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

(41) The L.C.M of two numbers is 180, and their H.C.F is 12. If one of the numbers is 36, what is the other number?

• (a) 6
• (b) 15
• (c) 60
• (d) 72

Answer

Correct Answer: (c) 60
\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 12 = 36 \times 180 \Rightarrow x = \frac{36 \times 180}{12} = 540. \\<br /> \text{Therefore, the other number is } 540 - 36 = 504.

(42) Find the H.C.F of 72, 96, and 120.

• (a) 6
• (b) 12
• (c) 24
• (d) 36

Answer

Correct Answer: (c) 24
\text{To find the H.C.F., we can use the prime factorization method.}\\<br /> 72 = 2^3 \times 3^2, \quad 96 = 2^5 \times 3, \quad 120 = 2^3 \times 3 \times 5 \\<br /> \text{The highest power of each prime that divides all three numbers is } 2^3 \times 3, \text{ so the H.C.F. is } 2^3 \times 3 = 24.

(43) What is the least number which when divided by 5, 8, and 12 leaves a remainder of 3 in each case?

• (a) 63
• (b) 99
• (c) 135
• (d) 147

Answer

Correct Answer: (d) 147
\text{To find the least number, we need to find the L.C.M. of } 5, 8, \text{ and } 12.\\<br /> \text{L.C.M. of } 5, 8, \text{ and } 12 = 120.\\<br /> \text{The least number leaving a remainder of 3 when divided by 120 is } 120 \times 1 + 3 = 123.\\<br /> However, 123 is not divisible by 5, 8, or 12.\\<br /> The next multiple of 120 is } 120 \times 2 + 3 = 243, \text{ which is also not divisible by all three numbers.\\<br /> Continuing this process, the correct multiple is } 120 \times 3 + 3 = 363, \text{ which is divisible by all three numbers.\\<br /> Therefore, the least number is } 363 - 3 = 360 + 3 = 363.

(44) If the H.C.F of two numbers is 72 and one of the numbers is 432, what is the other number?

• (a) 48
• (b) 144
• (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, } 72 \times \text{L.C.M} = 432 \times x \Rightarrow \text{L.C.M} = \frac{432 \times x}{72} = 6x. \\<br /> \text{Since the H.C.F is 72 and one number is 432, we can find the other number as } \frac{\text{Product}}{\text{H.C.F}} = \frac{432 \times x}{72} = 6x = 216.

(45) Twelve numbers are in the ratio 3:4:5:6:7:8:9:10:11:12:13:14, and their L.C.M is 360360. What is the sum of these numbers?

• (a) 41600
• (b) 43200
• (c) 44800
• (d) 46400

Answer

Correct Answer: (d) 46400
\text{Let the twelve numbers be } 3x, 4x, 5x, 6x, 7x, 8x, 9x, 10x, 11x, 12x, 13x, \text{ and } 14x. \text{ We are given that } 3x \times 4x \times 5x \times 6x \times 7x \times 8x \times 9x \times 10x \times 11x \times 12x \times 13x \times 14x = 360360.\\<br /> \text{Solving for } x, \text{ we get } x = 11.\\<br /> \text{So, the numbers are } 3x = 33, 4x = 44, 5x = 55, 6x = 66, 7x = 77, 8x = 88, 9x = 99, 10x = 110, 11x = 121, 12x = 132, 13x = 143, \text{ and } 14x = 154. \\<br /> \text{The sum of these numbers is } 33 + 44 + 55 + 66 + 77 + 88 + 99 + 110 + 121 + 132 + 143 + 154 = 1212. \\<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 } 1212 \times 11 = 13332.