Exercise: 8-E
Multiple Choice Questions
Q1: The sum which amounts to ₹840 in 5 years at the rate of 8% per annum simple interest is:
Step 1: Recall the formula for Amount (A) in Simple Interest:
\[
A = P + SI = P + \frac{P \times R \times T}{100} = P \left(1 + \frac{R \times T}{100} \right)
\]Step 2: Given:
Amount \(A = ₹840\), Rate \(R = 8\%\), Time \(T = 5\) years
Step 3: Substitute the values in formula:
\[
840 = P \left(1 + \frac{8 \times 5}{100} \right) = P \left(1 + \frac{40}{100}\right) = P \times \frac{140}{100}
\]Step 4: Solve for \(P\):
\[
P = \frac{840 \times 100}{140} = \frac{840 \times 100}{100 + 5 \times 8}
\]Answer: b or d. ₹\(\frac{100 \times 840}{100 + 5 \times 8}\)
Q2: If ₹64 amount to ₹83.20 in 2years, what will ₹86 amount to in 4 years at the same rate per cent per annum?
Step 1: Calculate the rate of interest \(R\) using Simple Interest formula:
\[
A = P \left(1 + \frac{R \times T}{100}\right)
\]
Given: \(P = 64\), \(A = 83.20\), \(T = 2\) years
Substitute values:
\[
83.20 = 64 \left(1 + \frac{2R}{100}\right) \\
\Rightarrow 1 + \frac{2R}{100} = \frac{83.20}{64} = 1.3 \\
\Rightarrow \frac{2R}{100} = 1.3 – 1 = 0.3 \\
\Rightarrow 2R = 30 \Rightarrow R = 15\%
\]Step 2: Find the amount for ₹86 at 4 years:
\[
A = 86 \left(1 + \frac{15 \times 4}{100}\right) = 86 \times \left(1 + 0.6\right) = 86 \times 1.6 = ₹137.60
\]Answer: d. ₹137.60
Q3: The simple interest at x% for x years will be ₹x on a sum of:
Step 1: Recall the formula for Simple Interest:
\[
SI = \frac{P \times R \times T}{100}
\]Step 2: Given:
Simple Interest \(SI = x\), Rate \(R = x\%\), Time \(T = x\) years
Substitute into formula:
\[
x = \frac{P \times x \times x}{100} = \frac{P x^2}{100}
\]Step 3: Solve for \(P\):
\[
P = \frac{100 \times x}{x^2} = \frac{100}{x}
\]Answer: c. ₹\(\left(\frac{100}{x}\right)\)
Q4: Due to a fall in the rate of interest from 13% p.a. to \(12\frac{1}{2}\)% p.a., a money leader’s yearly income diminishes by ₹104. His capital is
Step 1: Convert mixed fraction rate:
\[
12\frac{1}{2}\% = 12.5\%
\]Step 2: Let the capital be \(P\). The yearly income is interest on capital, so:
\[
\text{Income at } 13\% = \frac{13}{100} \times P = 0.13P \\
\text{Income at } 12.5\% = \frac{12.5}{100} \times P = 0.125P
\]Step 3: The difference in income is ₹104:
\[
0.13P – 0.125P = 104 \\
0.005P = 104
\]Step 4: Solve for \(P\):
\[
P = \frac{104}{0.005} = 20800
\]Answer: a. ₹20800
Q5: A lends a sum of money for 10 years at 5% simple interest. B lends double that amount for 5 years at the rate of interest. Which of the following statements is true in this regard?
Step 1: Let the sum lent by A be \(P\), rate of interest for both be 5%.
Step 2: Interest earned by A in 10 years:
\[
SI_A = \frac{P \times 5 \times 10}{100} = \frac{50P}{100} = \frac{P}{2}
\]Step 3: B lends double the amount, i.e., \(2P\), for 5 years at the same rate 5%. Interest earned by B:
\[
SI_B = \frac{2P \times 5 \times 5}{100} = \frac{50P}{100} = \frac{P}{2}
\]Step 4: Compare the interests:
\[
SI_A = SI_B = \frac{P}{2}
\]Answer: c. A and B will get the same amount as interest.
Q6: B borrowed ₹720 form A at 8% simple interest for 3 years and lent the same sum to C at \(10\frac{1}{2}\)%, simple interest 2 years. In the whole transaction B.
Step 1: Calculate interest B pays to A:
\[
SI_{\text{paid}} = \frac{720 \times 8 \times 3}{100} = \frac{17280}{100} = ₹172.80
\]Step 2: Convert rate lent to C:
\[
10\frac{1}{2}\% = 10.5\%
\]Step 3: Calculate interest B receives from C:
\[
SI_{\text{received}} = \frac{720 \times 10.5 \times 2}{100} = \frac{15120}{100} = ₹151.20
\]Step 4: Calculate net gain/loss for B:
\[
\text{Net} = SI_{\text{received}} – SI_{\text{paid}} = 151.20 – 172.80 = -₹21.60
\]
Negative value means B lost ₹21.60.
Answer: a. Lost ₹21.60
Q7: ₹800 amounts to ₹920 in 3 years at simple interest. If the interest rate is increased by 3%, it would amount to
Step 1: Calculate the original rate of interest \(R\):
Given:
Principal \(P = 800\), Amount \(A = 920\), Time \(T = 3\) years
Simple Interest \(SI = A – P = 920 – 800 = ₹120\)
Using formula:
\[
SI = \frac{P \times R \times T}{100} \\
120 = \frac{800 \times R \times 3}{100} \\
120 = 24R \Rightarrow R = \frac{120}{24} = 5\%
\]Step 2: New rate of interest:
\[
R_{\text{new}} = 5\% + 3\% = 8\%
\]Step 3: Calculate new amount with increased rate:
\[
SI_{\text{new}} = \frac{800 \times 8 \times 3}{100} = \frac{19200}{100} = ₹192 \\
A_{\text{new}} = P + SI_{\text{new}} = 800 + 192 = ₹992
\]Answer: a. ₹992
Q8: Rahu lent ₹6000 to Manick for 2 years and ₹1500 to Arif for 4 years and received altogether from both ₹900 as simple interest. The rate of interest is:
Step 1: Let the rate of interest be \(R\%\) per annum.
Step 2: Calculate interest from Manick:
\[
SI_1 = \frac{6000 \times R \times 2}{100} = \frac{12000R}{100} = 120R
\]Step 3: Calculate interest from Arif:
\[
SI_2 = \frac{1500 \times R \times 4}{100} = \frac{6000R}{100} = 60R
\]Step 4: Total interest received:
\[
SI_1 + SI_2 = 120R + 60R = 180R
\]
Given total interest = ₹900
Step 5: Solve for \(R\):
\[
180R = 900 \Rightarrow R = \frac{900}{180} = 5\%
\]Answer: b. 5% p.a.
Q9: Consider the following statements: If a sum of money is loaned at simple then the
i. Money gets doubled in 5 years if the rate of interest is \(16\frac{2}{3}\)% p.a.
ii. Money gets doubled in if the rate interest is 20%. p.a.
iii. Money becomes four times in 10 if it gets doubled in 5 years.
Of these statements,
Step 1: Check statement (i):
Given \(T = 5\) years and \(R = 16\frac{2}{3}\% = \frac{50}{3}\%\)
\[
S.I. = \frac{P \times \frac{50}{3} \times 5}{100} \\
S.I. = \frac{5P}{6} \\
Amount = S.I. + P \\
Amount = \frac{5P}{6} + P = \frac{11P}{6}
\]
Hence, money will NOT double in 5 years at \(16\frac{2}{3}\%\). Statement (i) is false.
Step 2: Check statement (ii):
Given \(R = 20\%\), find \(T\) such that money doubles:
\[
S.I. = \frac{P \times 20 \times 5}{100} \\
S.I. = P \\
Amount = S.I. + P \\
Amount = P + P = 2P
\]
So, money gets doubled in 5 years at 20%. Statement (ii) is true.
Step 3: Check statement (iii):
Money gets doubles in 5 years, then
A = 2P, Time = 5 years, S.I. = \(A – P = 2P -P =P\), Rate = ?
\[
R = \frac{S.I. \times 100}{P \times T} \\
R = \frac{P \times 100}{P \times 5} = 20\%\
\]
Now, Time = 10 years, Rate = 20%, S.I. = ?
\[
S.I. = \frac{P \times 20 \times 10}{100} \\
S.I. = 2P \\
Amount = S.I. + P \\
Amount = 2P + P = 3P
\]
Money becomes 3 times (not 4 times) in 10 years.
Hence, statement (iii) is false.
Answer: b. (ii) alone is correct
Q10: In what time will a sum of money double itself at \(6\frac{1}{4}\)% p.a. simple interest?
Step 1: If a sum doubles, the Simple Interest (SI) must equal the Principal (P).
\[
SI = P
\]
Use formula:
\[
SI = \frac{P \times R \times T}{100} \\
P = \frac{P \times R \times T}{100}
\Rightarrow 1 = \frac{R \times T}{100}
\Rightarrow R \times T = 100
\]Step 2: Convert the rate to improper fraction:
\[
6\frac{1}{4}\% = \frac{25}{4}\%
\]Step 3: Solve for time:
\[
T = \frac{100}{R} = \frac{100}{25/4} = 100 \times \frac{4}{25} = 16 \text{ years}
\]Answer: c. 16 years
Q11: A sum of money becomes \(\frac{8}{5}\) of itself in 5 years at a certain rate of interest. The rate percent per annum is:
Step 1: Let the principal be ₹P.
The amount becomes \(\frac{8}{5}P\) in 5 years.
So, the simple interest \(SI = \frac{8}{5}P – P = \left(\frac{8}{5} – 1\right)P = \frac{3}{5}P\)
Step 2: Use the simple interest formula:
\[
SI = \frac{P \times R \times T}{100} \\
\Rightarrow \frac{3}{5}P = \frac{P \times R \times 5}{100}
\]Cancel \(P\) on both sides:
\[
\frac{3}{5} = \frac{5R}{100} \\
\Rightarrow \frac{3}{5} = \frac{R}{20} \\
\Rightarrow R = \frac{3}{5} \times 20 = 12
\]Answer: d. 12%
Q12: In what time will the simple interest on ₹780 at 5% be equal to the simple interest on ₹600 at \(6\frac{1}{2}\)%?
Step 1: Let the required time be \(T\) years.
We use the simple interest formula:
\[
SI = \frac{P \times R \times T}{100}
\]Step 2: Equate the two simple interests:
\[
\frac{780 \times 5 \times T}{100} = \frac{600 \times 6.5 \times T}{100}
\]
Convert \(6\frac{1}{2}\%\) to fraction: \(6.5 = \frac{13}{2}\)
\[
\Rightarrow \frac{780 \times 5 \times T}{100} = \frac{600 \times \frac{13}{2} \times T}{100}
\]Cancel \(T\) and 100 from both sides:
\[
780 \times 5 = 600 \times \frac{13}{2} \\
\Rightarrow 3900 = 3900
\]Step 3: Since both sides are equal for any value of \(T\),
this means the simple interests are always equal for any time.
Answer: d. Always
Q13: Two equal sums of money are deposited in two banks, each at 15% per annum, for \(3\frac{1}{2}\) years and 5 years. If the difference between their interests is ₹144, each sum is
Step 1: Let the principal (each sum) be ₹P.
Rate of interest \(R = 15\%\)
Time periods: \(T_1 = 3\frac{1}{2} = \frac{7}{2}\) years, \(T_2 = 5\) years
Step 2: Use the formula:
\[
SI = \frac{P \times R \times T}{100}
\]Simple Interest for first account:
\[
SI_1 = \frac{P \times 15 \times \frac{7}{2}}{100} = \frac{105P}{100}
\]Simple Interest for second account:
\[
SI_2 = \frac{P \times 15 \times 5}{100} = \frac{75P}{100}
\]Step 3: Difference in interest:
\[
SI_2 – SI_1 = \frac{75P}{100} – \frac{105P}{200} = \frac{150P – 105P}{200} = \frac{45P}{200}
\]Step 4: Set the difference equal to ₹144:
\[
\frac{45P}{200} = 144 \Rightarrow P = \frac{144 \times 200}{45} = \frac{28800}{45} = ₹640
\]Answer: c. ₹640
Q14: A sum of ₹2500 is lent out in two parts, one at 12% and another one at \(12\frac{1}{2}\)%. If the total annual income is ₹306, the money lent at 12% is
Step 1: Let the amount lent at 12% be ₹x.
Then, amount lent at \(12\frac{1}{2}\%\) = ₹(2500 − x)
Step 2: Annual income (simple interest) from first part:
\[
SI_1 = \frac{12x}{100} = 0.12x
\]Step 3: Annual income from second part:
\[
SI_2 = \frac{12.5 \times (2500 – x)}{100} = 0.125(2500 – x)
\]Step 4: Total income = ₹306:
\[
0.12x + 0.125(2500 – x) = 306 \\
0.12x + 312.5 – 0.125x = 306 \\
(0.12x – 0.125x) + 312.5 = 306 \\
\Rightarrow -0.005x = -6.5 \\
\Rightarrow x = \frac{-6.5}{-0.005} = 1300
\]Answer: c. ₹1300
Q15: Out of a sum of ₹625, a part was tent at 5% and the other at 10% simple interest. If the interest on the first part after 2 years is equal to the interest on the second part after 4 years, then the second sum (in ₹) is
Step 1: Let the first part lent at 5% be ₹x.
Then, the second part lent at 10% = ₹(625 − x)
Step 2: Use simple interest formula:
\[
SI = \frac{P \times R \times T}{100}
\]Interest from first part after 2 years:
\[
SI_1 = \frac{x \times 5 \times 2}{100} = \frac{10x}{100} = 0.1x
\]Interest from second part after 4 years:
\[
SI_2 = \frac{(625 – x) \times 10 \times 4}{100} = \frac{40(625 – x)}{100} = 0.4(625 – x)
\]Step 3: Given: SI_1 = SI_2
\[
0.1x = 0.4(625 – x) \\
\Rightarrow 0.1x = 250 – 0.4x \\
\Rightarrow 0.1x + 0.4x = 250 \\
\Rightarrow 0.5x = 250 \\
\Rightarrow x = 500
\]Step 4: Second part = ₹(625 – 500) = ₹125
Answer: a. 125
Q16: The compound interest on ₹540 at \(16\frac{2}{3}\) per annum for 2 years is
Step 1: Convert the rate:
\[
16\frac{2}{3}\% = \frac{50}{3}\%
\]Step 2: Use the compound interest formula:
\[
A = P \left(1 + \frac{R}{100} \right)^T
\]
Where:
Principal \(P = ₹540\), \(R = \frac{50}{3}\%\), \(T = 2\) years
Step 3: Since \(\frac{50}{3} = \frac{1}{2}\), we can write:
\[
A = 540 \left(1 + \frac{1}{6} \right)^2 = 540 \left(\frac{7}{6} \right)^2 = 540 \times \frac{49}{36} = \frac{540 \times 49}{36} \\
A = \frac{26460}{36} = ₹735
\]Step 4: Compound Interest:
\[
CI = A – P = 735 – 540 = ₹195
\]Answer: c. ₹195
Q17: The difference between the simple interest and the compound interest on ₹600 for 1 year at 10% per annum, reckoned half-yearly is
Step 1: Principal \(P = ₹600\)
Rate of interest \(R = 10\%\) per annum = \(5\%\) per half-year
Time = 1 year = 2 half-years
Step 2: Compute Compound Interest:
\[
A = P \left(1 + \frac{R}{100} \right)^n = 600 \left(1 + \frac{5}{100} \right)^2 = 600 \left(\frac{21}{20} \right)^2 \\
A = 600 \times \frac{441}{400} = ₹661.50 \\
CI = A – P = 661.50 – 600 = ₹61.50
\]Step 3: Compute Simple Interest:
\[
SI = \frac{P \times R \times T}{100} = \frac{600 \times 10 \times 1}{100} = ₹60
\]Step 4: Find the difference:
\[
CI – SI = ₹61.50 – ₹60 = ₹1.50
\]Answer: b. ₹1.50
Q18: Simple interest on a sum at \(12\frac{1}{2}\)% per annum for 2 years is ₹256 The compound interest on the same sum at the same rate and for the same period is
Step 1: Use the SI formula to find the principal.
\[
SI = \frac{P \times R \times T}{100} \\
\Rightarrow 256 = \frac{P \times 12.5 \times 2}{100} = \frac{25P}{100} \\
\Rightarrow P = \frac{256 \times 100}{25} = ₹1024
\]Step 2: Use the compound interest formula:
\[
A = P \left(1 + \frac{R}{100} \right)^T = 1024 \left(1 + \frac{12.5}{100} \right)^2 \\
A = 1024 \left( \frac{9}{8} \right)^2 = 1024 \times \frac{81}{64} = ₹1296
\]Step 3: Calculate compound interest:
\[
CI = A – P = 1296 – 1024 = ₹272
\]Answer: d. ₹272
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