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.

Answer:

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?

Answer:

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?

Answer:

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?

Answer:

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?

Answer:

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?

Answer:

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?

Answer:

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

Answer:

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?

Answer:

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?

Answer:

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?

Answer:

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?

Answer:

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?

Answer:

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?

Answer:

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?

Answer:

\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

Answer:

\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.

Answer:

\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?

Answer:

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.

Answer:

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

Answer:

\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