The DAV Class 7 Maths Solutions and DAV Class 7 Maths Chapter 5 Worksheet 5 Solutions of Application of Percentage offer comprehensive answers to textbook questions.
DAV Class 7 Maths Ch 5 WS 5 Solutions
Question 1.
Find the unknown quantity in each of the following.
Answer:
(i) Here P = ?, R = 12%, T = 2\(\frac{1}{2}\) = \(\frac{5}{2}\) years, S.I = ₹ 1200, A = ?
S.I = \(\frac{P \times R \times T}{100}\)
⇒ 1200 = \(\frac{P \times 12 \times 5}{100 \times 2}\)
⇒ P = \(\frac{1200 \times 100 \times 2}{12 \times 5}\)
⇒ P = ₹ 4000
A = P + S.I.
= ₹ 4000 + ₹ 1200 = ₹ 5200
Hence P = ₹ 4000, A = ₹ 5200
(ii) Here P = ?, R = 3.5%, T = 2 years, A = ₹ 535, S.I. = ?
S.I = \(\frac{P \times R \times T}{100}\)
⇒ S.I = \(\frac{P \times 3.5 \times 2}{100}\)
⇒ S.I = \(\frac{7 P}{100}\)
A = P + S.I
⇒ 535 = P + \(\frac{7 P}{100}\)
⇒ 535 = \(\frac{107 \mathrm{P}}{100}\)
P = \(\frac{535 \times 100}{107}\) = 500
S.I. = A – P
⇒ S.I = ₹ 535 – ₹ 500
= ₹ 35
Hence P = ₹ 500, S.I = ₹ 35
(iii) Here P = ?, R = 4%, T = 3 years, S.I = ₹ 120, A = ?
S.I = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
120 = \(\frac{P \times 4 \times 3}{100}\)
P = \(\frac{120 \times 100}{4 \times 3}\) = ₹ 1000
A = P + S.I = ₹ 1000 + ₹ 120 = ₹ 1120
Hence P = ₹ 1000 and A = ₹ 1120
(iv) Here P = ₹ 450, R = ?, T = 3\(\frac{1}{2}=\frac{7}{2}\)years, S.I = ₹ 189, A = ?
S.I = \(\frac{P \times R \times T}{100}\)
⇒ 189 = \(\frac{450 \times R \times 7}{100 \times 2}\)
⇒ R = \(\frac{189 \times 100 \times 2}{450 \times 7}\) = 12%
A = P + S.I = ₹ 450 + ₹ 189
= ₹ 639
Hence R = 12% and A = ₹ 639
(v) Here P = 850, R = 6%, T = ?, S.I = ₹ 178.50, A = ?
S.I = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
⇒ 178.50 = \(\frac{850 \times 6 \times T}{100}\)
⇒ T = \(\frac{178.50 \times 100}{580 \times 6}\)
⇒ T = \(\frac{7}{2}\) = 3 \(\frac{1}{2}\) years
A = P + S.I
= ₹ 850 + ₹ 178.50
= ₹ 1028.50
Hence T = 3\(\frac{1}{2}\) years and A = ₹ 1028.50
(vi) Here P = ₹ 5400, R = ?, T = 3 years, S.I = ? A = ₹ 6210
S.I = A – P
S.I = 6210 – 5400
= 810
S.I = \(\frac{P \times R \times T}{100}\)
⇒ 810 = \(\frac{5400 \times \mathrm{R} \times 3}{100}\)
⇒ R = \(\frac{810 \times 100}{5400 \times 3}\)
⇒R = 4%
Hence R = 4% and S.I = ₹ 810
Question 2.
Find the sum of money that amounts to ₹ 5850 in 6 years at 5% P.a.
Answer:
Here, A = ₹ 5850, T = 6 years, R = 5% P.a., P = ?
Hence the sum = ₹ 4500
Question 3.
What sum will earn an interest oft 480 in 2y years at the rate of 3% p.a. simple interest?
Answer:
Here, S.I. = ₹ 480, T = 2\(\frac{1}{2}=\frac{5}{2}\) years, R = 3%, P = ?
S.I = \(\frac{P \times R \times T}{100}\)
480 = \(\frac{P \times 3 \times 5}{100 \times 2}\)
⇒ P = \(\frac{480 \times 100 \times 2}{3 \times 5}\)
⇒ P = ₹ 6400
Hence the sum = ₹ 6400
Question 4.
Mr. Mehta borrowed a sum of money at 8% P.a. If he paid ₹ 640 as interest after 5\(\frac{1}{3}\) years (5 years 4 months). Find the sum borrowed by him.
Answer:
Here R = 8%, S.I. = ₹ 640, T = 5\(\frac{1}{3}=\frac{16}{3}\) years, P = ?
S.I = \(\frac{P \times R \times T}{100}\)
640 = \(\frac{\mathrm{P} \times 8 \times 16}{100 \times 3}\)
⇒ P = \(\frac{640 \times 100 \times 3}{8 \times 16}\)
⇒ P = ₹ 1500
Hence the sum of money borrowed was ₹ 1500
Question 5.
Simple interest on a sum of money is \(\frac{9}{16}\) of the sum. If the rate is 4\(\frac{1}{2}\)% p.a., find the time.
Answer:
Let the principal be ₹ x
S.I = ₹ \(\frac{9}{16}\) x
R = 4\(\frac{1}{2}\)% = \(\frac{9}{2}\)% P.a., T = ?
S.I = \(\frac{P \times R \times T}{100}\)
\(\frac{9}{16} x=\frac{x \times 9 \times \mathrm{T}}{100 \times 2}\)
T = \(\frac{9}{16} x \times \frac{100 \times 2}{x \times 9}=\frac{25}{2}\) = 12\(\frac{1}{2}\) years
Question 6.
At what rate percent p.a. will a sum of money double itself in 8 years ?
Answer:
Let the principal be ₹ x
Amount = ₹ 2x
S.I. = A – P
= ₹ 2x – ₹ x
= ₹ x
T = 8 years, R = ?
S.I = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
x = \(\frac{x \times \mathrm{R} \times 8}{100}\)
R = \(\frac{100 \times x}{x \times 8}=\frac{25}{2}\)% or 12\(\frac{1}{2}\)% P.a
Question 7.
Rahul borrowed ₹ 50,000 from a bank on 1st March 2002 and paid ₹ 53150 on 6th October 2002. Find the rate of interest charged by the bank.
Answer:
T = March = 30 days
April = 30 days
May = 31 days
June = 30 days
July = 31 days
Aug. = 31 days
Sept. = 30 days
Oct. = 6 days
Total = 219 days
= \(\frac{219}{365}=\frac{3}{5}\) years
S.I. = A – P
= ₹ 53150 – ₹ 50,000
= ₹ 3150
S.I = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
3150 = \(\frac{50,000 \times \mathrm{R} \times 3}{100 \times 5}\)
R = \(\frac{3150 \times 100 \times 5}{50,000 \times 3}=\frac{21}{2}\) = 10\(\frac{1}{2}\)% P.a.
Question 8.
In what time will a sum of money double itself at 15% p.a. 1
Answer:
Let the sum be ₹ x
∴ Amount = ₹ 2x
S.I. = A – P = ₹ 2x – ₹ x = ₹ x
S.I = \(\frac{P \times R \times T}{100}\)
⇒ x = \(\frac{x \times 15 \times T}{100}\)
⇒ T = \(\frac{100 \times x}{15 \times x}=\frac{20}{3}\) = 6\(\frac{2}{3}\) years
Question 9.
In what time will the simple interest on ₹ 400 at 10% P.a. be the same as the simple interest on ₹ 1000 for 4 years at 4% P.a.?
Answer:
S.I on ₹ 400 = \(\frac{P \times R \times T}{100}=\frac{400 \times 10 \times T}{100}\) = ₹ 40 T
S.I on ₹ 1000 = \(\frac{1000 \times 4 \times 4}{100}\) = ₹ 160
40 T = 160
∴ T = \(\frac{160}{40}\) = 4 years
Question 10.
Mr. Jane donates ₹ 1 lakh to a school and the interest on it is to be used for awarding five scholarships of equal value. If the value of each scholarship is ₹ 1,500, find the rate of interest.
Answer:
Here, Principal = ₹ 1,00,000
Value of one scholarship = ₹ 1500
∴ Value of five scholarships of equal value = ₹ 5 × 1500 = ₹ 7500 = S.I.
Time = 1 year
Now, S.I = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
⇒ 7500 = \(\frac{100000 \times \mathrm{R} \times 1}{100}\)
⇒ R = \(\frac{7500 \times 100}{100000}\)
⇒ R = \(\frac{7500}{1000}=\frac{75}{10}\) = 7.5%
So, rate of interest is 7.5% per annum.
DAV Class 7 Maths Chapter 5 Value Based Questions
Question 1.
In a survey it was found that out of 125 people in a park, 12% jog, 16% do yoga and rest prefer to walk.
(i) Find the number of people who prefer to walk.
(ii) Discuss the importance of exercises like jogging, yoga, walking, etc.
Answer:
(i) Total number of people in park = 125
Number of people who prefer to walk = [100 – (12 + 16)]% = (100 -28)%
= 72% of the total number of people
= \(\frac{72}{100}\) × 125 = \(\frac{72}{4}\) × 5
= 18 × 5 = 90
So, 90 people prefer to walk.
(ii) Importance of Exercises: All these exercises keep us active, fit, physically and mentally alert.
Question 2.
The library teacher of a school keeps record of students of each class about their reading habits. The results of Class-Vll students for the month of April are as follows :
| Books read | Percentage |
| 0 | 14 |
| 1-3 | 28 |
| 4-6 | 26 |
| More than 6 | 32 |
(i) If total number of students ofClass-VII is 500, how many students had read books in the month of April ?
(ii) What is the importance of reading books?
Solution:
(i) Total number of students in class = 500
∴ Number of students who read books = (28 + 26 + 32)% = 86% of the total students
= \(\frac{86}{100}\) × 500 = 86 × 5 = 430
So, 430 students read books in the month of April.
(ii) Importance of Reading Books: Reading books increases vocabulary, improves thought process etc.