Question

A two digit number is obtained by either multiplying the sum of the digits by 8 and then subtracting 5 or by multiplying the difference of the digits by 16 and then adding 3.Find the number.​

Answer 1

Answer:-

Let the number be (10x + y).

Given:

Case - 1:-

It is obtained when multiplying the sum of the digits by 8 and subtracting 5 from the result.

That is,

→ 8(x + y) - 5 = 10x + y

→ 8x + 8y - 5 = 10x + y

→ 8x - 10x + 8y - y = 5

→ 7y - 2x = 5

→ 7y = (5 + 2x)

→ y = (5 + 2x)/7 -- equation (1)

Case - 2:-

It is obtained by multiplying the difference of the digits by 16 and adding 3.

→ 16(x - y) + 3 = 10x + y

→ 16x - 16y + 3 = 10x + y

→ 16x - 10x - 16y - y = - 3

→ 6x - 17y = - 3

On Putting the value of "y" we get,

→ 6x - 17(5 + 2x)/7 = - 3

→ (42x - 85 - 34x)/7 = - 3

→ 8x - 85 = (- 3)(7)

→ 8x = - 21 + 85

→ 8x = 64

→ x = 64/8

→ x = 8

Substitute "X" value in equation (1).

→ y = (5 + 2x)/7

→ y = 5 + 2(8)/7

→ y = 21/7

→ y = 3

Therefore, the number (10x + y) = 83.

Answer 2

Hey !

Given,

It's a 2-digit No.

Let the Unit digit of the no. be x

and Tens digit be y

No. is obtained by adding the digits with their place value

=> No. = 10(y) + 1(x)

Sum of the digits = x+y

As per the question,

No. can obtained by either multiplying the sum of the digits by 8 and then subtracting 5

=> 10y+x = 8(x+y) -5 ---------[1]

No. can also be obtained by multiplying the difference of the digits by 16 and then adding 3

Difference of the digits :

Here, to make sure result is positive, we assume that x>y and proceed with the difference as y-x

=> 10y+x = 16(y-x)+3 --------[2]



From (1) => 10y+x = 8x + 8y -5

=> 0 = 8x + 8y - 5 - 10y - x

=> 7x -2y -5 = 0 ---------------[3]

From (2)=> 10y+x = 16y-16x+3

=> 10y+x-16y+16x-3=0

=> 17x-6y- 3 =0------------------[4]



Now we have 2 unknown x and y,

To solve them we have 2 equations [3] and [4], which is enough to proceed further

Do 3×[3]

=> 3(7x-2y-5=0) => 21x - 6y -15 = 0

Subtract [4] from above

=> 21x-6y-15 - (17x -6y-3) = 0

=> 4x -12 = 0

=> 4x = 12

=> x= 3



Substitute x in [3]

=> 7(3)-2y-5 = 0

=> 21- 2y -5 =0

=> 2y = 16

=> y = 8

•°• No. = 10y +x =10(8)+3 = 83



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