Question

In what proportion must a number be
added so that Ye of the first part and
1/3 of the second part are together
equal to 1 of the original number?​

Answer 1

Answer:

Let the number to added be x

Let the 1st part be p

Second part is x-p

1/4 0f the 1st part is p/4

2/3 of the 2nd part is (x-p)×2/3

p/4 +(x--p) ×2/3 = x/2

(x-p) ×2/3 =x/2 --p/4

2/3 x-'--x/2 = 2p/3 --p/4

(4x--3x)/6 =(8p- 3p) /12

x/6=5p /12

2x=5p

x/p=5/2

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Step-by-step explanation:

Let the number = x

1st part be p, so the 2nd part = x-p

p/4 + 2(x-p)/3 = x/2

( 3p + 8x - 8p ) / 12 = x/2

(8x - 5p) = 12 * x/2 = 6x

8x - 6x = 5p

x = 5p/2

The 2nd part = x-p = 5p/2 - p = 3p/2

The proportion 1st part : 2nd part = p : 3p/2 = 2p : 3p = 2 : 3