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A vendor has 120 kg rice of one kind, 160 kg of another kind and 210 kg of a third kind. He wants to sell the rice by filling the three kinds of rice in bags of equal capacity. What should be the greatest capacity of such a bag?

A24 kg

B15 kg

C12 kg

D10 kg

Answer:

D. 10 kg

Read Explanation:

The greatest capacity of such a bag should be 10 kg.

To find the greatest capacity of a bag that can measure out all three quantities of rice exactly, you need to find the Highest Common Factor (HCF) or Greatest Common Divisor (GCD) of 120, 160, and 210.

  1. Find the prime factors of each number:

    • 120=2×2×2×3×5=23×3×5120 = 2 \times 2 \times 2 \times 3 \times 5 = 2^3 \times 3 \times 5

    • 160=2×2×2×2×2×5=25×5160 = 2 \times 2 \times 2 \times 2 \times 2 \times 5 = 2^5 \times 5

    • 210=2×3×5×7210 = 2 \times 3 \times 5 \times 7

  2. Identify the common factors:

    • The prime factors common to all three numbers are 2 and 5.

  3. Multiply the lowest powers of the common factors:

    • HCF=2×5=10\text{HCF} = 2 \times 5 = 10

Therefore, the maximum capacity of each bag must be 10 kg.


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