Question

Find the numbers which are in the ratio 3:2:4 such that the sum of the first and the second numbers added to the difference of the third and the second numbers is 21.
(A) 12, 8, 16
(C) 6, 4,8
(C) 9,6, 24
(D) 9,6, 12​

Answer 1

AnswEr:-

→Let number be 3x,2x,4x.

⇒3x+2x+4x-2x=21

⇒7x=21

⇒ x=3

Numbers are,

⇒  3*3=9

⇒2*3=6

⇒4*3=12 .

Hence 9,6,12 (option D) is the ans.

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Step-by-step explanation:

Answer 2

The numbers are 9 , 6 , 12 [ The correct option is (D) 9 , 6 , 12 ]

Given :

Three numbers are in the ratio 3 : 2 : 4 such that the sum of the first and the second numbers added to the difference of the third and the second numbers is 21.

To find :

The numbers are

(A) 12 , 8 , 16

(C) 6 , 4 , 8

(C) 9 , 6 , 24

(D) 9 , 6 , 12

Solution :

Step 1 of 2 :

Form the equation to find the numbers

Here it is given that three numbers are in the ratio 3 : 2 : 4

Let the numbers are 3x , 2x , 4x

Now it is given that sum of the first and the second numbers added to the difference of the third and the second numbers is 21

So by the given condition

$\displaystyle \sf (3x + 2x) + (4x - 2x) = 21$

Step 2 of 2 :

Find the numbers

$\displaystyle \sf (3x + 2x) + (4x - 2x) = 21$

$\displaystyle \sf{ \implies }5x + 2x = 21$

$\displaystyle \sf{ \implies }7x = 21$

$\displaystyle \sf{ \implies }x = \frac{21}{7}$

$\displaystyle \sf{ \implies }x = 3$

First number = 3x = 3 × 3 = 9

Second number = 2x = 2 × 3 = 6

Third number = 4x = 4 × 3 = 12

So the numbers are 9 , 6 , 12

Hence the correct option is (D) 9 , 6 , 12

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