Question

in a two digit number the units digit is 2 more than that of tens digit. the sum of the digits is 27 less than the number. find the product of the digits of the number.
a. 24.
b. 15.
c. 8.
d. 35.​

Answer 1

Answer:

c)

Step-by-step explanation:

Let the units digit be X

And tens digits be Y

Number=10Y+X

Units digit exceeds tens digit by 2

Hence X−Y=2

X=2+Y⟹(i)

And (10Y+X)(X+Y)=144

10XY+10Y

2

+X

2

+XY=144

11XY+10Y

2

+X

2

=144⟹(ii)

Substitute X from (i) in (ii) and get

11(2+Y)Y+10Y

2

+(2+Y)

2

=144

22Y+11Y

2

+10Y

2

+4+4Y+Y

2

=144

22X+11X

2

+40+40X+10X

2

=144

22Y

2

+26Y−140=0

11Y

2

+13Y−70=0

11Y

2

+35Y−22Y−70=0

Y(11Y+35)−2(11Y+35)=0

(y−2)(11Y+35)=0

Now either (Y−2)=0

Y=2

Or 11Y+35=0

Y=

11

−35

​

Since Y is non negative

Hence Y=2 and X=2+Y=2+2=4

Hence required number =4

Answer 2

Answer:

The correct answer of this question is $15$.

Step-by-step explanation:

Given - In two digit number the units digit is 2 more than that of tens digit and the sum of the digits is 27 less than the number.

To Find - Find the product of the digits of the number.

According tot he question ,

Let the unit digit be   $x$

and the tens digit be $x - 2$

Sum of digits is  $x + x - 2 = 2x - 2$

The number is -

$10 ( x - 2 ) + x = 10x - 20 + x\\= 11x - 20$

Now, according to the question,

$11x -20 - 27 = 2x - 2\\11x - 2x = -2 + 47 \\9x = 45 \\x = 5$

the tens digit -  x - 2

put the value of x in equation

$5 - 2 = 3$

So, the product of the digits is  $3$ × $5 = 15$

$15$ is the product of the digits of the number.

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