NIMCET Quantitative Aptitude Previous Year Questions (PYQs) – Page 1 of 4
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A sum of money was divided into 2 parts in the ratio 2:5. First part was invested for 2
years at the annual interest rate of 20% compounded annually. At what rate of simple
interest per annum the second part must be invested for 2 years, so that the interest
earned in both cases is the same?
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Solution
Ratio = $2:5 \;\Rightarrow\; \tfrac{2S}{7}, \tfrac{5S}{7}$
CI on $\tfrac{2S}{7}$ for 2 yrs @ $20\% =$ $0.44 \times \tfrac{2S}{7} = \tfrac{0.88S}{7}$
SI on $\tfrac{5S}{7}$ for 2 yrs @ $R\% =$ $\dfrac{5S}{7} \times \dfrac{R \times 2}{100} = \dfrac{RS}{70}$
Equating ⇒ $\tfrac{0.88S}{7} = \dfrac{RS}{70}$
$\therefore \; R = 8.8\% \; \text{p.a.}$
CI on $\tfrac{2S}{7}$ for 2 yrs @ $20\% =$ $0.44 \times \tfrac{2S}{7} = \tfrac{0.88S}{7}$
SI on $\tfrac{5S}{7}$ for 2 yrs @ $R\% =$ $\dfrac{5S}{7} \times \dfrac{R \times 2}{100} = \dfrac{RS}{70}$
Equating ⇒ $\tfrac{0.88S}{7} = \dfrac{RS}{70}$
$\therefore \; R = 8.8\% \; \text{p.a.}$
A number which, when increased by 15% becomes 161. What is the original number?
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Solution
Let the original number = x
Increased by 15% ⇒ New value = 1.15x
Given: 1.15x = 161
x = 161 ÷ 1.15
x = 140
✅ Final Answer: The original number is 140.
A set of consecutive positive integers beginning with
1 is written on the blackboard. A student came along
and erased one number. The average of the remaining
numbers is $35\frac{7}{15}$. What was the number erased?
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Solution
In a city, 40.1% of the adults are illiterate while
85.1% of the children are literate. If the ratio of the
adults to that of the children is 2:3, then what percent
of the population is literate?
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Solution
Let total population be $100$ (as given).
Ratio of adults : children $= 2 : 3$
So, total parts $= 2 + 3 = 5$
Adults $= \dfrac{2}{5} \times 100 = 40$
Children $= \dfrac{3}{5} \times 100 = 60$
Now use literacy data:
For adults:
Illiterate adults $= 40.1%$
So, literate adults $= 100% - 40.1% = 59.9%$
Literate adults $= 59.9%$ of $40$
$= \dfrac{59.9}{100} \times 40 = 23.96$
For children:
Literate children $= 85.1%$ of $60$
$= \dfrac{85.1}{100} \times 60 = 51.06$
Total literate population
$= 23.96 + 51.06 = 75.02$
Since total population is $100$,
Literate percentage $= 75.02% \approx 75%$.
A runs $1 \frac{2}{3}$ times as fast as B. If A gives B a start of
80m, how far must the winning post be so that A and
B might reach it at the same time?
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Solution
Mr. Kumar drives to work at an average speed of 48
Km/hr. The time taken to cover the first 60% of the
distance is 10 minutes more than the time taken to
cover the remaining distance. How far is his office?
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Solution
Two pipes A and B can fill the cistern in 37.5
minutes and 45 minutes respectively. Both pipes are
opened. The cistern will be filled in just half an hour,
if the B is turned off after
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Solution
In a shower, 5 cm of rainfalls. The volume of water
that falls on 1.5 hectares of ground is:
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Solution
In a competitive examination in Maharashtra state
9% candidates got selected from the total appeared
candidates. Tripura state had an equal number of
candidates appeared and 12% candidates got selected
with 102 more candidates got selected than
Maharashtra state. What was the number of
candidates appeared from each state?
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Solution
Let the number of candidates appeared from each state be $x$.
Maharashtra selected candidates
$= 9\%$ of $x = \dfrac{9}{100}x$
Tripura selected candidates
$= 12\%$ of $x = \dfrac{12}{100}x$
Given that Tripura has $102$ more selected candidates than Maharashtra:
\[
\dfrac{12}{100}x - \dfrac{9}{100}x = 102
\]
\[
\dfrac{3}{100}x = 102
\]
\[
x = 102 \times \dfrac{100}{3} = 3400
\]
Thus, the number of candidates appeared from each state is
3400.
Shiva gave 40% of his monthly salary to his mother
from the remaining he used 7% for electronic gadgets
and 23% he kept aside for his monthly expenses. The
remaining amount he transferred to his friend’s
account. The sum of the amount he gave to his
mother and he transferred to his friend account was
41000. What was Shiva’s monthly salary?
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Solution
Let Shiva's monthly salary be $x$.
He gave $40%$ of his salary to his mother:
Amount to mother $= 40%$ of $x = 0.40x$.
Remaining salary after giving to mother:
$= x - 0.40x = 0.60x$.
From this remaining $0.60x$:
He used $7%$ for electronic gadgets:
Amount for gadgets $= 7%$ of $0.60x = 0.07 \times 0.60x = 0.042x$.
He kept $23%$ for monthly expenses:
Amount for expenses $= 23%$ of $0.60x = 0.23 \times 0.60x = 0.138x$.
Total used from the remaining on gadgets and expenses:
$0.042x + 0.138x = 0.18x$.
So the amount transferred to his friend's account:
$= 0.60x - 0.18x = 0.42x$.
Now, sum of the amount given to his mother and transferred to his friend is given as $41000$:
Amount to mother $+ $ amount to friend $= 0.40x + 0.42x = 0.82x$.
So,
$0.82x = 41000$
$x = \dfrac{41000}{0.82} = 50000$.
Therefore, Shiva's monthly salary is 50000.
He gave $40%$ of his salary to his mother:
Amount to mother $= 40%$ of $x = 0.40x$.
Remaining salary after giving to mother:
$= x - 0.40x = 0.60x$.
From this remaining $0.60x$:
He used $7%$ for electronic gadgets:
Amount for gadgets $= 7%$ of $0.60x = 0.07 \times 0.60x = 0.042x$.
He kept $23%$ for monthly expenses:
Amount for expenses $= 23%$ of $0.60x = 0.23 \times 0.60x = 0.138x$.
Total used from the remaining on gadgets and expenses:
$0.042x + 0.138x = 0.18x$.
So the amount transferred to his friend's account:
$= 0.60x - 0.18x = 0.42x$.
Now, sum of the amount given to his mother and transferred to his friend is given as $41000$:
Amount to mother $+ $ amount to friend $= 0.40x + 0.42x = 0.82x$.
So,
$0.82x = 41000$
$x = \dfrac{41000}{0.82} = 50000$.
Therefore, Shiva's monthly salary is 50000.
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