MCA NIMCET 2016 Previous Year Question Paper and Solution with PDF

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MCA NIMCET Previous Year Questions (PYQs)

MCA NIMCET 2016 PYQ

Qus : 1 MCA NIMCET PYQ 2016

While cement is the basic raw material for producing cement tiles and cement paint which are used extensively in building construction. The main consumers of white cement are, therefore, cement tile and cement paint manufacturing units. These consumers, mostly in the small scale sector, are today facing a major crisis because of a significant increase in the price of white cement during a short period. The present actual licensed production capacity of white and grey cement in the country is approximately 3.5 lakh tonnes. The average demand is 2–2.5 lakh tonnes. This means that there is idle capacity to the tune of one lakh tonnes or more. The price rise is, therefore, not a phenomenon arising out of inadequate production capacity but evidently because of artificial scarcity created by the manufacturers in their self-interest.

The main reason for the continuing spurt in cement price is its decontrol. As it is, there is stiff competition in the cement paint and tile manufacturing business. Any further price revision at this stage is bound to have a severe adverse impact on the market conditions. The government should take adequate steps to ensure that suitable controls are brought in. Else it should allow import of cement.

What is the crisis being faced by the cement tile manufacturers as described in the passage?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 2 MCA NIMCET PYQ 2016

The main reason for the continuing spurt in cement price is its decontrol. As it is, there is stiff competition in the cement paint and tile manufacturing business. Any further price revision at this stage is bound to have a severe adverse impact on the market conditions. The government should take adequate steps to ensure that suitable controls are brought in. Else it should allow import of cement.

Which of the following words has the same meaning as the word ‘artificial’ as used in the passage?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 3 MCA NIMCET PYQ 2016

The main reason for the continuing spurt in cement price is its decontrol. As it is, there is stiff competition in the cement paint and tile manufacturing business. Any further price revision at this stage is bound to have a severe adverse impact on the market conditions. The government should take adequate steps to ensure that suitable controls are brought in. Else it should allow import of cement.

Which of the following words has the opposite meaning as the word ‘basic’ as used in the passage?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 4 MCA NIMCET PYQ 2016

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 5 MCA NIMCET PYQ 2016

Which of the following words means ‘Theatrical’?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

“Histrionic” means overly dramatic or theatrical in behavior.

Qus : 6 MCA NIMCET PYQ 2016

Identify the word which is different from the rest of the words:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Indisputable → something that cannot be questioned or doubted (certain, proven)

Uncertain → not sure, doubtful Dubious → doubtful or suspicious

Doubtful → uncertain, not convinced

? Key Idea: Options (2), (3), (4) all express doubt/uncertainty, while Indisputable expresses certainty.

Qus : 7 MCA NIMCET PYQ 2016

Choose the word that accurately signifies a person who makes money by starting or running business:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Entrepreneur → a person who starts or runs a business to make money ✅

Antrepreneur → incorrect spelling

Andrapreneur → not a valid word

Entrapranour → incorrect spelling

Only option (4) is the correct spelling and correct meaning.

Qus : 8 MCA NIMCET PYQ 2016

Which of the following is correct phrase to describe a group of insects?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

A swarm of insects → correct collective noun for insects ✅

A flock → used for birds

A school / shoal → used for fish

Hence, the correct phrase is “a swarm of insects.”

Qus : 9 MCA NIMCET PYQ 2016

Which of the following has closest meaning to the word ‘REPUTATION’?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Reputation means the way a person or thing is regarded or known by others — how widely they are talked about or perceived publicly.

Fame is the closest match, as it means being known or recognized by many people based on what others say/think about you.

Character refers to a person's inner moral qualities — not quite the same as reputation

Respect is admiration earned, not necessarily public perception Honour is a moral virtue, more of an internal quality

So Fame (3) is the best answer. ✓

Qus : 10 MCA NIMCET PYQ 2016

The meaning of the word “EGRESS” is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Egress means a way out or the act of going out/leaving. It is commonly used in architecture, safety regulations, and law to refer to an exit point.

Entrance is the opposite — a way in Double and Program have no relation to the word

So Exit (2) is correct. ✓

Qus : 11 MCA NIMCET PYQ 2016

Choose the answer which best expresses the meaning of the idiom/phrase ‘Elbow room’?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 12 MCA NIMCET PYQ 2016

Select the pair that best expresses a relationship similar to that expressed in SCALE:TONE.

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 13 MCA NIMCET PYQ 2016

Choose the answer which best expresses the meaning of the idiom/phrase “to burn a hole in the pocket”.

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 14 MCA NIMCET PYQ 2016

Choose the correct alternative to fill the blank My window look ............ the garden.

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 15 MCA NIMCET PYQ 2016

Fill in the blank with suitable article. ............ darkest cloud has a silver lining.

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 16 MCA NIMCET PYQ 2016

Fill in the blank with appropriate adjective. This steak is completely ..............., it is cold and tough.

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 17 MCA NIMCET PYQ 2016

Fill in the blank with appropriate preposition. We have been looking for a new flat ............ ages.

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 18 MCA NIMCET PYQ 2016

Fill in the blank with appropriate verb. Where is he? He should ............ home hours ago

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 19 MCA NIMCET PYQ 2016

Change the following sentence into passive sentence They studied Mathematics last year

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 20 MCA NIMCET PYQ 2016

Fill in the blank with appropriate verb

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 21 MCA NIMCET PYQ 2016

If a twelve sided regular polygon is inscribed in a circle of radius 3 centimeters, then the length of each side of the polygon is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 22 MCA NIMCET PYQ 2016

If C is the midpoint of AB and P is any point outside AB, then

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 23 MCA NIMCET PYQ 2016

The average marks of boys in class is 52 and that of

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 24 MCA NIMCET PYQ 2016

The number of 5 people groups that can be selected from 9 people when two particular persons are not to be in the same group is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 25 MCA NIMCET PYQ 2016

The solution set of equation log_x2 log_2x 2 = log_4x 2 is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 26 MCA NIMCET PYQ 2016

The equation of a circle with diameters are 2x – 3y + 12 = 0 and x + 4y – 5 = 0 and area of 154 sq. units is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 27 MCA NIMCET PYQ 2016

is equal to

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 28 MCA NIMCET PYQ 2016

, and

are vectors of magnitude α then the magnitude of the vector

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 29 MCA NIMCET PYQ 2016

A box contains 2 blue caps, 4 red caps, 5 green caps and 1 yellow cap. If four caps are picked at random, the probability that none of them is green is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 30 MCA NIMCET PYQ 2016

The line 3x + 5y = k touches the ellipse 16 x2 + 25 y2 = 400, if k is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 31 MCA NIMCET PYQ 2016

A, B, C, D, E, F and G are members of a family consisting of four adults, three children, three male and four female. Out of the children F and G are girls. A and D are brothers and A is a doctor. E is an engineer married to one of the brothers and has two children. B is married to D and G is their child. Who is C?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 32 MCA NIMCET PYQ 2016

P, Q, R and S are four logical statements such that if P is true, then Q is true; if Q is true, then R is true; and if S is true, then at least one of Q and R is false. Then it follows that

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 33 MCA NIMCET PYQ 2016

Eight friends J, K, L, M, N, O, P and Q live on eight different floors of a building but not necessarily in the same order. The lowermost floor of the building is numbered one, the one above that is numbered two and so on until the topmost floor is numbered eight.

J lives on the floor numbered six.
Only one person lives between J and L.
O lives on the floor immediately below L.
Only one person lives between O and P.
O lives above P.
K lives on an even numbered floor but not on the floor numbered two.
Two persons live between K and Q.
Q does not live on the lower most floor.
N lives on one of the floors above Q.

Which of the following is true about M?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 34 MCA NIMCET PYQ 2016

J lives on the floor numbered six.
Only one person lives between J and L.
O lives on the floor immediately below L.
Only one person lives between O and P.
O lives above P.
K lives on an even numbered floor but not on the floor numbered two.
Two persons live between K and Q.
Q does not live on the lower most floor.
N lives on one of the floors above Q.

Who amongst the following lives on the floor number eight?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 35 MCA NIMCET PYQ 2016

J lives on the floor numbered six.
Only one person lives between J and L.
O lives on the floor immediately below L.
Only one person lives between O and P.
O lives above P.
K lives on an even numbered floor but not on the floor numbered two.
Two persons live between K and Q.
Q does not live on the lower most floor.
N lives on one of the floors above Q.

Three of the following four are alike in a certain way based on the given arrangement and thus form a group. Which of the following does not belong to the group?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 36 MCA NIMCET PYQ 2016

J lives on the floor numbered six.
Only one person lives between J and L.
O lives on the floor immediately below L.
Only one person lives between O and P.
O lives above P.
K lives on an even numbered floor but not on the floor numbered two.
Two persons live between K and Q.
Q does not live on the lower most floor.
N lives on one of the floors above Q.

If P and L interchange their places, who will live between P and M?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 37 MCA NIMCET PYQ 2016

If the English word ‘EXAMINATION' is coded as 56149512965, then the word ‘GOVERNMENT' can be coded as :

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 38 MCA NIMCET PYQ 2016

A family has several children. Each boy in this family has as many sisters as brothers, but each girl has twice as many brothers as sisters. How many brothers and sisters are there?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 39 MCA NIMCET PYQ 2016

In a certain code language ‘134' means ‘good and tasty', ‘478' means ‘see good pictures' and ‘729' means ‘pictures are faint'. Which of the following numerical symbols stand for ‘see'?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 40 MCA NIMCET PYQ 2016

A team must be selected from the ten probable players A, B, C, D, E, F, G, H, I and J. Of these, A, C, E and J are forwards, B, G and H are point guards and D, F and I are defenders.

The team must have at least one forward, one point guard and one defender.
If the team includes J, it must also include F.
The team must include E or B, but not both.
If the team includes G, it must also include F.
The team must include exactly one among C, G and I.
C and F cannot be members of the same team.
D and H cannot be members of the same team.
The team must include both A and D or neither of them.
There is no restriction on the number of members in the team.

What could be the maximum size of the team that includes G?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 41 MCA NIMCET PYQ 2016

A team must be selected from the ten probable players A, B, C, D, E, F, G, H, I and J. Of these, A, C, E and J are forwards, B, G and H are point guards and D, F and I are defenders.

The team must have at least one forward, one point guard and one defender.
If the team includes J, it must also include F.
The team must include E or B, but not both.
If the team includes G, it must also include F.
The team must include exactly one among C, G and I.
C and F cannot be members of the same team.
D and H cannot be members of the same team.
The team must include both A and D or neither of them.
There is no restriction on the number of members in the team.

What would be the size of the largest possible team?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 42 MCA NIMCET PYQ 2016

A team must be selected from the ten probable players A, B, C, D, E, F, G, H, I and J. Of these, A, C, E and J are forwards, B, G and H are point guards and D, F and I are defenders.

The team must have at least one forward, one point guard and one defender.
If the team includes J, it must also include F.
The team must include E or B, but not both.
If the team includes G, it must also include F.
The team must include exactly one among C, G and I.
C and F cannot be members of the same team.
D and H cannot be members of the same team.
The team must include both A and D or neither of them.
There is no restriction on the number of members in the team.

Which of the following cannot be included in a team of size 6?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 43 MCA NIMCET PYQ 2016

A Group of 630 children are seated in n rows for a group photo session. Each row contains three less children than the row in front of it. Which one of the following number of rows is not possible?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Solution (AP method):

Total children = 630. Let rows = n, first row = a, common difference d = -3.

Sum of n terms: $S_n=\frac{n}{2}[2a+(n-1)d]=630$

$\Rightarrow 1260 = n\,[2a-3(n-1)] \;\Rightarrow\; 2a=\frac{1260}{n}+3(n-1)$

$\Rightarrow a=\frac{1}{2}\left(\frac{1260}{n}+3n-3\right)$. For a valid arrangement, a must be a positive integer and last row $a-3(n-1)>0$.

n = 3: $a=\tfrac{1}{2}(420+9-3)=\tfrac{1}{2}\cdot 426=213$ ✔️ integer (valid)
n = 4: $a=\tfrac{1}{2}(315+12-3)=\tfrac{1}{2}\cdot 324=162$ ✔️ integer (valid)
n = 5: $a=\tfrac{1}{2}(252+15-3)=\tfrac{1}{2}\cdot 264=132$ ✔️ integer (valid)
n = 6: $a=\tfrac{1}{2}(210+18-3)=\tfrac{1}{2}\cdot 225=112.5$ ❌ not integer (invalid)

✅ Final Answer: 6 rows is not possible.

Qus : 44 MCA NIMCET PYQ 2016

405 sweets were distributed equally among a group of children such that the number of sweets received by each child is one-fifth of the number of children. The number of children in the group is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 45 MCA NIMCET PYQ 2016

The number of common terms in the two sequences 17, 21, 25, ..........., 817 and 16, 21, 26, ..........., 851 is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Find the number of common terms in:
Series 1: $17, 21, 25, \ldots, 817$
Series 2: $16, 21, 26, \ldots, 851$

Step 1: Identify both APs
Series 1: $a_1 = 17,\ d_1 = 4$
Series 2: $a_2 = 16,\ d_2 = 5$

Step 2: First common term
By observation, $21$ is the first common term (appears in both series).

Step 3: Find common difference of new AP
Common terms will themselves form an AP.

Common difference $= \text{LCM}(d_1, d_2) = \text{LCM}(4, 5) = 20$

So common terms form AP: $21, 41, 61, \ldots$
with $a = 21,\ d = 20$

Step 4: Find the last common term
Last term must be $\leq$ smaller of the two last terms:
$\min(817, 851) = 817$

$T_n = a + (n-1)d \leq 817$
$21 + (n-1) \times 20 \leq 817$
$(n-1) \times 20 \leq 796$
$n - 1 \leq 39.8$
$n \leq 40.8$

So $n = 40$

Qus : 46 MCA NIMCET PYQ 2016

Fact 1 : Most stuffed toys are stuffed with beans.

Fact 2 : There are stuffed bears and stuffed tigers.

Fact 3 : Some chairs are stuffed with beans.

If the above statements are fact, which of the following statements must also be fact?

Only children's chairs are stuffed with beans.
All stuffed tigers are stuffed with beans.
Stuffed monkeys are not stuffed with beans.

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Statement 1 –

❌ Fact 3 says some chairs are stuffed with beans, but never specifies only children's chairs.

Statement 2 –

❌ Fact 1 says most (not all) stuffed toys have beans, so we can't say all stuffed tigers do.

Statement 3 –

❌ Fact 1 says most stuffed toys are stuffed with beans, so stuffed monkeys likely are — not aren't.

∴ Answer = (d)

Qus : 47 MCA NIMCET PYQ 2016

Eleven students, A, B, C, D, E, F, G, H, I, J and K are sitting in the first row of the class facing the teacher.
D who is to the immediate left of F is second to the right of C
A is second to the right of E, Who is at one of the ends.
J is the immediate neighbour of A and B and third to the left of G.
H is to the immediate left of D and third to the right of I.

Who is sitting in the middle of the row?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 48 MCA NIMCET PYQ 2016

Eleven students, A, B, C, D, E, F, G, H, I, J and K are sitting in the first row of the class facing the teacher.
D who is to the immediate left of F is second to the right of C
A is second to the right of E, Who is at one of the ends.
J is the immediate neighbour of A and B and third to the left of G.
H is to the immediate left of D and third to the right of I.

If E and D, C and B, A and H and K and F interchange their positions, which of the following pairs of students are sitting at the ends?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 49 MCA NIMCET PYQ 2016

Eleven students, A, B, C, D, E, F, G, H, I, J and K are sitting in the first row of the class facing the teacher.
D who is to the immediate left of F is second to the right of C
A is second to the right of E, Who is at one of the ends.
J is the immediate neighbour of A and B and third to the left of G.
H is to the immediate left of D and third to the right of I.

Which of the following groups of friends is sitting to the right of G?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 50 MCA NIMCET PYQ 2016

The day after the day after tomorrow is four days after Monday. What day is it today?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 51 MCA NIMCET PYQ 2016

A boy is asked to put in a basket one mango when ordered ‘One', one orange when ordered ‘Two', one apple when ordered “Three” and is asked to take out from the basket one mango and an orange when ordered “Four”. A sequence of orders is given as 12332142314223314113234.

How many total oranges were in the basket at the end of the above sequence?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 52 MCA NIMCET PYQ 2016

How many total fruits will be in the basket at the end of the above order sequence?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 53 MCA NIMCET PYQ 2016

How many pairs of letters are there in the word NECESSARY which have as many letters between them in the word as there are between them in the alphabet and in the same order

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 54 MCA NIMCET PYQ 2016

What is the missing number in the series 4, 7, 11, 18, 29, 47, __________, 123, 199?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 55 MCA NIMCET PYQ 2016

A, B, C, D, E and F are six members in a family in which there are two married couples.
D is brother of F.
Both D and F are lighter than B.
B is mother of D and lighter than E
C a lady, is neither heaviest nor lightest in the family.
E is lighter than C.
The grandfather in the family is the heaviest.

Which of the following is a pair of married couples?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 56 MCA NIMCET PYQ 2016

A, B, C, D, E and F are six members in a family in which there are two married couples.
D is brother of F.
Both D and F are lighter than B.
B is mother of D and lighter than E
C a lady, is neither heaviest nor lightest in the family.
E is lighter than C.
The grandfather in the family is the heaviest.

Who among the following will be in the second place if all the members in the family are arranged in descending order of their weights?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 57 MCA NIMCET PYQ 2016

A, B, C, D, E and F are six members in a family in which there are two married couples.
D is brother of F.
Both D and F are lighter than B.
B is mother of D and lighter than E
C a lady, is neither heaviest nor lightest in the family.
E is lighter than C.
The grandfather in the family is the heaviest.

How is C related to D?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 58 MCA NIMCET PYQ 2016

Raman was born on March 5, 1970. Lakshman was born 25 days before Raman. The year when they took birth, Republic Day fell on Monday. What is the day of birth of Lakshman?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Given:

Raman was born on March 5, 1970.
Lakshman was born 25 days before Raman.
January 26, 1970 was a Monday.

Step 1: Calculate Lakshman's date of birth.

Raman: March 5, 1970
Subtract 25 days → Lakshman's DOB = February 8, 1970

Step 2: Find the day of the week for February 8, 1970

We know January 26, 1970 = Monday
From Jan 26 to Feb 8 = 13 days
13 % 7 = 6 → Add 6 days to Monday
Monday + 6 days = Sunday → So February 1 = Sunday
Then, February 8 = Sunday + 7 = next Sunday = Sunday

✅ Final Answer: Lakshman was born on Sunday, February 8, 1970

Qus : 59 MCA NIMCET PYQ 2016

A clock is set right at 5 a.m. The clock looses 16 minutes in 24 hours. What will be the correct time when the clock indicates 10 p.m. on the 3rd day?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 60 MCA NIMCET PYQ 2016

There are six houses P, Q, R, S, T and U, three on either side of a road.
The houses are of different colours – red, blue, green, orange, yellow and white.
All the houses are of different heights.
T, the tallest house is exactly opposite to the red coloured house.
The shortest house is exactly opposite to the green coloured house.
U, the orange coloured house is located between P and S.
R, the yellow coloured house is exactly opposite to P.
Q, the green coloured house is exactly opposite to U.
P, the white coloured house is taller than R, but shorter than S and Q.

Which is the second largest house?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 61 MCA NIMCET PYQ 2016

There are six houses P, Q, R, S, T and U, three on either side of a road.
The houses are of different colours – red, blue, green, orange, yellow and white.
All the houses are of different heights.
T, the tallest house is exactly opposite to the red coloured house.
The shortest house is exactly opposite to the green coloured house.
U, the orange coloured house is located between P and S.
R, the yellow coloured house is exactly opposite to P.
Q, the green coloured house is exactly opposite to U.
P, the white coloured house is taller than R, but shorter than S and Q.

Which is the second shortest house?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 62 MCA NIMCET PYQ 2016

There are six houses P, Q, R, S, T and U, three on either side of a road.
The houses are of different colours – red, blue, green, orange, yellow and white.
All the houses are of different heights.
T, the tallest house is exactly opposite to the red coloured house.
The shortest house is exactly opposite to the green coloured house.
U, the orange coloured house is located between P and S.
R, the yellow coloured house is exactly opposite to P.
Q, the green coloured house is exactly opposite to U.
P, the white coloured house is taller than R, but shorter than S and Q.

What is the colored of the tallest house?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 63 MCA NIMCET PYQ 2016

Unscramble the letters in the following words and find the odd one.

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 64 MCA NIMCET PYQ 2016

A bus starts from its depot filled to seating capacity. It stops at a point A where 1/6th of the passengers alight and 10 board the bus. At point B, 1/5th of the passengers alight and 3 board the bus. At point C which is the last stop, all the 55 passengers alight. The capacity of the bus is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Let the number of children = n

Total sweets = 405

Each child gets = $ \frac{405}{n} $ sweets

Given: Each child’s sweets = $ \tfrac{1}{5} $ of the number of children

⇒ $ \frac{405}{n} = \frac{n}{5} $

Cross multiply:

$ 405 \times 5 = n^2 $
⇒ $ n^2 = 2025 $
⇒ $ n = \sqrt{2025} = 45 $

✅ Final Answer: The number of children = 45

Qus : 65 MCA NIMCET PYQ 2016

Kha-kha is an obscure island which is inhabited by two types of people : the “Yes” type and the “No” type. Native of type ‘Yes' ask only questions the right answer to which is ‘Yes' while those of type ‘No' ask only questions the right answer to which is ‘No'. For example the ‘Yes' type will ask questions like “Is 2 plus 2 equal to 4?” While the “No” type will ask questions like “Is 2 plus 2 equal to 5?”. The following question is based on your visit to the island Kha-Kha.

Kevin and Kumar are brothers from the island. Kumar asks you. Is at least one of us is of type ‘No' ? You can conclude that

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Logic Puzzle: Yes and No Types on Kha-Kha Island

Given:

“Yes” type always asks questions where the truthful answer is Yes.
“No” type always asks questions where the truthful answer is No.

Scenario:

Kumar asks: "Is at least one of us of type 'No'?"

Analyze:

Let’s consider possibilities:

Assume Kumar is “Yes” type:

Question must have truthful answer = “Yes”.
So, “At least one of us is ‘No’” is true.
Therefore, Kevin or Kumar is “No” type.
Since Kumar is “Yes” type, Kevin must be “No” type.

Assume Kumar is “No” type:

Question must have truthful answer = “No”.
So, “At least one of us is ‘No’” is false.
That means neither Kumar nor Kevin is “No” type.
This contradicts Kumar being “No” type.
So, this is impossible.

Conclusion:

Kumar is a “Yes” type, and Kevin is a “No” type.

Qus : 66 MCA NIMCET PYQ 2016

A group of six friends are sitting around a hexgonal table, each one at one corner of the hexagon.
Ram is sitting opposite to Ramesh.
Jyoti is sitting next to Seema.
Neeta is sitting opposite to Seema but not next to Ram.
Amrit has a person sitting between Ramesh and himself.

If Neeta sits to the right of Amrit, then who is sitting to the left of Amrit?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 67 MCA NIMCET PYQ 2016

A group of six friends are sitting around a hexgonal table, each one at one corner of the hexagon.
Ram is sitting opposite to Ramesh.
Jyoti is sitting next to Seema.
Neeta is sitting opposite to Seema but not next to Ram.
Amrit has a person sitting between Ramesh and himself.

Who is sitting between Amrit and Ramesh ?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 68 MCA NIMCET PYQ 2016

A group of six friends are sitting around a hexgonal table, each one at one corner of the hexagon.
Ram is sitting opposite to Ramesh.
Jyoti is sitting next to Seema.
Neeta is sitting opposite to Seema but not next to Ram.
Amrit has a person sitting between Ramesh and himself.

Who is sitting opposite to Jyoti?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 69 MCA NIMCET PYQ 2016

A group of six friends are sitting around a hexgonal table, each one at one corner of the hexagon.
Ram is sitting opposite to Ramesh.
Jyoti is sitting next to Seema.
Neeta is sitting opposite to Seema but not next to Ram.
Amrit has a person sitting between Ramesh and himself.

If Seema and Jyoti mutually interchange their positions, then who will be sitting opposite to Neeta?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 70 MCA NIMCET PYQ 2016

If all the 6's are replaced by 9's, then the algebraic sum of all the numbers from 1 to 100 (both inclusive)

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 71 MCA NIMCET PYQ 2016

If $X={4^n-3n-1,; n\in N}$ and $Y={9n-9,; n\in N}$, then $X\cup Y$ is equal to

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 72 MCA NIMCET PYQ 2016

$\int \left(\dfrac{\log x - 1}{1 + (\log x)^2}\right)^2 dx$ is equal to

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 73 MCA NIMCET PYQ 2016

Which of the following statements is False ?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 74 MCA NIMCET PYQ 2016

The volume of the parallelepiped determined by $u = i + 2j - k,; v = -2i + 3k$ and $w = 7j - 4k$

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 75 MCA NIMCET PYQ 2016

The vector perpendicular to the plane passing through $(1,-1,0),; (2,1,-1)$ and $(-1,1,2)$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 76 MCA NIMCET PYQ 2016

If $2x^2 + 7xy + 3y^2 + 8x + 14y + \lambda = 0$ represents a pair of straight lines, the value of $\lambda$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 77 MCA NIMCET PYQ 2016

The area of region bounded by the lines $y = |x - 1|$ and $y = 3 - |x|$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 78 MCA NIMCET PYQ 2016

In a triangle $ABC$, $a = 4$, $b = 3$, $\angle BAC = 60^\circ$, then the equation for which $c$ is the root is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 79 MCA NIMCET PYQ 2016

If $\cos\theta = \dfrac{5}{13},; \dfrac{3\pi}{2} < \theta < 2\pi$, then $\tan 2\theta$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 80 MCA NIMCET PYQ 2016

An experiment has $10$ equally likely outcomes. Let $A$ and $B$ be two non-empty events of the experiment. If $A$ consists of $4$ outcomes, the number of outcomes that $B$ must have so that $A$ and $B$ are independent is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 81 MCA NIMCET PYQ 2016

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three non-zero vectors, no two of which are collinear, if the vector $\vec{a}+2\vec{b}$ is collinear with $\vec{c}$ and $\vec{b}+3\vec{c}$ is collinear with $\vec{a}$, then $\vec{a}+2\vec{b}+6\vec{c}$ is equal to

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 82 MCA NIMCET PYQ 2016

The value of a, for which the sum of the square of the roots of the equation $x^2 - (a - 2)x - (a +1) = 0$, assumes the least value is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

The given equation is:

$x^2 - (a - 2)x - (a + 1) = 0$

Let the roots be $\alpha$ and $\beta$.

$\alpha + \beta = a - 2$

$\alpha\beta = -(a + 1)$

Let $f(a)$ represent the sum of the squares of the roots:

$f(a) = \alpha^2 + \beta^2$

$f(a) = (\alpha + \beta)^2 - 2\alpha\beta$

Substitute the expressions: $f(a) = (a - 2)^2 - 2\big(-(a + 1)\big)$

$f(a) = (a^2 - 4a + 4) + (2a + 2)$

$f(a) = a^2 - 2a + 6$

for mininum

$f'(a)=0$

$2a-2=0$

$a=1$

Qus : 83 MCA NIMCET PYQ 2016

For any two events $A$ and $B$, the probability that atleast one of them occur is $0.6$. If $A$ and $B$ occur simultaneously with a probability $0.3$ then $P(A')+P(B')$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 84 MCA NIMCET PYQ 2016

Two finite sets $A$ and $B$ are having $m$ and $n$ elements. The total number of subsets of the first set is $56$ more than the total number of subsets of the second set. The value of $m$ and $n$ are

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 85 MCA NIMCET PYQ 2016

The probability that $A$ speaks truth is $4/5$ while this probability for $B$ is $3/4$. The probability that they contradict each other when asked to speak on a fact is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 86 MCA NIMCET PYQ 2016

The 2's complement representation of the number (–100)₁₀ in an 8 bit computer is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 87 MCA NIMCET PYQ 2016

The sum of the expression $\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+\cdots+\dfrac{1}{\sqrt{80}+\sqrt{81}}$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 88 MCA NIMCET PYQ 2016

The number of terms in the product of sums canonical form of

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 89 MCA NIMCET PYQ 2016

Consider the function $f$ defined by $f(x)=\begin{cases} x^2-1, & x<3 \ 2ax, & x\geq 3 \end{cases}$ for all real number $x$. If $f$ is continuous at $x=3$, then value of $a$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 90 MCA NIMCET PYQ 2016

Find the odd man out :

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 91 MCA NIMCET PYQ 2016

Three house are available in a locality. Three persons apply for the houses. Each applies for one house without consulting other. The probability that all the three apply for the same house is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 92 MCA NIMCET PYQ 2016

Consider a hard disk with 16 recording surfaces (0-15) having 16384 cylinders (0-16383) and each cylinder contains 64 sectors(0-63). Data storage capacity in each sector is 512 bytes. Data are organized cyclinderwise and the addressing format is . A file of size 42797KB is stored in the disk and the starting disk location of the file is <1200,9,40>. What is cylinder number of the last sector of the file, if it is stored in a contiguous manner?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 93 MCA NIMCET PYQ 2016

Five horse are in a race. Mr. A selects two of the horses at random and bets on them. The probability that Mr. A selected winning horse is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 94 MCA NIMCET PYQ 2016

Consider the following min term expression for F.

F(P,Q,R,S) = ∑ (0, 2, 5, 7, 8, 10, 13, 15)

The minterms 2, 7, 8 and 13 are ‘do not care' terms.

The minimal sum of products form for F is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 95 MCA NIMCET PYQ 2016

If $3^x = 4^{x-1}$, then $x=$

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 96 MCA NIMCET PYQ 2016

Consider the equation (43)_x= (y3)₈ where x and y are unknown. The number of possible solutions is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 97 MCA NIMCET PYQ 2016

The matrix $A$ has $x$ rows and $(x+5)$ columns and the matrix $B$ has $y$ rows and $(11-y)$ columns. If both the matrices $AB$ and $BA$ exist, then value of $x$ and $y$ are

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 98 MCA NIMCET PYQ 2016

Subtract (1010)₂ from (1101)₂using first complement

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 99 MCA NIMCET PYQ 2016

A circus artist is climbing a $20$ m long rope, which is tightly streached and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is $30^\circ$

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 100 MCA NIMCET PYQ 2016

A hard disk has a rotational speed of 6000 rpm. Its average latency time is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 101 MCA NIMCET PYQ 2016

There are n equally spaced points 1,2,...,n marked on the circumference of a circle. If the point 15 is directly opposite to the point 49, then the total number of points is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Given: $n$ equally spaced points $1, 2, \ldots, n$ marked on circumference of a circle
Point $15$ is directly opposite to point $49$
Since the points are equally spaced on a circle, two points are directly opposite if they are diametrically opposite
$\therefore$ The number of points between $15$ and $49$ on each semicircle must be equal
Points between $15$ and $49$ (going from $15$ to $49$):
$16, 17, 18, \ldots, 48$
Number of points $= 48 - 16 + 1 = 33$
Since both semicircles must have equal number of points between the opposite points:
Points on other semicircle (from $49$ to $15$) must also be $33$
$\therefore$ Total number of points $= 2 + 33 + 33 = 68$

Qus : 102 MCA NIMCET PYQ 2016

The Boolean expression represented by the following Venn diagram is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 103 MCA NIMCET PYQ 2016

Let S={1,2,....,n}. The number of possible pairs of the form (A,B) with $A \subseteq B$ for subsets $A,B$ of $S$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Given: $S = \{1, 2, \ldots, n\}$
We need to find number of pairs $(A, B)$ such that $A \subseteq B \subseteq S$
Method: Element-wise analysis
For each element $i \in S$, there are three possibilities:
$i \notin A$ and $i \notin B$
$i \notin A$ and $i \in B$
$i \in A$ and $i \in B$
Note: The case $i \in A$ and $i \notin B$ is not possible since $A \subseteq B$
So each element independently has exactly $3$ choices
Since there are $n$ elements in $S$:
Total number of pairs $= \underbrace{3 \times 3 \times \cdots \times 3}_{n \text{ times}}$
$\therefore \boxed{3^n}$

Qus : 104 MCA NIMCET PYQ 2016

The range of n-bit signed magnitude representation is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

For an $n$-bit Signed Magnitude Representation:
The leftmost bit (MSB) is the sign bit:
$0 \rightarrow$ Positive
$1 \rightarrow$ Negative
Remaining $(n-1)$ bits represent the magnitude
Maximum positive value:
Sign bit $= 0$, remaining $(n-1)$ bits all $1$s
$\Rightarrow +(2^{n-1} - 1)$
Maximum negative value:
Sign bit $= 1$, remaining $(n-1)$ bits all $1$s
$\Rightarrow -(2^{n-1} - 1)$
Note: $+0$ and $-0$ both exist in signed magnitude representation
$\therefore$ Range of $n$-bit Signed Magnitude Representation is:
$\therefore \boxed{-(2^{n-1}-1) \leq x \leq +(2^{n-1}-1)}$

Qus : 105 MCA NIMCET PYQ 2016

Sum of the roots of the equation $4^x - 3(2^{x+3}) + 128 = 0$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Given: $4^x - 3(2^{x+3}) + 128 = 0$
Writing in terms of $2^x$:
$(2^x)^2 - 3 \cdot 2^x \cdot 2^3 + 128 = 0$
$(2^x)^2 - 24(2^x) + 128 = 0$
Let $t = 2^x$
$\Rightarrow t^2 - 24t + 128 = 0$
$\Rightarrow (t - 8)(t - 16) = 0$
$\Rightarrow t = 8$ or $t = 16$
When $t = 8$:
$2^x = 8 = 2^3 \Rightarrow x = 3$
When $t = 16$:
$2^x = 16 = 2^4 \Rightarrow x = 4$
Sum of roots $= 3 + 4$
$\therefore \boxed{\text{Sum of roots} = 7}$

Qus : 106 MCA NIMCET PYQ 2016

If the sum of the slopes of the lines given by $x^2 - 2cxy - 7y^2 = 0$ is four times their product, then the value of $c$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Given: $x^2 - 2cxy - 7y^2 = 0$
Comparing with $ax^2 + 2hxy + by^2 = 0$:
$a = 1,\ h = -c,\ b = -7$
By direct formula:
$m_1 + m_2 = \dfrac{-2h}{b} = \dfrac{-2(-c)}{-7} = \dfrac{-2c}{7}$
$m_1 \cdot m_2 = \dfrac{a}{b} = \dfrac{1}{-7} = \dfrac{-1}{7}$
Given condition: $m_1 + m_2 = 4(m_1 \cdot m_2)$
$\Rightarrow \dfrac{-2c}{7} = 4 \times \dfrac{-1}{7}$
$\Rightarrow -2c = -4$
$\Rightarrow c = 2$
$\therefore \boxed{c = 2}$

Qus : 107 MCA NIMCET PYQ 2016

The system of equations

x + y + 2z = a

x + z = b

2x + y + 3z = c

has a solution if

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Given system of equations:
$x + y + 2z = a$
$x + z = b$
$2x + y + 3z = c$
Writing in matrix form $[A|B]$:
$\left[\begin{array}{ccc|c} 1 & 1 & 2 & a \\ 1 & 0 & 1 & b \\ 2 & 1 & 3 & c \end{array}\right]$
Applying $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - 2R_1$:
$\left[\begin{array}{ccc|c} 1 & 1 & 2 & a \\ 0 & -1 & -1 & b-a \\ 0 & -1 & -1 & c-2a \end{array}\right]$
Applying $R_3 \rightarrow R_3 - R_2$:
$\left[\begin{array}{ccc|c} 1 & 1 & 2 & a \\ 0 & -1 & -1 & b-a \\ 0 & 0 & 0 & c-2a-(b-a) \end{array}\right]$
$\left[\begin{array}{ccc|c} 1 & 1 & 2 & a \\ 0 & -1 & -1 & b-a \\ 0 & 0 & 0 & c-a-b \end{array}\right]$
For system to have a solution:
$\rho(A) = \rho(A|B)$
$\Rightarrow c - a - b = 0$
$\therefore \boxed{a + b = c}$

Qus : 108 MCA NIMCET PYQ 2016

Let $f(x) = x^2 - bx + c$, b is an odd positive integer. If f(x)=0 has two prime numbers as roots and b + c = 35, then the global minimum value of f(x) is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Given: $f(x) = x^2 - bx + c$
where $b$ is an odd positive integer and $b + c = 35$
Let $\alpha, \beta$ be two prime roots of $f(x) = 0$
By Vieta's formulae:
$\alpha + \beta = b$
$\alpha \beta = c$
Now, $b + c = 35$
$\Rightarrow (\alpha + \beta) + \alpha\beta = 35$
Since $b$ is odd and $\alpha + \beta = b$ is odd
$\Rightarrow$ one of $\alpha, \beta$ must be even prime
$\therefore \alpha = 2$ (only even prime)
$\Rightarrow 2 + \beta + 2\beta = 35$
$\Rightarrow 3\beta = 33$
$\Rightarrow \beta = 11$
$\therefore b = \alpha + \beta = 2 + 11 = 13$
$\therefore c = \alpha\beta = 2 \times 11 = 22$
Verification: $b + c = 13 + 22 = 35$ ✓
So $f(x) = x^2 - 13x + 22$
Global minimum occurs at $x = \dfrac{b}{2} = \dfrac{13}{2}$
Minimum value $= f\left(\dfrac{13}{2}\right) = \left(\dfrac{13}{2}\right)^2 - 13\left(\dfrac{13}{2}\right) + 22$
$= \dfrac{169}{4} - \dfrac{169}{2} + 22$
$= \dfrac{169 - 338 + 88}{4}$
$= \dfrac{-81}{4}$
$\therefore \boxed{\text{Global Minimum} = -\dfrac{81}{4}}$

Qus : 109 MCA NIMCET PYQ 2016

The vertex of the parabola whose focus is (-1,1) and directrix is 4x + 3y - 24 = 0 is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Given: Focus $S(-1, 1)$ and directrix $4x + 3y - 24 = 0$
The vertex is the midpoint of focus and foot of perpendicular from focus to directrix
Foot of perpendicular from $S(-1, 1)$ to $4x + 3y - 24 = 0$:
$\dfrac{x + 1}{4} = \dfrac{y - 1}{3} = -\dfrac{4(-1) + 3(1) - 24}{4^2 + 3^2}$
$= -\dfrac{-4 + 3 - 24}{16 + 9}$
$= -\dfrac{-25}{25}$
$= 1$
$\therefore x + 1 = 4 \Rightarrow x = 3$
$\therefore y - 1 = 3 \Rightarrow y = 4$
$\therefore$ Foot of perpendicular $Z = (3, 4)$
Since vertex $V$ is midpoint of $SZ$:
$V = \left(\dfrac{-1 + 3}{2},\ \dfrac{1 + 4}{2}\right)$
$V = \left(\dfrac{2}{2},\ \dfrac{5}{2}\right)$
$\therefore \boxed{V = \left(1,\ \dfrac{5}{2}\right)}$

Qus : 110 MCA NIMCET PYQ 2016

The number of points in $(-\infty,\infty)$, for which $x^2 - x\sin x - \cos x = 0$ is

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Solution

Let $f(x) = x^2 - x\sin x - \cos x$
Differentiating with respect to $x$:
$f'(x) = 2x - \sin x - x\cos x + \sin x$
$f'(x) = 2x - x\cos x$
$f'(x) = x(2 - \cos x)$
Since $-1 \leq \cos x \leq 1$ $\Rightarrow$ $2 - \cos x \geq 1 > 0$ always
$\therefore$ Sign of $f'(x)$ depends on sign of $x$
$f'(x) < 0$ for $x < 0$ and $f'(x) > 0$ for $x > 0$
$\therefore f(x)$ has minimum at $x = 0$
Minimum value $= f(0) = 0 - 0 - \cos 0 = -1 < 0$
Also, $f(-x) = (-x)^2 - (-x)\sin(-x) - \cos(-x)$
$= x^2 - x\sin x - \cos x = f(x)$
$\therefore f(x)$ is an even function
As $x \rightarrow \pm\infty$, $f(x) \rightarrow +\infty$
Since $f(0) = -1 < 0$ is minimum and $f(x) \rightarrow +\infty$ on both sides
$\therefore$ Graph of $f(x)$ crosses $x$-axis at exactly two points
One in $(-\infty, 0)$ and one in $(0, \infty)$
$\therefore$ Total number of points $= \boxed{2}$

Qus : 111 MCA NIMCET PYQ 2016

There are $4$ books on fairy tales, $5$ novels and $3$ plays. In how many ways can they be arranged in the order books on fairy tales, novels and then plays so that books of same category are put together

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Given:
$4$ books on fairy tales, $5$ novels, $3$ plays
The order of arrangement is fixed as:
Fairy Tales $\rightarrow$ Novels $\rightarrow$ Plays
Step 1: Arrange books within each category
Fairy tales books can be arranged among themselves $= 4!= 24$ ways
Novels can be arranged among themselves $= 5! = 120$ ways
Plays can be arranged among themselves $= 3! = 6$ ways
Step 2: Total number of arrangements
$= 4! \times 5! \times 3!$
$= 24 \times 120 \times 6$
$\therefore \boxed{= 17280 \text{ ways}}$

Qus : 112 MCA NIMCET PYQ 2016

Suppose a population $A$ has $100$ observations $101,102,\ldots,200$ and another population $B$ has $100$ observations $151,152,\ldots,250$. If $V_A$ and $V_B$ represent variance of the two populations respectively, then $\dfrac{V_A}{V_B}$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Population $A$: $101, 102, \ldots, 200$
Population $B$: $151, 152, \ldots, 250$
Using Property:
Variance is independent of change of origin
i.e., if each observation is increased or decreased by a constant, variance remains unchanged
Here, each observation of $B$ = each observation of $A$ $+ 50$
$\Rightarrow$ Population $B$ is obtained by adding $50$ to each observation of population $A$
$\Rightarrow V_B = V_A$
$\therefore \dfrac{V_A}{V_B} = \boxed{1}$

Qus : 113 MCA NIMCET PYQ 2016

If $\vec{a}, \vec{b}$ are vector such that $|\vec{a}+\vec{b}|=\sqrt{29}$ and $\vec{a}\times (2\hat{i}+3\hat{j}+4\hat{k}) = (2\hat{i}+3\hat{j}+4\hat{k})\times \vec{b}$, then a possible value of $(\vec{a}+\vec{b})\cdot(-7\hat{i}+2\hat{j}+3\hat{k})$ is

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Solution

Qus : 114 MCA NIMCET PYQ 2016

Let $x_1,x_2,\ldots,x_n$ be $n$ observations such that $\sum x_i^2 = 400$ and $\sum x_i = 80$. Then a possible value of $n$ amongst the following is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Given: $\sum x_i^2 = 400$ and $\sum x_i = 80$
Using the inequality:
$\sum x_i^2 \geq \dfrac{(\sum x_i)^2}{n}$
This follows from Cauchy-Schwarz inequality:
$n \cdot \sum x_i^2 \geq \left(\sum x_i\right)^2$
$\Rightarrow n \times 400 \geq (80)^2$
$\Rightarrow 400n \geq 6400$
$\Rightarrow n \geq \dfrac{6400}{400}$
$\Rightarrow n \geq 16$
Also, Variance $\geq 0$:
$\dfrac{\sum x_i^2}{n} - \left(\dfrac{\sum x_i}{n}\right)^2 \geq 0$
$\Rightarrow \dfrac{400}{n} - \dfrac{6400}{n^2} \geq 0$
$\Rightarrow \dfrac{400n - 6400}{n^2} \geq 0$
$\Rightarrow n \geq 16$
$\therefore$ A possible value of $n$ is $\boxed{n = 20}$

Qus : 115 MCA NIMCET PYQ 2016

Area of the greatest rectangle that can be inscribed in the ellipse is

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Solution

Given ellipse: $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$
**Setting up the Rectangle:**
Let $P(a\cos\theta,\ b\sin\theta)$ be a point on the ellipse
Then the rectangle has:
Length $= 2a\cos\theta$
Breadth $= 2b\sin\theta$
**Area of Rectangle:**
$A = 2a\cos\theta \times 2b\sin\theta$
$A = 4ab\sin\theta\cos\theta$
$A = 2ab\sin 2\theta$
**Maximizing the Area:**
$A$ is maximum when $\sin 2\theta$ is maximum
$\Rightarrow \sin 2\theta = 1$
$\Rightarrow 2\theta = 90^\circ$
$\Rightarrow \theta = 45^\circ$
**Maximum Area:**
$A_{max} = 2ab \times 1$
$\therefore \boxed{A_{max} = 2ab}$

Qus : 116 MCA NIMCET PYQ 2016

. Two common tangents to the circles $x^2 + y^2 = 2a^2$ and parabola $y^2 = 8ax$ are

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Given circle: $x^2 + y^2 = 2a^2$
Given parabola: $y^2 = 8ax$
**Equation of tangent to parabola:**
For parabola $y^2 = 8ax$, tangent is:
$y = mx + \dfrac{2a}{m}$ ...(i)
**Condition for tangent to circle:**
For line $y = mx + \dfrac{2a}{m}$ to be tangent to circle $x^2 + y^2 = 2a^2$,
distance from centre $(0,0)$ = radius $= \sqrt{2}\ a$
$\Rightarrow \dfrac{\left|\dfrac{2a}{m}\right|}{\sqrt{1 + m^2}} = \sqrt{2}\ a$
$\Rightarrow \dfrac{2a}{|m|\sqrt{1+m^2}} = \sqrt{2}\ a$
$\Rightarrow \dfrac{2}{|m|\sqrt{1+m^2}} = \sqrt{2}$
$\Rightarrow |m|\sqrt{1+m^2} = \sqrt{2}$
Squaring both sides:
$\Rightarrow m^2(1+m^2) = 2$
$\Rightarrow m^4 + m^2 - 2 = 0$
$\Rightarrow (m^2 + 2)(m^2 - 1) = 0$
$\Rightarrow m^2 = 1 \quad (\because m^2 = -2 \text{ is not possible})$
$\Rightarrow m = \pm 1$
**Substituting in (i):**
When $m = 1$: $y = x + 2a$
When $m = -1$: $y = -x - 2a$
$\therefore$ The two common tangents are:
$\therefore \boxed{y = x + 2a \quad \text{and} \quad y = -(x + 2a)}$

Qus : 117 MCA NIMCET PYQ 2016

If $a_1,a_2,\ldots,a_n$ are in A.P. and $a_1=0$ then the value of $\left(\dfrac{a_3}{a_2}+\dfrac{a_4}{a_3}+\cdots+\dfrac{a_n}{a_{n-1}}\right)-a_2\left(\dfrac{1}{a_2}+\dfrac{1}{a_3}+\cdots+\dfrac{1}{a_{n-2}}\right)$ is equal to

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Solution

Qus : 118 MCA NIMCET PYQ 2016

The value of $\cos 20^\circ + \cos 100^\circ + \cos 140^\circ$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

We need to find: $\cos 20^\circ + \cos 100^\circ + \cos 140^\circ$
Using sum-to-product formula on first two terms:
$\cos 20^\circ + \cos 100^\circ = 2\cos\left(\dfrac{20^\circ + 100^\circ}{2}\right)\cos\left(\dfrac{100^\circ - 20^\circ}{2}\right)$
$= 2\cos 60^\circ \cos 40^\circ$
$= 2 \times \dfrac{1}{2} \times \cos 40^\circ$
$= \cos 40^\circ$
So the expression becomes:
$\cos 40^\circ + \cos 140^\circ$
Again applying sum-to-product formula:
$= 2\cos\left(\dfrac{40^\circ + 140^\circ}{2}\right)\cos\left(\dfrac{140^\circ - 40^\circ}{2}\right)$
$= 2\cos 90^\circ \cos 50^\circ$
$= 2 \times 0 \times \cos 50^\circ$
$= 0$
$\therefore \boxed{\cos 20^\circ + \cos 100^\circ + \cos 140^\circ = 0}$

Qus : 119 MCA NIMCET PYQ 2016

The permutations of ${a,b,c,d,e,f,g}$ are listed in lexicographic order. Which of the following permutations are just before and just after the permutation $bacdefg$ ?

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Solution

Qus : 120 MCA NIMCET PYQ 2016

The foci of the ellipse $\dfrac{x^2}{16} + \dfrac{y^2}{b^2} = 1$ and the hyperbola $\dfrac{x^2}{144} - \dfrac{y^2}{81} = \dfrac{1}{25}$ coincide. Then the value of $b^2$ is

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Solution

Given ellipse: $\dfrac{x^2}{16} + \dfrac{y^2}{b^2} = 1$ ...(i)
Given hyperbola: $\dfrac{x^2}{144} - \dfrac{y^2}{81} = \dfrac{1}{25}$ ...(ii)
**For Hyperbola:**
Rewriting (ii) in standard form:
$\dfrac{x^2}{\dfrac{144}{25}} - \dfrac{y^2}{\dfrac{81}{25}} = 1$
Here, $a^2 = \dfrac{144}{25}$ and $b^2 = \dfrac{81}{25}$
For hyperbola, $c^2 = a^2 + b^2$
$\Rightarrow c^2 = \dfrac{144}{25} + \dfrac{81}{25} = \dfrac{225}{25} = 9$
$\therefore c = 3$
So, foci of hyperbola $= (\pm 3,\ 0)$
**For Ellipse:**
Here, $a^2 = 16$
For ellipse, $c^2 = a^2 - b^2$
$\Rightarrow c^2 = 16 - b^2$
Since foci of ellipse and hyperbola coincide:
$\Rightarrow 16 - b^2 = 9$
$\Rightarrow b^2 = 16 - 9$
$\therefore \boxed{b^2 = 7}$

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✅ Final Answer: 6 rows is not possible.

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Given:

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Let the number of children = n

✅ Final Answer: The number of children = 45

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Logic Puzzle: Yes and No Types on Kha-Kha Island

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