The units digit of a two-digit number is 5 more than the tens digit. If the digits are reversed and the new nu

Question

The units digit of a two-digit number is 5 more than the tens digit. If the digits are reversed and the new number is divided by the original, the quotient is 2 and the remainder is 7. What is the original number?

Answer 1

first digit be a,
second digit be b.

ab

b = a + 5 --------------(1)

(10b + a) = 2*(10a + b) + 7 ----------------(2)


subst (1) into (2)

10a + 50 + a = 2*(10a + a + 5) + 7

11a + 50 = 22a + 17

11a = 33

a = 3 (ans)

subst into (1),

b = 3 +5 = 8 (ans)


Therefore ab = 38

Answer 2

Howdy! :)

I sure hope you're doing well. :) As for your question, remember that for a number such as 25, beyond the fact that it is obviously 20 + 5, the 20 itself can be thought of as 10x2. Now, to simplify solving the question, we need to assign variables (letters of the alphabet) to both digits. So:

Let the number be XY, which I will re-write as 10X + Y.

When reversed, the number becomes YX, which I'll re-write as 10Y + X.

We are given that Y = X + 5 (1)

We are also given that (10Y + X)/(10X + Y) = 2 + 7/(10X + Y) (2)

To solve, we substitute (1), namely (X + 5), into (2). We get:

[10(X + 5) + X]/[10X + (X + 5)] = 2 + 7/[10X + (X + 5)]

(10X + 50 + X)/(11X + 5) = 2 + 7/(11X + 5)

(11X + 50)/(11X + 5) = 2 + 7/(11X + 5)

Now, we multiply both sides of the equation by (11X + 5) so as neither side of the equation is in the form of a fraction. We get:

(11X + 50)(11X + 5)/(11X + 5) = 2(11X + 5) + 7(11X + 5)/(11X + 5)

11X + 50 = 22X + 10 + 7

11X + 50 = 22X + 17

Now, we subtract 22X & 50 from both sides of the equation so as to group the variable expressions on the left side (out of convenience, since grouping them on the right works too!) and the numbers on the right. We get:

11X -- 22X + 50 -- 50 = 22X -- 22X + 17 -- 50

--11X = --33

X = --33/--11

X = 3

So Y = X + 5

Y = 3 + 5

Y = 8

Therefore, the original number was 38. When reversed, it becomes 83. And 83 divided by 38 yields:

......2
....-----
38| 83
...-- 76
-----------
........7

That confirms the information given in the question!


I sure hope that helped and was clear as it can be for you to understand the solution as easily as possible. :) Good luck, take care and have a great day. :)

Cheers! :)

Answer 3

let the units digit be xand hence the ten's digit is x-5
The original no. is 10(x-5)+x
=11x-50
If the digits are reversed,the no. becomes
10x+(x-5)
=11x-5
By the problem,
(11x-5)/911x-50)=quotient of 2 and remainder of 7
=>11x-5=2(11x-50)+7
=>11x-5=22x-93
=>11x-22x=-93+5
=>x=8
and y=x-5=3
therefore,the original no. is 38

Answer 4

There are 4 possibilities for the unique extensive sort: 40 9, 38, 27, sixteen opposite each and each to get: ninety 4, 80 3, seventy two, sixty one Divide each and each new by skill of each and every unique: 1R45, 2R7, 2R18, 3R13 feels like the unique is 38

Answer 5

Only 16, 27, 38 and 49 satisfy the first condition and out of them, only 38 satisfies the second condition: 83/38 is 2 7/38

Answer 6

38

83/38 = 2 rem. 7

Trial and error. There's only 4 numbers it could possibly be from the start -
16, 27, 38, 49

Answer 7

OMG You're talkin' that scary "math" stuff, aren't you?

Ackkk! I never go there, if at all possible! I'm sorry....

It really feaks with my psyche....

Good luck, to ya, though.....maybe one of the "mathematician" people can figure this out.....I generally try to avoid them, myself....cause they scare me and I don't understand the way their brains work..... LOLOL