CAT Interest Questions

Let us assume the amount invested to be $$P$$, and the rate of interest to be $$r$$. 
Value of the investment after 3 years will be, $$P\left(1+r\right)^3$$
Value of the investment after 5 years will be, $$P\left(1+r\right)^5$$

$$\left(1+r\right)^2=\frac{36}{25}$$
$$\left(1+r\right)^2=1.44$$
$$r=0.2$$

We need to find the value of n for which $$4000\left(1+r\right)^n>20000$$
$$\left(1+r\right)^n>5$$
$$\left(1.2\right)^n>5$$
We see that, 
$$1.2^8=4.2999$$
$$1.2^9=5.15$$

Hence it takes 9 years to grow to over 20,000. 

Let's take the amount invested by Sunil to be X. 

The amount received by Anil at the end of 6 years would be $$22000\left(1+\frac{4}{2\times\ 100}\right)^{6\times\ 2}=22000\left(1.02\right)^{12}$$

The amount received by Sunil at the end of 5 years would be $$X\left(1.02\right)^{10}$$
In the 6th year, Sunil invests this at a simple interest of 10%, giving him an interest of $$X\left(1.02\right)^{10}\times\ 0.1$$
Giving the total amount with him at the end of 6 years to be $$X\left(1.02\right)^{10}\times\ \left(1+0.1\right)$$

Equating the final amount with Sunil and Anil, we get:
$$X\left(1.02\right)^{10}\times\ \left(1.1\right)=22000\left(1.02\right)^{12}$$
$$X=\frac{22000\left(1.02\right)^2}{1.1}=20808$$

Therefore, Option D is the correct answer. 

We are told that, 10000 is deposited in bank A for a certain number of years at a simple interest of 5% per annum. 
Let us say that the number of years is x
Total value of the deposit after x years is, $$10000\left(1+x\left(0.05\right)\right)$$

On maturity, the total amount received is deposited in bank B for another 5 years at a simple interest of 6% per annum
Here we know the years and the interest rate, 
$$10000\left(1+x\left(0.05\right)\right)\left(1+5\left(0.06\right)\right)$$
$$10000\left(1+\left(0.05\right)x\right)\left(1.3\right)$$

Interest received from Bank A is $$\left(x\left(0.05\right)\right)10000$$
Interest received from Bank B is $$0.3\left(10000\left(1+x\left(0.05\right)\right)\right)$$

This ratio is given to be 10:13. 

$$\dfrac{x\left(0.05\right)}{0.3\left(1+x\left(0.05\right)\right)}=\dfrac{10}{13}$$

$$0.65x=3+0.15x$$
$$0.5x=3$$
$$x=6$$

Hence the number of years the money was invested in Bank A is 6 years. 

It is given that Anil borrows Rs 2 lakhs at an interest rate of 8% per annum, compounded half-yearly. It is also known that he repays Rs 10320 at the end of the first year and closes the loan by paying the outstanding amount at the end of the third year.

The total amount at the end of the first year is: $$200000\times\ \frac{104}{100}\times\ \frac{104}{100}=216320$$

He repays 10320 rupees at the end of the first year, which implies the amount that remains unpaid at the end of the first year is 206000 rupees.

This unpaid amount will accrue interest for another two years.

Hence, the final amount at the end of three years is $$206000\times\ \frac{104}{100}\times\ \frac{104}{100}\times\ \frac{104}{100}\times\ \frac{104}{100}=240990.86$$

Hence, the accrued interest in these two years is (240990.86-206000) = 34990.86 rupees.

Hence, the total interest accrued over the three years = (34990.86+16320) = 51311 rupees.

The correct option is B

Anil invested 22000 for 6 years at 4% interest compounded half-yearly

=> Amount =$$22000\left(1.02\right)^{12}$$

Let Sunil invest 'S' rupees for 5 years at 4% C.I. half-yearly and 10% S.I. for 1 additional year

=> Amount = $$S\left(1.02\right)^{10}\left(1.1\right)$$

Given that the both amounts are equal

=> $$22000\left(1.02\right)^{12}=S\left(1.02\right)^{10}\left(1.1\right)$$

=> $$S=\dfrac{22000\left(1.02\right)^2}{1.1}=20808$$

Let the total investment me 15x and the no. of years required be T years

$$\frac{\left(3x\times6\times T\right)}{100}+\frac{\left(5x\times10\times T\right)}{100}+\frac{\left(7x\times1\times T\right)}{100}\ge15x$$

or, $$\frac{75xT}{100}\ge15x$$

or, $$T\ge20$$

So minimum value of T is 20 years

It is given,

$$\ \frac{\ 5.5\times1\times8000}{100}+\ \frac{\ 5.6\times1\times5000}{100}+\ \frac{\ x\times1\times7000}{100}=\frac{5}{100}\times20000$$

$$440+280+70x=1000$$

x = 4%

Interest = $$\ \frac{\ 20000\times4\times1}{100}$$ = Rs 800

The answer is option B.

Let the savings invested in first part and second part be 'x' and 'y', respectively.

It is given,

$$\ \frac{\ x\times15\times4}{100}=\ \frac{\ y\times12\times3}{100}$$

60x = 36y

5x = 3y

Required percentage = $$\frac{x}{x+y}\times100=\frac{3}{3+5}\times100=37.5\%$$

The answer is option A.

Let the principal amount be P and the interest rate be r.

Then $$P\left(1+r\right)^2-P\left(1+r\right)=806.25$$ -(1)

$$P\left(1+r\right)^3-P\left(1+r\right)^2=866.72$$ -(2)

Dividing (2) by (1), we get:

$$\frac{\left(P\left(1+r\right)^3-P\left(1+r\right)^2\right)}{P\left(1+r\right)^2-P\left(1+r\right)}=\frac{866.72}{806.25}$$

$$\frac{\left(\left(1+r\right)^2-1-r\right)}{1+r-1}=1.075$$

$$\frac{r^2+r}{r}=1.075$$

r=0.075 or 7.5%

$$\frac{\left(Interest\ accrued\ in\ 4th\ yr\right)}{Interest\ accrued\ in\ 3rd\ yr}=\frac{X}{866.72}$$

$$\frac{\left(P\left(1+r\right)^4-P\left(1+r\right)^3\right)}{P\left(1+r\right)^3-P\left(1+r\right)^2}=\frac{X}{866.72}$$

Dividing numerator and denominator by $$P\left(1+r\right)^2$$

$$\frac{r^2+2r+1-1-r}{1+r-1}=\frac{X}{866.72}$$

$$r+1=\frac{X}{866.72}$$

$$X=1.075\times\ 866.72=931.72$$

Bank A: 6% p.a. 1/2 yearly (CI)

Bank B: x% p.a (SI)

Bank C: 2x% p.a (SI)

Let Raju invest Rs P in bank B for t years. Hence, Rupa invests Rs 10,000 in bank C for 2t years.

Now, 

$$P\left(\frac{x}{100}\right)t\ =\ P\left(1+\frac{3}{100}\right)^2-P$$

$$\left(\frac{x}{100}\right)t\ =\ 1.0609-1$$

$$\left(\frac{x}{100}\right)t\ =\ 0.0609$$

We need to calculate

SI = $$10000\times\ 2t\times\ \left(\frac{2x}{100}\right)=40000\left(\frac{x}{100}\right)t=40000\times\ 0.0609=2436$$

For two years the compound interest is $$\frac{PR(1)}{100}+\frac{PR(1)}{100}\left(1+\frac{PR(1)}{100}\right)$$

For three years the simple interest is $$\frac{9PR}{100}$$

Now R(1)= 5% and R=3%

Hence $$\frac{5P}{100}+\frac{5P}{100}\left(1.05\right)-\frac{9P}{100}=1125$$

$$\frac{-4P}{100}+\frac{5.25P}{100}=1125$$

$$\frac{1.25P}{100}=1125$$

Solving we get P= 90000

Given, 

Rate of interest = 10%

Since it is compounded half-yearly, R=5%

n=3

We know, A = $$P\left(1+\frac{R}{100}\right)^{^n}$$

18522 = $$P\left(1+0.05\right)^3$$

=> P = 16000

Let their individual Amounts be equal after 't' years. Let their initial investments amount to $$A_V$$ and $$A_J$$ ; 

$$A_V\ =10,000\left(1+\frac{5t}{100}\right)$$ and $$A_J\ =8,000\left(1+\frac{10\left(t-2\right)}{100}\right)$$

Equating both: $$10,000\left(1+\frac{5t}{100}\right)\ =8,000\left(1+\frac{10\left(t-2\right)}{100}\right)$$

On simplifying both sides, we get: $$15t\ =\ 180\ ;\ t\ =\ 12$$

The amount with Amal at the end of 1 year = 12000*1.08+10000*1.03*1.03=23569

Interest received by Amal = 23569-22000=1569

Let the amount invested by Bimal = 100b

Interest received by Bimal = 100b*7.5*1/100=7.5b

It is given that the amount of interest received by both of them is the same 

7.5b=1569

b=209.2

Amount invested by Bimal = 100b = 20920

Assuming the amount invested in the ratio 2:1 was 200x and 100x, then the fixed deposit investment = 1500000-300x

Hence, the interest = 200x*4/100 = 8x and 100x*3/100=3x

Interest from the fixed deposit = (1500000-300x)*6/100 = 90000-18x

Hence the total interest  = 90000-18x+8x+3x=90000-7x =76000

=> 7x=14000   => x=2000

Hence, the fixed deposit investment = 1500000-300*2000 = 900000 = 9 lakhs

We have to equate the installments and the amount due either at the time of borrowing or at the time when the entire loan is repaid. Let us bring all values to the time frame in which all the dues get settled, i.e, by the end of 2 years. 

John borrowed Rs. 2,10,000 from the bank at 10% per annum. This loan will amount to 2,10,000*1.1*1.1 = Rs.2,54,100 by the end of 2 years. 
Let the amount paid as installment every year be Rs.x.

John would pay the first installment by the end of the first year. Therefore, we have to calculate the interest on this amount from the end of the first year to the end of the second year. The loan will get settled the moment the second installment is paid. 

=> 1.1x + x = 2,54,100
2.1x = 2,54,100
=> x = Rs. 1,21,000.

Therefore, 121000 is the correct answer. 

Amount of interest paid by Ishan to Gopal if the borrowed amount is Rs. (X+Y) = $$\dfrac{10}{100}*(X+Y)$$ = 0.1(X+Y)

Gopal also borrowed Rs. X from Ankit at 8% per annum. Therefore, he has to return Ankit Rs. 0.08X as the interest amount on borrowed sum. 

Hence, the interest retained by gopal = 0.1(X+Y) - 0.08X = 0.02X + 0.1Y   ... (1)

It is given that the net interest retained by Gopal is the same as that accrued to Ankit.

Therefore, 0.08X = 0.02X + 0.1Y 

$$\Rightarrow$$ X = (5/3)Y   ... (2)

Amount of interest paid by Ishan to Gopal if the borrowed amount is Rs. (X+2Y) = $$\dfrac{10}{100}*(X+2Y)$$ = 0.1X+0.2Y

In this case the amount of interest retained by Gopal = 0.1X+0.2Y - 0.08X = 0.02X + 0.2Y   ... (3)

It is given that the interest retained by Gopal increased by Rs. 150 in the second case. 

$$\Rightarrow$$ (0.02X + 0.2Y) - (0.02X + 0.1Y) = 150

$$\Rightarrow$$ Y = Rs. 1500

By substituting value of Y in equation (2), we can say that X = Rs. 2500

Therefore, (X+Y) = Rs. 4000.

As we know, formulae of compound interest for 2 years  will be:
$$P(1+\frac{r}{100})^{2}$$ = 625  (Where r is rate, P is principal amount)
For 3 years:
$$P(1+\frac{r}{100})^{3}$$ = 675
Dividing above two equations we will get r=8%