Question

four amount money are such that A is 20% of B and B is 20% more than C and C is 1/3 time more than D then find 20% of 2A/B+C?

Answer 1

Answer:

14/425

Step-by-step explanation:

C = D + D/3 = 4D/3

B = 2D/3 + 1/5 * 4D/3 = 14D/15

A = 1/5 * 14D/15 = 14D/75

A/B = 15/75 = 1/5

B/C = 14D/4D * 3/15 = 7/10

A : B : C = 7 : 35 : 50

20% of [2(7)/(35+50)

= 1/5 * 14/85 = 14/425

Answer 2

Given: The four amounts of money are such that A is 20% of B and B is 20% more than C and C is 1/3 time more than D.

Let amount D be $x$.

According to the given statement,

$C=\frac{1}{3}D$

$C=\frac{1}{3}x$

$B=120\%of C$

$B=1.20\times \frac{1}{3}x$

$B=0.40\times x$

$A=20\%of B$

$A=0.20\times 0.4x\\A=0.08x$

Then,

$20\% of (2\frac{A}{B+C})=0.20 (2\frac{0.08x}{0.4x+\frac{x}{3}})$

$20\% of (2\frac{A}{B+C})=0.20 (\frac{0.16x}{0.4x+0.33x})$

$20\% of (2\frac{A}{B+C})=0.20 (\frac{0.16x}{0.73x})$

$20\% of (2\frac{A}{B+C})=0.0438$

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