there are two 2 digit numbers such that the tens digit of the first no. is 3/2 times the tens digit of the second no., while the sum of the two no. is 158. which of the following can be the difference between them ?
Answers
Let the numbers be ab and cd
a = 1.5c
10a+b+10c+d = 158
Combination of a and c can be a=9 and c=6
So the two numbers are 9b and 6d. Maximum difference could be 38.
Given:
There are two 2 digit numbers such that the tens digit of the first number is 3/2 times the tens digit of the second number, while the sum of the two number is 158.
To Find:
What is possible difference of two number?
Solution:
Let:
First number = X Y
Where X is ten digit and Y is unit digit of first number:
Second number = A B
Where A is ten digit and B is unit digit of first number:
Given that the tens digit of the first number is 3/2 times the tens digit of the second number:
...1)
Given that the sum of the two number is 158, means:
10 X + Y + 10 A + B = 158 ...2)
From equation 1) and equation 2):
10× 1.5A + Y + 10A + B = 158
15 A + Y + 10 A + B = 158
25 A + Y + B = 158 ...3)
Where Y and B is whole number from zero to nine, means:
0 ≤ Y or B ≤ 9 ...4)
Equation 3) is only valid, when A is equal to:
A = 6
and
X = 1.5 A = 1.5× 6 =9
X = 9
Then:
Y + B = 158 - 25× 6
Y + B = 8 ...5)
Now possible value of Y and B may be:
Y B
7 1
6 2
5 3
4 4
Now numbers and their difference may be:
First Number Second Number Difference Of Numbers
XY AB XY - AB =
97 61 97 - 61 = 36
96 62 96 - 62 = 34
95 63 95 - 63 = 32
94 64 94 - 64 = 30
It is clear that possible difference may be 36, 34, 32 or 30 . So option C is correct.