Problems on H.C.F and L.C.M

(6) The L.C.M of two numbers is 45, and their H.C.F is 5. If one of the numbers is 15, what is the other number?

• (a) 9
• (b) 75
• (c) 25
• (d) 30

Answer

Correct Answer: (b) 75
\text{Let one of the numbers be } x. \text{ We know that } \text{L.C.M} \times \text{H.C.F} = \text{Product of the two numbers.}\\<br /> \text{So, } x \times 5 = 15 \times 45 \Rightarrow x = \frac{15 \times 45}{5} = 75.

(7) Find the H.C.F of 72, 108, and 144.

• (a) 6
• (b) 12
• (c) 24
• (d) 36

Answer

Correct Answer: (d) 36
\text{To find the H.C.F., we can use the prime factorization method.}\\<br /> 72 = 2^3 \times 3^2, \quad 108 = 2^2 \times 3^3, \quad 144 = 2^4 \times 3^2 \\<br /> \text{The highest power of each prime that divides all three numbers is } 2^2 \times 3^2, \text{ so the H.C.F. is } 2^2 \times 3^2 = 36.

(8) What is the least number which when divided by 4, 6, and 9 leaves a remainder of 2 in each case?

• (a) 28
• (b) 74
• (c) 86
• (d) 102

Answer

Correct Answer: (b) 74
\text{To find the least number, we need to find the L.C.M. of } 4, 6, \text{ and } 9.\\<br /> \text{L.C.M. of } 4, 6, \text{ and } 9 = 36.\\<br /> \text{The least number leaving a remainder of 2 when divided by 36 is } 36 \times 2 + 2 = 74.

(9) If the H.C.F of two numbers is 9 and one of the numbers is 45, what is the other number?

• (a) 18
• (b) 81
• (c) 27
• (d) 36

Answer

Correct Answer: (b) 81
\text{Let one of the numbers be } x. \text{ We know that } \text{H.C.F} \times \text{L.C.M} = \text{Product of the two numbers.}\\<br /> \text{So, } 9 \times \text{L.C.M} = 45 \times x \Rightarrow \text{L.C.M} = \frac{45 \times x}{9} = 5x. \\<br /> \text{Since the H.C.F is 9 and one number is 45, we can find the other number as } \frac{\text{Product}}{\text{H.C.F}} = \frac{45 \times x}{9} = 5x = 81.

(10) Three numbers are in the ratio 3:4:5, and their L.C.M is 120. What is the sum of these numbers?

• (a) 72
• (b) 96
• (c) 120
• (d) 144

Answer

Correct Answer: (b) 96
\text{Let the three numbers be } 3x, 4x, \text{ and } 5x. \text{ We are given that } 3x \times 4x \times 5x = 120.\\<br /> \text{Solving for } x, \text{ we get } x = 2.\\<br /> \text{So, the numbers are } 3x = 6, 4x = 8, \text{ and } 5x = 10. \\<br /> \text{The sum of these numbers is } 6 + 8 + 10 = 24.