Friday, September 18, 2020

A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is:

A. RS. 698
B. RS. 700
C. RS. 523
D. RS. 600

Solution:

\[\left. \begin{array} { l } { \text { S.I. for } 1 \text { year } = \text { Rs. } ( 854 – 815 ) = \text { Rs. } 39 } \\ { \text { S.I. for } 3 \text { years } = Rs. ( 39 \times 3 ) = Rs \text { . } 117 \text { . } } \\ { \therefore \text { Principal } = Rs \text { . } ( 815 – 117 ) = Rs .698 } \end{array} \right.\]

Mr. Thomas invested an amount of Rs. 13,900 divided in two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be Rs. 3508, what was the amount invested in Scheme B?

A. RS. 9134
B. RS. 5000
C. RS. 6400
D. RS. 6000

Solution:

Let the sum invested in Scheme A be Rs. x and that in Scheme B be Rs. (13900 – x).

\[\left. \begin{array} { l } { \text { Then, } ( \frac { x \times 14 \times 2 } { 100 } ) + ( \frac { ( 13900 – x ) \times 11 \times 2 } { 100 } ) = 3508 } \\ { \Rightarrow 28 x – 22 x = 350800 – ( 13900 \times 22 ) } \\ { \Rightarrow 6 x = 45000 } \\ { \Rightarrow x = 7500 } \\ { \text { So, sum invested in Scheme } B:}\\
{= Rs . ( 13900 – 7500 ) = Rs .6400 } \end{array} \right.\]

A sum fetched a total simple interest of Rs. 4016.25 at the rate of 9 p.c.p.a. in 5 years. What is the sum?

A. RS. 7252
B. RS. 8925
C. RS. 6748
D. RS. 5621

Solution:

\[\left. \begin{array}{l}{ \text { Principal } = Rs . ( \frac { 100 \times 4016.25 } { 9 \times 5 } ) }\\{ = Rs \cdot ( \frac { 401625 } { 45 } ) }\\{ = Rs .8925 }\end{array} \right.\]

How much time will it take for an amount of Rs. 450 to yield Rs. 81 as interest at 4.5% per annum of simple interest?

A. 14 years
B. 10 years
C. 5 years
D. 4 years

Solution:

\[\text { Time } = ( \frac { 100 \times 81 } { 450 \times 4.5 } ) \text { years } = 4 \text { years }\]

Reena took a loan of Rs. 1200 with simple interest for as many years as the rate of interest. If she paid Rs. 432 as interest at the end of the loan period, what was the rate of interest?

A. 6
B. 12
C. 2
D. 1

Solution:

\[\left. \begin{array} { l } { \text { Let rate } = R \% \text { and time } = R \text { years. } } \\ { \text { Then, } ( \frac { 1200 \times R \times R } { 100 } ) = 432 } \\ { \Rightarrow 12 R ^ { 2 } = 432 } \\ { \Rightarrow R ^ { 2 } = 36 } \\ { \Rightarrow R = 6 } \end{array} \right.\]

A sum of Rs. 12,500 amounts to Rs. 15,500 in 4 years at the rate of simple interest. What is the rate of interest?

A. 10%
B. 6%
C. 2%
D. 8%

Solution:

\[\left. \begin{array} { l } { \text { S.I. } = R s . ( 15500 – 12500 ) = R s .3000 } \\ { \text { Rate } = ( \frac { 100 \times 3000 } { 12500 \times 4 } ) \% = 6 \% } \end{array} \right.\]

An automobile financier claims to be lending money at simple interest, but he includes the interest every six months for calculating the principal. If he is charging an interest of 10%, the effective rate of interest becomes:

A.30%
B. 41.3%
C. 10.25%
D. 9.31%

Solution:

\[\left. \begin{array} { l } { \text { Let the sum be Rs. } 100 . \text { Then, } } \\ { \text { S.I. for first } 6 \text { months } = Rs \text { . } ( \frac { 100 \times 10 \times 1 } { 100 \times 2 } )}\\
{= Rs .5 } \\ { \text { S.I. for last } 6 \text { months } = Rs \text { . } ( \frac { 105 \times 10 \times 1 } { 100 \times 2 } ) }\\
{= Rs .5 .25 } \\ { \text { So, amount at the end of } 1 \text { year :}}\\
{= Rs . ( 100 + 5 + 5.25 ) = Rs .110 .25 } \\ { \therefore \text { Effective rate } = ( 110.25 – 100 ) = 10.25 \% } \end{array} \right.\]

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