RS Aggarwal Class 7 Solutions Simple Interest

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 per cent 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

A Man moves on a straight horizontal road with a block of 2kg in his hand. If he covers a distance of 20m with constant velocity of 1m/s. Find work done by man on block during motion.