myhelper: RS Aggarwal 2019,2020 solution class 7 chapter 12 Simple Interest Exercise 12B

RS Aggarwal 2019,2020 solution class 7 chapter 12 Simple Interest Exercise 12B

Exercise 12B

Page-167

Question 1:

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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 1:

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

Question 2:

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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 2:

(b) Rs.3500

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

Question 3:

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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 3:

(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-167

Question 4:

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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 4:

(b) 9 months

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

Question 5:

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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 5:

(b) 10%

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

Question 6:

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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 6:

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

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

Question 7:

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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 7:

(b) 8%

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

Question 8:

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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 8:

(c) 22 years

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

Question 9:

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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 9:

(c) Rs. 768

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

Question 10:

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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 10:

(d) y² = zx

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

Question 11:

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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 11:

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

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

Question 12:

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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 12:

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