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

(16) The L.C.M of two numbers is 72, and their H.C.F is 6. If one of the numbers is 18, what is the other number?

• (a) 8
• (b) 10
• (c) 12
• (d) 14

Answer

Correct Answer: (c) 12
\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 72 \Rightarrow x = \frac{18 \times 72}{6} = 216. \\<br /> \text{Therefore, the other number is } 216 - 18 = 198.

(17) Find the H.C.F of 54, 72, and 90.

• (a) 6
• (b) 18
• (c) 27
• (d) 36

Answer

Correct Answer: (b) 18
\text{To find the H.C.F., we can use the prime factorization method.}\\<br /> 54 = 2 \times 3^3, \quad 72 = 2^3 \times 3^2, \quad 90 = 2 \times 3^2 \times 5 \\<br /> \text{The highest power of each prime that divides all three numbers is } 2 \times 3^2, \text{ so the H.C.F. is } 2 \times 3^2 = 18.

(18) What is the least number which when divided by 7, 9, and 12 leaves a remainder of 5 in each case?

• (a) 101
• (b) 112
• (c) 125
• (d) 144

Answer

Correct Answer: (a) 101
\text{To find the least number, we need to find the L.C.M. of } 7, 9, \text{ and } 12.\\<br /> \text{L.C.M. of } 7, 9, \text{ and } 12 = 252.\\<br /> \text{The least number leaving a remainder of 5 when divided by 252 is } 252 \times 0 + 5 = 5.\\<br /> However, 5 is not divisible by 7, 9, or 12.\\<br /> The next multiple of 252 is } 252 \times 1 + 5 = 257, \text{ which is also not divisible by all three numbers.\\<br /> Continuing this process, the correct multiple is } 252 \times 2 + 5 = 509, \text{ which is divisible by all three numbers.\\<br /> Therefore, the least number is } 509 - 5 = 504 = 101.

(19) If the H.C.F of two numbers is 20 and one of the numbers is 100, what is the other number?

• (a) 40
• (b) 200
• (c) 120
• (d) 160

Answer

Correct Answer: (b) 200
\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, } 20 \times \text{L.C.M} = 100 \times x \Rightarrow \text{L.C.M} = \frac{100 \times x}{20} = 5x. \\<br /> \text{Since the H.C.F is 20 and one number is 100, we can find the other number as } \frac{\text{Product}}{\text{H.C.F}} = \frac{100 \times x}{20} = 5x = 200.

(20) Five numbers are in the ratio 2:3:4:5:6, and their L.C.M is 3600. What is the sum of these numbers?

• (a) 7200
• (b) 9000
• (c) 10800
• (d) 12600

Answer

Correct Answer: (c) 10800
\text{Let the five numbers be } 2x, 3x, 4x, 5x, \text{ and } 6x. \text{ We are given that } 2x \times 3x \times 4x \times 5x \times 6x = 3600.\\<br /> \text{Solving for } x, \text{ we get } x = 10.\\<br /> \text{So, the numbers are } 2x = 20, 3x = 30, 4x = 40, 5x = 50, \text{ and } 6x = 60. \\<br /> \text{The sum of these numbers is } 20 + 30 + 40 + 50 + 60 = 200. \\<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 } 200 \times 10 = 2000.