Ans .
48,900
Women = \(\frac{43}{83} \times 311250\) = 161250
Men = 311250 – 161250 = 150000
∴ Total number of literate persons
= \(\frac{161250 \times 8}{100}\) + 150000\(\times \frac{24}{100}\)
= 12900 + 36000 = 48900
Ans .
500
7x – 5x = 200
\(→\) 2x = 200 \(→\) x = 100
∴ Price of a pair of shoes
= 5x = 5 × 100 = 500
Ans .
4 : 3
\(x \times \frac{15}{100}\) = \(y \times \frac{20}{100}\)
\(→\) x × 15 = y × 20
\(→\) \(\frac{x}{y}\) = \(\frac{20}{15}\) = \(\frac{4}{3}\) = 4 : 3
Ans .
72
Boys in school = 2x
Girls = 3x
Students who are not scholarship holders :
Boys \(→\) \(\frac{2x \times 75}{100}\) = \(\frac{6x}{4}\)
Girls \(→\) \(\frac{3x \times 70}{100}\) = \(\frac{21x}{10}\)
Total students who do not hold scholarship = \(\frac{6x}{4}\) + \(\frac{21x}{10}\)
= \(\frac{72x}{20}\) = \(\frac{18x}{5}\)
∴ Required percentage = \(\frac{\frac{18x}{5}}{5x} \times 100\) = 72%
Ans .
4 : 3
Numbers \(→\) A and B
∴ \(\frac{A \times 5}{100}\) + \(\frac{B \times 4}{100}\)
= \(\frac{2}{3}\) \((\frac{A \times 6}{100} + \frac{B \times 8}{100})\)
\(→\) 5A + 4B = \(\frac{12A + 16B}{3}\)
\(→\) 15A + 12B = 12A + 16B
\(→\) 15A – 12A = 16B – 12B
\(→\) 3A = 4B
\(→\) \(\frac{A}{B}\) = \(\frac{4}{3}\)
Ans .
3 : 4
Acid – I
60
60 – 40 = 20
Acid – II
25
40 – 25 = 15
∴ Required ratio = 15 : 20 = 3 : 4
Ans .
3 : 5
50% of x = 30% of y
\(→\)\(\frac{x \times 50}{100}\) = \(\frac{y \times 30}{100}\)
\(→\)\(\frac{x}{y}\) = \(\frac{30}{50}\) = \(\frac{3}{5}\)
Ans .
23 : 27
Boys in the village = 3x
Girls in the village = 2x
Villagers who appeared in the examination
= \(\frac{3x \times 30}{100}\) + \(\frac{2x \times 70}{100}\)
= \(\frac{9x}{10}\) + \(\frac{14x}{10}\)
\(\frac{23x}{10}\)
Villagers who did not appear in the examination
= \(\frac{3x \times 70}{100}\) + \(\frac{2x \times 30}{100}\)
= \(\frac{21x}{10}\) + \(\frac{6x}{10}\)
\(\frac{27x}{10}\)
Required ratio = \(\frac{23x}{10}\) : \(\frac{27x}{10}\)
= 23 : 27
Ans .
1 : 4
C.P. of 1 litre of milk= Rs. 100
∴ Mixture sold for Rs. 125 = \(\frac{125}{100}\) = \(\frac{5}{4}\) litre
∴ Quantity of water = \(\frac{5}{4}\) - 1 = \(\frac{1}{4}\) litre
∴ Required ratio = \(\frac{1}{4}\) : 1 = 1 : 4
Ans .
75%
Percentage of syrup = \(\frac{3}{4} \times 100\) = 75%
Ans .
12
Let the numbers be 5x and 4x respectively
According to the question ,
\(5x \times \frac{40}{100}\) = 12
\(→\) 2x = 12 \(→\) x = 6
∴ 4x 50% = 4 × 6 × \(\frac{1}{2}\) = 12
Ans .
9 : 2
According to the question,
\(x \times \frac{10}{100}\) = \(3 \times y \times \frac{15}{100}\)
\(→\) 10x = 45y
\(→\) \(\frac{x}{y}\) = \(\frac{45}{10}\) = \(\frac{9}{2}\)
Ans .
110%
Required per cent = \(\frac{11}{10} \times 100\) = 110%
Ans .
78
Let the number of students
in school be 100.
Boys \(→\) 60
Girls \(→\) 40
Students who do not hold scholarship :
Boys \(→\) \(\frac{60 \times 80}{100}\) = 48
Girls \(→\) \(\frac{40 \times 75}{100}\) = 30
Required answer = 48 + 30
= 78 i.e., 78%
Ans .
5 : 7
According to the question,
A × 35% = B × 25%
\(→\) \(\frac{A}{B}\) =\(\frac{25}{35}\) =\(\frac{5}{7}\)
Ans .
150 litres
Glycerine in mixture = 40 litres
Water = 10 litres
Let x litres of pure glycerine is mixed with the mixture.
∴ \(\frac{40 + x}{50+x}\) = \(\frac{95}{100}\) = \(\frac{19}{20}\)
\(→\) 800 + 20x = 950 + 19x
\(→\) x = 950 – 800 = 150 litres
Ans .
\( 33 \frac{1}{3}\)%
Alcohol in original solution = \(\frac{40}{100} \times 5\) = 2 litres
Water in original solution = 3 litres
On adding 1 litre water, water becomes 4 litres.
Now, 6 litres of solution contains 2 litres of alcohol.
∴ 100 litres of solution contains = \(\frac{2}{6} \times 100\)
= \(\frac{100}{3}\) = \( 33 \frac{1}{3}\)% alcohol
Ans .
10.5%
In 12 litres salt solution,
Salt = \(\frac{7 \times 12}{100}\) = 0.84 units
Water = \(\frac{93 \times 12}{100}\) = 1116. units
After evaporation,
Percentage of salt = \(\frac{0.84}{8} \times 100\) = 10.5%
Ans .
20 litres
In 60 litres of solution, Water
= \(\frac{60 \times 20}{100}\) = 12 litres
On adding x litres of water, \(\frac{12 + x}{60+x} \times 100\) = 40
\(→\) 60 + 5x = 120 + 2x
\(→\) 3x = 60
\(→\) x = 20 litres
Ans .
100 gm
Sugar in original solution
= \(\frac{75 \times 30}{100}\) = 22.5 gm
Let x gm of sugar be mixed.
∴\(\frac{22.5 + x}{75 + x} \times 100\) = 70
\(→\) 2250 + 100x = 75 × 70 + 70x
\(→\) 2250 + 100x = 5250 + 70x
\(→\) 30x = 5250 – 2250 = 3000
\(→\) x = \(\frac{3000}{30}\) = 100 gm
Ans .
60%
In 30% alcohol solution,
Alcohol = \(\frac{30}{100} \times 6\) = 1.8 litres
Water = 4.2 litres
On mixing 1 litre of pure alcohol,
Percentage of water = \(\frac{4.2}{7} \times 100\) = 60%
Ans .
266.67
In 4 kg of ore, iron = 0.9 kg.
∴ Quantity of ore for 60 kg of iron = \(\frac{60 \times 4}{0.9}\) = 266.67 kg
Ans .
60 ml
Let x ml of water be added.
∴ \(\frac{20 + x}{100 + x} \times 100\) = 50
\(→\) 40 + 2x = 100 + x
\(→\) x = 60 ml
Ans .
1000 ml
In 1 litre i.e. 1000 ml of mixture,
Alcohol = 700 ml.
Water = 300 ml.
Let x ml of alcohol is mixed.
\(\frac{300}{1000 + x} \times 100\) = 15
\(→\) 1000 + x = 2000
\(→\) x = 1000 ml.
Ans .
\( 24 \frac{2}{7}\)%
In 10 litres of first type of liquid,
Water = \(\frac{1}{5} \times 10\) = 2 litres
In 4 litres of second type of liquid,
Water = \(4 \times \frac{35}{100}\) = \(\frac{7}{5}\) litres
Total amount of water = \(2 + \frac{7}{5}\) = \(\frac{17}{5}\) litres
Required percentage = \(\frac{\frac{17}{5}}{14} \times 100\) = \(\frac{170}{7}\) = \( 24 \frac{2}{7}\)%
Ans .
5 litres
Water content in 40 litres of mixture
= \(40 \times \frac{10}{100}\) = 4 litres
∴ Milk content = 40 – 4 = 36 litres
Let x litres of water is mixed.
Then,\(\frac{4 +x }{40+x}\) = \(\frac{20}{100}\) = \(\frac{1}{5}\)
\(→\) 20 + 5x = 40 + x
\(→\) 4x = 20 \(→\) x = 5 litres
Ans .
100 ml
Alcohol = \(\frac{15}{100} \times 400\)ml = 60 ml.
Water = 340 ml.
Let x ml of alcohol be added.
Then,\(\frac{60 + x}{400 + x} \times 100\) = 32
\(\frac{60 + x}{400 + x}\) = \(\frac{32}{100}\)= \(\frac{8}{25}\)
or 1500 + 25x = 3200 + 8x
or 17x = 1700
or x = 100 ml
Ans .
150 gm
Initial quantity of gold = \(\frac{50 \times 80}{100}\) = 40 gm
Let 'x ' gm be mixed.
(40+x) = \((50 + x) \times \frac{95}{100}\)
(40+x) = \((50 + x) \times \frac{19}{20}\)
\(→\) 800 + 20x = 950 + 19x
\(→\) x = 150 gm
Ans .
40 litres
In 200 litres of mixture,
Quantity of milk = \(\frac{85}{100} \times 200\) = 170 litres
Quantity of water = 30 litres
Let the quantity of additional milk added be x litres.
According to the question
\(\frac{170+x}{200+x} \times 100\) = 87.5
\(→\) (170 +x) × 100 = 17500 + 87.5x
\(→\) 100x – 87.5x = 17500 – 17000
\(→\) 12.5x = 500
\(→\) x = \(\frac{500}{12.5}\) = 40 litres
Ans .
1 : 3
Let x litres of first mixture is mixed with y litres of the second mixture.
According to the question,
\(\frac{x \times \frac{30}{100} + y \times \frac{50}{100}}{x \times \frac{70}{100} + y \times \frac{50}{100}}\) = \(\frac{45}{55}\)
\(\frac{0.3x + 0.5y}{0.7x + 0.5y}\) = \(\frac{9}{11}\)
\(→\) 6.3x + 4.5y = 3.3x + 5.5y
\(→\) 6.3x – 3.3x = 5.5y – 4.5y
\(→\) 3x = y
\(→\) \(\frac{x}{y}\) = 1:3
Ans .
2 : 3
∴ Required ratio = 10 : 15 = 2 : 3
Ans .
\( 16 \frac{2}{3}\)%
Alcohol = \(15 \times \frac{1}{5}\) = 3 litres
Water = \(15 \times \frac{4}{5}\) = 12 litres
∴ Required percentage
= \(\frac{3}{15 + 3} \times 100\)
= \(\frac{50}{3}\) = \( 16 \frac{2}{3}\)%
Ans .
575 kg
∵ 12 kg copper is contained
in 100 kg of alloy
69 kg copper is contained in
\(\frac{100}{12} \times 69\)= 575 kg of alloy
Ans .
2 : 3
∴ Required ratio = \(\frac{1}{10}\) : \(\frac{3}{20}\) = 2 : 3
Ans .
\( 16 \frac{2}{3}\)%
In 20 litres of mixture,
Alcohol \(→\) \(\frac{20 \times 20}{100}\) = 4 litres
Water \(→\) 20 - 4 = 16 litres
On adding 4 litres of water,
Quantity of water \(→\) 16 + 4 = 20 litres
Quantity of mixture = 24 litres
∴ Required per cent = \(\frac{4}{24} \times 100\) = \(\frac{50}{3}\) = \( 16 \frac{2}{3}\)%
Ans .
60 gram
In 300 gm of solution,
Sugar = \(\frac{300 \times 40}{100}\) = 120 gm.
Let x gm of sugar be mixed.
According to the question,
\(\frac{120 + x}{300 + x}\) = \(\frac{1}{2}\)
\(→\) 240 + 2x = 300 + x
\(→\) 2x – x = 300 – 240
\(→\) x = 60 gm.
Ans .
45
Quantity of sugar in the solution
= \(\frac{3 \times 60}{100}\) = 1.8 units
On adding 1 litre of water,
∴ Required percent = \(\frac{1.8}{4} \times 100\) = 45%
Ans .
16%
In 32 litres of solution,
Alcohol = \(\frac{32 \times 20}{100}\) = 6.4 litres
Water = 32 – 6.4 = 25.6 litres
On adding 8 litres of water,
Required percent = \(\frac{6.4}{40} \times 100\) = 16%
Ans .
\( 16 \frac{2}{3}\)%
Required percentage decrease
= \(\frac{Increase}{100 + Increase} \times 100\)
= \(\frac{20}{100 + 20} \times 100\)
= \(\frac{100}{6} \) = \( 16 \frac{2}{3}\)%
Ans .
\( 9 \frac{1}{11}\)%
Required answer
= \(\frac{10}{100 + 10} \times 100\)
= \(\frac{10}{110} \times 100\)
\(\frac{100}{11}\)% = \( 9 \frac{1}{11}\)%
Ans .
20%
Required reduction in consumption
= \(\frac{x}{100 + x} \times 100\)%
where x = 25
= \(\frac{25}{100 + 25} \times 100\)% = 20%
Ans .
\( 16 \frac{2}{3}\)%
Reduction in consumption = \(\frac{R}{100 + R} \times 100\)% = \(\frac{20}{120} \times 100\)% = \(\frac{50}{3}\)% = \( 16 \frac{2}{3}\)%
Ans .
4% decrease
Let the CP of each article = 100 and consumption = 100 units
Initial expenditure = (100 × 100) = 10000
New price of article = 80
Consumption = 120 units
Expenditure = (120 × 80) = 9600
Decrease = (10000 – 9600) = 400
∴ Percentage decrease = \(\frac{400 \times 100}{10000}\) = 4%
Ans .
\( 13 \frac{1}{23}\)%
If the price of a commodity increases by R%, then reduction
in consumption, not to increase the expenditure is given by
\(\frac{R}{100 + R} \times 100\) %
=\(\frac{15}{100 + 15} \times 100\) %
=\(\frac{15}{115} \times 100\) %
=\(\frac{300}{23}\)
\( 13 \frac{1}{23}\)%
Ans .
\(\frac{1}{3}\)
Required fractional decrease = \(\frac{R}{100 + R}\) = \(\frac{50}{100 + 50}\) = \(\frac{1}{3}\)
Ans .
20%
Percentage decrease = \(\frac{25}{125} \times 100\) = 20%
Ans .
\( 66 \frac{2}{3}\)%
Required increase percent = \(\frac{40}{100 - 40} \times 100\) = \(\frac{40}{60} \times 100\) = \( 66 \frac{2}{3}\)%
Ans .
\( 16 \frac{2}{3}\)%
Required percentage decrease = \(\frac{20}{100+20} \times 100\) = \(\frac{50}{3}\) = \( 16 \frac{2}{3}\)%
Ans .
20%
Percentage increase =
\(\frac{7.50 -6 }{6} \times 100\) = 25%
Percentage decrease in consumption
= \(\frac{25}{125} \times 100\) = 20%
Ans .
4% decrease
Percentage effect = 20 - 20 + \(\frac{20 \times -20}{100}\) % = – 4% Negative sign shows decrease
Ans .
37.5%
Required percentage = \(\frac{60}{160} \times 100\) = \(\frac{300}{8}\) = \(\frac{75}{2}\) = 37.5%
Ans .
20%
Required per cent = \(\frac{25 \times 100}{125}\) = 25%
Ans .
\( 16 \frac{2}{3}\)%
Percentage decrease in the consumption
of petrol
= \(\frac{20}{100 + 20} \times 100\)%
= \(\frac{50}{3}\) = \( 16 \frac{2}{3}\)%
Ans .
44.4%
Total candidates
= 1000 + 800 = 1800
The candidates who are passed
= \(1000 \times \frac{60}{10}\) + \(800 \times \frac{50}{100}\)
= 600 + 400 = 1000
The number of candidates who failed = 1800 – 1000 = 800
∴ Required percent = \(\frac{800}{1800} \times 100\) = 44.4%
Ans .
22%
Let the maximum marks be x.
According to question,
20% of x + 5 = 30% of x –20
\(→\) (30 –20)% of x = 25
\(→\) x = \(\frac{25 \times 100}{10}\) = 250
∴ Passing marks = 20% of 250 + 5 = 55
∴ % Passing marks = \(\frac{55}{250} \times 100\) = 22%
Ans .
600
According to question,
40% \(→\) 220 + 20 or 40% \(→\) 240
∴ 100% \(→\) \(\frac{240}{40} \times 100\) = 600
Ans .
500
Let the total marks be x.
According to the question,
25% of x + 40 = 33% of x
\(→\) (33 – 25)% of x = 40
\(→\) 8% of x = 40
\(→\) x = \(\frac{40 \times 100}{8}\) = 500
Ans .
42, 33
Let the marks obtained by first student be x.
∴ Marks obtained by second student = x + 9
Sum of their marks = 2x + 9
As given,
x + 9 = 56% of (2x + 9)
\(→\) x + 9 = \(\frac{56}{100} \times (2x+9)\)
\(→\) x + 9 = \(\frac{15}{25} \times (2x+9)\)
\(→\) 25x + 225 = 28x + 126
\(→\) 3x = 225 – 126
\(→\) x = \(\frac{99}{3}\) = 33
Ans .
90
Let marks obtained by Supriyo = x
∴ \(\frac{9x}{10}\) = 81 \(→\) x = \(\frac{81 \times 10}{9}\) = 90
Ans .
250
Let the maximum marks be x.
According to the question,
40% of x = 90 + 10
\(→\) x = \(\frac{100 \times 100}{40}\) = 250
Ans .
17%
Students passed only in Math = 65 – 30 = 35%
Students passed only in Physics = 48 – 30 = 18%
∴ Total passing % = 35 + 18 + 30 = 83%
∴ Failed = 100 – 83 = 17%
Ans .
250
Let the number of students
in the class be 100.
∴ Number of students in Biology = 72 and number of students in Maths = 44.
∴ Number of students opting for both subjects = 72 + 44 – 100 = 16
∵ When 16 students opt for both subjects, total number of students = 100
∴ When 40 students opt for both subjects,
total number of students = \(\frac{100}{16} \times 40\) = 250
Ans .
500
Let the maximum marks be x.
∴ \(\frac{x \times 33}{100}\) = 125 + 40 = 165
\(→\) x = \(\frac{165 \times 100}{33}\) = 500
Ans .
550
Let maximum marks be x,
then,
\(\frac{36 \times x}{100}\) = 113 + 85 = 198
\(→\) x = \(\frac{198 \times 100}{36}\) = 550
Ans .
67%
46% of 500 = \(\frac{500 \times 46}{100}\) = 230
32% of 300 = \(\frac{300 \times 32}{100}\) = 96
Required marks = 230 – 96 = 134
Let x% of 200 = 134
\(\frac{200 \times x}{100}\) = 134
2x = 134
x = \(\frac{134}{2}\) = 67%
Ans .
80%
A = 360;
B = \(\frac{360 \times 100}{90}\) = 400
C = \(\frac{400 \times 100}{125}\) = 320
D = \(\frac{320 \times 100}{80}\) = 400
\ Required percentage = \(\frac{400}{500} \times 100\) = 80%
Ans .
54·5%
Failed candidates = \(\frac{1100 \times 50}{100}\) + \(\frac{900 \times 60}{100}\)
= 550 + 540 = 1090
∴ Required percentage = \(\frac{1090}{2000} \times 100\) = 54.5%
Ans .
450
Successful boys in English or Maths or both = 80 + 85 – 75 = 90%
Unsuccessful boys = 10%
∴ Total number of boys = \(\frac{100}{10} \times 45\) = 450
Ans .
75%
25% of students pass in at least one subject
i.e.; they pass in one or both subjects.
∴ % of students who don’t pass or fail in both subjects = (100 – 25)% = 75%
Ans .
10%
The percentage of students
who pass in one or two or both subjects = 60 + 70 – 40 = 90
∴ Percentage of failed students = 100 – 90 = 10%
Ans .
240
Let total number of candidates = 100
70 candidates passed in English and 30 failed in it.
80 candidates passed in Maths and 20 failed in it.
10 candidates failed in English and Maths both.
∴ Out of 30 failed in English, 10 failed in Maths also.
∴ 30 – 10 = 20 failed in English alone.
Similarly, 20 – 10 = 10 failed in Maths alone.
∴ Total number of failures = 20 + 10 + 10 = 40
∴ 100 – 40 = 60 candidates passed in both subjects.
Now, if 60 candidates pass, total strength = 100
∴ For 144 candidates, total strength = \(\frac{100}{60} \times 144\) = 240
Ans .
25%
Difference of percentages of maximum marks obtained by two candidates = 32% – 20% = 12%
Difference of scores between two candidates = 30 +42 = 72
∴ 12% of maximum marks = 72
∴ Maximum marks = \(\frac{72 \times 100}{12}\) = 600
∴ Pass marks = 20% of 600 + 30 = 120 + 30 = 150
∴ Required percentage = \(\frac{150}{600} \times 100\) = 25%
Ans .
32.8%
Total number of students = 640 + 360 = 1000
Number of successful boys = 60% of 640 = 384
Number of successful girls = 80% of 360 = 288
Total number of successful students = 384 + 288 = 672
Number of unsuccessful students = 1000 – 672 = 328
\ Required percentage = \(\frac{328 \times 100}{1000}\) = 32.8%
Ans .
44%
Let total number of students = 100
Number of failures in Maths = 34
Number of failures in English = 42
Number of failures in both subjects = 20
Number of failures in Maths or
English or both = 34 + 42 – 20 = 56
Number of students who passed in both subjects = 100 – 56 = 44
The required percentage = 44%
Ans .
120
Difference of percentage = (40 – 30)% = 10%
Difference of marks = 6 + 6= 12
∵ 10% of total marks = 12
Total marks =\(\frac{12 \times 100}{10}\) = 120
Ans .
23%
Let the total number of students = 100
∴ Number of students who failed in Hindi or English or both = 52 + 42 – 17 = 77
∴ Number of students who passed in both subjects = 100 – 77 = 23%
∴ Required percentage = 23
Ans .
62%
Let total number of students = 100
∴ 27 students speak none of the two languages.
It means only 73 students speak either Hindi or English or both.
Let x students speak both languages.
∴ 73 = 70 – x + x + 65 – x
\(→\) x = 70 + 65 – 73 = 62%
Ans .
600
Clearly, 75 candidates qualify
∴ 75% of appearing candidates = 450
∴ Number of appearing candidates = \(\frac{450 \times 100}{75}\) = 600
Ans .
\( 68 \frac{4}{7}\)%
Number of students passed in first year = 75
Number of students passed in second year = \(\frac{60 \times 75}{100}\) = 45
Total number of passed students = 75+45=120
Total number of appeared students =175
∴ Required percentage = \(\frac{120}{175} \times 100\) = \( 68 \frac{4}{7}\)%
Ans .
600
Let the maximum marks be x.
According to the question,
35% of x = 200 +10
\(\frac{35x}{100}\) = 210
\(→\) x = \(\frac{210 \times 100}{35}\) = 600
Ans .
50
Let the full marks in that examination were x.
According to the question
\(\frac{30x}{100}\)+ 5 = \(\frac{40x}{100}\) - 10
\(\frac{3x}{10}\) - \(\frac{4x}{10}\) = - 10 - 5
∴ x = 150
∴ Minimum pass marks
\(\frac{30}{100} \times 150\) + 5 = 50
Ans .
5000
Let the total number of candidates = x
∴ Number of candidates passed in English = 0.6x
Number of candidates passed in Maths = 0.7x
Number of candidates failed in both subjects = 0.2x
Number of candidates passed in atleast one subject = x – 0.2x = 0.8x
∴ 0.6 x + 0.7x – 2500 = 0.8 x
\(→\) 1.3x – 0.8x = 2500
\(→\) 0.5x = 2500
\(→\) x = \(\frac{2500}{0.5}\) = 5000
Ans .
500
Let the maximum marks in the
examination = x.
According to the question,
\(\frac{40x}{100}\) - \(\frac{30x}{100}\) = 50
\(→\) \(\frac{10x}{100}\) = 50
\(→\) x = \(\frac{50 \times 100}{10}\) = 500
Ans .
8%
Let total candidates be 'x'
Percentage of the candidates passing in English or Mathematics or both
= n(E) + n(M) – n (E \({\displaystyle \cap }\) M)
= 80 + 85 – 73 = 92
\(→\) Percentage of candidates who failed in both the subjects
= 100 – 92 = 8 or 8%
Ans .
50%
Percentage of students who failed in Maths or English or both = (25 + 35 – 10)% = 50%
∴ Required percentage = (100 – 50)% = 50%
Ans .
3700
If the total number of students be x, then
7% of x = 259
\(\frac{x \times 7}{100}\) = 259
x = \(\frac{259 \times 100}{7}\) = 3700
Ans .
200
If the total number of students be x, then
x = \(\frac{90x}{100}\) + \(\frac{85x}{100}\) - 150
\(→\) 100x = 90x + 85x – 15000
\(→\) 175x – 100x = 15000
\(→\) 75x = 15000
\(→\) x = 200
Ans .
\( 88 \frac{2}{3}\)%
Required percentage = \(\frac{40 \times 100 + 50 \times 90 + 60 \times 80}{40 + 50 + 60}\) = \( 88 \frac{2}{3}\)%
Ans .
450
If D gets 100 marks, then
Marks obtained by C = 125
Marks obtained by B = \(\frac{125 \times 90}{100}\)
Marks obtained by A = \(\frac{125 \times 90}{100} \times \frac{125}{100}\)
∵ 100 = \(\frac{125 \times 125 \times 90}{10000}\)
∵ 320 = \(\frac{125 \times 125 \times 90 \times 320}{1000000}\) = 450
∵
Ans .
\( 68 \frac{4}{7}\)%
Total examinees = 80 + 60 = 140
Total successful examinees = \(\frac{80 \times 60}{100}\) + \(\frac{60 \times 80}{100}\)
= 48 + 48 = 96.
∴ Required percent = \(\frac{96}{140} \times 100\) = \(\frac{480}{7}\) = \( 68 \frac{4}{7}\)%
Ans .
78% of all students
a = 19%, b = 10%, c = 7%
Passed students in both the subjects
= 100 – (a + b – c)
= 100 – (19 + 10 – 7)
= 100 – 22 = 78%
Ans .
68
Successful students in both classes
= \(\frac{20 \times 80}{100}\) + \(\frac{30 \times 60}{100}\)
= 16 + 18 = 34
\ Required percentage = \(\frac{34}{50} \times 100\) = 68%
Ans .
400
Failures in English = 100 – 75 = 25
Failures in Maths = 100–60 = 40
Failures in both subjects = 25
Failures in English only = 25 – 25 = 0
Failures n Maths only = 40 – 25 = 15
Failures in one or both subjects = 25 + 15 = 40
Percentage of successfuls = 100 – 40 = 60
Let total students be x
\(x \times \frac{60}{100}\) = 240
\(→\) x = \(\frac{240 \times 100}{60}\)= 400
Ans .
400
Maximum marks in the examination = x (let)
\(\frac{40x}{100}\) - \(\frac{30x}{100}\) = 12 + 28
\(\frac{10x}{100}\) \(→\) x = 40 × 10 = 400
\(\frac{10x}{100}\) \(→\) x = 40 × 10 = 400
Ans .
70
Total marks scored in all three
subjects = \(\frac{300 \times 70}{100}\) = 210
∴ Marks scored in third subject
= 210 – 60 – 80 = 70
Ans .
625
Let total marks in the exam be x.
According to the question,
\(\frac{x \times 36}{100}\) = 190 + 35 = 225
\(→\) x × 36 = 225 × 100
\(→\) x = \(\frac{225 \times 100}{36}\) = 625
Ans .
25%
Let the full marks of exam be x.
According to the question,
\(\frac{x \times 32}{100}\) - \(\frac{x \times 20}{100}\) = 30 + 42
\(→\) \(\frac{12x}{100}\) = 72
\(→\) x = \(\frac{72 \times 100}{12}\) = 600
\ Minimum marks to pass = \(\frac{600 \times 20}{100}\) + 30 = 120 + 30 = 150
\ Required percentage = \(\frac{150}{600} \times 100\) = 25%
Ans .
30000
Percentage of students who
pass in one or two or both subjects = 73 + 70 – 64 = 79%
∴ Unsuccessful students
\(→\) 100 – 79 = 21%
If the total number of examinees be x, then 21% of x = 6300
\(→\) \(x \times \frac{21}{100}\) = 6300
\(→\) x = \(\frac{6300 \times 100}{21}\) = 30000
Ans .
12.5%
Let the number of students with less than 75% attendance = y
Total students in school = x
According to the question,
\(\frac{x}{10}\) + \(\frac{y}{5}\) = y
\(\frac{x}{10}\) = y - \(\frac{y}{5}\) = \(\frac{4y}{5}\)
\(\frac{x}{2}\) = 4y
\(\frac{y}{x}\) = \(\frac{1}{8}\)
\(\frac{y}{x} \times 100\) = \(\frac{100}{8}\) = 12.5%
Ans .
250
Number of students who wear spectacles = \(\frac{1400 \times 25}{100}\) = 350
∴ Girls who wear spectacles = 1 -\(\frac{2}{7}\) of 350
= \(350 \times \frac{5}{7}\) = 250