Three numbers which are co-prime to each other are such that the product of the first two is $77$ and that of the last two is $143.$ The sum of the three numbers is

31

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49

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23

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15

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Solution

The correct option is A $31$
Since the numbers are co-prime, they contain only $1$ as the common factor.
Also, the given two products have the middle number in common.
So, middle number $=$ H.C.F. of $77$ and $143 = 11$
Now, the first number $= \frac{77}{11} = 7$
and the third number $= \frac{143}{11} = 13$
$∴$ Required sum $= 7 + 11 + 13 = 31$

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Three numbers which are co-prime to each other are such that the product of the first two is 77 and that of the last two is 143. The sum of the three numbers is

The correct option is A 31Since the numbers are co-prime, they contain only 1 as the common factor. Also, the given two products have the middle number in common. So, middle number = H.C.F. of 77 and 143=11 Now, the first number =7711=7 and the third number =14311=13 ∴ Required sum =7+11+13=31

Three numbers which are co-prime to each other are such that the product of the first two is $77$ and that of the last two is $143.$ The sum of the three numbers is