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Aptitude question and answer on Simple interest



9
If an amount Rs. 8000 becomes to Rs. 9200 in 3 years at simple interest. If the rate of interest is increased by 3%, it would amount to
A.    Rs.9920
B.    Rs.10560
C.    Rs.11120
D.    Rs.11820

Answer : A.  Rs.9920

Explanation:
        Principal = Rs. 8000,
                S.I. = Rs. 1200,
               Time = 3years 
\(\dot{..}\)            Rate = \(\bigg({100 \times 1200 \over 8000 \times 3} \bigg)\)%
                       =5 %
        New rate  =8 %, Principal = Rs. 8000, Time = 3 years
                  S.I = Rs. \(\bigg({8000 \times 8 \times3 \over 100} \bigg)\)
                       =Rs. 1920
\(\dot{..}\) New amount = Rs. (8000+1920)
                       =9920

10
Rajnish took a loan for 7 years at the rate of 6% p.a. S.I. if the total interest paid was Rs.2100, The principal was
A.    Rs.4400
B.    Rs.4800
C.    Rs.5000
D.    Rs.5200

Answer : C.  Rs.5000

Explanation:
Principaal = Rs. \(\Bigg( {2100 \times 100 \over 7 \times6} \Bigg)\)
               =Rs. \(\Bigg( {300 \times 100 \over 6} \Bigg)\)
               =Rs. 5000

11
Mr. Santosh finds that an increase in the rate of interest from \(4{7 \over8}\)% to \(7{3 \over 4}\)%, his yearly income diminishes by Rs.615. His capital is
A.    Rs.260000
B.    Rs.246000
C.    Rs.238000
D.    Rs.224000

Answer : B.   Rs.246000

Explanation:
Let capital be Rs. x. Then

\({x \times 8 \times 1 \over 100}-x{31 \over 4} \times {1 \over 100} =615\)

or   \(32x-31x=61500 \times 4\)

\(\dot{..}\)      x= 246000.

12
A sum of money was lent at simple interest at 11% p.a. for \(3{1 \over 2}\)years and \(4{1 \over 2}\) years respectively. If the difference in interests for periods was Rs. 5500, then the sum is
A.    Rs.50050
B.    Rs.55000
C.    Rs.50000
D.    Rs.50500

Answer : C.  Rs.50000

Explanation:
Let the sum Rs.x. Then 
\(\Bigg( x\times 11 \times {9 \over 2} \times {1 \over 100} -x \times 11 \times {7 \over 2} \times {1 \over 100}\Bigg)\)= 5500
or            \({22x \over 200 }=5500\)
or           \(11x=550000\)
or              \(x=50000.\)

13
How much should money lender lend at simple rate of interest of 15% in order to have Rs. 3234 at the end of \(1{1 \over 2}\) years ?
A.    Rs. 1640
B.    Rs.2620
C.    Rs. 2610
D.    Rs.2640

Answer : D.  Rs.2640

Explanation:
Let the required money be x. Then 

        \(\Bigg( x+{x \times 15 \over 100} \times {3 \over 2} \Bigg)=3234\)

or  \(49x \over 40\)=3234

or  \(x = \bigg({3234 \times 40 \over 49 }\bigg)=2640\)

14
What annual payment will discharge a debit of Rs. 1740 due in 5 years, the rate being 8% per annum ?
A.    Rs. 360
B.    Rs. 300
C.    Rs. 320
D.    Rs. 340

Answer : B.   Rs. 300

Explanation:
Let annual instalment be Rs. x . Then

\(\bigg[x+\bigg({x \times 4 \times 8 \over100}\bigg) \bigg ]+ \bigg[x+\bigg({x \times 3 \times 8 \over100}\bigg) \bigg ]+\)

\(\bigg[x+\bigg({x \times 2 \times 8 \over100}\bigg) \bigg ]+\bigg[x+\bigg({x \times 1 \times 8 \over100}\bigg) \bigg ]+x\)=1740

Or  \({33x \over25}+{31x \over25}+{29x \over25}+{27x \over25}+x=1740\)

Or  \(x=300\)



15
A sum of Rs. 1000 is lent to be returned in 11 monthly installments of Rs. 100 each , interest being simple. The rate of interest is 
A.    \(9{1\over 11}\%\)
B.    \(10\%\)
C.    \(11\%\)
D.    \(21{9\over 11}\%\)

Answer : D.  \(21{9\over 11}\%\)

Explanation:
Rs. 1000+S.I. on Rs. 1000 for 11 months  

= Rs. 1000+S.I on Rs. 100 for (1+2+3+4+....+10) months 

Rs. 1000 S.I on Rs. 100. for \(110 \over 2\) months

= Rs. 1000+S.I.on Rs. 100 for 55 months

S.I. on Rs. 100 for 55 months = Rs. 100.

\(\dot{..}\)         Rate = \(\bigg({1000 \times 100 \times 12 \over 100 \times 55 } \bigg) \% = 21{9\over 11}\%\)
 

16
Rajan Lent Rs. 1200 to Rakesh for 3 years at a certain rate of ssimple interest and Rs. 1000 to Mukesh for the same time at same rate. If he gets Rs. 50 More from Rakessh thaan Mukesh, Then the rate percent is.
A.    \(8{1\over 3}\%\)
B.    \(6{2\over3}\%\)
C.    \(10{1\over 3} \%\)
D.    \(9{3\over2}\%\)

Answer : A.  \(8{1\over 3}\%\)

Explanation:
\({1200 \times R \times 3 \over 100 }-{1000 \times R \times 3 \over 100 }=50\)
or   \(6R=50\)
or   \(R=8{1 \over 3}\%\)