NIMCET 2009 Previous Year Questions (PYQs) – Page 6 of 12
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Statements:
All jewels are rings.
Some rings are necklaces.
Some cakes are jewels.
Conclusions:
I. Some necklaces are jewels.
II. Some rings are cakes.
III. No jewel is necklace.
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Solution
From “Some cakes are jewels” and “All jewels are rings” ⇒ those cakes are rings ⇒ “Some rings are cakes” is definitely true ⇒ Conclusion II follows.
From “Some rings are necklaces” and “All jewels are rings” we cannot be sure that the necklaces are jewels, so I does not follow.
“No jewel is necklace” is also not forced; it might be true or false, so III does not follow.
So, only II follows.
Statements:
All actors are writers.
Some writers are dancers.
All poets are writers.
Conclusions:
I. All actors are poets.
II. Some dancers are writers.
III. Some dancers are actors.
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Solution
From “Some writers are dancers” we can say some dancers are writers ⇒ Conclusion II follows.
Nothing links “actors” and “poets”, so I does not follow.
We are never told that any of the dancers are actors, so III does not follow.
So, only II follows.
Statements:
Some trees are branches.
All buds are branches.
All flowers are trees.
Conclusions:
I. Some branches are buds.
II. Some trees are flowers.
III. Some buds are trees.
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Solution
From “All buds are branches” and assuming buds exist, we get some branches are buds ⇒ I follows.
From “All flowers are trees” and assuming flowers exist, we get some trees are flowers ⇒ II follows.
There is no link given between buds and trees, so we cannot say “some buds are trees” ⇒ III does not follow.
So, only I and II follow.
Statements:
Some pots are eatables.
All eatables are drinks.
No banana is pot.
Conclusions:
I. Some pots are drinks.
II. All eatables are pots.
III. Some drinks are eatables.
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Solution
Check Conclusion I: “Some pots are drinks.”
We know:
Some pots are eatables.
All eatables are drinks.
So those eatables (which are pots) are definitely drinks.
Hence Some pots are drinks → TRUE
✔ Conclusion I follows.
Check Conclusion II: “All eatables are pots.”
We only know:
Some pots are eatables
But not
All eatables are pots
So this cannot be concluded.
❌ Conclusion II does NOT follow.
Check Conclusion III: “Some drinks are eatables.”
Given:
All eatables are drinks
This directly means:
Eatables ⊂ Drinks → So definitely some drinks are eatables.
✔ Conclusion III follows.
How many 5s are there in the number series each of which is immediately followed by 4 but not immediately preceded by 6?
Series: 456 656 455 455 654 456 456 5454
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Solution
We check every 5 followed by 4 and not preceded by 6.
Only the two 5s inside 5454 satisfy this condition.
You are in the land of logic where there are 3 types of rabbits.
Blue rabbits always tell the truth, green rabbits sometimes tell the truth and red rabbits never tell the truth.
A rabbit says to you: “I always lie.” What colour is the rabbit?
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Solution
A red rabbit cannot say “I always lie” because that would be true.
A blue rabbit cannot say it because that would be false.
A green rabbit can make a false statement.
How many pairs of letters are there in the word PRISON such that the number of letters between them in the word is the same as in the English alphabet?
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Solution
Valid pairs are: P–R, P–S, and R–S.
Total = 3 pairs.
If A1 = {3}, A2 = {5, 7, 9}, A3 = {11, 13, 15, 17, 19}, A4 = {21, 23, 25, 27, 29, 31, 33} and so on, what is the average of the numbers of the set A20?
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Solution
Given
$A_1={3}$,
$A_2={5,7,9}$,
$A_3={11,13,15,17,19}$,
$A_4={21,23,25,27,29,31,33}$, …
Observation:
• $A_k$ contains $2k-1$ numbers
• All elements are consecutive odd numbers
• So the average of $A_k$ is simply its middle term
First term of $A_k$:
$3,5,11,21,\dots$
This follows
$a_k = 3 + 2(k-1)^2$
Middle term of $A_k$:
$\text{Average} = a_k + (2k-2)$
So,
$\text{Average} = 3 + 2(k-1)^2 + 2(k-1)$
$\text{Average} = 3 + 2k(k-1)$
For $k=20$:
$\text{Average} = 3 + 2\times20\times19$
$\text{Average} = 3 + 760$
$\boxed{763}$
Identify the number of triangles in the given figure.
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Solution
The figure is a square divided into 16 smaller squares, with both diagonals and all midlines drawn.
Counting all small, medium, and large triangles step-by-step gives a total of 48 triangles.
Reena met Natasha’s children. Rahul said,
“The square of my age plus the cube of my sister’s age is 7148.”
Preeti said,
“The square of my age plus the cube of his age is 5274.”
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Solution
Let Rahul = R, Preeti = P
Check option 4:
R = 17, P = 19
17² + 19³ = 289 + 6859 = 7148 ✔
19² + 17³ = 361 + 4913 = 5274 ✔
This matches both statements.
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