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101. A litre of pure alcohol is added to 8 litre of 40% alcohol solution. Then what will be the percentage of water in the solution ?
Sol:

8AlcoholWater4.83.2 +140%

∴ required % = \(\frac{{4.8}}{{8 + 1}} \times 100\)
= \(\frac{{480}}{9}\)
= \(\frac{{160}}{3}\)
= \( \boldsymbol{53\frac{1}{3}} \)%       Answer

102. 40 litre mixture of milk and water contains 10% of milk, the milk to be added to make the milk content 20% in the new mixture will be how much ?
Sol:

40MilkWater364 +10%

∴ 20% = \(\frac{{4 + x}}{{40 + x}}\)
\(\frac{1}{5} = \frac{{4 + x}}{{40 + x}}\)
40 + x = 20 + 5x
4x = 20
x = 5     Answer

103. In what ratio must 20% of alcohol be mixed with 60% of alcohol to get a mixture of 40% strength alcohol.
Sol:

Alcohol I Alcohol II 20% 60% 40% 20% = 20% = ∴ required ratio = Answer

104. In a class the average score of boys in an exam is 63 and that of girls is 61 & the average score for the whole class is 61.8 then find the % of boys.
Sol:

Boys Girls 63 61 61.8 0.8 1.22 : 3

∴ required % of boys = \(\frac{2}{{3 + 2}} \times 100\)
40%      Answer

105. The population of a village was 8000. In a year with the increase in population of males by 8% and that of females by 5% the population of the village became 8458. What was the number of males in the village before increase ?
Sol:
% increase in population = \(\frac{{8458 – 8000}}{{8000}} \times 100\)
= \(\frac{{229}}{{40}}\)%
now using alligation method

malefemale85malefemale3202002292991

∴ required population of male = \(\frac{{29}}{{29 + 91}} \times 8000\)
= \(\frac{{29 \times 200}}{3}\)
= \(1933.3\bar 3\)
≈ 1934    Answer

106. In an examination 62% students failed in maths and 82% students failed in english while 52% students failed in both the subjects. If 48 students passed in both the subjects then find the total number of students appeared in the exam.
Sol:

Fail inmathFail inenglish105230

∴ total % of failed student = 10 + 52 + 30 = 92%
∴ total % of passed student = 100 – 92 = 8%
Given 8% ⟶ 48
          1% ⟶ 6
∴ total number of students = 100% = 100 × 6 = 600      Answer

107. In a group every person takes either tea or cold drink or both. If 72% takes tea & 44% takes cold drink and there are 160 persons who take both tea & cold drink. Then find total number of person in the group.
Sol:
persons taking either tea or cold drink = 72 + 44 = 116%
& total persons = 100%
Hence 16% of persons take tea and cold drink both.

561628

Given 16% ⟶ 160
            1% ⟶ 10
∴ total number of persons in the group = (56 + 16 + 28) × 10
1000     Answer

108. In an examination 80% students passed in Egnlish, 70% in Hindi while 15% failed in both the subjects. If 390 students passed in both the subjects. Then find the total number of stuents who appeared in the exam.
Sol:
passed in English = 80%
∴ failed in English = 20%
passed in Hindi = 70%
∴ failed in Hindi = 30%
& failed in both Hindi & English = 15%

Fail inEnglishFail inHindi51515failed in both

∴ total failed students = 5 + 15 + 15 = 35%
∴ total passed students = 100 – 35 = 65%
Given that 65% ⟶ 390
                     1% ⟶ 6
∴ total students = 100% = 600         Answer

109. Student A scores 30 marks in an examination out of 50 while another student B scores 20 marks out of 60. Who has performed better ?
Sol:
% of marks obtained by A = \(\frac{{30}}{{50}} \times 100\) = 60%
% of marks obtained by B = \(\frac{{20}}{{60}} \times 100\) = \(33\frac{1}{3}\)%
Hence A has performed better than B    Answer

110. The length and breadth of a rectangle are changed by +20% and -10% respectively. What is the % change in the area of the rectangle.
Sol:
Method(1):
net % change = \(20 – 10 – \frac{{20 \times 10}}{{100}}\)
= 10 – 2
+8%     Answer

Method(2):Length × Breadth = Area5 106 95054+20% = -10% = = +8% Answer