RS Aggarwal 2018 Solutions for Class 7 Math Chapter 12 - Simple Interest

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RS Aggarwal 2018 Solutions for Class 7 Math Chapter 12 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 RS Aggarwal 2018 Book of class 7 Math Chapter 12 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RS Aggarwal 2018 Solutions. All RS Aggarwal 2018 Solutions for class 7 Math are prepared by experts and are 100% accurate.

Page No 165:

Question 1:

Find the simple interest and the amount when:

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

Answer:

$\begin{array}{l} P = Rs . 6400, R = 6%, T = 2 years \\ S . I . = \frac{P×R×T}{100} = \frac{6400 \times 6 \times 2}{100} \\ = Rs. 768 \\ Amount=P+S.I . \end{array} =6400+768 = Rs. 7168$

Page No 165:

Question 2:

Find the simple interest and the amount when:

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

Answer:

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

$S.I. = \frac{P×R×T}{100} = \frac{2650×8×5}{100×2}$

$\begin{array}{l} = Rs . 530 \\ Amount = P + S . I . \end{array} =2650+530 = Rs . 3180$

Page No 165:

Question 3:

Find the simple interest and the amount when:

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

Answer:

$\begin{array}{l} P = Rs . 1500, R = 12 %, T = 3 + \frac{3}{12} = \frac{13}{4} years \\ S .I . = \frac{P×R×T}{100} = \frac{1500×12×13}{100×4} \\ = Rs . 585 \\ Amount=P+S .I . \end{array} =1500+585 =Rs . 2085$

Page No 165:

Question 4:

Find the simple interest and the amount when:

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

Answer:

$P = Rs . 9600 R = 7 \frac{1}{2} % T = 5 months = \frac{5}{12} years S . I . = \frac{P \times R \times T}{100} = \frac{9600 \times 15 \times 5}{100 \times 2 \times 12} = Rs . 300 Amount = P + S . I . = 9600 + 300 = Rs . 9900$

Page No 165:

Question 5:

Find the simple interest and the amount when:

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

Answer:

$\begin{array}{l} P = Rs . 5000, R = 9 %, T = 146 days = \frac{146}{365} years \\ S .I . = \frac{P×R×T}{100} = \frac{5000×9×146}{100×365} \\ =Rs . 180 \\ Amount=P+S .I . \end{array} =5000+180 =Rs . 5180$

Page No 165:

Question 6:

Find the time when:

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

Answer:

$\begin{array}{l} P = Rs . 64 00, S . I. = Rs . 1152, R = 6 % \\ T = \frac{S .I.×100}{P×R} = \frac{1152 \times 100}{6400 \times 6} \\ = \frac{1152}{384} \end{array} = 3 years$

Page No 165:

Question 7:

Find the time when:

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

Answer:

$\begin{array}{l} P = Rs . 9540, S . I. = Rs . 190 8, R = 8 % \\ T = \frac{S .I.×100}{P×R} = \frac{1908×100}{9540×8} \\ = \frac{10}{4} \\ =2 \frac{1}{2} years \end{array}$

Page No 165:

Question 8:

Find the time when:

Principal = Rs 5000, amount = Rs 6450 and rate = 12% p.a.

Answer:

$\begin{array}{l} P = Rs . 5 000, A=Rs . 645 0, R = 12 % \\ S . I. = A - P \\ = 6450 - 5000 \\ = Rs . 145 0 \\ T = \frac{S .I×100}{P×R} = \frac{1450×100}{5000×12} \\ = \frac{29}{12} \\ = 2 \frac{5}{12} \\ =2 years 5 months \end{array}$

Page No 166:

Question 9:

Find the rate when:

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

Answer:

$\begin{array}{l} P = Rs . 8250, S . I. = Rs . 1100, T = 2 years \\ R= \frac{S .I.×100}{P×T} = \frac{1100×100}{8250×2} \\ = \frac{1100}{165} =6 .67 % \end{array}$

Page No 166:

Question 10:

Find the rate when:

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

Answer:

$\begin{array}{l} P= Rs . 5200 , S .I.=Rs . 975 [ T=2 \frac{1}{2} years= \frac{5}{2} years] \\ R= \frac{S .I.×100}{P×T} = \frac{975×100×2}{5200×5} \\ = \frac{195}{26} \end{array} =7 .5%$

Page No 166:

Question 11:

Find the rate when:

Principal = Rs 3560, amount = Rs 4521.20 and time = 3 years.

Answer:

$\begin{array}{l} P = Rs . 356 0, A = Rs . 4521 . 20, T = 3 years \\ S . I. = A - P = 4521.20 - 3560 \\ = Rs . 961 . 20 \\ R = \frac{S .I.×100}{P×T} = \frac{961 .20×100}{3560×3} \\ = \frac{96120×100}{100×3560×3} \\ =9% \end{array}$

Page No 166:

Question 12:

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?

Answer:

$\begin{array}{l} P = Rs 6 000, R = 12 %, T= 3 years 8 months = 3 \frac{8}{12} = \frac{44}{12} years \\ S .I. = \frac{P×R×T}{100} = \frac{6000×12×44}{100×12} = Rs 2640 \\ A = P + S . I. \\ = 6000 + 2640 \\ = Rs 864 0 \end{array}$

Page No 166:

Question 13:

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?

Answer:

$\begin{array}{l} P = Rs . 126 00 R = 15 % T = 3 years \\ S .I.= \frac{P×R×T}{100} = \frac{12600×15×3}{100} \\ =Rs . 5670 \\ A = Rs . 126 00 + Rs . 567 0 = Rs . 1827 0 \\ Hari had to pay Rs . 1827 0 to the money lender, but he paid Rs . 7 070 and a goat . \\ ∴ Cost of the goat = Rs . 1827 0 - Rs . 7 070 \end{array} = Rs . 112 00$

Page No 166:

Question 14:

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

Answer:

$\begin{array}{l} Let the sum be Rs. P. \\ S .I. = Rs . 829 . 50, T = 3 years, R = 10 % \\ Now, P= \frac{S .I×100}{R×T} \\ = \frac{829 .50×100}{10×3} \\ = \frac{8295}{3} \end{array} = 2765 Hence, the sum is Rs . 2765 .$

Page No 166:

Question 15:

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

Answer:

$\begin{array}{l} Let the required sum be Rs. x . \\ A = Rs . 392 0, R = 7 \frac{1}{2} %, T = 3 years \\ Now, \\ Now, S .I .= \frac{P×R×T}{100} = \frac{x \times 15 \times 3}{2 \times 100} = \frac{9 x}{40} \\ A = P + S . I. \\ = x + \frac{9 x}{40} = \frac{40 x + 9 x}{40} = \frac{49 x}{40} \\ But the amount is Rs . 392 0 . \\ = > \frac{49 x}{40} = 3920 \\ = > x = \frac{3920 \times 40}{49} = \frac{156800}{49} = 3200 \\ Hence, the required sum is Rs . 32 00 . \end{array}$

Page No 166:

Question 16:

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?

Answer:

$\begin{array}{l} Given: R=11%, T=2 years 3 months = 2+ \frac{3}{12} = \frac{27}{12} years \\ Let the required sum be Rs. x . \\ S .I.= \frac{P×R×T}{100} = \frac{x \times 11 \times 2 7^{9}}{100 \times 1 2_{4}} = \frac{99 x}{400} \\ A = P + S . I. \\ = x + \frac{99 x}{400} = \frac{400 x + 99 x}{400} = \frac{499 x}{400} \\ But the amount is Rs . 4491. \\ = > \frac{499 x}{400} = 4491 \\ = > x = \frac{4491 \times 400}{499} = \frac{1796400}{499} = 3600 \\ Hence, the required sum is Rs . 3600. \\ ∴ S .I .= \frac{P \times R \times T}{100} = \frac{3600 \times 11 \times 3}{100} = Rs . 1188 \\ ∴ Amount=P+S .I.=3600+1188 \\ =Rs . 4788 \end{array}$

Page No 166:

Question 17:

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?

Answer:

$\begin{array}{l} Let the required sum be Rs. x . \\ S .I.= \frac{P×R×T}{100} = \frac{x \times 8 \times 2}{100} = \frac{16 x}{100} \\ A = P + S . I. \\ = x + \frac{16 x}{100} = \frac{100 x + 16 x}{100} = \frac{116 x}{100} \\ But the amount is Rs . 12122. \\ = > \frac{116 x}{100} = 12122 \\ = > x = \frac{12122 \times 100}{116} = 10450 \\ Now, S .I.= \frac{P×R×T}{100} = \frac{10450 \times 9^{3} \times 3 2^{8}}{100 \times 1 2_{4_{1}}} = Rs . 2508 \\ ∴ A=P+S .I. \\ = Rs . 10450 + Rs . 2508 \\ = Rs . 12958 \end{array}$

Page No 166:

Question 18:

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

Answer:

$\begin{array}{l} P = Rs . 36 00 A = Rs . 4734 T= 3 \frac{1}{2} = \frac{7}{2} years \\ S . I. = A - P \\ = 4734 - 3600 \\ = Rs . 1134 \\ R= \frac{S .I.×100}{P×T} \end{array} = \frac{1134×100×2}{3600×7} =9%$

Page No 166:

Question 19:

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?

Answer:

$\begin{array}{l} P = Rs . 64 0, A = Rs . 768, T= 2 years 6 months = \frac{5}{2} years \\ S . I. = A - P \\ = 768 - 640 \\ = Rs . 128 \\ R= \frac{S .I.×100}{P×T} = \frac{128×100×2}{640×5} =8% \\ P = Rs . 85 0, R = 8 %, T = 3 years \\ ∴ S .I. = \frac{P×R×T}{100} = \frac{850×8×3}{100} = \frac{2040}{10} =Rs . 204 \\ ∴ A = P + S . I. \\ = 850 + 204 \\ = Rs . 1054 \end{array}$

Page No 166:

Question 20:

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

Answer:

$\begin{array}{l} P = Rs . 56 00, A = Rs . 672 0, R = 8 % \\ S . I. = A - P \\ = 6720 - 5600 \\ = Rs . 112 0 \\ T= \frac{S .I.×100}{P×R} \end{array} = \frac{1120×100}{5600×8} = \frac{1120}{448} =2 \frac{1}{2} years$

Page No 166:

Question 21:

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.

Answer:

$\begin{array}{l} Let the sum be Rs. x . \\ Amount= \frac{8 x}{5} \\ ∴ S .I .=A - P= \frac{8 x}{5} - x \\ = \frac{3 x}{5} \\ Let the rate be R % . \\ S .I.= \frac{P×R×T}{100} \\ = > \frac{3 x}{5} = \frac{x \times R \times 5^{1}}{10 0_{20}} \\ = > 3 x \times 20 = R \times x \times 5 \\ = > R = \frac{3 \times x \times 2 0^{4}}{x \times 5} = 12 \end{array} Hence, the rate of interest is 12 % .$

Page No 166:

Question 22:

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 per annum.

Answer:

$\begin{array}{l} Amount in 3 years = (Principal + S . I . for 3 years) = Rs . 837 \\ Amount in 2 years = (Principal + S . I. for 2 years) = Rs . 783 \\ On subtracting : \\ S . I. for 1 year = (837 - 783) = Rs . 54 \\ S . I. for 2 years= (\frac{54}{1} \times 2) = Rs . 108 \\ ∴ Sum = Amount for 2 years - S . I. for 2 years \\ = 783 - 108 \\ = Rs . 675 \\ P = Rs . 675, S . I. = Rs . 1 0 8 and T = 2 years \\ R = \frac{S .I . \times 100}{P \times T} \\ = \frac{108 \times 10 0^{5 0^{2}}}{67 5_{27} \times 2_{1}} \\ = 8 % \end{array}$

Page No 166:

Question 23:

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

Answer:

$\begin{array}{l} Amount in 5 years = (Principal + S . I . for 5 years) = Rs . 5475 \\ Amount in 3 years = (Principal + S . I. for 3 years) = Rs . 4745 \\ On subtracting : \\ S . I. for 2 years = (5475 - 4745) = Rs . 73 0 \\ S . I. for 3 years= (\frac{730}{2} \times 3) = Rs . 1095 \\ ∴ Sum = Amount for 3 years - S . I. for 3 years \\ = 4745 - 1095 \\ = Rs . 3650 \\ P=Rs . 3650, S .I .=Rs . 1095, T=3 years \\ R= \frac{S .I . \times 100}{P \times T} \\ = \frac{1095 \times 100}{3650 \times 3} \\ = 10% \end{array}$

Page No 166:

Question 24:

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.

Answer:

$\begin{array}{l} Let the first part be Rs. x . \\ Second part = (3000 - x) \\ ∴ S . I. on x at 8 % per annum for 4 years = \frac{x \times 8 \times 4^{2^{1}}}{10 0_{5 0_{25}}} = \frac{8 x}{25} \\ S . I. on (3000 - x) at 9 % per annum = \frac{(3000 - x) \times 9 \times 2^{1}}{10 0_{50}} \\ = \frac{27000 - 9 x}{50} \\ ∴ \frac{8 x}{25} = \frac{27000 - 9 x}{50} \\ = > 8 x = \frac{(27000 - 9 x) \times 2 5^{1}}{5 0_{2}} \\ = > 16 x = 27000 - 9 x \\ = > 16 x + 9 x = 27000 \\ = > x = \frac{2700 0^{1080}}{2 5_{1}} = 1080 \\ ∴ F irst part = Rs .1 080 \end{array} Second part = (3000 - 1080) = Rs . 192 0$

Page No 166:

Question 25:

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 annum income is Rs 333.

Answer:

$\begin{array}{l} Let the first part be Rs. x . \\ Second part = (3600 - x) \\ ∴ S . I. on x at 9 % per annum for 1 years = \frac{x \times 9 \times 1}{100} = \frac{9 x}{100} \\ And, S . I. on (3600 - x) at 10 % per annum = \frac{(3600 - x) \times 1 \times 1 0^{1}}{100} = \frac{3600 - x}{10} \\ ∴ \frac{9 x}{100} + \frac{3600 - x}{10} = 333 \\ = > \frac{9 x + 36000 - 10 x}{100} = 333 \\ = > - x + 36000 = 33300 \\ = > - x = 33300 - 36000 \\ = > - x = - 2700 \\ = > x = 2700 \\ First part = Rs . 2700 \end{array} S econd part = (3600 - 2700) = Rs . 900$

Page No 166:

Question 1:

Mark (✓) against the correct answer
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

Answer:

$\begin{array}{l} (a) Rs . 125 \end{array} Principal = Rs . 6250 \begin{array}{l} Simple Interest = 4 % per annum \\ Time = 6 months = \frac{1}{2} years \\ Simple Interest= \frac{P×R×T}{100} \\ Simple Interest= \frac{6250×4×1}{100×2} \\ Simple Interest= \frac{250}{2} = Rs . 125 \end{array}$

Page No 166:

Question 2:

Mark (✓) against the correct answer
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

Answer:

(b) Rs.3500

$\begin{array}{l} Amount =Rs . 3605 \\ Time = \frac{219}{365} days= \frac{219}{365} days \\ Rate=5% per annum \\ Amount = Sum + \frac{Sum×Rate×Time}{100} \\ Amount = Sum (1+ \frac{Rate×Time}{100}) \\ Sum= \frac{3605}{1+ \frac{5}{100} × \frac{219}{365}} = \frac{3605×36500}{37595} \\ Sum= Rs . 3500 \end{array}$

Page No 166:

Question 3:

Mark (✓) against the correct answer
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%

Answer:

(c) 8%

$\begin{array}{l} Let the sum be Rs. x . \\ Rate of interest = r % \\ Time=2 \frac{1}{2} years= \frac{5}{2} years \\ Amount= \frac{6}{5} × Sum \\ Rate=? \\ Amount = \frac{6}{5} ×Sum \\ Principal + S .I . = Amount \\ Principal+ \frac{Principal \times Rate×Time}{100} = \frac{6}{5} ×Principal \\ => x + \frac{x r \times 5}{100×2} = \frac{6}{5} x \\ => x (1 + \frac{5 r}{100×2}) = \frac{6}{5} x \\ => 1 + \frac{r}{40} = \frac{6}{5} \\ = > r = 40 \times \frac{1}{5} \\ = > r = 8 \\ So, the rate of interest is 8%. \end{array}$

Page No 167:

Question 4:

Mark (✓) against the correct answer
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

Answer:

(b) 9 months

$\begin{array}{l} 4 . (b) \\ Let the time be t years . \\ Principal = Rs . 8000 \\ Amount = Rs . 8360 \\ Rate = 6 % per annum \\ Amount = Principal (1 + \frac{Rate×Time}{100}) \\ \frac{8360}{8000} = 1 + \frac{6 \times t}{100} \\ = > \frac{8360}{8000} - 1 = \frac{6 t}{100} \\ = > t = (\frac{8360 - 8000}{8000}) \times \frac{100}{6} \\ = \frac{360}{8000} \times \frac{100}{6} \\ = \frac{6}{8} \times 12 months \\ = 9 months \end{array}$

Page No 167:

Question 5:

Mark (✓) against the correct answer
At what rate per cent annum simple interest will a sum double itself in 10 years?

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

Answer:

(b) 10%

$\begin{array}{l} Let the sum be Rs. x and the rate be r % . \\ A/Q: \\ Amount =2 x \end{array} \Rightarrow P + S . I . = 2 x \Rightarrow P + \frac{P \times R \times T}{100} = 2 x \begin{array}{l} => x (1+ \frac{r \times 10}{100})=2 x \\ = > \frac{100 + 10 r}{100} = 2 \\ = > 10 r = 200 - 100 \end{array} \Rightarrow 10 r = 100 \Rightarrow r = \frac{100}{10} \Rightarrow r = 10$

Page No 167:

Question 6:

Mark (✓) against the correct answer
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) $Rs (\frac{100}{x^{2}})$

Answer:

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

$\begin{array}{l} Simple Interest=Rs . x \\ Rate= x % per annum \\ Time = x years \\ Simple Interest= \frac{Principal×Rate×Time}{100} \\ = > x = \frac{Principal \times x \times x}{100} \\ = > Principal = Rs . \frac{100}{x} \end{array}$

Page No 167:

Question 7:

Mark (✓) against the correct answer
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) $12 \frac{1}{2} %$

Answer:

(b) 8%

$\begin{array}{l} Time=5 years \\ Simple interest= \frac{2}{5} P \\ = > \frac{P×Rate×Time}{100} = \frac{2}{5} P \\ = > \frac{Rate×5}{100} = \frac{2}{5} \end{array} \Rightarrow R a t e = \frac{2 \times 100}{5 \times 5} =>Rate=8%$

Page No 167:

Question 8:

Mark (✓) against the correct answer
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

Answer:

(c) 22 years

$R 1 = 12 % R_{2} = 10 % P_{1} = R s . 8000 P_{2} = R s . 9100 Let their amount s be equal in T years . {Amount}_{1} = S . I ._{1} + P_{1} = \frac{P_{1} \times R_{1} \times T}{100} + P_{1} = \frac{8000 \times 12 \times T}{100} + 8000 = 960 T + 8000 {Amount}_{2} = S . I ._{2} + P_{2} = \frac{P_{2} \times R_{2} \times T}{100} + P_{2} = \frac{9100 \times 10 \times T}{100} + 9100 = 910 T + 9100 {Amount}_{1} = {Amount}_{2} \Rightarrow 960 T + 8000 = 910 T + 9100 \Rightarrow 960 T - 910 T = 9100 - 8000 \Rightarrow 50 T = 1100 \Rightarrow T = 22 Hence, a fter 22 years their amounts will be equal .$

Page No 167:

Question 9:

Mark (✓) against the correct answer
A sum of Rs 600 amounts of 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

Answer:

(c) Rs. 768

$\begin{array}{l} Let the rate be R %. \end{array} S . I . = A - P = 720 - 600 = Rs . 120 T ime = 4 years R = \frac{100 \times S I}{P \times T} R = \frac{100 \times 120}{600 \times 4} = 5 Rate of interest = 5 % Now, R = (5 + 2) % = 7 % S . I . = \frac{P \times R \times T}{100} = \frac{600 \times 7 \times 4}{100} = Rs . 168 Amount = SI + P = 600 + 168 = Rs . 768$

Page No 167:

Question 10:

Mark (✓) against the correct answer
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) z² = xy
(c) x² = yz
(d) y² = zx

Answer:

(d) y² = zx

$\begin{array}{l} y = S.I. on x = \frac{x \times R×T}{100} . . . (i) \end{array} z = S.I. on y = \frac{y \times R×T}{100} . . . (ii) Dividing equation (i) by (ii) : \Rightarrow \frac{y}{z} = (\frac{x \times R×T}{100} \times \frac{100}{y \times R×T}) \Rightarrow \frac{y}{z} = \frac{x}{y} \Rightarrow y^{2} = xz$

Page No 167:

Question 11:

Mark (✓) against the correct answer
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

Answer:

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

$\begin{array}{l} Rate=10% per annum \\ Simple Interest=0 .125 × Principal \\ => \frac{Principal×Rate×Time}{100} =0 .125×Principal \\ => \frac{Time}{10} =0 .125 \\ =>Time=1 .25=1 \frac{1}{4} years \end{array}$

Page No 167:

Question 12:

Mark (✓) against the correct answer
At which sum will the 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

Answer:

(b) Rs 2400
$\begin{array}{l} Rate=3 \frac{3}{4} % per annum \\ = \frac{15}{4} % per annum \\ Time=2 \frac{1}{3} years \\ = \frac{7}{3} years \\ S .I. = \frac{P× \frac{15}{4} × \frac{7}{3}}{100} \\ =>P= \frac{210×100}{(\frac{15}{4} × \frac{7}{3})} \\ =>P=600×4 \\ =>P= Rs 2400 \end{array}$

Page No 168:

Question 1:

Find the simple interest on Rs 6300 at 8% per annum for 8 months.

Answer:

$\begin{array}{l} P = Rs 63 00, R = 8 %, T = \frac{8}{12} years \\ ∴ S .I. = \frac{P×R×T}{100} = \frac{6300 \times 8 \times {8^{21}}^{4^{2}}}{100 \times 1 2_{6_{3_{1}}}} = Rs 336 \end{array}$

Page No 168:

Question 2:

What sum will amount to Rs 6600 in 2 years at 10% per annum simple interest?

Answer:

$\begin{array}{l} Let the sum be Rs x . \\ S .I.= \frac{P×R×T}{100} = \frac{x \times 10 \times 2^{1}}{10 0_{5}} = \frac{x}{5} \\ Now, A = P + S . I \\ A= x + \frac{x}{5} = \frac{6 x}{5} \\ But the amount is Rs 66 00 . \\ ∴ \frac{6 x}{5} = 6600 \\ = > x = \frac{660 0^{1100} \times 5}{6} = 5500 \\ Hence, the required sum is Rs 55 00 . \end{array}$

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Question 3:

At what rate per cent per annum simple interest will Rs 3625 amount to Rs 4495 in 2 years?

Answer:

$\begin{array}{l} P = Rs 3625, A = Rs 4495, T = 2 years \\ ∴ S . I. = A ― P \\ = 4495 ― 3625 \\ = Rs 87 0 \\ ∴ S .I.= \frac{P×R×T}{100} \\ = > 870 = \frac{3625 \times R \times 2}{100} \\ = > R = \frac{870 \times 100}{3625 \times 2} \\ = > R = \frac{87000}{7250} = 12 % \end{array}$

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Question 4:

In what time will Rs 3600 amount to Rs 4410 at 9% per annum simple interest?

Answer:

$\begin{array}{l} P = Rs . 36 00, A = Rs . 441 0, R = 9 % \\ S . I. = A - P \\ = 4410 - 3600 \\ = Rs . 81 0 \\ ∴ T= \frac{S .I.×100}{P×R} \end{array} = \frac{810×100}{3600×9} = \frac{90}{36} =2 \frac{1}{2} years$

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Question 5:

At what rate per cent per annum simple interest will a sum double itself in 12 years?

Answer:

$\begin{array}{l} Let the sum be Rs x . \\ Amount = Rs 2 x \\ ∴ S . I. = (2 x - x) = R s x \\ Time = 1 2 years \\ ∴ P = x, S . I . = x, T = 1 2 years \\ ∴ R= \frac{100×S .I.}{P×T} = \frac{100 \times x}{x \times 12} = 8.3 % \end{array}$

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Question 6:

A sum of money becomes $\frac{4}{3}$ of itself in 6 years at a certain rate of simple interest. Find the rate of interest.

Answer:

$\begin{array}{l} Let the sum be Rs x . \\ Amount = \frac{4}{3} x \\ S .I .=A - P= \frac{4}{3} x - x \\ = \frac{x}{3} \\ Let the rate be R % . \\ S .I.= \frac{P×R×T}{100} \\ = > \frac{x}{3} = \frac{x \times R×6}{100} \\ = > R = \frac{x \times 100}{x \times 6 \times 3} = \frac{100}{18} = 5.55 \end{array} Hence, the rate of interest is 5.55 % .$

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Question 7:

Mark (✓) against the correct answer
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) 9%
(d) 12%

Answer:

(c) 9%
$\begin{array}{l} Let the sum be Rs x . \\ A = \frac{49 x}{40} \\ We know: \\ A=P+S .I . \\ ∴ S .I. =A - P \\ = (\frac{49 x}{40} - x) \\ = \frac{49 x - 40 x}{40} \\ = \frac{9 x}{40} \\ Let the rate be R % per annum . \\ S .I.= \frac{P \times R \times T}{100} \\ = > \frac{9 x}{40} = \frac{x \times R \times 5^{1}}{10 0_{20} \times 2} \\ =>R= \frac{9 \times 2 0^{1} \times 2^{1}}{4 0_{2_{1}}} \\ =>R=9 \end{array}$

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Question 8:

Mark (✓) against the correct answer
A sum amounts to Rs 3626 in 219 days at 6% per annum simple interest. The sum is

(a) Rs 3000
(b) Rs 3200
(c) Rs 3500
(d) Rs 3600

Answer:

(c) 3500
$\begin{array}{l} A=Rs 3626, R=6%, T=219 days= \frac{219}{365} years \\ Let the required sum be Rs x . \\ S .I.= \frac{P×R×T}{100} = \frac{x \times 6 \times 219}{100 \times 365} = \frac{1314 x}{36500} \\ A = P + S.I. \\ = x + \frac{1314 x}{36500} \\ = \frac{36500 x + 1314 x}{36500} \\ = \frac{37814 x}{36500} \\ But the amount is Rs 3626. \\ ∴ \frac{37814 x}{36500} = 3626 \\ x = \frac{362 6^{259} \times 36500}{3781 4_{2701}} = 3500 \\ The required sum is Rs 35 00 . \end{array}$

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Question 9:

Mark (✓) against the correct answer
In what time will Rs 6000 amount to Rs 6360 at 8% per annum simple interest?

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

Answer:

(a) 9 months

$\begin{array}{l} P = Rs 6 000, A = Rs 636 0, R = 8 % \\ S . I. = A - P \\ = 6360 - 6000 \\ = Rs . 36 0 \\ T= \frac{S .I.×100}{P×T} = \frac{360×100}{6000×8} \\ = \frac{6}{8} years \\ = \frac{6}{8} ×12 months \\ = \frac{72}{8} \\ =9 months \end{array}$

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Question 10:

Mark (✓) against the correct answer
The simple interest on a sum for 5 yars is $\frac{3}{5}$ of the sum. The rate per cent per annum is
(a) 8%
(b) 10%
(c) 12%
(d) $12 \frac{1}{2} %$

Answer:

$(c) 12 % \begin{array}{l} Let the sum be Rs x . \\ S .I .=Rs \frac{3}{5} \\ Let the rate be R % per annum . \\ S .I.= \frac{P×R×T}{100} = \frac{x ×R×5}{100} = \frac{R x}{20} \\ ∴ \frac{3 x}{5} = \frac{R x}{20} \\ = > R = \frac{3 x \times 20}{5 x} = 12 % \end{array}$

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Question 11:

Mark (✓) against the correct answer
The simple interest at x% per annum for x years will be Rs x on a sum of

(a) Rs x
(b) Rs 10 x
(c) Rs 100 x
(d) Rs $(\frac{100}{x})$

Answer:

$(d) Rs \frac{100}{x} \begin{array}{l} Let the sum be Rs P. \\ Time= x years \\ Rate of interest, r = x % per annum \\ S .I .=Rs x \\ Simple Interest = \frac{P \times R \times T}{100} \\ => x = \frac{P \times x \times x}{100} \\ =>P= \frac{x \times 100}{x \times x} \\ ∴ P= (\frac{100}{x}) \end{array}$

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Question 12:

Mark (✓) against the correct answer
At what rate per cent per annum simple interest will a sum double itself in 10 years?

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

Answer:

$(b) 10% \begin{array}{l} Let the sum be Rs x . \\ Amount=Rs 2 x \\ Time=10 years \\ ∴ S .I .=A - P=2 x - x \\ =Rs x \\ ∴ S .I.= \frac{P \times R \times T}{100} \\ = > x = \frac{x \times R \times 10}{100} \\ = > R= \frac{x \times 100}{x \times 10} \end{array} = > R = 10$

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Question 13:

Fill in the blanks.

(i) $P = \frac{100 \times (. . . . . .)}{R \times T}$
(ii) $R = \frac{100 \times SI}{(. . . . . .) \times T}$
(iii) At (......)% per annum simple interest a sum doubles itself in 10 years.
(iv) At simple interest a sum becomes $\frac{6}{5}$ of itself in $2 \frac{1}{2}$ years. The rate of interest is (......)% per annum.

Answer:

$\begin{array}{l} (i) P = \frac{100 \times S . I .}{R×T} \\ (ii) R= \frac{100×S . I .}{(P)×T} \end{array} (iii) 10 % Let the sum be Rs x . Amount = Rs 2 x S . I . = A - P = 2 x - x = Rs x Let the rate of inetrset be R % . Time = 10 yrs S . I . = \frac{P \times R \times T}{100} \Rightarrow x = \frac{x \times R \times 10}{100} \Rightarrow R = \frac{100}{10} \Rightarrow R = 10 (i v) 8 % Let the sum be Rs x . Amount = Rs \frac{6}{5} x S . I . = A - P = \frac{6}{5} x - x = Rs \frac{x}{5} Let the rate of inetrset be R % . Time = 2.5 yrs S . I . = \frac{P \times R \times T}{100} \Rightarrow \frac{x}{5} = \frac{x \times R \times 2.5}{100} \Rightarrow R = \frac{100}{2.5 \times 5} \Rightarrow R = 8$

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Question 14:

Write 'T' for true and 'F' for false

(i) Simple interest on Rs x for x years is Rs x. Then, the rate of interest is x% per annum.
(ii) Rate = $\frac{100 \times SI}{P \times T}$ .
(iii) A sum doubles itself at simple interest at 10% per annum in 10 years.
(iv) Simple interest on Rs 1000 at 5% per annum for 73 days is Rs 10.

Answer:

$(i) False S . I . = \frac{P \times R \times T}{100} \Rightarrow x = \frac{x \times R \times x}{100} \Rightarrow R = \frac{100 \times x}{x \times x} \Rightarrow R = \frac{100}{x} % (ii) True (iii) True Let the sum be Rs x . S . I . = \frac{P \times R \times T}{100} S . I . = \frac{x \times 10 \times 10}{100} S . I . = Rs x Amount = S . I . + P = x + x = Rs 2 x (iv) True P = Rs 1000, R = 5 %, T = 73 days = \frac{73}{365} yrs S . I . = \frac{P \times R \times T}{100} = \frac{1000 \times 5 \times 73}{100 \times 365} = Rs 10$

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