Solve the following Objective Questions on Percentages.
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 ?
The sum of 15% of a positive number and 10% of the same number is 70. What is twice of that number?
Therefore, Double of the given number = 280 × 2 = 560
In order to pass in an exam a student is required to get 975 marks out of the aggregate marks. Priya got 870 marks and was declared failed by 7 percent. What are the maximum aggregate marks a student can get in the examination?
(a) 1500
(b) 1000
(c) 1200
(d) Cannot be determined
(e) None of these
Solution: (a)
Minimum marks to pass = 975
Priya failed by 975 – 870 = 105 marks
Maximum marks = \(\displaystyle \frac{{105}}{7}\times 100=1500\)
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.
\(\displaystyle \begin{array}{l}\begin{array}{*{20}{l}} \begin{array}{l}Let\text{ }the\text{ }no.\text{ }of\text{ }persons\text{ }in\text{ }the\text{ }city\text{ }be\text{ }100x\\So,\text{ }n\left( A \right)\text{ }=\text{ }42x\text{ },\text{ }\left[ {n\left( A \right)=\text{ }people\text{ }that\text{ }read\text{ }newspaper\text{ A}} \right]\end{array} \\ {n\left( B \right)\text{ }=\text{ }68x,\text{ }\left[ {n\left( B \right)\text{ }=\text{ }people\text{ }that\text{ }read\text{ }newspaper\text{ B}} \right]} \\ {n\left( C \right)\text{ }=\text{ }51x\text{ }\left[ {n\left( C \right)=\text{ }people\text{ }that\text{ }read\text{ }newspaper\text{ C}} \right]} \end{array}\\n\left( {A\cap B} \right)=30x\left[ {n\left( {A\cap B} \right)=\text{ }people\text{ }that\text{ }read\text{ }newspaper\text{ A and B}} \right]\\n\left( {B\cap C} \right)=28x\left[ {n\left( {B\cap C} \right)=\text{ }people\text{ }that\text{ }read\text{ }newspaper\text{ B and C}} \right]\\n\left( {A\cap C} \right)=36x\left[ {n\left( {A\cap C} \right)=\text{ }people\text{ }that\text{ }read\text{ }newspaper\text{ A and C}} \right]\\n(A\cap B\cap C)=people\text{ }that\text{ }read\text{ }newspaper\text{ A, B and C}\\n(A\cup B\cup C)=82x[people\text{ who donot }read\text{ any }newspaper\text{ A, B and C }\!\!]\!\!\text{ }\\\text{(100-18=82)}\\\Rightarrow n\left( {A\cup B\cup C} \right)=n(A)+n(b)+n(c)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C)\\Putting\text{ }the\text{ }above\text{ }values,\text{ }we\text{ }have\\82x=42x+68x+51x-30x-28x-36x+n(A\cap B\cap C)\\\Rightarrow n(A\cap B\cap C)=82x-67x=15x\\Therefore\text{ }the\text{ }number\text{ }of\text{ }people\text{ }reading\text{ }all\text{ }the\text{ }three\text{ }newspapers\text{ }are\text{ }15x\text{ }peopleout\text{ }of\text{ }100x\text{ }which\text{ }is\text{ }15\%\text{ }of\text{ }the\text{ }total\text{ }people\end{array}\)
A salesgirl’s terms were changed from a flat commission of 5% on all her sales to a fixed salary of ₹ 1000 plus 2.5% commission on all sales exceeding ₹4000. If her remuneration as per the new scheme was ₹600 more than that by the previous scheme, her total sales was
Mr Akbar salary increases every year by 10% in June. If there is no other increase or reduction in the salary and his salary in June 2022 was ₹22,385, what was his salary in June 2020?
Salary in June 2020 = \(\displaystyle \frac{{22385}}{{1.1\times 1.1}}=18500\)
An HR Company employees 4800 people, out of which 45 percent are males and 60 percent of the males are either 25 years or older. How many males are employed in that HR Company who are younger than 25 years?
(a) 2640
(b) 2160
(c) 1296
(d) 864
(e) None of these
Solution: (d)
Total Number of employees = 4800
45% of these 4800 are male
Therefore, total male = \(\displaystyle \frac{{45}}{{100}}\times 4800=2160\)
60% of these 2160 are 25years or older
Therefore, total male who are 25 years or older=\(\displaystyle \frac{{60}}{{100}}\times 2160=1296\)
The number of males less that 25 years=2160 – 1296 = 864
(or)
Direct alternate method,
Required number = \(\displaystyle 4800\times \frac{{45}}{{100}}\times \frac{{40}}{{100}}=864\)
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 ?
a) ₹ 11,10,400
(b) ₹13,24,800
(c) ₹ 23,36,800
(d) ₹ 1,52,200
(e) None of these
Solution: (b)
Ravina’s monthly income = \(\displaystyle 32000\times \frac{{100+15}}{{100}}=32000\times \frac{{115}}{{100}}=36800\)
Ramola’s annual income = \(\displaystyle 36800\times 3\times 12\)
= ₹ 1324800
In an Entrance Examination Ritu scored 56 percent marks, Smita scored 92 percent marks and Rina scored 634 marks. The maximum marks of the examination are 875. What are the average marks scored by all the three girls together?
(a) 1929
(b) 815
(c) 690
(d) 643
(e) None of these
Solution: (d)
Marks scored by Ritu = \(\displaystyle 875\times \frac{{56}}{{100}}=490\)
Marks scored by Smita = \(\displaystyle 875\times \frac{{92}}{{100}}=805\)
Therefore, Average marks scored by all the three together =\(\displaystyle \frac{{490+805+634}}{3}=\frac{{1929}}{3}=643\)
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 \(\displaystyle \frac{x}{y}\)
