Application Based Problem on Percentage

**Category : **7th Class

The following are the points to remember to solve the problem related to variation in the price of an article.

- If the price of an article increases by \[x\text{ }%\]then the reduction in consumption so that expenditure remains unaffected is \[\left( \frac{x}{100+x}\times 100 \right)%\]
- If the price of an article decreases by \[x\text{ }%\] then the increase in consumption so that expenditure remains unaffected is \[\left( \frac{x}{100-x}\times 100 \right)%\]

**In the new budget, the price of petrol increased by 25 %. By how much % person should reduce his consumption so that his expenditure is not affected **

(a) 10%

(b) 20%

(c) 25%

(d) 30%

(e) None of these

**Answer:** (b)

**Explanation**

Reduction in consumption \[\left( \frac{25}{100+25}\times 100 \right)%=20%\]

** Due to reduction of \[6\frac{1}{4}%\] in the price of sugar, a man is able to buy 1 kg more sugar for Rs 120. The reduced rate of sugar is: **

(a) Rs 8 per kg

(b) Rs 6.5 per kg

(c) Rs 7.5 per kg

(d) Rs 9 per kg

(e) None of these

**Answer:** (c)

** Explanation**

Suppose original rate of sugar \[Rs\,x\] per kg.

Reduced rate \[=\left[ \left( 100-\frac{25}{4} \right)\times \frac{1}{100}\times x \right]=\frac{15x}{16}\]

According to question, \[\frac{120}{\frac{15x}{16}}-\frac{120}{x}=1\]

\[\Rightarrow \frac{128}{x}-\frac{120}{x}=1\Rightarrow x=8\]

Reduced rate \[=\frac{15}{16}\times 8=Rs\,7.5\] per kg

**Problem Based on the Population of a Locality**

Suppose the present population of a locality be 'A' and let it increases by \[x\text{ }%\]per annum then

- Population after y years \[=A{{\left( 1+\frac{x}{100} \right)}^{y}}\]
- Population before y years \[\frac{A}{{{\left( 1+\frac{x}{100} \right)}^{y}}}\]

**The population of a town is 176400. If it increases at the rate of 5 % per annum then what was its population 2 years ago? **

(a) 166400

(b) 154600

(c) 160000

(d) 166000

(e) None of these

**Answer:** (c)

**Explanation **

According to question

Population of city before y years, if it increases at the rate of \[x%=\frac{A}{{{\left( 1+\frac{x}{100} \right)}^{y}}}\]

\[\therefore \]The population if city \[=\frac{176400}{{{\left( 1+\frac{5}{100} \right)}^{2}}}=176400\left( \frac{20}{21}\times \frac{20}{21} \right)=160000\]

**In a certain year the population of London is 200000. If it increases at the rate of 6.5 % per annum then what will be its population after 2 years? **

(a) 226845

(b) 228645

(c) 224685

(d) 228465

(e) None of these

**Answer:** (a)

**Explanation**

Population after 2 years

\[=200000{{\left( 1+\frac{6.5}{100} \right)}^{2}}=226845\]

**Match the following: **

(i) | \[\frac{1}{2}\]is what percent of \[\frac{1}{3}\] | (a) | \[4%\] |

(ii) | What percent of 7 is 84? | (b) | \[150%\] |

(iii) | What percent of 6.5 liter is 130 ml | (c) | \[2%\] |

(iv) | 5 is what percent of 125 | (d) | \[1200%\] |

**Which one of the following options is correct? **

(a) i - b, ii - d, iii - a, iv - c

(b) i - b ii - c, iii ? d, iv - a

(c) i-c, ii-d, iii-b, iv-a

(d) i - b,ii - d, iii - c, iv-a

(e) None of these

**Answer:** (d)

**Stephen's mathematics test had 85 problems, which contains 20 algebra, 30 statistics and 35 geometry problems. He answered 70 % of algebra, 40 % of the statistics and 60 % of geometry problems correctly. He did not pass the test because he got less than 60 % of the problem correct. How many more questions he would have need to answer correctly to earn 60% passing grade? **

(a) 5

(b) 4

(c) 1

(d) 3

(e) None of these

**Answer:** (b)

**Explanation **

Number of questions attempted correctly \[=(70\text{ }%20+40\text{ }%\text{ }of30+60\text{ }%\text{ }of35)\]of \[=(14+12+21)=47\]

Questions to be answered correctly to obtained \[60%=60%\]of\[~85=51\]

Required number of question\[~=(51-47)=4\]

**The tax on an electronic goods is \[15\text{ }%\] and is increased by \[10\text{ }%.\] What the percent of total tax on the electronic goods? **

(a) \[26.5\text{ }%\]increase

(b) \[23\text{ }%\]decrease

(c) \[25\text{ }%\] increase

(d) \[27\text{ }%\] decrease

(e) None of these

**Answer: (**a)

**If the duty on an article be reduced by 40 % of its present amount, then by how much percent must the consumption be increased in order than the revenue may remain unaltered? **

(a)\[\text{50 }%\]

(b) \[166\frac{2}{3}%\]

(c)\[40%\]

(d)\[20%\]

(e) None of these

**Answer:** (b)

**What is the total number of candidates appear in an examination, if \[31%\] is fail and the number of passed candidates are 247 more than the number of fail candidates? **

(a) 650

(b) 750

(c) 800

(d) 900

(e) None of these

**Answer:** (a)

- Percent means parts per hundred and we use the symbol \[%\] to represent it.
- Percent can be converted into decimals, fractions, ratio and vice versa.
- If the price of an article increases \[by\text{ }x%\]then the reduction in consumption so that expenditure remains un effected is \[\left( \frac{x}{100+x}\times 100 \right)%\]
- If the price of an article decreases by x % then the increase in consumption so that expenditure remains un effected is \[\left( \frac{x}{100-x}\times 100 \right)%\]

- In the 14th century, Madhava of Sangamagrama, the founder of the so-called Kerala School of Mathematics, found the Madhava-Leibniz series and using 21 terms, computed the value of \[\pi \] as 3.14159265359.
- In the 12th century, Bhskara II lived in southern India and wrote extensively on all then known branches of mathematic. His work contains mathematical objects equivalent or approximately equivalent to infinitesimals, derivatives, the mean value theorem and the derivative of the sine function.

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