Question Detail

What is the present worth of Rs. 132 due in 2 years at 5% simple interest per annum

Answer: Option B

Explanation:

Let the present worth be Rs.x
Then,S.I.= Rs.(132 - x)

=› (x*5*2/100) = 132 - x

=› 10x = 13200 - 100x

=› 110x = 13200

x= 120

1. In how many years Rs 150 will produce the same interest at 8% as Rs. 800 produce in 3 years at 9/2%

Answer: Option B

Explanation:

Clue:
Firstly we need to calculate the SI with prinical 800,Time 3 years and Rate 9/2%, it will be Rs. 108

Then we can get the Time as

Time = (100*108)/(150*8) = 9

2. There was simple interest of Rs. 4016.25 on a principal amount at the rate of 9%p.a. in 5 years. Find the principal amount

Rs 7925
Rs 8925
Rs 7926
Rs 7925

Answer: Option B

Explanation:

\begin{aligned}
P = \frac{S.I. * 100}{R*T}
\end{aligned}

So by putting values from our question we can get the answer

\begin{aligned}
P = \frac{4016.25 * 100}{9*5} \\
= 8925
\end{aligned}

3. We have total amount Rs. 2379, now divide this amount in three parts so that their sum become equal after 2, 3 and 4 years respectively. If rate of interest is 5% per annum then first part will be ?

Answer: Option B

Explanation:

Lets assume that three parts are x, y and z.
Simple Interest, R = 5%

From question we can conclude that, x + interest (on x) for 2 years = y + interest (on y) for 3 years = z + interest (on y) for 4 years

\begin{aligned}
\left( x + \frac{x*5*2}{100} \right) = \left( y + \frac{y*5*3}{100} \right) = \left( z + \frac{z*5*4}{100} \right)\\
\left( x + \frac{x}{10} \right) = \left( y + \frac{3y}{20} \right) = \left( z + \frac{z}{5} \right) \\
=> \frac{11x}{10} = \frac{23y}{20} = \frac{6z}{5} \\

\text{lets assume k = }\frac{11x}{10} = \frac{23y}{20} = \frac{6z}{5} \\
\text{then }x = \frac{10k}{11} \\
y = \frac{20k}{23}\\
z = \frac{5k}{6}\\
\text{we know x+y+z = 2379}\\
=> \frac{10k}{11} + \frac{20k}{23} + \frac{5k}{6} = 2379\\
\text{10k*23*6+20k*11*6+5k*11*23=2379*11*23*6}\\
\text{1380k+1320k+1265k=2379*11*23*6}\\
\text{3965k=2379*11*23*6}\\
k = \frac{2379*11*23*6}{3965}\\
\text{by putting value of k we can get x} \\
x = \frac{10k}{11} \\
=>x = \frac{10}{11}*\frac{2379*11*23*6}{3965}\\
=>x = \frac{10*2379*23*6}{3965}\\
= \frac{2*2379*23*6}{793}\\
= 2 * 3 * 23 * 6 = 828
\end{aligned}

4. At 5% per annum simple interest, Rahul borrowed Rs. 500. What amount will he pay to clear the debt after 4 years

Answer: Option D

Explanation:

We need to calculate the total amount to be paid by him after 4 years, So it will be Principal + simple interest.

So,
\begin{aligned}
=> 500 + \frac{500*5*4}{100}
=> Rs. 600
\end{aligned}

5. Rs. 800 becomes Rs. 956 in 3 years at a certain rate of simple interest. If the rate of interest is increased by 4%, what amount will Rs. 800 become in 3 years.

Rs 1052
Rs 1152
Rs 1252
Rs 1352

Answer: Option A

Explanation:

S.I. = 956 - 800 = Rs 156

\begin{aligned}
R = \frac{156*100}{800*3} \\
R = 6\frac{1}{2}\% \\

\text{ New Rate = }6\frac{1}{2}+4 \\
= \frac{21}{2} \% \\

\text{ New S.I. = }800\times\frac{21}{2}\times{3}{100} \\
= 252
\end{aligned}

Now amount will be 800 + 252 = 1052