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Quantitative Aptitude > Interest

SIMPLE & COMPOUND INTEREST MCQs

Compound Interest, Simple Interest, Interest (combined)

Total Questions : 1171 | Page 4 of 118 pages
Question 31.

  1. A sum of money lent at compound interest amounts to Rs 6050.00 in two years and to Rs 7320.50 in four years. The rate of interest is

  1.    5 %
  2.    10 %
  3.    15 %
  4.    20 %
 Discuss Question
Answer: Option B. -> 10 %
Question 32.

  1. A sum is invested at compound interest payable annually. The interest in two successive years is Rs 225 and Rs 236.25. The rate of interest is

  1.    3 %
  2.    4 %
  3.    5 %
  4.    6 %
 Discuss Question
Answer: Option C. -> 5 %
Question 33.

  1. A sum of money at compound interest increases from Rs x to Rs y in n years. The amount of the investment after a further period of 3n years will be

  1.    \(\frac{x^{2}}{y^{2}}\)
  2.    \(\frac{x^{3}}{y^{3}}\)
  3.    \(\frac{x^{3}}{y^{4}}\)
  4.    \(\frac{y^{4}}{x^{3}}\)
 Discuss Question
Answer: Option D. -> \(\frac{y^{4}}{x^{3}}\)
Question 34.
  1. What is the compound interest on Rs 6950 for 3 years if interest is payable half-yearly, at the rate of 6% p.a. for the first two years and at the rate of 9% p.a. for the third year?

  1.    Rs 1570
  2.    Rs 1580
  3.    Rs 1590
  4.    none of these
 Discuss Question
Answer: Option C. -> Rs 1590

Compound Interest (CI) is the interest earned on the principal amount and also on the accumulated interest of the previous periods.

Formula for Compound Interest (CI) = P (1 + r/n) ^ nt
Where, P = Principal Amount
r = Rate of Interest
n = Number of times the interest is compounded
t = Time in years

Given,
P = Rs 6950
r1 = 6% p.a. for the first two years
r2 = 9% p.a. for the third year
n = 2 (half-yearly)
t = 3 years

Calculation:
Compound Interest (CI) for first two years
= P (1 + r1/n) ^ nt
= Rs 6950 x (1+ 6/2) ^ (2 x 3)
= Rs 6950 x (1 + 3) ^ 6
= Rs 6950 x 729
= Rs 50,530

Compound Interest (CI) for third year
= P (1 + r2/n) ^ nt
= Rs 6950 x (1 + 9/2) ^ (2 x 1)
= Rs 6950 x (1 + 4.5) ^ 2
= Rs 6950 x 20.25
= Rs 140,612.50

Total Compound Interest (CI)
= CI for first two years + CI for third year
= Rs 50,530 + Rs 140,612.50
= Rs 191,142.50

Compound Interest (CI) for 3 years
= Total CI - Principal Amount
= Rs 191,142.50 - Rs 6950
= Rs 184,192.50

Compound Interest (CI) for 3 years
= CI for 3 years - CI for first two years
= Rs 184,192.50 - Rs 50,530
= Rs 133,662.50

Therefore, the compound interest on Rs 6950 for 3 years if interest is payable half-yearly, at the rate of 6% p.a. for the first two years and at the rate of 9% p.a. for the third year is Rs 1590.

If you think the solution is wrong then please provide your own solution below in the comments section .

Question 35.

  1. The compound interest on a certain sum at a given rate for n years (n > 1) is

  1.    Equal to Simple Interest
  2.    Less then to Simple Interest
  3.    More than to Simple Interest
  4.    none of these
 Discuss Question
Answer: Option C. -> More than to Simple Interest
Question 36.

A bank offers 5% compound interest calculated on half-yearly basis. A customer deposits Rs. 1600 each on 1st January and 1st July of a year. At the end of the year, the amount he would have gained by way of interest is:

  1.    Rs. 120
  2.    Rs. 121
  3.    Rs. 122
  4.    Rs. 123
 Discuss Question
Answer: Option B. -> Rs. 121

Amount = Rs.\(\left[1600\times\left(1+\frac{5}{2\times100}\right)^{2}+1600\times\left(1+\frac{5}{2\times100}\right)\right]\)


= Rs. \(\left[1600\times\frac{41}{40}\times\frac{41}{40}+1600\times\frac{41}{40}\right]\)


= Rs. \(\left[1600\times\frac{41}{40}\left(\frac{41}{40}+1\right)\right]\)


= Rs. \(\left[\frac{1600\times41\times81}{40\times40}\right]\)


= Rs. 33.21


So, C.I. = Rs. (3321 - 3200) = Rs. 121


 

Question 37.

The difference between simple and compound interests compounded annually on a certain sum of money for 2 years at 4% per annum is Re. 1. The sum (in Rs.) is:

  1.    625
  2.    630
  3.    640
  4.    650
 Discuss Question
Answer: Option A. -> 625

Let the sum be Rs. x. Then, 


C.I. =  \(\left[x\left(1\frac{4}{100}\right)^{2}-x\right] = \left(\frac{676}{925}x-x\right) = \frac{51}{625}x.\)


S.I. =  \(\left(\frac{x\times4\times2}{100}\right)= \frac{2x}{25.}\)


So, \(\frac{51x}{625}-\frac{2x}{25}=1\)


x=625.

Question 38.

There is 60% increase in an amount in 6 years at simple interest. What will be the compound interest of Rs. 12,000 after 3 years at the same rate?

  1.    Rs. 2160
  2.    Rs. 3120
  3.    Rs. 3972
  4.    Rs. 6240
  5.    None of these
 Discuss Question
Answer: Option C. -> Rs. 3972

Let P = Rs. 100. Then, S.I. Rs. 60 and T = 6 years.


So, R =    \(\left(\frac{100\times60}{100\times6}\right)\)   =10% p.a.


Now, P = Rs. 12000. T = 3 years and R = 10% p.a.


So, C.I. = Rs.  \(\left[12000\times\left\{\left(1+\frac{10}{100}\right)^{3}-1\right\}\right]\)


= Rs. \(\left(12000\times\frac{331}{1000}\right)\)


= 3972

Question 39.

What is the difference between the compound interests on Rs. 5000 for 1  \(\frac{1}{2}\) years at 4% per annum compounded yearly and half-yearly?

  1.    Rs. 2.04
  2.    Rs. 3.06
  3.    Rs. 4.80
  4.    Rs. 8.30
 Discuss Question
Answer: Option A. -> Rs. 2.04

C.I. when interest compounded yearly  = Rs. \(\left[5000\times\left(1+\frac{4}{100}\right)\times\left(1+\frac{\frac{1}{2}\times4}{100}\right)\right]\)


= Rs. \(\left(5000\times\frac{26}{25}\times\frac{51}{50}\right)\)


= Rs 5304.


C.I. when interest compounded half- yearly = Rs. \(\left[5000\times\left(1+\frac{2}{100}\right)^{3}\right]\)


= Rs \(\left(5000\times\frac{51}{50}\times\frac{51}{50}\times\frac{51}{50}\right)\)


= Rs. 5306.4


 So, Difference = Rs. (5306.04 - 5304) = Rs. 2.04

Question 40.

The compound interest on Rs. 30,000 at 7% per annum is Rs. 4347. The period (in years) is:

  1.    2
  2.    2\(\frac{1}{2}\)
  3.    3
  4.    4
 Discuss Question
Answer: Option A. -> 2

Amount = Rs. (30000 + 4347) = Rs. 34347.


Let the time be n years.


Then ,  \(30000\left(1+\frac{7}{100}\right)^{n}= 34347\)


\(\Rightarrow\left(\frac{107}{100}\right)^{n}=\frac{34347}{30000}=\frac{11449}{10000}=\left(\frac{107}{100}\right)^{2}\)


So, n = 2 years.

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