CAT 2018 Quant Question: Sequence & Series
for every positive integer n ≥ 2. 
then k equals (TITA)
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Answer:
Hence, the answer is 24
CAT 2018 Quant Question: Mensuration
From a rectangle ABCD of area 768 sq cm, a semicircular part with diameter AB and area 72π sq cm is removed. The perimeter of the leftover portion, in cm, is
A. 88 + 12π
B. 80 + 16π
C. 86 + 8π
D. 82 + 24π
Answer:
Given that ABCD be the rectangle of area of 768 sq cm. A semicircular part with diameter AB and area 72π sq cm is removed
From the area given we can find the radius r as
The diameter AB is 24 since r = 12
So we have a rectangle ABCD
AB × BC = 768
24 × BC = 768
BC = 
BC = 32
Now we have to find the perimeter of the left over portion (shaded area)
Perimeter = AD + DC + BC + 
perimeter of the circle
Perimeter = 32 + 24 + 32 + 
(r = 12 cm)
Perimeter = 88 + 12π cm
The perimeter of the leftover portion, in cm, is 88 + 12π
Hence, the answer is 88 + 12π cm
Choice A is the correct answer.
CAT 2018 Quant Question: Number Theory
If N and x are positive integers such that 
then the largest possible x is (TITA)
CAT 2018 Quant Question: Geometry
A chord of length 5 cm subtends an angle of 60° at the centre of a circle. The length, in cm, of a chord that subtends an angle of 120° at the centre of the same circle is
A. 2π
B. 5√3
C. 6√2
D. 8
Answer:
Given that a chord of length 5 cm subtends an angle of 60° at the centre of a circle.
We have to find the length of a chord that subtends an angle of 120° at the centre of the same circle
The length of a chord that subtends an angle of 120° at the centre of the same circle is 5√3
Hence, the answer is 5√3
Choice B is the correct answer.
CAT 2018 Quant Question: Logarithm
CAT 2018 Quant Question: Permutation & Combination
In a tournament, there are 43 junior level and 51 senior level participants. Each pair of juniors play one match. Each pair of seniors play one match. There is no junior versus senior match. The number of girl versus girl matches in junior level is 153, while the number of boy versus boy matches in senior level is 276. The number of matches a boy plays against a girl is (TITA)
Answer:
Given that in a tournament there are 43 junior level and 51 senior level participants, let us assume that there are n number of girls and m number of boys
The number of girl versus girl matches in junior level is 153
n(n-1) = 306 = 18 × 17
There are 18 girls and 43 -18 = 25 boys.
m(m-1) = 552 = 276 × 2 = 138 × 4 = 69 × 8 = 23 × 24
In senior level there are 24 boys and 27 girls
The number of matches a boy plays against a girl has to be found
⟹ 25 × 18 + 24 × 27
⟹ 450 + 648
⟹ 1098
The number of matches a boy plays against a girl is 1098.
Hence, the answer is 1098
CAT 2018 Quant Question: Mixtures
A 20% ethanol solution is mixed with another ethanol solution, say, S of unknown concentration in the proportion 1:3 by volume. This mixture is then mixed with an equal volume of 20% ethanol solution. If the resultant mixture is a 31.25% ethanol solution, then the unknown concentration of S is
A. 50%
B. 55%
C. 48%
D. 52%
Answer:
Given that 20% ethanol solution is mixed with another ethanol solution, say, S of unknown concentration in the proportion 1:3 by volume
This M is then mixed with an equal volume of 20% ethanol solution. If the resultant mixture is a 31.25% ethanol solution ,then the unknown concentration of S has to be found
Equal quantities of mixture and 20% ethanol solution are mixed in equal ratio to get 31.25%
The mixture and 20% ethanol solution are mixed in equal ratio to get 31.25%. This 31.25% is 11.25% more than this 20%. This mixture must be 11.25% more than 31.25% so mixture is equal to 42.5%
This mixture 42.5% is mixed in the ratio 1:3
Using allegations we can find this 22.5% which should be in the ratio 1:3 so the other one is 7.5%
S = 42.5 + 7.5
S = 50%
Here to find the solution, do the second thing first and then go to the first thing
The second one is mixed in the ratio 1 : 1 so that the final thing should be bang in the middle
Hence, the answer is 50%
Choice A is the correct answer.
CAT 2018 Quant Question: Mensuration
The area of a rectangle and the square of its perimeter are in the ratio 1 : 25. Then the lengths of the shorter and longer sides of the rectangle are in the ratio
A. 3 : 8
B. 2 : 9
C. 1 : 4
D. 1 : 3
Answer:
Given that the area of a rectangle and the square of its perimeter are in the ratio 1: 25, we have to find the ratio of the lengths of the shorter and longer sides of the rectangle i.e. b : l
The area of the rectangle = lb
⟹ 4l[l- 4b] – b [l- 4b] = 0
⟹ [l - 4b][4l - b] = 0
By this we can find that the ratio of the shorter and longer sides of the rectangle b : l is in the ratio 1: 4
In other way ,if we want to deal with only one variable and do not want to factorize divide this
Now x can be found by simplifying it
Hence, the answer is 1 : 4
Choice C is the correct answer.
CAT 2018 Quant Question: Number Theory | Inequalities
The smallest integer n for which 
holds, is closest to
A. 33
B. 39
C. 37
D. 35
Answer:
And similarly n cannot be less than 38 so we can eliminate the options less than 38 and so the only possible value n could take is 39
Hence, the answer is 39
Choice B is the correct answer.
CAT 2018 Quant Question: Polynomials | Inequalities
The smallest integer n such that 
32n - 28 ＞ 0 is (TITA)
Answer:
For this we can assume some values for n, such that n = 10 , 9 , 8 ,… so on and can find the smallest integer
Let us now assume that n = 10
1000 – 1100 + 320 - 28 > 0
192 > 0 Hence this n = 10 works out
Similarly n = 9 , 8 also works so we can try out with n = 8 and if it works we can try for lower numbers otherwise we can try with 9
512 – 704 + 256 – 28 > 0
36 > 0
Since n = 8 works lets check for n = 7
343 – 539 + 224 – 28 = 0
Hence the smallest integer was found to be 8
The other way of thinking about it can be by substituting it with smaller numbers and checking if something can factorize it
(n – 2)(n - 2)(n - 7) > 0
This will be greater than 0 when n = 7 and when n is between 2 and 7 it will be negative
When n = 1 ,
1 - 11 + 32 - 28 = -6 so this doesn’t work
Hence the smallest integer which will work is n = 8
Hence, the answer is 8