Question

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​

Answer 1

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 2

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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