Question

The sum of a two-digit number and the number obtained by interchanging its digits is 110. If the original number exceeds six times the sum of its digits by 4, find the original number. Solve the equations by the
cross-multiplication method
​

Answer 1

Answer:

The sum of a two digit number and the number formed by interchanging its digits is 110. If 10 is subtracted from the first number, the new number is 4 more than 5 times the sum of its digits in the first number. Find the first number

ANSWER

let the unit place digit be x and tens place digit be y

then the two-digit number will be 10y + x

and the number formed by interchanging the unit place and tens place digits will be 10x + y

according to the first condition given in the qs i.e, the sum of two numbers is 110 that is

10y + x + 10x + y = 110

=> 11x + 11y = 110

divide the above equation by 11 we get

x + y = 10

x = 10 - y ....(i)

now according to the second equation,

if 10 is subtracted from the first number i.e, the new number is 10y + x - 10

given that the new number is 4 more than 5 time the sum of its digits in the first number i.e

the sum of its digits in the first number is x + y, now 5 times of its, 5(x + y), and now 4 more that is, 4 + 5(x + y)

therefore new number = 4 + 5(x + y)

10y + x - 10 = 4 +5(x + y)

10y - 5y + x = 4 +10 +5x

5y = 14 + 4x.....(ii)

substitute the value of x from eq(i) to eq (ii)

we get , 5y = 14 + 4(10 - y)

5y = 14 + 40 - 4y

y = 6

and from eq(i)

x = 4

then the first number 10y + x = 10x6 + 4 = 64

first number is 64.

Answer 2

Answer:

The original no. will be 64 or 46

Step-by-step explanation:

According to first statement

∴10x + y + 10x + y = 110

∴11x + 11y = 110

Dividing by 11

∴x + y = 10   ---(1)

Now, according to second statement

10x+y=6(x+y) + 4   or   10y+x=6(y+x) +4

First for 10x+y=6(x+y) + 4

∴10x+y=6x+6y+4

∴10x-6x=6y-y+4

∴4x=5y+4

∴4x-5y=4   ---(2)

From (1)

x+y=10

Solving (1) and (2) by cross multiplication method

a1= 1                             a2= 4

b1= 1                            b2= -5

c1= -10                         c2= -4

∴  $\frac{x}{b1c2-b2c1} = \frac{y}{c1a2-c2a1} = \frac{1}{a1b2-a2b1}$

∴   $\frac{x}{(1)(-4) - (-5)(-10)} = \frac{y}{(-10)(4)-(-4)(1)} = \frac{1}{(1)(-5)-(4)(1)}$

∴ $\frac{x}{-4-50} = \frac{y}{-40+4} = \frac{1}{-5-4}$

∴ $\frac{x}{-54} = \frac{y}{-36} = \frac{1}{-9}$

Now, taking

∴  $\frac{x}{-54} = \frac{1}{-9}$

∴ -9x=-54

∴ $x=\frac{-54}{-9}$

∴ x= 6

Now,

∴ $\frac{y}{-36} = \frac{1}{-9}$

∴$y=\frac{-36}{-9}$

∴y=4

Therefore, the number is 10x+y=64

Similarly if we take

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

4y-5x=1

we get the answer

Hope this gives your answer