CRE 3 - Successive % change

If A is increased by 20%, i.e. A₁ = 1.2A

∴ A₁² = (1.2A)² = 1.44A²

∴ A² increases by 44%.

Alternately,

Percentage increase in A² will be same as 2 successive percentage increases of 20% each,

i.e. 20 + 20 + (20 × 20)/100 = 40 + 4 = 44%.

Hence, option (b).

When 8% of X is added to X, we get:

X + 0.08X

Now, again 8% of this result is added to the result.

⇒ (X + 0.08X) + 0.08(X + 0.08X)

⇒ X + 0.08X + 0.08X + 0.0064X

⇒ X + 0.16X + 0.0064X

⇒ 1.1664X

Alternately,

Here we have to successively increase X twice by 8%.

Final value of X = X × 1.08 × 1.08 = 1.1664X

Hence, option (b).

Let initial price be P and consumption be C.

Initial expenditure (E) = P × C   ...(1)

New price = 1.2P and new consumption = 0.7C

New expenditure (E') = 1.2P × 0.7C = 0.84PC = 0.84E (From (1))

Hence, % change in expenditure = $\frac{0.84E-E}{E}$ × 100 = -16%.

Hence, expenditure reduces by 16%.

Alternately,
Expenditure is first increased by 20% (due to price) and then successively reduced by 30% (due to consumption)

∴ Net % change = 20 – 30 + $\frac{20\times -30}{100}$ = -10 – 6 = -16%

Hence, option (a).

Let the original salary of man be x.

After increase by 10% of x, the new salary is 1.1x

A deduction of new salary by 10% means

0.9 (1.1 x) = 0.99x

⇒ 1% decrease

Alternately,
Net change = 10 - 10 + $\frac{10\times -10}{100}$ = -1%

Alternately,
When a number is increased by p% and then successively decreased by p% (or vice-versa), the net % change is $-\frac{p^2}{100}\%$.

Here, net % change = $-\frac{{10}^2}{100}\%$ = -1%

Hence, option (c).

Let P be the price/unit (Rs./ unit), E be the expenditure (Rs.) and C be the consumption in units.

Then, E = P × C (Rs./unit . unit)

Now,

Price has reduced by 40% ⇒ New price = 0.6P.

Expenditure remains same i.e. E

Let new consumption be C’

⇒ New expenditure = E = P' × C'

⇒ P × C = 0.6P × C'

⇒ C' = C/0.6 = 5C/3 = 1.667C 

⇒ C' is $66\frac{2}{3}$% more than C.

Alternately,
Price becomes $\left(1-\frac{40}{100}\right)$ = $\frac{3}{5}$ times.

∴ For Expenditure to remain constant, Consumption should become $\frac{5}{3}$ times

∴ % change in consumption = $\left(\frac{5}{3}-1\right)\times 100$ = $66\frac{2}{3}$%

Hence, option (a).

Net %change = 50 – 50 + $\frac{50\times -50}{100}$ = -25 = -25%

Alternately,

When a quantity is increased by p% and then successively decreased by p%, net change is $-\frac{p^2}{100}$.

Here, net change = -50²/100 = -25%.

Hence, option (a).

Let the original sum be x

After 5% reduction on x, we have, 0.95 x

Further 10% reduction on available 0.95 x means

0.9 (0.95 x) = 0.855 x = 171

x = 171/0.855 = Rs. 200

Hence, option (c).

Let the length and breadth of rectangle be L and B. So area = LB

New Length = 1.2L

New Breadth = 0.9B

Hence new area = 1.2L × 0.9B = 1.08 LB

Percentage change in area = (1.08 LB - LB)/LB × 100 = 8% increase.

Hence, option (e).

Let p be the price/unit (Rs./ unit) q be the quantity sold (units)

Then sales = pq (Rs./Unit . Unit)

After price and quantity changes, new price = 1.5p and new quantity = 0.8q

Total sales after these changes = 1.5p × 0.8q

Hence, % effect on sales = $\frac{1.5p\times 0.8q}{pq}$ = 1.2

⇒ Sales increased by 20%.

Alternately,

Net change = 50 - 20 + $\frac{50\times -20}{100}$ = 20

Hence, option (a).

Let Length and Breadth be L and B respectively. So area = LB

Length becomes 0.8 L

Breadth become 0.8 B

New Area = 0.8 L × 0.8B = 0.64 LB

Percentage decrease in area = $\frac{LB-0.64 LB}{LB}$ × 100 = 36%.

Alternately,

Net change = -20 -20 + $\frac{-20\times -20}{100}$ = -36%

Hence, option (d).

Let p be the original price/unit, consumption q = 18 quintals 

And expenditure e = p × 18 quintals 

10% reduction in price means that new price = 0.9p

We are now required to find consumption x, such that 0.9px = p × 18

⇒ x = 18/0.9 = 20 quintal.

Alternately,

Let the % change in consumption be C%.

Net change in expenditure = 0 = -10 + C + $\frac{-10\times C}{100}$

Solving this we get C = 11.11%

Hence, new consumption = 18$\left(1+\frac{11.11}{100}\right)$ = 18$\left(1+\frac{1}{9}\right)$ = 18 × $\frac{10}{9}$ = 20 quintals.

Alternately,
Price becomes (1 - 10/100) = 9/10 times.

For expenditure to remain constant, consumption should become 10/9 times.

∴ consumption at lower price = 18 × 10/9 = 20 quintals.

Hence, option (a).

Let the price of milk be = P and consumption = C.

New price of Milk = 1.5P and new consumption = C'.

New expenditure = initial expenditure

P × C = P' × C'

⇒ P × C = 1.5P × C'

⇒ C' = C/1.5 = 2C/3

Hence, new consumption is 2/3^rd, i.e. reduces by 1/3^rd = 33.33% reduction.

Alternately,
Price becomes (1 + 50/100) = 3/2 times

For expenditure to remain constant, consumption should become 2/3 times.

∴ % change in consumption = (2/3 - 1) × 100% = 33.33%

Hence, option (e).

Initially let the price of sugar be Rs. 'p' and quantity consumed be 'q'.

⇒ Expenditure = 60 = p × q   ...(1)

New price = 0.8p and new quantity = q + 6

⇒ Expenditure = 60 = 0.8p × (q + 6)   ...(2)

From (1) and (2)

p × q = 0.8p × (q + 6)

⇒ q = 0.8q + 4.8

⇒ 0.2q = 4.8

⇒ q = 24 kgs.

∴ New quantity = 24 + 6 = 30 kgs and

New price = 60/30 = Rs. 2/kg

Alternately,
Because price reduces by 20%, he saves 20% of 60 = Rs. 12

Since he saves Rs. 12 he buys 6 kg more.

So new increased price of 6 kg. = 12 Rs.

Increased price of 1 kg = 12/6 = 2 Rs.

Hence, option (a).

Given population, p = 35000

Let male population be m

Then, female population would be (35,000 – m)

From the problem,

1.06m + 1.04(35000 – m) = 36760

1.06m + 36400 – 1.04 m = 36760

⇒ 0.02m = 360

⇒ m = 360/0.02 = 18,000

Hence, option (d).

Let original salary = Rs. 100

Final salary = 100 × 0.7 × 1.3 = 91

So percentage decrease = (100 - 91)/100 × 100 = 9%    

Alternately,

When a quantity is increased and subsequently decreased by same p%, net % change = -p²/100.

Here, net % change in salary = - 30²/100 = -9%

Hence, option (c).

10% Increase for 2 consecutive years = 10 + 10 + (10 × 10)/100 = 21%

New population after 2 years = 2500 × 1.21 = 3025

Hence, option (c).

Let C be the % change in consumption.

Net % change in expenditure = 0 = 20 + C + (20 × C)/100

⇒ -20 = C + C/5

⇒ -20 = 6C/5

⇒ C = -100/6% or 16.67% reduction.

Hence, option (a).

Let the original price of gold/unit is p Rs./unit

Let q be the quantity purchased in units

Let e be the expenditure incurred

Thus, e = p × q

When price increases by 30% i.e., 1.3p

Then new quantity purchased q’ units be such that,

1.3pq’ = p × q 

q' = q/1.3 ≈ 77% of q.

∴ reduction in q must be 23% = $23\frac{1}{13}\%$

Hence, option (b).

Let the monthly rent of property be r.

Investment on property = 10 lakhs

Money on repairs (Annually) = 12.5/100 × r × 12

Annual taxes = Rs. 3320

Thus, expenditure recovered from rent, annually = 12 × 12.5r/100 + 3320 = 1.5r + 3320

Savings from rent (Annually) = 10/100 × 10,00,000 = 1,00,000

Annual income - Annual Expenditure = Annual Savings

Thus, 12r – (1.5r + 3320) = 1,00,000

or, 10.5r = 96,680

or, r = Rs. 9207.6

Hence, option (d).

Let the original price = P and consumption = C

New price = P' and consumption = C'

Expenditure increases by 10%,

⇒ 1.1 × P × C = P' × C'

⇒ 1.1 × P × C = 0.9P × C'

⇒ C' = 1.1C/0.9 = 11/9 × C

∴ % change in consumption = (11/9 - 1) × 100% = 22.22%

Hence, consumption increases by 22.22%.

Hence, option (b).