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Three fractions, x,y and z, are such that `x succ y succ z`. When the smallest of them is divided by the greatest, the result is `9/16`, which exceeds y by 0.0625. If `x+y+z =1 13/24`, then the value of x+z is :
तीन भिन्न x, y और z इस प्रकार है कि `x succ y succ z` है | जब इनमें से सबसे छोटे को सबसे बड़े से विभाजित किया जाता है, तो परिणाम `9/16` आता है जो y से 0.0625 अधिक है | यदि `x+y+z =1 13/24` है, तो x+z का मान ज्ञात करें |

A

`(7)/(8)`

B

1

C

`(25)/(24)`

D

`(7)/(6)`

Text Solution

Verified by Experts

The correct Answer is:
C
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