Ans .
1218
Percentage of boys = 60%
∴ Percentage of girls = 40%
Boys : Girls = 60 : 40 = 3 : 2
Number of girls = 812
∴ Number of boys = \(\frac{3}{2} \times 812\) = 1218
Ans .
75.5%
Marks scored by A :
First subject \(→\) \(\frac{900 \times 72}{100}\) = 648
Second subject \(→\) \(\frac{700 \times 80}{100}\) = 560
Total marks scored = 648 + 560 = 1208
Total maximum marks = 900 + 700 = 1600
\ Required per cent = \(\frac{1208}{1600} \times 100\) = 75.5%
Ans .
40%
Percentage of failures either in 1 subject or both subjects = (35 + 45 – 20)% = 60%
∴ Percentage of the successful = (100– 60)% = 40%
Ans .
80%
Total marks of 50 students = 50 × 70 = 3500
Total marks of 25 students = 25 × 60 = 1500
Total marks of 24 students = 24 × 80 = 1920
∴ Marks obtained by last student = 3500 – 1500 – 1920
= 80 i.e, 80%
Ans .
42 and 33
Let marks obtained by the first student be x.
∴ Marks obtained by the second student = x – 9
According to the question,
x = 56% of (x + x – 9)
x = \(\frac{(2x - 9) \times 56}{100}\)
\(→\) 100x = 112x – 504
\(→\) 112x – 100x = 504
\(→\) 12x = 504
\(→\) x = \(\frac{504}{12}\) = 42
∴ Marks obtained by the second student = 42 – 9 = 33
Ans .
360 marks
Maximum marks of examination = x (let)
According to the question,
25% of x = 47 + 43
\(\frac{x \times 25}{100}\) = 90
\(\frac{x}{4}\) = 90
\(→\) x = 4 × 90 = 360
Ans .
4% decreased
Change in his salary = 20 - 20 - \(\frac{20 \times 20}{100}\) %
= -\(\frac{400}{100}\)% = -4%
i.e. 4% decrease
Ans .
8% increase
Net % change = A + B + \(\frac{AB}{100}\) %
Here, A = 20%, B = – 10%
∴ Net % change
= 20 - 10 - \(\frac{200}{100}\)
= 10 – 2= 8%
+ ve sign shows increase
Ans .
1% decrease
Required change = \(\frac{(10)^2}{100}\) %decrease
= 1% decrease
Ans .
19%
A single equivalent reduction to reduction series of x%, y%
= x + y - \(\frac{xy}{100}\) %
= 10 + 10 - \(\frac{10 \times 10}{100}\) %
= 20 - 1 % = 19%
Ans .
10% decrease
The net change in price = -25 +20 - \(\frac{-25 \times 20}{100}\) % = (–25 + 20 – 5)% = – 10% Negative sign shows decrease.
Ans .
\( 2 \frac{1}{2}\)% decrease
Let the price of the article be 100 and the daily sale be 100
units.
∴ Revenue day = 100 × 100 = 10000
New receipts = 75 × 130 = 9750
Decrease = (10000 – 9750) = 250
∴ % decrease = \(\frac{2500}{10000} \times 100\) = \( 2 \frac{1}{2}\)% decrease
Ans .
21%
Single equivalent percentage increase in price = 10 + 10 + \(\frac{10 \times 10}{100}\) % = 21%
Ans .
25
Effective increase percentage
= 10 + 20 + \(\frac{10 \times 20}{100}\) % = 32%
\(\frac{x \times 132}{100}\) = 33
\(→\) x = \(\frac{33 \times 100}{132}\) = 25
Ans .
44%
Effective percentage increase = 20 + 20 + \(\frac{20 \times 20}{100}\) % = 44%
Ans .
1% decrease
Net change
= 10 - 10 - \(\frac{10 \times 10}{100}\) %
= –1% = 1% decrease
Ans .
32%
Required percentage increase = 10 + 20 + \(\frac{10 \times 20}{100}\) % = 32%
Ans .
44% increase
Required effect
= 80 - 20 - \(\frac{80 \times 20}{100}\) % = 44%
Positive sign shows increase.
Ans .
decreased by 1%
Net effect = 10 - 10 - \(\frac{10 \times 10}{100}\) % = -1%
Negative sign shows decrease
Ans .
10% decrease
Required per cent effect = 20 - 25 - \(\frac{20 \times 25}{100}\) % = -10%
Ans .
21
Percentage effect = 10 + 10 + \(\frac{10 \times 10}{100}\) % = 21%
Ans .
\(\frac{8}{15}\)
Original fraction = \(\frac{x}{y}\)
∴ \(\frac{\frac{120x}{100}x}{y \times \frac{80}{100}}\) = \(\frac{4}{5}\)
∴ \(\frac{120x}{80y}\) = \(\frac{4}{5}\)
∴ \(\frac{6x}{4y}\) = \(\frac{4}{5}\)
∴ \(\frac{x}{y}\) = \(\frac{8}{15}\)
Ans .
\(\frac{95}{48}\)
Let original fraction be
\(\frac{x}{y}\)
According to the question,
\(\frac{\frac{120}{100}x}{\frac{95y}{100}}\) = \frac{5}{2}
\(\frac{120x}{95y}\) = \frac{5}{2}
\(\frac{x}{y}\) = \frac{95}{48}
Ans .
decreased by \( 6 \frac{1}{4}\)%
Effective value
= x + y + \(\frac{x \times y}{100}\) %
= 25 - 25 - \(\frac{25 \times 25}{100}\) % = decreased by \( 6 \frac{1}{4}\)%
Ans .
240
Larger number = x and smaller
number = 520 – x
\(\frac{96x}{100}\) = \(\frac{520-x}{100} \times 112\)
\(→\) 96x = 520 × 112 – 112x
\(→\) 112x + 96x = 520 × 112
\(→\) 208x = 520 × 112
\(→\) x = \(\frac{520 \times 112}{208}\) = 280
∴ Smaller number
= 520 – 280 = 240
Ans .
400
If the original price of article
be x, then
\(x \times \frac{80}{100}\times \frac{130}{100}\) = 416
\(→\) x = \(\frac{416 \times 100\times 100}{80 \times 130}\) = 400
Ans .
80
If the number be x, then \(x \times \frac{245}{200}\) = 98
\(→\) x = \(\frac{98 \times 200}{245}\) = 80
Ans .
\( 11 \frac{1}{9}\)%
Let the original price be 100
New price after 10% decrease = 90
In order to restore the price to its original value,
it must be increased by 10% increase = \(\frac{10}{90} \times 100\) = \(\frac{100}{9}\) = \( 11 \frac{1}{9}\)%
Ans .
25%
Clearly, 75% of the number = 225
∴ Number = \(\frac{225 \times 100}{75}\) = 300
Again, 125% of 300 = 375
Hence, the number should be increased by 25%
Ans .
\( 30 \frac{5}{9}\)%
Let the number be 100.
After 20% increase, number = 120
After 20% increase of 120, number = \(120 \times \frac{120}{100}\) = 144
∴ Per cent decrease
= \(\frac{44}{144} \times 100\)
= \(\frac{275}{9}\) = \( 30 \frac{5}{9}\)%
Ans .
\(\frac{25}{4}\)%
Let the original number of employees be 100 and wages per
head be 100.
Total wages = (100 × 100) = 10000
New number of employees = 125
New wages per head = 75
Total new wages = (125 × 75) = 9375
Decrease = (10000 – 9375) = 625
∴ Percentage decrease = \(\frac{625}{10000} \times 100\)
= \(\frac{625}{100}\) = \(\frac{25}{4}\)%
Ans .
5000
Let original number be x
\(\frac{90}{100}x \times \frac{110}{100}\) = x - 50
\(\frac{99x}{100}\) = x-50
x - \(\frac{99x}{100}\) = 50
\(\frac{x}{100}\) = 50
\(→\) x = 5000
Ans .
5%
Let the income be x and the rate of income tax be y %
According to the question,
\(\frac{xy \times 1.19}{100}\) - \(\frac{xy}{100}\) = \((x - \frac{xy}{100}) \times \frac{1}{100}\)
\(→\) 1.19 xy – xy = - \(\frac{xy}{100}\)
\(→\) 0.19 y = - \(\frac{y}{100}\)
\(→\) \(\frac{y}{100}\) + 0.19y = 1 \(→\) y\(\frac{1+19}{100}\) = 1
\(→\) = \(\frac{100}{20}\) = 5%
Ans .
50%
Man’s income = 100 (let)
Expenditure = 75
Savings = 25
New income = \(\frac{100 \times 120}{100}\) = 120
New expenditure = \(\frac{75 \times 110}{100}\) = 82.5
Savings = 120 – 82.5 = 37.5
Increase in savings = 37.5 – 25 = 12.5
\ Increase per cent = \(\frac{12.5}{25} \times 100\) = 50%
Ans .
33.1%
Single equivalent increase for 10% and 10% = 10 + 10 + \(\frac{10 \times 10}{100}\) % = 21%
Again, single equivalent increase for 21% and 10% = 21 + 10 + \(\frac{21 \times 10}{100}\) % = 33.1%
Ans .
increased by 8.9%
Increase in first year = 10%
Decrease in 2nd year = 10%
Effective result = 10 - 10 - \(\frac{10 \times 10}{100}\) %
= –1%
Increase in 3rd year = 10%
∴ Effective result = 10 - 1 - \(\frac{10 \times 1}{100}\)%
= (9 – 0.1)% = 8.9% (increase)
Ans .
80
Let the number be x.
∴ (20 + 25)% of x = 36
\(\frac{45x}{100}\) = 36
\(→\) x = \(\frac{36 \times 100}{45}\) = 80
Ans .
500
Effective percentage = -20 + 20 - \(\frac{20 \times 20}{100}\) % = -4%
If the number be x, then 4% of x = 20
\(x \times \frac{4}{100}\) = 20
x = \(\frac{20 \times 100}{4}\) = 500
Ans .
\(\frac{100x}{100+x}\)%
\(\frac{B \times 30}{100}\) \(→\) increased value \(→\) P + \(\frac{Px}{100}\)
= P\(\frac{100+x}{100}\)
∴ Required answer =\(\frac{x}{100+x} \times 100\)
Ans .
28%
Effective percentage decrease = x + y + \(\frac{x \times y}{100}\) %
= - 10 - 20 + \(\frac{-10 \times -20}{100}\) %
= (– 30 + 2)% = –28%
Ans .
0
Cost of edible oil = 100 per kg.
Consumption = 1 kg.
Again, New price = 125 per kg.
Consumption = 0.8 kg.
Expenditure = Rs. (125 × 0.8) = Rs. 100
Percentage effect = 25 - 20 - \(\frac{20 \times 25}{100}\) % = 0%
Ans .
27500
Let the total number of votes be 100.
Number of uncast votes = 8
∴ Number of votes polled = 92
Number of votes obtained by the winner = 48
∴ Number of votes obtained by the loser = 92 – 48 = 44
If the difference of win be 4 votes, total voters = 100
∴ When the difference be 1100 votes, total voters
= \(\frac{100}{4} \times 1100\) = 27500
Ans .
16800
Let the total number of voters enrolled be x.
Number of votes polled = = 75% of x = \(\frac{3x}{4}\)
Number of valid votes = \(\frac{3x}{4}\) - \(\frac{2}{100} \times \frac{3x}{4}\)
= \(\frac{3x}{4}\) - \(\frac{2}{100} \times \frac{3x}{4}\)
= \(\frac{147x}{200}\)
Now, 75% of = \(\frac{147x}{200}\) = 9261
or \(\frac{3}{4}\) of \(\frac{147x}{200}\) = 9261
or x = \(\frac{9261 \times 4 \times 200}{3 \times 147}\) = 16800
Ans .
42,000
Difference of percentage of
votes = 60% – 40% = 20%
∴ 20% of total votes = 14000
∴ 60% of total votes
= \(\frac{14000}{20} \times 60\) = 42000
Ans .
52%
Let total employees = 100
∴ Required percentage = \(\frac{40 \times 40}{100}\) + \(\frac{60 \times 60}{100}\)
= 16 + 36 = 52%
Ans .
1490
Let votes polled = x
\(x \times \frac{60-40}{100}\) = 298
\(→\) x ×\(\frac{1}{5}\) = 298
\(→\) x = 298 × 5 = 1490
Ans .
56056
Number of valid votes = \(104000 \times \frac{98}{100}\) = 101920
∴ Valid votes received by the candidate = \(\frac{101920 \times 55}{100}\) = 56056
Ans .
8000
If the number of votes polled be x, then \(\frac{x \times 20}{100}\) = 1600
\(→\) x = \(\frac{1600 \times 100}{20}\) = 8000
Ans .
8640
Vote percentage of third candidate
= 100 – 40 – 36 = 24%
∴ Votes got by third candidate = \(\frac{36000 \times 24}{100}\) = 8640
Ans .
3,00,000
Total votes polled = x
∴ (57 – 43)% of x = 42000
\(x \times \frac{14}{100}\) = 42000
x = \(\frac{42000 \times 100}{14}\) = 300000
Ans .
700
Total number of votes polled = x
\(\frac{x \times 84}{100}\) - \(\frac{x \times 16}{100}\) = 476
\(\frac{68x}{100}\) = 476
\(→\) x = \(\frac{476 \times 100}{68}\) = 700
Ans .
30000
Number of valid votes = x (let)
\ (62 – 38)% of x = 7200
\(x \times \frac{24}{100}\) = 7200
x = \(\frac{7200 \times 100}{24}\) = 30000
Ans .
600
Total number of votes polled = x (let)
According to the question
\(\frac{x \times 62}{100}\) - \(\frac{x \times (100 - 62)}{100}\) = 144
\(\frac{x \times 62}{100}\) - \(\frac{x \times 38}{100}\) = 144
x = 600
Ans .
6200
Total voters in the list = x
Votes got by the winner = \(\frac{47x}{100}\)
Votes got by the loser = x – \(\frac{x}{10}\) - 60 - \(\frac{47x}{100}\)
= \(\frac{9x}{10}\) - 60 - \(\frac{47x}{100}\)
= \(\frac{43x}{100}\) - 60
According to the question,
\(\frac{47x}{100}\) - \(\frac{43x}{100}\) + 60 = 308
x = 6200
Ans .
1490
Total votes polled = x
According to the question,
(60 – 40)% of x = 298
\(→\) \(x \times \frac{20}{100}\) = 298
\(→\) \(\frac{x}{5}\) = 298
\(→\) x = 298 × 5 = 1490
Ans .
217800
Required population after two years = = 180000 \((1+ \frac{10}{100})^2\)
= \(180000 \times \frac{11}{10}\times \frac{11}{100}\) = 217800
Ans .
48.8%
If the present worth of the equipment be 100, then its price after 3 years
= \(100 \times (\frac{80}{100})^3\) = 51.2
∴ Depriciation = 48.8%
Ans .
68921
P = P0 \((1 + \frac{R}{100})^T\)
P = 64000 \((1 + \frac{5}{200})^3\)
P = 64000 \(( \frac{41}{40})^3\)
P = \(\frac{64000 \times 41 \times 41 \times 41}{40 \times 40 \times 40}\)
P = 68921
Ans .
10000
Suppose the value of property two years ago was x
According to question
∴ x\((1-\frac{10}{100})^2\) = 8100
∴ x\((\frac{90}{100})^2\) = 8100
\(→\) x = 10000
Ans .
56,700
Let the present population be P.
∴ P = 62500\((1 - \frac{4}{100})^2\)
P = 57600
Ans .
54,080
Required population = \(( 1 + \frac{4}{100})^2\) = 54,080
Ans .
Let the man’s annual salary in 2006 be x.
∴ \(\frac{110x}{100}\) = 880000
x = = 800000
Ans .
62500
Population of the village two years ago = \(\frac{P}{(1+\frac{R}{100})^2}\)
= \(\frac{67600}{(1+\frac{4}{100})^2}\)
= 62500
Ans .
1,80,500
A = \(P (1-\frac{R}{100})^T\)
A = \(200000 (1-\frac{5}{100})^2\)
= 1,80,500
Ans .
4,80,000
Value of the property 3 years ago
= \(\frac{P}{(1-\frac{R}{100})^T}\)
= \(\frac{411540}{(1-\frac{5}{100})^3}\)
= 480000
Ans .
85184
Population of town = \(P (1-\frac{R}{100})^T\)
= \(P (1-\frac{R}{100})^T\)
= \(64000 (1-\frac{10}{100})^3\)
= 85184
Ans .
6400
Present population = \(10000 (1-\frac{20}{100})^2\) = 6400
Ans .
5
Required percent = \(\frac{1}{4} \times 3\) + \(\frac{2}{3} \times 5\) + \( (1-\frac{1}{4}-\frac{2}{3}) \times 11\) = 5%
Ans .
3,150
Required price of the machine = = \(6250 (1-\frac{10}{100})(1-\frac{20}{100})(1-\frac{30}{100})\) = 3150
Ans .
40,500
Required value = \(50000 (1-\frac{10}{100})^2\) = 40,500
Ans .
1000
If the price of machine 3years ago be x. then 729 = \(x (1-\frac{10}{100})^3\) 729 = \(x \times (\frac{9}{10})^3\) \(→\) x = 1000
Ans .
1720
Required Raman’s salary = \(\frac{100}{100+5} \times 1806\) = \(\frac{100}{105} \times 1806\) = 1720
Ans .
4000
∴ Men : Women = 1 : 1
∴ Number of men = \(\frac{1}{2} \times 8000\) = 4000
Ans .
5600
Let the number of males = x
\ Number of females= 9800 – x
According to the question,
\(x \times \frac{108}{100}\) + \((9800-x) \times \frac{105}{100}\) = 10858
\(→\) 108 x + 9800 ×105 – 105x
= 1045800
\(→\) 3x + 1029000 = 1045800
\(→\) 3x = 1045800 – 1029000
= 16800
\(→\) = \(\frac{16800}{3}\) = 5600
Ans .
5120
Population in the beginning of the year = \(\frac{10000}{ (1+\frac{25}{100})^3}\) = 5120
Ans .
111%
If the number of men be 100, then Number of women = 90
∴ Required per cent = \(\frac{100}{90} \times 100\) = 111%
Ans .
562432
Required population = \(500000 (1+\frac{4}{100})^3\) = 562432
Ans .
4000
If the population of village two years ago be P0
, then 4410 = \(P0 (1+\frac{1}{20})^2\)
P0 = 4000
Ans .
19,950
Value of TV after one year = 21000 × (100 – 5)% = \(\frac{21000 \times 95}{100}\) = = Rs. 19950
Ans .
72.8 %
Single equivalent increase for 20% and 20% = 20 + 20 + \(\frac{20 \times 20}{100}\) % = 44%
Single equivalent increase for 44% and 20% = 44 + 20 + \(\frac{44 \times 20}{100}\) % = = 72.8%
Ans .
32.5
Population of town = 1000
Males \(→\) 600
Females \(→\) 400
Literate males = \(\frac{600 \times 20}{100}\) = 20
Total literate inhabitants = \(\frac{1000 \times 25}{100}\) = 250
∴ Literate females = 250 – 120 = 130
∴ Required percent = \(\frac{130}{400} \times 100\) = 32.5%
Ans .
10%
If the rate of increase per annum be R%, then
A = \(P (1-\frac{R}{100})^T\)
48400 = \(40000 (1+\frac{R}{100})^2\)
R = 10% per annum
Ans .
Rs. 10,000
A = \(P (1-\frac{R}{100})^3\)
7290 = \(P (1-\frac{10}{100})^3\)
P = Rs. 10000
Ans .
8000
9261 = \(P0 (1+\frac{5}{100})^3\)
P0 = 8000
Ans .
6125
Original population of village = x (let)
According to the question,
\(x \times \frac{95}{100} \times \frac{80}{100}\) = 4655
x = 6125
Ans .
9600
In the village,
Females = 3000
Males = 9000 – 3000 = 6000
After respective increases,
Population of village = \(3000 \times \frac{105}{100}\) + \(\frac{6000 \times 107.5}{100}\)
= 3150 + 6450 = 9600
Ans .
60
Let the population of the city be 100.
Total illiterate people = 40
Poor people = 60
Rich people = 40
Illiterate rich people = \(\frac{40 \times 10}{100}\) = 4
∴ Illiterate poor people = 40 – 4 = 36
∴ Required per cent = \(\frac{36}{60} \times 100\)% = 60%
Ans .
31200
Population of city after two years = \(P (1+\frac{R1}{100})(1+\frac{R1}{100})\) = \(20000 (1+\frac{20}{100})(1+\frac{30}{100})\) = \(20000 \times \frac{120}{100} \times \frac{130}{100}\) = 31200
Ans .
4100
Let the total population of village be x.
According to the question,
\(\frac{x \times 14}{100}\) = 574
\(→\) x = \(\frac{574 \times 100}{14}\) = 4100
Ans .
13.50
10 per cent of 837 = \(\frac{10}{100} \times 837\) = 83.7
∴ Reduced per kg price = \(\frac{83.7}{62}\) = 13.50
Ans .
18
Let the original price of sugar = x /kg.
Reduced price of sugar = 80% of x
= \(\frac{x \times 80}{100}\) = \(\frac{4 \times x}{5}\)
∴ \(\frac{36}{\frac{4x}{5}}\) - \(\frac{36}{x}\) = \(\frac{1}{2}\)
\(→\) x = 9 × 2 = 18/Kg
Ans .
18
Reduced price of 6.2kg of sugar
= 10% of 1116
= 111.6
∴ Reduced price per kg = \(\frac{111.6}{62}\) = 18
Ans .
30
Let the original price of sugar be x/kg.
∴ New price = \(\frac{9x}{10}\)
\(\frac{270}{\frac{9x}{10}}\) - \(\frac{270}{x}\) = 1
\(→\) x = 30 /kg
Ans .
12.96
Let the original price of apples be x/ dozen
∴ New price = \(\frac{4x}{5}\) /dozen
\(\frac{54}{\frac{4x}{5}}\) - \(\frac{54}{x}\) = \(\frac{10}{2}\)
\(\frac{4x}{5}\) = 12.96
Ans .
36
The original price of 1 egg = x
Present price = \(\frac{3}{2}x\)
\(\frac{24}{x}\) - \(\frac{24}{\frac{3x}{2}}\) = 4
\(→\) x = 2
∴ Present price of eggs per dozen = \(12 \times \frac{3}{2} \times 2\) = 36
Ans .
16
Original price of wheat = x /kg.
New price of wheat = \(\frac{4x}{5}\)
\(\frac{320}{\frac{4x}{5}}\) - \(\frac{320}{x}\) = 5
\(→\) x = 16
Ans .
28.80
Original rate = x per egg
New rate = \(\frac{6x}{5}\) per egg
\(\frac{24}{x}\) - \(\frac{24 \times 5}{6x}\) = 2
x = 2
∴ New rate = \(\frac{12}{5}\) per egg.
∴ Rate per dozen of eggs = \(\frac{12}{5} \times 12\)= 28.80
Ans .
15
Let original price of rice per kg = x (let)
∴ New price of rice per kg = \(\frac{3x}{4}\)
\(\frac{600}{\frac{3x}{4}}\) - \(\frac{600}{x}\) = 10
\(→\) x = 20
∴ New price = \(\frac{3 \times 20}{4}\) = 15/kg
Ans .
20%
Percentage decrease = \(\frac{0.25}{1.25} \times 100\) = 20%
Ans .
30
If the number be x, then x – 15 = \(\frac{4x}{5}\)
\(→\) 5x – 75 = 4x \(→\) x = 75
∴ 40% of 75 = \(\frac{75 \times 40}{100}\) = 30