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

Let the strength of the school $x$

$\% \hspace{2pt} of \hspace{2pt} Boys = (100-36) \% = 64\%$

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

or $x = 2250$

Therefore the strength of the school = 2250 students

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

Let the monthly salary  be $x Rs.$

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

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

Or  $x = Rs. 12500$  Monthly Salary

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

Let the Maximum Marks  $x$

Then  $\frac{40}{100} \times x = 178+32 \hspace{2pt} or \hspace{2pt} x = 525$  Maximum Marks

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

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

% of students attended the school = 92%

Then, $\frac{92}{100} \times x = 1633 \hspace{2pt} or \hspace{2pt} x = 1775$ students

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

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

Let the original Price be $x Rs.$

Then $x(1+\frac{14}{100}) = 1995 \hspace{2pt} or \hspace{2pt} x = 1750 Rs.$ Original Price

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

Let the original price be $x Rs.$

Then $x(1-\frac{6}{100}) = 1551 \hspace{2pt} or \hspace{2pt} x = 1650 Rs.$ Original Price

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

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

Then $x(1-\frac{15}{100}) = 52.7 \hspace{2pt} or \hspace{2pt} x =62 kg.$  Original weight.

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

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

Then

$\frac{58}{100}x- \frac{42}{100}x= 2650 \\ \\ 0.16x - 2560 \\ \\ or x = 16000 \hspace{2pt} votes$

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

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

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

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

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

Worst performance is in English

Best performance is in Math.

$Aggregate = \frac{total \hspace{2pt}marks \hspace{2pt}obtained}{total \hspace{2pt}marks} \times 100$

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

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

Let the price of the article be $x$

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

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

Then   $= 1.25x(1-\frac{y}{100}) = x$

$1.25(1-\frac{y}{100}) = 1 \hspace{2pt} or \hspace{2pt} y = 20$

Therefore the new price must be decreased by 20%

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

Let the value of the article be $x$

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

Let us increase the price by $y\%$

Then $0.9x(1+\frac{y}{100}) = x$

$90+0.9y=100 \hspace{2pt} or \hspace{2pt} y = 11.11\%$

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

Let the price of the tea $x$

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

Let the consumption be decreased by $y\%$

Therefore $1.2x(1-\frac{y}{100}) = x \hspace{2pt} or \hspace{2pt} y = 16.67\%$

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

Let the man have $x Rs.$

Money given to his elder son  $0.35 x$

Money left $= (1- 0.35 x)$

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

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

Hence $0.39x=11700 \hspace{2pt} or \hspace{2pt} 30,000 Rs.$

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

Let the initial population be $x Rs.$

People killed $=0.05x$

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

Hence   $x-0.05x-0.076x=43700 \hspace{2pt} or \hspace{2pt} x = 50,000$

The initial population =50,000 people

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

Let the height of tower B  $x$

Height of tower A $= 0.8x$

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

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

Percentage of student failed in English $=30\%$

Percentage of students failed in G.K.$=35\%$

Percentage of students failed in both $=27\%$

Hence

Let the total number of candidates be $x$

% of students failed in English $=3\%$

% of students failed in G.K. $=8\%$

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

Therefore $0.63x = 310 \hspace{2pt} or \hspace{2pt} x = 500$ Students

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

Let the value of car = 65000 Rs.

Value of car after 1 year $= (1-0.1) \times 65000 \hspace{2pt} Rs =0.9 \times 65000 \hspace{2pt} Rs =585000 \hspace{2pt} Rs.$

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

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

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

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

Let the number be  $x$

The right answer is  $= \frac{5}{3}x$

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

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

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

Let the number of voters be $x$

No. of voters who did not cast vote $0.1x$

No. of votes found invalid $= 0.9x \times 0.1=0.09x$

No of votes Successful candidate got $= \frac{54}{100} \times (x-0.1x-0.09x) = \frac{54}{100} \times 0.81x$

No of votes the other candidate got $= \frac{46}{100} \times 0.81x$

Hence $(\frac{54}{100} -\frac{46}{100}) \times 0.81x = 1620 \hspace{2pt} or \hspace{2pt} x = 25,000$  votes