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This set contains Percentage Questions with Solutions — Set 11 (Q101 to Q110) covering a mix of question types and difficulty levels — from basic to advanced — exactly as asked in real competitive exams.

Solutions are written in a simple, step-by-step notebook style for easy self-study and quick understanding. Each solution is broken down step by step so even the toughest question feels easy. These questions are hand-picked for students preparing for SSC CGL, SSC CHSL, CAT, Bank PO, Bank Clerk, UPSC CSAT, Railway RRB, AMCAT, eLitmus, TCS NQT and all campus placement aptitude tests. International students preparing for GRE, GMAT, SAT, ACT, MAT and all Numerical Reasoning Tests will find these equally useful.

✏️ Attempt each question on your own first — then check the solution below.

Percentage Questions 101 to 110 with Solutions

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 ?

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 ?

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

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.

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 ?

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.

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.

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.

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 ?

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.

Solutions — Percentage Questions 101 to 110

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

 Well done on completing Set 11!

Continue practising with Percentage Questions 111 to 120 → Set 12 or revisit the Percentage Concept Page to strengthen your formulas and tricks before moving ahead.

Consistent practice is the key to mastering percentage for SSC CGL, SSC CHSL, CAT, Bank PO, Bank Clerk, UPSC CSAT, Railway RRB, AMCAT, eLitmus, TCS NQT and international exams including GRE, GMAT, SAT, ACT, MAT and all Numerical Reasoning Tests. Want to understand the concept better? Read about Percentage on Wikipedia before attempting the next set.

This page is part of our complete series of percentage questions with solutions for competitive exams — covering every question type from basic to advanced so you can build speed, accuracy and confidence. Practising these questions regularly will also strengthen your core percentage concept before your exam day.

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