If the prime factorisation of a and b are a-22- 3 - x and b-2 - 32- y - LCM of a- b-22- 32- 52 and HCF of a- b-2 - 3 - 5- which of the following is true-A- x y-25 B- x-y-30C- x-y-0D- x-5

The correct option is B x + y = 30Since product of two numbers = LCM × HCF (2^2.3.x) × (2.3^2.y) =(2× 3× 5) × (2^2× 3^2× 5^2) which gives xy = 5 × 5^2 So, x an ...

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Byju's Answer

Standard IX

Mathematics

Laws of Exponents for Real Numbers

If the prime ...

Question

If the prime factorisation of a and b are $a = 2^{2} \times 3 \times x$ and $b = 2 \times 3^{2} \times y$ ; LCM of (a, b) = $2^{2} \times 3^{2} \times 5^{2}$ and HCF of (a, b) = $2 \times 3 \times 5$ , which of the following is true?

xy = 25

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x + y = 30

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x – y = 0

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x = 5

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Solution

The correct option is B x + y = 30
Since product of two numbers = LCM $\times$ HCF
$(2^{2} .3 . x) \times ({2.3}^{2} . y) = (2 \times 3 \times 5) \times (2^{2} \times 3^{2} \times 5^{2})$ which gives $x y = 5 \times 5^{2}$
So, x and y can alternate between the values 5 and 25 and we can not specify respective values. But, we can say that the sum is always 25 + 5 = 30.

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If the prime factorisation of a and b are a=22×3×x and b=2×32×y ; LCM of (a, b) =22×32×52 and HCF of (a, b) = 2×3×5, which of the following is true?

The correct option is B x + y = 30Since product of two numbers = LCM × HCF (22.3.x)×(2.32.y)=(2×3×5)×(22×32×52) which gives xy=5×52 So, x and y can alternate between the values 5 and 25 and we can not specify respective values. But, we can say that the sum is always 25 + 5 = 30.

If the prime factorisation of a and b are $a = 2^{2} \times 3 \times x$ and $b = 2 \times 3^{2} \times y$ ; LCM of (a, b) = $2^{2} \times 3^{2} \times 5^{2}$ and HCF of (a, b) = $2 \times 3 \times 5$ , which of the following is true?