Question

82. Three numbers A, B and Care in the ratio 1:2:3
Their average is 600. If A is increased by 10% and B
is decreased by 20%, then to get the average
increased by 5%, C will be increased by
(a) 90 (b) 100 (C) 150 (d) 180​

Answer 1

Answer:

Option d (180)

Given:

Three numbers A,B and C in the ratio 1:2:3

A and B are increased by 10% and 20% respectively

To find:

How much by C increased

Solution:-

Let A,B and C be x,2x and 3x respectively

$average = \frac{x + 2x + 3x}{3} = 600$

$\frac{6x}{3} = 600$

$6x = 1800$

$x = 300$

A=300,B=600 and C=900

A increased by 10%

$300 \times (\frac{1 + 10}{100})$

$300 \times \frac{110}{100} = 330$

B decreased by 20%

$600 \times (\frac{1 - 20}{100})$

$600 \times \frac{80}{100} = 480$

Now

Average increased by 5%

$600 \times (\frac{1 + 5}{100})$

$600 \times \frac{150}{100} =630$

•New average=630

Total average

$\frac{330 + 480 + c}{3} = 630$

$810 +c = 630 \times 3$

$810 + c = 1890$

$c = 1890 - 810$

$= 1080$

C increased by 180

As 1080-900=180

Answer 2

Answer:

$\huge{\pink{\green{\sf{Increase=180}}}}$

Step-by-step explanation:

$\huge{\pink{\green{\underline{\red{\sf{SOLUTION-}}}}}}$

▪In the given question given about three number ratio's and their average after that increase in first number decrease in second number and increase in average.

▪So, we have to find how much increase in third number.

$\underline \bold{Given : } \\ \implies Ratio \: of \: A : B : C= 1 : 2 : 3 \\ \implies Average = 600 \\ \\ \underline \bold{To \: Find : } \\ \implies Third \: number\:increase = ?$

▪According to given question :

$\implies Average = \frac{A + B + C}{3} \\ \implies 600 = \frac{x + 2x + 3x}{3} \\ \implies 1800 = 6x \\ \bold{\implies x = 300 }\\ \\ \bold{\implies A= x = 300} \\ \bold{\implies B= 2x = 2 \times 300 = 600} \\ \bold{\implies C =3 x = 3 \times 300 = 900}$

▪Increase % of A

$\implies Increased = 300 \times \frac{10}{100} \\ \implies Increased =30 \\ \bold{\implies A_{1} =300 + 30= 330} \\ \\ \bold{For \: B_{1}} \implies \\ \implies Decrease = 600 \times \frac{20}{100} \\ \implies Decrease = 120 \\ \bold {\implies B_{1} = 600 - 120 = 480} \\ \\ \bold {For \: New \: Average} \implies \\ \implies average \: Increase = 600 \times \frac{5}{100} \\ \implies average \: Increase = 30 \\ \bold{\implies new \: Average = 600 + 30 = 630 }\\ \\ \implies New \: Average = \frac{ A_{1} +B_{1} +C_{1}}{3} \\ \implies 630 = \frac{330 + 480 +C_{1}}{3} \\ \implies 810 + C_{1} = 1890 \\ \implies C_{1} = 1890 - 810 \\ \implies C_{1} = 1080 \\ \\ \implies Increase = C_{1} - C\\ \implies Increase = 1080 - 900 \\ \bold{\implies Increase = 180}$

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