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

a)
46

b)
48

c)
64

d)
84

Correct answer is option 'C'. Can you explain this answer?

Question

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

a)
46
b)
48
c)
64
d)
84

Correct answer is option 'C'. Can you explain this answer?

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

OPTION D -11 AND 5

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

From where you got this question

Answer

Explanation:

Given Equation:
2x^2 + xy - 3y^2 + x + ay - 10 = (2x + 3y + b)(x - y - 2)

Expand the Right-hand side:
(2x + 3y + b)(x - y - 2) = 2x^2 - 2xy - 4x + 3xy - 3y^2 - 6y + bx - by - 2b
= 2x^2 + xy - 3y^2 - 4x - 6y + bx - by - 2b

Compare coefficients of like terms on both sides:
For x^2 terms: 2 = 2 (Coefficients of x^2 on both sides)
For xy terms: 1 = 1 (Coefficients of xy on both sides)
For y^2 terms: -3 = -3 (Coefficients of y^2 on both sides)
For x terms: 1 = -4 + b (Coefficients of x on both sides)
For y terms: a = -6 - b (Coefficients of y on both sides)
Constant terms: -10 = -2b (Constants on both sides)

Solve the equations:
From x terms: b = 1
From y terms: a = -6 - 1 = -7
From constant terms: -10 = -2(1) => -10 = -2
This is not possible, so there might be a mistake in the given question.
Therefore, the correct values of a and b cannot be determined from the given equation.

Answer

11

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