XAT Interest Questions

Let the amount Jose borrowed be 'x'. The rate of interest is 10%

Interest occurred on 'x' is = $$\ \frac{\ x\times1\times\ \ 0.1\ }{5}+\ \frac{\ 2x\ \times2\times\ \ 0.1}{5}\ +\ \ \frac{2x\times\ 3\times\ 0.1\ }{5}$$

                                         = $$\ \frac{\ 1.1x\ }{5}$$

The invested amount doubled at the end of the fourth year, i.e. 2x.

Jose's profit after paying principal and interest amounts to his friend at the end of the fourth year is Rs 97500.

i.e., $$\ 2x-x-\frac{\ 1.1x\ }{5}=97500$$

       $$\frac{\ 3.9x}{5}=97500$$

       $$\ \ \ \ x=125000$$

The amount that Jose borrowed is Rs 1,25,000.

Option (D) is correct.

Let the total amount be 3x

Case 1: 

Smaller amount = x, rate of interest = 10

Larger amount = 2x, rate of interest = 5

Total amount received at the end of two years( smaller amount) = $$x\left(1+\frac{10}{100}\right)^2\ =\ 1.21x$$. CI = 0.21x

Total amount received at the end of two years( larger amount) = $$2x\left(1+\frac{5}{100}\right)^2\ =\ 2.205x$$   CI = 0.205x

Given, 0.21x + 0.205x = 830

=> x = 2000

2x= 4000

Case 2:

Smaller amount = x, rate of interest = 20

Larger amount = 2x, rate of interest = 10

Total amount received at the end of two years( smaller amount) = $$x\left(1+\frac{20}{100}\right)^2\ =\ 1.44x$$. CI = 0.44x

Total amount received at the end of two years( larger amount) = $$2x\left(1+\frac{10}{100}\right)^2\ =\ 2.42x$$ CI = 0.42x

Given, 0.44x+0.42x = 830

=> x = 965.11 which is not valid since it should be greater than 1000

Amount on which interest will be charged = 19200 - 4800 = 14400

The total amount paid will be equal to the sum of all monthly instalments. Therefore, we have

$$14400*k^{5a} = I (k^{4a} +k^{3a}+k^{2a}+k^{a}+1 )$$   .....(1)

where, k = $$1+\frac{12}{100}$$ & a = $$\frac{1}{12}$$

We know that, $$k^{5a}-1 = (k-1)(k^{4a} +k^{3a}+k^{2a}+k^{a}+1)$$

=>$$k^{4a} +k^{3a}+k^{2a}+k^{a}+1$$ = $$\frac{k^{5a}-1}{k-1}$$ ....(2)

Substituting in equation (1) we get

I = $$14400*k^{5a}*\left[\frac{k-1}{k^{5a}-1}\right]$$ ....(3)

On substituting the values of k and a in equation (3) we get
I $$\approx$$ 2965

Hence, option B.

Let the principal amount = $$P$$ and rate of interest = $$r \%$$

Interest accumulated from 2004 to 2007 is Rs.10,000 and from 2004 to 2010 is Rs.25,000

Using, $$C.I. = P[(1 + \frac{R}{100})^T - 1]$$

=> $$P[(1 + \frac{r}{100})^3 - 1] = 10,000$$ ----------Eqn(I)

and $$P[(1 + \frac{r}{100})^6 - 1] = 25,000$$ -----------Eqn(II)

Dividing eqn(II) from (I), we get : 

=> $$\frac{P[(1 + \frac{r}{100})^6 - 1]}{P[(1 + \frac{r}{100})^3 - 1]} = \frac{5}{2}$$

Let $$(1 + \frac{r}{100})^3 = x$$

=> $$\frac{x^2 - 1}{x - 1} = \frac{5}{2}$$

=> $$2x^2 - 5x + 3 = 0$$

=> $$(2x - 3) (x - 1) = 0$$

=> $$x = \frac{3}{2} , 1$$    $$(x \neq 1)$$ because then, r = 0

=> $$(1 + \frac{r}{100})^3 = \frac{3}{2}$$

Substituting it in eqn(I)

=> $$P[\frac{3}{2} - 1] = 10,000$$

=> $$P = 10,000 \times 2 = 20,000$$

Rs. 15 lakhs is to be divided equally.

In the case of the younger son,
Principal = Rs. 750000, time = 9 years and Interest = Rs. 1350000
Rate of interest = $$\dfrac{1350000 * 100}{750000 * 9}$$ = 20%

In the case of the elder son, 
Principal = Rs. 750000, time = 6 years and Interest = Rs. 1350000
Rate of interest = $$\dfrac{1350000 * 100}{750000 * 6}$$ = 30%

Hence, option E is the correct answer.

Amount man gets after 1 year

= $$6000 + (\frac{6000 \times 5 \times 1}{100}) - 1200$$

= $$6000 + 300 - 1200 = 5100$$

$$\therefore$$ Amount at the beginning of third year, i.e. after 2 years

= $$5100 + (\frac{5100 \times 5 \times 1}{100}) - 1200$$

= $$5100 + 255 - 1200 = 4155$$