GRE Percentage

Percentage. We utter this word daily in our real life. You will absolutely get one or more GRE percentage problem in your exam. GRE percentage problems are slightly more tricky than our real life problems. By going through our chapter you will learn how to attack on a GRE percentage problem.

The concept of percentage mainly applies to ratios, & the percentage value of ratio is arrived at by multiplying by 100 the decimal value of the ratio.

For example, a student scores 20 marks out of a maximum possible 30 marks. His marks can then  be denoted as 20  out of 30

= \((\frac { 20 }{ 30 } )\quad or\quad (\frac { 20 }{ 30 } )\quad \times \quad 100%\quad =\quad 66.66%\)

GRE PERCENTAGE PROBLEMS  

Problem 1)   A hostess at an art gallery makes $100 for each exhibit that she works. She also receives \(2\frac { 1 }{ 2 } %\)  of the art sales. If she earned $900 for a single exhibit, how much were the art sales?

Sol.  Money earned by hostess of the art sales = $800

\((Art\quad Sales)\quad (\frac { 100 }{ 10000 } \times \frac { 5 }{ 2 } )\quad =\quad 800\\ (\frac { 1 }{ 100 } )(Art\quad Sales)(\frac { 5 }{ 2 } )\quad =\quad 800\\ Art\quad Sales\quad =\quad (800\quad \times \quad 100)(\frac { 2 }{ 5 } )\quad =\quad (800)(20)(2)\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad (800)(40)\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad 32000/-\)

 

Problem 2) Gasoline at a certain station has increased from $2.98 to $3.07. If it then decreases by half the percent of the percent increase, what is the new price?

Sol.  Gasoline increased from $2.98 to $3.67.

Let’s percentage increase be x.

\(2.98(1+\frac { x }{ 100 } )\quad =\quad 3.07\\ \quad \quad \quad \quad \quad \frac { x }{ 100 } =\quad (\frac { 3.07 }{ 2.98 } -1)\\ \quad \quad \quad \quad \quad \quad \quad x\quad =\quad (\frac { 3.07 }{ 2.98 } -1)(100)\)

New Price \(=\quad (1-\frac { 50 }{ 100 } (\frac { 3.07 }{ 2.98 } -1))\\ =\quad 3.07(1-0.015)\\ =\quad { (3.02395) }^{ Ans }\)

 

Problem 3)  Which of the following is the largest number

a) 20% of 200  b) 7% of 500      c) 1300% of 3    d) 700% of 9

Sol.  700% of 9 = 63 is the highest number.

(d) is correct.

 

Problem 4) What is 20% of 50% of 75% of 70?

Sol. 

\((\frac { 20 }{ 100 } )(\frac { 50 }{ 100 } )(\frac { 75 }{ 100 } \times 70)\\ =\quad \frac { 1 }{ 5 } \times \frac { 1 }{ 2 } \times \frac { 3 }{ 4 } \times 70\\ =\quad { 5.25 }^{ Ans }\)

 

Problem 5) The length, breadth and height of a room in the shape of a cuboid are increased by 10%, 20% and 50% respectively. Find the percentage change in the volume of the cuboid?

Sol.

\(lbh\quad =\quad V\\ l(\frac { 110 }{ 100 } )b(\frac { 120 }{ 100 } )h(\frac { 150 }{ 100 } )\quad =\quad V(1+\frac { x }{ 100 } )\\ (100\quad +\quad x)\quad =\quad \frac { (110)(120)(150) }{ 10000 } \quad =\quad 198\\ x\quad =\quad 98%\quad\)

 

Problem 6) The salary if A it is 30% more than of Varun. Find by what percentage is the salary of Varun less than that of Amit?

a) 12%     b) 23.07%      c) 21.23%      d) 27.27%

Sol.

Let Salary of Varun is x% less than that of Amit.

\( Salary\quad of\quad Amit\quad =\quad (Salary\quad of\quad Varun)(\frac { 130 }{ 100 } )\\ Salary\quad of\quad Varun\quad =\quad (Salary\quad of\quad Amit)(1-\frac { x }{ 100 } )\\ (1-\frac { x }{ 100 } )\quad =\quad (\frac { 100 }{ 130 } )\\ 1\quad -\quad \frac { 100 }{ 130 } \quad =\quad \frac { x }{ 100 } \\ x\quad =\quad 100(\frac { 30 }{ 130 } )\quad =\quad 23.07%\)

(b) is correct.

 

Problem 7) In the recent, climate conference in New York, out of 700 men, 500 women, 800 children present inside the building premises, 20% of the men, 40% of the women & 10% of the children were Indians. Find the percentage of people who were not indian.

a) 73%     b) 77%      c) 79%       d) 83%

Sol.  Percentage =

\((\frac { 700(\frac { 80 }{ 100 } )\quad +\quad 500(\frac { 60 }{ 100 } )\quad +\quad 800(\frac { 90 }{ 100 } ) }{ 700\quad +\quad 500\quad +\quad 800 } )\times 100\\ =(\frac { 560\quad +\quad 300\quad +\quad 720 }{ 2000 } )\times 100\\ =79%\)

(c) is correct.

 

Problem 8) Out of the total production of iron from hematite. An ore of iron, 20% of the ore gets wasted, & out of the remaining ore, only 25% is the pure iron. If the pure iron obtained in a year from a mine of hematite was 80,000kg, then the quantity of hematite mined from that mine in the year is?

a) 5,000 kg     b) 4,00,000 kg     c) 4,50,000 kg     d) None of these

Sol.  Let Quantity of hematite mixed from that mine in the year is x.

\(x(\frac { 80 }{ 100 } )(\frac { 25 }{ 100 } )\quad =\quad 80,000\\ x\quad =\quad (\frac { 80,000\times 10000 }{ 80\times 25 } )\\ =\quad 4,00,000\quad Kg\)

(b) is correct.

 

Problem 9) A man buys a truck for ¥2,50,000. The annual repair cost comes to 2% of the price of the purchase. Besides, he has to pay an annual tax of ¥2,000. At what monthly rent must be rent out the truck to get a return of 15% on his net investment of the first year?

a) 3,350     b) 2,500       c) 4,000     d) 3,212.50

Sol. Monthly rent =

\((2,50,000(1\quad +\quad \frac { 2 }{ 100 } )\quad +\quad 2000)\times (\frac { 15 }{ 100 } )\times (\frac { 1 }{ 12 } )\\ =\quad (2,50,000(\frac { 102 }{ 100 } )\quad +\quad 2000)\times (\frac { 15 }{ 1200 } )\\ =\quad (3212.5)\)

(d) is correct.

 

Problem 10)   \({ (\frac { 4 }{ 5 } ) }^{ th }\) of the voters in Bellary promised to vote for Sonia & the rest promised to vote for Sushi. Of these voters, 10% of the voters who had promised to vote for Sonia, didn’t vote on the election day, while 20% of the voters who had promised to vote for Sushi didn’t vote on the election day. What is the total number of voters polled if Sonia got 216 votes?

a) 200     b) 300        c) 264         d) 100

Sol.  Sonia got 216 votes. Let voters be x.

\(216\quad =\quad (\frac { 4 }{ 5 } )x(\frac { 90 }{ 100 } )\\ x\quad =\quad (\frac { 216\times 5\times 100 }{ 4\times 90 } )\)

Total number of votes polled =

\(=\quad x\quad -\quad x(\frac { 4 }{ 5 } )(\frac { 10 }{ 100 } )\quad -\quad x(\frac { 1 }{ 5 } )(\frac { 20 }{ 100 } )\\ =\quad (\frac { 216 }{ 4\times 90 } )(500-40-20)\\ =\quad (\frac { 216 }{ 4\times 90 } )\times (440)\\ =\quad (\frac { 216\times 44 }{ 4\times 9 } )\\ =\quad (\frac { 216\times (11) }{ 9 } )\\ =\quad 264\)

(c) is correct.

 

Problem 11) Ravana spends 30% of his salary on house rent, 30% of the rest he spends on his children’s education & 24% of the total salary he spends on clothes. After his expenditure, he is lift with ₹2500. What is Ravana’s salary?

a) 11,494.25   b) 20,000    c) 10,000     d) 15,000

Sol.  Let salary be x.

\(2500\quad =\quad x(1-\frac { 30 }{ 100 } -\frac { 70 }{ 100 } (\frac { 30 }{ 100 } )-\frac { 24 }{ 100 } )\\ 2500\quad =\quad x(1-\frac { 30 }{ 100 } -\frac { 21 }{ 100 } -\frac { 24 }{ 100 } )\\ 2500\quad =\quad x(1-\frac { 75 }{ 100 } )\\ 2500\quad =\quad x(\frac { 25 }{ 100 } )\\ x\quad =\quad 10,000\)

(c) is correct.

 

Problem 12)  The entrance ticket at the Minerva theatre in London is worth £ 250. When the price of the ticket was lowered, the sale of tickets increased by 50% while the collection recorded a decrease of 17.5% . Find the deduction in the ticket price.

a) 150    b) 112.5           c) 105               d) 120

Sol.  Let number of ticket sold previously be x.

Number of ticket sold now be y.

\(y\quad =\quad x(\frac { 150 }{ 100 } )\)

Let new price be P.

\(250(x)(1-\frac { 17.5 }{ 100 } )\quad =\quad yp\quad =\quad (\frac { 150 }{ 100 } )xp\\ (250)(1-\frac { 17.5 }{ 100 } )\quad =\quad (\frac { 150 }{ 100 } )p\\ p\quad =\quad \frac { 25\times (825) }{ 150 }\)

Deduction =

\(250\quad -\quad \frac { 25(825) }{ 150 } \\ =\quad 25(10\quad -\quad \frac { 825 }{ 150 } )\\ =\quad 112.5\)

 (b) is correct.

 

Problem 13)  In 1970, company X had 2,000 employees, 15% of whom were women & 10% of these women were executives, what was the percent increase in the number of women executives from 1970 to 2012?

Sol.  No. of women executives in 1970 =

\(=\quad (\frac { 15 }{ 100 } \times 2000)\quad \times \quad (\frac { 10 }{ 100 } )\\ =\quad 30\)

No. of women executives in 2012 =

\(=\quad (12000)\times ()\times ()\\ =\quad 12\times 45\times 4\\ =\quad 12\times 180\\ =\quad 2160\)

Percent increase in the number of women executive from 1970 to 2012 =

\(=\quad (\frac { Difference }{ Original } )\times 100\\ =\quad (\frac { 2130 }{ 30 } \times 700)\\ =\quad 7100%\)

 

Problem 14) Raymond borrowed $450 at 0% interest. If he pays back 0.5% of the total amt. every 7 days, beginning exactly 7 days after the loan was disbursed & has thus far paid back $18, with the most recent payment made today. How many days ago did he borrow the money?

Sol.  0.5% of the total =

\((\frac { 5 }{ 1000 } \times 450)\\ =\quad \frac { 45 }{ 20 } \\ =\quad 2.25\)

2.25$ × 8 = 18$

8 × 7 = 56

He borrowed money 56 days ago.

 

Problem 15) If Ken’s salary were 20% higher, it would be 20% less than Lorena’s. If Lorena’s salary is $60,000. What is Ken’s salary?

Sol.  Let Ken’s salary be x

\(x\quad \times \quad (\frac { 120 }{ 100 } )\quad =\quad 60.000\quad \times \quad (\frac { 80 }{ 100 } )\\ x\quad =\quad 60000\quad \times \quad \frac { 80 }{ 100 } \quad \times \quad \frac { 100 }{ 120 } \\ =\quad 60,000\quad \times \quad \frac { 4 }{ 6 } \\ =\quad 40,000\)

Ken’ s Salary is  $40,000.

 

Problem 16) Aloysius spends 50% of his income on rent, utilities & insurance & 20% on food. If he spends 30% of the remainder on video games & has no other expenditures. What percent of his income is lift after all of the expenditures?

Sol.  After spending 70% money on rent, utilities, insurance & food. He spends 30% of his remaining 30% salary on video games. After all expenditure 70% of his 30% salary remains with him.

Let x% of his income left after all of the expenditures.

\((\frac { 70 }{ 100 } )\times (\frac { 30 }{ 100 } \times Salary)\quad =\quad \frac { x }{ 100 } \times Salary\\ =\quad 21%\)

 

Problem 17) Roselba’s annual income exceeds twice Jane’s annual income & both pay the same positive percent of their respective incomes in transportation fees.

Quantity A:       The annual amt. Jane pays in transportation fees.

Quantity B:       Half the annual amt. Roselba pays in transportation fees.

a) Quantity A is greater

b) Quantity B is greater

c) The two quantities are equal

d) The relationship cannot be determined from the information given.

Sol.  The correct answer is “Quantity B is greater”. Roselba’s income is more than twice as great as Jane’s income. If both pay the same percent of income in transportation fees, that means Roselba must pay more than twice as much as Jane in transportation fees.

Quantity B is greater.

 

Problem 18) At the end of April, the price of fuel was 40% greater than the price at the beginning of the month. At the end of May, the price of fuel was 30% greater than of the price at the end of April.

Quantity A:       The price increase in April.

Quantity B:       The price increase in May.

a) Quantity A is greater.

b) Quantity B is greater.

c) The two quantities are equal.

d) The relationship cannot be determined from the information given.

Sol. Let price of fuel at the starting of April be x.

Price increase in April = 0.40x

Price of fuel increases to 1.40x

Price increase in May =

\(=\quad (1.40x)\times (\frac { 30 }{ 100 } )\\ =\quad 0.42x\)

“Quantity B is greater” price  increase in May is greater than price increase in April.

 

Problem 19) 75% of all the boys & 48% of all the girls at Smith high school take civics. If there are 20% flower boys then there are girls in the school. What percent of all the students take civics?

Sol.  No. of boys took civics =

\((\frac { 75 }{ 100 } \times (Total\quad number\quad of\quad boys))\)

No. of girls took civics =

\(\frac { 48 }{ 100 } \times (Total\quad No.\quad of\quad girls)\)

No. of boys =

\((\frac { 80 }{ 100 } \times (No.\quad of\quad girls))\)

Percent of all the students who took civics =

\((\frac { \frac { 75 }{ 100 } \times (Total\quad no.\quad of\quad boys)\quad +\quad \frac { 48 }{ 100 } \times (Total\quad no.\quad of\quad girls) }{ Total\quad no.\quad of\quad boys\quad +\quad Total\quad no.\quad of\quad girls } )\times 100\\ =\quad (\frac { \frac { 75 }{ 100 } \times (\frac { 80 }{ 100 } \times Total\quad no.\quad of\quad girls)\quad +\quad \frac { 48 }{ 100 } \times (Total\quad no.\quad of\quad girls) }{ \frac { 80 }{ 100 } \times Total\quad no.\quad of\quad girls\quad +\quad Total\quad no.\quad of\quad girls } )\times 100\\ =\quad (\frac { \frac { 48 }{ 100 } +(\frac { 75 }{ 100 } \times \frac { 80 }{ 100 } ) }{ 1\quad +\quad \frac { 80 }{ 100 } } )\times 100\\ =\quad (\frac { \frac { 48 }{ 100 } \quad +\quad (\frac { 60 }{ 100 } ) }{ \frac { 180 }{ 100 } } )\times 100\\ =\quad (\frac { 108 }{ 180 } \times 100)\\ =\quad 60%\\ =\quad ()\)

 

60% of all students took civics.

 

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