Quantitative Aptitude

Simple and Compound Interest

3,394 Questions
All Simple interest and its inverse forms 768 Compound interest with annual compounding 742 Simple-interest growth and split sums 626 Difference between compound and simple interest 479 Half-yearly and quarterly compounding 252 Changing rates and fractional years 201 Installments (CI based) 190 Ratio of principal to interest 118 Population growth and depreciation 16

Simple and Compound Interest Questions

Multiple choice

Suppose $P_1, \, P_2$ are principles, $T_1, \, T_2$ are times periods and $R_1,\, R_2$ are rates of interest in two cases. If $P_1\,:\, P_2\,=\, 2:\,3,\, T_1\,:\, T_2\, =\, 3\, :\, 5$ and $R_1\,:\,R_2\,=\, 5\, :\, 8$ then find the percent with which SI second case $(I_2)$ is more than the S.I. in I case $(I_1)$ ?

  1. 100 %

  2. 200 %

  3. 300 %

  4. 400 %
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

I1 = P1*R1*T1/100. I2 = P2*R2*T2/100. I2/I1 = (P2/P1) * (R2/R1) * (T2/T1) = (3/2) * (8/5) * (5/3) = 4. I2 = 4 * I1. The increase is (I2 - I1) / I1 = (4I1 - I1) / I1 = 3 = 300%.

Multiple choice

A sum of Rs. 1,550 was partly lent at 5% and 8% p.a for three years. Interest earned for Rs. 1,550 which was partly lent at 5% and 8% p.a. was Rs. 300. The ratio of the money lent at 5% to that lent at 8% is

  1. 8 : 5

  2. 5 : 8

  3. 16 : 15

  4. 15 : 16
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let x be the amount at 5% and (1550-x) at 8%. Interest = (x*5*3)/100 + ((1550-x)*8*3)/100 = 300. 15x + 37200 - 24x = 30000. 9x = 7200, x = 800. The other part is 750. Ratio 800:750 = 16:15.

Multiple choice

A man invests a certain sum of money at $6\%\ p.a.$ simple interest and another sum at $7\%\ p.a.$ simple interest. His income from interest after $2$ years was $Rs.\ 354$. One-fourth of the first sum is equal to one-fifth of the second sum. The total sum invested was
  1. $Rs.\ 1500$
  2. $Rs.\ 1200$
  3. $Rs.\ 2700$
  4. $Rs.\ 5400$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let sums be S1 and S2. S1/4 = S2/5 => S2 = 1.25*S1. Interest: (S1 * 0.06 * 2) + (S2 * 0.07 * 2) = 354. 0.12*S1 + 0.14*(1.25*S1) = 354. 0.12*S1 + 0.175*S1 = 354. 0.295*S1 = 354. S1 = 1200. S2 = 1500. Total = 2700.

Multiple choice

Ravi borrowed Rs45000 from Sahil A part of the sum is borrowed at 10% pa under simple interest for 4 years and the remaining part at 11% pa under simple interest for $\displaystyle 4\frac{1}{2}$ years If the total interest earned by Sahil is Rs 19900 then find the sum borrowed at each rate

  1. 10000, 35000

  2. 25000, 20000

  3. 15000, 30000

  4. 5000, 40000
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let x be the sum at 10% for 4 years, and (45000-x) at 11% for 4.5 years. Interest = (x * 0.1 * 4) + ((45000-x) * 0.11 * 4.5) = 19900. Solving gives x = 25000.

Multiple choice

Q and R borrowed Rs.26000 and Rs. 25000 respectively for a period of 2 years Q paid simple interest at the rate of 2% p.a. while R paid compound interest at the same rate compounded annually Who paid more interest and by how much ?

  1. R paid more than Q by Rs.40

  2. R paid more than Q by Rs.30

  3. Q paid more than R by Rs.40

  4. Q paid more than R by Rs.30
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Q interest = 26000 * 0.02 * 2 = 1040. R interest = 25000 * ((1.02)^2 - 1) = 25000 * 0.0404 = 1010. Q paid more by 1040 - 1010 = 30.

Multiple choice

The simple interest on a certain sum of money for $3$ years at $8\%$ per annum is half the compound interest on $Rs.\,4000$ for $2$ years at $10\%$ per annum. The sum placed on simple interest is .............. .
  1. $Rs.\,1550$
  2. $Rs.\,1650$
  3. $Rs.\,1750$
  4. $Rs.\,2000$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Compound interest on 4000 for 2 years at 10% is 4000 * (1.1^2 - 1) = 840. Simple interest is half of this, 420. Using SI = (P * R * T) / 100: 420 = (P * 8 * 3) / 100 => 42000 = 24P => P = 1750.