A person wanted to withdraw X rupees and Y paise from the bank. The cashier made a mistake and gave him Y rupees and X paise instead. Neither the person nor the cashier noticed the error. After spending 20 paise, the person counted the money and found he had exactly double the amount he originally wanted to withdraw. Find X and Y. (1 Rupee = 100 Paise)

Question

Adhir Mishra has three children: Urmila, Raghu and Sumit. Sumit married Roma, the eldest daughter of Mr. and Mrs. Mohan. The Mohans married their youngest daughter to the eldest son of Mr. and Mrs. Sharma and they had two children Sandeep and Shaifali. The Mohans have two more children, Roshan and Bimla, both elder to Sheila. Sohan and Shivendar are sons of Sumit and Roma. Leela is the daughter of Sandeep.

Answer 1

We can solve this without testing the options, by treating it as a straightforward pair of linear equations problem.

Let the amount the person wanted be W = 100X + Y paise (X rupees, Y paise), and the amount the cashier actually paid be R = 100Y + X paise (the digits swapped).
The condition "after spending 20 paise he had double what he wanted" gives: R - 20 = 2W, that is, (100Y + X) - 20 = 2(100X + Y).
Expanding: 100Y + X - 20 = 200X + 2Y, which rearranges to 98Y = 199X + 20.
Since X and Y must both be whole numbers between 0 and 99 (paise cannot exceed 99), we look for integer solutions of 98Y = 199X + 20. Rearranging for Y: Y = (199X + 20)/98.
For Y to come out as a whole number, 199X + 20 must be exactly divisible by 98. Testing X = 26: 199(26) + 20 = 5194, and 5194 / 98 = 53 exactly. So X = 26 gives Y = 53, both valid paise values (less than 100).
Verification: wanted amount = 2653 paise, wrongly paid amount = 5326 paise, after spending 20 paise he has 5306 paise, and 2 x 2653 = 5306. The condition holds perfectly.

So X = 26 and Y = 53, confirming option (2). \[\boxed{X=26,\ Y=53}\]

Answer 2

We can solve this without testing the options, by treating it as a straightforward pair of linear equations problem.

Let the amount the person wanted be W = 100X + Y paise (X rupees, Y paise), and the amount the cashier actually paid be R = 100Y + X paise (the digits swapped).
The condition "after spending 20 paise he had double what he wanted" gives: R - 20 = 2W, that is, (100Y + X) - 20 = 2(100X + Y).
Expanding: 100Y + X - 20 = 200X + 2Y, which rearranges to 98Y = 199X + 20.
Since X and Y must both be whole numbers between 0 and 99 (paise cannot exceed 99), we look for integer solutions of 98Y = 199X + 20. Rearranging for Y: Y = (199X + 20)/98.
For Y to come out as a whole number, 199X + 20 must be exactly divisible by 98. Testing X = 26: 199(26) + 20 = 5194, and 5194 / 98 = 53 exactly. So X = 26 gives Y = 53, both valid paise values (less than 100).
Verification: wanted amount = 2653 paise, wrongly paid amount = 5326 paise, after spending 20 paise he has 5306 paise, and 2 x 2653 = 5306. The condition holds perfectly.

So X = 26 and Y = 53, confirming option (2). \[\boxed{X=26,\ Y=53}\]

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The Correct Option is B

Solution and Explanation

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