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 %
