NIMCET 2020 Previous Year Questions (PYQs) – Page 4 of 12
Study Stuff for MCA Examinations
Solution
Solution
Solution
Average Age of Anu and Dhanu
Let the ages be:
Anu = \(A\), Bhanu = \(B\), Chanu = \(C\), Dhanu = \(D\)
Given:
1) \(A + B = 10 + B + C + D \Rightarrow A = 10 + C + D\)
2) Average age of Chanu and Dhanu = 19
\(\Rightarrow \frac{C + D}{2} = 19 \Rightarrow C + D = 38\)
3) Dhanu is 10 years elder than Chanu:
\(D = C + 10\)
Substitute \(D\) in \(C + D = 38\):
\[
C + (C + 10) = 38 \Rightarrow 2C = 28 \Rightarrow C = 14
\]
\[
D = 14 + 10 = 24
\]
Now, \(A = 10 + C + D = 10 + 14 + 24 = 48\)
Average age of Anu and Dhanu:
\[
\frac{A + D}{2} = \frac{48 + 24}{2} = \frac{72}{2} = 36
\]
Answer: The average age of Anu and Dhanu is 36 years.
Solution
Solution
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.
Solution
Solution
Solution
Solution
Let the total number of students = N
We know,
- 78% students were successful → 22% students failed.
- Number of failures = 176.
So,
22% of N = 176
⇒ 22N / 100 = 176
⇒ N = (176 × 100) / 22 = 800
Total students appeared = 800
Now, 34% got first class:
First class students = 34% of 800
= (34 × 800) / 100
= 272
✅ Final Answer: 272 students got first class.
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Commented Feb 06, 2021
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