Latest m4maths Aptitude Question SOLUTION: A fruit-seller sells half of his fruits plus 4 oranges; again he sells half of what remains plus 4 oranges; and so again a third time. His store is now exhausted; How many did he

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(#M40019732) M4MATHS question fruit-seller Keep an EYE

A fruit-seller sells half of his fruits plus 4 oranges; again he sells half of what remains plus 4 oranges; and so again a third time. His store is now exhausted; How many did he start with?

Asked In m4maths suman (10 years ago)

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shiv kumar garg  (10 years ago)

let f is the total no. of fruit then in first time he sold=f/2+4 and remaining are f/2-4
second time he sold (f/2-4)/2+4 and remaining are [(f/2-4)/2]/2-4;
so total fruits are first time +second time+third time=total
(f/2+4)+(f/2-4)/2+4+[(f/2-4)/2]/2-4=f
after solving
f=56.

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Prathyusha BSN  (10 years ago)

the oranges are also a part of all fruits.
so everytym we take half of remaining fruits, we have to subtract the extra four oranges given.
therefore, x/2 +4 + (x-8)/4 +4 + (x-24)/8 +4 = x..
solving, 8x=7x+96-40
so, x=56... solved...

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pramod dusad  (10 years ago)

let the total fruits is x.
so after first time remaining fruits is = x/2 - 4
= (x-8)/2

after second time remaining fruits is = ((x-8)/2)/2 -4
= (x-8)/4 - 4
= (x-8-16)/4
= (x-24)/4

after third time remaining fruits is = ((x-24)/4)/2 - 4
= (x-24)/8 - 4
= (x-24-32)/8
= (x-56)/8

and given that after third time store is exhausted so

(x-56)/8 = 0
so ans is x=56

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