Percentage — Part 2 MCQs with Explanations

Let the fraction be $$\frac{{\text{x}}}{{\text{y}}}$$ When fraction is squared its numerator is reduced by $$33\frac{1}{3}$$ and denominator is reduced by 20% $$\eqalign{ & {\text{According}}\,{\text{to}}\,{\text{question,}} \cr & {\left( {\frac{x}{y}} \right)^2} \times \frac{{33\left( {\frac{1}{3}} \right)\% }}{{20\% }} = 2\left( {\frac{x}{y}} \right) \cr & {\text{Or}},\,{\left( {\frac{x}{y}} \right)^2} \times \frac{{\left( {\frac{2}{3}} \right)}}{{\left( {\frac{1}{5}} \right)}} = 2\left( {\frac{x}{y}} \right) \cr & {\text{Or}},\,\frac{x}{y} = \frac{3}{5} \cr & {\text{Sum of numerator and denominator is}} \cr & \left( {x + y} \right) = 3 + 5 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 8 \cr} $$

Let the percentage marks in QA =(10a + b)% Let the percentage marks in DI = (10b + a)% Let the percentage marks in VA = x% Now, according to the question, we have, $$\eqalign{ & \frac{{\left( {10{\text{a}} + {\text{b}}} \right) + {\text{x}} + \left( {10{\text{b}} + {\text{a}}} \right)}}{3} = {\text{x}} \cr & 11{\text{a}} + 11{\text{b}} + {\text{x}} = 3{\text{x}} \cr & {\text{or, x}} = \frac{{11\left( {{\text{a}} + {\text{b}}} \right)}}{2} \cr} $$ Clearly, we can see that the percentage of the VA section will be a multiple of 11 So, required answer will be 66

Let the initial price = Rs. 10000p Price after first increment = 10000p + 100xp Price after first decrement = 10000p + 100xp - (100px + px 2 ) = 10000p - px 2 Now, total decrement, px 2 = 21025 . . . . . (1) Price after second increment, = 10000p - px 2 + 100xp - $$\frac{{{\text{p}}{{\text{x}}^3}}}{{100}}$$ Price after second increment, = 10000p - p 2 + 100xp - $$\frac{{{{\text{p}}^3}}}{{100}}$$ - 100xp + $$\frac{{{\text{p}}{{\text{x}}^3}}}{{100}}$$ - px 2 + $$\frac{{{\text{p}}{{\text{x}}^4}}}{{10000}}$$ = 10000p - 2px 2 + $$\frac{{{\text{p}}{{\text{x}}^2}}}{{10000}}$$ = 484416 . . . . . . (2) On solving equation (1) and (2), We get x = 20 Substituting back we get, p = 5,25,625

No. of Machines Output Manuf. cost Est. cost Total cost Profit 12 48, 000 24, 000 10, 000 34, 000 14, 000 11 44, 000 22, 000 10, 000 32, 000 12, 000 Profit, = Output - total cost = 44000 - 32000 = Rs. 12000 Initial value of share holders, = 14000 $$ \times \frac{{10}}{{100}}$$ = Rs. 1400 New value of share holders, = 12000 $$ \times \frac{{10}}{{100}}$$ = Rs. 1200 Decrease in Share holder value = 1400 - 1200 = 200 percentage decrease in the value of shareholders is : $$\eqalign{ & = \frac{{200 \times 100}}{{1400}} \cr & = 14.28\% \cr} $$

Let the price of wheat is x per kg. Then price of wheat will be 5x per kg. Expenditure on rice = 25 × x = 25x Expenditure of wheat = 9 × 5x = 45x Total cost, 25x + 45x = 350 70x = 350 x = 5 Hence, price of Rice = Rs. 5 per kg. Price of wheat = 25 per kg. Now, price of wheat = 25 ---- 20% ↑----> Rs. 30 per kg. Let the new amount of rice is N kg, then N*5 + 9*30 = 350 N = 16 kg. % decrease in the amount of rice $$\eqalign{ & = \frac{{\left( {25 - 16} \right) \times 100}}{{25}} \cr & = 36\% \cr} $$

Let the initial cost of raw material be 100. So, initial labor cost was 25 and net cost was 125 Now, 15% increment in raw materials cost and labor cost has gone up to 30% from 25 % Raw material cost = 115 And Labor cost = (115 × 30%) = 34.5 So, New net cost, = 115 + 34.5 = 149.5 Difference of labor cost = 149.5 - 125 = 24.5 % reduction = $$\frac{{24.5 \times 100}}{{149.5}}$$ = 17%(approx.)

$$\eqalign{ & {\text{His bonus}}, \cr & = \frac{{ {20 \times 1000000} }}{{100}} \cr & = 2\, \text{lakh} \cr & {\text{Total}}\,{\text{profit}} = {\text{Net}}\,{\text{profit}} + \frac{{ {10 \times {\text{net}}\,{\text{profit}}} }}{{100}} \cr & 1.32\,\text{lakh} = {\text{Net}}\,{\text{profit}} \times \left[ {1 + {\frac{{10}}{{100}}} } \right] \cr & {\text{Net}}\,{\text{profit}} = \frac{{132000}}{{1.1}} = 120000 \cr & {\text{Commission}}, \cr & = \left( {{\text{Total}}\,{\text{profit}} - {\text{Net}}\,{\text{profit}}} \right) \cr & = 132000 - 120000 \cr & = 12000 \cr & {\text{Hence}}, {\text{his}}\,{\text{total}}\,{\text{earnings}} \cr & = 2\,\text{lakh} + 12000 \cr & = Rs.\,212000 \cr} $$

There is no need of the number of goats given i.e. 34,398. Initially, let there be 100 goats. Then 100 == 40% ↑==> 140 == 30%↓(declined to 70%) ==> 98 == 30%↑ ==> 127.4 == 10%↓(sold) ==> 114.66 Hence, % increase = 14.66% [As 100 becomes 114.66]

Total employees = 1600 Female employees, 60% of 1600 $$ = \frac{{60 \times 1600}}{{100}} = 960$$ Then male employees = 640 50% of male are computer literate, = 320 male computer literate 62% of total employees are computer literate, $$ = \frac{{62 \times 1600}}{{100}} = 992$$ computer literate Thus, Female computer literate = 992 - 320 = 672 Alternatively : Let 60% employees are female and 40% are male Then, 20% of male are computer literate and 42% are female computer literate Female computer literate $$ = \frac{{1600 \times 42}}{{100}} = 672$$

Price after third depreciation, 100 ==25%↓ ==> 75 == 20%↓==>60 == 15% ↓ ==> 51 The price will be, = Rs. 5,10,000 Alternatively : 1000000 × 0.75 × 0.80 × 0.85 = Rs. 5,10,000

100 === 25% ↑===> 125 ===> 20% ↓===> 100 So there is no change in the area of rectangle Alternatively : Let area be 1 Area of rectangle, = l × b = 0.8 × 1.25 = 1

Cost of fresh mangoes + Cost of packaging = Total cost Let initial Cost of fresh, mangoes = 100 Then, packaging cost = 40 Thus, Initial total cost = 100 + 40 = 140 After increasing in cost of fresh mangoes 30%, Cost of fresh mangoes = 130 And cost of packing go down by 50 % so, Cost of packing = 20 Now Total cost = 130 + 20 = 150 Increased cost = 150 - 140 = 10 % increased = $$ = \frac{{10 \times 100}}{{140}} = 7.14\% $$

Given, 220% of X = 44 Or, X = 20 Thus, 44% of 20 $$ = \frac{{44 \times 20}}{{100}} = 8.8$$

Let initially the quantity and rate be 100 each Quantity × rate = Expenditure 100 × 100 = 10000 Now, Increase in price is 25% and new quantity is 70% of original Quantity × rate = Expenditure 70 × 125 = 8750 Decreased expenditure, = 10000 - 8750 = 1250 % decrease $$ = \frac{{1250 \times 100}}{{10000}} = 12.5\% $$

$$\eqalign{ & {\text{Daily}}\,{\text{supply}}, \cr & = \left( {100 - z} \right)\% \,{\text{of}}\,y \cr & = \frac{{ {\left( {100 - z} \right)y} }}{{100}} \cr & {\text{Thus,}}\,{\text{required}}\,{\text{number}}\,{\text{of}}\,{\text{days}} \cr & = {\frac{{ {100x} }}{{\left( {100 - z} \right)y}}} \cr} $$

Let the initial revenue be 100 100 === 50% ↑ (Ticket up) ===> 150 === 20% ↓ (Visitors down) ===> 120 There is 20% increase in the revenue

Result Physics Chemistry Failed 35% 45% Passed 65% 55% Now, 55% of 600 passed in Chemistry = $$\frac{{55 \times 600}}{{100}}$$ = 330 From question, 40% of those who passed in Chemistry(330 Students) also passed in physics = $$\frac{{40 \times 330}}{{100}}$$ = 132, so, 132 students passed in both Therefore % who passed in both = $$\frac{{132 \times 100}}{{600}} = 22\% $$ Passed in both subject = 22% of total students % of students who passed in either Physics or Chemistry or both, = (65 + 55) - 22 = 98% Thus, percentage of students who failed in both subjects = 2% Number of students who failed = 2% of 600 = 12

Copper : Aluminum = 7 : 4 Let Copper and Aluminum in the weapon be 7x and 4x respectively Given, Aluminum in weapon = 12 kg So, → 4x = 12 → x = 3 Copper = 7x = 7 × 3 = 21 Kg. Total alloy in the weapon = 12 + 21 = 33 kg But 12% alloy get destroyed in making the weapon, i.e. 88% alloy is used in the weapon, so, → 88 % alloy = 33 kg → 100 % alloy = 37.5 kg

Let the smaller number be x and larger number be y According to the question, 80% of x + 4 = 40% of y 4y - 8x = 40 y - 2x = 10 --------- (1) y - x = 85 ----------- (2) By using (1) and (2), we get x = 75 y = 160 x + y = 235

Actual result = 10x By mistake it has been divided by 10 = $$\frac{x}{{10}}$$ %Change = $$ {\frac{{10x - \left( {\frac{{10}}{x}} \right)}}{{10x}}} $$ × 100 = 99 = -99% Since, actual value is greater than the wrong value. Alternatively, Let, x = 10 Actual result = 10 × 10 = 100 Wrong result = $$\frac{{10}}{{10}}$$ = 1 Change = (1 - 100) = -99 % Change = - 99%

Let the number of people be x who has been asked for the donations People already solicited = 60% of x = 0.6x Remaining people = 40% of x = 0.4x Amount collected from the people solicited, = 600 × 0.6x = 360x 360x = 75% of the amount collected Remaining amount = 25% = 120x Thus, Average donations from remaining people, $$\eqalign{ & = \frac{{120{\text{x}}}}{{0.4{\text{x}}}} \cr & = 300 \cr} $$

Female employees = 45% Male employees = 55% Difference = 55 - 45 = 10% → 10% = 72 → 1% = $$\frac{{72}}{{10}}$$ → 100 = 7.2 × 100 = 720

100 == 25% ↑ ==> 125 == x%↓ ==> 100 $${\text{x}}\% = \frac{{25 \times 100}}{{100}} = 25\% $$ We don't need the actual value to calculate the answer

Day 1 2 3 Initial mangoes x (let) 0.45x 0.2025x Sales 0.5x 0.225x 0.10125x Over night remaining 0.5x 0.225x 0.10125x Rotten 0.05x 0.0225x 0.010125x Next day stock 0.45x 0.2025x Thus, Total mangoes rotten amount is ⇒ 0.05x + 0.0225x + 0.010125x = 0.082625x = 1983x = 24,000

Non-defective products M 1 = 25 × 0.98 = 24.5% Non-defective products M 2 = 35 × 0.96 = 33.6% Non-defective products M 3 = 40 × 0.95 = 38% Percentage of non-defective products = 24.5 + 33.6 + 38 = 96.1%

$$\eqalign{ & {\text{Let the positive number be x}}. \cr & {\text{According}}\,{\text{to}}\,{\text{the}}\,{\text{question}}, \cr & x \times x = x + \frac{{\left( {x \times 2000} \right)}}{{100}} \cr & {x^2} = x + 20x \cr & {x^2} - 21x = 0 \cr & {\text{Either}}, \cr & x = 0,\,\,or \cr & x = 21 \cr & {\text{21}}\,{\text{is}}\,{\text{the}}\,{\text{possible}}\,{\text{value}}{\text{.}} \cr & {\text{Then}}\,{\text{square}}\,{\text{of}}\,21\,{\text{is}}\,441 \cr} $$

Let the bike's initial cost be x and then car's initial cost be 5x After the increase, Bike price = 1.2x Car price = 5.75x Initial total cost of 5 cars and 10 bikes, = 25x + 10x = 35x New cost, = 28.75x + 12x = 40.75x Change in cost = (40.75x - 35x) = 5.75x % change = $$\frac{{5.75{\text{x}} \times 100}}{{35}} = 16\frac{3}{7}\% $$

The average weight of the bars given to Brennan (light) < the average weight of the bars given to Blair < the average weight of the bars given to Maya (heavy). Let the total weight of all the bars be X. The weight of the bars given to Brennan, = 45% of X = 0.45X The weight of the bars given to Maya, = 26% of X = 0.26X The weight of the bars given to Claire = rest = 29% of X = 0.29X The average weight of the bars given to Brennan, $$ = \frac{{{\text{Weight}}}}{{{\text{Number}}\,{\text{of}}\,{\text{bars}}}} = \frac{{0.45{\text{X}}}}{{24}}$$ The average weight of the bars given to Maya = Weight / number of bars = $$\frac{{0.26{\text{X}}}}{{13}}$$ Similarly, if the number of bars given to Blair = B, then the average weight of the bars given to Blair = $$\frac{{0.29{\text{X}}}}{{\text{B}}}$$ As, the average weight of the bars given to Brennan (light) < the average weight of the bars given to Blair < the average weight of the bars given to Maya So , Option (B) is the right answer

$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{original}}\,{\text{price}} = y, \cr & {\text{After}}\,{\text{first}}\,{\text{change,}}\,{\text{it}}\,{\text{becomes}}, \cr & y \times \left( {1 + {\frac{x}{{100}}} } \right) \cr & {\text{After}}\,{\text{second}}\,{\text{change,}}\,{\text{it}}\,{\text{becomes}} \cr & y \times \left( {1 + {\frac{x}{{100}}} } \right)\left( {1 - {\frac{x}{{100}}} } \right) \cr & = y\left( {1 - {{\left( {\frac{x}{{100}}} \right)}^2}} \right) \cr & {\text{Thus}}, \cr & {x^2} \times y = {10^6} - - - - \left( 1 \right) \cr & {x^2} = \frac{{{{10}^6}}}{y} \cr & {\text{Now}}, \cr & y{\left( {1 - {\frac{{{{10}^6}}}{{10000y}}} } \right)^2} \cr & = 2304\left( {{\text{similar}}\,{\text{to}}\,{\text{above}}} \right) \cr & y{\left( {1 - \frac{{100}}{y}} \right)^2} = 2304 \cr & y = 2500 \cr} $$

Year Value 2000 100 (Let) 2001 110 2002 121 2003 133.1 2004 119.79 2005 107.811 2006 97.0299 $$\eqalign{ & {\text{Now}}, \cr & \frac{{ {\left( {100 - 97.0299} \right) \times 100} }}{{100}} \approx 3 \cr} $$

20 litres juice contain 10% Tomato, i.e. 20L juice = $$\frac{{20 \times 10}}{{100}}$$ = 2L Tomato Tomato puree contains 80% of water and 20% tomato. This 80% tomato = 2L (which is contained by 100 puree) So, Now this 2 L Consist 80% in puree Thus, total puree will be $$\frac{2}{{.8}} = 2.5{\text{L}}$$

If we take different values of x then, Let x = 150, then error% $$\eqalign{ & = \frac{{ {\frac{{ {200 - 100} }}{{100}}} }}{{100}} \cr & = 100\% \cr} $$ Again if x = 100, then error% $$\eqalign{ & = \frac{{ {\left( {150 - 50} \right) \times 100} }}{{100}} \cr & = 200\% \cr} $$ If we take different value of x, we get different answer so we can't determine it.

Total student = 100 Boys = 60 Girls = 40 Boys who plays hockey = 40% = 24 There is no information about boys who play badminton. Girls who plays Badminton = 75% = 30 No girls plays hockey. Since, we do not have information that whether the rest of the boys are playing badminton or not. So, we cannot determine the total no. of student who don't play any game.

$$\eqalign{ & {\text{Let the required number of pages be }}x. \cr & 30 \times 25 \times 35 = x \times 30 \times 28 \cr & x = 31.25 \approx 32 \cr & \% \,{\text{increase}}\,{\text{in}}\,{\text{number}}\,{\text{of}}\,{\text{pages}}, \cr & = {\frac{2}{{30}}} \times 100 \cr & = 6.66\% \cr} $$

$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{fraction}}\,{\text{be}}\,\frac{{100x}}{{100y}} \cr & {\text{Now}}\,{\text{according}}\,{\text{to}}\,{\text{the}}\,{\text{question}}, \cr & {\left( {\frac{{100x}}{{100y}}} \right)^2} = \frac{{125{x^2}}}{{80{y^2}}} = \frac{{25{x^2}}}{{16{y^2}}} \cr & \frac{{25{x^2}}}{{16{y^2}}} = \frac{5}{8}\left( {\frac{{100x}}{{100y}}} \right) \cr & {\frac{{100x}}{{100y}}} = \frac{2}{5} \cr & {\text{hence}}, \cr & {\text{product of numerator and denominator}} \cr & = 2 \times 5 = 10 \cr} $$

Let X be the number which is added to 80 80% of X = 0.8X Now, 80 + 0.8X = X 0.2X = 80 X = $$\frac{{80}}{{0.2}} = 400$$

$$\eqalign{ & {\text{Reduction}} \cr & = {\frac{7}{{107}}} \times 2568 \cr & = 168 \cr} $$

Since, Run scored = Over × Run rate If overs is reduced by 25%, run rate will go up by 33.33%. Hence, Australia could have scored any number of runs.

In 2000, the market share was 40%, 30% and 30%, means the ratio is 4 : 3 : 3 In 2001, a new product (A) enters and has 10% market share, 90% of the remaining market is shared by the previous 3. Now, divide 90% in the ratio 4 : 3 : 3 , i.e 36%, 27%, 27%. Now the ratio is 36 : 27 : 27 : 10 In 2002, another new product (B) enters and has 10% market share, now the remaining 90% market share is distributed in the ratio, 36 : 27 : 27 : 10 and hence remaining 32.4%, 24.3%, 24.3%, 9% Thus, the market share of Margo in 2002 is 32.4%

$$\eqalign{ & {\text{Let the total number of applicants be x}}. \cr & {\text{Number of eligible candidates}} \cr & = {\text{ }}95\% {\text{ }}of{\text{ }}x \cr & {\text{Eligible candidates of other categories}}, \cr & = 15\% \,of\,\left( {95\% {\text{ }}of{\text{ }}x} \right) \cr & = {\frac{{15}}{{100}}} \times {\frac{{95}}{{100}}} \times x \cr & = \frac{{57}}{{400}}x \cr & or,\left( {\frac{{57}}{{400}}} \right)x \cr & x = \frac{{ {4275 \times 400} }}{{57}} \cr & \,\,\,\,\,\, = 30000 \cr} $$