NIMCET 2027 2016 Previous Year Questions (PYQs) – Page 5 of 12
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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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Solution
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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Solution
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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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.
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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Solution
The number of common terms in the two sequences 17, 21, 25, ..........., 817 and 16, 21, 26, ..........., 851 is
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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 $n = 40$
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 $n = 40$
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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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)
- 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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Solution
- 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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Solution
- 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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The day after the day after tomorrow is four days after Monday. What day is it today?
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Solution
NIMCET 2027
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