Test Level 2: Simple & Compound Interest - 1 - CAT MCQ

Total money pay by A to B(after end of the 2 year) = 20,000 + 10,000 (1 + 10/100) = 24,200 + 11,000 = Rs. 35,200

 Let the amount be Rs. x and Rs. y for the younger and elder son, respectively.
x + y = Rs. 18,750 ... (I)
Applying simple interest :
 x (1 + 6 × 5/100) = y (1 + 4 × 5/100) [ We know: S.I = (P×R×T)/100]
 x/12 = y/13 = 750
 x = Rs. 9,000
y = Rs. 9,750
Hence, amount allotted to elder son = Rs. 9,750

Let principle be P, amount be 8P and time(t) be 3 years compounded annually.


r = 100%

At 6.25%, money becomes 6 times. i.e; SI = 6 - 1 = 5 times.
SI and rate have inverse relation.
So, (11 - 1)/r = (6 - 1)/6.25;
r = {10 x (6.25/5)}%
= 12.5%

Let principal = P, rate = r and time = n

⇒ 2^n/2 ≥ 7
If n = 5, 2^5/2 < 7
If n = 6, 2³ > 7
So, minimum 6 years are required.

Simple interest after 1 month = (1000 × 10 × 1)/(100 × 12)
Simple interest after 2 months = (1000 × 10 × 2)/(100 × 12)
Simple interest after 3 months = (1000 × 10 × 3)/(100 × 12)
Similarly, simple interest after 12 months = (1000 × 10 × 12)/(100 × 12)
Required amount = 1000 × 12 + 1000 × 
= Rs. 12650

5000 
5 + 25 r₁/100 = 6 + 30r₂/100
So, we cannot find r₁/r₂ from this.

SI at 11% for 2 years on a sum of Rs. 1000 = Rs. 220.
Let the rate at which the interest is compounded annually be R.
Since CI is twice the SI,

⇒ R = 20%
Required rate = 20%

Amount = Principal
The present value of Rs. 11,000, which is paid after 1 year, is 
The present value of Rs. 60,500, which is paid after 2 years, is 
So, borrowed amount = 10,000 + 50,000 = Rs. 60,000

CI - SI = P  (For 2 years)
7.35 = P 
Or, P = = Rs. 1500