Question

In a two digit number , the units digit is 3 more than the ten's digit. The sum of the digit is 18 less than the original number. The products of digit is ___________

Answer 1

Answer:

B

36

10's digit 1's digit

x x+3

The number = 10x+x+3=11x+3

Interchanging the digits

10'sdigit 1's digit

x+3 x

Then the number =10x+30+x=11x+30

so

11x+3

11x+30

=

4

7

44x+120=77x+21

33x=99

x=3

∴ The number =36

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

• In a two digit number , the units digit is 3 more than the ten's digit.

Let assume that

• Digit at tens place be x.

So,

• Digit at unit place is x + 3

So,

Number formed = 10 × x + 1 × (x + 3) = 10x + x + 3 = 11x + 3

According to statement

The sum of the digit is 18 less than the original number.

$\rm :\longmapsto\:x + (3 + x) = 11x + 3 - 18$

$\rm :\longmapsto\:x + 3 + x = 11x- 15$

$\rm :\longmapsto\:2x + 3 = 11x- 15$

$\rm :\longmapsto\:2x - 11x = - 3- 15$

$\rm :\longmapsto\: - 9x = - 18$

$\bf\implies \:x = 2$

Thus,

• Digit at tens place, x = 2

• Digit at ones place, x + 3 2 + 3 = 5

So,

• Product of digits = 2 × 5 = 10

Verification

Now,

➢ Digit at tens place = 2

➢ Digit at ones place = 5

➢ So, Number formed = 25

➢ Now, Sum of digits = 2 + 5 = 7

➢ Now, 25 - 7 = 18 implies In a two digit number, the sum of the digit is 18 less than the original number.