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a two digit number is such that the tens digit exceeds twice the units digit by 2 and the number obtained by inter changing the digits is 5 more than three times the sum of the digits.find the two digit number.

 Let the number be 10x+y.

As per question equation x= 2y+2 (equation i)

Second statement. number obtained by reversing the digits = 10y+x.

3 times the sum of the digits = 3(x+y)

As per questin 10y+x = 3(x+y)+5 Equation II

we will replace x in equation II by putting its value as 2y+2 given in equation 1

now 10y+2y+2 = 3(2y+2+y)+5

12y+2 = 3( 3y+2) +5

12y+2 = 9y+ 6 + 5

12y-9y = 11-2

3y = 9

so y = 3

After putting the value of y in equation I x would be 3*2+2 = 8

SO the number is 10*8+3 = 83.

Cheers.

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solution follows as:

Let the digit at the ten's place and unit's place bexandyrespectively.

∴ Original number = 10x+y

and Reverse number = 10y+x

According to the question,

x= 2y+ 2 ..........(i)

and 10y+x= 3 (x+y) + 5

⇒ 10y+x= 3x+ 3y+ 5

⇒ 7y= 2x+5 ..................(ii)

∴ on solving (i) and (ii), we have

⇒7y= 2(2y+ 2) + 5

⇒7y= 4y+ 4 + 5

⇒ 3y= 9

⇒y= 3

Put y= 3 in equation (i)

x= 2 × 3 + 2 = 8

∴Original number = 10x+y= 10 × 8 + 3 = 80 + 3 = 83

Thus, the two digit number is 83.

  • 17

Let the no be 10y+2y+2=9y+6+5 12y+2=9y+11 3y=9 y=3 the no 10 x8+3=83

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