Question

DO QUESTION NO 1 4 7 9

Answer 1

1) let one be X and second be X+1 then
1/x+1/(x+1)=23/132

solve u get quadratic equation solve it.

And ask only one question is one because it is very hadd

Answer 2

4) Sum of Reciprocals is..
Let no. 1 is 'x'. Therefore it's Reciprocal is 1/x. Since we are given sum of consecutive nos. So let no. 2 be 'x + 1' and it's Reciprocal is 1/x+1.

So, there sum is...
1/x + 1/x+1 = 23/132
==> (2x + 1)/x^2 +x = 23/132 (taking L.C.M.)

==> 132(2x + 1) = 23(x^2 + x)
==> 264x + 132 = 23x^2 + 23x
==> 23x^2 +23x - 264x -132 = 0
==> 23x^2 -241x - 132 = 0
==> 23x^2 - 253x + 12x - 132 = 0
==> 23x ( x - 11) + 12 ( x - 11) = 0
==> (23x + 12)(x - 11) = 0
So we get x = 11 because x = -12/23 can not be our answer as it is negative.

So, the required numbers are 11 and 12. You can answer by the following way :--

Reciprocal of 11 = 1/11
Reciprocal of 12 = 1/12

So,
1/11 + 1/12 = 23/132
Therefore x = 11 is correct and numbers are 11 and 12.
I hope i cleared your doubt regarding this question..

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