Percentage — Part 1 MCQs with Explanations

$$\eqalign{ & {\text{Required}}\,{\text{Percentage}} \cr & = \frac{{ {1.14 \times 100} }}{{1.9}} \cr & = 60\% \cr} $$

Students passed in English = 80% Students passed in Math's = 85% Students passed in both subjects = 73% Then, number of students passed in at least one subject = (80+85)-73 = 92%. [The percentage of students passed in English and Maths individually, have already included the percentage of students passed in both subjects. So, We are subtracting percentage of students who have passed in both subjects to find out percentage of students at least passed in one subject.] Thus, students failed in both subjects = 100-92 = 8%.

$$\eqalign{ & {\text{As}}\,{\text{we}}\,{\text{know}},\,1\% = \frac{1}{{100}} \cr & {\text{Hence}},\, \cr & \frac{1}{2}\% = {\frac{1}{2} \times \frac{1}{{100}}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \frac{1}{{200}} \cr & \,\,\,\,\,\,\,\,\,\,\, = 0.005 \cr} $$

Let the initial price of the commodity be 100. After 50% increase in price, It will become, 100 ------50% increase----> 150. Now, we have to reduce the consumption to keep expenditure 100. Increase in price= 150 - 100 = 50 We have to reduce the consumption, $$\eqalign{ & = \frac{{50}}{{150}} \times 100 \cr & = \frac{1}{3}{\text{ or 33}}{\text{.33}}\% \cr} $$ Other Method: Here, we use, Final product constant graphic. 100 ==50% up== 150===33.33% down===>100 Consumption Reduce = 33.33% = $$\frac{1}{3}$$

Here we can use the compound interest based formula, $$\eqalign{ & {\text{Population after }}n{\text{ years}} \cr & = P \times {\left[ {1 + {\frac{r}{{100}}} } \right]^n} \cr & {\text{Population after 2 years}} \cr & = 50000 \times {\left[ {1 + {\frac{4}{{100}}} } \right]^2} \cr} $$ Population after 2 years = 54080 Alternatively, we can use, net percentage change graphic as well, 50,000------4%↑---→ 52,000---- 4%↑---→ 54,080. Then, population after 2 years= 54,080. In this calculation, we need to find 1% of 50,000 first, which is easily calculated by dividing 50,000 by 100.

As A and B are fixed, C is any point on AB, so if AC is increases then CB decreases. A ________3 cm_________ C _____2 cm____ B Then, solution can be visualized as, Increase in AC 6% = $$\frac{{106 \times 3}}{{100}} = 3.18\,{\text{cm}}{\text{.}}$$ Decrease in CB = 0.18 cm % decrease = $$\frac{{0.18}}{2} \times 100 = 9\% $$ Alternatively, AC = 3 Cm. BC = 2 Cm. Increase in AC by 6%, then New, AC = 3 + 6% of 3 = 3 + 0.18 = 3.18 cm. 0.18 cm increase in AC means 0.18 cm decrease in BC as already mentioned AB as the fixed point. So, % decrease in BC, $$\eqalign{ & = \frac{{{\text{Actual}}\,{\text{Decrease}}\,{\text{in}}\,{\text{BC}}}}{{{\text{Original BC}}}} \times 100 \cr & = \frac{{0.18}}{2} \times 100 = 9\% \cr} $$

Initial Cost = Rs. 75 After 20% increase in the cost, it becomes, (75 + 20% of 75) = Rs. 90 Now, Cost is decreased by 20%, So cost will become, (90 - 20% of 90) = Rs. 72 So, present cost is Rs. 72

Let the initial expenses on Sugar was Rs. 100. Now, Price of Sugar rises 25%. So, to buy same amount of Sugar, they need to expense, = (100 + 25% of 100) = Rs. 125. But, They want to keep expenses on Sugar, so they have to cut Rs. 25 in the expenses to keep it to Rs. 100. Now, % decrease in Consumption, $$\frac{{25}}{{125}} \times 100 = 20\% $$ Mind Calculation Method; 100 === 25%↑ ===> → 125 === X%↓ ===> 100 Here, X = $$\frac{{25}}{{125}} \times 100 = 20\% $$

Let the Original length of the rectangle be 20 unit and breadth be 10 unit. Then Original Area = length *breadth = 20*10 = 200 Square unit. 40% decrease in each side, then Length = (20 - 40% of 20) = 12 unit. Breadth = (10 - 40% of 10) = 6 unit. Now, Area = 12 × 6 = 72 Square unit. Decrease in area = 200 - 72 = 128 square unit. % Decrease in Area = $$\frac{{128}}{{200}} \times 100 = 64\% $$ Mind Calculation Method: Let the original area be 100 square unit. 100 === 40%↓(decrease in length) ==⇒ 60 === 40%↓ === (decrease in breadth) ===> 36 Diminished in area = 100 - 36 = 64%

Let the initial expenditure on the commodity be Rs. 100. Now, the price decreases by 20%, Current Price = (100 - 20% of 100) = Rs. 80. Same time due to decrements in price 20% consumption has been increased. So, Current expenses on commodity = (80 + 20% of 80)= Rs. 96. Here, the initial expenditure was Rs. 100 which became 96 at the end, it means there is 4% decrements in the expenditure of the commodity. Mind Calculation Method: 100 === 20%↓(Decrements in Price) ===> 80 === 20%↑(Increment in Consumption) ===> 96. Thus, there is a decrements of 4%

Let B's Salary is Rs. 100. Then, A's Salary = (100 + 25% of 100) = Rs. 125 Difference between A's Salary and B's Salary = 125 - 100 = Rs. 25 % Difference (lower) = $$\frac{{25}}{{125}} \times 100 = 20\% $$ Mind Calculation Method: 100(B salary) === 25%↑ ===> 125(A salary) === 20%↓ ===> 100 (B salary) B's salary is 20% lower than A's

Net percentage increase in Population = (2.5 - 0.5) = 2% each year. Let the Original Population of the town be 100. Population of Town after 1 year = (100 + 2% of 100) = 102. Population of the town after 2nd year = (102 + 2% of 102 ) = 104.04 Now, % increase in population = $$\frac{{4.04}}{{100}} \times 100 = 4.04\% $$ Mind Calculation Method: 100 == 2% Up(1 st year) ==> 102 == 2%Up(2 nd year) ==> 104.04 % population increase in 2 years = 4.04%.

Winner gets 65% of valid votes and loser gets 35% of votes Difference between this two = 2748 (65-35)% = 2748 30% = 2748 Total number of voters, 100% = $$\frac{{2748 \times 100}}{{30}}$$ = 9160

$$\eqalign{ & {\text{Let the present population of the town be }}P \cr & {\text{Using compound interest formula}} \cr & {\text{Then}}, \cr & P = x\left[ {1 + \left( {\frac{R}{{100}}} \right)} \right] - - - \,\left( i \right) \cr & {\text{And}}\,y = P\left[ {1 + \left( {\frac{R}{{100}}} \right)} \right] \cr & = P \times \frac{P}{x} - - - - \,\left( {ii} \right) \cr & {P^2} = xy; \cr & {\text{Hence}},\,P = \sqrt {xy} \cr} $$

Narayan's saving and rent = 1000 + 1800 = Rs. 2800 Let his monthly income be Rs. 100 30% of his income he spent on education i.e. Rs. 30 Remaining = 100 - 30 = 70 50% of remaining on food = $$\frac{{70 \times 50}}{{100}} = {\text{Rs}}{\text{. 35}}$$ Now, that 35 must be equal to his saving and rent i.e. 35 = 2800 then, 1 = $$\frac{{2800}}{{35}}$$ 100 = $$\frac{{2800 \times 100}}{{35}} = {\text{Rs}}{\text{. 8000}}$$ So, his income = Rs. 8000

Let Q = 10. Then, P = 60. Q is 50 less than P. Q, % less than P = $$\frac{{50}}{{60}} \times 100 = 83.33\% $$ Alternative Method 10 (Q) == (6 times greater) ==> 60(P) == x%↓(Less than Q) ==> 10 (Q) Now, $${\text{x}} = \frac{{50 \times 100}}{{60}} = 83.33\% $$

Let the third number be 100. Then, 1 st number = 130 2 nd number = 140 % 1 st number to the 2 nd number $$\eqalign{ & = \frac{{130 \times 100}}{{140}} \cr & = \frac{{650}}{7} \cr & = 92\frac{6}{7}\% \cr} $$

$$\eqalign{ & {\text{Let number of men in the population be }}x \cr & {\text{Number of women}} = \left( {35000 - x} \right) \cr & {\text{Increase in the number of men}} \cr & = 6\% \,of\,x = \frac{{6x}}{{100}} \cr & {\text{Increase in the number of women}} \cr & = \left( {3500 - x} \right) \times \frac{4}{{100}} \cr & {\text{Increase in whole population}} \cr & = 36760 - 35000 = 1760 \cr & {\text{Now}}, \cr & \frac{{6x}}{{100}} + \left[ {\left( {35000 - x} \right) \times \frac{4}{{100}}} \right] = 1760 \cr & \left[ {\left( {6x - 4x} \right) + 35000 \times \frac{4}{{100}}} \right] = 1760 \cr & 2x + 35000 \times 4 = 1760 \times 100 \cr & 2x = 176000 - 35000 \times 4 \cr & x = 18000 \cr & {\text{Number}}\,{\text{of}}\,{\text{men}} = 18000 \cr & {\text{Number}}\,{\text{of}}\,{\text{women}} \cr & = 35000 - 18000 \cr & = 17000 \cr} $$

Let each side of the cuboid be 10 unit initially. Initial Volume of the cuboid, = length * breadth * height = 10 × 10 × 10 = 1000 cubic unit. After increment dimensions become, Length = (10 + 10% of 10) = 11 unit. Breadth = (10 + 20% of 10) = 12 unit. Height = (10 + 50% of 10) = 15 unit. Now, present volume = 11 × 12 × 15 = 1980 cubic unit. Increase in volume = 1980 - 1000 = 980 cubic unit. % increase in volume = $$\frac{{980}}{{1000}} \times 100 = 98\% $$ Mind Calculation Method: 100 == 50%↑(height effects) ==> 150 == 20%↑(breadth) ==> 180 == 10%↑(length effects) ==> 198 Change in volume = 98% [We can take net percentage change in any order]

Let Rs. 100 be spend on rice initially for 20 kg. As the price falls by 20%, new price for 20 kg rice,= (100 - 20% of 100) = 80 New price of rice = $$\frac{{80}}{{20}}$$ = Rs. 4 per kg. Rice can bought now at = $$\frac{{100}}{{4}}$$ = 25 kg.

Let the number be x According to the question, 91 - $$\frac{{30{\text{x}}}}{{100}}$$ = x 9100 - 30x = 100x Or, 9100 = 130x Or, x = $$\frac{{9100}}{{130}}$$ Hence, x = 70

100000 == 10%↑(1 st year) ==> 110000 == 10%↑(2 nd year) ==> 121000 Population at starting of 3 rd year = 121000 [By using net percentage change method, we have saved our self from cumber some formulas on population which is based on compound interest].

Let Ashu's salary = 100; Ashu's salary after rise = 125 Then Vicky's salary = 175 Vicky's salary after rise of 40% = 245 [As 10% of Vicky's salary is 17.5 then 40% = 17.5 × 4 = 70] Difference between Vicky's salary and Ashu's salary = 245 - 125 = 120 % more Vicky's salary than Ashu's = $$\frac{{120 \times 100}}{{125}} = 96\% $$

Let there is 100 kg of ore. 25% ore contains 90% off Iron that means 25 kg contains; $$\frac{{25 \times 90}}{{100}} = 22.5\,{\text{kg}}\,{\text{iron}}$$ 22.5 kg Iron contains 100 kg of ore. Then, 1 kg of iron contains = $$\frac{{25}}{{100}}{\text{kg}}\,{\text{ore}}$$ Hence, 60 kg iron contains = $$\frac{{100 \times 60}}{{22.5}}$$ = 266.66 kg ore

Number of Indians men present there = $$\frac{{700 \times 20}}{{100}} = 140$$ Indian women = $$\frac{{500 \times 40}}{{100}} = 200$$ Indian children = $$\frac{{800 \times 10}}{{100}} = 80$$ Total member present in climate conference = 700 + 500 + 800 = 2000 Total Indian = 200 + 140 + 80 = 420 Hence, % of Indian present there = $$\frac{{420 \times 100}}{{2000}} = 21\% $$ % of people who were not Indian = 100 - 21 = 79%

Let 100 kg of hematite be obtained then 20% of it get wasted that means 80 kg of ore remains. Pure iron = 25% of remaining ore = $$\frac{{80 \times 25}}{{100}} = 20\,{\text{kg}}$$ 20 kg pure Iron is obtained from 100 of hematite 1 kg pure Iron is obtained from = $$\frac{{100}}{{20}}$$ hematite Then, 80000 kg pure Iron is obtained from =$$\,\frac{{100}}{{20}} \times 80000 = \,$$ $$400000\,{\text{kg}}$$ hematite.

The total cost for the year = 250000 + 2% of 2500000 + 2000 = Rs. 257000 For getting return of 15% he must earn = $$\frac{{257000 \times 15}}{{100}}$$ = Rs. 38550 per year Then, Monthly Rent = $$\frac{{38550}}{{12}}$$ = Rs. 3212.5

Let Ram's salary be x He spends on rent = 30% of x = $$\frac{{30{\text{x}}}}{{100}}$$ He spends on education = 30% from rest of the salary = $$\frac{{30 \times 70{\text{x}}}}{{100 \times 100}} = \frac{{21{\text{x}}}}{{100}}$$ He Spends on clothes = 24% of total salary = $$\frac{{24{\text{x}}}}{{100}}$$ Saving = 2500 Salary of ram = x $$\eqalign{ & \frac{{30{\text{x}}}}{{100}} + \frac{{21{\text{x}}}}{{100}} + \frac{{24{\text{x}}}}{{100}} + 2500 = {\text{x}} \cr & {\text{or,}}\,\frac{{75x}}{{100}} = {\text{x}} - 2500 \cr & {\text{or,}}\,75{\text{x}} = 100{\text{x}} - 250000 \cr & {\text{or, }}100{\text{x}} - 75{\text{x}} = 250000 \cr & {\text{or, x}} = \frac{{250000}}{{25}} \cr & {\text{or, x}} = 10000 \cr & {\text{Ram's}}\,{\text{salary = Rs}}{\text{.}}\,10000 \cr} $$

Let x be the page required when report is retyped Now, we can use work equivalence method 20 × 55 × 65 = 70 × 65 × x Or, x = $$\frac{{20 \times 55 \times 65}}{{70 \times 65}}$$ Or, x = 15.70 = 16 pages. Hence, % reduction in pages = $$\left( {20 - 16} \right) \times \frac{{100}}{{20}}$$ $$\, = \,$$ $$20\% $$

Let initial price of Maruti Car be Rs. 100 As price increases 30%, price of car will become, (100 + 30% of 100) = Rs. 130 Due to increase in price, sales is down by 20%. It means, it is going make 20% less revenue as expected after increment of price So, New revenue = (130 - 20% of 130) = Rs. 104 The initial revenue was Rs. 100 which becomes Rs. 104 at the end. It means there is 4% increment in the total revenue Mind Calculation Method: 100 == 30%↑(price effect) ==> 130 == 20%↓(sales effects) ==> 104 Hence, 4% rises

Let us assume there are 100 students in the institute. Then, number of boys = 60 And, number of girls = 40 Further, 15% of boys get fee waiver = 9 boys 7.5% of girls get fee waiver = 3 girls Total = 12 students who gets fee waiver But, here given 90 students are getting fee waiver. So we compare 12 = 90 So, 1 = $$\frac{{90}}{{12}}$$ = 7.5 Now number of students who are not getting fee waiver = 51 boys and 37 girls 50% concession = 25.5 boys and 18.5 girls (i.e. total 44) Hence, required students = 44 × 7.5 = 330

We solve it through options. Choosing options for trial depends on mental thought. 100 == 12% up ==> 112 == 12% up ==> 125.44 == 12% Up ==> 140.49 So, answer is 12%

Let the marks obtained in five subjects be 6x, 7x, 8x, 9x and 10x. Total marks obtained = 40x Max. Marks of the five subjects = $$\frac{{40{\text{x}}}}{{0.6}}$$ [40x is 60% of total marks] Max. Marks in each subject = $$\frac{{40{\text{x}}}}{{0.6 \times 5}}$$ = 13.33x Hence, % of each subject = $$\frac{{6{\text{x}} \times 100}}{{13.33}}$$ = 45.01% Or, $$\frac{{7{\text{x}} \times 100}}{{13.33}}$$ = 52.51 In same way other percentage are 60.01%, 67.52%, 75.01%. Hence, number of subjects in which he gets more than 50% marks = 4

Let length, breadth and height of the room be 3, 2, 1 unit respectively. Area of walls = 2(l + b) × h = 2(3 + 2) × 1 = 10 sq. unit. Now, length, breadth and height of room will become 6, 1 and $$\frac{1}{2}$$ respectively. Now, area of walls = $$2\left( {6 + 1} \right) \times \frac{1}{2}$$ = 7 sq. unit. % decrease in the area of walls = $$\left( {10 - 7} \right) \times \frac{{100}}{{10}}$$ = 30%

Let the number of bacteria in the 1 st generation be x, then number of bacteria in 2 nd , 3 rd , 4 th . . . . . Generation would be $$8\left( {\frac{{\text{x}}}{2}} \right),\,8\left( {\frac{{4{\text{x}}}}{2}} \right),\,8\left( {\frac{{16{\text{x}}}}{2}} \right)$$ . . . . And so on. As x, 4x, 16x, 64x . . . . . it is in GP with common ratio 4 Hence, 7th term of GP, x(4) 6 = 4096 or, x = 1 or 1 million.

Increase in the price of sugar = (8 + 2) = 10% Hence, price of sugar on, Jan. 1, 1996 $$\eqalign{ & = \frac{{20 \times 110 \times 110}}{{100 \times 100}} \cr & = {\text{Rs}}{\text{.}}\,24.20 \cr} $$

Passed in none = 5% Passed in all = 5% Passed in four = 20% of 90% = 18% Passed in one = 25% of 90% = 22.5% Passed in two = 24.5% Passed in three = (100 - 5 - 5 - 22.5 - 24.5 - 18) = 25% But given 300 students passed in three Hence, 25% = 300 So, 100% = 1200 1200 students must have appeared

Time lost over two weeks = 25% a week time(given that $$\frac{1}{2}$$ % clock loses in first week and in the second week it gains $$\frac{1}{4}$$ % on true time) A week = 168 hours Hence, clock lost = 0.42 hours = 25.2 minutes or 25 minute 12 seconds Thus, correct time = 11 : 34 : 48

$$\eqalign{ & {\text{Shrinking of cloth}}, \cr & = {\frac{{ {36 - 33} }}{{36}}} \times 100 \cr & = \frac{{100}}{{12}}\% \cr & {\text{Second time the strip shrinks,}} \cr & = \frac{{ {48 \times 100} }}{{1200}} \cr & = 4\,\text{inches} \cr & {\text{hence,}}\,{\text{the}}\,{\text{cloth}}\,{\text{remains}} \cr & = 48 - 4 \cr & = 44 \cr} $$

$$\eqalign{ & {\text{According}}\,{\text{to}}\,{\text{question,}} \cr & {\frac{{XY}}{{100}}} + {\frac{{YX}}{{100}}} \cr & = \frac{{2XY}}{{100}} \cr & = 2\% \,of\,XY \cr} $$