DAV Class 7 Maths Chapter 5 Worksheet 5

Application of Percentage Worksheet 5

1. Find the unknown quantity in each of the following:

(i) Principal = ?
Rate of Interest = 12%
Time = $2 \frac{1}{2}$ years
Simple Interest = ₹1200
Amount = ?

Solution

$\begin{aligned} S . I & = ₹ 1200 \\ R & = 12 % \\ T & = 2 \frac{1}{2} ⟹ \frac{5}{2} y e a r s \\ P & = \frac{S . I \times 100}{R \times T} \\ = \frac{1200 \times 100}{12 \times \frac{5}{2}} \\ = \frac{1200^{100} \times 100^{20} \times 2}{12_{1} \times 5_{1}} \\ = 100 \times 20 \times 2 \\ P & = ₹ 4000 \\ A & = P + S . I \\ = 4000 + 1200 \\ A & = ₹ 5200 \end{aligned}$

Answer $P = ₹ 4000, A m o u n t = ₹ 5200$

(ii) Principal = ?
Rate of Interest = 3.5%
Time = 2 years
Simple Interest = ?
Amount = ₹ 535

Solution

$\begin{aligned} R & = 3.5 % \\ T & = 2 y e a r s \\ A & = ₹ 535 \\ P + S . I & = 535 \\ P + \frac{P \times R \times T}{100} & = 535 \\ P + \frac{P \times 3.5 \times 2}{100} & = 535 \\ P + \frac{P \times 7}{100} & = 535 \\ P + \frac{7 P}{100} & = 535 \\ \frac{100 P + 7 P}{100} & = 535 \\ \frac{107 P}{100} & = 535 \\ \frac{107}{100} \times P & = 535 \\ P & = \frac{535^{5} \times 100}{107_{1}} \\ P & = 5 \times 100 \\ P & = ₹ 500 \\ A & = ₹ 535 \\ S . I & = A - P \\ S . I & = 535 - 500 \\ S . I & = ₹ 35 \end{aligned}$

Answer $P = ₹ 500, S . I = ₹ 35$

(iii) Principal = ?
Rate of Interest = 4%
Time = 3 years
Simple Interest = ₹120
Amount = ?

Solution

$\begin{aligned} S . I & = ₹ 120 \\ R & = 4 % \\ T & = 3 years \\ P & = \frac{S . I \times 100}{R \times T} \\ = \frac{{120^{30}}^{10} \times 100}{4_{1} \times 3_{1}} \\ = 10 \times 100 \\ P & = ₹ 1000 \\ A & = P + S . I \\ = 1000 + 120 \\ A & = ₹ 1120 \end{aligned}$

Answer $P = ₹ 1000, A m o u n t = ₹ 1120$

(iv) Principal = ₹450
Rate of Interest = ?
Time = $3 \frac{1}{2}$ years
Simple Interest = ₹189
Amount = ?

Solution

$\begin{aligned} P & = ₹ 450 \\ S . I & = ₹ 189 \\ T & = 3 \frac{1}{2} ⟹ \frac{7}{2} years \\ R & = \frac{S . I \times 100}{P \times T} \\ = \frac{189 \times 100}{450 \times \frac{7}{2}} \\ = \frac{{189^{21}}^{3} \times 100^{2} \times 2}{{450_{9}}_{1} \times 7_{1}} \\ = 3 \times 2 \times 2 \\ R & = 12 % \\ A & = P + S . I \\ = 450 + 189 \\ A & = ₹ 639 \end{aligned}$

Answer $R = 12 %, A m o u n t = ₹ 639$

(v) Principal = ₹850
Rate of Interest = 6%
Time = ?
Simple Interest = ₹178.50
Amount = ?

Solution

$\begin{aligned} P & = ₹ 850 \\ S . I & = ₹ 178.50 \\ R & = 6 % \\ T & = \frac{S . I \times 100}{P \times R} \\ = \frac{178.50 \times 100}{850 \times 6} \\ = \frac{{17850^{21}}^{7}}{850_{1} \times 6^{2}} \\ = \frac{7}{2} \\ T & = 3 \frac{1}{2} years \\ A & = P + S . I \\ = 850 + 178.50 \\ A & = ₹ 1028.50 \end{aligned}$

Answer $T = 3 \frac{1}{2} years, A m o u n t = ₹ 1028.50$

(vi) Principal = ₹5,400
Rate of Interest = ?
Time = 3 years
Simple Interest = ?
Amount = ₹6,210

Solution

$\begin{aligned} P & = ₹ 5400 \\ A & = ₹ 6210 \\ S . I & = A - P \\ = 6210 - 5400 \\ S . I & = ₹ 810 \\ T & = 3 years \\ R & = \frac{S . I \times 100}{P \times T} \\ = \frac{{810^{270}}^{5} \times 100^{1}}{{5400_{54}}_{1} \times 3^{1}} \\ R & = 5 % \end{aligned}$

Answer $S . I = ₹ 810, R = 5 %$

2. Find the sum of money that amounts to ₹5,850 in six years at 5% per annum.

Solution

$\begin{aligned} P & = ? \\ R & = 5 % \\ T & = 6 years \\ A & = ₹ 5850 \\ P + S . I & = 5850 \\ P + \frac{P \times R \times T}{100} & = 5850 \\ P + \frac{P \times 5 \times 6}{100} & = 5850 \\ P + \frac{P \times 30^{3}}{100_{10}} & = 5850 \\ P + \frac{3 P}{10} & = 5850 \\ \frac{10 P + 3 P}{10} & = 5850 \\ \frac{13 P}{10} & = 5850 \\ \frac{13}{10} \times P & = 5850 \\ P & = \frac{5850^{450} \times 10}{13_{1}} \\ P & = 450 \times 10 \\ P & = ₹ 4500 \end{aligned}$

Answer Sum of money $(P r i n c i p a l) = ₹ 4500$

3. What sum will earn an interest of ₹480 in $2 \frac{1}{2}$ years at the rate of 3% per annum simple interest?

Solution

$\begin{aligned} P & = ? \\ S . I & = ₹ 480 \\ R & = 3 % \\ T & = 2 \frac{1}{2} years ⟹ \frac{5}{2} years \\ P & = \frac{S . I \times 100}{R \times T} \\ = \frac{480 \times 100}{3 \times \frac{5}{2}} \\ = \frac{480^{160} \times 100^{20} \times 2}{3_{1} \times 5_{1}} \\ = 160 \times 20 \times 2 \\ = 3200 \times 2 \\ P & = ₹ 6400 \end{aligned}$

Answer Sum $(P r i n c i p a l) = ₹ 6400$

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.

Solution

$\begin{aligned} P & = ? \\ S . I & = ₹ 640 \\ R & = 8 % \\ T & = 5 \frac{1}{3} years ⟹ \frac{16}{3} years \\ P & = \frac{S . I \times 100}{R \times T} \\ = \frac{640 \times 100}{8 \times \frac{16}{3}} \\ = \frac{{640^{80}}^{5} \times 100 \times 3}{8_{1} \times 16_{1}} \\ = 5 \times 100 \times 3 \\ P & = ₹ 1500 \end{aligned}$

Answer Sum borrowed $(P r i n c i p a l) = ₹ 1500$

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.

Solution

$\begin{aligned} Let P & = P \\ S . I & = \frac{9}{16} \times P \\ R & = 4 \frac{1}{2} % ⟹ \frac{9}{2} % \\ T & = \frac{S . I \times 100}{P \times R} \\ = \frac{(\frac{9}{16} \times P) \times 100}{P \times \frac{9}{2}} \\ = \frac{9 \times P \times 100^{25} \times 2^{1}}{{16_{8}}_{2} \times P \times 9} \\ = \frac{25}{2} \\ T & = 12 \frac{1}{2} years \end{aligned}$

Answer Time $= 12 \frac{1}{2} y e a r s$

6. At what rate percent p.a. will a sum of money double itself in 8 years?

Solution

$\begin{aligned} Let P & = P \\ Amount (A) & = 2 P \\ S . I & = A - P \\ = 2 P - P \\ = P \\ T & = 8 years \\ R & = \frac{S . I \times 100}{P \times T} \\ = \frac{P \times 100^{25}}{P \times 8_{2}} \\ = \frac{25}{2} \\ R & = 12 \frac{1}{2} % \end{aligned}$

Answer Rate of interest $(R) = 12 \frac{1}{2} %$

7. Rahul borrowed ₹ 50,000 from a bank on 1st March 2002 and paid ₹ 53,150 on 6th October 2002. Find the rate of interest charged by the bank.

Solution

$\begin{aligned} P & = ₹ 50000 \\ A & = ₹ 53150 \\ S . I & = A - P \\ = 53150 - 50000 \\ S . I & = ₹ 3150 \\ T & = From 1st March to 6th October \\ M a r c h & = 31 d a y s \\ A p r i l & = 30 d a y s \\ M a y & = 31 d a y s \\ J u n e & = 30 d a y s \\ J u l y & = 31 d a y s \\ A u g u s t & = 31 d a y s \\ O c t o b e r & = 5 d a y s \\ T o t a l & = 219 d a y s \\ = \frac{219^{3}}{365_{5}} \\ T & = \frac{3}{5} y e a r s \\ R & = \frac{S.I \times 100}{P \times T} \\ R & = \frac{3150 \times 100}{50000 \times \frac{3}{5}} \\ R & = \frac{3150^{1050} \times 100^{1} \times 5^{1}}{{50000_{500}}_{100} \times 3_{1}} \\ R & = \frac{1050}{100} \\ R & = 10.5 % \end{aligned}$

Answer Rate of Interest $(R) = 10.5 %$

8. In what time will a sum of money double itself at 15% p.a.?

Solution

$\begin{aligned} Let P & = P \\ Amount (A) & = 2 P \\ S . I & = A - P \\ = 2 P - P \\ S . I & = P \\ R & = 15 % \\ T & = \frac{S . I \times 100}{P \times R} \\ = \frac{P \times 100^{20}}{P \times 15_{3}} \\ = \frac{20}{3} \\ T & = 6 \frac{2}{3} years \\ T & = 6 years + \frac{2}{3} years \\ T & = 6 years + \frac{2}{3_{1}} \times 12^{4} \\ T & = 6 years 8 m o n t h s \end{aligned}$

Answer Time $(T) = 6 years 8 m o n t h s$

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.?

Solution

$\begin{aligned} S t e p - 1 \\ P & = ₹ 1000 \\ T & = 4 y e a r s \\ R & = 4 % \\ S . I & = \frac{1000^{10} \times 4 \times 4}{100_{1}} \\ = 10 \times 16 \\ S . I & = ₹ 160 \\ S t e p - 2 \\ S . I & = ₹ 160 \\ P & = 400 \\ R & = 10 % \\ T & = \frac{S . I \times 100}{P \times R} \\ = \frac{160^{4} \times 100}{400_{1} \times 10} \\ T & = 4 years \end{aligned}$

Answer Time required $(T) = 4 years$

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.

Solution

$\begin{aligned} P & = ₹ 1, 00, 000 \\ Value of 1 scholarship & = ₹1500 \\ Value of 5 scholarship & = 5 \times 1500 \\ = ₹7,500 \\ S . I & = ₹7,500 \\ T & = 1 year \\ R & = \frac{S . I \times 100}{P \times T} \\ = \frac{7500 \times 100}{100000 \times 1} \\ = \frac{75}{10} \\ = 7.5 % \end{aligned}$

Answer Rate of interest $(R) = 7.5 %$