NCVT ITI Time Table 2020 (Released) | Download NCVT ITI Revised Time Table July-August Semester and Annual Date Sheet

NCVT ITI Time Table 2020

NCVT ITI Time Table 2020:  NCVT Under the Guidelines of Government of India’s Ministry of Skill Development and Entrepreneurship, Directorate General of Training (DGT) will be conducting the examination for AITT (All India Trading Test) under the apprenticeship Training Scheme (ATS) in June and July 2020. In this article, we are discussing the NCVT ITI 2020 Exam Dates. The NCVT Exam dates 2020 for the semester and Annual Exams are given below and the Revised Schedule of ITI NCVT Exam time Table 2020 NCVT is discussed below.

NCVT MIS 2020 is the administration information system in a computerized database, of details regarding the enrollments and candidates for both the engineering and non-engineering skill development courses offered in several Industrial Training Institutes (ITIs). ITI Exam date 2020 Maharashtra is in June 2020.

ITI NCVT MIS Time Table 2020 (Annual Exam)

Date, Date & Time Year Trade Duration (Years) Subject
1st August 2020 (Thursday) 10:00 AM 1 2 Engineering-Employability Skills – (Employability Skills)
1st August 2020 (Thursday) 02:30 PM 1 1 Engineering-Employability Skills – (Employability Skills)
30th July 2020 (Tuesday) 10:00 AM 1 2 Engineering-Engineering Drawing – (Engineering Drawing)
30th July 2020 (Tuesday) 2:30 PM 1 1 Engineering-Engineering Drawing – (Engineering Drawing)
22nd July 2020 (Monday) 9:30 AM 1 1 Engineering-Trade Practical – (Trade Practical)
22nd July 2020 (Monday) 9:30 AM 1 2 Engineering-Trade Practical – (Trade Practical)
31st July 2020 (Wednesday) 10:00 AM 1 2 Engineering-Trade Theory – (Trade Theory)
31st July 2020 (Wednesday) 02:30 AM 1 1 Engineering-Trade Theory – (Trade Theory)
02nd August 2020 (Friday) 10:00 AM 1 2 Engineering-Workshop Calculation and Science – (Workshop Calculation and Science)
02nd August 2020 (Friday) 02:30 AM 1 1 Engineering-Workshop Calculation and Science – (Workshop Calculation and Science)
01st August 2020 (Thursday) 10:00 AM 1 2 Non-Engineering-Employability Skills – (Employability Skills)
01st August 2020 (Thursday) 02:30 AM 1 1 Non-Engineering-Employability Skills – (Employability Skills)
22nd July (Monday) 9:30 AM 1 1 Non-Engineering-Trade Practical – (Trade Practical)
22nd July (Monday) 9:30 AM 1 2 Non-Engineering-Trade Practical-(Trade Practical)
31st July 2020 (Wednesday) 10:00 AM 1 2 Non-Engineering-Trade Theory-(Trade Theory)
 31st July 2020 (Wednesday) 02:30 PM 1 1 Non-Engineering-Trade Theory-(Trade Theory)
30th July 2020 (Tuesday) 10:00 AM 2 2 Engineering-Engineering Drawing-(Engineering Drawing)
22nd July 2020 (Monday) 9:30 AM 2 2 Engineering-Trade Practical-(Trade Practical)
31st July 2020
(Wednesday) 10:00 AM
2 2 Engineering-Trade Theory-(Trade Theory)
02nd August 2020 (Friday) 10:00 AM 2 2 Engineering-Workshop Calculation and Science-(Workshop Calculation and Science)
22nd July 2020 (Monday) 9:30 AM 2 2 Non-Engineering-Trade Practical-(Trade Practical)
31st July 2020 (Wednesday) 10:00 AM 2 2 Non-Engineering-Trade Theory- (Trade Theory)

ITI Time Table 2020 NCVT (Semester Exam)

Day, Date & Time Semester Trade Duration (Years) Subject
22nd July 2020 (Monday) 9:30 AM 1 1 Engineering-Practical – (Trade Practical)
22nd July 2020 (Monday) 9:30 AM 1 2 Engineering-Practical- (Trade Practical)
22nd July 2020 (Monday) 9:30 AM 1 1 Non-Engineering-Practical- (Trade Practical)
22nd July 2020 (Monday) 9:30 AM 1 2 Non-Engineering-Practical- (Trade Practical)
23rd July 2020 (Tuesday) 9:30 AM 2 1 Engineering-Practical- (Trade Practical)
23rd July 2020 (Tuesday) 9:30 AM 2 2 Engineering-Practical- (Trade Practical)
23rd July 2020 (Tuesday) 9:30 AM 2 1 Non-Engineering-Practical- (Trade Practical)
23rd July 2020 (Tuesday) 9:30 AM 2 2 Non-Engineering-Practical- (Trade Practical)
24th July 2020 (Wednesday) 9:30 AM 3 2 Engineering-Practical- (Trade Practical)
24th July 2020 (Wednesday) 9:30 AM 3 2 Non-Engineering-Practical- (Trade Practical)
25th July 2020 (Thursday) 9:30 AM 4 2 Engineering-Practical- (Trade Practical)
25th July 2020 (Thursday) 9:30 AM 4 2 Non-Engineering-Practical- (Trade Practical)
3rd August (Saturday) 10:00 AM 2 1 Engineering-Paper-III- (Engg. Drawing)
3rd August (Saturday) 10:00 AM 2 2 Engineering-Paper-III- (Engg. Drawing)
3rd August (Saturday) 02:30 AM 3 2 Engineering-Paper-III- (Engg. Drawing)
05th August 2020 (Monday) 10:00 AM 4 2 Engineering-Paper-III- (Engg. Drawing)
06th August 2020 (Tuesday) 2:30 PM 3 2 Engineering-Paper-I- (Trade Theory)
06th August 2020 (Tuesday) 2:30 PM 3 2 Non-Engineering-Paper-I- (Trade Theory)
06th August 2020 (Tuesday) 10:00 AM 4 2 Engineering-Paper-I- (Trade Theory)
06th August 2020 (Tuesday) 10:00 AM 4 2 Non-Engineering-Paper-I- (Trade Theory)
07th August 2020 (Wednesday) 2:30 PM 3 2 Engineering-Paper-II- (Workshop Cal. & Sci.)
07th August 2020 (Wednesday) 10:00 AM 4 2 Engineering-Paper-II- (Workshop Cal. & Sci.)
08th August 2020 2:30 PM (Thursday) 1 1 Engineering-Paper-II- (Part-A Workshop Calculation & Science) (Part-B Engineering Drawing)
08th August 2020 2:30 PM (Thursday) 1 2 Engineering-Paper-II- (Part-A Workshop Calculation & Science) (Part-B Engineering Drawing)
08th August 2020 2:30 PM (Thursday) 1 1 Engineering-Paper-II- (Workshop Calculation & Science)
08th August 2020 2:30 PM (Thursday) 1 2 Engineering-Paper-II- (Workshop Calculation & Science)
08th August 2020 10:00 AM (Thursday) 2 1 Engineering-Paper-II- (Part-A Workshop Cal. & Sci.) (Part-B Employability Skills)
08th August 2020 10:00 AM (Thursday) 2 2 Engineering-Paper-II- (Part-A Workshop Cal. & Sci.) (Part-B Employability Skills)
08th August 2020 (Thursday) 10:00 AM 2 1 Non-Engineering-Paper-II- (Employability Skills)
08th August 2020 (Thursday) 10:00 AM 2 2 Non-Engineering-Paper-II- (Employability Skills)
9th August 2020 (Friday) 2:30 PM 1 1 Engineering-Paper-I- (Part-A Trade Theory) (Part-B Employability Skills)
9th August 2020 (Friday) 2:30 PM 1 2 Engineering-Paper-I- (Part-A Trade Theory) (Part-B Employability Skills)
9th August 2020 (Friday) 2:30 PM 1 1 Non-Engineering-Paper-I- (Part-A Trade Theory) (Part-B Employability Skills)
9th August 2020 (Friday) 2:30 PM 1 2 Non-Engineering-Paper-I- (Part-A Trade Theory) (Part-B Employability Skills)
Friday 09th August 2020 10:00 AM 2 1 Engineering-Paper-I- (Trade Theory)
Friday 09th August 2020 10:00 AM 2 2 Engineering-Paper-I- (Trade Theory)
Friday 09th August 2020 10:00 AM 2 1 Non-Engineering-Paper-I- (Trade Theory)
Friday 09th August 2020 10:00 AM 2 2 Non-Engineering-Paper-I- (Trade Theory)

Academic Calendar for NCVT ITI Programmes 2020

Task Name Dates (Tentative)
Generate Question Paper Indent 01-May-2020
Record Attendance and Session Marks Quarter 2 01-Jun-2020
Record Trainees’ Exam Fee Status 01-Jun-2020
Assign Bank/Police Station 01-Jun-2020
Generate Hall Ticket 01-Jun-2020
Print Hall Ticket 01-Jun-2020
Assign Examination Center 01-Jun-2020
AITT Exams 24-Jul-2020
Record Admitted Trainee 01-Aug-2020
Transfer of Trainee 01-Aug-2020
OMR Collection 17-Aug-2020
Capture Practical and Engineering Drawing Marks 17-Aug-2020
OMR Check 01-Sep-2020
OMR Upload 01-Oct-2020
Record Attendance and Sessional Marks Quarter 1 01-Oct-2020
Result Declaration 06-Oct-2020
Print Mark sheet 01-Nov-2020
Generate Question Paper Indent 01-Nov-2020
Record Attendance and Sessional Marks Quarter 1 19-Nov-2020
Record Attendance and Sessional Marks Quarter 2 19-Nov-2020
Record Trainees’ Exam Fee Status 19-Nov-2020
Generate Hall Ticket 17-Dec-2020
Print Hall Ticket 17-Dec-2020
Assign Examination Center 17-Dec-2020
AITT Exams 15-Jan-2020
OMR Collection 01-Feb-2020
Capture Practical and Engineering Drawing Marks 01-Feb-2020
OMR Check 05-Feb-2020
OMR Upload 16-Feb-2020
Result Declaration 01-Mar-2020
Print Mark sheet 10-Mar-2020
Generate Certificate 10-Mar-2020

Previous NCVT ITI Time Table 2020

Subject Date Time
First Semester
Trade Practical 18-July-2020 to 19-July-2020 9:30 AM
Paper-I (Trade Theory and Employability Skills) for 2 year Course 31-July-2020 10:30 AM
Paper-I (Trade Theory and Employability Skills) for 1 year Course 31-July-2020 10:30 AM
Paper-II (Workshop Calculation & Science and Engineering Drawing )for 2 year Course 01-Aug-2020 10:30 AM
Paper-II (Workshop Calculation & Science and Engineering Drawing )for 1 year Course 01-Aug-2020 10:30 AM
Second Semester
Trade Practical 20-July-2020 to 22-July-2020 9:30 AM
Paper-I (Trade Theory) for 2 year Course 03-Aug-2020 10:30 AM
Paper-I (Trade Theory) for 1 year Course 03-Aug-2020 02:30 PM
Paper-II (Workshop Calculation & Science and Employability Skills) for 2 year Course 04-Aug-2020 10:30 AM
Paper-II (Workshop Calculation & Science and Employability Skills) for 1 year Course 04-Aug-2020 02:30 PM
Paper-III (Engineering Drawing ) for 2 year Course 05-Aug-2020 09:30 AM
Paper-III (Engineering Drawing) for 1 year Course 05-Aug-2020 02:30 PM
Third Semester
Trade Practical 24-July-2020 to 25-July-2020 9:30 AM
Paper-I (Trade Theory) 31-July-2020 02:30 PM
Paper-II (Workshop Calculation & Science) 01-Aug-2020 02:30 PM
Paper-III (Engineering Drawing) 02-Aug-2020 09:30 AM
Fourth Semester
Trade Practical 26-July-2020 to 28-July-2020 9:30 AM
Paper-I (Trade Theory) 08-Aug-2020 10:30 AM
Paper-II (Workshop Calculation & Science) 09-Aug-2020 10:30 AM
Paper-III (Engineering Drawing ) 10-Aug-2020 09:30 AM

State-Wise NCVT ITI Time Table 2020

ITI is a government setup for studnets of engineering and non-engineering technical feilds. Students can get into this course directly after the completing of Class X or Class XII. After the conclusion of ITI students need to undergo ATS (Apprenticeship Training Scheme) to get into eligible for National NCVT Certificate. The major objective of this initiative is to provide skilled labour and manpower to the fast-emerging industry of our economy.

State Wise ITI Exam Details

NCVT ITI Admit Card 2020

NCVT ITI Admit Card will be available in June 2020 and students need to download their NCVT MIS Admit Card from the official website of ATS. NCVT Hall Ticket is an important document and candidates need to bring them during the examination.

Students will be able to download NCVT ITI Admit Card from the following steps

  • Step – 1: Visit the official website of NCVT.org
  • Step – 2: Now search for Admit Card Download Portal
  • Step – 3: Now enter the required credentials.
  • Step – 4: Now ATS admit card will display on the screen.
  • Step – 5: Now download the Admit Card and take a print out for future reference.

NCVT ITI Result 2020

NCVT ITI MIS Result 2020, All India Trades Test (AITT) under the apprenticeship Training Scheme (ATS) under the guidelines of Government of India’s Ministry of Skill Development and Entrepreneurship, Directorate General of Training (DGT) will be held from January 2020. AITT 2020 will be held in both offline and online depending upon the nature of the exam. Results of AITT will be released by SAA on the official website of NCVT MIS Login in the month of April 2020. ITI Annual Exam Result Date is 26th April 2020.

Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Practice Set 4.2

Maharashtra State Board Class 10 Maths Solutions Chapter 4 Financial Planning Practice Set 4.2

Practice Set 4.2 Financial Planning Question 1. ‘Chetana Store’ paid total GST of ₹ 1,00,500 at the time of purchase and collected GST ₹ 1,22,500 at the time of sale during 1st of July 2017 to 31st July 2017. Find the GST payable by Chetana Stores.
Answer:
Output tax (Tax collected at the time of sale)
= ₹ 1,22,500
Input tax (Tax paid at the time of purchase)
= ₹ 1,00,500
ITC (Input Tax credit) = ₹ 1,00,500.
GST payable = Output tax – ITC
= 1,22,500 – 1,00,500
= ₹ 22,000
GST payable by Chetana stores is ₹ 22,000.

Financial Planning Practice Set 4.2 Question 2. Nazama is a proprietor of a firm, registered under GST. She has paid GST of ₹ 12,500 on purchase and collected ₹ 14,750 on sale. What is the amount of ITC to be claimed? What is the amount of GST payable?
Solution:
Output tax = ₹ 14,750
Input tax = ₹ 12,500
∴ ITC for Nazama = ₹ 12,500.
∴ GST payable = Output tax – ITC
= 14750 – 12500
= ₹ 2250
∴ Amount of ITC to be claimed is ₹ 12,500 and amount of GST payable is ₹ 2250.

Financial Planning 10th Practice Set 4.2 Question 3. Amir Enterprise purchased chocolate sauce bottles and paid GST of ₹ 3800. He sold those bottles to Akbari Bros, and collected GST of ₹ 4100. Mayank Food Corner purchased these bottles from Akbari Bros, and paid GST of ₹ 4500. Find the amount of GST payable at every stage of trading and hence find payable CGST and SGST.
Solution:
For Amir Enterprise:
Output tax = ₹ 4100
Input tax = ₹ 3800
ITC for Amir enterprise = ₹ 3800.
∴ GST payable = Output tax – ITC
= 4100 – 3800
= ₹ 300
For Akbari Bros.:
Output tax = ₹ 4500
Input tax = ₹ 4100
ITC for Akbari Bros = ₹ 4100.
GST payable = Output tax – ITC
= 4500 – 4100 = ₹ 400
∴ Statement of GST payable at every stage of trading:
Practice Set 4.2 Algebra 10th

Financial Planning Class 10 Practice Set 4.2 Question 4. Malik Gas Agency (Chandigarh Union Territory) purchased some gas cylinders for industrial use for ₹ 24,500, and sold them to the local customers for ₹ 26,500. Find the GST to be paid at the rate of 5% and hence the CGST and UTGST to be paid for this transaction, (for Union Territories there is UTGST instead of SGST.)
Solution:
For Malik Gas Agency:
Output tax = 5% of 26500
= \(\frac { 5 }{ 100 } \) × 26500
= ₹ 1325
Input tax = 5% of 24500
= \(\frac { 5 }{ 100 } \) × 24500
= ₹ 1225
ITC for Malik Gas Agency = ₹ 1225.
∴ GST payable = Output tax – ITC
= 1325 – 1225
= ₹ 100
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Practice Set 4.2 2
∴ CGST = UTGST = ₹ 50
∴ The GST to be paid at the rate of 5% is ₹ 100 and hence, CGST and UTGST paid for the transaction is ₹ 50 each.

Question 5.
M/s Beauty Products paid 18% GST on cosmetics worth ₹ 6000 and sold to a customer for ₹ 10,000. What are the amounts of CGST and SGST shown in the tax invoice issued?
Solution:
Output tax = 18% of 10,000
= \(\frac { 18 }{ 100 } \) × 10,000
= ₹ 1800
M/S Beauty Products Paid 18
∴ Amount of CGST and SGST shown in the tax invoice issued is ₹ 900 each.

Question 6.
Prepare Business to Consumer (B2C) tax invoice using given information. Write the name of the supplier, address, state, Date, Invoice number, GSTIN etc. as per your choice.
Supplier: M/s ______ Address _______ State _______ Date _______ Invoice No. _______ GSTIN _______
Particulars
Rate of Mobile Battery ₹ 200 Rate of GST 12% HSN 8507 1 PC
Rate of Headphone ₹750 Rate of GST 18% HSN 8518 1 Pc
Solution:
Rate of Mobile Battery = ₹200
CGST = 6% of 200
= \(\frac { 6 }{ 100 } \) × 200
= ₹ 12
∴ CGST = SGST = ₹ 12

Rate of Headphone = ₹ 750
COST = 9% of 750
= \(\frac { 9 }{ 100 } \) × 750
= ₹ 67.5
∴ CGST = SGST = ₹ 67.5
Algebra Practice Set 4.2

Question 7.
Prepare Business to Business (B2B) Tax Invoice as per the details given below, name of the supplier, address, Date etc. as per your choice.
Supplier – Name, Address, State, GSTIN, Invoice No., Date
Recipient – Name, Address, State, GSTIN,
Items:
i. Pencil boxes 100, HSN – 3924, Rate – ₹ 20, GST 12%
ii. Jigsaw Puzzles 50, HSN 9503, Rate – ₹ 100 GST 12%.
Solution:
Cost of 100 Pencil boxes
= 20 × 1oo
= ₹ 2000
CGST = 6% of 2000
= \(\frac { 6 }{ 100 } \) × 2000
= ₹ 120
∴ CGST = SGST = ₹ 120

Cost of 50 Jigsaw Puzzles = 100 × 50
= ₹ 5000
CGST = 6% of 5000
= \(\frac { 6 }{ 100 } \) × 5000
= ₹ 300
CGST – SGST = ₹ 300
Malik Gas Agency Purchased

Question 1.
Suppose a manufacturer sold a cycle for a taxable value of ₹ 4000 to the wholesaler. Wholesaler sold it to the retailer for ₹ 4800 (taxable value). Retailer sold it to a customer for ₹ 5200 (taxable value). Rate of GST is 12%. Complete the following activity to find the payable CGST and SGST at each stage of trading. (Textbook pg. no. 92)
Solution:
Suppose a Manufacturer Sold a Cycle
GST payable by manufacturer = ₹ 480
Output tax of wholesaler
= 12% of 4800 = \(\frac { 12 }{ 100 } \) × 4800 = ₹ 576
∴ GST payable by wholesaler
= Output tax – Input tax
= 576 – 480
= ₹ 96
Output tax of retailer = 12% of 5200
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Practice Set 4.2 7

Practice Set 4.2 Question 2. Suppose in the month of July the output tax of a trader is equal to the input tax, then what is his payable GST?(Textbook pg. no. 93)
Answer:
Here, output tax is same as input tax.
∴ Trader payable GST will be zero.

Question 3.
Suppose in the month of July output tax of a trader is less than the input tax then how to compute his GST? (Textbook pg. no. 93)
Answer:
If output tax of a trader in a particular month is less than his input tax, then he won’t be able to get entire credit for his input tax. In such a case his balance credit will be carried forward and adjusted against the subsequent transactions.

Maharashtra Board Class 10 Maths Solutions

Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Practice Set 4.3

Maharashtra State Board Class 10 Maths Solutions Chapter 4 Financial Planning Practice Set 4.3

Practice Set 4.3 Financial Planning Question 1. Complete the following table by writing suitable numbers and words.
Practice Set 4.3 Algebra 10th
Solution:
i. Here, share is at par.
∴ MV = FV
∴ MV = ₹ 100

ii. Here, Premium = ₹ 500, MV = ₹ 575
∴ FV + Premium = MV
∴ FV + 500 = 575
∴ FV = 575 – 500
∴ FV = ₹ 75

iii. Here, FV = ₹ 10, MV = ₹ 5
∴ FV > MV
Share is at discount.
FV – Discount = MV
∴ 10 – Discount = 5
∴ 10 – 5 = Discount
₹ Discount = ₹ 5
Math Part 1 Practice Set 4.3

Practice Set 4.3 Question 2. Mr. Amol purchased 50 shares of Face value ₹ 100 when the Market value of the share was ₹ 80. Company had given 20% dividend. Find the rate of return on investment.
Solution:
Here, MV = ₹ 80, FV = ₹ 100,
Number of shares = 50, Rate of dividend = 20%
∴ Sum invested = Number of shares × MV
= 50 × 80
= ₹ 4000

Dividend per share = 20% of FV
= \(\frac { 20 }{ 100 } \) × 100 = ₹ 20
∴ Total dividend of 50 shares = 50 × 20
= ₹ 1000
Financial Planning Practice
∴ Rate of return on investment is 25%.

Question 3.
Joseph purchased following shares, Find his total investment.
Company A : 200 shares, FV = ₹ 2, Premium = ₹ 18.
Company B : 45 shares, MV = ₹ 500
Company C : 1 share, MV = ₹ 10,540
Solution:
For company A:
FV = ₹ 2, premium = ₹ 18,
Number of shares = 200
∴ MV = FV+ Premium
= 2 + 18
= ₹ 20
Sum invested = Number of shares × MV
= 200 × 20
= ₹ 14000

For company B:
MV = ₹ 500, Number of shares = 45
Sum invested = Number of shares × MV
= 45 × 500 = ₹ 22,500

For company C:
MV = ₹ 10,540, Number of shares = 1
∴ Sum invested = Number of shares × MV
= 1 × 10540
= ₹ 10,540
∴ Total investment of Joseph
= Investment for company A + Investment for company B + Investment for company C
= 4000 + 22,500 + 10,540
= ₹ 37040
∴ Total investment done by Joseph is ₹ 37,040.

Question 4.
Smt. Deshpande purchased shares of FV ₹ 5 at a premium of ₹ 20. How many shares will she get for ₹ 20,000?
Solution:
Here, FV = ₹ 5, Premium = ₹ 20,
Sum invested = ₹ 20,000
∴ MV = FV + Premium
= 5 + 20
∴ MV = ₹ 25
Now, sum invested = Number of shares × MV
Mr Amol Purchased 50 Shares
∴ Smt. Deshpande got 800 shares for ₹ 20,000.

Question 5.
Shri Shantilal has purchased 150 shares of FV ₹ 100, for MV of ₹ 120. Company has paid dividend at 7%. Find the rate of return on his investment.
Solution:
Here, FV = ₹ 100, MV = ₹ 120
Dividend = 7%, Number of shares = 150
∴ Sum invested = Number of shares × MV
= 150 × 120 = ₹ 18000
Dividend per share = 7% of FV
= \(\frac { 7 }{ 100 } \) × 100 = ₹ 7
∴ Total dividend of 150 shares
= 150 × 7 = ₹ 1050
Algebra Practice Set 4.3
∴ Rate of return on investment is 5.83%.

4.3 Class 10 Question 6. If the face value of both the shares is same, then which investment out of the following is more profitable?
Company A : dividend 16%, MV = ₹ 80,
Company B : dividend 20%, MV = ₹ 120.
Solution:
Let the face value of share be ₹ x.
For company A:
MV = ₹ 80, Dividend = 16%
Dividend = 16% of FV
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Practice Set 4.3 6
∴ Rate of return of company A is more.
∴ Investment in company A is profitable.

Question 1.
Smita has invested ₹ 12,000 and purchased shares of FV ₹ 10 at a premium of ₹ 2. Find the number of shares she purchased. Complete the given activity to get the answer. (Textbook pg. no. 101.)
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Practice Set 4.3 7

Maharashtra Board Class 10 Maths Solutions

Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3

Maharashtra State Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3

Practice Set 5.3 Geometry Class 10 Question 1. Angles made by the line with the positive direction of X-axis are given. Find the slope of these lines.
i. 45°
ii. 60°
iii. 90°
Solution:
i. Angle made with the positive direction of
X-axis (θ) = 45°
Slope of the line (m) = tan θ
∴ m = tan 45° = 1
∴ The slope of the line is 1.

ii. Angle made with the positive direction of X-axis (θ) = 60°
Slope of the line (m) = tan θ
∴ m = tan 60° = \(\sqrt { 3 }\)
∴ The slope of the line is \(\sqrt { 3 }\).

iii. Angle made with the positive direction of
X-axis (θ) = 90°
Slope of the line (m) = tan θ
∴ m = tan 90°
But, the value of tan 90° is not defined.
∴ The slope of the line cannot be determined.

Practice Set 5.3 Geometry Question 2. Find the slopes of the lines passing through the given points.
i. A (2, 3), B (4, 7)
ii. P(-3, 1), Q (5, -2)
iii. C (5, -2), D (7, 3)
iv. L (-2, -3), M (-6, -8)
v. E (-4, -2), F (6, 3)
vi. T (0, -3), s (0,4)
Solution:
i. A (x1, y1) = A (2, 3) and B (x2, y2) = B (4, 7)
Here, x1 = 2, x2 = 4, y1 = 3, y2 = 7
Practice Set 5.3
∴ The slope of line AB is 2.

ii. P (x1, y1) = P (-3, 1) and Q (x2, y2) = Q (5, -2)
Here, x1 = -3, x2 = 5, y1 = 1, y2 = -2
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 2
∴ The slope of line PQ is \(\frac { -3 }{ 8 } \)

iii. C (x1, y1) = C (5, -2) and D (x2, y2) = D (7, 3)
Here, x1 = 5, x2 = 7, y1 = -2, y2 = 3
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 3
∴ The slope of line CD is \(\frac { 5 }{ 2 } \)

iv. L (x1, y1) = L (-2, -3) and M (x2,y2) = M (-6, -8)
Here, x1 = -2, x2 = – 6, y1 = – 3, y2 = – 8
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 4
∴ The slope of line LM is \(\frac { 5 }{ 4 } \)

v. E (x1, y1) = E (-4, -2) and F (x2, y2) = F (6, 3)
Here,x1 = -4, x2 = 6, y1 = -2, y2 = 3
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 5
∴ The slope of line EF is \(\frac { 1 }{ 2 } \).

vi. T (x1, y1) = T (0, -3) and S (x2, y2) = S (0, 4)
Here, x1 = 0, x2 = 0, y1 = -3, y2 = 4
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 6
∴ The slope of line TS cannot be determined.

5.3.5 Practice Question 3. Determine whether the following points are collinear.
i. A (-1, -1), B (0, 1), C (1, 3)
ii. D (- 2, -3), E (1, 0), F (2, 1)
iii. L (2, 5), M (3, 3), N (5, 1)
iv. P (2, -5), Q (1, -3), R (-2, 3)
v. R (1, -4), S (-2, 2), T (-3,4)
vi. A(-4,4),K[-2,\(\frac { 5 }{ 2 } \)], N (4,-2)
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 7
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 8
∴ slope of line AB = slope of line BC
∴ line AB || line BC
Also, point B is common to both the lines.
∴ Both lines are the same.
∴ Points A, B and C are collinear.

Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 9
∴ slope of line DE = slope of line EF
∴ line DE || line EF
Also, point E is common to both the lines.
∴ Both lines are the same.
∴ Points D, E and F are collinear.

Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 10
∴ slope of line LM ≠ slope of line MN
∴ Points L, M and N are not collinear.

Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 11
∴ slope of line PQ = slope of line QR
∴ line PQ || line QR
Also, point Q is common to both the lines.
∴ Both lines are the same.
∴ Points P, Q and R are collinear.

Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 12
∴ slope of line RS = slope of line ST
∴ line RS || line ST
Also, point S is common to both the lines.
∴ Both lines are the same.
∴ Points R, S and T are collinear.

Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 13
∴ slope of line AK = slope of line KN
∴ line AK || line KN
Also, point K is common to both the lines.
∴ Both lines are the same.
∴ Points A, K and N are collinear.

Practice Set 5.3 Geometry 9th Standard Question 4. If A (1, -1), B (0,4), C (-5,3) are vertices of a triangle, then find the slope of each side.
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 14
∴ The slopes of the sides AB, BC and AC are -5, \(\frac { 1 }{ 5 } \) and \(\frac { -2 }{ 3 } \) respectively.

Geometry 5.3 Question 5. Show that A (-4, -7), B (-1, 2), C (8, 5) and D (5, -4) are the vertices of a parallelogram.
Proof:
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 15
5.3.10 Practice Questions
∴ Slope of side AB = Slope of side CD … [From (i) and (iii)]
∴ side AB || side CD
Slope of side BC = Slope of side AD … [From (ii) and (iv)]
∴ side BC || side AD
Both the pairs of opposite sides of ꠸ABCD are parallel.
꠸ABCD is a parallelogram.
Points A(-4, -7), B(-1, 2), C(8, 5) and D(5, -4) are the vertices of a parallelogram.

Question 6.
Find k, if R (1, -1), S (-2, k) and slope of line RS is -2.
Solution:
R(x1, y1) = R (1, -1), S (x2, y2) = S (-2, k)
Here, x1 = 1, x2 = -2, y1 = -1, y2 = k
5.3 Practice a Geometry
But, slope of line RS is -2. … [Given]
∴ -2 = \(\frac { k+1 }{ -3 } \)
∴ k + 1 = 6
∴ k = 6 – 1
∴ k = 5

5.3 Class 10 Question 7. Find k, if B (k, -5), C (1, 2) and slope of the line is 7.
Solution:
B(x1, y1) = B (k, -5), C (x2, y2) = C (1, 2)
Here, x1 = k, x2 = 1, y1 = -5, y2 = 2
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.3 18
But, slope of line BC is 7. …[Given]
∴ 7 = \(\frac { 7 }{ 1-k } \)
∴ 7(1 – k) = 7
∴ 1 – k = \(\frac { 7 }{ 7 } \)
∴ 1 – k = 1
∴ k = 0

Question 8.
Find k, if PQ || RS and P (2, 4), Q (3, 6), R (3,1), S (5, k).
Solution:
Class 10 Math 5.3
But, line PQ || line RS … [Given]
∴ Slope of line PQ = Slope of line RS
∴ 2 = \(\frac { k-1 }{ 2 } \)
∴ 4 = k – 1
∴ k = 4 + 1
∴ k = 5

Maharashtra Board Class 10 Maths Solutions

Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1

Maharashtra State Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1

Practice Set 5.1 Geometry Class 10 Question 1. Find the distance between each of the following pairs of points.
i. A (2, 3), B (4,1)
ii. P (-5, 7), Q (-1, 3)
iii. R (0, -3), S (0,\(\frac { 5 }{ 2 } \))
iv. L (5, -8), M (-7, -3)
v. T (-3, 6), R (9, -10)
vi. W(\(\frac { -7 }{ 2 } \),4), X(11, 4)
Solution:
i. Let A (x1, y1) and B (x2, y2) be the given points.
∴ x1 = 2, y1 = 3, x2 = 4, y2 = 1
By distance formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 1
∴ d(A, B) = 2\(\sqrt { 2 }\) units
∴ The distance between the points A and B is 2\(\sqrt { 2 }\) units.

ii. Let P (x1, y1 ) and Q (x2, y2) be the given points.
∴ x1 = -5, y1 = 7, x2 = -1, y2 = 3
By distance formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 2
∴ d(P, Q) = 4\(\sqrt { 2 }\) units
∴ The distance between the points P and Q is 4\(\sqrt { 2 }\) units.

iii. Let R (x1, y1) and S (x2, y2) be the given points.
∴ x1 = 0, y1 = -3, x2 = 0, y2 = \(\frac { 5 }{ 2 } \)
By distance formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 3
∴ d(R, S) = -y units
∴ The distance between the points R and S is \(\frac { 11 }{ 2 } \) units.

iv. Let L (x1, y1) and M (x2, y2) be the given points.
∴ x1 = 5, y1 = -8, x2 = -7, y2 = -3
By distance formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 4
∴ d(L, M) = 13 units
∴ The distance between the points L and M is 13 units.

v. Let T (x1,y1) and R (x2, y2) be the given points.
∴ x1 = -3, y1 = 6,x2 = 9,y2 = -10
By distance formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 5
∴ d(T, R) = 20 units
∴ The distance between the points T and R 20 units.

vi. Let W (x1, y1) and X (x2, y2) be the given points.
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 6
∴ d(W, X) = \(\frac { 29 }{ 2 } \) units
∴ The distance between the points W and X is \(\frac { 29 }{ 2 } \) units.

Practice Set 5.1 Geometry 10th Question 2. Determine whether the points are collinear.
i. A (1, -3), B (2, -5), C (-4, 7)
ii. L (-2, 3), M (1, -3), N (5, 4)
iii. R (0, 3), D (2, 1), S (3, -1)
iv. P (-2, 3), Q (1, 2), R (4, 1)
Solution:
i. By distance formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 7
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 8
∴ d(A, B) = \(\sqrt { 5 }\) …(i)
On adding (i) and (iii),
d(A, B) + d(A, C)= \(\sqrt { 5 }\) + 5\(\sqrt { 5 }\) = 6\(\sqrt { 5 }\)
∴ d(A, B) + d(A, C) = d(B, C) … [From (ii)]
∴ Points A, B and C are collinear.

ii. By distance formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 9
On adding (i) and (iii),
d(L, M) + d(L, N) = 3\(\sqrt { 5 }\) + 5\(\sqrt { 2 }\) ≠ \(\sqrt { 65 }\)
∴ d(L, M) + d(L, N) ≠ d(M, N) … [From (ii)]
∴ Points L, M and N are not collinear.

iii. By distance formula,
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 10
On adding (i) and (ii),
∴ d(R, D) + d(D, S) = \(\sqrt { 8 }\) + \(\sqrt { 5 }\) ≠ 5
∴ d(R, D) + d(D, S) ≠ d(R, S) … [From (iii)]
∴ Points R, D and S are not collinear.

iv. By distance formula,
5.1 Practice a Geometry Answers
On adding (i) and (ii),
d(P, Q) + d(Q, R) = \(\sqrt { 10 }\) + \(\sqrt { 10 }\) = 2\(\sqrt { 10 }\)
∴ d(P, Q) + d(Q, R) = d(P, R) … [From (iii)]
∴ Points P, Q and R are collinear.

Coordinate Geometry Class 10 Practice Set 5.1 Question 3. Find the point on the X-axis which is equidistant from A (-3,4) and B (1, -4).
Solution:
Let point C be on the X-axis which is equidistant from points A and B.
Point C lies on X-axis.
∴ its y co-ordinate is 0.
Let C = (x, 0)
C is equidistant from points A and B.
∴ AC = BC
Coordinate Geometry Practice Set 5.1
∴ (x + 3)2 + (-4)2 = (x- 1)2 + 42
∴ x2 + 6x + 9 + 16 = x2 – 2x + 1 + 16
∴ 8x = – 8
∴ x = – \(\frac { 8 }{ 8 } \) = -1
∴ The point on X-axis which is equidistant from points A and B is (-1,0).

10th Geometry Practice Set 5.1 Question 4. Verify that points P (-2, 2), Q (2, 2) and R (2, 7) are vertices of a right angled triangle.
Solution:
Distance between two points
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 13
Consider, PQ2 + QR2 = 42 + 52 = 16 + 25 = 41 … [From (i) and (ii)]
∴ PR2 = PQ2 + QR2 … [From (iii)]
∴ ∆PQR is a right angled triangle. … [Converse of Pythagoras theorem]
∴ Points P, Q and R are the vertices of a right angled triangle.

Question 5.
Show that points P (2, -2), Q (7, 3), R (11, -1) and S (6, -6) are vertices of a parallelogram.
Proof:
Distance between two points
10th Class Geometry Practice Set 5.1
PQ = RS … [From (i) and (iii)]
QR = PS … [From (ii) and (iv)]
A quadrilateral is a parallelogram, if both the pairs of its opposite sides are congruent.
∴ □ PQRS is a parallelogram.
∴ Points P, Q, R and S are the vertices of a parallelogram.

Question 6.
Show that points A (-4, -7), B (-1, 2), C (8, 5) and D (5, -4) are vertices of rhombus ABCD.
Proof:
Distance between two points
5.1.10 Practice Questions
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 16
∴ AB = BC = CD = AD …[From (i), (ii), (iii) and (iv)]
In a quadrilateral, if all the sides are equal, then it is a rhombus.
∴ □ ABCD is a rhombus.
∴ Points A, B, C and D are the vertices of rhombus ABCD.

Practice Set 5.1 Question 7. Find x if distance between points L (x, 7) and M (1,15) is 10.
Solution:
X1 = x, y1 = 7, x2 = 1, y2 = 15
By distance formula,
Geometry Practice Set 5.1
∴ 1 – x = ± 6
∴ 1 – x = 6 or l – x = -6
∴ x = – 5 or x = 7
∴ The value of x is – 5 or 7.

Geometry 5.1 Question 8. Show that the points A (1, 2), B (1, 6), C (1 + 2\(\sqrt { 3 }\), 4) are vertices of an equilateral triangle.
Proof:
Distance between two points
Maharashtra Board Class 10 Maths Solutions Chapter 5 Co-ordinate Geometry Practice Set 5.1 18
∴ AB = BC = AC … [From (i), (ii) and (iii)]
∴ ∆ABC is an equilateral triangle.
∴ Points A, B and C are the vertices of an equilateral triangle.

Maharashtra Board Class 10 Maths Chapter 5 Coordinate Geometry Intext Questions and Activities

Question 1.
In the figure, seg AB || Y-axis and seg CB || X-axis. Co-ordinates of points A and C are given. To find AC, fill in the boxes given below. (Textbook pa. no. 102)
Maths Part 2 Practice Set 5.1
Solution:
In ∆ABC, ∠B = 900
∴ (AB)2 + (BC)2 = [(Ac)2 …(i) … [Pythagoras theorem]
seg CB || X-axis
∴ y co-ordinate of B = 2
seg BA || Y-axis
∴ x co-ordinate of B = 2
∴ co-ordinate of B is (2, 2) = (x1,y1)
co-ordinate of A is (2, 3) = (x2, Y2)
Since, AB || to Y-axis,
d(A, B) = Y2 – Y1
d(A,B) = 3 – 2 = 1
co-ordinate of C is (-2,2) = (x1,y1)
co-ordinate of B is (2, 2) = (x2, y2)
Since, BC || to X-axis,
d(B, C) = x2 – x1
d(B,C) = 2 – -2 = 4
∴ AC2 = 12 + 42 …[From (i)]
= 1 + 16 = 17
∴ AC = \(\sqrt { 17 }\) units …[Taking square root of both sides]

Maharashtra Board Class 10 Maths Solutions

Maharashtra Board Class 9 Maths Solutions Chapter 5 Linear Equations in Two Variables Practice Set 5.2

Maharashtra State Board Class 9 Maths Solutions Chapter 5 Linear Equations in Two Variables Practice Set 5.2

Linear equations in two variables class 9 practice set 5.2 Question 1. In an envelope there are some ₹5 notes and some ₹10 notes. Total amount of these notes together is ₹350. Number of ₹5 notes are less by 10 than twice the number of ₹10 notes. Then find the number of ₹5 and ₹10 notes.
Solution:
Let the number of ₹5 notes be ‘x’ and the number of ₹10 notes be ‘y’
Total amount of x notes of ₹ 5 = ₹ 5x
Total amount ofy notes of ₹ 10 = ₹ 10y
∴ Total amount = 5x + 10y
According to the first condition,
total amount of the notes together is ₹350.
∴ 5x + 10y = 350 …(i)
According to the second condition,
Number of ₹ 5 notes are less by 10 than twice the number of ₹ 10 notes.
∴ x = 2y – 10
∴ x – 2y = -10 …..(ii)
Multiplying equation (ii) by 5,
5x – 10y = -50 …(iii)
Adding equations (i) and (iii),
5x + 10y =350
+ 5x – 10y = -50
10x =300
∴ x = \(\frac { 300 }{ 10 }\)
∴ x = 30
Substituting x = 30 in equation (ii),
x – 2y = -10
30 – 2y = -10
∴ 30 + 10 = 2y
∴ 40 = 2y
∴ y = \(\frac { 40 }{ 2 }\)
∴ y = 20
There are 30 notes of ₹ 5 and 20 notes of ₹ 10 in the envelope.

Practice set 5.2 algebra class 9 Question 2. The denominator of a fraction is 1 less than twice its numerator. If 1 is added to numerator and denominator respectively, the ratio of numerator to denominator is 3 : 5. Find the fraction.
Solution:
Let the numerator of the fraction be ‘x’ and its denominator be ‘y’.
Then, the required fraction is \(\frac { x }{ y }\) .
According to the first condition,
the denominator is 1 less than twice its numerator.
∴ y = 2x – 1
∴ 2x – y = 1 …(i)
According to the second condition,
if 1 is added to the numerator and the denominator, the ratio of numerator to denominator is 3 : 5.
∴ \(\frac { x+1 }{ y+1 }\) = \(\frac { 3 }{ 5 }\)
∴ y + 1 = 5
∴ 5(x + 1) = 3(y + 1)
∴ 5x + 5 = 3y + 3
∴ 5x – 3y = 3 – 5
∴ 5x – 3y = -2 ……(ii)
Multiplying equation (i) by 3,
6x – 3y = 3 …(iii)
Subtracting equation (ii) from (iii),
Practice set 5.2 algebra 9th
Substituting x = 5 in equation (i),
∴ 2x – y = 1
∴ 2(5) – y = 1
∴ 10 – y = 1
∴ y= 10 – 1 =9
∴ The required fraction is \(\frac { 5 }{ 9 }\).

9th class maths part 1 practice set 5.2 Question 3. The sum of ages of Priyanka and Deepika is 34 years. Priyanka is elder to Deepika by 6 years. Then find their present ages.
Solution:
Let the present age of Priyanka be ‘x’ years and that of Deepika be ‘y’ years.
According to the first condition,
Priyanka’s age + Deepika’s age = 34 years
∴ x + y = 34 …(i)
According to the second condition,
Priyanka is elder to Deepika by 6 years.
∴ x =y + 6
∴ x – y = 6 …..(ii)
Adding equations (i) and (ii),
Maharashtra Board Class 9 Maths Solutions Chapter 5 Linear Equations in Two Variables Practice Set 5.2 3
∴ x = 20
Substituting x = 20 in equation (i),
x + y = 34
∴ 20 + y = 34
∴ y = 34 -20= 14
∴ The present age of Priyanka is 20 years and that of Deepika is 14 years.

9th algebra practice set 5.2 Question 4. The total number of lions and peacocks in a certain zoo is 50. The total number of their legs is 140. Then find the number of lions and peacocks in the zoo.
Solution:
Let the number of lions in the zoo be ‘x’ and the number of peacocks be ‘y’.
According to the first condition,
the total number of lions and peacocks is 50.
∴ x + y = 50 …(i)
Lion has 4 legs and Peacock has 2 legs.
According to the second condition,
the total number of their legs is 140.
∴ 4x + 2y = 140
Dividing both sides by 2,
2x + y = 70 …(ii)
Subtracting equation (i) from (ii),
Maharashtra Board Class 9 Maths Solutions Chapter 5 Linear Equations in Two Variables Practice Set 5.2 4
Substituting x = 20 in equation (i),
x + y = 50
∴ 20 + y = 50
∴ y = 50 – 20 = 30
∴ The number of lions and peacocks in the zoo are 20 and 30 respectively.

Linear equations in two variables class 9 Maharashtra board Question 5. Sanjay gets fixed monthly income. Every year there is a certain increment in his salary. After 4 years, his monthly salary was ₹ 4500 and after 10 years his monthly salary became ₹ 5400, then find his original salary and yearly increment.
Solution:
Let the original salary of Sanjay be ₹ ‘x’ and his yearly increment be ₹ ‘y’.
According to the first condition, after 4 years his monthly salary was ₹ 4500
∴ x + 4y = 4500 …..(i)
According to the second condition,
after 10 years his monthly salary became ₹ 5400
∴ x + 10y = 5400 …(ii)
Subtracting equation (i) from (ii),
Maharashtra Board Class 9 Maths Solutions Chapter 5 Linear Equations in Two Variables Practice Set 5.2 5
∴ y = 150
Substituting y = 150 in equation (i),
x + 4y = 4500
∴ x +4(150) = 4500
∴ x + 600 = 4500
∴ x = 4500 – 600 = 3900
∴ The original salary of Sanjay is ₹ 3900 and his yearly increment is ₹ 150.

9th class algebra practice set 5.2 Question 6. The price of 3 chairs and 2 tables is ₹ 4500 and price of 5 chairs and 3 tables is ₹ 7000, then find the price of 2 chairs and 2 tables.
Solution:
Let the price of one chair be ₹ ‘x’ and that of one table be ₹ ‘y’.
According to the first condition,
the price of 3 chairs and 2 tables is ₹ 4500
∴ 3x + 2y = 4500 ,..(i)
According to the second condition, the price of 5 chairs and 3 tables is ? 7000
∴ 5x + 3y = 7000 …(ii)
Multiplying equation (i) by 3,
9x + 6y = 13500 ….(iii)
Multiplying equation (ii) by 2,
10x + 6y= 14000 …(iv)
Subtracting equation (iii) from (iv),
linear equations in two variables class 9
Substituting x = 500 in equation (i),
3x + 2y = 4500
∴ 3(500)+ 2y = 4500
∴ 1500 + 2y = 4500
∴ 2y = 4500- 1500
∴ 2y = 3000
∴ y = \(\frac { 3000 }{ 2 }\)
∴ y = 1500
∴ Price of 2 chairs and 2 tables = 2x + 2y
= 2(500)+ 2(1500)
= 1000 + 3000 = ₹ 4000
∴ The price of 2 chairs and 2 tables is ₹ 4000.

Question 7.
The sum of the digits in a two-digit number is 9. The number obtained by interchanging the digits exceeds the original number by 27. Find the two-digit number.
Solution:
Let the digit in unit’s place be ‘x’ and the digit in ten’s place be ‘y’.
linear equations in two variables class 9 problem set 5
According to the first condition.
the sum of the digits in a two-digit number is 9
x + y = 9 …(i)
According to the second condition,
the number obtained by interchanging the digits exceeds the original number by 27
∴ 10x + y = 10y + x + 27
∴ 10x – x + y – 10y = 27
∴ 9x – 9y = 27
Dividing both sides by 9,
x – y = 3 …….(ii)
Adding equations (i) and (ii),
Maharashtra Board Class 9 Maths Solutions Chapter 5 Linear Equations in Two Variables Practice Set 5.2 7a
∴ x = 6
Substituting x = 6 in equation (i),
x + y = 9
∴ 6 + y = 9
∴ y = 9 – 6 = 3
∴ Original number = 10y + x = 10(3)+ 6
= 30 + 6 = 36
∴ The two digit number is 36.

practice set 5.2 Question 8. In ∆ABC, the measure of ∠A is equal to the sum of the measures of ∠B and ∠C. Also the ratio of measures of ∠B and ∠C is 4 : 5. Then find the measures of angles of the triangle.
Solution:
Let the measure of ∠B be ‘x°’ and that of ∠C be ‘y°’.
According to the first condition,
m∠A = m∠B + m∠C
∴ m∠A = x° + y°
In AABC,
m∠A + m∠B + m∠C = 180° …[Sum of the measures of the angles of a triangle is 180°]
∴ x + y + x + y = 180 ,
∴ 2x + 2y = 180
Dividing both sides by 2,
x + y = 90 …(i)
According to the second condition,
the ratio of the measures of ∠B and ∠C is 4 : 5.
∴ \(\frac { x }{ y }\) = \(\frac { 4 }{ 5 }\)
∴ 5x = 4y
∴ 5x – 4y = 0 …….(ii)
Multiplying equation (i) by 4,
4x + 4y = 360 …(iii)
Adding equations (ii) and (iii),
Maharashtra Board Class 9 Maths Solutions Chapter 5 Linear Equations in Two Variables Practice Set 5.2 7b
∴ x = 40
Substituting x = 40 in equation (i),
x + y = 90
∴ 40 + y = 90
∴ y = 90 – 40
∴ y = 50
∴ m∠A = x° + y° = 40° + 50° = 90°
∴ The measures of ∠A, ∠B and ∠C are 90°, 40°, and 50° respectively.

std 9 algebra practice set 5.2 Question 9. Divide a rope of length 560 cm into 2 parts such that twice the length of the smaller part is equal to \(\frac { 1 }{ 3 }\) of the larger part. Then find the length of the larger part.
Solution:
Let the length of the smaller part of the rope be ‘x’ cm and that of the larger part be ‘y’ cm.
According to the first condition,
total length of the rope is 560 cm.
∴ x + y = 560 …(i)
Twice the length of the smaller part = 2x
\(\frac { 1 }{ 3 }\)rd length of the larger part = \(\frac { 1 }{ 3 }\)y
According to the second condition,
2x = \(\frac { 1 }{ 3 }\) 3
∴ 6x = y
∴ 6x – y = 0 ……(ii)
Adding equations (i) and (ii),
Maharashtra Board Class 9 Maths Solutions Chapter 5 Linear Equations in Two Variables Practice Set 5.2 8
∴ x = 80
Substituting x = 80 in equation (ii),
6x – y = 0
∴ 6(80) – y = 0
∴ 480 – y = 0
∴ y = 480
∴ The length of the larger part of the rope is 480 cm.

Question 10.
In a competitive examination, there were 60 questions. The correct answer would carry 2 marks, and for incorrect answer 1 mark would be subtracted. Yashwant had attempted all the questions and he got total 90 marks. Then how many questions he got wrong?
Solution:
Let us suppose that Yashwant got ‘x’ questions right and ‘y’ questions wrong.
According to the first condition, total number of questions in the examination are 60.
∴ x + y = 60 …(i)
Yashwant got 2 marks for each correct answer and 1 mark was deducted for each wrong answer.
∴ He got 2x – y marks.
According to the second condition,
he got 90 marks.
2x – y = 90 … (ii)
Adding equations (i) and (ii),
Maharashtra Board Class 9 Maths Solutions Chapter 5 Linear Equations in Two Variables Practice Set 5.2 8a
∴ x = 50
Substituting x = 50 in equation (i),
50 + y = 60
∴ y = 60 – 50 = 10
∴ Yashwant got 10 questions wrong.

Question 1.
The population of a certain town was 50,000. In a year, male population was increased by 5% and female population was increased by 3%. Now the populatin
became 52020. Then what was the number of males and females in the previous year? (Textbook pg. no. 89)
Solution:
Step 1: Read the given word problem carefully and try to understand it.

Step 2: Make assumptions using two variables x and y.
Let the number of males in previous year be
‘x’ and the number of females be ‘y’.

Step 3: From the given information, form mathematical statements using the above variables.
According to the first condition,
the total population of town was 50,000.
∴ x + y = 50000 …(i)
Male population increased by 5%.
∴ Number of males = x + 5% of x , 5
Maharashtra Board Class 9 Maths Solutions Chapter 5 Linear Equations in Two Variables Practice Set 5.2 9
Female population increased by 3%.
∴ Number of females = y + 3% of y
Maharashtra Board Class 9 Maths Solutions Chapter 5 Linear Equations in Two Variables Practice Set 5.2 9a
According to the second condition,
in a year population became 52020
∴ \(\frac{105}{100} x+\frac{103}{100} y=52020\)
∴ 105 x + 103 y = 5202000 …(ii)
Multiplying equation (i) by 103,
103 x + 103 y = 5150000 …(iii)

Step 4: Here, we use elimination method.
Subtracting equation (iii) from (ii),
Maharashtra Board Class 9 Maths Solutions Chapter 5 Linear Equations in Two Variables Practice Set 5.2 9b
∴ x = 26000
Substituting x = 26000 in equation (i),
∴ 26000 + y = 50000
∴ y = 50000 – 26000
∴ y = 24000
∴ Number of males = x = 26000
∴ Number of females = y = 24000

Step 5: Write the answer.
The number of males and females in the previous year were 26,000 and 24,000 respectively.

Step 6: Verify your result using smart check.

Maharashtra Board Class 9 Maths Solutions

 

Tamilnadu Board Class 10 English Solutions Poem Chapter 1 Life

Tamilnadu State Board Class 10 English Solutions Poem Chapter 1 Life

Life Textual Questions

A. Read the following lines from the poem and answer the questions that follows.

Life Poem by Henry Van Dyke Line by Line Explanation Question 1. Let me but live my life from year to year,
With forward face and unreluctant soul;
(a) Whom does the word ‘me’ refer to?
(b) What kind of life does the poet want to lead?
Answer:
(a) ‘Me’ refers to the poet, Henry Van Dyke.
(b) The poet wants to live a joyous life with plenty of positivity towards the future.

2. Not hurrying to, nor turning from the goal;
Not mourning for the things that disappear
(a) Why do you think the poet is not in a hurry?
(b) What should one not mourn for?
Answer:
(a) The poet wants to move towards his goal without hurrying or turning away from it.
(b) One must not mourn for the things lost in the past.

Life Poem Figure of Speech by Henry Van Dyke Question 3. In the dim past, nor holding back in fear
From what the future veils; but with a whole
And happy heart, that pays its toll
To Youth and Age, and travels on with cheer.
(a) What does the poet mean by the phrase ‘in the dim past’?
(b) Is the poet afraid of future?
(c) How can one travel on with cheer?
Answer:
(a) ‘In the dim past’ means the sad days of the past.
(b) No, the poet is not afraid of future.
(c) One can travel on with cheer by retaining the pleasures of childhood.

So Let the Way Wind Up the Hill or Down Figure of Speech Question 4. So let the way wind up the hill or down,
O’er rough or smooth, the journey will be joy:
Still seeking what I sought when but a boy,
New friendship, high adventure, and a crown
(a) How is the way of life?
(b) How should be the journey of life?
(c) What did the poet seek as a boy?
Answer:
(a) The way of life could go up the hill or down, rough or smooth.
(b) The journey of life must be joyful.
(c) The poet seeks new friends and high adventure.

English Poems about Life Question 5. My heart will keep the courage of the quest,
And hope the road’s last turn will be the best.
(a) What kind of quest does the poet seek here?
(b) What is the poet’s hope?
Answer:
(a) The quest is to seek a purposeful life with courage and determination.
(b) The poet hopes for a beautiful life with a clear sense of purpose.

6. In the dim past, nor holding back in fear
From what the future veils; but with a whole
And happy heart, that pays its toll
To Youth and Age, and travels on with cheer.
(a) Identify the rhyming words of the given lines.
Answer:
fear – cheer; whole – toll.

Let Me But Live My Life from Year to Year Poem Question 7. Let me but live my life from year to year,
With forward face and unreluctant soul;
Not hurrying to, nor turning from the goal;
Not mourning for the things that disappear
(a) Identify the rhyme scheme of the given lines.
Answer:
The rhyming scheme: a b b a.

Poetic Comprehension – Additional

Poem on Life In English Question 1. Let me but live my life from year to year,
With forward face and unreluctant soul;
Not hurrying to, nor turning from the goal;
Not mourning for the things that disappear
(a) How does the poet want to live his life?
(b) What does he say about past events?
Answer:
(a) He wants to live his life happily with forwarding momentum and positivity.
(b) He tells us not to worry about the past things that disappeared.

Poetic Devices – Additional

English Poem about Life Question 1. With forward face and unreluctant soul;
Not hurrying to, nor turning from the goal;
(a) What literary device is used here?
(b) Pick out the alliterated words in these lines.
Answer:
(a) Couplet is used in this poem. A couplet is two lines of verse that are joined by a rhyme.
(b) The alliterated words: Forward – face; (1st line), Not-nor; (2nd line)

Poem about Life Journey Question 2. “So let the way wind up the hill or down”
(a) What is the figure of speech used here?
Answer:
Personification. The way is personified as a human. It takes us up the hill or down.

3. “My heart will keep the courage of the quest,
And hope the road’s last turn will be the best”.
(a) What is the literary device used here?
Answer:
Couplet is the literary device used here.

B. Answer the following question in about 80-100 words.

1. Describe the journey of life as depicted in the poem by Henry Van Dyke.
Answer:
In this poem, life is described not as an entity, but as an experience. One should live with courage and dedication. Life should be lived without hurry and with a clear sense of purpose that drives the mind and soul. The poet encourages us to let go of all that has been lost in the past as well as the uncertainty the future holds. He tells us to embrace the present with the happiness which nourishes the young and the old. Happiness gives us nourishment on this journey with a smile on our face. Whatever situation life throws at us, it is the journey that should be joyous, for it teaches us to grow and live. Our imagination should have the innocence and fearlessness of childhood. We should seek out new friendships, new adventures and new experiences which enrich us. He encourages us to have faith and determination in our hearts, as we take on this beautiful journey. We should have eternal hope that our story ends joyfully.

Paragraph Questions & Answers Additional

English Poetry on Life Question 1.
What is the message of the poem “Life”?
Answer:
Henry Van Dyke was a visionary American author. His poem “Life” describes life taking on life in its truest form, an adventure. The poet wants to live his life looking ahead, willing to do something. He neither wants to hurry nor move away from his goal. He does not want to mourn the things he has lost, not hold back for fear of the future. He instead prefers to live his life with a whole and happy heart which cheerfully travels from youth to old age. Therefore, it does not matter to him whether the path goes up or down the hill, rough or smooth, the journey will be joyful. He will continue to seek what he wanted as a boy – new friendship, high adventure and a crown (prize). His heart will remain courageous and pursue his desires. He hopes that every turn in his life’s journey will be the best.

Question 2.
What does the poet narrate to us in this poem?
Answer:
In this poem, the poet narrates to us about how he would want to look forward with a happy and cheerful mind without worrying too much about the future. He thinks that he should live life to the fullest by retaining the child-like innocence and pleasures of life. Finally, he hopes that his life will be meaningful and that better things will happen to him in the future which lies ahead of him.

English Poem on Life Question 3.
What are the main concepts and ideas in the poem ‘Life’?
Answer:
The poet wants to live his life looking ahead, willing to do something. He neither wants to hurry nor move away from his goal. He does not want to mourn the things he has lost, not hold back for fear of the future. He instead prefers to live his life with a whole and happy heart which cheerfully travels from youth to old age. Therefore, it does not matter to him whether the path goes up or down the hill, rough or smooth, the journey will be joyful. He will continue to seek what he wanted as a boy – new friendship, (vii) high adventure and a crown (prize). His heart will remain courageous and pursue his desires. He hopes that every turn in his life’s journey will be the best.

C. Based on your understanding of the poem, complete the following passage by using the phrases given in the box.

youth to old age up or down the hill to hurry nor move away
high adventure joyful mourn looking ahead

The poet wants to live his life (i) ________, willing to do something. He neither wants (ii) ________ from his goal. He does not want to (iii) ________ the things he has lost, not hold back for fear of the future. He instead prefers to live his life with a whole and happy heart which cheerfully travels from (iv) ________. Therefore, it does not matter to him whether the path goes (v)________, rough or smooth, the journey will be (vi)________. He will continue to seek what he wanted as a boy – new friendship, (vii) _______ ________ and a crown (prize). His heart will remain courageous and pursue his desires. He hopes that every turn in his life’s journey will be the best.
Answer:
(i) looking ahead
(ii) to hurry nor move away
(iii) mourn
(iv) youth to old age
(v) up or down the hill
(vi) joyful
(vii) high adventure

Life By Henry Van Dyke

Henry Van Dyke (1852 – 1933) was born in Pennsylvania, USA. A nature lover and avid reader, he earned degrees from Princeton, then served as a Presbyterian minister for more than 20 years. (He was considered one of the best preachers in New York City). He eventually returned to Princeton, where he spent nearly 20 years as a professor of English, with a bit of service as the U.S. Ambassador to Luxembourg and the Netherlands in between. A writer whose talent extended to many different genres, Henry’s best-known works are probably the lyrics of the hymn “Joyful, Joyful, We Adore Thee” and the two Christmas stories, “The Other Wise Man” and “The First Christmas Tree”.

Henry Van Dyke is a visionary American author. His poem “Life” describes life taking on life in its truest form, an adventure. This poem is beautiful and inspiring but also idealistic. One cannot help but be charmed by his childlike hope and absolute faith in the abilities of a warm heart and an able mind.

Life Key points

  • Life is an experience.
  • To be lived with courage.
  • One should not worry about the uncertain future.
  • Happiness nourishes life with extra energy.
  • Imaginations to be fearless and pure.
  • New friendships, new adventures, new explorations to enrich us.
  • To always hope for a joyous future with determination and faith.

Life Summary

In this poem, life is described not as an entity, but as an experience. One should live with courage and dedication. Life should be lived without hurry and with a clear sense of purpose that drives the mind and soul. The poet encourages us to let go of all that has been lost in the past as well as the uncertainty the future holds. He tells us to embrace the present with the happiness which nourishes the young and the old. Happiness gives us. nourishment on this journey with a smile on our face.

Whatever situation life throws at us, it is the journey that should be joyous, for it teaches us to grow and live. Our imagination should have the innocence and fearlessness of childhood. We should seek out new friendships, new adventures and new experiences which enrich us. He encourages us to have faith and determination in our hearts, as we take on this beautiful journey. We should have eternal hope that our story ends joyfully.

Life Explanation of Poetic Lines

Line No. 1 – 2
Let me but live my life from year to year,
With forward face and unreluctant soul;
Explanation:
The poet is giving advice to the readers from his own personal experiences. The poet wants to live his life happily as it comes with forward momentum and optimism. He wants to live his life every year with a happy heart.

Line No. 3 – 4
Not hurrying to, nor turning from the goal;
Not mourning for the things that disappear
Explanation:
The poet does not like to hurry in any matters or situations. He doesn’t want to turn away from his aims. What he wanted to achieve, he will do it with confidence and hope. He does not like to feel sad and keep on worrying about the things that have passed away.

Line No. 5 – 6
In the dim past, nor holding back in
fear From what the future veils;
Explanation:
He doesn’t want to think about his dull past and hold back in fear about the uncertainty of his future. He wanted to look forward with a happy and cheerful mind. He doesn’t want to worry too much about his future.

Line No. 6 – 8
but with a whole.
And happy heart, that pays its toll
To Youth and Age, and travels on with
cheer.
Explanation:
He thinks that he should live life to the fullest by retaining the child-like innocence and pleasures of life. He likes to travel his journey of life cheerfully.

Line No. 9 -10
So let the way wind up the hill or down,
O’er rough or smooth, the journey will be joy:
Explanation:
Though his way goes up in the path of a mountain or down the valley, he would enjoy his journey. No matter whatever happens if his path is difficult or easy, he will make it a joyful journey.

Line No. 11 – 12
Still seeking what I sought when but a boy,
New friendship, high adventure, and a crown,
Explanation:
He is searching to find what he was searching for when he was a boy – his new adventures, his hopes, his new friendship and his new experiences which enrich him.

Line No. 13 – 14
My heart will keep the courage of the quest,
And hope the road’s last turn will be the best.
Explanation:
Life is described not as an entity but as an experience. One should live with courage, dedication and a clear sense of purpose that drives the mind and soul. The poet encourages us to have faith and determination in our hearts, as we take on this beautiful journey.

Life Glossary

crown (n) – a prize or position offered for being the best
dim – dark
goal – aim
mourning (v) – feeling or expressing great sadness
quest (n) – a long search for something that is difficult to find
seeking – hunting
sought – desired
toll – payment
unreluctant (adj.) – willing to do something (This form is generally not used but the poet has coined it for emphasis)
veils (v) – to hide or cover something so that you cannot see it clearly or understand it
wind – curve

Tamilnadu Board Class 10 English Solutions

Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Problem Set 4B

Maharashtra State Board Class 10 Maths Solutions Chapter 4 Financial Planning Problem Set 4B

Financial Planning Class 10 Problem Set 4b Question 1. Write the correct alternative for the following questions.

i. If the Face Value of a share is ₹ 100 and Market value is ₹ 75, then which of the following statement is correct?
(A) The share is at premium of ₹ 175
(B) The share is at discount of ₹ 25
(C) The share is at premium of ₹ 25
(D) The share is at discount of ₹ 75
Answer:
(B)

ii. What is the amount of dividend received per share of face value ₹ 10 if dividend declared is 50%.
(A) ₹ 50
(B) ₹ 5
(C) ₹ 500
(D) ₹ 100
Answer:
Dividend = 10 × \(\frac { 50 }{ 100 } \) = ₹ 5
(B)

iii. The NAV of a unit in mutual fund scheme is ₹ 10.65, then find the amount required to buy 500 such units.
(A) 5325
(B) 5235
(C) 532500
(D) 53250
Answer:
(A)

iv. Rate of GST on brokerage is _______
(A) 5%
(B) 12%
(C) 18%
(D) 28%
Answer:
(C)

v. To find the cost of one share at the time of buying the amount of Brokerage and GST is to be ______ MV of share.
(A) added to
(B) subtracted from
(C) Multiplied with
(D) divided by
Answer:
(A)

Problem Set 4b Algebra Class 10 Question 2. Find the purchase price of a share of FV ₹ 100 if it is at premium of ₹ 30. The brokerage rate is 0.3%.
Solution:
Here, Face Value of share = ₹ 100,
premium = ₹ 30, brokerage = 0.3%
MV = FV + Premium
= 100 + 30
= ₹ 130
Brokerage = 0.3% of MV
= \(\frac { 0.3 }{ 100 } \) × 130 = ₹ 0.39
Purchase price of a share = MV + Brokerage
= 130 + 0.39
= ₹ 130.39
Purchase price of a share is ₹ 130.39.

Question 3.
Prashant bought 50 shares of FV ₹ 100, having MV ₹ 180. Company gave 40% dividend on the shares. Find the rate of return on investment.
Solution:
Here, Number of shares = 50, FV = ₹ 100,
MV = ₹ 180, rate of dividend = 40%
∴ Sum invested = Number of shares × MV
= 50 × 180
= ₹ 9000
Dividend per share = 40% of FV
= \(\frac { 40 }{ 100 } \) × 100
Dividend = ₹ 40
∴ Total dividend on 50 shares = 50 × 40
= ₹ 2000
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Problem Set 4B 1
∴ Rate of return on investment is 22.2%.

Question 4.
Find the amount received when 300 shares of FV ₹ 100, were sold at a discount of ₹ 30.
Solution:
Here, FV = ₹ 100, number of shares = 300,
discount = ₹ 30
MV of 1 share = FV – Discount
= 100 – 30 = ₹ 70
∴ MV of 300 shares = 300 × 70
= ₹ 21,000
∴ Amount received is ₹ 21,000.

Question 5.
Find the number of shares received when ₹ 60,000 was invested in the shares of FV ₹ 100 and MV ₹ 120.
Solution:
Here, FV = ₹ 100, MV = ₹ 120,
Sum invested = ₹ 60,000
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Problem Set 4B 2
∴ Number of shares received were 500.

Question 6.
Smt. Mita Agrawal invested ₹ 10,200 when MV of the share is ₹ 100. She sold 60 shares when the MV was ₹ 125 and sold remaining shares when the MV was ₹ 90. She paid 0.1% brokerage for each trading. Find whether she made profit or loss? and how much?
Solution:
For purchasing shares:
Here, sum invested = ₹ 10,200, MV = ₹ 100
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Problem Set 4B 3
For selling shares:
60 shares sold at MV of ₹ 125.
∴ MV of 60 shares = 125 × 60
= ₹ 7500
Brokerage = \(\frac { 0.1 }{ 100 } \) × 7500 = ₹ 7.5
∴ Sale value of 60 shares = 7500 – 7.5 = ₹ 7492.5
Now, remaining shares = 102 – 60 = 42
But 42 shares sold at MV of ₹ 90.
∴ MV of 42 shares = 42 × 90 = ₹ 3780
∴ Brokerage = \(\frac { 0.1 }{ 100 } \) × 3780 = ₹ 3.78
∴ Sale value of 42 shares = 3780 – 3.78 = ₹ 3776.22
Total sale value = 7492.5 + 3776.22 = ₹ 11268.72
Since, Purchase value < Sale value
∴ Profit is gained.
∴ Profit = Sale value – Purchase value
= 11268.72 – 10210.2
= ₹ 1058.52
∴ Smt. Mita Agrawal gained a profit of ₹ 1058.52.

Problem Set 4b Algebra Question 7. Market value of shares and dividend declared by the two companies is given below.
Face value is same and it is 7 100 for both the shares. Investment in which company is more profitable?
i. Company A – ₹ 132,12%
ii Company B – ₹ 144,16%
Solution:
For company A:
FV = ₹ 100, MV = ₹ 132,
Rate of dividend = 12%
Dividend = 12% of FV
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Problem Set 4B 4
∴ Rate of return of company B is more.
∴ Investment in company B is more profitable.

10th Algebra Problem Set 4b Question 8. Shri. Aditya Sanghavi invested ₹ 50,118 in shares of FV ₹ 100, when the market value is ₹ 50. Rate of brokerage is 0.2% and Rate of GST on brokerage is 18%, then How many shares were purchased for ₹ 50,118?
Solution:
Here, FV = ₹ 100, MV = ₹ 50
Purchase value of shares = ₹ 50118,
Rate of brokerage = 0.2%, Rate of GST = 18%
Brokerage = 0.2% of MV
Financial Planning Class 9 Problem Set
Financial Planning Problems
∴ 1000 shares were purchased for ₹ 50,118.

Financial Planning Problem Set 4b Question 9. Shri. Batliwala sold shares of ₹ 30,350 and purchased shares of ₹ 69,650 in a day. He paid brokerage at the rate of 0.1% on sale and purchase. 18% GST was charged on brokerage. Find his total expenditure on brokerage and tax.
Solution:
Total amount = sale value + Purchase value
= 30350 + 69650
= ₹ 1,00,000
Rate of Brokerage = 0.1 %
Brokerage = 0.1 % of 1,00,000
= \(\frac { 0.1 }{ 100 } \) × 1,00,000
= ₹ 100
Rate of GST = 18%
∴ GST = 18 % of brokerage
= \(\frac { 18 }{ 100 } \) × 100
∴ GST = ₹ 18
Total expenditure on brokerage and tax
= 100 + 18 = ₹ 118
∴ Total expenditure on brokerage and tax is ₹ 118.

Alternate Method:
Brokerage = 0.1 %, GST = 18%
At the time of selling shares:
Total sale amount of shares = ₹ 30,350
Brokerage = 0.1% of 30,350
Financial Planning Problems and Solutions
For purchasing shares:
Total purchase amount of shares = ₹ 69,650
Brokerage = 0.1% of 69,650
= \(\frac { 0.1 }{ 100 } \) × 69650
= ₹ 69.65
GST = 18% of 69.65
= \(\frac { 18 }{ 100 } \) × 69.65
= ₹ 12.537
∴ Total expenditure on brokerage and tax = Brokerage and tax on selling + Brokerage and tax on purchasing
= (30.35 + 5.463) + (69.65 + 12.537)
= ₹ 118
∴ Total expenditure on brokerage and tax is ₹ 118.

Problem Set 4 Algebra 10th Question 10. Sint. Aruna Thakkar purchased 100 shares of FV 100 when the MV is ₹ 1200. She paid brokerage at the rate of 0.3% and 18% GST on brokerage. Find the following –
i. Net amount paid for 100 shares.
ii. Brokerage paid on sum invested.
iii. GST paid on brokerage.
iv. Total amount paid for 100 shares.
Solution:
Here, FV = ₹ 100,
Number of shares = 100, MV = ₹ 1200
Brokerage = 0.3%, GST = 18%
i. Sum invested = Number of shares × MV
= 100 × 1200 = ₹ 1,20,000
∴ Net amount paid for 100 shares is ₹ 1,20,000.

ii. Brokerage = 0.3% of sum invested
= \(\frac { 0.3 }{ 100 } \) × 1,20,000 = ₹ 360
∴ Brokerage paid on sum invested is ₹ 360.

iii. GST = 18% of brokerage
= \(\frac { 18 }{ 100 } \) × 360 = ₹ 64.80
∴ GST paid on brokerage is ₹ 64.80.

iv. Total amount paid for 100 shares
= Sum invested + Brokerage + GST
= 1,20,000 + 360 + 64.80
= ₹ 1,20,424.80
∴ Total amount paid for 100 shares is ₹ 1,20,424.80.

Algebra Financial Planning Question 11. Smt. Anagha Doshi purchased 22 shares of FV ₹ 100 for Market Value of ₹ 660. Find the sum invested. After taking 20% dividend, she sold all the shares when market value was ₹ 650. She paid 0.1% brokerage for each trading done. Find the percent of profit or loss in the share trading. (Write your answer to the nearest integer)
Solution:
For purchasing shares:
Here, FV = ₹ 100, MV = ₹ 660, Number of shares = 22, rate of brokerage = 0.1%
Sum invested = MV × Number of shares
= 660 × 22
= ₹ 14,520
Brokerage = 0.1 % of sum invested
= \(\frac { 0.1 }{ 100 } \) × 14520 = ₹ 14.52
∴ Amount invested for 22 shares
= Sum invested + Brokerage
= 14520 + 14.52
= ₹ 14534.52
For dividend:
Rate of dividend = 20%
∴ Dividend per share = 20 % of FV
Financial Planning Class
∴ Percentage of profit in the share trading is 1 % (nearest integer).

Alternate Method:
For purchasing share:
Here, FV = ₹ 100, MV = ₹ 660, Number of shares = 22, rate of brokerage = 0.1%
Sum invested = MV × Number of shares
= 660 × 22
= ₹ 14,520
Brokerage = 0.1 % of MV
= \(\frac { 0.1 }{ 100 } \) × 660 = ₹ 0.66
Amount invested for 1 share = 660 + 0.66
= ₹ 660.66
For dividend:
Rate of dividend = 20%
Dividend = 20% of FV = \(\frac { 20 }{ 100 } \) × 100 = ₹ 20
For selling share:
MV = ₹ 650, rate of brokerage = 0.1%
Brokerage = 0.1 % of MV
= \(\frac { 0.1 }{ 100 } \) × 650 = ₹ 0.65 100
Amount received after selling 1 share
= 650 – 0.65 = 649.35
∴ Amount received including divided
= selling price of 1 share + dividend per share
= 649.35 + 20
= ₹ 669.35
Since, income > Amount invested
∴ Profit is gained.
∴ profit = 669.35 – 660.66 = ₹ 8.69
Profit Percentage = \(\frac { 8.69 }{ 660.66 } \) × 100= 1.31%
∴ Percentage of profit in the share trading is 1 % (nearest integer).

Maharashtra Board Class 10 Maths Solutions

Maharashtra Board Class 9 Maths Solutions Chapter 4 Ratio and Proportion Practice Set 4.2

Maharashtra Board Class 9 Maths Solutions Chapter 4 Ratio and Proportion Practice Set 4.2

Ratio and Proportion 9th Class Practice Set 4.2 Question 1. Using the property \(\frac { a }{ b }\) = \(\frac { ak }{ bk }\), fill in the blanks by substituting proper numbers in the following.
Maharashtra Board Class 9 Maths Solutions Chapter 4 Ratio and Proportion Practice Set 4.2 1
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 4 Ratio and Proportion Practice Set 4.2 1a
Maharashtra Board Class 9 Maths Solutions Chapter 4 Ratio and Proportion Practice Set 4.2 1b

Practice Set 4.2 Algebra 9th Question 2. Find the following ratios.
i. The ratio of radius to circumference of the circle.
ii. The ratio of circumference of circle with radius r to its area.
iii. The ratio of diagonal of a square to its side, if the length of side is 7 cm.
iv. The lengths of sides of a rectangle are 5 cm and 3.5 cm. Find the ratio of numbers denoting its perimeter to area.
Solution:
i. Let the radius of circle be r.
then, its circumference = 2πr
Ratio of radius to circumference of the circle
Maharashtra Board Class 9 Maths Solutions Chapter 4 Ratio and Proportion Practice Set 4.2 2
The ratio of radius to circumference of the circle is 1 : 2π.

ii. Let the radius of the circle is r.
∴ circumference = 2πr and area = πr2
Ratio of circumference to the area of circle
Maharashtra Board Class 9 Maths Solutions Chapter 4 Ratio and Proportion Practice Set 4.2 2a
∴ The ratio of circumference of circle with radius r to its area is 2 : r.

iii. Length of side of square = 7 cm
∴ Diagonal of square = √2 x side
= √2 x 7
= 7 √2 cm
Ratio of diagonal of a square to its side
Maharashtra Board Class 9 Maths Solutions Chapter 4 Ratio and Proportion Practice Set 4.2 2b
∴ The ratio of diagonal of a square to its side is √2 : 1.

iv. Length of rectangle = (l) = 5 cm,
Breadth of rectangle = (b) = 3.5 cm
Perimeter of the rectangle = 2(l + b)
= 2(5 + 3.5)
= 2 x 8.5
= 17 cm
Area of the rectangle = l x b
= 5 x 3.5
= 17.5 cm2
Ratio of numbers denoting perimeter to the area of rectangle
Maharashtra Board Class 9 Maths Solutions Chapter 4 Ratio and Proportion Practice Set 4.2 2c
∴ Ratio of numbers denoting perimeter to the area of rectangle is 34 : 35.

Ratio and Proportion Practice Set 4.2 Question 3. Compare the following
Maharashtra Board Class 9 Maths Solutions Chapter 4 Ratio and Proportion Practice Set 4.2 3
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 4 Ratio and Proportion Practice Set 4.2 3a
If The Product Of Two Numbers Is 360 and Their Ratio Is 10 Is To 9 Then Find The Numbers
Ratio and Proportion Class 9 Practice Set 4.1
Maharashtra Board Class 9 Maths Solutions Chapter 4 Ratio and Proportion Practice Set 4.2 3d

Ratio and Proportion Class 9 Practice Set 4.2 Question 4. Solve.
ABCD is a parallelogram. The ratio of ∠A and ∠B of this parallelogram is 5 : 4. FInd the measure of ∠B. [2 Marksl
Solution:
Ratio of ∠A and ∠B for given parallelogram is 5 : 4
Let the common multiple be x.
9th Class Algebra Practice Set 4.2
m∠A = 5x°and m∠B=4x°
Now, m∠A + m∠B = 180° …[Adjacent angles of a parallelogram arc supplementary]
∴ 5x° + 4x°= 180°
∴ 9x° = 180°
∴ x° = 20°
∴ m∠B=4x°= 4 x 20° = 80°
∴ The measure of ∠B is 800.

ii. The ratio of present ages of Albert and Salim is 5 : 9. Five years hence ratio of their ages will be 3 : 5. Find their present ages.
Solution:
The ratio of present ages of Albert and Salim is 5 : 9
Let the common multiple be x.
∴ Present age of Albert = 5x years and
Present age of Salim = 9x years
After 5 years,
Albert’s age = (5x + 5) years and
Salim’s age = (9x + 5) years
According to the given condition,
Five years hence ratio of their ages will be 3 : 5
\(\frac{5 x+5}{9 x+5}=\frac{3}{5}\)
∴ 5(5x + 5) = 3(9x + 5)
∴ 25x + 25 = 27x + 15
∴ 25 – 15 = 27 x – 25 x
∴ 10 = 2x
∴ x = 5
∴ Present age of Albert = 5x = 5 x 5 = 25 years
Present age of Salim = 9x = 9 x 5 = 45 years
∴ The present ages of Albert and Salim are 25 years and 45 years respectively.

iii. The ratio of length and breadth of a rectangle is 3 : 1, and its perimeter is 36 cm. Find the length and breadth of the rectangle.
Solution:
The ratio of length and breadth of a rectangle is 3 : 1
Let the common multiple be x.
Length of the rectangle (l) = 3x cm
and Breadth of the rectangle (b) = x cm
Given, perimeter of the rectangle = 36 cm
Since, Perimeter of the rectangle = 2(l + b)
∴ 36 = 2(3x + x)
∴ 36 = 2(4x)
∴ 36 = 8x
∴ \(x=\frac{36}{8}=\frac{9}{2}=4.5\)
Length of the rectangle = 3x = 3 x 4.5 = 13.5 cm
∴ The length of the rectangle is 13.5 cm and its breadth is 4.5 cm.

iv. The ratio of two numbers is 31 : 23 and their sum is 216. Find these numbers.
Solution:
The ratio of two numbers is 31 : 23
Let the common multiple be x.
∴ First number = 31x and
Second number = 23x
According to the given condition,
Sum of the numbers is 216
∴ 31x + 23x = 216
∴ 54x = 216
∴ x = 4
∴ First number = 31x = 31 x 4 = 124
Second number = 23x = 23 x 4 = 92
∴ The two numbers are 124 and 92.

v. If the product of two numbers is 360 and their ratio is 10 : 9, then find the numbers.
Solution:
Ratio of two numbers is 10 : 9
Let the common multiple be x.
∴ First number = 10x and
Second number = 9x
According to the given condition,
Product of two numbers is 360
∴ (10x) (9x) = 360
∴ 90x2 = 360
∴ x2 = 4
∴ x = 2 …. [Taking positive square root on both sides]
∴ First number = 10x = 10x2 = 20
Second number = 9x = 9x2 = 18
∴ The two numbers are 20 and 18.

9th Class Maths Part 1 Practice Set 4.2 Question 5. If a : b = 3 : 1 and b : c = 5 : 1, then find the value of [3 Marks each]
Maharashtra Board Class 9 Maths Solutions Chapter 4 Ratio and Proportion Practice Set 4.2 5
Solution:
Given, a : b = 3 : 1
∴ \(\frac { a }{ b }\) = \(\frac { 3 }{ 1 }\)
∴ a = 3b ….(i)
and b : c = 5 : 1
∴ \(\frac { b }{ c }\) = \(\frac { 5 }{ 1 }\)
b = 5c …..(ii)
Substituting (ii) in (i),
we get a = 3(5c)
∴ a = 15c …(iii)
Ratio and Proportion Class 9 Practice Set 4.3

Ratio and Proportion 9th Class Practice Set 4.1 Question 6. If \(\sqrt{0.04 \times 0.4 \times a}=0.4 \times 0.04 \times \sqrt{b}\) , then find the ratio \(\frac { a }{ b }\).
Solution:
\(\sqrt{0.04 \times 0.4 \times a}=0.4 \times 0.04 \times \sqrt{b}\) … [Given]
∴ 0.04 x 0.4 x a = (0.4)2 x (0.04)2 x b … [Squaring both sides]
Ratio and Proportion Practice Set 4.1

9th Algebra Practice Set 4.2 Question 7. (x + 3) : (x + 11) = (x – 2) : (x + 1), then find the value of x.
Solution:
(x + 3) : (x + 11) = (x- 2) : (x+ 1)
\(\quad \frac{x+3}{x+11}=\frac{x-2}{x+1}\)
∴ (x + 3)(x +1) = (x – 2)(x + 11)
∴ x(x +1) + 3(x + 1) = x(x + 11) – 2(x + 11)
∴ x2 + x + 3x + 3 = x2 + 1 lx – 2x – 22
∴ x2 + 4x + 3 = x2 + 9x – 22
∴ 4x + 3 = 9x – 22
∴ 3 + 22 = 9x – 4x
∴ 25 = 5x
∴ x = 5

Maharashtra Board Class 9 Maths Solutions

Tamilnadu Board Class 9 English Solutions Prose Chapter 6 From Zero to Infinity

Tamilnadu State Board Class 9 English Solutions Prose Chapter 6 From Zero to Infinity

From Zero to Infinity In-Text Questions

From Zero to Infinity Summary Question. What was the reaction of the classmates to Ramanujan’s question?
Answer:
The classmates laughed at Ramanujan’s question.

From Zero to Infinity Mind Map Question. What did the Indian mathematician Bhaskara prove?
Answer:
The Indian mathematician Bhaskara proved that zero divided by zero is infinity.

Question.
Where did Ramanujan get S.L. Loney’s book on Trigonometry?
Answer:
Ramanujan got Loney’s “Trigonometry” book from a college library.

From Zero to Infinity 9th Standard English Question. Where did Ramanujan do his mathematical problems?
Answer:
Ramanujan did his mathematical problems on loose sheets of paper or on a slate.

Question.
What were the subjects neglected by Ramanujan in college?
Answer:
History, English, Physiology were the subjects neglected by Ramanujan in college.

From Zero to Infinity Biography of Srinivasa Ramanujan Question. Which University granted him a fellowship of H75 a month?
Answer:
University of Madras granted him a fellowship of? 75 a month.

From Zero to Infinity Lesson Plan Question. What did Ramanujan send to G.H. Hardy?
Answer:
Ramanujan sent a letter in which he set out 120 theorems and formulae to G.H Hardy.

Question.
Who discovered a rare mathematical genius in Ramanujan?
Answer:
G.H Hardy and his colleague J.E. Littlewood discovered a rare mathematical genius in Ramanujan.

A. Answer the following questions in a sentence or two.

From Zero to Infinity Lesson Summary Question 1.
Why did the students laugh at Ramanujan?
Answer:
The students laughed at Ramanujan because he asked if no banana was distributed among no one, would every one get one banana.

Question 2.
Why did the teacher compliment Ramanujan?
Answer:
The teacher complimented Ramanuj for asking a question that took centuries for mathematicians to answer.

Question 3.
Question What did Ramanujan do after reading the book on Trigonometry?
Answer:
After reading the book on Trigonometry, Ramanujan began his own research. He came forth with many mathematical theorems and formulae not given in the book

Question 4.
What disappointed Ramanujan’s father?
Answer:
Ramanujan failed twice in his first year arts examination in college as he neglected other subjects such as History, English and Physiology. This disappointed his father.

Question 5.
How did Ramanujan manage his paper crisis?
Answer:
Ramanujan needed about 2,000 sheets of paper every month. He started using even scraps of paper he found lying on the streets. Sometimes he used a red pen to write over what was written in blue ink.

Question 6.
Why were Ramanujan’s application for jobs rejected?
Answer:
Ramanujan would show his frayed notebooks to every officers. But no one could understand what was written in the notebooks. So, his applications for jobs were rejected.

Question 7.
Why was Ramanujan sent back to India?
Answer:
While Ramanujan continued his research work, Tuberculosis, then an incurable disease, was devouring him. So, he was sent back to India.

Short Questions and Answers : Additional

Question 1.
What was the teacher doing?
Answer:
The teacher was solving questions on division. He drew three bananas on the blackboard and started to solve the problems.

Question 2.
What was the most significant turn In Ramanujam’s life?
Answer:
The most significant turn came two years later, when one of his senior friends showed him Synopsis of Elementary Results in Pure Applied Mathematics by George Shoo bridge Carr. This book triggered the mathematical genius in him.

Question 3.
What did Ramanujam do, before he went abroad? .
Answer:
Before Ramanujan went abroad, he had filled three notebooks, which later became famous as ‘Ramanujan’s Frayed Notebooks’.

Question 4.
Why did Ramanujan’s father think that his son was mad?
Ans :
When Ramanujan’s father found his son always scribbling numbers and not doing much else, he thought Ramanujan had gone mad.

Question 5.
Who was impressed by his notebooks?
Answer:
The Director of Madras Port Trust, Francis Spring was impressed by his
notebooks and gave him a clerical job on a monthly salary of? 25.

Question 6.
How did Ramanujan find himself at Cambridge?
Answer:
Ramanujan found himself a stranger at Cambridge. The cold was hard to bear and being a vegetarian, he had to cook his own food. However, he continued his research in Mathematics with determination.

Question 7.
What did Ramanujan do to forget his agonising pain?
Answer:
Ramanujan continued to play with numbers even on his death-bed to forget his agonising pain.

Question 8.
What was Ramanujan, beside being a mathematician?
Answer:
Beside being a mathematician, Ramanujan was an astrologer of repute and a good speaker.

Question 9.
While at research work, what disease was devouring him?
Answer:
While Ramanujan continued his research work, Tuberculosis, then an incurable disease was devouring him.

Question 10.
On what subjects did Ramanujan used to give lectures?
Answer:
Ramanujan used to give lectures on subjects like ‘God, Zero and Infinity’.

B. Answer the following questions in about 80 -100 words.

Question 1.
Describe the life of Srinivasa Ramanujan in India.
Answer:
Ramanujan was born in Erode in Tamilnadu on December 22,1887. From early childhood, it was evident that he was a prodigy. Senior students used to get his assistance in solving math problems. At the age of 13, he began his own research on Trigonometry. The book “Elementary Results in Pure Applied Mathematics” by George Shoobridge Carr triggered the genius in Ramanujan. He used to do problems on loose sheets and enter the results in notebooks which are now famous as “Ramanujan’s Frayed Notebooks”.

Although Ramanujan secured a first class in Mathematics in the matriculation examination and was awarded the Subramanyan Scholarship, he failed twice in his first year arts examination in college as he neglected other subjects such as History, English and Physiology. He searched for job for food and papers to do calculations. The Director of Madras Port Trust gave a clerical job to Ramanujan on a monthly salary of Rupees 25.

Question 2.
Narrate the association of Ramanujan with G.H. Hardy.
Answer:
Ramanujan sent a letter to the great Mathematician G.H. Hardy of Cambridge University, in which he set out 120 theorems and formulae which included the Reimann Series. Hardy and his colleague Littlewood realized that they had discovered a rare mathematical genius. They invited him to Britain.Despite the cold weather and food, Ramanujan continued his research with determination in the company of Hardy and Littlewood. Hardy found an unsystematic mathematician in Ramanujan due to his lack of formal education. Ramanujan’s achievements include the Hardy-Ramanujan-Littlewood circle method in number theory.

Paragraph Questions and Answers : Additional

Question 1.
Narrate the incident, which took place in the class of Ramanujan.
Answer:
The arithmetic class was in progress. The teacher was solving questions on division. He drew three bananas on the black board and asked the students if they have three bananas and three boys, how many each will get. A smart boy in the front row replied that each one would get one. He then proceeded to ask if 1000 bananas are distributed among 1000 boys would each one get one banana. A boy sitting in one comer raised his hand and stood up. The teacher stopped his explanation and waited for the boy to speak. He asked the teacher if no banana is distributed among no one, will everyone still get one banana.

There was a roar of laughter in the class. The teacher asked the ‘ students to stop laughing and explained to the boys, what he was asking. It was if zero banana is divided among zero, would each one get one? The answer would be ‘no’. Mathematically, each would get an infinite number of bananas. The boy had asked a question that had taken mathematicians several centuries to answer.

Question 2.
Why were Ramanujan’s applications for Jobs turned down?
Answer:
Ramanujan began to look for a job, as he had to find money not only for food, but for papers as well to do his calculations. He needed about 2000 sheets of paper every month. He visited offices and showed everyone his frayed notebooks and told them that he knew mathematics and he could do a clerical job. He behaved in an unpleasant way and was not neatly dressed. No one could understand what was written in the notebooks and so his applications for jobs were turned down.

C. Match the words with correct Synonym and Antonym from the table.
Answer:
From Zero to Infinity Tamil

Listening

D. Listen to the anecdote “Two Geniuses” and narrate it in your own words.
Answer:

Narration of “Two Geniuses”

There’s a story about how Dr. Albert Einstein was travelling to Universities in his car, delivering lectures on his theory of relativity. During one tired journey, his driver Hans remarked “Dr. Einstein, I have heard you deliver that lecture about 30 times. ,t I know it by heart and bet I could give it myself.”

“Well, I’ll give you the chance”, said the Dr. “They don’t know me at the next University, so when we get there, I’ll put on your cap, and you introduce yourself as Dr. Einstein and give the lecture.”

The driver delivered Einstein’s lecture without any mistakes. When he finished, he started to leave, but one of the professor stopped him and asked a complex question filled with mathematical equations and formulae. The driver thought fast. “The answer to that problem is so simple,” he said,” I’m surprised you have to ask me. In fact, to show you just how simple it is, I’m going to ask my driver Hans to come up here and answer your question”.

Speaking

E. Divide the students into groups of five and conduct a group discussion on the topic “Importance of Mathematics in Our Everyday Life” The teacher will act as a moderator.
Answer:

Group Discussion on Importance of Mathematics

Teacher : Good morning students! We have just learnt the life of the great mathematician Ramanujan. Now let’s have a group discussion on “Importance of Mathematics in Our Everyday Life”. Divide yourselves into groups of five.

Harsha (Group A) : The importance of maths in everyday life. Mathematics is a methodical application of matter. It is so said because the subject makes a man methodical or systematic. Mathematics makes our life orderly and prevents chaos.

Varsha (Group B) : In Hebrew, it’s root is “thinking.” They tell us that mathematics gives us the critical ability to learn and think logically in any field of endeavor. The skills of learning today are more important than knowledge, which is so readily available on the Internet.

Yusuf (Group C) : Math is an important part of our lives, because in the future you will get a job that deals with math. Math is pretty much in everything you do, really. Math is important because it is the most widely used subject in the world. Every career uses some sort of math.

Adhira (Group D) : Maths improves problem-solving abilities. Teaches clearer logical reasoning. Sharpens concentration and observance. Develops confidence and self-esteem.

Danny (Group E) : Knowing basic math principles keeps you from having to carry around a calculator because good use of math allows you to do many calculations in your head.

Reading

F. Answer the following questions based on the given passage.

From Zero to Infinity Question 1. What made John Shepherd-Barron to come up with the idea of ATM?
Answer:
It was then John’s habit to withdraw money on a Saturday, but on this particular weekend he had arrived one minute late and found the bank doors locked against him. This made him to come up with idea of ATM.

Question 2.
When and where was the first ATM installed?
Answer:
The first ATM was installed at a branch in the North London suburb of Enfield on June 27, 1967.

Question 3.
Who was the first person to withdraw cash from the ATM?
Answer:
The first person to withdraw cash from the ATM was Reg Varney, a celebrity resident of Enfield known for his part in the number of popular television series.

Question 4.
Why did Shepherd-Barron reduce the PIN number from six digits to four?
Answer:
Shepherd-Barron’s wife said that she could only remember four figures, because of her, four figures became the world standard.

Question 5.
Which theory of Ramanujan helps the ATMs to dispense cash?
Answer:
Ramanujan’s Partition theory helps the ATMs to dispense cash.

Writing

G. Paragraph Writing

Question 1.
Write a paragraph of 100-120 words about a memorable anecdote / incident of your life.
Answer:

A memorable anecdote/incident in my life

I was then a student of class four. One day, I was left at home with my grandmother. It was in the afternoon, my grandmother was taking a nap. I was a very restless one. The toys soon bored me and I looked around for something new. The unique thing which caught my attention was my Grandma’s spectacles.

I put it on my nose just in the style of my Grandma and looked around. Soon my eyes got tired. As I felt pain in my eyes, I removed the specs and threw them away. They struck the wall and landed on the ground broken. Now I got worried and afraid. I started trying to repair it. As I was holding these glass pieces I felt a severe pain in the middle finger of my right hand, I looked at it .

Blood was trickling down from a deep cut in my finger. I started crying loudly. On hearing my loud wailing my Grandma woke up. She hurriedly came out of her room, took a quick glance at my adventure and detecting the source of my trouble, she pressed her hand on my cut finger for sometime and then she took me to the doctor for bandaging. I was very much afraid of punishment but my Grandma forgave me although she had to suffer difficulty in seeing until the glasses were repaired. However I was naturally punished as I could neither eat my meals nor do my home work for three days.

Question 2.
Write a paragraph of 100-120 words about your favourite personality.
Answer:

My favourite personality

There are many people all around the world who are very famous and celebrities. But my favourite personality is my father. My father is my hero. He is kind, polite and really friendly to everyone. He is a teacher by profession and is very good in teaching. He is always ready to help and support the needy and helpless. He is a God fearing person and always teaches us to remember the God’s gifts and God’s love for the world.

I am so proud to have a father like him. He is a simple man with kind rules. He is handsome, my favourite and my ideal man. He is my friend and always ready to encourage, appreciate me for success and always ready to help me wherever I need a friend or a support of my father. I am proud of my father and wish him good health forever.

Grammar

Connectors

A. Complete the following sentences using appropriate Connectors from the box.

moreover
although
meanwhile
therefore
because
as long as
thus
above all
for instance
except

1. She felt cold _________ she was wearing a winter coat.
Answer:
although
2. This restaurant has some of the best chefs in the town.__________ their service is excellent.
Answer:
Moreover

3. I’m not going to the party tonight __________ I didn’t get an invitation.
Answer:
because

4. You can set the table. __________, I’ll start making dinner.
Answer:
Meanwhile

5. I can play quite a few instruments __________ , the flute, the guitar and the piano.
Answer:
For instance

6. The store was out of chocolate chips; __________ they would need to make a different type of cookies.
Answer:
therefore

7. The stores are open daily __________ Sundays.
Answer:
except

8. I’ll stay __________ you need me.
Answer:
as long as

9. This detergent is highly concentrated and __________ you will need to dilute it.
Answer:
thus

10. It was the thing he prized __________ .
Answer:
above all

Active Voice and Passive Voice

B. Convert the following active sentences into passive sentences by supplying an appropriate passive verb form.

Question 1.
She will not recognize us. / We__________ by her.
(a) will not recognize
(b) will not being recognized
(c) will not be recognized
Answer:
(c) will not be recognized

Question 2.
They didn’t invite me, but I went anyway. /I __________ but I went anyway.
(a) wasn’t invited
(b) wasn’t being invited
(c) wasn’t inviting
Answer:
(a) wasn’t invited

Question 3.
They broke up the table for firewood. / The table __________ up for firewood.
(a) broke
(b) had broken
(c) was broken
Answer:
(c) was broken

Question 4.
She has won the first prize. / The first prize __________ by her.
(a) has won
(b) has been won
(c) had been won
Answer:
(b) has been won

Question 5.
A friend of mine is repairing the car. / The car __________ by a friend of mine.
(a) is repairing
(b) is repaired
(c) is being repaired
Answer:
(c) is being repaired

Question 6.
Begin the work tomorrow. / Let the work __________ tomorrow.
(a) be begun
(b) begin
(c) is beginning
Answer:
(a) be begun

Question 7.
They speak English in New Zealand. / English __________ in New Zealand.
(a) is speaking
(b) is spoken
(c) is being spoken
Answer:
(b) is spoken

Question 8.
His attitude shocked me. /I __________ by his attitude.
(a) had shocked
(b) had been shocked
(c) was shocked
Answer:
(c) was shocked

Question 9.
She had already sent the parcel. / The parcel __________ by her.
(a) has already been sent
(b) had already been sent
(c) was already sent
Answer:
(b) had already been sent

Question 10.
Her silence worries me. /I __________ her silence.
(a) am worrying by
(b) am worried by
(c) have worried by
Answer:
(b) am worried by

C. Match the following Active voice sentences with Passive voice.
From Zero to Infinity Lesson In Tamil
Answer:

  1. (e)
  2. (c)
  3. (d)
  4. (b)
  5. (a)

D. Change the following into passive voice.

Zero to Infinity Solutions Question 1. Stanley will inform you later.
Answer:
You will be informed by Stanley later.

Question 2.
People speak Portuguese in Brazil.
Answer:
Portuguese is spoken by people in Brazil.

Question 3.
My grandfather built this house in 1943.
Answer:
This house was built by my grandfather in 1943.

Question 4.
Do not hurt the animals.
Answer:
You are warned not to hurt the animals.

Question 5.
You must not drop litter in the streets.
Answer:
You are warned not to drop litter in the streets.

Question 6.
Carry it home.
Answer:
Let it be carried to home.

Question 7.
They are decorating the wall.
Answer:
The wall is being decorated by them.

Question 8.
He has already mended the TV set.
Answer:
The TV set has already been mended by him.

Project

E. Make a scrapbook of ‘famous Biographies’ by collecting at least five biographies of famous scientists, mathematicians, inventors, artists etc., of your choice. You may also collect the pictures related to their achievements, inventions etc.
Ramanujan Zero
(To be done by the student)

From Zero to Infinity Textual Activities

Warm Up

Question.
Solve this
From Zero to Infinity Theme
Answer:
Tamilnadu Board Class 9 English Solutions Prose Chapter 6 From Zero to Infinity - 10

Question.
Did you enjoy solving this?
Answer:
Yes. I enjoyed solving this puzzle

Question.
Was it easy or hard to solve?
Answer:
Yes. It was easy.

Question.
Do you like Mathematics? Give reasons.
Answer:
Yes. I like Mathematics as it is exciting.

Synonyms & Antonyms : Additional Questions

I. Choose the correct Synonym for the underlined words.

Question 1.
The teacher complimented the boy who had asked that absurd question.
(a) clever
(b) silly
(c) wise
Answer:
(b) silly

Question 2.
Senior students used to go to his dingy house.
(a) dark and dirty place
(b) bright place
(c) attractive place
Answer:
(a) dark and dirty place

Question 3.
But Ramanujan was ignorant of the work of the German mathematician George. F. Riemann.
(a) conscious
(b) educated
(c) unaware
Answer:
(c) unaware

Question 4.
His father was a petty clerk in a cloth shop.
(a) significant
(b) insignificant
(c) important
Answer:
(b) insignificant

Question 5.
Unkempt and uncouth, he would visit offices.
(a) neat
(b) clean
(c) not neat
Answer:
(c) not neat

Another Type of Exercise

Choose the appropriate Synonym of the underlined words.

1. Ramanujan’s father was a petty (i) clerk in a cloth shop. From early childhood it was evident that he was a prodigy (ii). Senior students used to go to his dingy (iii) house to get their difficulties in mathematics solved. At the age of 13, Ramanujan was lent a book on advanced (iv) trigonometry written by S.L. Loney. Not only did he master this rather (v) difficult book but also began his own research.

i) (a) insignificant
(b) significant
(c) royal
(d) major
Answer:
(a) insignificant

ii) (a) imbecile
(b) genius
(c) normality
(d) regularity
Answer:
(b) genius

iii) (a) dark
(b) gloomy
(c) bright and clean
(d) dull
Answer:
(c) bright and clean

iv) (a) basic
(b) middle
(c) primitive
(d) progressive
Answer:
(d) progressive

v) (a) quite
(b) extremely
(c) insignificantly
(d) violently
Answer:
(a) quite

2. Although Ramanujan secured (i) a first class in mathematics in the matriculation examination and was awarded (ii) the Subramanyan Scholarship, he failed twice in his first-year arts examination in college, as he neglected (iii) other subjects such as History, English and Physiology. This disappointed (iv) his father. When he found the boy always scribbling numbers and not doing much else, he thought Ramanujan had gone mad (v)

i) (a) missed
(b) gave up
(c) obtained
(d) lost
Answer:
(d) lost

ii) (a) granted
(b) denied
(c) refused
(d) disallowed
Answer:
(a) granted

iii) (a) included
(b) abandoned
(c) cherished
(d) completed
Answer:
(b) abandoned

iv) (a) distressed
(b) charmed
(c) contented
(d) comforted
Answer:
(a) distressed

v) (a) wise
(b) rational
(c) sane
(d) insane
Answer:
(d) insane

3. Unkempt (i) and uncouth (ii), he would visit offices, showing everyone his frayed (iii) notebooks and telling them that he knew mathematics and could do a clerical job. But no one could understand what was written in the notebooks and his applications for jobs were turned down (iv). Luckily for him, he at last found someone who was impressed (v) by his notebooks.

i) (a) neat
(b) messy
(c) tidy
(d) trim
Answer:
(b) messy

ii) (a) rude
(b) decent
(c) civilized
(d) polite
Answer:
(a) rude

iii) (a) neat
(b) perfect
(c) worn out
(d) smart
Answer:
(c) worn out

iv) (a) processed
(b) accepted
(c) considered
(d) rejected
Answer:
(d) rejected

v) (a) apathetic
(b) amazed
(c) callous
(d) heedless
Answer:
(b) amazed

II. Choose the correct Antonym for the underlined words.

Question 1.
The arithmetic class was in progress.
(a) advance
(b) decline
(c) movement
Answer:
(b) decline

Question 2.
He could forget much of the hardship, he had to endure.
(a) suffer
(b) combat
(c) agree
Answer:
(b) combat

Question 3.
He came forth with many mathematical theorems.
(a) backward
(b) forward
(c) onward
Answer:
(a) backward

Question 4.
The teacher was solving questions on division.
(a) addition
(b) subtraction
(c) multiplication
Answer:
(c) multiplication

Question 5.
There was a roar of laughter in the class.
(a) cry
(c) giggle
(c) smile
Answer:
(a) cry

Another Type of Exercise

Choose the appropriate Antonyms of the underlined words.

1. Everyone laughed again. The boys understood the trick (i), arithmetic had played upon them. What they could not understand was why the teacher later complimented (ii) the boy who had asked that absurd (iii) question. The boy who asked the intriguing (iv) question was Srinivasa Ramanujan. Throughout his life, he was always ahead (v) of his mathematics teachers.

i) (a) deceive
(b) cheat
(c) honesty
(d) fool
Answer:
(c) honesty

ii) (a) commended
(b) abused
(c) praised
(d) congratulated
Answer:
(b) abused

iii) (a) logical
(b) ridiculous
(c) senseless
(d) foolish
Answer:
(a) logical

iv) (a) interesting
(b) alluring
(c) boring
(d) appealing
Answer:
(c) boring

v) (a) in advance
(b) backward
(c) forward
(d) onward
Answer:
(b) backward

2. In Ramanujan, Hardy found an unsystematic mathematician, similar to one who knows the Pythagoras theorem but does not know what a congruent (i) triangle means. Several discrepancies in his research could be attributed (ii) to his lack of formal education. Ramanujan played with numbers, as a child would with a toy. It was sheer genius (iii) that led him to mathematical “truths”. The task of proving them, so important (iv) in Science, he left to lesser mortals (v).

i) (a) concurring
(b) harmonious
(c) agreeable
(d) disagreeable
Answer:
(d) disagreeable

ii) (a) connected
(b) unrelated
(c) credited
(d) associated
Answer:
(b) unrelated

iii) (a) ignorance
(b) ability
(c) flair
(d) aptitude
Answer:
(a) ignorance

iv)(a) critical
(b) unimportant
(c) crucial
(d) imperative
Answer:
(b) unimportant

v) (a) humans
(b) characters
(c) individuals
(d) immortals
Answer:
(d) immortals

3. While Ramanujan continued his research work, Tuberculosis, then an incurable disease, was devouring (i) him. Ramanujan was sent back to India and when he disembarked (ii), his friends found him pale, exhausted (iii) and emaciated (iv). To forget the agonizing (v) pain, he continued to play with numbers even on his death bed.

i) (a) consuming
(b) eating
(c) absorbing
(d) avoiding
Answer:
(c) hated

ii) (a) alighted
(b) landed
(c) embarked
(d) decended
Answer:
(b) boredom

iii) (a) tired
(b) energetic
(c) worn out
(d) fatigued
Answer:
(d) inconveniently

iv) (a) lean
(b)chubby
(c) thin
(d) skinny
Answer:
(a) unlikely

v) (a) painful
(b) severe
(c) pleasant
(d) distressing
Answer:
(c) go

MCQs – Additional

Complete the following by choosing the correct answer from the options given.

Question 1.
The teacher was solving questions on _______________.
(a) division
(b) Mathematics
(c) multiplication
(d) addition
Answer:
(a) division

Question 2.
There was a roar of laughter in the _______________.
(a) room
(b) corridor
(c) class
(d) hall
Answer:
(c) class

Question 3.
“Mathematically, each will get an _______________ number of bananas!”
(a) equal
(b) unlimited
(c) infinite
(d) extra
Answer:
(c) infinite

Question 4.
The boys understood the trick, _______________ had played upon them.
(a) mathematics
(b) the teacher
(c) the student
(d) arithmetic
Answer:
(d) arithmetic

Question 5.
The boy who _______________the intriguing question was Srinivasa Ramanujan.
(a) answered
(b) asked
(c) put forward
(d) enquired
Answer:
(b) asked

Question 6.
His father was a petty clerk in a _______________ shop.
(a) grocery
(b) toy
(c) cloth
(d) jewellery
Answer:
(c) cloth

Question 7.
He needed about _______________ sheets of paper every month,
(a) 1000
(b) 2000
(c) 1500
(d)3000
Answer:
(b) 2000

Question 8.
On March 17,1914, he sailed for _______________.
(a) Britain
(b) London
(c) Germany
(d) France
Answer:
(a) Britain

Question 9.
Ramanujan found himself a stranger at _______________.
(a) Cambridge
(c) Yale
(b) Oxford
(d) London
Answer:
(a) Cambridge

Question 10.
He was _______________ the Indian to receive distinguished fellowship,
(a) oldest
(b) first
(c) second
(d) youngest
Answer:
(d) youngest

Question 11.
A smart boy in the front row replied _______________.
(a) all will get two
(b) each will get one
(c) no one will get anything
Answer:
(b) each will get one

Question 12.
_______________ the teacher said loudlv.
(a) Quiet
(b) Be silent
(c) sit down
Answer:
(a) Quiet

Question 13.
‘Uncouth’ is behaving in a _______________.
(a) pleasant way
(b) unpleasant way
(c) careless way
Answer:
(b) unpleasant way

Question 14.
The teacher stopped and _______________.
(a) asked the boy to answer
(b) waited for the boy to speak
(c) enquired the boy
Answer:
(b) waited for the boy to speak

Question 15.
He used to lecture on subjects like _______________.
(a) astrologies and mathematics
(b) health and diseases
(c) ‘God, Zero and infinity’
Answer:
(c) ‘God, Zero and infinity’

From Zero to Infinity About the Author

Srinivasa Ramanujan (22 December 1887 – 26 April 1920) was an Indian mathematician who lived during the British rule in India. Though he had almost no formal training in pure Mathematics, he made substantial contributions to mathematical analysis, number theory, infinite series, and continued fractions, including solutions to mathematical problems considered to be unsolvable. He was born on 22 December 1887 into a Tamil Brahmin Iyengar family in Erode, Madras Presidency (now Tamil Nadu). The Man Who Knew Infinity: A Life of the Genius Ramanujan is a biography of Ramanujan, written in 1991 by Robert Kanigel and published by Washington Square Press.
Tamilnadu Board Class 9 English Solutions Prose Chapter 6 From Zero to Infinity - 1

From Zero to Infinity Summary

This lesson ‘From Zero to Infinity’ is a biography of Srinivasa Ramanujan. He was an Indian h Mathematician, who lived during the British rule in India. His knowledge of Mathematics was extraordinary.

One day, when the arithmetic class was in progress, his teacher was solving questions on , division. When the teacher asked the students if there were three students and three bananas, how many each students would get, a smart boy replied that each one would get one banana. Then he proceeded to ask if 1000 bananas are distributed among 1000 boys, would each one get one banana? f Ramanujan stood up and asked his teacher, if no banana is distributed among no one, would everyone still get one banana? All the other students laughed at this silly question.

But the teacher understood his question and explained to the students what he had asked. It was if zero banana is divided among zero, would each one get one? The answer would be ‘no’. Mathematically, each would get an infinite number of bananas. The boy had asked a question that had taken mathematicians several centuries to answer.

Ramanujan was bom in Erode in Tamilnadu on December 22,1%87. From early childhood, it was evident that he was a prodigy. Senior students used to get his assistance in solving math problems. At the age of 13, he began his own research on Trigonometry. The book “Synopsis of Elementary Results in Pure Applied Mathematics” by George Shoobridge Carr triggered the genius in Ramanuj an.

He used to do problems on loose sheets and enter the results in notebooks which are now famous as “Ramanujan’s Frayed Notebooks”. Although Ramanujan secured a first class in Mathematics in the matriculation examination and was awarded the Subramanyan Scholarship, he failed twice in his first year arts examination in college as he neglected other subjects such as History, English and Physiology. He searched for job for food and papers to do calculations. The Director of Madras Port Tmst gave a clerical job to Ramanujan on a monthly salary of Rupees 25.

Ramanujan sent a letter to the great Mathematician G. H. Hardy of Cambridge University, in which he set out 120 theorems and formulae, which included the Reimann Series. Hardy and his colleague Littlewood realized that they had discovered a rare mathematical genius. They invited him to Britain.

Despite the cold weather and food, Ramanujan continued his research with determination ‘ in the company of Hardy and Littlewood. Hardy found an unsystematic mathematician in Ramanuj an due to his lack of formal education. Ramanujan’s achievements include the Hardy-Ramanujan- Littlewood circle method in number theory.

As Ramanujan was suffering from Tuberculosis, he was sent back to India. Even on his death bed, he continued to play with numbers, so that he could forget his agonising pain. Ramanujan was also an astrologer and a good speaker. He gave lectures on subjects like “God, Zero and Infinity”.

From Zero to Infinity Mind Map

Tamilnadu Board Class 9 English Solutions Prose Chapter 6 From Zero to Infinity - 2

From Zero to Infinity Glossary

Tamilnadu Board Class 9 English Solutions Prose Chapter 6 From Zero to Infinity - 3

Synonyms
Tamilnadu Board Class 9 English Solutions Prose Chapter 6 From Zero to Infinity - 4

Antonyms
Tamilnadu Board Class 9 English Solutions Prose Chapter 6 From Zero to Infinity - 5

Tamilnadu Board Class 9 English Solutions