A two-digit number is seven times the sum of its digits. The number formed by reversing the digits is 6 more than half of the original number. Find the difference of the digits of the given number.

Question

A two-digit number is seven times the sum of its digits. The number formed by reversing the digits is 6 more than half of the original number. Find the difference of the digits of the given number. (A) 2 (B) 3 (C) 4 (D) 5

Tardigrade Admin · Accepted Answer

Let the two-digit number be 10 x+&gamma; Given, 10 x+y=7(x+&gamma;) ⇒ 3 x=6 &gamma; ⇒ x=2 &gamma; Also, 10 y+x=(1/2)(10 x+y)+6 ⇒-8 x+19 y=12 ⇒-16 y+19 y=12 (∵ text from Eq. (1)) ⇒ y=4 text and x=8 ∴ x-y=4

Q. A two-digit number is seven times the sum of its digits. The number formed by reversing the digits is 6 more than half of the original number. Find the difference of the digits of the given number.

2

3

4

5

Let the two-digit number be $10 x + γ$
Given, $10 x + y = 7 (x + γ)$
$\Rightarrow 3 x = 6 γ \Rightarrow x = 2 γ$
Also, $10 y + x = \frac{1}{2} (10 x + y) + 6$
$\Rightarrow - 8 x + 19 y = 12$
$\Rightarrow - 16 y + 19 y = 12 (∵ from Eq. (1))$
$\Rightarrow y = 4 and x = 8$
$∴ x - y = 4$

Q. A two-digit number is seven times the sum of its digits. The number formed by reversing the digits is 6 more than half of the original number. Find the difference of the digits of the given number.

2

3

4

5

Let the two-digit number be 10x+γ Given, 10x+y=7(x+γ) ⇒3x=6γ⇒x=2γ Also, 10y+x=21​(10x+y)+6 ⇒−8x+19y=12 ⇒−16y+19y=12(∵ from Eq. (1)) ⇒y=4 and x=8 ∴x−y=4

Let the two-digit number be $10 x + γ$
Given, $10 x + y = 7 (x + γ)$
$\Rightarrow 3 x = 6 γ \Rightarrow x = 2 γ$
Also, $10 y + x = \frac{1}{2} (10 x + y) + 6$
$\Rightarrow - 8 x + 19 y = 12$
$\Rightarrow - 16 y + 19 y = 12 (∵ from Eq. (1))$
$\Rightarrow y = 4 and x = 8$
$∴ x - y = 4$