Rd Sharma 2018 Solutions for Class 7 Math Chapter 13 Simple Interest are provided here with simple step-by-step explanations. These solutions for Simple Interest are extremely popular among Class 7 students for Math Simple Interest Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2018 Book of Class 7 Math Chapter 13 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Rd Sharma 2018 Solutions. All Rd Sharma 2018 Solutions for class Class 7 Math are prepared by experts and are 100% accurate.

#### Page No 13.7:

#### Question 1:

Find the simple interest, when:

(i) Principal = Rs 2000, Rate of Interest = 5% per annum and Time = 5 years.

(ii) Principal = Rs 500, Rate of Interest = 12.5% per annum and Time = 4 years.

(iii) Principal = Rs 4500, Rate of Interest = 4% per annum and Time = months.

(iv) Principal = Rs 12000, Rate of Interest = 18% per annum and Time = 4 months.

(v) Principal = Rs 1000, Rate of Interest = 10% per annum and Time = 73 days.

#### Answer:

(i) Principal (*P*) = Rs 2000

Rate of interest (*R*) = 5% p.a.

Time (*T*) = 5 years

Simple interest = $\frac{P\times R\times T}{100}=\frac{2000\times 5\times 5}{100}=\mathrm{Rs}500$

(ii) Principal (*P*) = Rs 500

Rate of interest (*R*) = 12.5% p.a.

Time (*T*) = 4 years

Simple interest = $\frac{P\times R\times T}{100}=\frac{500\times 12.5\times 4}{100}=\mathrm{Rs}250$

(iii) Principal (*P*) = Rs 4500

Rate of interest (*R*) = 4% p.a.

Time (*T*) = 6 months *T *=$\frac{6}{12}=\frac{1}{2}$ year (1 year = 12 months)

Simple interest = $\frac{P\times R\times T}{100}=\frac{4500\times 4\times {\displaystyle \frac{1}{2}}}{100}=\frac{4500\times 4\times 1}{100\times 2}=\mathrm{Rs}90$

(iv) Principal (*P*) = Rs 12000

Rate of interest (*R*) = 18% p.a.

$Time\; (T)\; =\; 4months=\frac{4}{12}=\frac{1}{3}\text{y}\mathrm{ear}$ (1 year = 12 months)

Simple interest = $\frac{P\times R\times T}{100}=\frac{12000\times 18\times 1}{100\times 3}=\mathrm{Rs}720$

(v) Principal (*P*) = Rs 1000

Rate of interest (*R*) = 10% p.a.

$\mathrm{Time\; (}T)\; =\; 73days=\frac{73}{365}\mathrm{year}$ (1 year = 365 days)

Simple interest = $\frac{P\times R\times T}{100}=\frac{1000\times 10\times 73}{100\times 365}=\mathrm{Rs}20$

#### Page No 13.8:

#### Question 2:

Find the interest on Rs 500 for a period of 4 years at the rate of 8% per annum. Also, find the amount to be paid at the end of the period.

#### Answer:

Principal amount (*P*) = Rs 500

Time period (*T*) = 4 years

Rate of interest (*R*) = 8% p.a.

Interest = $\frac{P\times R\times T}{100}=\frac{500\times 4\times 8}{100}=\mathrm{Rs}160$

Total amount paid = Principal amount + Interest = Rs 500 + 160

= Rs 660

#### Page No 13.8:

#### Question 3:

A sum of Rs 400 is lent at the rate of 5% per annum. Find the interest at the end of 2 years.

#### Answer:

Principal amount (*P*) = Rs 400

Time period (*T*) = 2 years

Rate of interest (*R*) = 5% p.a.

Interest paid after 2 years$=\frac{P\times R\times T}{100}=\frac{400\times 5\times 2}{100}=\mathrm{Rs}40$

#### Page No 13.8:

#### Question 4:

A sum of Rs 400 is lent for 3 years at the rate of 6% per annum. Find the interest.

#### Answer:

Principal amount (*P*) = Rs 400

Time period (*T*) = 3 years

Rate of interest (*R*) = 6% p.a.

Interest after 3 years = $\frac{P\times R\times T}{100}=\frac{400\times 6\times 3}{100}$ = Rs 72

#### Page No 13.8:

#### Question 5:

A person deposits Rs 25000 in a firm who pays an interest at the rate of 20% per annum. Calculate the income he gets from it annually.

#### Answer:

Principal amount (*P*) = Rs 25000

Time period (*T*) = 1 year

Rate of interest (*R*) = 20% p.a.

Annual interest = $\frac{P\times R\times T}{100}=\frac{25000\times 20\times 1}{100}$ = Rs 5000

#### Page No 13.8:

#### Question 6:

A man borrowed Rs 8000 from a bank at 8% per annum. Find the amount he has to pay after $4\frac{1}{2}$ years.

#### Answer:

Principal amount (*P*) = Rs 8000

Time period (*T*) = $4\frac{1}{2}=\frac{9}{2}\mathrm{years}$

Rate of interest (*R*) = 8% p.a.

Interest = $\frac{P\times R\times T}{100}=\frac{8000\times 8\times 9}{100\times 2}$ = Rs 2880

Total amount paid after $4\frac{1}{2}$ years = Principal amount + Interest = Rs 8000 + Rs 2880

= Rs 10880

#### Page No 13.8:

#### Question 7:

Rakesh lent out Rs 8000 for 5 years at 15% per annum and borrowed Rs 6000 for 3 years at 12% per annum. How much did he gain or lose?

#### Answer:

Principal amount lent out by Rakesh (*P*) = Rs 8000

Time period (*T*) = 5 years

Rate of interest (*R*) = 15% p.a.

Interest = $\frac{P\times R\times T}{100}=\frac{8000\times 15\times 5}{100}$ = Rs 6000

Principal amount borrowed by Rakesh (*P*) = Rs 6000

Time period (*T*) = 3 years

Rate of interest (*R*) = 12% p.a.

Interest = $\frac{P\times R\times T}{100}=\frac{6000\times 12\times 3}{100}$ = Rs 2160

Amount gained by Rakesh = Rs 6000 − Rs 2160 = Rs 3840

#### Page No 13.8:

#### Question 8:

Anita deposits Rs 1000 in a savings bank account. The bank pays interest at the rate of 5% per annum. What amount can Anita get after one year?

#### Answer:

Principal amount (*P*) = Rs 1000

Time period (*T*) = 1 year

Rate of interest (*R*) = 5% p.a.

Interest = $\frac{P\times R\times T}{100}=\frac{1000\times 5\times 1}{100}$ = Rs 50

Total amount paid after 1 year = Principal amount + Interest = Rs 1000 + Rs 50

= Rs 1050

#### Page No 13.8:

#### Question 9:

Nalini borrowed Rs 550 from her friend at 8% per annum. She returned the amount after 6 months. How much did she pay?

#### Answer:

Principal amount (*P*) = Rs 550

Time period (*T*) = 6 months = $\frac{6}{12}=\frac{1}{2}$ year (1 year = 12 months)

Rate of interest (*R*) = 8% p.a.

Interest = $\frac{P\times R\times T}{100}=\frac{550\times 8\times 1}{100\times 2}=\mathrm{Rs}22$

Total amount paid after 6 months = Principal amount + Interest = Rs 550 + Rs 22

= Rs 572

#### Page No 13.8:

#### Question 10:

Rohit borowed Rs 600000 from a bank at 9% per annum for 2 years. He lent this sum of money to Rohan at 10% per annum for 2 years. How much did Rohit earn from this transaction?

#### Answer:

Principal amount lent out by Rohit (P) = Rs. 60000

Time period (T) = 2 years

Rate of interest (R) = 10% p.a.

Interest = $\frac{P\times R\times T}{100}=\; Rs.\frac{60000\times 10\times 2}{100}=\; Rs.\; 12000$

Principal amount borrowed by Rohit from the bank (P) = Rs. 60000

Time period (T) = 2 years

Rate of interest (R) = 9% p.a.

Interest = $\frac{P\times R\times T}{100}=\; Rs.\frac{60000\times 9\times 2}{100}=\; Rs.\; 10800$

Amount gained by Rohit = Rs. 12000 - 10800 = Rs. 1200

#### Page No 13.8:

#### Question 11:

Romesh borrowed Rs 2000 at 2% per annum and Rs 1000 at 5% per annum. He cleared his debt after 2 years by giving Rs 2800 and a watch. What is the cost of the watch?

#### Answer:

Principal amount borrowed by Romesh (P) = Rs. 2000

Time period (T) = 2 years

Rate of interest (R) = 2% p.a.

Interest = $\frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100}=\mathrm{Rs}.\frac{2000\times 2\times 2}{100}=\mathrm{Rs}.80$

Principal amount borrowed by Romesh (P) = Rs. 1000

Time period (T) = 2 years

Rate of interest (R) = 5% p.a.

Interest = $\frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100}=\mathrm{Rs}.\frac{1000\times 5\times 2}{100}=\mathrm{Rs}.100$

Total amount that he will have to return = Rs. 2000 + 1000 + 80 + 100 = Rs. 3180

Amount repaid = Rs. 2800

Value of the watch = Rs. 3180 - 2800 = Rs. 380

#### Page No 13.8:

#### Question 12:

Mr Garg lent Rs 15000 to his friend. He charged 15% per annum on Rs 12500 and 18% on the rest. How much interest does he earn in 3 years?

#### Answer:

Principal amount (*P*) = Rs 12500

Time period (*T*) = 3 years

Rate of interest (*R*) = 15% p.a.

Interest = $\frac{P\times R\times T}{100}=\frac{12500\times 15\times 3}{100}$ = Rs 5625

Rest of the amount lent = Rs 15000 − Rs 12500 = Rs 2500

Rate of interest = 18 % p.a.

Time period = 3 years

Interest = $\frac{P\times R\times T}{100}=\frac{2500\times 18\times 3}{100}$ = Rs 1350

Total interest earned = Rs 5625 + Rs 1350 = Rs 6975

#### Page No 13.8:

#### Question 13:

Shikha deposited Rs 2000 in a bank which pays 6% simple interest. She withdrew Rs 700 at the end of first year. What will be her balance after 3 years?

#### Answer:

Principal amount deposited (*P*) = Rs 2000

Time period (*T*) = 1 year

Rate of interest (*R*) = 6% p.a.

Interest after 1 year = $\frac{P\times R\times T}{100}=\frac{2000\times 6\times 1}{100}=\mathrm{Rs}120$

So amount after 1 year = Principal amount + Interest = 2000 + 120 = Rs 2120

After 1 year, amount withdrawn = Rs 700

Principal amount left (*P*_{1}) = Rs 2120 − Rs 700 = Rs 1420

Time period (*T*) = 2 years

Rate of interest (*R*) = 6% p.a.

Interest after 2 years = $\frac{{P}_{1}\times R\times T}{100}=\frac{1420\times 6\times 2}{100}=\mathrm{Rs}170.40$

Total amount after 3 years = Rs 1420 + Rs 170.40 = Rs 1590.40

#### Page No 13.8:

#### Question 14:

Reema took a loan of Rs 8000 from a money lender, who charged interest at the rate of 18% per annum. After 2 years, Reema paid him Rs 10400 and wrist watch to clear the debt. What is the price of the watch?

#### Answer:

Principal amount (*P*) = Rs 8,000

Rate of interest (*R*) = 18%

Time period (*T*) = 2 years

Interest after 2 years = $\frac{P\times R\times T}{100}=\frac{8000\times 18\times 2}{100}$ = Rs 2,880

Total amount payable by Reema after 2 years = Rs 8,000 + Rs 2,880 = Rs 10,880

Amount paid = Rs 10,400

Value of the watch = Rs 10,880 − Rs 10,400 = Rs 480

#### Page No 13.8:

#### Question 15:

Mr Sharma deposited Rs 20000 as a fixed deposit in a bank at 10% per annual. If 30% is deducted as income tax on the interest earned, find his annual income.

#### Answer:

Amount deposit (*P*) = Rs 20,000

Rate of interest (*R*) = 10% p.a.

Time period (*T*) = 1 year

Interest after 1 year = $\frac{P\times R\times T}{100}=\frac{20000\times 10\times 1}{100}$ = Rs 2,000

Amount deducted as income tax = $30\%\mathrm{of}\mathrm{Rs}2,000=\frac{30\times 2000}{100}=\mathrm{Rs}600$

Annual interest after tax deduction = Rs 2,000 − Rs 600 = Rs 1,400

#### Page No 13.8:

#### Question 1:

If the simple interest on a certain sum for 2 years at the rate of 5% per annum is â‚¹4000, then the sum is

(a) â‚¹46,000

(b) â‚¹44,000

(c) â‚¹40,000

(d) â‚¹48,000

#### Answer:

We know, $I=\frac{P\times T\times R}{100}$

It is given that,*T* = 2 years*R* = 5%*I* = â‚¹4000

Then,

$4000=\frac{P\times 5\times 2}{100}\phantom{\rule{0ex}{0ex}}\Rightarrow 4000=\frac{10P}{100}\phantom{\rule{0ex}{0ex}}\Rightarrow P=40000$

Thus, *P* = â‚¹40,000

Hence, the correct option is (c).

#### Page No 13.9:

#### Question 2:

In how many years will a certain sum become 3 times itself at 25% per annum under simple interest?

(a) 5

(b) 8

(c) 12

(d) 6

#### Answer:

Amount = 3 times the sum = 3*P*

Simple interest (*I*) = Amount − Sum = 3*P* − *P* = 2*P*

Let the sum (*P*) be* x*.

Then, simple interest (*I*) = 2*x*

Rate (*R*) = 25%

Time = *T*

$I=\frac{P\times R\times T}{100}\phantom{\rule{0ex}{0ex}}\Rightarrow T=\frac{100\times I}{P\times R}\phantom{\rule{0ex}{0ex}}=\frac{100\times 2x}{x\times 25}\phantom{\rule{0ex}{0ex}}=4\times 2\phantom{\rule{0ex}{0ex}}=8\mathrm{years}$

Hence, the correct option is (b).

#### Page No 13.9:

#### Question 3:

The amount on â‚¹25,000 at 8% per annum for 6 years under simple interest is

(a) â‚¹35,000

(b) â‚¹37,000

(c) â‚¹45,000

(d) â‚¹47,000

#### Answer:

It is given that,

Sum (*P*) = â‚¹25,000

Rate (*R*) = 8%

Time (*T*) = 6 years

$I=\frac{P\times R\times T}{100}\phantom{\rule{0ex}{0ex}}=\frac{25000\times 8\times 6}{100}\phantom{\rule{0ex}{0ex}}=12000$

Therefore, simple interest (*I*) = â‚¹12,000

Now, Amount = *P* + *I* = â‚¹25,000 + â‚¹12,000 = â‚¹37,000

Hence, the correct option is (b).

#### Page No 13.9:

#### Question 4:

The simple interest for â‚¹1500 at 8% per annum for 3 years is

(a) â‚¹400

(b) â‚¹360

(c) â‚¹450

(d) â‚¹500

#### Answer:

It is given that,

Sum (*P*) = â‚¹1500

Rate (*R*) = 8%

Time (*T*) = 3 years

$I=\frac{P\times R\times T}{100}\phantom{\rule{0ex}{0ex}}=\frac{1500\times 8\times 3}{100}\phantom{\rule{0ex}{0ex}}=360$

Therefore, simple interest (*I*) = â‚¹360

Hence, the correct option is (b).

#### Page No 13.9:

#### Question 5:

The difference between the interest obtained for â‚¹1000 at 12% per annum for 3 years and that for â‚¹1500 at 8% per annum for $1\frac{1}{2}$ years is

(a) â‚¹360

(b) â‚¹300

(c) â‚¹180

(d) â‚¹200

#### Answer:

It is given that,

Sum (*P*_{1}) = â‚¹1000

Rate (*R*_{1}) = 12%

Time (*T*_{1}) = 3 years

${I}_{1}=\frac{{P}_{1}\times {R}_{1}\times {T}_{1}}{100}\phantom{\rule{0ex}{0ex}}=\frac{1000\times 12\times 3}{100}\phantom{\rule{0ex}{0ex}}=360....\left(1\right)$

Sum (*P*_{2}) = â‚¹1500

Rate (*R*_{2}) = 8%

Time (*T*_{2}) = $1\frac{1}{2}$ years = $\frac{3}{2}$ years

${I}_{2}=\frac{{P}_{2}\times {R}_{2}\times {T}_{2}}{100}\phantom{\rule{0ex}{0ex}}=\frac{1500\times 8\times 3}{100\times 2}\phantom{\rule{0ex}{0ex}}=180....\left(2\right)$

Subtracting (2) from (1), we get

${I}_{2}-{I}_{1}=360-180=180$

Hence, the correct option is (c).

#### Page No 13.9:

#### Question 6:

Which of the following yields maximum interest for 2 years?

(a) â‚¹1500 at 8% per annum

(b) â‚¹1000 at 11% per annum

(c) â‚¹2000 at 5% per annum

(d) â‚¹900 at 20% per annum

#### Answer:

(a) It is given that,

Sum (*P*_{1}) = â‚¹1500

Rate (*R*_{1}) = 8%

Time (*T*_{1}) = 2 years

${I}_{1}=\frac{{P}_{1}\times {R}_{1}\times {T}_{1}}{100}\phantom{\rule{0ex}{0ex}}=\frac{1500\times 8\times 2}{100}\phantom{\rule{0ex}{0ex}}=240....\left(1\right)$

(b) It is given that,

Sum (*P*_{2}) = â‚¹1000

Rate (*R*_{2}) = 11%

Time (*T*_{2}) = 2 years

${I}_{2}=\frac{{P}_{2}\times {R}_{2}\times {T}_{2}}{100}\phantom{\rule{0ex}{0ex}}=\frac{1000\times 11\times 2}{100}\phantom{\rule{0ex}{0ex}}=220....\left(2\right)$

(c) It is given that,

Sum (*P*_{3}) = â‚¹2000

Rate (*R*_{3}) = 5%

Time (*T*_{3}) = 2 years

${I}_{3}=\frac{{P}_{3}\times {R}_{3}\times {T}_{3}}{100}\phantom{\rule{0ex}{0ex}}=\frac{2000\times 5\times 2}{100}\phantom{\rule{0ex}{0ex}}=200....\left(3\right)$

(d) It is given that,

Sum (*P*_{4}) = â‚¹900

Rate (*R*_{4}) = 20%

Time (*T*_{4}) = 2 years

${I}_{4}=\frac{{P}_{4}\times {R}_{4}\times {T}_{4}}{100}\phantom{\rule{0ex}{0ex}}=\frac{900\times 20\times 2}{100}\phantom{\rule{0ex}{0ex}}=360....\left(4\right)$

From (1), (2), (3) and (4),

â‚¹900 at 20% per annum yields maximum interest for 2 years.

â€‹

Hence, the correct option is (d).

#### Page No 13.9:

#### Question 7:

If a sum of â‚¹3000 is lent out at 3% per annum for 20 years under simple interest, then the amount at the end of 20^{th} year is

(a) â‚¹1800

(b) â‚¹1080

(c) â‚¹3600

(d) â‚¹4800

#### Answer:

It is given that,

Sum (*P*) = â‚¹3000

Rate (*R*) = 3%

Time (*T*) = 20 years

$I=\frac{P\times R\times T}{100}\phantom{\rule{0ex}{0ex}}=\frac{3000\times 3\times 20}{100}\phantom{\rule{0ex}{0ex}}=1800$

Amount = *I* + *P* = â‚¹1800 + â‚¹3000 = â‚¹4800

Hence, the correct option is (d).

#### Page No 13.9:

#### Question 8:

If a sum of â‚¹2000 is lent out at 2% per annum for 10 years under simple interest, then the amount is

(a) â‚¹1400

(b) â‚¹2400

(c) â‚¹200

(d) â‚¹1500

#### Answer:

It is given that,

Sum (*P*) = â‚¹2000

Rate (*R*) = 2%

Time (*T*) = 10 years

$I=\frac{P\times R\times T}{100}\phantom{\rule{0ex}{0ex}}=\frac{2000\times 2\times 10}{100}\phantom{\rule{0ex}{0ex}}=400$

Amount = *I* + *P* = â‚¹400 + â‚¹2000 = â‚¹2400

Hence, the correct option is (b).

#### Page No 13.9:

#### Question 9:

If interest on â‚¹*x* for 2 years at *R*% per annum is â‚¹80, the interest on â‚¹2*x* for one year at *R*% per annum is

(a) â‚¹160

(b) â‚¹40

(c) â‚¹80

(d) â‚¹120

#### Answer:

It is given that,

Sum (*P*_{1}) = â‚¹*x*

Rate (*R*_{1}) = *R*%

Time (*T*_{1}) = 2 years

Interest (*I*_{1}) = â‚¹80

${I}_{1}=\frac{{P}_{1}\times {R}_{1}\times {T}_{1}}{100}\phantom{\rule{0ex}{0ex}}\Rightarrow 80=\frac{x\times R\times 2}{100}\phantom{\rule{0ex}{0ex}}=\frac{2Rx}{100}...\left(1\right)$

Now,

Sum (*P*_{2}) = â‚¹2*x*

Rate (*R*_{2}) = *R*%

Time (*T*_{2}) = 1 year

${I}_{2}=\frac{{P}_{2}\times {R}_{2}\times {T}_{2}}{100}\phantom{\rule{0ex}{0ex}}=\frac{2x\times R\times 1}{100}\phantom{\rule{0ex}{0ex}}=\frac{2Rx}{100}\phantom{\rule{0ex}{0ex}}=80\left[\mathrm{From}\left(1\right)\right]$

Therefore, *I*_{2} = â‚¹80

Hence, the correct option is (c).

#### Page No 13.9:

#### Question 10:

At simple interest a sum becomes $\frac{49}{40}$ of itself in $2\frac{1}{2}$ years. The rate of interest per annum is

(a) 7%

(b) 8%

(c) 12%

(d) 9%

#### Answer:

Amount = $\frac{49}{40}$ times the sum = $\frac{49}{40}$*P*

Simple interest (*I*) = Amount − Sum = $\frac{49}{40}$*P* − *P* = $\frac{9}{40}$*P*

Let the sum (*P*) be* x*.

Then, simple interest (*I*) = $\frac{9}{40}$*x*

Rate (*R*) = *R*%

Time (*T*) = $2\frac{1}{2}$ years = $\frac{5}{2}$ years

$I=\frac{P\times R\times T}{100}\phantom{\rule{0ex}{0ex}}\Rightarrow R=\frac{100\times I}{P\times T}\phantom{\rule{0ex}{0ex}}=\frac{100\times {\displaystyle \frac{9}{40}}x}{x\times {\displaystyle \frac{5}{2}}}\phantom{\rule{0ex}{0ex}}=\frac{45}{5}\phantom{\rule{0ex}{0ex}}=9\%$

Hence, the correct option is (d).

#### Page No 13.9:

#### Question 11:

At what rate percent per annum simple interest will a sum double itself in 10 years?

(a) 8%

(b) 10%

(c) 12%

(d) $12\frac{1}{2}$%

#### Answer:

Amount = 2 times the sum = 2*P*

Simple interest (*I*) = Amount − Sum = 2*P* − *P* = *P*

Let the sum (*P*) be* x*.

Then, simple interest (*I*) = *x*

Rate (*R*) = *R*%

Time (*T*) = 10 years

$I=\frac{P\times R\times T}{100}\phantom{\rule{0ex}{0ex}}\Rightarrow R=\frac{100\times I}{P\times T}\phantom{\rule{0ex}{0ex}}=\frac{100\times x}{x\times {\displaystyle 10}}\phantom{\rule{0ex}{0ex}}=10\%$

Hence, the correct option is (b).

#### Page No 13.9:

#### Question 12:

In what time will a sum of â‚¹8000 amount to â‚¹8360 at 6% per annum simple interest?

(a) 8 months

(b) 9 months

(c) $1\frac{1}{4}$ months

(d) $1\frac{1}{2}$ years

#### Answer:

It is given that,

Amount = â‚¹8360

Sum = â‚¹8000

Simple interest (*I*) = Amount − Sum = â‚¹8360 − â‚¹8000 = â‚¹360

Also,

Rate (*R*) = 6%

Time (*T*) = *T* years

$I=\frac{P\times R\times T}{100}\phantom{\rule{0ex}{0ex}}\Rightarrow T=\frac{100\times I}{P\times R}\phantom{\rule{0ex}{0ex}}=\frac{100\times 360}{8000\times 6}\phantom{\rule{0ex}{0ex}}=\frac{3}{4}\mathrm{years}\phantom{\rule{0ex}{0ex}}=\frac{3}{4}\times 12\mathrm{months}\phantom{\rule{0ex}{0ex}}=9\mathrm{months}$

Hence, the correct option is (b).

#### Page No 13.9:

#### Question 13:

If *a*, *b* and *c* are three sums of money such that *b* is the simple interest on* a* and* c* is the simple interest on *b* for the same time and same rate. Which of the following is correct?

(a) *abc* = 1

(b) *c*^{2} = *ab*

(c) *b*^{2} = *ac*

(d) *a*^{2} = *bc*

#### Answer:

It is given that,

Simple interest (*I*_{1}) = *b*

Sum (*P*_{1}) = *a*

Rate (*R*_{1}) = *R*%

Time (*T*_{1}) = *T* years

Now,

${I}_{1}=\frac{{P}_{1}\times {R}_{1}\times {T}_{1}}{100}\phantom{\rule{0ex}{0ex}}\Rightarrow b=\frac{a\times R\times T}{100}\phantom{\rule{0ex}{0ex}}\Rightarrow R\times T=\frac{100b}{a}....\left(1\right)$

Also,

Simple interest (*I*_{2}) = *c*

Sum (*P*_{2}) = *b*

Rate (*R*_{2}) = *R*%

Time (*T*_{2}) = *T* years

Now,

${I}_{2}=\frac{{P}_{2}\times {R}_{2}\times {T}_{2}}{100}\phantom{\rule{0ex}{0ex}}\Rightarrow c=\frac{b\times R\times T}{100}\phantom{\rule{0ex}{0ex}}\Rightarrow R\times T=\frac{100c}{b}....\left(2\right)$

On equating (1) and (2), we get

$\frac{100b}{a}=\frac{100c}{b}\phantom{\rule{0ex}{0ex}}\Rightarrow {b}^{2}=ac$

Hence, the correct option is (c).

#### Page No 13.9:

#### Question 14:

The simple interest at* R*% per annum for *n *years will be â‚¹*n* on a sum of

(a) â‚¹*n*

(b) â‚¹100*n*

(c) â‚¹$\frac{100}{n}$

(d) â‚¹$\frac{100}{{n}^{2}}$

#### Answer:

It is given that,

Simple interest (*I*) = â‚¹*n*

Rate (*R*) = *R*%

Time (*T*) = *n* years

$I=\frac{P\times R\times T}{100}\phantom{\rule{0ex}{0ex}}\Rightarrow P=\frac{100\times I}{R\times T}\phantom{\rule{0ex}{0ex}}=\frac{100\times n}{R\times n}\phantom{\rule{0ex}{0ex}}=\frac{100}{R}$

Hence, the correct option is (c).

#### Page No 13.9:

#### Question 15:

The simple interest on a certain sum is $\frac{16}{25}$ of the sum. If the rate percent per annum and the time are numerically equal, then the rate percent is

(a) 8%

(b) 4%

(c) 6%

(d) 12%

#### Answer:

Let the sum (*P*) be â‚¹*x*

Then, the simple interest (*I*) = â‚¹$\frac{16}{25}$*x*

Also,

Rate (*R*) = *R*%

Time (*T*) = *R* years (âˆµ the rate percent per annum and the time are numerically equal)

$I=\frac{P\times R\times T}{100}\phantom{\rule{0ex}{0ex}}\Rightarrow R=\frac{100\times I}{P\times T}\phantom{\rule{0ex}{0ex}}\Rightarrow R=\frac{100\times {\displaystyle \frac{16}{25}}x}{x\times R}\phantom{\rule{0ex}{0ex}}\Rightarrow R\times R=\frac{64x}{x}\phantom{\rule{0ex}{0ex}}\Rightarrow R\times R=8\times 8\phantom{\rule{0ex}{0ex}}\Rightarrow R=8\%$

Hence, the correct option is (a).

#### Page No 13.9:

#### Question 16:

At which rate percent per annum simple interest will a sum triple itself in 16 years?

(a) 12%

(b) 10.5%

(c) 11.5%

(d) 12.5%

#### Answer:

Amount = 3 times the sum = 3*P*

Simple interest (*I*) = Amount − Sum = 3*P* − *P* = 2*P*

Let the sum (*P*) be* x*.

Then, simple interest (*I*) = 2*x*

Rate (*R*) = *R*%

Time (*T*) = 16 years

$I=\frac{P\times R\times T}{100}\phantom{\rule{0ex}{0ex}}\Rightarrow R=\frac{100\times I}{P\times T}\phantom{\rule{0ex}{0ex}}=\frac{100\times 2x}{x\times {\displaystyle 16}}\phantom{\rule{0ex}{0ex}}=12.5\%$

Hence, the correct option is (d).

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