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Sum of two numbers is 384. HCF of the numbers is 48. The difference of the numbers is = ?

1100
2192
3288
4336

Answer:- 1

Explanation:-

Solution:

HCF = 48
Let number are 48x and 48y respectively

\begin{aligned} \Rightarrow 48 x + 48 y = 384 \\ (x + y) = \frac{384}{48} = 8 \end{aligned}

$\eqalign{ & \Rightarrow 48x + 48y = 384 \cr & \left( {x + y} \right) = \frac{{384}}{{48}} = 8 \cr}$ So, possible pairs of co - prime number are (1, 7) (3, 5)
numbers are (48, 336) or (144, 240)
∴ difference between numbers is = 336 - 48
= 288 and 240 - 144 = 96

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So, possible pairs of co - prime number are (1, 7) (3, 5)
numbers are (48, 336) or (144, 240)
∴ difference between numbers is = 336 - 48
= 288 and 240 - 144 = 96 ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

HCF = 48
Let number are 48x and 48y respectively

\begin{aligned} \Rightarrow 48 x + 48 y = 384 \\ (x + y) = \frac{384}{48} = 8 \end{aligned}

$\eqalign{ & \Rightarrow 48x + 48y = 384 \cr & \left( {x + y} \right) = \frac{{384}}{{48}} = 8 \cr}$ So, possible pairs of co - prime number are (1, 7) (3, 5)
numbers are (48, 336) or (144, 240)
∴ difference between numbers is = 336 - 48
= 288 and 240 - 144 = 96 ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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