Simple and Compound Interest Questions with Solution (Basic)

Solution:

Option C

If Diff. between SI & CI for 2 years is Rs. x, then Principal = x (100/r)² P =  50 x (100x100)/(10x10) → P = 5000.

Solution:

Option D

If Diff. between SI & CI for 2 years is Rs. x, Principal = x (100/r)² P = 162 x  (100 x100)/(18 x18) → P = 5000.

Solution:

Option E

SI = Rs. 200 = Rs. 100 + Rs. 100

CI = Rs. 208 = Rs. 100 + Rs. 108 (In first year SI & CI are equal.) Therefore, Rs.8 Gap is because of the interest of 1^st year interest. Rs. 8 is interest on Rs. 100, R= (8x100)/(100x1) % ,R = 8%

Solution:

Option B

Principal = x (100/r)² where x is the difference in CI & SI for 2 years. ∴ P =  25(100/10)² = 2500.

Solution:

Option A

SI for 3 years = 225

→ SI for 1 year = 75

∴ CI for 1 year = 75.

So CI for 2^nd year = 90 and SI for 2^nd year = 75. Difference = 15  ∴ Rate of interest = (15/75)x100 = 20 %

Solution:

Option B

SI for 3 years = 150

→ SI for 1 years = 75

∴ CI for 1 year = 75.

So CI for 2^nd year = 80 and SI for 2^nd year = 75. Difference = 5  ∴ Rate of interest = (5/75) x 100 = 20/3 %.

Solution:

Option C

Let, the capital of Tulsi = 4.

∴ Capital of Mihir = 9.

4(1+(50/100))³ = [9+(27R)/100]R

50/3 %

Solution:

Option D

Let us assume P= Rs. 1000. SI= Rs. 300, CI = Rs. 331

Difference = 331 - 300 = Rs. 31.

Applying the unitary method, the difference Rs. 372 is 12 times Rs. 31. Therefore, the principle will also be 12 times of Rs. 1000 i.e. Rs. 12000.

Solution:

Option B

Let, the capital of Sahil = 6.

∴ Capital of Chaya = 7

6(1+(20/100))³ = [7+(14R)/100]R

= 11 5/7%

Solution:

Option A

Let us assume P= Rs. 1000, SI = Rs. 600, CI = Rs. 728

Difference= 728 – 600 = Rs. 128

Applying the unitary method, the difference Rs. 640 is 5 times Rs. 128. Therefore, the principle will also be 5 times of Rs. 1000 i.e. Rs. 5000.

Solution:

Option A

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

let the time be n years

Then,30000(1+7/100)^n = 34347

(107/100)^n = 34347/30000 = 11449/10000 = (107/100)^2 n = 2 years

Solution:

Option B

Let the sum be Rs.P.then

P(1+R/100)^3=6690…(i)

and P(1+R/100)^6=10035…(ii)

On dividing,we get (1+R/100)^3=10025/6690=3/2.

Substituting this value in (i),we get:

P*(3/2)=6690 or P=(6690*2/3)=4460

Hence, the sum is rs.4460.

Solution:

Option D

I → (P*R*5)/100 = P → R = 20

II → P(1 + R/100)² – P - (P*R*2)/100 = 400 → PR² → 4000000 III → (P*R*1)/100 = 2000 → PR = 200000

PR²/ PR = 4000000 / 200000 → R = 20

Thus I only or (II and III) give answer.

Solution:

Option B

Time = 2 years 4 months = 2(4/12) years = 2(1/3) years.

Amount = Rs'. [8000 X (1+(15/100))^2 X (1+((1/3)*15)/100)] =Rs. [8000 * (23/20) * (23/20) * (21/20)]

= Rs. 11109. .

:. C.I. = Rs. (11109 - 8000) = Rs. 3109.

Solution:

.Option B

Let sum=Rs.x

C.I. when compounded half yearly = [x(1+10/100)⁴ – x] = 4641/10000 C.I. when compounded annually = [x(20/100)² – x] = 11/25

4641x/10000 – 11x/25 = 482

→ x = 20000

Solution:

Option D

Principal = Rs. 10000; Rate = 2% per half-year; Time = 2 years = 4 half years.

Amount == Rs. 10824.32.

Rs [10000*(1+2/100)⁴]

C.I. = Rs. (10824.32 - 10000) = Rs. 824.32

Solution:

Option B

Principal = Rs. 16000; Time = 9 months =3 quarters;

Rate = 20% per annum = 5% per quarter.

Amount = Rs. [16000 x (1+(5/100))3] = Rs. 18522

= Rs. (18522 - 16000) = Rs. 2522

Solution:

Option B

Let rate = R% and time = R years.

Then, (122*R*R)/100 = 432

12R²=432

R²=36

R=6

Solution:

Option D

Rate of growth of population r = 10%

Present population = 26620

Let population 3 years ago be x

Applying the formula for population growth which is similar to compound interest, 26620 = x(1+r/100)³

26620 = x(1+10/100)³

26620 = x(110/100)³

X = 26620 (100/110)³

X = 20,000

The population 3 years ago was 20,000

Solution:

Option B

Difference between depreciation in value between the first and second years Rs.  4,000 - Rs. 3,600 = Rs. 400.

⇒ Depreciation of one year on Rs. 4,000 = Rs. 400

⇒ Rate of depreciation = 400/4000 * 100 = 10%