Successive Percentage Change MCQ Quiz - Objective Question with Answer for Successive Percentage Change - Download Free PDF

Given:

Sachin's initial salary = 100 (assume as base for simplicity)

Reduction percentage = 16%

Subsequent increase percentage = 10%

Formula used:

Final Salary = Initial Salary × (1 - Reduction Percentage) × (1 + Increase Percentage)

Percentage decrease = ((Initial Salary - Final Salary) / Initial Salary) × 100

Calculation:

Final Salary = 100 × (1 - 0.16) × (1 + 0.10)

⇒ Final Salary = 100 × 0.84 × 1.10

⇒ Final Salary = 92.4

Percentage decrease = ((100 - 92.4) / 100) × 100

⇒ Percentage decrease = (7.6 / 100) × 100

⇒ Percentage decrease = 7.6%

∴ The correct answer is option (2).

Given:

Initial value of building (P_building) = ₹31,25,000

Depreciation rate of building (r_building) = 4%

Time (t_building) = 3 years

Initial value of land (P_land) = ₹32,00,000

Appreciation rate of land (r_land) = 5%

Time (t_land) = 3 years

Formula used:

For depreciation: A = P × (1 - r/100)^t

For appreciation: A = P × (1 + r/100)^t

Calculations:

For building value after 3 years (x):

A_building = P_building × (1 - r_building/100)^t_building

⇒ A_building = 31,25,000 × (1 - 4/100)³

⇒ A_building = 31,25,000 × (0.96)³

⇒ A_building = ₹27,64,800

For land value after 3 years (y):

A_land = P_land × (1 + r_land/100)^t_land

⇒ A_land = 32,00,000 × (1 + 5/100)³

⇒ A_land = 32,00,000 × (1.05)³

⇒ A_land = ₹37,04,400

Difference (y - x):

y - x = A_land - A_building

⇒ y - x = 37,04,400 - 27,64,800 = ₹9,39,600

∴ The correct answer is option (2).

Given:

The price of an article was increased by x%. Later the new price was decreased by x%.

Latest price = ₹1

Calculation:

Let the original price be P. After increasing by x%, the price becomes:

P × (1 + x/100)

After decreasing by x%, the new price becomes:

P × (1 + x/100) × (1 - x/100)

We are given that the latest price = ₹1, so:

P × (1 + x/100) × (1 - x/100) = 1

⇒ P × (1 - x²/100²) = 1

⇒ P = 1 / (1 - x²/100²)

⇒ P = 100² / (100² - x²)

⇒ P = ₹(10000) / ((100 + x)(100 - x))

∴ The correct answer is option (1).

Given:

Initial salary reduction = 10%

Then increased by = 10%

Formula used:

Net change = -x% + x% - (x × x) ÷ 100

Where x = same percentage increase and decrease

Calculation:

⇒ Net change = -10 + 10 - (10 × 10) ÷ 100

⇒ Net change = 0 - 100 ÷ 100

⇒ Net change = -1%

∴ Ramesh's ultimate loss is 1%.

Given:

Initial Salary = \(S\)

First increase = 20%

Then decrease = 20%

Formula Used:

Final Salary = Initial Salary × (1 + Increase%) × (1 - Decrease%)

Calculation:

Initial Salary = \(S\)

After 20% increase:

\(S_{\text{new}} = S × (1 + \frac{20}{100})\)

\(S_{\text{new}} = S × 1.20\)

After 20% decrease on new salary:

\(S_{\text{final}} = S_{\text{new}} × (1 - \frac{20}{100})\)

\(S_{\text{final}} = S × 1.20 × 0.80\)

\(S_{\text{final}} = S × 0.96\)

Change in Salary = Final Salary - Initial Salary

Change in Salary = \(S × 0.96 - S\)

Change in Salary = \(S(0.96 - 1)\)

Change in Salary = \(S(-0.04)\)

Percentage change in Salary = \(\frac{-0.04S}{S} × 100\)

Percentage change in Salary = -4%

The correct answer is option 1: 4% decreased.

Given:

During the first year, the strength of school increased by 12%

It decreases by 12% in the second year and in the third year it increase by 10%

Strength at the end of third year is nearly 10842.

Calculation:

Given:

30% increased in 2019

15% increased i. 2020

40% decreased in 2021

Mask manufactured in 2021 = 179400

Formula used:

Increased % = 100 + increased

Decreased %  = 100 - decreased

Calculations:

⇒ Mask manufactured in 2019 = X × (100 + 30) = X × 130%

⇒ Mask manufactured in 2020 = (130%X) × (100 + 15) = 115% (130% of X)

⇒ Mask manufactured in 2021 = 115% (130% of X) × (100 - 40) = 60% ( 115%( 130% of X))

But according to the question,

⇒ Manufactured in 2021 = 179400

So,

⇒ 60%{115%(130% of X)} = 179400

⇒ \(\frac{X × 60×115×130}{100×100×100}= 179400\)

⇒ X =  \(\frac{179400 × 100×100×100}{60×130×115} \) 

⇒ X = 200000

⇒ Hence, The manufacture  of mask in 2018 is 200000

Shortcut Trick

So, 897 unit → 179400

then, 1000 unit → 179400/897 × 1000 = 200000

Given:

The cost of apples is increased by 20% and then decreased by 20%.

Calculation:

Let the price of apples be Rs. 100

After increase it is 100 × 120%

⇒ Rs. 120

After decrease it is 120 × 80%

⇒ Rs. 96

So, decrease = 100 - 96

⇒ Rs. 4

% of decrease = (4/100) × 100

⇒ 4%

∴ The net percentage decrease is 4%.

Shortcut Trick

Given:

Increment in income for the first month = 30%

Decrement in income for the second month = 30%

Formula used:

Effective percentage change in salary = Inc. - dec. - (inc × dec)/100

Where, Inc = increment%; dec = decrement%

Calculation:

Effective percentage change in salary = Inc. - dec. - (inc × dec)/100

⇒ 30 - 30 - (30 × 30)/100

⇒ 0 - 9 = - 9%

Note: (- Ve) sign denotes the decrement in salary

∴ The correct option is 4.

Given:

Initial Salary = \(S\)

First increase = 20%

Then decrease = 20%

Formula Used:

Final Salary = Initial Salary × (1 + Increase%) × (1 - Decrease%)

Calculation:

Initial Salary = \(S\)

After 20% increase:

\(S_{\text{new}} = S × (1 + \frac{20}{100})\)

\(S_{\text{new}} = S × 1.20\)

After 20% decrease on new salary:

\(S_{\text{final}} = S_{\text{new}} × (1 - \frac{20}{100})\)

\(S_{\text{final}} = S × 1.20 × 0.80\)

\(S_{\text{final}} = S × 0.96\)

Change in Salary = Final Salary - Initial Salary

Change in Salary = \(S × 0.96 - S\)

Change in Salary = \(S(0.96 - 1)\)

Change in Salary = \(S(-0.04)\)

Percentage change in Salary = \(\frac{-0.04S}{S} × 100\)

Percentage change in Salary = -4%

The correct answer is option 1: 4% decreased.

Given:

Initial value of building (P_building) = ₹31,25,000

Depreciation rate of building (r_building) = 4%

Time (t_building) = 3 years

Initial value of land (P_land) = ₹32,00,000

Appreciation rate of land (r_land) = 5%

Time (t_land) = 3 years

Formula used:

For depreciation: A = P × (1 - r/100)^t

For appreciation: A = P × (1 + r/100)^t

Calculations:

For building value after 3 years (x):

A_building = P_building × (1 - r_building/100)^t_building

⇒ A_building = 31,25,000 × (1 - 4/100)³

⇒ A_building = 31,25,000 × (0.96)³

⇒ A_building = ₹27,64,800

For land value after 3 years (y):

A_land = P_land × (1 + r_land/100)^t_land

⇒ A_land = 32,00,000 × (1 + 5/100)³

⇒ A_land = 32,00,000 × (1.05)³

⇒ A_land = ₹37,04,400

Difference (y - x):

y - x = A_land - A_building

⇒ y - x = 37,04,400 - 27,64,800 = ₹9,39,600

∴ The correct answer is option (2).

Given:

Initial salary reduction = 10%

Then increased by = 10%

Formula used:

Net change = -x% + x% - (x × x) ÷ 100

Where x = same percentage increase and decrease

Calculation:

⇒ Net change = -10 + 10 - (10 × 10) ÷ 100

⇒ Net change = 0 - 100 ÷ 100

⇒ Net change = -1%

∴ Ramesh's ultimate loss is 1%.

Given:

During the first year, the strength of school increased by 12%

It decreases by 12% in the second year and in the third year it increase by 10%

Strength at the end of third year is nearly 10842.

Calculation:

Given:

A number is increased by 12%, then by 23%, and then decreased by 34%.

Formula Used:

For successive percentage changes:

Final Value = Initial Value × (1 + Percentage Change/100)ⁿ

Calculation:

Let the initial number be 100 (for simplicity).

After a 12% increase:

New Value = 100 × (1 + 12/100)

⇒ New Value = 100 × 1.12

⇒ New Value = 112

After a 23% increase:

New Value = 112 × (1 + 23/100)

⇒ New Value = 112 × 1.23

⇒ New Value = 137.76

After a 34% decrease:

New Value = 137.76 × (1 - 34/100)

⇒ New Value = 137.76 × 0.66

⇒ New Value = 90.9216

Net change in value = Final Value - Initial Value

Net change in value = 90.9216 - 100

Net change in value = -9.0784

Net percentage change = (Net change in value / Initial Value) × 100

Net percentage change = (-9.0784 / 100) × 100

Net percentage change = -9.0784%

The net decrease percent in the original number is appro×imately 9%.

Given:

30% increased in 2019

15% increased i. 2020

40% decreased in 2021

Mask manufactured in 2021 = 179400

Formula used:

Increased % = 100 + increased

Decreased %  = 100 - decreased

Calculations:

⇒ Mask manufactured in 2019 = X × (100 + 30) = X × 130%

⇒ Mask manufactured in 2020 = (130%X) × (100 + 15) = 115% (130% of X)

⇒ Mask manufactured in 2021 = 115% (130% of X) × (100 - 40) = 60% ( 115%( 130% of X))

But according to the question,

⇒ Manufactured in 2021 = 179400

So,

⇒ 60%{115%(130% of X)} = 179400

⇒ \(\frac{X × 60×115×130}{100×100×100}= 179400\)

⇒ X =  \(\frac{179400 × 100×100×100}{60×130×115} \) 

⇒ X = 200000

⇒ Hence, The manufacture  of mask in 2018 is 200000

Shortcut Trick

So, 897 unit → 179400

then, 1000 unit → 179400/897 × 1000 = 200000