Question 1: $\displaystyle 36\%$ of the students in a school is girls. If the number of boys is $\displaystyle 1440,$ find the total strength of the school.

Let the strength of the school $\displaystyle x$

$\displaystyle \% \text{ of Boys } = (100-36) \% = 64\%$

$\displaystyle \text{Therefore } \frac{64}{100} \times x = 1440$

$\displaystyle \text{or } x = 2250$

Therefore the strength of the school $\displaystyle = 2250$ students

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Question 2: Geeta saves $\displaystyle 18\%$ of her monthly salary. If she spends Rs. $\displaystyle 10250$ per month, what in her monthly salary?

Let the monthly salary be $\displaystyle x \text{ Rs. }$

Geeta spends $\displaystyle = (100-18) \% = 82\%$ of her salary

$\displaystyle \text{Then } \frac{72}{100} \times x = 10250$

$\displaystyle \text{Or } x = Rs. 12500$  Monthly Salary

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Question 3: In an Examination, a student secures $\displaystyle 40\%$ mark to pass. Raul gets $\displaystyle 178$ marks and fails by $\displaystyle 32$ marks. What are the maximum marks?

Let the Maximum Marks $\displaystyle x$

$\displaystyle \text{Then } \frac{40}{100} \times x = 178+32 \text{ or } x = 525$ Maximum Marks

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Question 4: $\displaystyle 8\%$ of the students in a school remained absent on a day. If $\displaystyle 1633$ attended the school on that day, how many remained absent?

Let the total number of students in school be $\displaystyle x$

$\displaystyle \%$ of students attended the school $\displaystyle = 92\%$

$\displaystyle \text{Then } \frac{92}{100} \times x = 1633 \text{ or } x = 1775$ students

Therefore the number of students absent $\displaystyle \frac{8}{100} \times 1775 = 142$

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Question 5: On increasing the price of the article by $\displaystyle 14\%$, it becomes Rs. $\displaystyle 1995$. What was its original Price?

Let the original Price be $\displaystyle x \text{ Rs. }$

$\displaystyle \text{Then } x(1+ \frac{14}{100} ) = 1995 \text{ or } x = 1750 Rs.$ Original Price

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Question 6: On decreasing the price of an article by $\displaystyle 6\%$, it becomes Rs. $\displaystyle 1551$. What was the original price?

Let the original price be $\displaystyle x \text{ Rs. }$

$\displaystyle \text{Then } x(1- \frac{6}{100} ) = 1551 \text{ or } x = 1650 Rs.$ Original Price

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Question 7: Reenu reduced her weight by $\displaystyle 15\%$. If she now weighs $\displaystyle 52.7$ kg, what was her original weight?

Let Reenu’s original weight be $\displaystyle x kg.$

$\displaystyle \text{Then } x(1- \frac{15}{100} ) = 52.7 \text{ or } x =62 kg.$ Original weight.

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Question 8: Two candidates contest an election. One of them secure $\displaystyle 58\%$ votes and won the election by a margin of $\displaystyle 2560$ votes how many votes were polled in all

Let the total no. of votes polled be $\displaystyle x$

Then

$\displaystyle \frac{58}{100} x- \frac{42}{100} x= 2650 \\ \\ 0.16x - 2560 \\ \\\text{ or } x = 16000 votes$

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Question 9: In an examination, Preety scored $\displaystyle 60$ out of $\displaystyle 75$ in sciences, $\displaystyle 84$ out of $\displaystyle 100$ in mathematics, $\displaystyle 36$ out of $\displaystyle 50$ in Hindi and $\displaystyle 30$ out of $\displaystyle 45$ in English

In which subject the performance is worst?

In which subject the performance is best?

What is her aggregate percentage of marks?

Let’s first calculate % marks in each subject

$\displaystyle \text{Science } \% = \frac{60}{75} \times 100=80\%$

$\displaystyle \text{Mathematics } \% = \frac{84}{100} \times 100=84\%$

$\displaystyle \text{Hindi } \% = \frac{35}{45} \times 100=72\%$

$\displaystyle \text{English } \% = \frac{30}{45} \times 100=66.67\%$

Worst performance is in English

Best performance is in Math.

$\displaystyle \text{Aggregate } = \frac{ \text{total marks obtained} } {\text{total marks}} \times 100$

$\displaystyle \text{Aggregate } = \frac{60+84+36+30}{75+100+50+45} \times 100 = 77.78\%$

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Question 10: The price of an article is increased by $\displaystyle 25\%$. By how much $\displaystyle \%$ must this new price be decreased to return to its formal value?

Let the price of the article be $\displaystyle x$

New price $\displaystyle = x(1+ \frac{25}{100} ) = 1.25x$

Let % to be decreased to restore it to formal value be $\displaystyle y$

$\displaystyle \text{Then } = 1.25x(1- \frac{y}{100} ) = x$

$\displaystyle 1.25(1- \frac{y}{100} ) = 1 \text{ or } y = 20$

Therefore the new price must be decreased by $20\%$

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Question 11: The price of an article is reduced by $\displaystyle 10\%$ by how much this $\displaystyle \%$ value be increased to restore it to its formal value?

Let the value of the article be $\displaystyle x$

$\displaystyle \text{New price } = x(1- \frac{10}{100} ) = 0.9x$

Let us increase the price by $\displaystyle y\%$

$\displaystyle \text{Then } 0.9x(1+ \frac{y}{100} ) = x$

$\displaystyle 90+0.9y=100 \text{ or } y = 11.11\%$

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Question 12: The price of the tea is increased by $\displaystyle 20\%$. By how much $\displaystyle \%$ a housewife should reduce the consumption of tea so as not to increase the expenditure on tea?

Let the price of the tea $\displaystyle x$

$\displaystyle \text{New price } = x(1+ \frac{20}{100} ) = 1.2x$

Let the consumption be decreased by $\displaystyle y\%$

$\displaystyle \text{Therefore } 1.2x(1- \frac{y}{100} ) = x \text{ or } y = 16.67\%$

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Question 13: A man gave $\displaystyle 35\%$ of his money to his elder son and $\displaystyle 40\%$ of the remainder to younger son. Now, he is left with Rs. $\displaystyle 11700$. How much money he had originally?

Let the man have $\displaystyle x \text{ Rs. }$

Money given to his elder son $\displaystyle 0.35 x$

Money left $\displaystyle = (1- 0.35 x)$

Money given to younger son $\displaystyle =0.40 \times 0.65x=0.26x$

Total money left $\displaystyle =x-0.35x-0.26x=0.39x$

Hence $\displaystyle 0.39x=11700 \text{ or } 30000 \text{ Rs. }$

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Question 14: $\displaystyle 5\%$ of the population of a town was killed in an earthquake and $\displaystyle 8\%$ of the remainder left the town. If the population of the town now is $\displaystyle 43700$, what was the population at the beginning?

Let the initial population be $\displaystyle x \text{ Rs. }$

People killed $\displaystyle =0.05x$

People left after earthquake $\displaystyle = (1-0.05)\times 0.08x=0.076x$

Hence $\displaystyle x-0.05x-0.076x=43700 \text{ or } x = 50,000$

The initial population $\displaystyle =50000$ people

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Question 15: A and B are two towers. The height of the tower A is $\displaystyle 20\%$ less than that of tower B. How much percent is B’s height more than that of A?

$\displaystyle \text{ Let the height of tower B } x$

$\displaystyle \text{ Height of tower A } = 0.8x$

$\displaystyle \text{ \% of B's height more than A } = \frac{0.2x}{0.8x} = 25\%$

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Question 16: In an examination, $\displaystyle 30\%$ of the candidates failed in English, $\displaystyle 35\%$ fail in G.K. and $\displaystyle 27\%$ failed in both the subjects. If $\displaystyle 310$ passed in both, how many candidates appeared in the examination

$\displaystyle \text{ Percentage of student failed in English } =30\%$

$\displaystyle \text{ Percentage of students failed in G.K. } =35\%$

$\displaystyle \text{ Percentage of students failed in both } =27\%$

Hence

$\displaystyle \text{ Let the total number of candidates be } x$

$\displaystyle \text{ \% of students failed in English } =3\%$

$\displaystyle \text{ \% of students failed in G.K. } =8\%$

$\displaystyle \text{ \% of students failed in one or both subjects } = 27 + 3+ 8 = 38\%$

$\displaystyle \text{Therefore } 0.63x = 310 \text{ or } x = 500$ Students

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Question 17: The value of a car depreciates annually by $\displaystyle 10\%$. If the present value of car be Rs. $\displaystyle 650000$, find the value of car after two years.

$\displaystyle \text{ Let the value of car } = 65000 \text{ Rs. }$

$\displaystyle \text{ Value of car after 1 year } = (1-0.1) \times 65000 Rs =0.9 \times 65000 Rs =585000 \text{ Rs. }$

$\displaystyle \text{ Value of car after 2 years } 0.9 \times 585000 Rs=526500 \text{ Rs. }$

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Question 18: The population of village increased by $\displaystyle 7\%$ every year. If the present population is $\displaystyle 80,000$ then what will be the population after $\displaystyle 2$ years?

Population after $\displaystyle 2$ years $\displaystyle =(1.07) \times (1.07) \times 80000=91592$

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Question 19: A student was asked to multiply a number by . He multiplied by instead. Find the $\displaystyle \%$ error in calculation.

Let the number be $\displaystyle x$

$\displaystyle \text{ The right answer is } = \frac{5}{3} x$

$\displaystyle \text{ Wrong answer } = \frac{3}{5} x$

$\displaystyle \% \text{ error } = ( \frac{\frac{5}{3}x - \frac{3}{5}x}{\frac{5}{3}x} ) \times 100 = \frac{(25-9)\times 3 \times 100}{15 \times 5} = 64\%$

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Question 20: In an election between two candidates, $\displaystyle 10\%$ of the votes did not cast their votes. $\displaystyle 10\%$ of the votes polled were found invalid. The successful candidate got $\displaystyle 54\%$ of the valid votes and won by the majority of $\displaystyle 1620$ votes. Find the number of votes enrolled on the voters list

$\displaystyle \text{ Let the number of voters be } x$
$\displaystyle \text{No. of voters who did not cast vote } 0.1x$
$\displaystyle \text{No. of votes found invalid } = 0.9x \times 0.1=0.09x$
$\displaystyle \text{No of votes Successful candidate got } = \frac{54}{100} \times (x-0.1x-0.09x) = \frac{54}{100} \times 0.81x$
$\displaystyle \text{No of votes the other candidate got } = \frac{46}{100} \times 0.81x$
$\displaystyle \text{Hence } ( \frac{54}{100} - \frac{46}{100} ) \times 0.81x = 1620 \text{ or } x = 25000$ votes