IPMAT 2022 - Quantitative Aptitude

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Previous Year Questions with Detailed Solutions
Practice IPMAT 2022 - Quantitative Aptitude previous year questions with step-by-step solutions from the IIM IPMAT exam. Every question includes the correct answer and a full worked explanation — no need to leave the page.

72

Aruna purchases a certain number of apples for INR 20 each and a certain number of mangoes for INR 25 each. If she sells all the apples at 1% profit and all the mangoes at 20% loss, overall she makes neither profit nor loss. Instead, if she sell all the apples at 20% loss and all the mangoes at 10% profit, overall she makes a loss of INR 150. Then the number of apples purchased by Aruna is_________.
50


Let P(X) denote power set of a set X. If A is the null set, then the number of elements in P(P(P(P(A)))) is_______.
16

The number – 16, 2x+3 – 22x-1 – 16, 22x-1 + 16 are in an arithmetic progression. Then x equals________.
3

When Geeta increases her speed from 12 km/hr to 20 km/hr. she takes one hour less than the usual time to cover the distance between her home and office. The distance between her home and office is______ km.
30

Given that f(x) = |x| = 2|x – 1| + |x – 2| + |x – 4| + |x – 6| + 2|x – 10|, x ∈ (– ∞, ∞) the minimum value of f(x) is_______.
26
f(x)= |x| + 2|x – 1| + |x – 2| + |x – 4| + |x – 6| +2|x – 10| x ∈ (– ∞, ∞)
Case 1 ? x ≤ 0
f(x)= –x – 2(x – 1) – (x – 2) – (x – 4) – (x – 6) – 2 (x – 10)
= – x – 2x + 2 – x + 2 – x + 4 – x + 6 – 2x + 20
= – 8x + 34
Minimum value =34
Case 2 ? 0 < x ≤ 1
f(x) = x – 2(x – 1) – (x – 2) – (x – 4) – (x – 6) – 2(x – 10)
= x – 2x + 2 – x + 2 – x + 4 – x + 6 – 2x + 20
= – 6x + 34
= 28
Case 3 ? 1 < x ≤ 2
f(x) = x – 2(x – 1) – (x – 2) – (x – 4) – (x – 6) – 2(x – 10)
= x + 2x – 2 – x + 2 – x + 4 – x + 6 – 2x + 20
= – 2x + 30
Minimum value at x = 2
= – 2(2) + 30
= 26
Case 4 ? 2 < x ≤ 4
f(x) = x + 2(x – 1) – (x – 2) – (x – 4) – (x – 6) – 2(x – 10)
= x + 2x – 2 – x – 2 – x + 4 – x + 6 – 2x + 20
= 26
Minimum value at f(x) is 26
Case 4 ? 4 < x ≤ 6
f(x) = x + 2(x – 1) + (x – 2) + (x – 4) – (x – 6) – 2(x – 10)
= x + 2x – 2 + x – 2 + x – 4 – x + 6 – 2x + 20
= 2x + 18
Minimum value at x = 4
= 26
Case 6 ? 6 < x ≤ 10
f(x) = x + 2(x – 1) + (x – 2) + x – 4 + x – 6 – 2x + 20
= x + 2x – 2 + x – 2 + x – 4 + x – 6 – 2x + 20
= 4x + 6
Minimum value at x = 6
= 24 + 6
= 30
Case 7 ? x > 10
= x + 2x – 2 + x – 2 + x – 4 + x – 6 + 2x – 20
= 8x – 34
Minimum value at x = 10
= 80 – 34
= 46
So the minimum value of f(x) in all cases combined = 26
The sum of the coefficients of all the terms in the expansion of (5x – 9)4 is _________.
256

The number of triangles that can be formed by choosing points from 7 points on a line and 5 points on another parallel line is__________.
175
A triangle can be formed by choosing 2 points from 1 line & one point from another line
Case 1 → choosing 2 points from line with 7 points
7c2 x 5c1
= 105 triangles
Case 1 → choosing 2 points from line with 5 points
= 5c2 × 7c1
= 10 × 7
= 105 triangles
Total triangles = 105 + 70 = 175 triangles
The area enclosed by 2|x| + 3|y| ≤ 6 is ________ sq. units.
12

The above big is graph of 2|x| + 3|y| = 6
Then Are enclosed by 2|x| + 3|y| ≤ 6
= Area enclosed by parallelogram ABCD
A new sequence is obtained from the sequence of positive integers (1, 2, 3…) by deleting all the perfect squares. Then the 2022nd term of the new sequence is________.
2067
S = (1, 2, 3, 4, 5, 6, 7 ……..)
If no number is removed then 2022nd number will be 2022 but all the squares are removed so check how many numbers are removed then a2 < 2022
So 44 no.’s are removed 442 = 1836
So 2022nd number will be 452 = 2025
2022
462 = 2116
44
--------
2066
But 452 will also be removed as 2025 < 2066 so we have to go to next number to get 2022nd number.
2022nd number = 2067

100

If
then the absolute value of the determinant of (A9 + A6 + A3 + A) is __________.
32

Let 50 distinct positive integers be chosen such that the highest among them is 100, and the average of the largest 25 integers among them exceeds the average of the remaining integers by 50. Then the maximum possible value of the sum of all the 50 integers is_________.
3150

The 3rd, 14th and 69th terms of an arithmetic progression form three distinct and consecutive terms of a geometric progression. If the next term of the geometric progression is the nth terms of the arithmetic progression, then n equals _____________.
344


Mrs. and Mr. Sharma, and Mrs. and Mr. Ahuja along with four other persons are to be seated at a round table for dinner. If Mrs. and Mr. Sharma are to be seated next to each other, and Mrs. and Mr. Ahuja are not to be seated next to each other, then the total number of seating arrangements is___________.
960
Total Arrangements where Mr & Mrs Sharma together & Mr & Mrs Ahuja not together = Total Cases where Mr & Mrs Sharma together - Arrangements where Mr & Mrs Sharma sit together & Mr & Mrs Ahuja sit together
For Mr & Mrs Sharma sitting together consider them one unit and arrange them within themselves later
Seating Arrangement in a circle = (n - 1)!
So there are 7 people => 6 people & 7th one Mr & Mrs Sharma
Total cases where Mr & Mrs Sharma
Together = 6! × 2!
Arrangements where Mr & Mrs Sharma
Together & Mr. & Mrs Ahuja Together = 5! x 2! x 2!
Cases where Mr & Mrs Sharma
Together & Mr & Mrs Ahuja no together = 6! x 2! (5! x 2! x 2!)
= 5! × 2! (6 – 2)
= 240 x 4
= 960
Alternate Method ?
Arrange Mr & Mrs Sharma along with 4 persons excluding Mr & Mrs Ahuja from the group.
= 4! X 2!
(Formula ? Arrangement in circular table = (∩ - 1)!
The spaces created b/w arrangements is 4 + 1 = 5
Select any 2 places & arrange Mr & Mrs Ahuja so they will never be together
= 4! X 2! X 5P2 = 960
When the square of the difference of two natural numbers is subtracted from the square of the sum of the same two numbers and the result is divided by four, we get
the product of the LCM and HCF of the two numbers

In a 400-metre race, Ashok beats Bipin and Chandan respectively by 15 seconds and 25 seconds. If Ashok beats Bipin by 150 metres, by how many metres does Bipin beat Chandan in the race?
80


The set of real values of x for which the inequality
holds is
[2, 81)

If f[x2 + f(y)] = xf(x) + y for all non – negative integers x and y, then the value of [f(0)]2 + f(0) equals ________.
0

In how many ways can the letters of the word MANAGEMENT be arranged such that no two vowels appear together ?
37800

If the five digit number abcde is divisible by 6, then which of the following number is not necessarily divisible by 6 ?
edcba
Given
abcde = 6x
Then e = 2n
i.e. e is divisible by 2
& a + b + c + d + e = 3m i.e. sum is divisible by 3
Then checking all options whether they are divisible by 6
(i) eee
Sum = 3e so number is divisible by 3
& e = 2n and e is unit digit
So eee is divisible by 6
(ii) e + d + c + b + a 3 m then no. is divisible by 3 but we cannot comment whether a is even or odd
So edcba is not necessarily divisible by 6.
Similarly checking other 2 options so they are divisible
Hence, answer is – edcba
The curve represented by the equation 
an ellipse with the foci on the y-axis





In a bowl containing 60 ml orange juice, 40 ml of water is poured. Thereafter, 100 ml of apple juice is poured to make a fruit punch. Madhu drinks 50 ml of this fruit punch and comments that the proportion of orange juice needs to be higher for better taste. How much orange juice should be poured into the fruit punch that remained, in order to bring up the level of orange juice to 50 percentage ?
60 ml

Ayesha is standing stop a vertical tower 200m high and observes a car moving away from the lower on a straight, horizontal road from the foot of the tower. At 11 : 00 AM, she observes the angle of depression of the car to be 45°. At 11 : 02 AM, she observes the angle of depression of the car to be 30°. The speed at which the car is moving is approximately.
4.39 km per hour

The set of all possible values of f(x) for which (81)x + (81)f(x) = 3 is
(−∞, 0.25)
(81)x + (81)f(x) = 3
ax > 0 for any value of x
So (81)f(x) can take value (0, 3) based on which x will adjust itself to satisfy above relation & it can’t be equal to 3 as (81)f(x)< 3
0 < (81)f(x) < 3
3−∞ < 34f(x) < 31
− ∞ < 4f(x) < 1
− ∞ < f(x) < 1/4
f(x) ∈ (− ∞, 0.25)
The number of four-digit integers which are greater than 1000 and divisible by both 2 and 3, but not by 5, is
1200
Four digit numbers divisible by both 2 & 3 would be numbers divisible by 6
Total numbers divisible by 6 & > 1000 = 1002, 1008, 1014 ….. 9996
an = a + (n – 1)d
9996 = 1002 + (n – 1) 6
8994/6 = (n – 1)
n = 1500
Number divisible 2, 3 & 5 = Numbers divisible by 30
Total numbers divisible by 30 = 1020, 1050 ….. 9990
an = a1 + (n – 1)d1 a1 = 1020 d1 = 30
9990 = 1020 + (n – 1) 30
8970/30 = n-1
n = 300
So number of 4 digit numbers greater than 1000
divisible by both 2 & 3 but not 5 = 1500 – 300 = 1200
Let A= {1, 2, 3} and B = {a, b}. Assuming all relations from set A to set B are equally likely, what is the probability that a relation from A to B is also a function ?
1/8


None of these

In a right-angled triangle ABC, the hypotenuse AC is of length 13 cm. A line drawn connecting the midpoints D and E of sides AB and AC is found to be 6 cm in length. The length of BC is
12 cm

In a room, there are n persons whose average height is 160 cm. If m more persons, whose average height is 172 cm, enter the room, then the average height of all persons in the room becomes 164 cm. then m : n is
1 : 2

If one of the factors of the number 3728173 is randomly chosen, then the probability that the chosen factor will be a perfect square is_____________.
5/36

The lengths of the sides of a triangle are x, 21 and 40, where x is the shortest side. A possible value of x is -
20
For a triangle sum of 2 sides > 3rd side
x + 21 > 40
x > 19
Ans is 20
A set of all possible values the function f(x) = x/|x| , where x ≠ 0 takes is
{1, – 1}

The sum of the first 15 terms in an arithmetic progression is 200, while the sum of the next 15 terms is 350. Then the common difference is
2/3

The cost of a piece of jewellery is proportional to the square of its weight. A piece of jewellery weighing 10 grams is INR 3600. The cost of a piece of jewellery of the same kind weighing 4 grams is
INR 576

The value of k for which the following lines x – y – 1 = 0, 2x + 3y – 12 = 0, 2x – 3y + k = 0 are concurrent is
0
For lines to be concurrent they must pass through a common point.
x – y – 1 = 0
2x + 3y – 12 = 0
2x – 3y + k = 0
x – y = 1 × 2
2x + 3y = 12
2x – 2y = 12
5y = 10
y = 2
x – y = 1
x – 2 = 1
x = 3
Then 2x – 3y + k = 0 also passes from (3, 2)
2 (3) – 3 (2) + K = 0
6 – 6 + K = 0
K = 0
The sum of the squares of all the roots of the equation x2 + |x + 4| + |x – 4| – 35 = 0
50

Let A and B be two sets such that the Cartesian product A × B consists of four elements. If two elements of A × B are (1, 4) and (4, 1), then
A × B = B × A
A × B has 4 elements
Then A has 2 elements & B has 2 elements
Because 2 values of A × B have different value for B
So elements of B are = {1, 4}
Then elements of A are = {1, 4}
Then A × B = {(1, 1), (1, 4), (4, 1), (4, 4)}
And B × A = {(1, 1), (1, 4), (4, 1), (4, 4)}
Hence A × B = B × A
Suppose a, b and c are integers such that a > b > c > 0, and
Then the value of the determinant of A
is negative
