CMAT Interest Questions

Initial amount = Rs 4000

Compounded amount after one year = $$4000\left(1+\frac{15}{100}\right)=4600$$

Final amount after one year = 4600 - 1500 = Rs 3100

Compounded amount after at the end of two years = $$3100\left(1+\frac{15}{100}\right)=3565$$

Final amount at the end of two years = 3565-1500 = Rs 2065

Compounded amount after at the end of three years = $$2065\left(1+\frac{15}{100}\right)=2374.75$$

Final additional amount at the end of three years = 2374.75-1500 = Rs 874.75

The answer is option D.

The difference between C.I and S.I for a period of 2 years is given by : P(R/100)^2 
So we get :
P(R/100)^2 = 32
5000(R/100)^2 = 32
(R/100)^2 = 32/5000 = 16/2500
We get 
R/100 = 4/50 
R = 8%

We have 
$$P\left(\frac{R}{100}\right)^2=\ 64$$
And 
$$P\left(1+\left(\frac{R}{100}\right)\right)^{^2}-p\ =\ 704$$
Now dividing both we get :
$$\frac{\left(\frac{R}{100}\right)^2}{\left(1+\frac{R}{100}\right)^2-1}\ =\ \frac{1}{11}$$
Solving we get R/100 = 5
Now substituting we get P = 64*25 = 1600

Principal sum to be borrower = Rs. 25,000

Amount when compounded annually = $$P(1+\frac{R}{100})^T$$

=> Amount to be paid after 3 years = $$25,000(1+\frac{3}{100})$$ $$(1+\frac{4}{100})$$ $$(1+\frac{5}{100})$$

= $$250\times103\times\frac{26}{25}\times\frac{21}{20}$$

= $$103\times13\times21=Rs.$$ $$28,119$$

=> Ans - (D)

If a sum of money becomes $$x$$ times in $$a$$ years, it will become $$y$$ times in $$b$$ years

=> $$(x)^{\frac{1}{a}}=(y)^{\frac{1}{b}}$$

Similarly, $$(8)^{\frac{1}{3}}=(16)^{\frac{1}{t}}$$

=> $$(2)^{\frac{3}{3}}=(2)^{\frac{4}{t}}$$

=> $$1=\frac{4}{t}$$

=> $$t=4$$ years

=> Ans - (A)

If the principle amount 'P' when compounded half-yearly at R% interest rate per annum for 'n' years, the new amount is P'.

then $$P' = P{[1 + \frac{R}{2\times 100}]}^{n}$$

Given P' = 6,632.55, P = 6,250 and R = 4%

$$\Rightarrow 6,632.55 = 6,250{[1 + \frac{4}{2\times 100}]}^{n}$$

$$\Rightarrow 1.061 = {1.02}^{n}$$

Taking logarithm on both sides we get,

n = log(1.061)$$\div$$log(1.02) = 3

Since n refers to half a year in this case, the number of years will be $$\frac{3}{2}$$ years.

Let say after a Time period (T) 'x'  years the amount Rs. 2 lakh gets doubled at a rate of interest (R) = 11.5% per annum

Initial principle amount (P) = 2,00,000 Rs.

Simple Interest (I)  for 'x' years is given by I = $$PTR\div 100$$ = $$200000\times x\times 11.5\div 100$$

                       => I = $$23000\times x$$.

Final Principle amount (P') = P + I = $$2\times P$$ (as given) 

                                               => I = $$2\times P - P$$ = P

                                              => $$23000\times x$$ = 200000

                                              => x = $$200000\div 23000$$

                                              => x = 8.695 = between 8 and 9.                                                              

If the  principle amount 'P' when compounded annually for 'n' years at 'R%" interest rate per annum becomes P'.

Then $$P' = P {[1 +\frac{R}{100}]}^{n}$$

Given P' = 2000, n = 2 years, R = 10% 

$$\Rightarrow P = P'\div {[1 +\frac{R}{100}]}^{n}$$

$$\Rightarrow P = 2000\div {[1 +\frac{10}{100}]}^{2}$$ 

$$\Rightarrow P = 2000\div 1.21$$

$$\Rightarrow P = 1653$$

Hence the initial amount P = Rs. 1,653.