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

(1) What is the smallest number that when divided by 20, 25, and 30 leaves a remainder of 7?

• (a) 357
• (b) 377
• (c) 427
• (d) 457

Answer

Correct Answer: (c) 427
L.C.M. \text{ of } 20, 25, \text{ and } 30 = 300. \text{The smallest number leaving a remainder of 7 is } 300 + 7 = 307. \text{However, 307 is not divisible by all three numbers. The next multiple is } 300 \times 2 + 7 = 607, \text{ which is also not divisible. The correct multiple that works for all is } 300 \times 3 + 7 = 907, \text{ which is also incorrect. Continuing this process, } 300 \times 4 + 7 = 1207, \text{ and subtracting the multiples of 20, 25, or 30 from 1207, we get the smallest number as } 427.

(2) If the H.C.F of two numbers is 12 and their L.C.M is 288, what is the product of the two numbers?

• (a) 3456
• (b) 2368
• (c) 4032
• (d) 5184

Answer

Correct Answer: (a) 3456
\text{The product of two numbers is equal to the product of their H.C.F. and L.C.M. Hence, Product } = H.C.F \times L.C.M = 12 \times 288 = 3456.

(3) Which of the following cannot be the H.C.F of two numbers whose product is 7680?

• (a) 32
• (b) 50
• (c) 64
• (d) 128

Answer

Correct Answer: (b) 50
\text{The H.C.F of two numbers must be a factor of their product. The product is 7680. The factors of 7680 include 32, 64, and 128 but not 50. Therefore, 50 cannot be the H.C.F.}

(4) Find the H.C.F of 48 and 64.

• (a) 12
• (b) 8
• (c) 16
• (d) 24

Answer

Correct Answer: (c) 16
\text{To find the H.C.F., we can use the prime factorization method.}\\<br /> 48 = 2^4 \times 3 \text{ and } 64 = 2^6 \\<br /> \text{The common prime factor is } 2^4, \text{ so the H.C.F. is } 2^4 = 16.

(5) Three bells ring at intervals of 12, 18, and 24 minutes, respectively. If they all ring together at 12:00 PM, at what time will they next ring together?

• (a) 1:00 PM
• (b) 2:00 PM
• (c) 2:24 PM
• (d) 3:00 PM

Answer

Correct Answer: (c) 2:24 PM
\text{To find the time at which they will next ring together, we need to find the L.C.M. of 12, 18, and 24.}\\<br /> \text{L.C.M. of } 12, 18, \text{ and } 24 = 72 \text{ minutes.}\\<br /> \text{Therefore, they will next ring together at 12:00 PM + 72 minutes = 2:24 PM.}