# RS Aggarwal Class 7 Solutions Chapter 12 - Simple Interest

## RS Aggarwal Class 7 Chapter 12 - Simple Interest Solutions Free PDF

The rate at which we lend or borrow money is called Simple Interest . The main concept behind this form of interest, is when a person borrows money from a lender they have to pay an extra amount, plus the principal amount. The Simple Interest formula is a culmination of Principal and Interest which can be mathematically represented as:

I= P x R x T

Here, R= Rate of Interest per annum as percent

I= Interest amount

P= Principal Amount

Learn RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest below:

Find the simple interest and the amount when:

Q1: Principal = Rs. 6400, rate = 6% p.a. and time = 2 years

Sol:

P = Rs. 6400, R = 6% , T = 2 years

SI = $\frac{P \times R \times T}{100} = \frac{6400 \times 6 \times 2}{100}$

= Rs. 768

Amount = P+ SI

= 6400 + 768

= Rs. 7168

Q2: Principal = Rs. 2650, rate = 8% p.a. and time = $2\frac{1}{2}$ years.

Sol:

P = Rs. 2650, R = 8% , T = $2\frac{1}{2} = \frac{5}{2}$ years

SI = $\frac{P \times R \times T}{100} = \frac{2650 \times 8 \times 5}{100 \times 2}$

= Rs. 530

Amount = P+ SI

= 2650 + 530

= Rs. 3180

Q3: Principal = Rs. 1500, rate = 12% p.a. and time = 3 years 3 months.

Sol:

P = Rs. 1500, R = 12% , T = $3 + \frac{3}{12} = \frac{13}{4}$ years

SI = $\frac{P \times R \times T}{100} = \frac{1500 \times 12 \times 13 }{100 \times 4}$

= Rs 585

Amount = P+ SI

= 1500 + 585

= Rs. 2085

Q4: Principal = Rs. 9600, rate = $7 \frac{1}{2}$% p.a. and time = 5 months.

Sol:

P = Rs. 9600, R = $7 \frac{1}{2}% = \frac{15}{2}$ , T = 5 months = $\frac{5}{12}$ years

SI = $\frac{P \times R \times T}{100} = \frac{9600 \times 15 \times 5 }{100 \times 2 \times 12}$

= Rs 300

Amount = P+ SI

= 9600 + 300

= Rs. 9900

Q5: Principal = Rs. 5000, rate = 9% p.a. and time = 146 days.

Sol:

P = Rs. 5000, R = 9% , T = 146 days = $\frac{146}{365}$ years

SI = $\frac{P \times R \times T}{100} = \frac{5000 \times 9 \times 146 }{100 \times 365 }$

= Rs 180

Amount = P+ SI

= 5000 + 180

= Rs. 5180

Find the time when:

Q6: Principal = Rs. 6400, SI = Rs. 1152 and rate = 6% p.a.

Sol:

P = Rs. 6400, SI = Rs. 1152, R = 6%

$T = \frac{SI \times 100}{P \times R} = \frac{1152 \times 100}{6400 \times 6}$

= $\frac{1152}{384}$

= 3 years

Q7: Principal = Rs. 9540, SI = Rs. 1908 and rate = 8% p.a.

Sol:

P = Rs. 9540, SI = Rs. 1908, R = 8%

$T = \frac{SI \times 100}{P \times R} = \frac{1908 \times 100}{9450 \times 8}$

= $\frac{10}{4}$

= $2\frac{1}{2}$ years

Q8: Principal = Rs. 5000, Amount = Rs. 6450 and rate = 12% p.a.

Sol:

P = Rs. 5000, A = Rs. 6450 , R = 12%

SI = A – P

= 6450 – 5000

= Rs. 1450

$T = \frac{SI \times 100}{P \times R} = \frac{1450 \times 100}{5000 \times 12}$

= $\frac{20}{12}$

= $2 \frac{5}{12}$ years

= 2 years 5 months

Find the rate when:

Q9: Principal = Rs. 8250, SI = Rs. 1100 and time = 2 years.

Sol:

P = Rs. 8250, SI = Rs. 1100, T = 2 years

$R = \frac{SI \times 100}{P \times T} = \frac{1100 \times 100}{8250 \times 2}$

= $\frac{1100}{165}$

= 6.67 %

Q10: Principal = Rs. 5200, SI = Rs. 975 and time = $2\frac{1}{2}$ years.

Sol:

P = Rs. 5200, SI = Rs. 975, T = $2\frac{1}{2} = \frac{5}{2}$ years

$R = \frac{SI \times 100}{P \times T} = \frac{975 \times 100 \times 2 }{5200 \times 5}$

= $\frac{195}{26}$

= 7.5 %

Q11: Principal = Rs. 3560, Amount = Rs. 4521.20 and time = 3 years.

Sol:

P = Rs. 3560, A = 4521.20, T = 3 years

SI = A – P = 4521.20 – 3560

= Rs. 961.20

$R = \frac{SI \times 100}{P \times T} = \frac{ 961.20 \times 100}{3560 \times 3}$

= 9 %

Q12: Shanta borrowed Rs. 6000 from the State Bank of India for 3 years 8 months at 12% per annum. What amount will clear off her debt?

Sol:

P = Rs. 6000, R = 12%, T = 3 years 8 months = $3\frac{8}{12} = \frac{44}{12}$ years

SI = $\frac{P \times R \times T}{100} = \frac{6000 \times 12 \times 44 }{100 \times 12 }$

= Rs 2640

Amount = P+ SI

= 6000 + 2640

= Rs. 8640

Q13: Hari borrowed Rs. 12600 from a moneylender at 15% per annum simple interest. After 3 years, he paid Rs. 7070 and gave a goat to clear off the debt. What is the cost of the goat?

Sol:

P = Rs. 12600, R = 15%, T = 3 years

SI = $\frac{P \times R \times T}{100} = \frac{12600 \times 15 \times 3 }{100}$

= Rs 5670

Amount = P+ SI

= 12600 + 5670

= Rs. 18270

Hari had to pay Rs. 18270 to the money lender, but he paid Rs. 7070 and a goat.

Therefore, the Cost of the goat = Rs. 18270 – Rs. 7070

= Rs. 11200

Q14: The simple interest on a certain sum of 3 years at 10% per annum is Rs. 829.50. Find the sum.

Sol:

Let the sum be Rs. P

SI = Rs. 829.50, T = 3 years, R = 10%

Now, $P = \frac{SI \times 100}{R \times T}$

= $P = \frac{829.50 \times 100}{10 \times 3}$

= $\frac{8295}{3}$

= 2765

Hence, the sum is Rs. 2765.

Q15: A sum when reckoned at $7 \frac{1}{2}$ % per annum amounts to Rs. 3920 in 3 years. Find the sum.

Sol:

Let the required sum be Rs. x

A = Rs.3920, R = $7 \frac{1}{2}$% , T = 3 years

Now,

SI = $\frac{P \times R \times T}{100} = \frac{x \times 15 \times 3 }{100 \times 2} = \frac{9x}{40}$

Amount = P+ SI

= x + $\frac{9x}{40} = \frac{40 x + 9x}{40} = \frac{49 x} {40}$

But the amount is Rs. 3920

$\Rightarrow \frac{49 x}{40} = 3920$ $\Rightarrow x = \frac{3920 \times 40}{49} = 3200$

Hence, the required sum is Rs. 3200.

Q16: A sum of money put at 11% per annum amounts to Rs. 4491 in 2 years 3 months. What will it amount to in 3 years at the same rate?

Sol:

Given:

R = 11%, T = 2 years 3 months = 2 + $\frac{3}{12} = \frac{27}{12}$ years

Let the required sum be Rs. x.

SI = $\frac{P \times R \times T}{100} = \frac{x \times 11 \times 27 }{100 \times 12} = \frac{99x}{400}$

Amount = P+ SI

= x + $\frac{99x}{400} = \frac{400 x + 99 x}{400} = \frac{499 x} {400}$

But the amount is Rs. 4491

$\Rightarrow \frac{499 x}{400} = 4491$ $\Rightarrow x = \frac{4491 \times 400}{499} = 3600$

Hence, the required sum is Rs. 3600.

Therefore SI for 3 years = $\frac{P \times R \times T}{100} = \frac{3600 \times 11 \times 3 }{100 } = Rs. 1188$

And Amount = P + SI = 3600 + 1188

= Rs. 4788

Q17: A sum of money invested at 8% per annum amounts to Rs. 12122 in 2 years. What will it amount to in 2 years 8 months at 9% per annum?

Sol:

Let the required sum be Rs. x

SI = $\frac{P \times R \times T}{100} = \frac{x \times 8 \times 2 }{100} = \frac{16x}{100}$

Amount = P+ SI

= x + $\frac{16x}{100} = \frac{100 x + 16 x}{100} = \frac{116 x} {100}$

But the amount is Rs. 12122

$\Rightarrow \frac{116 x}{100} = 12122$ $\Rightarrow x = \frac{12122 \times 100}{116} = 10450$

Amount in 2 years 8 months at 9% per annum will be:

SI = $\frac{P \times R \times T}{100} = \frac{10450 \times 9 \times 32 }{100 \times 12} = Rs. 2508$

Amount = P+ SI

= Rs. 10450 + Rs. 2508

= Rs. 12958

Q18: At what rate per cent annum will Rs. 3600 amount to Rs. 4734 in $3\frac{1}{2}$ years?

Sol:

P = Rs. 3600, A = Rs. 4734, T = $3\frac{1}{2} = \frac{7}{2}$ years

SI = A – P

= 4734 – 3600

= Rs. 1134

R = $\frac{SI \times 100}{P \times T}$

= $\frac{1134 \times 100 \times 2}{3600 \times 7}$

= 9%

Q19: If Rs. 640 amounts to Rs. 768 in 2 years 6 months, what will Rs. 850 amount to in 3 years at the same rate per cent per annum?

Sol:

P = Rs. 60, A = 768, T = 2 years 6 months = $\frac{5}{2}$ years

SI = A – P

= 768 – 640

= Rs. 128

R = $\frac{SI \times 100}{P \times T}$

= $\frac{128 \times 100 \times 2}{640 \times 5}$

= 8%

P = Rs. 850, R = 8%, T = 3 years

Therefore, SI = $\frac{P \times R \times T}{100} = \frac{800 \times 8 \times 3 }{100} = \frac{2040}{10}$

= Rs. 204

Amount = P+ SI

= 850 + 204

= Rs. 1054

Q20: In what time will Rs. 5600 amount to Rs. 6720 at 8% per annum?

Sol:

P = Rs. 5600, A = 6720, R = 8%

SI = A – P

= 6720 – 5600

= Rs. 1120

T = $\frac{SI \times 100}{P \times R}$

= $\frac{1120 \times 100}{5600 \times 8}$

= $\frac{1120}{448}$

= $2 \frac{1}{2}$ years

Q21: A sum of money becomes $\frac{8}{5}$ of itself in 5 years at a certain rate of simple interest. Find the rate of interest.

Sol:

Let the sum be Rs. x.

Amount = $\frac{8x}{5}$

Therefore, SI = A – P = $\frac{8x}{5}$ – x

= $\frac{3x}{5}$

Let the rate be R%.

SI = $\frac{P \times R \times T}{100}$ $\Rightarrow \frac{3x}{5} = \frac{x \times R \times 5 }{100}$ $\Rightarrow 3 x \times 20 = R \times x \times 5$ $\Rightarrow R = 12$

Hence the rate of interest is 12%.

Q22: A sum of money lent at simple interest amounts to Rs. 783 in 2 years and to Rs. 837 in 3 years. Find the sum and the rate percent annum.

Sol:

Amount in 3 years = (Principal + SI for 3 years) = Rs. 837

Amount in 2 years= ( Principal + SI for 2 years) = Rs. 783

On subtracting :

SI for 1 year = ( 837 – 783) = Rs. 54

SI for 2 years = $\left ( \frac{54}{1} \times 2 \right )$ = Rs. 108

Therefore, Sum = Amount for 2 years – SI for 2 years

= 783 – 108

= Rs. 675

P = Rs. 675, SI = Rs. 108 and T = 2 years

R = $\frac{SI \times 100}{P \times T}$

= $\frac{ 108 \times 100}{675 \times 2}$

= 8%

Q23: A sum of money lent at simple interest amounts to Rs. 4745 in 3 years and to Rs. 5475 in 5 years. Find the sum and the rate per cent per annum.

Sol:

Amount in 5 years = ( Principal + SI for 5 years) = Rs. 5475

Amount in 3 years= ( Principal + SI for 3 years) = Rs. 4745

On subtracting :

SI for 2 year = (5475 – 4745) = Rs. 730

SI for 2 years = $\left ( \frac{730}{2} \times 3 \right )$ = Rs. 1095

Therefore, Sum = Amount for 3 years – SI for 3 years

= 4745 – 1095

= Rs. 3650

P = Rs. 3650, SI = Rs. 1095 and T = 3 years

R = $\frac{SI \times 100}{P \times T}$

= $\frac{ 1095 \times 100}{3650 \times 3}$

= 10%

Q24: Divide Rs. 3000 into two parts such that the simple interest on the first part for 4 years at 8% per annum is equal to the simple interest on the second part for 2 years at 9% per annum.

Sol:

Let the first part be Rs. x.

Second part = (3000 – x)

Therefore, SI on x at 8% per annum for 4 years = $\frac{x \times 8 \times 4}{100} = \frac{8x}{25}$

SI on ( 3000 – x) at 9% per annum = $\frac{(3000 – x) \times 9 \times 2}{100} = \frac{27000 – 9x}{50}$

Therefore, $\frac{8x}{25} = \frac{27000 – 9x}{50}$ $\Rightarrow 8x = \frac{(27000 – 9x) \times 25}{50}$ $\Rightarrow 16x = 27000 – 9x$ $\Rightarrow 25x = 27000$ $\Rightarrow x = 1080$

Therefore First part = Rs. 1080

Second Part = (3000 – 1080) = Rs. 1920

Q25: Divide Rs. 3600 into two parts such that if one part be lent at 9% per annum and the other at 10% per annum, the total annual income is Rs. 333.

Sol:

Let the first part be Rs. x.

Second part = (3600 – x)

Therefore, SI on x at 9% per annum for 1 years = $\frac{x \times 9 \times 1}{100} = \frac{9x}{100}$

SI on ( 3600 – x) at 10% per annum = $\frac{(3600 – x) \times 10 \times 1}{100} = \frac{3600 – x}{10}$

Therefore, $\frac{9x}{100} + \frac{3600 – x}{100} = 333$ $\Rightarrow \frac{9x + 36000 – 10x}{100} = 333$ $\Rightarrow -x + 36000 = 33300$ $\Rightarrow -x = -2700$ $\Rightarrow x = 2700$

Therefore First part = Rs. 2700

Second Part = (3600 – 2700) = Rs. 900

Exercise 12B

Mark the correct answer in each of the following:

Q1: The simple interest on Rs. 6250 at 4% per annum for 6 months is

(a) Rs. 125                  (b) Rs. 150                  (c) Rs. 175               (d) Rs. 135

Sol:

(a) Rs. 125

Principal = Rs. 6250

Rate = 4% per annum

Time = 6 months = $\frac{1}{2}$ years

SI = $\frac{P \times R \times T}{100}$ $= \frac{6250 \times 4 \times 1}{100 \times 2}$ $= \frac{250 }{2}$

= Rs. 125

Q2: A sum amounts to Rs. 3605 in 219 days at 5% per annum. The sum is

(a) Rs. 3250               (b) Rs. 3500                (c) Rs. 3400            (d) Rs. 3550

Sol:

(b) Rs. 3500

Amount = Rs. 3605

Time = 219 days = $\frac{219}{365}$ years

Rate = 5% per annum

Amount = Sum + $\frac{Sum \times Rate \times Time}{100}$

= Sum$\left ( 1 + \frac{Rate \times Time}{100} \right )$

Sum = $\frac{3605}{1 + \frac{5}{100} \times \frac{219}{365}} = \frac{3605 \times 36500}{37595}$

Sum = Rs. 3500

Q3: At simple interest a sum becomes $\frac{6}{5}$ of itself in $2 \frac{1}{2}$ years. The rate of interest per annum is

(a) 6%              (b) $7 \frac{1}{2}$ %               (c) 8%              (d) 9%

Sol:

(c) 8%

Le the sum be Rs. x.

Rate of interest = r%

Time = $2\frac{1}{2}$ years = $\frac{5}{2}$ years

Amount = $\frac{6}{5} \times Sum$

Rate = ?

Principal + SI = Amount

Principal + $\frac{Principal \times Rate \times Time}{100} = \frac{6}{5} \times Principal$ $\Rightarrow \frac{x r \times 5}{100 \times 2} = \frac{6}{5} \times x$ $\Rightarrow x \left ( 1 + \frac{ 5r }{100 \times 2} \right ) = \frac{6}{5} \times x$ $\Rightarrow 1 + \frac{r}{40} = \frac{6}{5}$ $\Rightarrow r = 40 \times \frac{1}{5}$ $\Rightarrow r = 8$

So, the rate of interest is 8%.

Q4: In what time will Rs. 8000 amount to Rs. 8360 at 6% per annum simple interest?

(a) 8 months                (b) 9 months                (c) $1 \frac{1}{4}$ years             (d) $1 \frac{1}{2}$ years

Sol:

(b) 9 months

Let the time be t years.

Principal = Rs. 8000

Amount = Rs. 8360

Rate = 6% per annum

Amount = Principal $\left ( 1 + \frac{Rate \times Time}{100} \right )$

8360 = 8000 $\times \left ( 1 + \frac{6 \times t}{100} \right )$ $\frac{8360}{8000} = \left ( 1 + \frac{6 \times t}{100} \right )$ $\Rightarrow \frac{8360}{8000} – 1 = \frac{6 t}{100}$ $\Rightarrow t = \left ( \frac{8360 – 8000}{8000} \right ) \times \frac{100}{6}$ $= \frac{360}{8000} \times \frac{100}{6}$ $= \frac{6}{8} \times 12$ months

= 9 months

Q5: At what rate per cent annum simple interest will a sum double itself in 10 years?

(a) 8%                      (b) 10%                      (c) 12%                       (d) $12 \frac{1}{2}$%

Sol:

(b) 10%

Let the sum be Rs. x and the rate be r%.

Amount = 2r

Principal + SI = Amount

P + SI = 2x

$\Rightarrow P + \frac{P \times R \times T}{100} = 2x$ $\Rightarrow x\left ( 1 + \frac{r \times 10}{100} \right ) = 2x$ $\Rightarrow \frac{100 + 10r}{100} = 2$ $\Rightarrow$ 10r = 200 – 100

$\Rightarrow$10 r = 100

$\Rightarrow r = \frac{100}{10} = 10$

Q6: The simple interest at x% per annum for x years will be Rs. x on a sum of

(a) Rs. x                      (b) Rs. 100x                    (c) Rs. $\frac{100}{x}$                      (d) Rs. $\frac{100}{x^{2}}$

Sol:

(c) Rs. $\frac{100}{x}$

Simple interest = Rs. x

Rate = x% per annum

Time = x years

Time = x years

SI = $\frac{P \times R \times T}{100}$ $\Rightarrow x = \frac{Principal \times x \times x}{100} = 2$ $\Rightarrow Principal = Rs. \frac{100}{x}$

Q7: The simple interest on a sum for 5 years is $\frac{2}{5}$ of the sum. The rate per cent per annum is

(a) 10%                   (b) 8%                        (c) 6%                  (d) $12 \frac{1}{2}$%

Sol:

(b) 8%

Time = 5 years

SI = $\frac{2}{5}$P

$\Rightarrow \frac{P \times R \times T}{100} = \frac{2}{5}$P

$\Rightarrow \frac{Rate \times 5}{100} = \frac{2}{5}$ $\Rightarrow Rate = \frac{2 \times 100}{5 \times 5}$

Rate = 8%

Q8: A borrows Rs. 8000 at 12% per annum simple interest and B borrows Rs. 9100 at 10% per annum simple interest. In how many years will their amounts be equal?

(a) 18 years             (b) 20 years              (c) 22 years               (d) 24 years

Sol:

(c) 22 years

$R_{1}$ = 12%

$R_{2}$ = 10%

$P_{1}$ = Rs. 8000

$P_{2}$ = Rs. 9100

Let their amounts be equal in T years.

$Amount_{1} = SI_{1} + P_{1}$ $= \frac{P_{1} \times R_{1} \times T }{100} + P_{1}$

= $= \frac{8000 \times 12 \times T }{100} + 8000$

= 960T + 8000

$Amount_{2} = SI_{2} + P_{2}$ $= \frac{P_{2} \times R_{2} \times T }{100} + P_{2}$

= $= \frac{9100 \times 10 \times T }{100} + 9100$

= 910T + 9100

$Amount_{1} = Amount_{2}$ $\Rightarrow$ 960T + 8000 = 910T + 9100

$\Rightarrow$ 960T – 910T = 9100 – 8000

$\Rightarrow$ 50T = 1100

$\Rightarrow$ T = 22

Hence, after 22 years their amounts will be equal.

Q9: A sum of Rs. 600 amounts to Rs. 720 in 4 years. What will it amount to if the rate of interest is increased by 2%?

(a) Rs. 724                (b) Rs. 648                    (c) Rs. 768               (d) Rs. 792

Sol:

(c) Rs. 768

Let the rate be R%.

SI = A – P

= 720 – 600

= Rs. 120

Time = 4 years

R = $\frac{100 \times SI }{P \times T}$

R = $\frac{100 \times 120}{600 \times 4 }$

= 5

Rate of interest = 5%

Now R = 5 + 2 = 7%

SI = $\frac{P \times R \times T}{100}$

= $\frac{600 \times 7 \times 4 }{100}$

= Rs. 168

Amount = SI  + P

= 600 + 168

= Rs. 768

Q10: x, y and z are three sums of money such that y is the simple interest on x and z is the simple interest on y for the same time and same rate. Which of the following is correct?

(a) xyz = 1                (b) $z^{2} = xy$                 (c) $x^{2} = yz$                   (d) $y^{2} = zx$

Sol:

(d) $y^{2} = zx$

y = SI on x = $\frac{x \times R \times T }{100}$  ……….(i)

Z = SI on y = $\frac{y \times R \times T }{100}$  ……….(ii)

Dividing equation (i) by (ii) :

$\Rightarrow \frac{y}{z} = \left ( \frac{x \times R \times T}{100} \times \frac{100}{y \times R \times T} \right )$ $\Rightarrow \frac{y}{z} = \frac{x}{y}$ $\Rightarrow y^{2} = zx$

Q11: In how much time would the simple interest on a certain sum be 0.125 \times  the principal at 10% per annum?

(a) $1\frac{1}{4}$ years                    (b) $1\frac{3}{4}$ years                (c) $2 \frac{1}{4}$ years                  (d) $2 \frac{3}{4}$ years

Sol:

(a) $1\frac{1}{4}$ years

Rate = 10% per annum

SI = 0.125 $\times$ Principal

$\Rightarrow \frac{Principal \times Rate \times Time}{100} = 0.125 \times Principal$ $\Rightarrow \frac{Time}{10} = 0.125$ $\Rightarrow Time = 1.25$ =  $1\frac{1}{4}$ years

Q12: At which sum will simple interest at the rate of $3 \frac{3}{4}$% per annum be Rs. 210 in $2 \frac{1}{3}$ years?

(a) Rs. 1580            (b) Rs. 2400            (c) Rs. 2800          (d) none of these

Sol:

(b) Rs. 2400

Rate = $3 \frac{3}{4}$% per annum

= $\frac{15}{4}$% per annum

Time = $2 \frac{1}{3}$ years

$\frac{7}{3}$ years

SI = $\frac{P \times \frac{15}{4} \times \frac{7}{3}}{100}$ $\Rightarrow P = \frac{210 \times 100}{\left ( \frac{15}{4} \times \frac{7}{3}\right )}$ $\Rightarrow P = 600 \times 4$ $\Rightarrow$ P = Rs. 2400

#### Practise This Question

Photosynthesis occurs with the help of a pigment present in chloroplasts called