In a two digit number, the digit in ten's place
exceeds the digit in unit's place by 4. The sum of the digits is 1/7 of the number, then
the digit at the unit's place is
(1) 5
(2) 4
(3) 3
(4) 1
Answers
Answer:
(2). 4
Step-by-step explanation:
So if x is the tens digit, and y is the ones digit, we can write two equations.
The first equation is x=y+4.
The second equation is 7(x+y)=10x+y.
We can plug in (y+4) for x in the second equation.
You would have 7[(y+4)+y]=10(y+4)+y. Expanding would be 7(2y+4)=10y+40+y. Further expanding and simplifying would get you 14y+28=11y+40. Moving all the terms with x to the left side and moving all the constants to the right side would give you 14y-11y=40-28. Simplifying would get you 3y=12. By dividing both sides by three, you get y=4.
Answer:
Let the digit in the units place =x
Let the digit in the tens place =x+4
∴ The number is 10(x+4)+x=10x+40+x=11x+40
Sum of the digits =x+x+4=2x+4
Given, 2x+4=
7
1
(11x+40)
⇒14x+28=11x+40
⇒3x=12
⇒x=4
∴ the number is 11×4+40=44+40=84.
Step-by-step explanation:
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