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Jay has with him a total of Rs 29 in 5-rupee and 2-rupee denominators. The number of 5-rupee notes in one-half of one less than the number of 2-rupee notes. How many 5-rupee notes and 2-rupee notes does Jay have respectively?









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Option(B) is correct

Let $x$ be the 5-rupee notes and $y$ be the number of 2-rupee notes.

\(5x+2y=29\)-------- (i)


\(2x-y=-1\)-------- (ii)

On solving both the equation we get:

$x =\textbf{3}$ and $y =\textbf{7}$

Edit: For an alternative approach making use of the option choices, check comment by Sravan Reddy.

(3) Comment(s)

Sravan Reddy

Hi Abhijeet,

They said "Number of 5-rupee notes in one-half of one less than the number of 2-rupee notes"

So this condition is satisfied by only 2 options which are (3,7) and (2,5). So you can quickly check these two options: (3 Rs.5 notes & 7 Rs.2 notes make up 29 || 2 Rs.5 notes and 5 Rs.2 notes make up only 20)

So, answer is (3,7). This doesn't apply to all problems obviously but I am used to look for shortcuts in exam as soon as I see the question. If I don't find any, I do the actual solution. These methods really helped me in my CAT.

P.S. I understood that you did not understand many of my methods, but please do not use terms like "fake, out of world, messing, etc."

Sravan Reddy

Kill it in 10 sec:

=> Possible options: (3,7) and (2,5).

By calculating only 3,7 adds up to 29


either yur logic is fake or out of this world..are u just messing with our brains? if anyone can clear this logic by 21st century kindly reply here