If the prime factorisation of a and b are a - 2-2- 3- x and b - 2- 3-2- y - LCM of -a- b- -2-2- 3-2- 5-2 and nbsp-HCF of -a- b- - 2 - 3 - 5- which of the following is true- -x-1-y-1-xy - 25x - y - 30x - y - 0x - 6

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

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Byju's Answer

Standard X

Mathematics

Product of Two Numbers = HCF X LCM of the Two 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? ( $x \neq 1, y \neq 1$ )

xy = 25

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x + y = 30

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x – y = 0

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x = 6

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Solution

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

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Similar questions

Q. Two positive integers a and b can be written as a = x³y² and b = xy³, where x and y are prime numbers. Find HCF(a, b) and LCM(a, b).

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\frac{3}{x + y} + \frac{2}{x - y} = 2 and \frac{9}{x + y} - \frac{4}{x - y} = 1

then
(a)

x = \frac{1}{2}, y = \frac{3}{2}

(b)

x = \frac{5}{2}, y = \frac{1}{2}

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x = \frac{3}{2}, y = \frac{1}{2}

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Q. If a and b are positive integers such that a = x³y² and b = xy³, where x, y are prime numbers, then HCF (a, b) = ?

(a) xy
(b) xy²
(c) x²y²
(d) x³y³

Q. 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? (

x \neq 1, y \neq 1

)

If the prime factorisation of a and b are

$a = 2^{2} \times 3 \times x$

$b = 2 \times 3^{2} \times y$ and

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? ( $x \neq 1, y \neq 1$ )

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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? (x≠1, y≠1)

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 Therefore either x = 5 and y = 25 or x = 25 and y = 5. We cannot 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? ( $x \neq 1, y \neq 1$ )