Simple Interest and Compound Interest - Practice Questions with Multiple Methods

Simple and Compound Interest

Example 1: A sum of money lent at simple interest amounts to Rs 880 in 2 years and to Rs 920 in 3 years. Find rate of interest?

Solution:  Method 1:  Amount after 2 years is Rs 880 and after 3 years is Rs 920.
It means S.I for one year = 920 – 880 = Rs 40
  ⇒ S.I for 2 years = 2 x 40 = Rs 80
Principal will be = Amount – S.I = 880 – 80 = Rs 800
  ⇒ Rate = (80x100/800x2) = 5%

 Method 2:  Direct formula: If a sum amounts to Rs A1 in t1 years and Rs A2 in t2 years at Simple rate of interest, then Rate per annum =     100 x [A1 - A2]/[A1 x t2 - A2 x t1]
  ⇒ Rate = 100 x [920 – 880]/[880 x 3 – 920 x 2] = 5%

Example 2: The sum of money that will give Rs 1 as interest per day at the rate of 5% per annum simple interest is?

Solution: S.I = PRT/100
  ⇒ 1 = (P x 5 x 1)/(100 x 365)
  ⇒ P = Rs 7300

Example 3: What annual installment will discharge a debt of Rs 6450 due in 4 years at 5% simple interest?

Solution:  Method 1:  Basic formula: Let I be the annual installment.
  ⇒ I +  (I x R x (N-1))/100  +  I + (I x R x (N-2))/100  + …. +  I + (I x R x (N-N))/100  = Z
  ⇒ I + Ix5x3/100 + I + Ix5x2/100 + I = 6450
  ⇒ I = Rs 1500

 Method 2:  Method of percentage:
Let the annual payment (for fourth year) = 100%
The amount at 5% for first year = 100% + rate% = 100% + 5% = 105%
The amount at 5% for second year = 100% + 10% = 110%
The amount at 5% for third year = 100% + 15% = 115%

These three years amounts together with the last annual payment will discharge a debt of Rs 6450.
  ⇒ (100 + 105 + 110 + 115)% = Rs 6450
  ⇒ 430% = Rs 6450
  ⇒ 100% = 6450 x 100/430 = Rs 1500
  ⇒ Annual payment = Rs 1500

 Method 3:  Direct formula: The annual payment that will discharge a debt of Rs Z due in T years at the rate of interest R% per annum is = (100Z/[100T + RT(T – 1)/2])
  ⇒ Annual payment = (100 x 6450/[100x4 + 5x4(4– 1)/2])
  ⇒ Annual payment = Rs 1500

Example 4: A sum of Rs 1550 was lent partly at 5% and partly at 8% simple interest. The total interest received after 3 years is Rs 300. The ratio of money lent at 5% to that at 8% is?

Solution:  Method 1:  Let the sum lent at 5% is Rs x and at 8% is Rs (1550 – x). 
  ⇒ x x 5 x 3/100 +  (1550 – x) x 8 x 3/100 = 300 
  ⇒ x = Rs 800
Sum lent at 8% = 1550 – 800 = Rs 750
  ⇒ Ratio = 800:750 = 16:15

 Method 2:  By Alligation:
First rate % for 3 years = 5 x 3 = 15%
Second rate % for 3 years = 8 x 3 = 24%
Overall rate of interest = 300 x 100/1550 = 600/31%

Ratio = 149:135 = 16:15

Example 5: A person invests money in three different schemes for 6 years, 10 years and 12 years at 10%, 12% and 15% respectively. At the completion of each scheme, he gets the same interest. The ratio of his investment is?

Solution: Ration of investments = 1/R1t1 : 1/R2t2 : 1/R3t3
  ⇒ 1/60 : 1/120 : 1/180 = 6:3:2

Example 6: A sum was invested on simple interest at a certain rate for 2 years. Had it been put at 3% higher rate, it would have fetched Rs 72 more. The sum is?

Solution: We know principal is always 100%.
Rate for 2 years will be = 3 x 2 = 6%
This rate fetch Rs 72 more, it means: 6% = 72
  ⇒ 100% = 72 x 100/6 = Rs 1200

Example 7: A sum of money doubles itself in 10 years at simple interest. What is the rate of interest?

Solution: If sum doubles, then S.I must be equal to principal. If sum trebles, then S.I must be double to principal. It means rate of interest always 1 less than the multiple number of principal.
  ⇒ Rate of interest = (Rate x Time)% = Multiple number of principal – 1
  ⇒ (R x 10)% = 2 – 1
  ⇒ R = 1/10 x 100 = 10%

Example 8: A man borrows Rs 21000 at 10% compound interest. How much he has to pay annually at the end of each year, to settle his loan in two years?

Solution:  Method 1:  Basic: If each installment be x, then:
Present worth of first installment will be P1 = x/(1+10/100) = 10x/11
Present worth of second installment P2 = x(1 + 10/100)2 = 1100x/121
  ⇒ P1 + P2 = Rs 21000
  ⇒ x = Rs 12100

 Method 2:  Direct Formula: Let value of installment is Rs I

⇒ I = 21000/[(100/(100+10))1 + (100/(100+10))2]
  ⇒ I = Rs 12100

 Method 3:  Second Formula: Let value of installment is Rs I

  ⇒ I = (21000 x 10/100)/[1 - (100/100+10)2]
  ⇒ I = Rs 12100

Example 9: A sum of money doubles itself in 4 years at compound interest. It will amount to 8 times itself at the same rate of interest in how many years?

Solution:  Method 1:  Basic: Let the sum be Rs x.
Sum becomes double in 4 years, it means amount = 2x
Amount = P(1 + R/100)2
  ⇒ 2x = x(1 + R/100)4
  ⇒ 2 = (1 + R/100)4
We need to cube both sides so that amount will be 8 times to the sum.
  ⇒ 8 = (1 + R/100)12
It means amount will be 8 times to the sum in 12 years

 Method 2:  Conceptual formula:
If a sum becomes x times in y years at C.I then it will be (x)n times in ny years.
  ⇒ Amount is 2 times in = 4 years
  ⇒ It will be 8 or (23) times in = 3 x 4 = 12 years

Example 10: A certain sum of money amounts to Rs 2420 in 2 years and Rs 2662 in 3 years at same rate of compound interest, compounded annually. The rate of interest per annum is?

Solution:  Method 1:  Basic: Amount = P(1 + R/100)n
  ⇒ 2420 = P(1 + R/100)2  ……..(1)
  ⇒ 2662 = P(1 + R/100)3  ……..(2)
On dividing equation (2) by equation (1) we get:
  ⇒ 2662/2420 = 1 + R/100
  ⇒ R = 10%

 Method 2:  Concept: When difference of year is 1 only, then:
We know that S.I and C.I for one year are same. It means for third year amount, amount of second year will become principal.
  ⇒ Interest = Amount – principal = 2662 – 2420 = Rs 242

S.I = PRT/100
  ⇒ 242 = 2420 x R x 1/100
  ⇒ R = 10%

Example 11: The compound interest on a certain sum of money at 5% per annum for 2 years is Rs 246. The simple interest on the same sum for 3 years at 6% per annum is?

Solution:  Method 1:  Method of percentage:
C.I for 2 years at 5% = 2 x 5 + (5)2/100 = 10.25% of sum or principal
S.I for 2 years at 5% = 10% of sum or principal

C.I for 2 years is given as = Rs 246
  ⇒ 10.25% of sum = Rs 246
  ⇒ 10% of sum = 246/10.25 x 10 = Rs 240
  ⇒ S.I on same sum for 2 years is Rs 240.

Rate of interest for 3 years at 6% will be = 3 x 6 = 18% of the sum.
  ⇒ 10% of sum = Rs 240
  ⇒ 18% of sum = Rs 240 X 18/10 = Rs 432
  ⇒ S.I on same sum for 3 years at 6% per annum is Rs 432.

 Method 2:  Direct Formula:
If S.I on certain sum for two years at same Rate% is x and C.I is y, then: y = x(1+ R/200)
  ⇒ 246 = x(1+ 5/200)
  ⇒ x = Rs 240
This simple interest is calculated on 5% rate of interest.

Now by basic formula: S.I = PRT/100
  ⇒ 240 = P x 5 x 2/100
  ⇒ P = 2400
S.I for 3 years on same sum at 6% rate of interest = 2400 x 6 x 3/100 = Rs 432

Example 12: The compound interest on a certain sum of money at a certain rate for 2 years is Rs 40.80 and the simple interest on the same sum is Rs 40 at the same rate and for the same time. Find the rate of interest and sum?

Solution:  Method 1:  Direct Formula:
For 2 years: Rate = 2 x Difference in C.I & S.I x 100/S.I
  ⇒ Rate = 2 x 0.8 x 100/40 = 4%

For 2 years: Sum = Difference between C.I & S.I x (100/R)2
  ⇒ Sum = 0.80 x (100/4)2 = Rs 500

 Method 2:  Concept: Difference between C.I & S.I = Rs 0.80
S.I for one year = 40/2 = Rs 20

We know that S.I and C.I are same for first year but C.I for second year includes rate of interest on S.I. It means difference between C.I & S.I is rate of interest on S.I.

  ⇒ Rate% of S.I = Difference between C.I and S.I
  ⇒ Rate% of 20 = 0.80
  ⇒ Rate = 4%