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The product of two numbers is 2028 and their HCF is 13. The number of such of pair is = ?

11
22
33
44

Answer:- 1

Explanation:-

Solution:

HCF = 13 (given)
Let number are 13x & 13y respectively

\begin{aligned} Also given \\ 13 x \times 13 y = 2028 \\ 13 \times 13 \times x y = 2028 \\ x y = \frac{2028}{13 \times 13} = 12 \\ ∴ Possible pairs are \\ = (1, 12) (3, 4) \end{aligned}

$\eqalign{ & {\text{Also given}} \cr & 13x \times 13y = 2028 \cr & 13 \times 13 \times xy = 2028 \cr & xy = \frac{{2028}}{{13 \times 13}} = 12 \cr & \therefore {\text{Possible pairs are}} \cr & {\text{ = }}\left( {1,12} \right)\left( {3,4} \right) \cr}$ Only two pair are possible

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Only two pair are possible ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

HCF = 13 (given)
Let number are 13x & 13y respectively

\begin{aligned} Also given \\ 13 x \times 13 y = 2028 \\ 13 \times 13 \times x y = 2028 \\ x y = \frac{2028}{13 \times 13} = 12 \\ ∴ Possible pairs are \\ = (1, 12) (3, 4) \end{aligned}

$\eqalign{ & {\text{Also given}} \cr & 13x \times 13y = 2028 \cr & 13 \times 13 \times xy = 2028 \cr & xy = \frac{{2028}}{{13 \times 13}} = 12 \cr & \therefore {\text{Possible pairs are}} \cr & {\text{ = }}\left( {1,12} \right)\left( {3,4} \right) \cr}$ Only two pair are possible ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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