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30 Jul 2021, 03:15
Kudos
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
5%
(low)
Question Stats:
91% (01:43) correct
9%
(02:19)
wrong
based on 45
sessions
History
Date
Time
Result
Not Attempted Yet
In a two digit number the digit in the unit’s place is twice the digit in the ten’s place and the number obtained by interchanging the digits is more than the original number by 27. What is 50% of the original number?
(A) 36
(B) 24
(C) 18
(D) 12
(E) 6
(A) 36
(B) 24
(C) 18
(D) 12
(E) 6
perskbot
Current Student
Joined: 13 Mar 2019
Last visit: 18 May 2022
Posts: 3
Own Kudos:
Given Kudos: 8
Location: United States (UT)
Concentration: Finance, Real Estate
Schools: Wharton '24 (A) Stanford '24 (A)
GMAT 1: 760 Q48 V47
GPA: 3.59
WE:Real Estate (Real Estate)
30 Jul 2021, 03:29
Kudos
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Possible number pairings:
12
24
36
48
If you flip the two numbers, the only option in which the difference is 27 is 36 (63-36 = 27)
36 * 50% = 18
Alternatively you could solve it using a more calculation intensive approach.
If x is the 10s digit and y is the 1s digit:
10x + y = 10y+x - 27
2x= y
Two equations and two variables is solvable.
x = 3, y = 6
10 * 3 + 6 = 36
Don't forget to take 50% of 36 to get 18
Posted from my mobile device
12
24
36
48
If you flip the two numbers, the only option in which the difference is 27 is 36 (63-36 = 27)
36 * 50% = 18
Alternatively you could solve it using a more calculation intensive approach.
If x is the 10s digit and y is the 1s digit:
10x + y = 10y+x - 27
2x= y
Two equations and two variables is solvable.
x = 3, y = 6
10 * 3 + 6 = 36
Don't forget to take 50% of 36 to get 18
Posted from my mobile device
30 Jul 2021, 08:08
Kudos
Bookmarks
Solution:
Let the 2-digit original number be \(10x+y\) where \(x\) and \(y\) are tens and units digit respectively.
We are given that the digit in the unit’s place is twice the digit in the ten’s place. This means we can write \(y=2x\).
So, the 2-digit number becomes \(10x+2x=12x\)
Secondly, we are given that number obtained by interchanging the digits is more than the original number by 27.
This means that \(10y+x\) (interchanging the digits) \(- 10x+y\) (original) \(= 27\)
\(⇒ 10y+x-10x-y=27\)
\(⇒ 9y-9x=27\)
\(⇒ 9(y-x)=27\)
\(⇒ y-x=\frac{27}{9}\)
\(⇒ y-x=3\)
\(⇒ 2x-x=3\)
\(⇒ x=3\)
So, original number \(= 12x=12\times 3=36\)
Thus, \(50\)% of the number \(= \frac{50}{100}\times 36=18\)
Hence the right answer is Option C.
Let the 2-digit original number be \(10x+y\) where \(x\) and \(y\) are tens and units digit respectively.
We are given that the digit in the unit’s place is twice the digit in the ten’s place. This means we can write \(y=2x\).
So, the 2-digit number becomes \(10x+2x=12x\)
Secondly, we are given that number obtained by interchanging the digits is more than the original number by 27.
This means that \(10y+x\) (interchanging the digits) \(- 10x+y\) (original) \(= 27\)
\(⇒ 10y+x-10x-y=27\)
\(⇒ 9y-9x=27\)
\(⇒ 9(y-x)=27\)
\(⇒ y-x=\frac{27}{9}\)
\(⇒ y-x=3\)
\(⇒ 2x-x=3\)
\(⇒ x=3\)
So, original number \(= 12x=12\times 3=36\)
Thus, \(50\)% of the number \(= \frac{50}{100}\times 36=18\)
Hence the right answer is Option C.
