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

(36) The L.C.M of two numbers is 120, and their H.C.F is 15. If one of the numbers is 45, what is the other number?

• (a) 20
• (b) 25
• (c) 80
• (d) 100

Answer

Correct Answer: (c) 80
\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 15 = 45 \times 120 \Rightarrow x = \frac{45 \times 120}{15} = 360. \\<br /> \text{Therefore, the other number is } 360 - 45 = 315.

(37) Find the H.C.F of 63, 84, and 105.

• (a) 7
• (b) 9
• (c) 14
• (d) 21

Answer

Correct Answer: (d) 21
\text{To find the H.C.F., we can use the prime factorization method.}\\<br /> 63 = 3^2 \times 7, \quad 84 = 2^2 \times 3 \times 7, \quad 105 = 3 \times 5 \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.

(38) What is the least number which when divided by 9, 12, and 15 leaves a remainder of 6 in each case?

• (a) 33
• (b) 45
• (c) 51
• (d) 66

Answer

Correct Answer: (d) 66
\text{To find the least number, we need to find the L.C.M. of } 9, 12, \text{ and } 15.\\<br /> \text{L.C.M. of } 9, 12, \text{ and } 15 = 180.\\<br /> \text{The least number leaving a remainder of 6 when divided by 180 is } 180 \times 0 + 6 = 6.\\<br /> However, 6 is not divisible by 9, 12, or 15.\\<br /> The next multiple of 180 is } 180 \times 1 + 6 = 186, \text{ which is also not divisible by all three numbers.\\<br /> Continuing this process, the correct multiple is } 180 \times 2 + 6 = 366, \text{ which is divisible by all three numbers.\\<br /> Therefore, the least number is } 366 - 6 = 360 + 6 = 366.

(39) If the H.C.F of two numbers is 48 and one of the numbers is 288, what is the other number?

• (a) 72
• (b) 96
• (c) 192
• (d) 240

Answer

Correct Answer: (c) 192
\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, } 48 \times \text{L.C.M} = 288 \times x \Rightarrow \text{L.C.M} = \frac{288 \times x}{48} = 6x. \\<br /> \text{Since the H.C.F is 48 and one number is 288, we can find the other number as } \frac{\text{Product}}{\text{H.C.F}} = \frac{288 \times x}{48} = 6x = 192.

(40) Ten numbers are in the ratio 2:3:4:5:6:7:8:9:10:11, and their L.C.M is 27720. What is the sum of these numbers?

• (a) 22770
• (b) 24990
• (c) 27180
• (d) 29460

Answer

Correct Answer: (d) 29460
\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 = 27720.\\<br /> \text{Solving for } x, \text{ we get } x = 10.\\<br /> \text{So, the numbers are } 2x = 20, 3x = 30, 4x = 40, 5x = 50, 6x = 60, 7x = 70, 8x = 80, 9x = 90, 10x = 100, \text{ and } 11x = 110. \\<br /> \text{The sum of these numbers is } 20 + 30 + 40 + 50 + 60 + 70 + 80 + 90 + 100 + 110 = 660. \\<br /> \text{To find the sum of the original numbers, we need to multiply by 10 (the value of } x), \text{ so the sum is } 660 \times 10 = 6600.