SChand Composite Mathematics Class 7 Chapter 9 Percentage and it's application Exercise 9E

  Exercise 9 E 

Question 1

Find the simple interest and the amount at simple interest in the following problems 

1. Rs 184 for 2 year at 5% p.a 

Sol: S.I = $\frac{P \times R \times T}{100}=\frac{184 \times 5 \times 2}{100}=₹ 18.40$

Amount = 184 +18.4 = Rs 202.40

2. Rs 600 for 5 years at $12 \frac{1}{2} \% p.a$

Sol: S.I = $\frac{P \times R \times T}{100}=\frac{600\times 5 \times 12 .5}{100 \times 10}$

= Rs 375

A = 600+375 = Rs 975

3. Rs 350 for 4 years at  6% p.a 

Sol: S.I= $\frac{350 \times 4 \times 6}{100}=Rs 84$

$A=350+84=Rs 434$

4. Rs 960 For 3 years at $5 \frac{1}{2} \% p \cdot q$

Sol: S.I= $\frac{960 \times 3 \times 5.5}{100 \times 10}$ 

= $\frac{96 \times 33}{20}=Rs 158.40$

$\begin{aligned} A &=P+S . I \\ &=960+158.40=\$ Rs 1118.40 \end{aligned}$

Question 5

Vineet deposited Rs 8500 in a finance company which pays 13% interest per year. Find the amount he will receive after 3 years . 

Sol: S.I = $\frac{P \times R \times T}{100}$

$\frac{8500 \times 18 \times 3}{100}=4590$

$\begin{aligned} A &=P+S \cdot I \cdot \\ &=8500+4590=Rs 13,090 \end{aligned}$

Question 6

A former borrowed Rs 3600 at 15% interest per annum. At the end of 4 year he cleared his account by paying Rs 4000 and a cow. Find the cost of the cow. 

Sol: S.I = $\frac{P \times R \times T}{100}=\frac{3600 \times 15 \times 4}{100}=2160$

Amount = 3600+ 2160 = Rs 5760

Paid 4000+ cow= Rs 5760

$\operatorname{cow}=5760-4000$

Cost of Cow $= Rs1,760 \mathrm{Ans}$

Question 7

Find the rate percent per annum if 

(i) Rs 350 yields Rs 105 interest in 6 years 

Sol: S.I = $\frac{P \times R \times T}{100} \Rightarrow 105=\frac{350 \times R \times 6}{100}$

R = $\frac{105 \times 100}{350 \times 6}$= $\frac{10}{2}=5 \%$

(ii) Rs 425 yields Rs 238 interest in 8 years 

Sol: S.I $\frac{P \times R \times T}{100} \Rightarrow 238=\frac{425 \times R \times 8}{100}$

=$\frac{238 \times 100}{425 \times 8}$= R = 7%

(iii) Rs 600 yield Rs 72 interest in 4 years 

Sol: S.I =$\frac{P \times R \times T}{100} \Rightarrow 72=\frac{600 \times R \times 4}{100}$

R= $\frac{72 \times 100}{600 \times 4} \quad \Rightarrow \quad R=3 \%$

(iv) Rs 700 yields Rs 105 interest in 3 years 

Sol: S.I = $\frac{P \times R \times T}{100}$ 

= $105=\frac{700 \times R \times 3}{100} \Rightarrow R= 5$

Question 8

Find in what time: 

(i) Rs 300 will yields Rs 60 interest at 5% p.a 

Sol: S.I =$\frac{P \times R \times T}{100} \Rightarrow 60$

=$\frac{300 \times 5 \times T}{100}$

T= $\frac{60}{15}$

=4 Years 

(ii) Rs 500 will yield Rs 105 interest at $3 \frac{1}{2} \%$ P.a 

Sol: $105=\frac{500 \times 3.5 \times T}{100}$

T = $\frac{10}{5 \times 3.5}$= 6 Years

(iii) Rs 750 will yield Rs 225 interest at 6 % p.a 

Sol: 225 = $\frac{750 \times 6 \times T}{100}$

=$T=\frac{225 \times 100}{750 \times 6}$

=5 Years 

(iv) Rs 150 will yield Rs 36 interest at 6% p.a 

Sol: 36 = $\frac{150 \times 6 \times 1}{100}$

T=  $\frac{36 \times 100}{150 \times 6}$

T= 4 Years answer 

Question 9

Ranbir donates Rs 5000 to a school, the interest of which is to be used for awarding 10 scholarship of equal value every years. IF the donation earns an interest of 11% per annum . find the value of each scholarship. 

Sol: Ranbir donates per scholarship = $\frac{5000}{10}=500$

Interest = $\frac{P \times R \times T}{100}=\frac{500 \times 11 \times 1}{100}$

= Rs 55 answer 

Question 10

In how many years will Rs 150 double itself at 4% S.I ? 

Sol: P = 150 ; A = 300; S.I = A- P = 300- 150 =Rs 150

S.I = $\frac{P \times R \times T}{100} \Rightarrow 150=\frac{150 \times 4 \times T}{100} \Rightarrow T=\frac{150 \times 100}{150 \times 4}$

T= 25 years 

Question 11

AT what rate of S.I will a sum of money double itself in 20 years 

Sol: Let p = x ; A = 2x ; S.I = 2x- x = x 

x= $\frac{4 \times R \times 20}{100} \Rightarrow R$ = 5 %