CAT 2022 QA Slot 1 | Previous Year CAT Paper
Ankita buys 4 kg cashews, 14 kg peanuts and 6 kg almonds when the cost of 7 kg cashews is the same as that of 30 kg peanuts or 9 kg almonds. She mixes all the three nuts and marks a price for the mixture in order to make a profit of ₹1752. She sells 4 kg of the mixture at this marked price and the remaining at a 20% discount on the marked price, thus making a total profit of ₹744. Then the amount, in rupees, that she had spent in buying almonds is
- A.
1176
- B.
1680
- C.
2520
- D.
1440
Answer: Option B
Explanation :
Let the price of a kg of peanuts, cashews and almonds is Rs. p, c and a.
Given, 7c = 30p = 9a
⇒ c : p : a = 90 : 21 : 70
Ankita mixes 4 kg cashews, 14 kg peanuts and 6 kg almonds, i.e., total 24 kgs.
Total cost price of the mixture = 4 × 90x + 14 × 21x + 6 × 70x = 1074x.
Cost price/kg = 44.75x/kg
She plans to sells this 24 kg mixture for a Profit of Rs. 1752 i.e, at a profit of 1752/24 = Rs. 73/kg.
She sells 4 kg of this mixture at the original selling price and hence earns a profit of 4 × 73 = Rs. 292.
Total profit she actually earns is Rs. 744.
⇒ Profit earned on remaining 20 kgs = 744 – 292 = Rs. 452.
⇒ Profit originally planned on remaining 20 kgs = 20 × 73 = Rs. 1460.
Cost of this 20 kg = 20 × 44.75x = 895x
⇒ (895x + 1460) × 0.8 = 895x + 452
⇒ 1460 × 0.8 – 452 = 895x × 0.2
⇒ 716 = 179x
⇒ x = 4
∴ Price of 6 kg of almonds = 6 × 70 × 4 = 1680
Hence, option (b).
Workspace:
In a village, the ratio of number of males to females is 5 : 4. The ratio of number of literate males to literate females is 2 : 3. The ratio of the number of illiterate males to illiterate females is 4 : 3. If 3600 males in the village are literate, then the total number of females in the village is
Answer: 43200
Explanation :
Number of male literates = 3600
=
⇒ =
⇒ Number of literate females = 5400
Let number of illiterate males and females be 4x and 3x respectively.
⇒ =
⇒ =
⇒ 14400 + 16x = 27000 + 15x
⇒ x = 12,600
⇒ Total females in the village = 5400 + 3x = 5400 + 37,800 = 43,200
Hence, 43200.
Workspace:
Pinky is standing in a queue at a ticket counter. Suppose the ratio of the number of persons standing ahead of Pinky to the number of persons standing behind her in the queue is 3 : 5. If the total number of persons in the queue is less than 300, then the maximum possible number of persons standing ahead of Pinky is
Answer: 111
Explanation :
Let the number of people standing in front and behind Pinky be 3x and 5x respectively.
∴ Total number of people in the line = 3x + 5x + 1 (Pinky herself) = 8x + 1
⇒ 8x + 1 < 300
⇒ x < 299/8 = 37.375
∴ Maximum value of x is 37.
⇒ Maximum number of people standing in front of Pinky = 3x = 111.
Hence, 111.
Workspace:
A trapezium ABCD has side AD parallel to BC. ∠BAD = 90°, BC = 3 cm and AD = 8 cm. If the perimeter of this trapezium is 36 cm, then its area, in sq. cm, is
Answer: 66
Explanation :
Let AB = x
In right ∆CED,CE = AB = x
⇒ CD =
In the given figure,
⇒ 3 + 8 + x + = 36
⇒ = 25 – x
⇒ x2 + 25 = 625 + x2 – 50x
⇒ 50x = 600
⇒ x = 12
Now, area of a trapezium = 1/2 × (sum of parallel sides) × height
= 1/2 × 11 × 12 = 66
Hence, 66.
Workspace:
Amal buys 110 kg of syrup and 120 kg of juice, syrup being 20% less costly than juice, per kg. He sells 10 kg of syrup at 10% profit and 20 kg of juice at 20% profit. Mixing the remaining juice and syrup, Amal sells the mixture at ₹ 308.32 per kg and makes an overall profit of 64%. Then, Amal’s cost price for syrup, in Rupees per kg, is
Answer: 160
Explanation :
Let the cost price of juice be Rs. 10x/kg, hence cost price of syrup = 8x/kg
10 kg syrup is sold at 10% profit i.e., at 8.8x/kg.
∴ Total selling price = Rs. 88x
20 kg of juice is sold at 20% profit i.e., at 12x/kg
∴ Total selling price = Rs. 240x
Remaining (110 + 120 – 30 =) 200 kg at Rs. 308.32/kg.
Total cost price of the mixture = 110 × 8x + 120 × 10x = 2080x
⇒ 2080x × 1.64 = 88x + 240x + 308.32 × 200
⇒ 2080x × 1.64 - 328x = 308.32 × 200
⇒ 3083.2x = 308.32 × 200
⇒ x = 20
Cost price of syrup = 8x = 160/kg.
Hence, 160.
Workspace:
Let a and b be natural numbers. If a2 + ab + a = 14 and b2 + ab + b = 28, then (2a + b) equals
- A.
9
- B.
10
- C.
8
- D.
7
Answer: Option C
Explanation :
Given a and b are natural numbers.
a2 + ab + a = 14
⇒ a(a + b + 1) = 14 = 1 × 14 or 2 × 7
Case 1: a = 1 and a + b + 1 = 14
⇒ b = 12
a = 1 and b = 12 does not satisfy b2 + ba + b = 28
Case 2: a = 2 and a + b + 1 = 7
⇒ b = 4
a = 2 and b = 4 satisfies b2 + ba + b = 28
∴ 2a + b = 4 + 4 = 8
Hence, option (c).
Workspace:
The number of ways of distributing 20 identical balloons among 4 children such that each child gets some balloons but no child gets an odd number of balloons, is
Answer: 84
Explanation :
Let the number of balloons each child gets is 2a, 2b, 2c and 2d respectively.
a, b, c, d > 0
⇒ 2a + 2b + 2c + 2d = 20
⇒ a + b + c + d = 10
Number of combinations of a, b, c and d such that a, b, c and d > 0 = (10-4)+4-1C4-1 = 9C3 = 84.
Hence, 84.
Workspace:
Let ABCD be a parallelogram such that the coordinates of its three vertices A, B, C are (1, 1), (3, 4) and (−2, 8), respectively. Then, the coordinates of the vertex D are
- A.
(4, 5)
- B.
(0, 11)
- C.
(-4, 5)
- D.
(-3, 4)
Answer: Option C
Explanation :
Let (x, y) be the coordinates of 4th vertex.
In a parallelogram, diagonals bisect each other.
∴ Intersection point of diagonals is same as the mid-point of two opposite vertices.
Sum of coordinates of two pair of opposite vertices is same.
A and C are opposite vertices while B and D are opposite vertices.
(1, 1) is opposite (-2, 8) while (3, 4) is opposite (x, y)
⇒ 1 - 2 = 3 + x ⇒ x = -4
⇒ 1 + 8 = 4 + y ⇒ y = 5
∴ (x, y) = (-4, 5)
Of the three possibilities only (-4, 5) is mentioned in options.
Hence, option (c).
Workspace:
Let A be the largest positive integer that divides all the numbers of the form 3k + 4k + 5k and B be the largest positive integer that divides all the numbesr of the form 4k + 3(4k) + 4k+2, where k is any positive integer. Then (A + B) equals
Answer: 82
Explanation :
Given, 3k + 4k + 5k
put k = 1, we have 3k + 4k + 5k = 3 + 4 + 5 = 12
put k = 2, we have 3k + 4k + 5k = 9 + 16 + 25 = 50
The only common factor between 12 and 50 is 2.
3k + 4k + 5k is always even for any value of k, and hence always divisible by 2.
∴ A = 2
Given, 4k + 3(4k) + 4k+2
= 4k(1 + 3 + 42)
= 4k × 20
For k ≥ 1, this number is always divisible by 4 × 20 = 80
∴ B = 80
⇒ A + B = 2 + 80 = 82.
Hence, 82.
Workspace:
Let a, b and c be non-zero real numbers such that b2 < 4ac, and f(x) = ax2 + bx + c. If the set S consists of all integers m such that f(m) < 0, then the set S must necessarily be
- A.
the empty set
- B.
the set of all integers
- C.
either the empty set or the set of all integers
- D.
the set of all positive integers
Answer: Option C
Explanation :
Since b2 < 4ac ⇒ D < 0, hence, the roots of f(x) are imaginary.
Case 1: a > 0
Since a > 0 and D < 0 ⇒ f(x) > 0
Hence, there is no value of m for which f(m) < 0
⇒ m is an empty set.
Case 2: a < 0
Since a < 0 and D < 0 ⇒ f(x) < 0
Hence, for all values of m, f(m) < 0
⇒ m can take any integral value.
Hence, option (c).
Workspace:
A mixture contains lemon juice and sugar syrup in equal proportion. If a new mixture is created by adding this mixture and sugar syrup in the ratio 1 : 3, then the ratio of lemon juice and sugar syrup in the new mixture is
- A.
1 : 4
- B.
1 : 6
- C.
1 : 5
- D.
1 : 7
Answer: Option D
Explanation :
Let 10 units of mixture is mixed with 30 units of sugar syrup.
Amount of sugar syrup in 10 units of mixture = 5 units
Amount of lemon juice in 10 units of mixture = 5 units
∴ Total amount of lemon juice in the final mixture = 5 units, and
Total amount of sugar syrup in the final mixture = 5 + 30 = 35 units
Ratio of lemon juice and sugar syrup in final mixture = 5 : 35 = 1 : 7.
Hence, option (d).
Workspace:
For natural numbers x, y and z, if xy + yz = 19 and yz + xz = 51, then the minimum possible value of xyz is
Answer: 34
Explanation :
x, y and z are natural numbers.
⇒ xy + yz = 19
⇒ y(x + z) = 19 = 1 × 19
∴ y = 1 and x + z = 19 [x + z ≥ 2, since they are natural numbers] …(1)
Also, yz + xz = 51
⇒ z(1 + x) = 51 = 3 × 17 or 1 × 51
Case 1: z = 3 and 1 + x = 17
⇒ x = 16
But then x + z = 19
Hence, this case is accepted.
⇒ xyz = 16 × 1 × 3 = 84
Case 2: z = 17 and 1 + x = 3
⇒ x = 2
But then x + z = 19
Hence, this case is accepted.
⇒ xyz = 2 × 1 × 17 = 34
Case 3: z = 1 and 1 + x = 51
⇒ x = 50
But then x + z = 51
Hence, this case is rejected.
Case 4: z = 51 and 1 + x = 1
⇒ x = 0
But x should be a natural number.
Hence, this case is rejected.
∴ Lowest possible value of xyz = 34 [Case 2].
Hence, 34.
Workspace:
For any natural number n, suppose the sum of the first n terms of an arithmetic progression is (n+ 2n2). If the nth term of the progression is divisible by 9, then the smallest possible value of n is
- A.
9
- B.
8
- C.
4
- D.
7
Answer: Option D
Explanation :
Sn = n + 2n2
Tn = Sn – Sn-1
⇒ Tn = n + 2n2 – [(n-1) + 2(n-1)2]
⇒ Tn = n + 2n2 – [n-1 + 2n2 – 4n + 2]
⇒ Tn = n + 2n2 – n + 1 - 2n2 + 4n – 2
⇒ Tn = 4n – 1
Tn is divisible by 9 when n = 7.
Hence, option (d).
Workspace:
The average weight of students in a class increases by 600 gm when some new students join the class. If the average weight of the new students is 3 kg more than the average weight of the original students, then the ratio of number of original students to the number of new students is
- A.
3 : 1
- B.
4 : 1
- C.
1 : 2
- D.
1 : 4
Answer: Option B
Explanation :
Let the number of students initially be I and the number of students joining be J.
Average of students initially be a kg, while those joining be (a + 3) kg such that final average of all students becomes (a + 0.6) kg
Initial Students Students Joining
I J
a a + 3
a + 0.6
2.4 0.6
⇒ I/J = 2.4/0.6 = 4/1
Hence, option (b).
Workspace:
The largest real value of a for which the equation |x + a| + |x - 1| = 2 has an infinite number of solutions for x is
- A.
-1
- B.
2
- C.
0
- D.
1
Answer: Option D
Explanation :
Given, |x + a| + |x - 1| = 2
⇒ Sum of distance of x from 1 and -a is 2.
This is possible when a = -3 or 1.
Hence, option (d).
Workspace:
All the vertices of a rectangle lie on a circle of radius R. If the perimeter of the rectangle is P, then the area of the rectangle is
- A.
- 2PR
- B.
- 2R2
- C.
-
- D.
- R2
Answer: Option B
Explanation :
Let the longer side of the rectangle be a and shorter side by b.
In the figure given, (2R)2 = a2 + b2
⇒ P = 2(a + b)
⇒ = a + b
Squaring both sides.
⇒ = a2 + b2 + 2ab
⇒ - 4R2 = 2ab
⇒ - 2R2 = ab
Area of the rectangle = a × b = ab =
Hence, option (b).
Workspace:
For any real number x, let [x] be the largest integer less than or equal to x. If = 25, then N is
Answer: 44
Explanation :
= 0 when < 1 ⇒ n < 20
= 1 when 1 ≤ < 2 ⇒ 20 ≤ n < 45
= + = 0 + (1 + 1 + 1 + … + 1 (25 times)) = 25
∴ N = 44
Hence, 44.
Workspace:
Trains A and B start traveling at the same time towards each other with constant speeds from stations X and Y, respectively. Train A reaches station Y in 10 minutes while train B takes 9 minutes to reach station X after meeting train A. Then the total time taken, in minutes, by train B to travel from station Y to station X is
- A.
12
- B.
15
- C.
10
- D.
6
Answer: Option B
Explanation :
Let the time taken to meet after starting be t mins.
Time taken by A to reach Y after meeting = (10 – t) mins
Time taken by B to reach X after meeting = 9 mins
⇒ t2 = (10 - t) × 9
⇒ t2 +9t – 90 = 0
⇒ (t + 15)(t - 6) = 0
⇒ t = 6 (-15 is rejected)
Total time taken by B to reach Y = 6 + 9 = 15 mins.
Hence, option (b).
Workspace:
The average of three integers is 13. When a natural number n is included, the average of these four integers remains an odd integer. The minimum possible value of n is:
- A.
5
- B.
1
- C.
3
- D.
4
Answer: Option A
Explanation :
Sum of the original 3 numbers = 3 × 13 = 39.
Now, = odd = 2k - 1
⇒ 39 + n = 8k – 4
⇒ 43 + n = 8k
For k to be an integer least possible natural value of n = 5
Hence, option (a).
Workspace:
In a class of 100 students, 73 like coffee, 80 like tea and 52 like lemonade. It may be possible that some students do not like any of these three drinks. Then the difference between the maximum and minimum possible number of students who like all the three drinks is
- A.
52
- B.
47
- C.
48
- D.
53
Answer: Option B
Explanation :
Maximum people who can like all three drinks is 52 since least number of people linking a particular drink is 52 for lemonade.
Let the number of students liking
exactly one drink = a
exactly two drinks = b
exactly three drinks = c
exactly no drink = n
⇒ a + 2b + 3c = 73 + 80 + 52 = 205
Even if a = c = 0 and b = 100, a + 2b + 3c = 200
∴ c cannot be zero.
If we shift one student from b to c, the sum will increase by 1.
Hence, to increase the sum by 5, we need to shift 5 students from b to c.
∴ Least possible value of c is 5.
⇒ The difference between the maximum and minimum possible number of students who like all the three drinks = 52 – 5 = 47
Hence, option (b).
Workspace:
Alex invested his savings in two parts. The simple interest earned on the first part at 15% per annum for 4 years is the same as the simple interest earned on the second part at 12% per annum for 3 years. Then, the percentage of his savings invested in the first part is
- A.
40%
- B.
62.5%
- C.
60%
- D.
37.5%
Answer: Option D
Explanation :
Let the amount invested in first and second parts is F and S respectively.
⇒ =
⇒ =
⇒ % amount invested in first part = × 100% = 37.5%
Hence, option (d).
Workspace:
Let 0 ≤ a ≤ x ≤ 100 and f(x) = |x - a| + |x - 100| + |x - a - 50|. Then the maximum value of f(x) becomes 100 when a is equal to
- A.
0
- B.
25
- C.
50
- D.
100
Answer: Option C
Explanation :
Since a ≤ x ≤ 100
⇒ f(x) = x - a + -(x – 100) + |x - a - 50|
⇒ f(x) = x - a - x + 100 + |x - a - 50|
⇒ f(x) = 100 – a + |x - a - 50| ≤ 100
Option (a): a = 0
⇒ f(x) = 100 – 0 + |x - 0 - 50|
⇒ f(x) = 100 + |x - 50|
Here, when x = 0, f(x) > 100.
∴ This option is rejected.
Option (b): a = 25
⇒ f(x) = 100 – 25 + |x - 25 - 50|
⇒ f(x) = 75 + |x - 75|
Here, when x = 25, f(x) > 100.
∴ This option is rejected.
Option (c): a = 50
⇒ f(x) = 100 – 50 + |x - 50 - 50|
⇒ f(x) = 50 + |x - 100|
For any value of x ≥ 50, f(x) ≤ 100.
∴ This option is correct.
Option (d): a = 100
⇒ f(x) = 100 – 100 + |x - 100 - 50|
⇒ f(x) = |x - 150|
The only value x can take in this case is 100.
⇒ f(x) = 50
Here, the maximum value of f(x) is not 100.
∴ This option is rejected.
Hence, option (c).
Workspace:
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