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

(26) The L.C.M of two numbers is 36, and their H.C.F is 6. If one of the numbers is 18, what is the other number?

• (a) 12
• (b) 18
• (c) 72
• (d) 144

Answer

Correct Answer: (c) 72
\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 6 = 18 \times 36 \Rightarrow x = \frac{18 \times 36}{6} = 108. \\<br /> \text{Therefore, the other number is } 108 - 18 = 90.

(27) Find the H.C.F of 36, 48, and 60.

• (a) 6
• (b) 12
• (c) 18
• (d) 24

Answer

Correct Answer: (b) 12
\text{To find the H.C.F., we can use the prime factorization method.}\\<br /> 36 = 2^2 \times 3^2, \quad 48 = 2^4 \times 3, \quad 60 = 2^2 \times 3 \times 5 \\<br /> \text{The highest power of each prime that divides all three numbers is } 2^2 \times 3, \text{ so the H.C.F. is } 2^2 \times 3 = 12.

(28) What is the least number which when divided by 5, 7, and 9 leaves a remainder of 3 in each case?

• (a) 18
• (b) 33
• (c) 45
• (d) 93

Answer

Correct Answer: (d) 93
\text{To find the least number, we need to find the L.C.M. of } 5, 7, \text{ and } 9.\\<br /> \text{L.C.M. of } 5, 7, \text{ and } 9 = 315.\\<br /> \text{The least number leaving a remainder of 3 when divided by 315 is } 315 \times 0 + 3 = 3.\\<br /> However, 3 is not divisible by 5, 7, or 9.\\<br /> The next multiple of 315 is } 315 \times 1 + 3 = 318, \text{ which is also not divisible by all three numbers.\\<br /> Continuing this process, the correct multiple is } 315 \times 2 + 3 = 633, \text{ which is divisible by all three numbers.\\<br /> Therefore, the least number is } 633 - 3 = 630 + 3 = 633.

(29) If the H.C.F of two numbers is 30 and one of the numbers is 150, what is the other number?

• (a) 50
• (b) 60
• (c) 70
• (d) 90

Answer

Correct Answer: (d) 90
\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, } 30 \times \text{L.C.M} = 150 \times x \Rightarrow \text{L.C.M} = \frac{150 \times x}{30} = 5x. \\<br /> \text{Since the H.C.F is 30 and one number is 150, we can find the other number as } \frac{\text{Product}}{\text{H.C.F}} = \frac{150 \times x}{30} = 5x = 90.

(30) Seven numbers are in the ratio 4:5:6:7:8:9:10, and their L.C.M is 2520. What is the sum of these numbers?

• (a) 2520
• (b) 2880
• (c) 3150
• (d) 3600

Answer

Correct Answer: (b) 2880
\text{Let the seven numbers be } 4x, 5x, 6x, 7x, 8x, 9x, \text{ and } 10x. \text{ We are given that } 4x \times 5x \times 6x \times 7x \times 8x \times 9x \times 10x = 2520.\\<br /> \text{Solving for } x, \text{ we get } x = 3.\\<br /> \text{So, the numbers are } 4x = 12, 5x = 15, 6x = 18, 7x = 21, 8x = 24, 9x = 27, \text{ and } 10x = 30. \\<br /> \text{The sum of these numbers is } 12 + 15 + 18 + 21 + 24 + 27 + 30 = 147. \\<br /> \text{To find the sum of the original numbers, we need to multiply by 3 (the value of } x), \text{ so the sum is } 147 \times 3 = 441.