Aptitude - Problems on Numbers - Discussion
Discussion Forum : Problems on Numbers - General Questions (Q.No. 4)
4.
The difference between a two-digit number and the number obtained by interchanging the digits is 36. What is the difference between the sum and the difference of the digits of the number if the ratio between the digits of the number is 1 : 2 ?
Answer: Option
Explanation:
Since the number is greater than the number obtained on reversing the digits, so the ten's digit is greater than the unit's digit.
Let ten's and unit's digits be 2x and x respectively.
Then, (10 x 2x + x) - (10x + 2x) = 36
9x = 36
x = 4.
Required difference = (2x + x) - (2x - x) = 2x = 8.
Discussion:
46 comments Page 2 of 5.
Nathi said:
2 decades ago
At first we should decide the digits of 10's & unit's be X & 2X respectively.
((10 x 2X) + X) - ((10 x X) + 2X) = 36
(20X + X) - (10X + 2X) = 36
21X - 12X = 36
9X = 36
X = 4
The difference between two digits = (2X+X)-(2X-x)
= 3X - X
= 2X
= 2(4)
= 8
((10 x 2X) + X) - ((10 x X) + 2X) = 36
(20X + X) - (10X + 2X) = 36
21X - 12X = 36
9X = 36
X = 4
The difference between two digits = (2X+X)-(2X-x)
= 3X - X
= 2X
= 2(4)
= 8
(2)
Jonsihay said:
1 decade ago
The Problem should be re-stated specially on the aspect of the ratio.
It should "NOT JUST BE" 'if the ratio between the digits of the number is 1:2. But it should be restated as 'if the ratio between the unit's and ten's digits of the number is 1:2.
Then if that's ok, I think reliably be solvable!
It should "NOT JUST BE" 'if the ratio between the digits of the number is 1:2. But it should be restated as 'if the ratio between the unit's and ten's digits of the number is 1:2.
Then if that's ok, I think reliably be solvable!
(1)
Kolkata said:
7 years ago
I am not understanding it, please someone explain me.
(1)
Md Parvez Islam said:
6 years ago
Thanks @Chinnadurai.
(1)
Sravanthi said:
2 decades ago
Hi I can't understand this problem. Any more explnation please.
(1)
Sikha said:
2 decades ago
It can't b (x+y) - (x-y).
Rather it could be (10x +y) - (10y + x).
Rather it could be (10x +y) - (10y + x).
(1)
Ramya said:
2 decades ago
Hi I could not understand this problem. Can anyone explain this more clearly please?
(1)
Ajeya said:
8 years ago
Lets assume n1=10a+b and n2 = a+10b where a and b are digits in a two digit number,given n1-n2 = 36,
So, 10a+b-a-10b = 36,
9a-9b = 36,
a-b = 4.
How can it be 8? Please help.
So, 10a+b-a-10b = 36,
9a-9b = 36,
a-b = 4.
How can it be 8? Please help.
Karthik said:
2 weeks ago
Let X = tens place, Y = units place.
(10x+y)-(10y+x) = 36,
9(x-y) = 36,
x-y = 4.
Given X:Y = 1:2.
Then y = x/2.
Put y = x/2 in eqn (x-y) = 4.
Then x = 8.
y = 4.
Then the sum of digits is (x+y) = 8 + 4 = 12.
The difference is (x-y) = 8-4 = 4.
Then the difference between those is 12 - 4 = 8.
(10x+y)-(10y+x) = 36,
9(x-y) = 36,
x-y = 4.
Given X:Y = 1:2.
Then y = x/2.
Put y = x/2 in eqn (x-y) = 4.
Then x = 8.
y = 4.
Then the sum of digits is (x+y) = 8 + 4 = 12.
The difference is (x-y) = 8-4 = 4.
Then the difference between those is 12 - 4 = 8.
Sarita said:
8 years ago
Thank you for the answer @Anju.
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