Question

Twenty-four is divided into two parts such that 7 times the first part added to 5 times the second part makes 146 . Find each part
han the original number. Find the number.​

Answer 1

Answer:

11 and 13

Step-by-step explanation:

Let the original numbers be x and y

from 1st condition,

x + y = 24 -----------(1)

from 2nd condition,

7x + 5y = 146 ----------(2)

multiplying equation (1) by 7 - equation (2),we get,

2y =22

y = 11

putting value of y in equation (1) we get,

x + 11 = 24

x = 24 - 11

x = 13

Answer 2

Question :

• Twenty-four is divided into two parts such that 7 times the first part added to 5 times the second part makes 146 . Find each part of number.

Answer :

• The numbers are 11 and 13.

Solution :

• Let , two parts be x and y.

Given,

• 7 times the first part added to 5 times the second part makes 146.

$\implies \: 7x + 5y = 146 - - (eq.1)$

• Sum of divided numbers is equal to 24

$\implies \: x + y = 24 \: - - (eq.2)$

Now , solve eq(1) and eq(2) :

$\implies7x + 5y = 146 \\ \: \: \: \: x + y = 24$

$eq \: 2 \times 5 \implies \: 5x + 5y = 120$

$\implies \: 7x + 5y = 146 \\ \: \: \: \: \: \: \: \: \: \: \: \: - 5x - 5y = - 100$

By adding both equations ,

$\implies \: (7x - 5x) = (146 - 100)$

$\implies \: 2x = 46$

$\implies \: x = \dfrac{46}{2}$

$\implies \: x = \cancel \dfrac{46}{2}$

$\implies \: x = 13.$

• The value of x is 13.

Now by substituting x value in eq(2) we get ,

$x + y = 24$

$13 + y = 24$

$y = 24 - 13$

$y = 11$

• The value of y is 11.

The required numbers are 11 & 13