model 2 net increase or decrease in % Section-Wise Topic Notes With Detailed Explanation And Example Questions

MOST IMPORTANT quantitative aptitude - 11 EXERCISES

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The following question based on percentage topic of quantitative aptitude

Questions : If the numerator of a fraction is increased by 20% and the denominator is decreased by 5%, the value of the new fraction becomes $5/2$. The original fraction is

(a) $3/18$

(b) $48/95$

(c) $95/48$

(d) $24/19$

The correct answers to the above question in:

Answer: (c)

Let original fraction be $x/y$

According to the question,

${120/100x}/{{95y}/100} = 5/2$

${120x}/{95y} = 5/2$

$x/y = 5/2 × 95/120 = 95/48$

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Read more net increase or decrease Based Quantitative Aptitude Questions and Answers

Question : 1

The price of an article is reduced by 25% but the daily sale of the article is increased by 30%. The net effect on the daily sale receipts is

a) 2$1/2$% decrease

b) 2% decrease

c) 2% increase

d) 2$1/2$% increase

Answer: (a)

Let the price of the article be Rs.100 and the daily sale be 100 units.

Revenue day = 100 × 100 = Rs.10000

New receipts = 75 × 130 = Rs.9750

Decrease = Rs.(10000 – 9750) = Rs.250

% decrease = $250/10000 × 100 = 2{1}/2%$

Using Rule 5,

If a number is decreased by a% and then it is increased by b%, then net increase or decrease per cent is $(-a + b - {ab}/100)$% (Negative sign for decrease; Positive sign for increase)

Required change = $(a - b - {ab}/100)$

= $(30 - 25 – {30 × 25}/100)%$

= $(5 - 15/2) = –2.5% = 2{1}/2%$ decrease.

Question : 2

A number is first decreased by 10% and then increased by 10%. The number so obtained is 50 less than the original number. The original number is

a) 5000

b) 5050

c) 5500

d) 5900

Answer: (a)

Let the original number be x.

$90/100x × 110/100 = x – 50$

${99x}/100 =x - 50$

$x - {99x}/100 = 50$

$x/100 = 50 ⇒ x = 5000$

Using Rule 3,

If an amount is increased by a% and then it is reduced by a% again, then percentage change will be a decrease of $a^2/100 %$

Let the number be x

Decrease % = $10^2/100% = 1%$

x – 1% of x =x – 50

$x/100 = 50$ ⇒ x = 5000

Question : 3

The number of employees working in a farm is increased by 25% and the wages per head are decreased by 25%. If it results in x % decrease in total wages, then the value of x is

a) 25%

b) $25/4$%

c) 20%

d) 0%

Answer: (b)

Let the original number of employees be 100 and wages per head be Rs.100.

Total wages = Rs.(100 × 100) = Rs.10000

New number of employees = 125

New wages per head = Rs.75

Total new wages = Rs.(125 × 75) = Rs.9375

Decrease = Rs.(10000 – 9375) = Rs.625

Percentage decrease =$625/10000 × 100$

=$625/100 = 25/4%$

Question : 4

The strength of a school increases and decreases in every alternate year by 10%. It started with increase in 2000. Then the strength of the school in 2003 as compared to that in 2000 was

a) decreased by 8.9%

b) decreased by 9.8%

c) increased by 9.8%

d) increased by 8.9%

Answer: (d)

Using Rule 4,

If a number is increased by a% and then it is decreased by b%, then the resultant change in percentage will be $(a - b - {ab}/100)$% (Negative for decrease, Positive for increase)

Increase in first year = 10%

Decrease in 2nd year = 10%

Effective result = $(10 - 10 - {10 × 10}/100)% = -1%$

Increase in 3rd year = 10%

Effective result = $(10 - 1 - {10 × 1}/100)$%

= (9 – 0.1)% = 8.9% (increase)

Question : 5

The sum of two numbers is 520. If the bigger number is decreased by 4% and the smaller number is increased by 12%, then the numbers obtained are equal. The smaller number is

a) 210

b) 300

c) 240

d) 280

Answer: (c)

Larger number = x and smaller number = 520 – x

${96x}/100 = {520 - x}/100 × 112$

96x = 520 × 112 – 112x

112x + 96x = 520 × 112

208x = 520 × 112

$x = {520 × 112}/208 = 280$

Smaller number = 520 – 280 = 240

Question : 6

A number is first decreased by 20%. The decreased number is then increased by 20%. The resulting number is less than the original number by 20. Then the original number is

a) 400

b) 600

c) 500

d) 200

Answer: (c)

Effective percentage = $(-20 + 20 - {20 × 20}/100)$= - 4%

If the number be x, then 4% of x = 20

$x × 4/100 = 20$

$x = {20 × 100}/4 = 500$

Using Rule 3,

If an amount is increased by a% and then it is reduced by a% again, then percentage change will be a decrease of $a^2/100%$

Let the number be x

Decrease % = $20^2/100$ = 4%

x – 4% of x = x – 20

${4x}/100 = + 20 ⇒ x = 500$

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GET percentage PRACTICE TEST EXERCISES

model 1 simple percentage questions

model 2 net increase or decrease in %

model 3 reducing & exceeding prices

model 4 x & y comparison

model 5 income & expenditure

model 6 consumption & remaining

model 7 marks scored in examinations

Model 8 population based

model 9 voters & election

model 10 percentage with allegations & mixture

model 11 percentage with ratios
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