Raju

The DAV Maths Class 4 Solutions and DAV Class 4 Maths Chapter 14 Worksheet 2 Solutions of Fun with Patterns offer comprehensive answers to textbook questions.

DAV Class 4 Maths Ch 14 WS 2 Solutions

Question 1.
Look for the rules and complete the following patterns.
(a) 52, 57, 62, 67, ______, ______, ______, ______
Answer:
72, 77, 82, 87 (Increases by 5)

(b) 1, 3, 5, 7, ______, ______, ______, ______
Answer:
9, 11, 13, 15 (Increases by 2)

(c) 30, 27, 24, 21, ______, ______, ______, ______
Answer:
18, 15, 12, 9 (Decreases by 3)

(d) 100, 90, 80, 70, ______, ______, ______, ______
Answer:
60, 50, 40, 30 (Decreases by 3)

(e) 1, 3, 7, 13, ______, ______, ______, ______
Answer:
21, 31, 43, 57

(f) 11, 22, 33, 44, ______, ______, ______, ______
Answer:
55, 66, 77, 88 (Increases by 11)

(g) 2, 4, 8, 16, 32, ______, ______, ______, ______
Answer:
64, 128. 256. 512 (each no. × 2)

(h) 99, 101, 103, 105, ______, ______, ______, ______
Answer:
107. 109. 111. 113 (Increases by 2)

(i) 9, 109, 209, ______, ______, ______, ______
Answer:
309. 409. 509, 609 (Increases by 100)

(j) 586, 686, 786, ______, ______, ______, ______
Answer:
886, 986, 1086, 1186 (Increases by 100)

DAV Class 4 Maths Chapter 14 Worksheet 2 Notes

Magic square
In this number box, no number comes twice in any row or column.

In this Magic box, no letter comes twice in any row or column.

Lollipop game
Question 1.

Write digit 1 to 7 in lollipop so that each row with 3 digit adds up to 12.
Answer:

The DAV Maths Class 4 Solutions and DAV Class 4 Maths Chapter 14 Worksheet 1 Solutions of Fun with Patterns offer comprehensive answers to textbook questions.

DAV Class 4 Maths Ch 14 WS 1 Solutions

Question 1.

Answer:

Answer:

Answer:

Answer:

Answer:

Question 2.
Create your own patterns in the space provided.

Answer:

DAV Class 4 Maths Chapter 14 Worksheet 1 Notes

We see patterns in many things around us.
Every pattern follow some rules of addition, subtraction, division or multiplication.

Name three more things in which you find some patterns.

• Furniture
• Tiles
• Bed sheet

All patterns follow same rules.

Two flowers, and one leaf

One upright and one downward triangle.

The DAV Books Solutions Class 5 Maths and DAV Class 5 Maths Chapter 14 Worksheet 1 Solutions of Bills offer comprehensive answers to textbook questions.

DAV Class 5 Maths Ch 14 Worksheet 1 Solutions

Question 1.
Who prepares a bill and for whom?
Solution:
The shopkeeper prepares a bill for the customer.

Question 2.
Name any three bills you receive at home. Do all of them look alike? If not, why?
Solution:
Water bill, electricity bill, and mobile bill. No, all of them are different because entries made in various bills are different.

Question 3.
Write three uses of a bill.
Solution:

1. A shopkeeper can keep track on day to day sales.
2. The customer can cross-check the prices of items purchased and their total cost.
3. In case of any problem, the customer can exchange the item easily by showing the bill.

Question 4.
Mr. Kumar bought the following items from Toys Comer, Kamla Nagar, Delhi on 13 March 2014. Prepare a bill for these items.
(a) One teddy bear at ₹ 75 per piece.
(b) Three toy cars at ₹ 45.50 per car.
(c) Two building blocks at ₹ 36 per block.
(d) One jigsaw puzzle for ₹ 85.50.
Solution:

Total amount = ₹ 369

Question 5.
Mrs. Kiran purchased the following items from New Style Garment Store, Karol Bagh, Delhi on 7 September 2014.
(a) Two frocks at ₹ 255.50 per frock.
(b) One salwar-kameez for ₹ 325.
(c) Three shirts at ₹ 125.50 per shirt.
(d) Four nightgowns at ₹ 105.50 per gown.
If Mrs. Kiran gives the shopkeeper ₹ 1,700, how much balance does she get back?
Solution:

Amount of purchase = ₹ 1,634.50
The total amount given by Mrs. Kiran = ₹ 1,700
Balance she get back = ₹ 65.50

DAV Class 5 Maths Chapter 14 Value Based Questions

During the X-mas festival, sweets and gifts are exchanged with friends and relatives. Julie went with her mother to buy sweets for distribution. They purchased 2 kg cake at ₹ 320 per kg, 30 pastries at ₹ 45 per piece, and 25 doughnuts at ₹ 50 per piece. The shopkeeper made a bill. Julie cross-checked the bill and saw that the shopkeeper had charged ₹ 100 less than the actual bill amount. She informed the shopkeeper. The shopkeeper praised Julie for her honesty and gave her a chocolate.

Question 1.
Calculate the actual bill amount.
Solution:

Question 2.
How many items were purchased?
Solution:
3 items

Question 3.
If Julie’s mother wants to buy a 3 kg cake, how much more should she pay?
Solution:
Cost of 3 kg cake = 3 × ₹ 320 = ₹ 960
Now, she should pay ₹ 320 more.

Question 4.
Suppose you find a wristwatch from the school playground, what will you do?
Solution:
I will ask from children playing in the playground if the watch belongs to them or not. If they say no, then I will give it to reception, who make an announcement in the morning assembly for lost property.

DAV Class 5 Maths Chapter 14 Worksheet 1 Notes

Bill:
The slip of paper on which a shopkeeper notes down the requirements of a customer (buyer) and calculates the total cost of items purchased is called a bill.

Uses of a Bill:

• A shopkeeper can keep track on day to day sales.
• The customer can cross-check the prices of items purchased and their total cost.
• In case of any problem, the customer can exchange the items easily by showing the bill.

We obtain gas bills, electricity bills, water bills, ration bills, etc. All of them look different as entries made in various bills are not exactly the same.

Preparing A Bill

• Step 1. Write the name of the shop, its location, phone number, and date of purchase of items.
• Step 2. Make five columns – item number, name of item, quantity, rate, and amount.
• Step 3. Enter the name of the item, rate, and quantity.
• Step 4. We calculate the amount for each item by multiplying the rate by quantity.
• Step 5. We find the total cost of items purchased.

DAV Class 5 Sanskrit Book Solutions: Students who are preparing for exams of class 5 Sanskrit can happily read our DAV Sanskrit Book Class 5 Solutions to easily understand the concept and know the pattern of answers to execute in the exam. Surbhi Sanskrit Book Class 5 DAV Solutions always provides you the accurate and up-to-date worksheets according to DAV Class 5 Sanskrit Book Pdf.

Do not look for help to ace your exam preparation, as we are here to provide DAV Class 5 Sanskrit Solutions along with free download. Students can happily prepare for the exam with our DAV Public School Class 5 Sanskrit Book Solutions Pdf and score good marks in the exam.

DAV Public School Class 5 Sanskrit Book Solutions – DAV Sanskrit Book Class 5 Solutions

Candidates who are looking for the Surbhi Sanskrit Book Class 5 Solutions Pdf DAV can look below and download the Pdf if they want or can study there to score good marks.

Also, Check:

• DAV Books Solutions Class 5

DAV Class 5 Sanskrit Solutions – Surbhi Sanskrit Book Class 5 DAV Solutions

• Chapter 1 मधुरा प्रभातवेला
• Chapter 2 मम परिवारः
• Chapter 3 वयम् पश्याम: जन्तुशालाम्
• Chapter 4 शाकहट्टम्
• Chapter 5 अस्माकम् प्रियमित्राणि (पक्षिणः)
• Chapter 6 ये फलानि खादन्ति ते सुखिनः वसन्ति
• Chapter 7 चित्रप्रदर्शनी
• Chapter 8 अभिनवः किम् किम् करोति?
• Chapter 9 विडालः कुत्र अस्ति?
• Chapter 10 आगच्छ! गायामः
• Chapter 11 हरिणम् मा मारय !
• Chapter 12 कथयन्तु! कस्य कः वर्णः?
• Chapter 13 कः चतुरः अस्ति?
• Chapter 14 कति वस्तूनि सन्ति?
• सङ्ख्या, इदम् अस्माकं शरीरम्

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1. Where to find DAV School Class 5 Sanskrit Book Solutions for all chapters?

Students can find all chapters of DAV Class 5 Sanskrit Solution online and offline at our learncram.com website.

2. Are these chapterwise DAV Books Solutions Class 5 Sanskrit free to download?

Yes, these chapterwise DAV Surbhi Sanskrit Book Class 5 Solutions are free to download and available in pdf format too.

3. What is the name of the DAV Class 5 Sanskrit Book Pdf?

The name of the DAV Sanskrit Book Class 5 Pdf is Surbhi (Praveshika).

Key Takeaways

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The DAV Books Solutions Class 5 Maths and DAV Class 5 Maths Chapter 13 Brain Teasers Solutions of Simple Interest offer comprehensive answers to textbook questions.

DAV Class 5 Maths Ch 13 Brain Teasers Solutions

Question 1.
Tick (✓) the correct answer.
(а) The extra money paid by a bank for every ₹ 100 after one year is called-
(i) Interest
(ii) Rate of interest
(iii) Amount
(iv) Principal
Solution:
(ii) Rate of Interest

(b) Mr Gupta borrowed ₹ 1,75,000 from a bank to buy a car. This amount is called-
(i) Interest
(ii) Rate of interest
(iii) Loan
(iv) Time
Solution:
(iii) Loan

(c) Simple interest for ₹ 1,000 at 10% for 9 months is-
(i) ₹ 75
(ii) ₹ 900
(iii) ₹ 90
(iv) ₹ 1,075
Solution:
(i) ₹ 75
SI = \(\frac{1000 \times 10 \times 9}{100 \times 12}\)
= \(\frac{900}{12}\)
= ₹ 75

(d) If amount = ₹ 1,645.50, simple interest = ₹ 95, principal is-
(i) ₹ 1,740.50
(ii) ₹ 1,550
(iii) ₹ 1,740
(iv) ₹ 1,550.50
Solution:
(iv) ₹ 1,550.50
A = SI + P
1,645.50 = 95 + P
1,645.50 – 95 = P
P = ₹ 1,550.50

(e) If loan = ₹ 1,75,000, Rate of interest = 4%, Time = 2 years. Then amount to be repaid is equal to-
(i) ₹ 1,89,000
(ii) ₹ 1,76,400
(iii) ₹ 1,98,000
(iv) ₹ 1,80,000
Solution:
(i) ₹ 1,89,000
SI = \(\frac{1,75,000 \times 4 \times 2}{100}\) = ₹ 14,000
A = P + I
= 1,75,000 + 14,000
= ₹ 1,89,000

Question 2.
Name the three factors that determine simple interest.
Solution:

1. Principal
2. Rate of Interest
3. Time

Question 3.
Find the amount if the principal is ₹ 625 and the interest is ₹ 55.
Solution:
A = P + I
= ₹ 625 + ₹ 55
= ₹ 680

Question 4.
Calculate the simple interest on ₹ 1,200 at 3\(\frac{1}{2}\)% interest per annum for six months.
Solution:
P = ₹ 1,200
R = 3\(\frac{1}{2}\)% = \(\frac{7}{2}\)%
T = six months = \(\frac{6}{12}\) = \(\frac{1}{2}\) year
SI = PRT
= \(\frac{1,200 \times 7 \times 1}{100 \times 2 \times 2}\)
= ₹ 21

Question 5.
Sohan deposited ₹ 6,000 in a bank at 7\(\frac{1}{2}\)% per annum for three years. Mohan deposited ₹ 6,000 in another bank at 11% per annum for 2\(\frac{1}{2}\) years. Who will get more simple interest?
Solution:
P = ₹ 6,000
R = 7\(\frac{1}{2}\)% = \(\frac{15}{2}\)% = \(\frac{15}{200}\) per annum
T = 3 years
SI got by Sohan = 6,000 × \(\frac{15}{200}\) × 3 = ₹ 1350
P = ₹ 6,000
R = 11% = \(\frac{11}{100}\) per annum
T = \(\frac{5}{2}\) years
SI got by Mohan = \(\frac{6,000 \times 11 \times 5}{100 \times 2}\) = ₹ 1650
Mohan will get more simple interest.

Question 6.
A farmer borrowed ₹ 2,400 at 12% interest per annum. At the end of 3 years, he repaid ₹ 1,200 and a cow for the balance amount. Find the cost of the cow.
Solution:
P = ₹ 2,400
R = 12%
T = 3 years
SI = 2,400 × \(\frac{12}{100}\) × 3
= ₹ 864
A = P + SI
= ₹ 2,400 + ₹ 864
= ₹ 3,264
Cost of cow = ₹ 3,264 – ₹ 1200 = ₹ 2,064
Cost of cow is ₹ 2,064.

Question 7.
Mr Kannan had deposited ₹ 2,500 in a bank at 12% per annum. After 3\(\frac{1}{2}\) years, he withdraws this amount. Out of this money, he want to buy a dressing table costing ₹ 4000. How much more money does he need?
Solution:
P = ₹ 2,500
R = 12% = latex]\frac{12}{100}[/latex] per annum
T = 3\(\frac{1}{2}\) years = \(\frac{7}{2}\) years
SI = 2,500 × \(\frac{12}{100} \times \frac{7}{2}\) = ₹ 1,050
A = P + I
= ₹ 2,500 + ₹ 1,050
= ₹ 3,550
Cost of table = ₹ 4,000
Amount he needs = ₹ 4,000 – ₹ 3,550 = ₹ 450
He needs ₹ 450 more.

Additional Questions

Question 1.
Calculate simple interest for the following:
(a) P = ₹ 4,000; R = 3% per annum; T = 2 years
Solution:
P = ₹ 4,000
R = 3% per annum = \(\frac{3}{100}\) per annum
T = 2 years
Simple Interest = PRT
= 4,000 × \(\frac{3}{100}\) × 2
= ₹ 240

(b) P = ₹ 1,200; R = 2\(\frac{1}{2}\)% per annum; T = 1\(\frac{1}{2}\) years
Solution:
P = ₹ 1,200
R = 2\(\frac{1}{2}\)% = \(\frac{5}{2}\)% per annum
T = 1\(\frac{1}{2}\) years = \(\frac{3}{2}\) years
Simple Interest = PRT
= 1200 × \(\frac{5}{200} \times \frac{3}{2}\)
= ₹ 45

(c) P = ₹ 1,040; R = 12% per annum; T = 9 months
Solution:
P = ₹ 1040
R = 12% per annum
T = \(\frac{9}{12}\) year
Simple Interest = PRT
= 1,040 × \(\frac{12}{100} \times \frac{9}{12}\)
= \(\frac{468}{5}\)
= ₹ 93.6

Question 2.
A man deposited ₹ 12,000 in a bank at an interest of 9% per annum. How much simple interest will he get at the end of 24 months?
Solution:
Principal = ₹ 12,000
R = 9% per annum = \(\frac{9}{100}\) per annum
T = 24 months = 2 years
SI = PRT
= 12000 × \(\frac{9}{100}\) × 2
= ₹ 2160
He will get a simple interest of ₹ 2160 at the end of 24 months.

Question 3.
Calculate the following:
(a) P = ₹ 1,200; SI = ?; Amount = ₹ 1,450
Solution:
A = P + SI
₹ 1,450 = ₹ 1,200 + SI
SI = ₹ 1,450 – ₹ 1,200 = ₹ 250

(b) P = ?; SI = ₹ 134; Amount = ₹ 3100
Solution:
A = P + SI
P = A – SI
= ₹ 3,100 – ₹ 134
= ₹ 2,966

(c) P = ₹ 1,50,000; SI = ?; Amount = ₹ 1,75,000
Solution:
A = P + SI
SI = A – P
= ₹ 1,75,000 – ₹ 1,50,000
= ₹ 25,000

Question 4.
Find the amount.
(a) P = ₹ 21,000; Interest = ₹ 350
Solution:
A = P + I
= ₹ 21,000 + ₹ 350
= ₹ 21,350

(b) P = ₹ 1,750; Interest = ₹ 250
Solution:
A = P + I
= ₹ 1,750 + ₹ 250
= ₹ 2,000

(c) P = ₹ 7,500; Interest = ₹ 200.50
Solution:
A = P + I
= ₹ 7,500 + ₹ 200.50
= ₹ 7,700.50

Question 5.
Maya deposited ₹ 3,500 in a bank at 14% interest per annum. She withdraws her money after 9 months. Out of this amount, she purchases a centre table worth ₹ 3,800. How much money is left with her?
Solution:
Principal = ₹ 3,500
R = 14% per annum
T = 9 months = \(\frac{9}{12}\) years
SI = PRT
= 3,500 × \(\frac{14}{100} \times \frac{9}{12}\)
= ₹ 367.5
A = P + I
= ₹ 3,500 + ₹ 367.5
= ₹ 3,867.5
Cost of centre table = ₹ 3,800
Money left with her = ₹ 3,867.5 – ₹ 3,800 = ₹ 67.5

Question 6.
Tick (✓) the correct answer.
(a) Mr Sharma borrowed ₹ 20,00,000 from the bank to buy a small truck. This amount is called
(i) Principal
(ii) Rate of Interest
(iii) Loan
(iv) Time
Solution:
(iii) Loan

(b) The extra money paid by the bank is called
(i) Principal
(ii) Rate of Interest
(iii) Loan
(iv) Interest
Solution:
(iv) Interest

(c) Extra money charged by the bank for every hundred rupees deposit is called
(i) Principal
(ii) Amount
(iii) Interest
(iv) Loan
Solution:
(iii) Interest

(d) The period of time for which money is kept in a bank is called
(i) Amount
(ii) Rate of Interest
(iii) Principal
(iv) Time period
Solution:
(iv) Time period

DAV Class 3 Maths Book Solutions Pdf: Ace up your exam preparation with the help of our DAV Class 3 Maths Solutions, It covers all the chapter’s study resources. Check out our free DAV Maths Book Class 3 Solutions online and download DAV public school maths worksheets with solutions in pdf format for exam preparation.

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DAV Public School Class 3 Maths Book Solutions – DAV Maths Book Class 3 Solutions

Here are the pdf formatted links that are useful and free to download chapterwise DAV Public School Class 3 Maths Solutions. It can also examine your knowledge with the help of the provided chapters worksheets & brain teasers practice tests.

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• DAV Books Solutions Class 3

DAV Class 3 Maths Book Pdf Chapter 1 Numbers up to 9999

• DAV Class 3 Maths Chapter 1 Worksheet 1
• DAV Class 3 Maths Chapter 1 Worksheet 2
• DAV Class 3 Maths Chapter 1 Worksheet 3
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• DAV Class 3 Maths Chapter 1 Brain Teasers

DAV Class 3 Maths Solutions Chapter 2 Addition

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DAV Class 3 Maths Worksheets Chapter 5 Division

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• DAV Class 3 Maths Chapter 5 Brain Teasers

DAV Class 3 Maths Book Solutions Pdf Chapter 6 Day, Date and Time

• DAV Class 3 Maths Chapter 6 Worksheet 1
• DAV Class 3 Maths Chapter 6 Worksheet 2
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Class 3 DAV Maths Book Pdf Download Chapter 7 Money

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Class 3 Maths DAV Book Chapter 8 Length

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Primary Mathematics Class 3 DAV Chapter 10 Capacity

• DAV Class 3 Maths Chapter 10 Worksheet 1
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DAV Maths Book Class 3 Solutions Chapter 11 Fraction

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The DAV Books Solutions Class 5 Maths and DAV Class 5 Maths Chapter 13 Worksheet 3 Solutions of Simple Interest offer comprehensive answers to textbook questions.

DAV Class 5 Maths Ch 13 Worksheet 3 Solutions

Question 1.
Find the amount for the following:

Solution:
(a) P = ₹ 1,500, I = ₹ 150
A = P + I
= ₹ 1500 + ₹ 150
= ₹ 1650

(b) P = ₹ 750, I = ₹ 35
A = P + I
= ₹ 750 + ₹ 35
= ₹ 785

(c) P = ₹ 15,000, I = ₹ 980
A = P + I
= ₹15,000 + ₹ 980
= ₹ 15,980

(d) P = ₹ 4500, I = ₹ 215
A = P + I
= ₹ 4,500 + ₹ 215
= ₹ 4,715

Question 2.
Fill in the blanks:

Solution:
(a) P = ₹ 800, SI = ?, A = ₹ 905
A = P + I
905 = 800 + I
905 – 800 = I
105 = I
I = ₹ 105

(b) SI = ₹ 75.50, A = ₹ 450, P = ?
A = SI + P
450 = 75.50 + P
450 – 75.50 = P
P = ₹ 374.50

(c) P = ₹ 25,000, SI = ?, A = ₹ 31,000
A = P + SI
31,000 = 25,000 + SI
31,000 – 25,000 = I
SI = ₹ 6,000

(d) SI = ₹ 515, A = ₹ 1,680, P = ?
A = P + SI
1,680 = P + 515
P = 1,680 – 515 = ₹ 1,165

Question 3.
Calculate the amount for the following:
(a) ₹ 4,000 at 9% per annum for 3 years.
Solution:
A = ?
P = ₹ 4,000
R = 9%
T = 3 years
SI = ?
SI = PRT
= 4,000 × 3 × \(\frac{9}{100}\)
= ₹ 1,080
A = P + SI
= ₹ 4,000 + ₹ 1,080
= ₹ 5,080

(b) ₹ 750 at 3\(\frac{1}{2}\)% per annum for 2 years.
Solution:
P = ₹ 750
R = 3\(\frac{1}{2}\)% = \(\frac{7}{2}\)% = \(\frac{7}{200}\)
T = 2 years
SI = ?
SI = PRT
= 750 × \(\frac{7}{200}\) × 2
= ₹ 52.5
A = P + SI
= ₹ 750 + ₹ 52.5
= ₹ 802.50

(c) ₹ 3,200 at 7\(\frac{1}{2}\)% per annum for 3\(\frac{1}{2}\) years.
Solution:
P = ₹ 3,200
R = 7\(\frac{1}{2}\)% = \(\frac{15}{2}\)% = \(\frac{15}{200}\)
T = 3\(\frac{1}{2}\) years = \(\frac{7}{2}\) years
SI = ?
SI = PRT
= 3,200 × \(\frac{15}{200} \times \frac{7}{2}\)
= ₹ 840
A = SI + P
= ₹ 840 + ₹ 3,200
= ₹ 4,040

(d) ₹ 900 at 5% per annum for 5\(\frac{1}{2}\) years.
Solution:
P = ₹ 900
R = 5%
T = 5\(\frac{1}{2}\) years = \(\frac{11}{2}\) years
SI = ?
SI = PRT
= 900 × \(\frac{5}{100} \times \frac{11}{2}\)
= ₹ 247.50
A = P + SI
= ₹ 900 + ₹ 247.50
= ₹ 1,147.50

Question 4.
Solve the following questions:
(a) Arun took a loan of ₹ 600 for 1\(\frac{1}{2}\) years with interest at the rate of 4% per annum. Find the interest he pays after 1\(\frac{1}{2}\) years . Also, calculate the amount he pays back after 1\(\frac{1}{2}\) years.
Solution:
Principal = ₹ 600
T = 1\(\frac{1}{2}\) years = \(\frac{3}{2}\) years
R = 4% = \(\frac{4}{100}\)
SI = P × R × T
= 600 × \(\frac{4}{100} \times \frac{3}{2}\)
= ₹ 36
A = ₹ 600 + ₹ 36 = ₹ 636
He pays interest of ₹ 36 and total amount ₹ 636 after 1\(\frac{1}{2}\) years.

(b) Amit deposited ₹ 4,800 in his account. The bank pays an interest of 6% per annum. What amount will Amit get back after three years?
Solution:
Principal = ₹ 4,800
R = 6% per annum
T = 3 years
SI = P × R × T
= 4,800 × \(\frac{6}{100}\) × 3
= ₹ 864
A = P + I
= ₹4,800 + ₹ 864
= ₹ 5,664
Amit will get back ₹ 5,664 after three years.

(c) A man borrowed ₹ 700 from his friend. He promised to return the amount after six months at an interest of 8% per annum. How much money will he pay back after six months?
Solution:
Principal = ₹ 700
R = 8%
T = 6 months = \(\frac{1}{2}\) year
SI = P × R × T
= 700 × \(\frac{8}{100} \times \frac{1}{2}\)
= ₹ 28
A = SI + P
= ₹ 700 + ₹ 28
= ₹ 728
He will pay back ₹ 728 after 6 months.

(d) Surender deposited ₹ 2,500 in a bank at 12% interest per annum. He withdraws his money after 2\(\frac{1}{2}\) years. Out of this amount, he buys an almirah for ₹ 3,000. How much money is left with him?
Solution:
P = ₹ 2,500
R = 12%
T = 2\(\frac{1}{2}\) = \(\frac{5}{2}\) years
SI = P × R × T
= 2,500 × \(\frac{12}{100} \times \frac{5}{2}\)
= ₹ 750
A = P + I
= ₹ 2,500 + ₹ 750
= ₹ 3,250
Cost of almirah = ₹ 3,000
Amount left = A – cost of almirah
= ₹ 3,250 – ₹ 3,000
= ₹ 250
₹ 250 is left with him.

DAV Class 5 Maths Chapter 13 Value Based Questions

Nakul’s father Mr. Gupta got transferred from Delhi to Mumbai. Mr. Gupta had deposited ₹ 55000 in the Punjab National Bank two years before at 8% per annum. He closed the account and with the interest money he purchased gifts for his friends and family members. Nakul also got two pairs of jeans, two shirts, and a video game as a gift. Nakul with the permission of his parents gifted one jean and one shirt to Rahul, the son of the milkman. Rahul was very happy and thanked Nakul for the gift.

Question 1.
How much money did Mr Gupta get from the bank?
Solution:
Principal = ₹ 55,000
T = 2 years
R = 8% = \(\frac{8}{100}\) per annum
SI = PRT
= 55,000 × \(\frac{8}{100}\) × 2
= ₹ 8,800
A = P + I
= ₹ 55000 + ₹ 8800
= ₹ 63,800
Mr. Gupta got ₹ 63,800 from the bank.

Question 2.
How much money was spent by Mr Gupta on gifts?
Solution:
Interest = ₹ 8,800
With interest money, he purchased a gift. So he purchased a shirt and jeans for ₹ 8,800.

Question 3.
How do you feel when you gift/donate anything to the needy?
Solution:
When we gift/donate anything to the needy we feel happy and self-satisfied.

DAV Class 5 Maths Chapter 13 Worksheet 3 Notes

Concept of Amount
The amount we get back after a certain period of time is the sum of Principal and Interest.

• Amount = Principal + Interest
• Principal = Amount – Interest
• Interest = Amount – Principal

Example 1.
Sarita deposited ₹ 3,000 in a bank at 8% interest per annum. What amount will she get back after 3\(\frac{1}{2}\) years?
Solution:
Principal = ₹ 3,000
R = 8% = \(\frac{8}{100}\)
Time = 3\(\frac{1}{2}\) years = \(\frac{7}{2}\) years
SI = P × R × T
= 3,000 × \(\frac{8}{100} \times \frac{7}{2}\)
= ₹ 840
Amount received at the end of 3\(\frac{1}{2}\) years = P + Interest
= ₹ 3,000 + ₹ 840
= ₹ 3,840