Question

Two numbers are such that if 7 is added to the first
number, a number twice the second number is
obtained. If 20 is added to the second number,
the number obtained is four times the first number.
Find the two numbers.

Answer 1

Answer:

$\large\bold\red{\frac{47}{7}}\: and\:\bold\red{\frac{48}{7}}$

Step-by-step explanation:

Let,

the first number be 'x'

and

the second numbers be 'y'

Now,

According to the Question,

if 7 is added to the first number, a number twice the second number is obtained.

Therefore,

$7 + x = 2y \\ \\ = > x = (2y - 7) \: \: \: \: \: \: \: \: \: \: \: \: ............(i)$

Also,

If 20 is added to the second number, the number obtained is four times the first number.

Therefore,

$y + 20 = 4x \\ \\ = > y + 20 = 4(2y - 7) \\ \\ = > y + 20 = 8y - 28 \\ \\ = > 8y - y = 28 + 20 \\ \\ = > 7y = 48 \\ \\ = > y = \frac{48}{7}$

Therefore,

$= > x = 2 \times \frac{48}{7} - 7 \\ \\ = \frac{96 - 49}{7} \\ \\ = \frac{47}{7}$

Hence,

the required numbers are

$\large\bold{\frac{47}{7}}\: and\:\bold{\frac{48}{7}}$