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What number should be subtracted from each of the numbers 23, 30, 57 and 78 so that the resultant numbers are in proportion ?

A4

B3

C6

D7

Answer:

C. 6

Read Explanation:

The number that should be subtracted is 6.

Let the number to be subtracted from each term be xx.

According to the problem, the resulting numbers (23x)(23-x), (30x)(30-x), (57x)(57-x), and (78x)(78-x) must be in proportion:
23x30x=57x78x\frac{23 - x}{30 - x} = \frac{57 - x}{78 - x}

Step 1: Cross-multiply the fractions
(23x)(78x)=(57x)(30x)(23 - x)(78 - x) = (57 - x)(30 - x)

Step 2: Expand both sides

  • Left side: 179423x78x+x2=1794101x+x21794 - 23x - 78x + x^2 = 1794 - 101x + x^2

  • Right side: 171057x30x+x2=171087x+x21710 - 57x - 30x + x^2 = 1710 - 87x + x^2

Now set them equal:
1794101x+x2=171087x+x21794 - 101x + x^2 = 1710 - 87x + x^2

Step 3: Simplify and solve for xx
Cancel x2x^2 from both sides:
1794101x=171087x1794 - 101x = 1710 - 87x

Rearrange the terms to bring xx to one side:
17941710=87x+101x1794 - 1710 = -87x + 101x
84=14x84 = 14x
x=8414=6x = \frac{84}{14} = 6


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