1. If the numerator of certain fraction is increased by 200% and the denominator is increased by 150% the new fraction thus formed is \(\displaystyle \frac{9}{{10}}\) . What is the original fraction?
(a)\(\displaystyle \frac{3}{4}\)
(b)\(\displaystyle \frac{1}{4}\)
(c)\(\displaystyle \frac{3}{5}\)
(d)\(\displaystyle \frac{2}{5}\)
(e)None of these
Solution: (a)
Let the original fraction be =\(\displaystyle \frac{x}{y}\)
Therefore, \(\displaystyle \frac{{x\times 300}}{{y\times 250}}=\frac{9}{{10}}\)
\(\displaystyle \Rightarrow \) \(\displaystyle \frac{{x\times 6}}{{y\times 5}}=\frac{9}{{10}}\)
\(\displaystyle \Rightarrow \)\(\displaystyle \frac{x}{y}=\frac{9}{{10}}\times \frac{5}{6}=\frac{3}{4}\)
2. If the numerator of a certain fractions increased by 100% and the denominator is increased by 200%; the new fraction thus formed is \(\displaystyle \frac{4}{{21}}\). What is the original fraction?
(a) \(\displaystyle \frac{2}{7}\)
(b) \(\displaystyle \frac{3}{7}\)
(c) \(\displaystyle \frac{2}{5}\)
(d) \(\displaystyle \frac{4}{7}\)
(e) None of these
Solution: (a)
Let the original fraction be = \(\displaystyle \frac{x}{y}\)
\(\displaystyle \Rightarrow \)\(\displaystyle \frac{{x\times 200}}{{y\times 300}}=\frac{4}{{21}} \)
\(\displaystyle \Rightarrow \)\(\displaystyle \frac{x}{y}=\frac{4}{{21}}\times \frac{3}{2}=\frac{2}{7}\)
3. The difference between 55% of a number and 14% of the same number is 8610. What is 85% of that number?
(a) 15850
(b) 17020
(c) 17850
(d) 18450
(e) None of these
Solution: (c)
Let the number be x. Now, according to the question,
(55 – 14) % of x = 8610
\(\displaystyle \Rightarrow \)\(\displaystyle \frac{{x\times 41}}{{100}}=8610\)
\(\displaystyle \Rightarrow \)\(\displaystyle x=\frac{{8610\times 100}}{{41}}=2100\)
\(\displaystyle \Rightarrow \)\(\displaystyle \frac{{21000\times 85}}{{100}}\)=17850
4. Animesh got 102 marks in Hindi, 118 marks in Science, 104 marks in Sanskrit, 114 marks in Maths and 96 marks in English. The maximum marks of each subject are 120. How much overall percentage of marks did Animesh get?
(a) 82
(b) 89
(c) 77
(d) 71
(e) None of these
Solution: (b)
Total marks obtained by Animesh
= 102 + 118 + 104 + 114 + 96 = 534
Total maximum marks = 120 × 5 = 600
Therefore, required percentage= \(\displaystyle \frac{{534}}{{600}}\times 100\) =89
5. Madhur got 101 marks in Hindi, 100 marks in Science, 96 marks in Sanskrit, 108 marks in Maths and 78 marks in English. If the maximum marks of each subject is equal and if Madhur scored 84 per cent marks in all the subjects together, what is the maximum marks of each subject?
(a) 110
(b) 120
(c) 115
(d) 100
(e) None of these
Solution: (c)
Total marks = 101 + 100 + 96 + 108 + 78 = 483
Since it is 84% of the total maximum marks,
Let total maximum marks be x
\(\displaystyle \frac{{84}}{{100}}x=483\)
x=\(\displaystyle \frac{{483\times 100}}{{84}}=575\)
Maximum marks of each subject = \(\displaystyle \frac{{575}}{5}\) = 115
Alternate method,
If each paper is of “x” marks, then total marks=5x
Given, total marks = 101 + 100 + 96 + 108 + 78 = 483
Mathur scored 84% marks in total, (which is 484 marks)
483 marks —— 5x marks
84%————–100%
Cross-multiply
\(\displaystyle \begin{array}{l}5x\times 84=483\times 100\\5x=\frac{{483\times 100}}{{84}}=575\\x=115\end{array}\)
6. In an examination, the maximum aggregate marks that a student can get is 1040. In order to pass the exam, a student is required to get 676 marks out of the aggregate marks. Mina got 624 marks. By what per cent did Mina fail in the exam?
(a) 5%
(b) 8%
(c) 7%
(d) Cannot be determined
(e) None of these
Solution: (a)
Minal failed by (676 – 624) = 52 marks
% marks =\(\displaystyle \frac{{52}}{{1040}}\times 100=5\%\)
7. In a town three newspapers A, B and C are published. 42% of the people in that town read A, 68% read B, 51% read C, 30% read A and B, 28% read B and C, 36% A and C and 18% do not read any paper. Find the % of population of town that reads all the three.
(or) Elaborated method,
\(\displaystyle \begin{array}{l}\begin{array}{*{20}{l}} \begin{array}{l}Let\text{ }the\text{ }no.\text{ }of\text{ }persons\text{ }in\text{ }the\text{ }city\text{ }be\text{ }100x\\So,\text{ }n\left( A \right)\text{ }=\text{ }42x\text{ },\text{ }\left[ {n\left( A \right)=\text{ }people\text{ }that\text{ }read\text{ }newspaper\text{ A}} \right]\end{array} \\ {n\left( B \right)\text{ }=\text{ }68x,\text{ }\left[ {n\left( B \right)\text{ }=\text{ }people\text{ }that\text{ }read\text{ }newspaper\text{ B}} \right]} \\ {n\left( C \right)\text{ }=\text{ }51x\text{ }\left[ {n\left( C \right)=\text{ }people\text{ }that\text{ }read\text{ }newspaper\text{ C}} \right]} \end{array}\\n\left( {A\cap B} \right)=30x\left[ {n\left( {A\cap B} \right)=\text{ }people\text{ }that\text{ }read\text{ }newspaper\text{ A and B}} \right]\\n\left( {B\cap C} \right)=28x\left[ {n\left( {B\cap C} \right)=\text{ }people\text{ }that\text{ }read\text{ }newspaper\text{ B and C}} \right]\\n\left( {A\cap C} \right)=36x\left[ {n\left( {A\cap C} \right)=\text{ }people\text{ }that\text{ }read\text{ }newspaper\text{ A and C}} \right]\\n(A\cap B\cap C)=people\text{ }that\text{ }read\text{ }newspaper\text{ A, B and C}\\n(A\cup B\cup C)=82x[people\text{ who donot }read\text{ any }newspaper\text{ A, B and C }\!\!]\!\!\text{ }\\\text{(100-18=82)}\\\Rightarrow n\left( {A\cup B\cup C} \right)=n(A)+n(b)+n(c)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C)\\Putting\text{ }the\text{ }above\text{ }values,\text{ }we\text{ }have\\82x=42x+68x+51x-30x-28x-36x+n(A\cap B\cap C)\\\Rightarrow n(A\cap B\cap C)=82x-67x=15x\\Therefore\text{ }the\text{ }number\text{ }of\text{ }people\text{ }reading\text{ }all\text{ }the\text{ }three\text{ }newspapers\text{ }are\text{ }15x\text{ }peopleout\text{ }of\text{ }100x\text{ }which\text{ }is\text{ }15\%\text{ }of\text{ }the\text{ }total\text{ }people\end{array}\)
8. Ramola’s monthly income is three times Ravina’s monthly income. Ravina’s monthly income is fifteen percent more than Ruchira’s monthly income. Ruchira’s monthly income is ₹ 32,000. What is Ramola’s annual income?
9. If the numerator of a fraction is increased by 300% and the denominator is increased by 200%, the resultant fraction is \(\displaystyle \frac{4}{15}\). What is the original fraction?
(a) \(\displaystyle \frac{3}{5}\)
(b) \(\displaystyle \frac{4}{5}\)
(c) \(\displaystyle \frac{2}{5}\)
(d) \(\displaystyle \frac{1}{5}\)
(e) None of these
Solution: (d)
Let the original fraction be \(\displaystyle \frac{x}{y}\)
According to the question
Therefore, \(\displaystyle \frac{{x\times 400}}{{y\times 300}}=\frac{4}{{15}}\)
\(\displaystyle \Rightarrow \)\(\displaystyle \frac{x}{y}=\frac{4}{{15}}\times \frac{3}{4}=\frac{1}{5}\)
10. A student was awarded certain marks in an examination. However, after re-evaluation, his marks were reduced by 40% of the marks that were originally awarded to him so that the new score now became 96. How many marks did the student lose after re-evaluation?
(a) 58
(b) 68
(c) 63
(d) 56
(e) 64
Solution: (e)
Let initial marks of student = x
After re-evaluation marks reduced by 40% of x
New score = 60% of x = 96
= \(\displaystyle \frac{{60}}{{100}}\times x=96\)
\(\displaystyle x=\frac{{96\times 100}}{{60}}\)
x=160
Marks lose = 160 – 96 = 64
