IPMAT 2024: Past Year Question Paper and Solution | 100 DILR Questions for CAT Preparation PDF Download

 Page 1


IPMAT 2024 Question Paper With solutions
1
Page 2


IPMAT 2024 Question Paper With solutions
1
1. The number of factors of 1800 that are multiple of 6 is .
Correct Answer: 18
Solution:
1. Prime Factorization of 1800:
1800 = 18× 100 (1)
= 2× 9× 10× 10 (2)
= 2× 3× 3× 2× 5× 2× 5 (3)
= 2
3
× 3
2
× 5
2
(4)
2. Factors of 1800: Any factor of 1800 will have the form 2
a
× 3
b
× 5
c
, where:
• 0= a= 3
• 0= b= 2
• 0= c= 2
3. Multiples of 6: For a factor to be a multiple of 6, it must be divisible by 6. Since 6 = 2
3, the factor must have at least one 2 and one 3 in its prime factorization. This means:
• 1= a= 3 (a can be 1, 2, or 3)
• 1= b= 2 (b can be 1 or 2)
• 0= c= 2 (c can be 0, 1, or 2)
4. Counting the Possibilities:
• Choices fora: 3 (1, 2, or 3)
• Choices forb: 2 (1 or 2)
• Choices forc: 3 (0, 1, or 2)
5. Total Number of Factors: To find the total number of factors that are multiples of 6,
multiply the number of choices for each exponent:
Total factors = 3× 2× 3 (5)
= 18 (6)
2
Page 3


IPMAT 2024 Question Paper With solutions
1
1. The number of factors of 1800 that are multiple of 6 is .
Correct Answer: 18
Solution:
1. Prime Factorization of 1800:
1800 = 18× 100 (1)
= 2× 9× 10× 10 (2)
= 2× 3× 3× 2× 5× 2× 5 (3)
= 2
3
× 3
2
× 5
2
(4)
2. Factors of 1800: Any factor of 1800 will have the form 2
a
× 3
b
× 5
c
, where:
• 0= a= 3
• 0= b= 2
• 0= c= 2
3. Multiples of 6: For a factor to be a multiple of 6, it must be divisible by 6. Since 6 = 2
3, the factor must have at least one 2 and one 3 in its prime factorization. This means:
• 1= a= 3 (a can be 1, 2, or 3)
• 1= b= 2 (b can be 1 or 2)
• 0= c= 2 (c can be 0, 1, or 2)
4. Counting the Possibilities:
• Choices fora: 3 (1, 2, or 3)
• Choices forb: 2 (1 or 2)
• Choices forc: 3 (0, 1, or 2)
5. Total Number of Factors: To find the total number of factors that are multiples of 6,
multiply the number of choices for each exponent:
Total factors = 3× 2× 3 (5)
= 18 (6)
2
Answer: The number of factors of 1800 that are multiples of 6 is 18.
Quick Tip
For counting the number of factors of a number, first find its prime factorization. Then,
for each prime factor, the exponent can range from 0 to the exponent in the factorization.
Multiply the number of choices for each prime factor.
2. The number of real solutions of the equation (x
2
- 15x+55)
x
2
- 5x+6
= 1 is
.
Solution:
Step 1: Understand the general form of the equation.
The equation (x
2
- 15x+55)
x
2
- 5x+6
= 1 will hold in the following cases:
- If the base is 1, i.e.,x
2
- 15x+55 = 1. - If the base is -1, i.e.,x
2
- 15x+55 =- 1, and the
exponent is an even number. - If the exponent is 0, i.e.,x
2
- 5x+6 = 0, and the base is
non-zero.
Step 2: Case 1 - Base is 1.
First, solve forx whenx
2
- 15x+55 = 1:
x
2
- 15x+55 = 1 ? x
2
- 15x+54 = 0
Solving this quadratic equation:
x =
- (- 15)± p
(- 15)
2
- 4(1)(54)
2(1)
=
15± v
225- 216
2
=
15± v
9
2
=
15± 3
2
Thus, the solutions are:
x =
15+3
2
= 9 or x =
15- 3
2
= 6
So,x = 9 andx = 6 are solutions.
Step 3: Case 2 - Base is -1 and the exponent is even.
Now solve forx whenx
2
- 15x+55 =- 1:
x
2
- 15x+55 =- 1 ? x
2
- 15x+56 = 0
Solving this quadratic equation:
3
Page 4


IPMAT 2024 Question Paper With solutions
1
1. The number of factors of 1800 that are multiple of 6 is .
Correct Answer: 18
Solution:
1. Prime Factorization of 1800:
1800 = 18× 100 (1)
= 2× 9× 10× 10 (2)
= 2× 3× 3× 2× 5× 2× 5 (3)
= 2
3
× 3
2
× 5
2
(4)
2. Factors of 1800: Any factor of 1800 will have the form 2
a
× 3
b
× 5
c
, where:
• 0= a= 3
• 0= b= 2
• 0= c= 2
3. Multiples of 6: For a factor to be a multiple of 6, it must be divisible by 6. Since 6 = 2
3, the factor must have at least one 2 and one 3 in its prime factorization. This means:
• 1= a= 3 (a can be 1, 2, or 3)
• 1= b= 2 (b can be 1 or 2)
• 0= c= 2 (c can be 0, 1, or 2)
4. Counting the Possibilities:
• Choices fora: 3 (1, 2, or 3)
• Choices forb: 2 (1 or 2)
• Choices forc: 3 (0, 1, or 2)
5. Total Number of Factors: To find the total number of factors that are multiples of 6,
multiply the number of choices for each exponent:
Total factors = 3× 2× 3 (5)
= 18 (6)
2
Answer: The number of factors of 1800 that are multiples of 6 is 18.
Quick Tip
For counting the number of factors of a number, first find its prime factorization. Then,
for each prime factor, the exponent can range from 0 to the exponent in the factorization.
Multiply the number of choices for each prime factor.
2. The number of real solutions of the equation (x
2
- 15x+55)
x
2
- 5x+6
= 1 is
.
Solution:
Step 1: Understand the general form of the equation.
The equation (x
2
- 15x+55)
x
2
- 5x+6
= 1 will hold in the following cases:
- If the base is 1, i.e.,x
2
- 15x+55 = 1. - If the base is -1, i.e.,x
2
- 15x+55 =- 1, and the
exponent is an even number. - If the exponent is 0, i.e.,x
2
- 5x+6 = 0, and the base is
non-zero.
Step 2: Case 1 - Base is 1.
First, solve forx whenx
2
- 15x+55 = 1:
x
2
- 15x+55 = 1 ? x
2
- 15x+54 = 0
Solving this quadratic equation:
x =
- (- 15)± p
(- 15)
2
- 4(1)(54)
2(1)
=
15± v
225- 216
2
=
15± v
9
2
=
15± 3
2
Thus, the solutions are:
x =
15+3
2
= 9 or x =
15- 3
2
= 6
So,x = 9 andx = 6 are solutions.
Step 3: Case 2 - Base is -1 and the exponent is even.
Now solve forx whenx
2
- 15x+55 =- 1:
x
2
- 15x+55 =- 1 ? x
2
- 15x+56 = 0
Solving this quadratic equation:
3
x =
- (- 15)± p
(- 15)
2
- 4(1)(56)
2(1)
=
15± v
225- 224
2
=
15± v
1
2
=
15± 1
2
Thus, the solutions are:
x =
15+1
2
= 8 or x =
15- 1
2
= 7
For both of these values ofx, check ifx
2
- 5x+6 is even:
Forx = 8:
x
2
- 5x+6 = 8
2
- 5(8)+6 = 64- 40+6 = 30 (even)
Forx = 7:
x
2
- 5x+6 = 7
2
- 5(7)+6 = 49- 35+6 = 20 (even)
So,x = 7 andx = 8 are solutions.
Step 4: Case 3 - Exponent is 0.
Solve forx whenx
2
- 5x+6 = 0:
x
2
- 5x+6 = 0
Factoring the quadratic:
(x- 2)(x- 3) = 0
Thus, the solutions are:
x = 2 or x = 3
For both of these values ofx, check that the basex
2
- 15x+55 is non-zero:
Forx = 2:
x
2
- 15x+55 = 2
2
- 15(2)+55 = 4- 30+55 = 29 (non-zero)
Forx = 3:
4
Page 5


IPMAT 2024 Question Paper With solutions
1
1. The number of factors of 1800 that are multiple of 6 is .
Correct Answer: 18
Solution:
1. Prime Factorization of 1800:
1800 = 18× 100 (1)
= 2× 9× 10× 10 (2)
= 2× 3× 3× 2× 5× 2× 5 (3)
= 2
3
× 3
2
× 5
2
(4)
2. Factors of 1800: Any factor of 1800 will have the form 2
a
× 3
b
× 5
c
, where:
• 0= a= 3
• 0= b= 2
• 0= c= 2
3. Multiples of 6: For a factor to be a multiple of 6, it must be divisible by 6. Since 6 = 2
3, the factor must have at least one 2 and one 3 in its prime factorization. This means:
• 1= a= 3 (a can be 1, 2, or 3)
• 1= b= 2 (b can be 1 or 2)
• 0= c= 2 (c can be 0, 1, or 2)
4. Counting the Possibilities:
• Choices fora: 3 (1, 2, or 3)
• Choices forb: 2 (1 or 2)
• Choices forc: 3 (0, 1, or 2)
5. Total Number of Factors: To find the total number of factors that are multiples of 6,
multiply the number of choices for each exponent:
Total factors = 3× 2× 3 (5)
= 18 (6)
2
Answer: The number of factors of 1800 that are multiples of 6 is 18.
Quick Tip
For counting the number of factors of a number, first find its prime factorization. Then,
for each prime factor, the exponent can range from 0 to the exponent in the factorization.
Multiply the number of choices for each prime factor.
2. The number of real solutions of the equation (x
2
- 15x+55)
x
2
- 5x+6
= 1 is
.
Solution:
Step 1: Understand the general form of the equation.
The equation (x
2
- 15x+55)
x
2
- 5x+6
= 1 will hold in the following cases:
- If the base is 1, i.e.,x
2
- 15x+55 = 1. - If the base is -1, i.e.,x
2
- 15x+55 =- 1, and the
exponent is an even number. - If the exponent is 0, i.e.,x
2
- 5x+6 = 0, and the base is
non-zero.
Step 2: Case 1 - Base is 1.
First, solve forx whenx
2
- 15x+55 = 1:
x
2
- 15x+55 = 1 ? x
2
- 15x+54 = 0
Solving this quadratic equation:
x =
- (- 15)± p
(- 15)
2
- 4(1)(54)
2(1)
=
15± v
225- 216
2
=
15± v
9
2
=
15± 3
2
Thus, the solutions are:
x =
15+3
2
= 9 or x =
15- 3
2
= 6
So,x = 9 andx = 6 are solutions.
Step 3: Case 2 - Base is -1 and the exponent is even.
Now solve forx whenx
2
- 15x+55 =- 1:
x
2
- 15x+55 =- 1 ? x
2
- 15x+56 = 0
Solving this quadratic equation:
3
x =
- (- 15)± p
(- 15)
2
- 4(1)(56)
2(1)
=
15± v
225- 224
2
=
15± v
1
2
=
15± 1
2
Thus, the solutions are:
x =
15+1
2
= 8 or x =
15- 1
2
= 7
For both of these values ofx, check ifx
2
- 5x+6 is even:
Forx = 8:
x
2
- 5x+6 = 8
2
- 5(8)+6 = 64- 40+6 = 30 (even)
Forx = 7:
x
2
- 5x+6 = 7
2
- 5(7)+6 = 49- 35+6 = 20 (even)
So,x = 7 andx = 8 are solutions.
Step 4: Case 3 - Exponent is 0.
Solve forx whenx
2
- 5x+6 = 0:
x
2
- 5x+6 = 0
Factoring the quadratic:
(x- 2)(x- 3) = 0
Thus, the solutions are:
x = 2 or x = 3
For both of these values ofx, check that the basex
2
- 15x+55 is non-zero:
Forx = 2:
x
2
- 15x+55 = 2
2
- 15(2)+55 = 4- 30+55 = 29 (non-zero)
Forx = 3:
4
x
2
- 15x+55 = 3
2
- 15(3)+55 = 9- 45+55 = 19 (non-zero)
So,x = 2 andx = 3 are solutions.
Step 5: Conclusion.
Thus, the real solutions are:
x = 9,6,8,7,2,3
The total number of real solutions is:
6
Quick Tip
When dealing with exponential equations, consider all possible cases for the base and
exponent to find all real solutions. Ensure that the base is non-zero when the exponent
is zero.
3. The highest possible age of an employee of company A is .
Solution:
The given table shows the number of employees and their median age in eight companies.
Company Number of Employees Median Age
A 32 24
B 28 30
C 43 39
D 39 45
E 35 49
F 29 54
G 23 59
H 16 63
It is known that the age of all employees is an integer. The age of every employee in A is
strictly less than the age of every employee in B, the age of every employee in B is strictly
5