Wednesday, October 21, 2020

Section 27 of the NHSO deals with ___________________.

A. Fitness certificate for transport vehicle
B. Changing the design and body of vehicle
C. Changing the ownership of vehicle
D. none of these

Show Answer
C. Changing the ownership of vehicle

234(333)257 345( )678

A. 233

B. 333

C. 133

D. None of these

Shared by: Fahad Hafeez

Show Answer
A. 233

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:

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

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