Daily Practice Questions CAT Previous Year Questions with Answer PDF

Q1: Let n be the least positive integer such that 168 is a factor of 1134ⁿ. If m is the least positive integer such that 1134ⁿ is a factor of 168^m, then m + n equals      [2023]
(a) 12
(b) 9
(c) 15
(d) 24
Ans: (c)
Sol: The prime factorizations of 168 and 1134 are as follows:
168 = 2³ × 3 × 7
1134 = 2 × 3⁴ × 7
Clearly, the smallest positive integral value of n, such that 168 is a factor of 1134ⁿ is 3.
1134ⁿ = 1134³ = 2³ × 3¹² × 7³
Clearly, the smallest positive integral value of m, such that 1134³ is a factor of 168^m is 12.
Therefore, m + n = 12 + 3 = 15 

Q2: The number of all natural numbers up to 1000 with non-repeating digits is      [2023]
(a) 585
(b) 504
(c) 648
(d) 738
Ans: (d)
Sol: Single digit numbers with non-repeating digits = 9
(The unit’s digit is non-zero)
Two digit numbers with non-repeating digits = 9 × 9
(The tenth’s digit is non-zero and the unit digit can be any digit except the tenth’s digit.)
Three digit numbers with non-repeating digits = 9 × 9 × 8
(The hundred’s digit is non-zero and the tenth’s digit can be any digit except the hundred’s digit and the unit digit can be any digit except the tenth’s digit.)
So, totally there are (9 + 9 × 9 + 9 × 9 × 8) = 738 natural numbers up to 1000 with non-repeating digits. 

Q3: For any natural numbers m, n , and k, such that k divides both m + 2n and 3m + 4n, k must be a common divisor of      [2023]
(a) 2m and 3n
(b) m and 2n
(c) 2m and n
(d) m and n
Ans: (b)
Sol: k divides m + 2n
So, k also divides 2(m + 2n) = 2m + 4n
It is given that k divides 3m + 4n
Which means, k should also divide (3m + 4n) – (2m + 4n)
∴ k divides m
Since k divides m and m + 2n
k should also divide (m + 2n) – m = 2n
Therefore, k divides m and 2n. 

Q4: The number of positive integers less than 50, having exactly two distinct factors other than 1 and itself, is      [2023]
Sol: A positive integer less than 50, having exactly two distinct factors other than 1 and itself, is either a perfect cube below 50 or an integer that is a product of exactly two distinct primes.
Case i)
Perfect cubes below 50 are 2³ and 3³. So, two numbers here
Case ii)
For the product of two primes to be below 50, the individual primes should be below 25.
(Because, the smallest prime is 2 and multiplying 2 with anything greater than or equal to 25 yields a number greater than or equal to 50.)
2, 3, 5, 7, 11, 13, 17, 19, 23 are prime numbers less than 25. 
2, 3, 5, 7 are the primes less than √50 any product of two numbers among them yields a product less than 50.
So, there are 4C₂ = 6 pairs here.
11, 13, 17, 19, 23 are the primes greater than √50, any product of two numbers among them yields a product greater than 50.
So, there are 0 pairs here.
Between the two lists 11 and 13 can pair with 2 and 3, while 17, 19, and 23 can only pair with 2.
So, there are 7 pairs here.
So, totally, there are 2 + 6 + 0 + 7 = 15 such numbers. 

Q5: For any real number x, let |x| be the largest integer less than or equal to x. If  then N is       [2022]
Sol: It is given,

For n = 1 to n = 19, value of function is zero.
For n = 20 to n = 44, value of function will be 1.
44 = 20 + n - 1
n = 25 which is equal to given value.
This implies N = 44

Q6: A donation box can receive only cheques of ₹100, ₹250, and ₹500. On one good day, the donation box was found to contain exactly 100 cheques amounting to a total sum of ₹15250. Then, the maximum possible number of cheques of ₹500 that the donation box may have contained, is
Sol: Let the number of 100 cheques, 250 cheques and 500 cheques be x, y and z respectively.
We need to find the maximum value of z.
x + y + z = 100 ...... (1)
100x + 250y + 500z = 15250
2x + 5y + 10z = 305 ...... (2)
2x + 2y + 2z = 200 ....... (1)
(2) - (1), we get
3y + 8z = 105
At z = 12, x = 3
Therefore, maximum value z can take is 12.

Q7: How many three-digit numbers are greater than 100 and increase by 198 when the three digits are arranged in the reverse order?      [2021]
Sol: Let the numbers be of the form 100a + 10b + c, where a, b, and c represent single digits.
Then (100c + 10b + a) - (100a+ 10b + c) = 198
99c - 99a = 198
c - a = 2.
Now, a can take the values 1-7. a cannot be zero as the initial number has 3 digits and cannot be 8 or 9 as then c would not be a single-digit number.
Thus, there can be 7 cases.
B can take the value of any digit from 0-9, as it does not affect the answer. Hence, the total cases will be 7× 10 = 70.

Q8: For a 4-digit number, the sum of its digits in the thousands, hundreds and tens places is 14, the sum of its digits in the hundreds, tens and units places is 15, and the tens place digit is 4 more than the units place digit. Then the highest possible 4-digit number satisfying the above conditions is      [2021]
Sol: Given the 4 digit number:
Considering the number in thousands digit is a number in the hundredth digit is b, number in tens digit is c, number in the units digit is d.
Let the number be abcd.
Given that a + b + c = 14. (1)
b + c + d = 15. (2)
c = d + 4. (3).
In order to find the maximum number which satisfies the condition, we need to have abcd such that a is maximum which is the digit in thousands place in order to maximize the value of the number. b, c, and d are less than 9 each as they are single-digit numbers.
Substituting (3) in (2) we have b + d + 4 + d = 15, b + 2 x d = 11. (4)
Subtracting (2) and (1) : (2) - (1) = d = a+1. (5)
Since c cannot be greater than 9 considering c to be the maximum value 9 the value of d is 5.
If d = 5, using d = a + 1, a = 4.
Hence the maximum value of a = 4 when c = 9, d = 5.
Substituting b + 2 x d = 11. b = 1.
The highest four-digit number satisfying the condition is 4195

Q9: Suppose one of the roots of the equation ax² - bx + c = 0 is  2+ √3 , Where a,b and c are rational numbers and a ≠ 0. If b = c³  then | a | equals.       [2021]
(a) 1
(b) 2
(c) 3
(d) 4
Ans: (b)
Sol: Given a, b, c are rational numbers.
Hence a, b, c are three numbers that can be written in the form of p/Q
Hence if one both the root is 2 + √3 and considering the other root to be x.
The sum of the roots and the product of the two roots must be rational numbers.
For this to happen the other root must be the conjugate of 2 + √3  so the product and the sum of the roots are rational numbers which are represented by: b/a, c/a
Hence the sum of the roots is 2 + √3 + 2 - √3 = 4
The product of the roots is 2 + √3 x  2 - √3 =1
b/a = 4, c/a = 1.
b = 4 x a, c= a.
Since b = c³
4 x a = a³ 
a² = 4.
a = 2 or -2.
|a| = 2

Q10: For a sequence of real numbers x₁ ,x₂ ,...x_n , If x₁ −x₂ +x₃ −....+(−1)ⁿ⁺¹ x_n = n² +2n for all natural numbers n, then the sum x₄₉ + x₅₀ equals       [2021]
(a) 200
(b) 2
(c) -200
(d) -2
Ans: (d)
Sol: Now as per the given series :
we get x₁ =1 + 2 =3
Now x₁ −x₂ = 8
So, x₂ = -5
Now x₁ − x₂ + x₃  = 15
So, x₃ = 7
So, we get
So x₄₉  = 99 and x₅₀ = −101
Therefore x₄₉ + x₅₀  = −2

Q11: For a real number x the condition ∣3x−20∣+∣3x−40∣ = 20 necessarily holds if       [2021]
(a) 10 < x < 15
(b) 9 < x < 14
(c) 7 < x < 12
(d) 6 < x < 11
Ans: (c)
Sol: Case 1 : x ≥ 40 / 3
we get 3x - 20  + 3x - 40 = 20
6x = 80

Case 2:
we get 3x - 20 + 40 - 3x = 20
we get 20 = 20
So, we get
Case 3:
we get 20 - 3x + 40 - 3x = 20
40 = 6x
x = 20/3
but this is not possible
so we get from case 1,2 and 3

Now looking at options
we can say only option C satisfies for all x .
Hence 7 < x < 12.

Q12: If n is a positive integer such that , then the smallest value of n is       [2021]
Sol: 

For minimum value of n,

1 + 2 + ... + n = 21
We can see that if n = 6, 1 + 2 + 3 + ... + 6 = 21.

Q13: If 3x + 2∣y∣ + y=7 and x + ∣x∣ + 3y = 1 then x + 2yx + 2y is:       [2021]
(a) -4/3
(b) 8/3
(c) 0
(d) 1
Ans: (c)
Sol: We need to check for all regions:
x >= 0, y >= 0
x >= 0, y < 0
x < 0, y >= 0
x < 0, y < 0
However, once we find out the answer for any one of the regions, we do not need to calculate for other regions since the options suggest that there will be a single answer.
Let us start with x >= 0, y >= 0,
3x + 3y = 7
2x + 3y = 1
Hence, x = 6 and y = -11/3
Since y > = 0, this is not satisfying the set of rules.
Next, let us test x >= 0, y < 0,
3x - y = 7
2x + 3y = 1
Hence, y = -1
x = 2.
This satisfies both the conditions. Hence, this is the correct point.
We need the value of x + 2y
x + 2y = 2 + 2(-1) = 2 - 2 = 0.

Q14: If a, b and c are positive integers such that ab = 432, bc = 96 and c < 9, then the smallest possible value of a + b + c is     [2020]
(a) 56
(b) 59
(c) 49
(d) 46
Ans: (d)
Sol: Given ab = 432, bc = 96 and c < 9
To find the minimum value for a + b + c, the two larger numbers should be as close as possible.
The closest combination whose product is 432 is 18 x 24 . For b = 24, we get c = 4 and a = 18 .
Hence the least value for a + b + c = 46 .

Q15: How many 3-digit numbers are there, for which the product of their digits is more than 2 but less than 7?      [2020]
Sol: The product of the digits of the three-digit numbers should be more than 2 and less than 7 .
Hence the possible numbers are as follows.

Q16: The mean of all 4 -digit even natural numbers of the form 'aabb', where a >0, is       [2020]
(a) 5050
(b) 4466
(c) 5544
(d) 4864 
Ans: (c)
Sol: The sum of possible even digit numbers in the form aabb is 1100 + 1122 + 1144 + 1166 + 1188 + 2200 + 2222 + 2288 + ......9900 + 9922 + 9988 i.e. (45 numbers)
⇒ 1100 + 1100 +1100+ 1100 + 1100 + 22 + 44 + 66 + 88 + 2200 + 2200 + 2200 + 2200 + 2200 +22 + 44 + 66 + 88 +… + 9900 + 9900 + 9900 + 990 + 9900 + 22 + 44 + 66 + 88
⇒ 5(1100 + 2200 + .....9900) + 9(22 + 44 + 66 + 88)5 x 1100(1 + 2 + .... 9) + 9 x  22(1 + 2 + 3 + 4)
⇒ 5500(45) + 45 ´ 44 = 45(5544)
Hence mean =5544 

Q17: Let A, B and C be three positive integers such that the sum of A and the mean of B and C is 5. In addition, the sum of B and the mean of A and C is 7. Then the sum of A and B is      [2020]
(a) 6
(b) 5
(c) 7
(d) 4
Ans: (a)
Sol: Given A, B and C are positive integers such that  



The least value for A=1 in which case B=5.
Hence A + B = 6

Q18: If x and y are positive real numbers satisfying x + y=102, then the minimum possible value of  is      [2020]
Sol: AM ≥ GM ≥ HM

Given x + y = 102
⇒
⇒
The minimum value of
=
= 2704

Q19: How many 4-digit numbers, each greater than 1000 and each having all four digits distinct, are there with 7 coming before 3?      [2020]
Sol:  Consider four blanks

7 is in thousand place, then 3 can be placed in any of the 3 places in 3 ways. Remaining two blanks can be filled with remaining eight digits in ⁸P₂ ways. The number of numbers that have 7 is in thousand place is 3  x ⁸P₂ = 168
Thousand place cannot be 0,7 and 3, it can be filled with remaining 7 digits in 7 ways. In remaining 3 blanks, 7 and 3 can be arranged in 3 ways. Fourth blank can be filled in 7 ways. The number of numbers that are formed where 7 and 3 is not in thousand place is 7 x 3 x 7 = 147 . Hence total required numbers = 168 + 147 = 315 .

Q20: In how many ways can a pair of integers (x , a) be chosen such that x² - 2| x| + | a - 2| = 0?      [2020]
(a) 4
(b) 5
(c) 6
(d) 7
Ans: (d)
Sol: x² - 2| x| + | a- 2 |= 0


If a > 2; | a - 2 |= a - 2

since x is integer 3 - a ≥ 0
a ≤ 3
The possible values of a is = 3
Then x = ± 1;
If a = 2, | x |=| 1 ± 1 |, ⇒ x = ± 2, 0
If a < 2, | a - 2 |= 2 - a

Since x is integer a -1 ≥ 0 ⇒ a ≥ 1
∴ The possible values of a is 1
If a = 1, | x |= 1 ⇒ x = ± 1
∴ The possible pairs =(-1,3), (1,3), (1,1), (-1,1), (2,2),(-2,2), (0,2) i.e., 7

Q21: The number of integers that satisfy the equality ( x² - 5x + 7 ) ^x+1 = 1 is      [2020]
(a) 2
(b) 3
(c) 5
(d) 4
Ans: (b)
Sol: ( x² - 5x + 7 ) ^x+1  = 1
We know, for ab = 1,
if -a = -1 then b is even.
-a = 1 then b is any number
-a > 0 then b = 0
Case 1: x + 1 = 0 Þ x = -1
Case 2: x² -5x+7 =1 ⇒ x2 -5x + 6 = 0 ⇒ x = 2 or 3
Case 3: x² -5x + 7 = -1 ⇒ x² -5x + 8 = 0
but x is not an integer
∴ The number of integers satisfies the equation is 3

Q22: The number of pairs of integers (x , y) satisfying x ≥ y ≥ -20 and 2 x + 5 y = 99 is      [2020]
Ans: 2 x + 5 y = 99
When y = - 19, x = 97; since x ≥ y ; the maximum value of y is 13 and corresponding value of x is 17.
We know that the solutions of y are in arithmetic progression with common difference 2.
Here a = - 19, d = 2, tn = 13
t_n = a + (n -1)d
-19 + (n - 1)(2) = 13
(n - 1)2 = 32 ⇒ n = 17
Hence number of pairs integers is 17

Q23: If x and y are non-negative integers such that x + 9 = z, y + 1 = z and x + y < z + 5, then the maximum possible value of 2x + y equals      [2020]
Sol: Given x + 9 = z = y + 1 and x + y < z + 5
⇒ (z -9) + (z - 1) < z + 5
⇒ z < 15
Hence the maximum value of z = 14, max of x = 5 and max of y = 13
Required answer, 2 x + y = 2 x 5 + 13 = 23

Q24: How many pairs (a, b) of positive integers are there such that a ≤ b and ab = 4²⁰¹⁷ ?      [2020]
(a) 2017
(b) 2019
(c) 2020
(d) 2018
Ans: (b)
Sol: Given, a × b = 4²⁰¹⁷ = 2⁴⁰³⁴
Since a × b = 4²⁰¹⁷ , is a perfect square the number of factors of 2⁴⁰³⁴ is odd.
Required answer, the number of values of

= 2018

Q25: If x₁ = -1 and x_m = x_{m +1} + (m + 1) for every positive integer $m,$ then x₁₀₀ equals       [2020]
(a) -5151
(b) -5150
(c) -5051
(d) -5050
Ans: (d)
Sol: x_{m + 1} = x_m - (m + 1)
x₂ = x₁ - 2 = -1 - 2 = -3
x₃ = x₂ - 3 = -1 - 2 - 3 = - 6
Similarly,

Hence,

= -5050

Q26: Let m and n be natural numbers such that n is even and . Then m - 2n equals      [2020]
(a) 3
(b) 4
(c) 1
(d) 2
Ans: (c)
Sol: Given,


⇒ n = 4
since 0.2 < n/m < 0.5 m and n = 4, m = 9
m - 2n = 9 - 2 x 4 = 1

Q27: Let N, x and y be positive integers such that N = x + y, 2 < x < 10 and 14 < y < 23. If N > 25, then how many distinct values are possible for N?      [2020]
Sol: Given, 2 < x < 10 and 14 < y < 23 Þ 17 < ( x + y ) < 32 i.e. 17 < N < 32
But N > 25 hence 25 < N < 32
N can take 6 distinct values. 

Q28: How many of the integers 1, 2, … , 120, are divisible by none of 2, 5 and 7?      [2020]
(a) 40
(b) 42
(c) 43
(d) 41
Ans: (d)
Sol: The required answer
=
= 41.14
Required answer is the integral part of 41.14 = 41