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02 Aug 2021, 03:45
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
68% (01:53) correct
32%
(02:05)
wrong
based on 93
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If the digit in the units place of a two-digit number is halved and the digit in the tens place is doubled, the number thus obtained is equal to the number obtained by interchanging the digits. Which of the following is definitely true?
(A) Digits in the units place and the tens place are equal.
(B) Sum of the digits is a two-digit number.
(C) Digit is the units place is half of the digit in the tens place.
(D) Digit in the units place is twice the digit in the tens place.
(E) None of these
(A) Digits in the units place and the tens place are equal.
(B) Sum of the digits is a two-digit number.
(C) Digit is the units place is half of the digit in the tens place.
(D) Digit in the units place is twice the digit in the tens place.
(E) None of these
Updated on: 02 Aug 2021, 04:28
Originally posted by rajatchopra1994 on 02 Aug 2021, 03:54.
Last edited by rajatchopra1994 on 02 Aug 2021, 04:28, edited 1 time in total.
Last edited by rajatchopra1994 on 02 Aug 2021, 04:28, edited 1 time in total.
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Explanation:
20x + y/2 = 10y + x
19x = 19y/2
38x = 19y
2x = y
IMO-C IMO-D
20x + y/2 = 10y + x
19x = 19y/2
38x = 19y
2x = y
02 Aug 2021, 04:23
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Solution:
Let us assume a two-digit number be \(= 10x+y\). Where \(x\) and \(y\) are ten's and unit's digit respectively.
We are forming a new number such that the digit in the unit's place is halved and the digit in the tens place is doubled. This means the new number \(= 10 (2x)+\frac{y}{2}\)
We are told that this new number = number obtained by interchanging the digits
So, we can say: \(10 (2x)+\frac{y}{2} = 10y+x\)
\(⇒ 20x+\frac{y}{2}=10y+x\)
\(⇒ 20x-x=10y-\frac{y}{2}\)
\(⇒ 19x=\frac{(20y-y)}{2}\)
\(⇒ 19x=\frac{19y}{2}\)
\(⇒ x=\frac{y}{2}\)
\(⇒ 2x=y\)
Thus we can say that the digit in the unit's place (\(y\)) is twice the digit in the tens place (\(x\)).
Hence the right answer is Option D.
Let us assume a two-digit number be \(= 10x+y\). Where \(x\) and \(y\) are ten's and unit's digit respectively.
We are forming a new number such that the digit in the unit's place is halved and the digit in the tens place is doubled. This means the new number \(= 10 (2x)+\frac{y}{2}\)
We are told that this new number = number obtained by interchanging the digits
So, we can say: \(10 (2x)+\frac{y}{2} = 10y+x\)
\(⇒ 20x+\frac{y}{2}=10y+x\)
\(⇒ 20x-x=10y-\frac{y}{2}\)
\(⇒ 19x=\frac{(20y-y)}{2}\)
\(⇒ 19x=\frac{19y}{2}\)
\(⇒ x=\frac{y}{2}\)
\(⇒ 2x=y\)
Thus we can say that the digit in the unit's place (\(y\)) is twice the digit in the tens place (\(x\)).
Hence the right answer is Option D.
