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

(46) The L.C.M of two numbers is 240, and their H.C.F is 8. If one of the numbers is 32, what is the other number?

• (a) 60
• (b) 72
• (c) 90
• (d) 120

Answer

Correct Answer: (c) 90
\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 8 = 32 \times 240 \Rightarrow x = \frac{32 \times 240}{8} = 960. \\<br /> \text{Therefore, the other number is } 960 - 32 = 928.

(47) Find the H.C.F of 84, 105, and 126.

• (a) 14
• (b) 21
• (c) 42
• (d) 63

Answer

Correct Answer: (c) 42
\text{To find the H.C.F., we can use the prime factorization method.}\\<br /> 84 = 2^2 \times 3 \times 7, \quad 105 = 3 \times 5 \times 7, \quad 126 = 2 \times 3^2 \times 7 \\<br /> \text{The highest power of each prime that divides all three numbers is } 3 \times 7, \text{ so the H.C.F. is } 3 \times 7 = 21.

(48) What is the least number which when divided by 7, 9, and 15 leaves a remainder of 4 in each case?

• (a) 109
• (b) 139
• (c) 169
• (d) 199

Answer

Correct Answer: (d) 199
\text{To find the least number, we need to find the L.C.M. of } 7, 9, \text{ and } 15.\\<br /> \text{L.C.M. of } 7, 9, \text{ and } 15 = 315.\\<br /> \text{The least number leaving a remainder of 4 when divided by 315 is } 315 \times 0 + 4 = 4.\\<br /> However, 4 is not divisible by 7, 9, or 15.\\<br /> The next multiple of 315 is } 315 \times 1 + 4 = 319, \text{ which is also not divisible by all three numbers.\\<br /> Continuing this process, the correct multiple is } 315 \times 2 + 4 = 634, \text{ which is divisible by all three numbers.\\<br /> Therefore, the least number is } 634 - 4 = 630 + 4 = 634.

(49) If the H.C.F of two numbers is 36 and one of the numbers is 144, what is the other number?

• (a) 72
• (b) 108
• (c) 216
• (d) 288

Answer

Correct Answer: (c) 216
\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, } 36 \times \text{L.C.M} = 144 \times x \Rightarrow \text{L.C.M} = \frac{144 \times x}{36} = 4x. \\<br /> \text{Since the H.C.F is 36 and one number is 144, we can find the other number as } \frac{\text{Product}}{\text{H.C.F}} = \frac{144 \times x}{36} = 4x = 216.

(50) Ten numbers are in the ratio 2:3:4:5:6:7:8:9:10:11, and their L.C.M is 360360. What is the sum of these numbers?

• (a) 39600
• (b) 41040
• (c) 42480
• (d) 43920

Answer

Correct Answer: (d) 43920
\text{Let the ten numbers be } 2x, 3x, 4x, 5x, 6x, 7x, 8x, 9x, 10x, \text{ and } 11x. \text{ We are given that } 2x \times 3x \times 4x \times 5x \times 6x \times 7x \times 8x \times 9x \times 10x \times 11x = 360360.\\<br /> \text{Solving for } x, \text{ we get } x = 11.\\<br /> \text{So, the numbers are } 2x = 22, 3x = 33, 4x = 44, 5x = 55, 6x = 66, 7x = 77, 8x = 88, 9x = 99, 10x = 110, \text{ and } 11x = 121. \\<br /> \text{The sum of these numbers is } 22 + 33 + 44 + 55 + 66 + 77 + 88 + 99 + 110 + 121 = 715. \\<br /> \text{To find the sum of the original numbers, we need to multiply by 11 (the value of } x), \text{ so the sum is } 715 \times 11 = 7865.