Question Detail

A sum of money amounts to Rs 9800 after 5 years and Rs 12005 after 8 years at the same rate of simple interest. The rate of interest per annum is

Answer: Option D

Explanation:

We can get SI of 3 years = 12005 - 9800 = 2205

SI for 5 years = (2205/3)*5 = 3675 [so that we can get principal amount after deducting SI]

Principal = 12005 - 3675 = 6125

So Rate = (100*3675)/(6125*5) = 12%

1. 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

2. 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}

3. 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

4. Sachin borrows Rs. 5000 for 2 years at 4% p.a. simple interest. He immediately lends money to Rahul at 25/4% p.a. for 2 years. Find the gain of one year by Sachin.

110.50
111.50
112.50
113.50

Answer: Option C

Explanation:

Two things need to give attention in this question, First we need to calculate gain for 1 year only.
Second, where we take money at some interest and lends at other, then we use to subtract each other to get result in this type of question. Lets solve this Simple Interest question now.

\begin{aligned}
\text{Gain in 2 year = } \\
[(5000 \times \frac{25}{4} \times \frac{2}{100})-(\frac{5000 \times 4 \times 2}{100})] \\
= (625 - 400) = 225 \\
\text{ So gain for 1 year = }\\
\frac{225}{2} = 112.50
\end{aligned}

5. Find the simple interest on the Rs. 2000 at 25/4% per annum for the period from 4th Feb 2005 to 18th April 2005

Rs 25
Rs 30
Rs 35
Rs 40

Answer: Option A

Explanation:

One thing which is tricky in this question is to calculate the number of days.
Always remember that the day on which money is deposited is not counted while the day on which money is withdrawn is counted.
So lets calculate the number of days now,
Time = (24+31+18) days = 73/365 years = 1/5 years
P = 2000
R = 25/4%

\begin{aligned}
\text{ S.I. = } = \frac{2000 \times 25 }{4 \times 5 \times 100} = 25
\end{aligned}