Question #9c419

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Answer 1 · 2016-12-03T08:30:24

58

Let in the two digit number the tens digit be x and units digit be y .
So the number is $10 x + y$ $" "10x+y$

By the 1st condition

$" "y-x=3.......(1)$

And by the2nd condition

$10x+y-4(x+y)=6$

$=>6x-3y=6$

$=>2x-y=2.....(2)$

Adding (1) and (2) we gwt

$x=5$

Inserting $x=5$ in (1) we get

$y-5=3=>y=8$

Hence the the two digit number is $10x+y=10xx5+8=58$

Answer 2 · 2016-12-03T08:34:51

See explanation.

If we denote the tens digit by $a$ and unit digit by $b$ then the question leads to the following equations:

Tens digit is 3 less than units digit: $a=b-3$ .

The original number is six more than four times than the sum of dygits: $10a+b-6=4*(a+b)$ .

If we solve the system we get:

${(a=b-3),(10a+b-6=4a+4b):}$

${(a=b-3),(10(b-3)+b-6=4(b-3)+4b):}$

Now from the second equation we calculate $b$ :

$10b-30+b-6=4b-12+4b$

$10b+b-8b=36-12$

$3b=24$

$b=8$

So: $a=8-3=5$

Answer: The number is 58.