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02 Aug 2021, 03:15
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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:
94% (01:30) correct
6%
(01:42)
wrong
based on 150
sessions
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The difference between a two-digit number and the number obtained by interchanging the digits is 18. The sum of the digits is 10 and the digit at the ten’s place is bigger than the digit at the unit’s place, What is the two-digit number?
(A) 28
(B) 46
(C) 64
(D) 73
(E) 82
(A) 28
(B) 46
(C) 64
(D) 73
(E) 82
02 Aug 2021, 03:20
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Explanation:
10x+y - 10y - x = 18
9x - 9y = 18
x-y = 2 --- Eqn 1
x+y = 10 --- Eqn 2
From Eqn 1 & 2,
2x = 12
x = 6
y = 4
number = 64
IMO-C
10x+y - 10y - x = 18
9x - 9y = 18
x-y = 2 --- Eqn 1
x+y = 10 --- Eqn 2
From Eqn 1 & 2,
2x = 12
x = 6
y = 4
number = 64
IMO-C
02 Aug 2021, 04:41
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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.
And the number obtained by interchanging the digits will be \(= 10y+x\).
We are given that their difference is \(18\). Thus we can say \(10x+y-(10y+x)=18\)
\(⇒ 10x+y-10y-x=18\)
\(⇒ 9x-9y=18\)
\(⇒ 9(x-y)=18\)
\(⇒ x-y=\frac{18}{9}\)
\(⇒ x-y=2\)..... \((i)\)
We are given that the sum of the digits is \(10\). Thus we can say \(x+y=10\)..... \((ii)\)
Adding \(eq(i)\) and \(eq(ii)\), we get: \(x-y+x+y=2+10\)
\(⇒ 2x=12\)
\(⇒ x=\frac{12}{2}\)
\(⇒ x=6\)
Plugging \(x=6\) in equation \((i)\), we get \(x-y=2\)
\(⇒ 6-y=2\)
\(⇒ y=6-2\)
\(⇒ y=4\)
We have both the digits now. Thus the number \(= 10x+y=10\times 6+4=64\)
Hence the right answer is Option C.
Let us assume a two-digit number be \(10x+y\). Where \(x\) and \(y\) are ten's and unit's digit respectively.
And the number obtained by interchanging the digits will be \(= 10y+x\).
We are given that their difference is \(18\). Thus we can say \(10x+y-(10y+x)=18\)
\(⇒ 10x+y-10y-x=18\)
\(⇒ 9x-9y=18\)
\(⇒ 9(x-y)=18\)
\(⇒ x-y=\frac{18}{9}\)
\(⇒ x-y=2\)..... \((i)\)
We are given that the sum of the digits is \(10\). Thus we can say \(x+y=10\)..... \((ii)\)
Adding \(eq(i)\) and \(eq(ii)\), we get: \(x-y+x+y=2+10\)
\(⇒ 2x=12\)
\(⇒ x=\frac{12}{2}\)
\(⇒ x=6\)
Plugging \(x=6\) in equation \((i)\), we get \(x-y=2\)
\(⇒ 6-y=2\)
\(⇒ y=6-2\)
\(⇒ y=4\)
We have both the digits now. Thus the number \(= 10x+y=10\times 6+4=64\)
Hence the right answer is Option C.
