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04 Aug 2021, 03:15
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B
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:
89% (01:03) correct
11%
(01:29)
wrong
based on 63
sessions
History
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Time
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Not Attempted Yet
The difference between a two-digit number and the number obtained by interchanging the two digits of the number is 9. If the sum of the two digits of the number is 15, then what is the original number?
(A) 89
(B) 87
(C) 67
(D) 65
(E) Cannot be determined.
(A) 89
(B) 87
(C) 67
(D) 65
(E) Cannot be determined.
04 Aug 2021, 03:40
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Explanation:
10x + y - 10y - x = 9
9x - 9y = 9
x-y = 1 -- Eqn 1
x+y = 15 -- Eqn 2
From eqn 1 and 2,
2x = 16
x=8
y=7
Number = 87
IMO-B
10x + y - 10y - x = 9
9x - 9y = 9
x-y = 1 -- Eqn 1
x+y = 15 -- Eqn 2
From eqn 1 and 2,
2x = 16
x=8
y=7
Number = 87
IMO-B
04 Aug 2021, 04:00
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Solution:
There is a methodical way and hit and trial method to solve this question.
Methodical Way:
Let us assume the \(2-\)digit number \(= 10x+y\). Where \(x\) and \(y\) are ten's and unit's digit respectively.
So. the number after interchanging the digit will be \(= 10y+x\)
We are given that difference between original and digit reversed number \(= 9\). Thus we can say:
\(10x+y-(10y+x)=9\)
\(⇒ 10x+y-10y-x=9\)
\(⇒ 9x-9y=9\)
\(⇒ 9(x-y)=9\)
\(⇒ x-y=1\).......\((i)\). This is our equation \(1\).
We are also given that sum of digits \(= 15\). Thus we can say:
\(⇒ x+y=15\).....\((ii)\). This is our equation \(2\).
On adding the 2 equations we get: \(x-y+x-y=1+15\)
\(⇒ 2x=16\)
\(⇒ x=\frac{16}{2}\)
\(⇒ x=8\)
Plugging \(x=8\) in equation 2 we get,
\(⇒ x+y=15\)
\(⇒ 8+y=15\)
\(⇒ y=7\)
Hence the 2-digit number \(= 87\)
Hit and Trial Method:
Since sum of digits \(= 15\), answer can be either Option B or E
Looking at the second scenario, the difference between original and digit reversed number \(= 9\). We see that \(87-78=9\)
Hence the right answer is Option B.
There is a methodical way and hit and trial method to solve this question.
Methodical Way:
Let us assume the \(2-\)digit number \(= 10x+y\). Where \(x\) and \(y\) are ten's and unit's digit respectively.
So. the number after interchanging the digit will be \(= 10y+x\)
We are given that difference between original and digit reversed number \(= 9\). Thus we can say:
\(10x+y-(10y+x)=9\)
\(⇒ 10x+y-10y-x=9\)
\(⇒ 9x-9y=9\)
\(⇒ 9(x-y)=9\)
\(⇒ x-y=1\).......\((i)\). This is our equation \(1\).
We are also given that sum of digits \(= 15\). Thus we can say:
\(⇒ x+y=15\).....\((ii)\). This is our equation \(2\).
On adding the 2 equations we get: \(x-y+x-y=1+15\)
\(⇒ 2x=16\)
\(⇒ x=\frac{16}{2}\)
\(⇒ x=8\)
Plugging \(x=8\) in equation 2 we get,
\(⇒ x+y=15\)
\(⇒ 8+y=15\)
\(⇒ y=7\)
Hence the 2-digit number \(= 87\)
Hit and Trial Method:
Since sum of digits \(= 15\), answer can be either Option B or E
Looking at the second scenario, the difference between original and digit reversed number \(= 9\). We see that \(87-78=9\)
Hence the right answer is Option B.
