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MCA NIMCET Previous Year Questions (PYQs)
MCA NIMCET Probability Critical Thinking PYQ
MCA NIMCET PYQ
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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ
In a class of 100 students, 55 passed in Mathematics and 67 passed in Physics.The number of students who passed in Physics only is:
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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ
Solution
Use the formula:$\text{Physics only} = n(P) - n(M \cap P)$
First find $n(M \cap P$) using:
$n(M) + n(P) - n(M \cap P) = 100$
So:
$55 + 67 - n(M \cap P) = 100$
$122 - n(M \cap P) = 100$
$n(M \cap P) = 22$
Now:
$\text{Physics only} = 67 - 22 = 45$
MCA NIMCET PYQ
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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2010 PYQ
Survey: $63%$ like cheese, $76%$ like apples. If $x%$ like both?
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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2010 PYQ
Solution
Using inclusion–exclusion:$63 + 76 - x \le 100$
$139 - x \le 100$
$x \ge 39$
Max value of $x$ cannot exceed minimum of both percentages
→ $x \le 63$
Thus $39 \le x \le 63$.
MCA NIMCET PYQ
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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2008 PYQ
A six-faced die is a biased one. It is thrice more likely to show an odd number than to show an even number. It is thrown twice. The probability that the sum of the numbers in the two throws is even is
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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2008 PYQ
Solution
Let $P(\text{even})=p$ and $P(\text{odd})=3p$ $p+3p=1 \Rightarrow p=\dfrac14$ So $P(\text{even})=\dfrac14,\quad P(\text{odd})=\dfrac34$ Sum is even when both outcomes are even or both are odd. $P=\left(\dfrac14\right)^2+\left(\dfrac34\right)^2=\dfrac1{16}+\dfrac9{16}=\dfrac{10}{16}=\dfrac58$ Answer: $\boxed{\dfrac58}$
MCA NIMCET PYQ
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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2008 PYQ
A letter is known to have come from either TATANAGAR or CALCUTTA. On the envelope, just two consecutive letters, TA, are visible. The probability that the letter has come from CALCUTTA is
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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2008 PYQ
Solution
TATANAGAR has $2$ occurrences of TA CALCUTTA has $3$ occurrences of TA Total occurrences $=5$ Required probability $=\dfrac{3}{5}$ This is not listed. Answer: $\boxed{\text{None of these}}$MCA NIMCET
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