GRE: Ratios and Proportions

If general, the ratio of a number x to a number y is defined as the quotient of the number x & y.

The numbers that form the ratio are called the terms of the ratio. The numerator of the ratio is called the antecedent & the denominator is called the consequent of the ratio.

GRE: Ratios and Proportions

Important properties:

  1. If

\(\frac { a }{ b } =\frac { c }{ d } =\frac { e }{ f } =\frac { g }{ h } =k\quad\)

, then

\(\frac { a }{ b } =\frac { c }{ d } =\frac { e }{ f } =\frac { g }{ h } =k=\frac { (a\quad +\quad c\quad +\quad e\quad +\quad g) }{ (b\quad +\quad d\quad +\quad f\quad +\quad h) }\)

2. If  \(\frac { { a }_{ 1 } }{ { b }_{ 1 } } ,\frac { { a }_{ 2 } }{ { b }_{ 2 } } ,\frac { { a }_{ 3 } }{ { b }_{ 3 } } ——-\frac { { a }_{ n } }{ { b }_{ n } }\) are unequal fractions.

Then the ratio:

\(\frac { { a }_{ 1 }\quad +\quad { a }_{ 2 }\quad +\quad { a }_{ 3 }\quad ——-{ a }_{ n } }{ { b }_{ 1 }\quad +\quad { b }_{ 2 }\quad +\quad { b }_{ 3 }——–{ b }_{ n } }\)

lies between the lowest & highest of these percentage.

GRE: Ratios and Proportions

Problem 1)   In a pet store, the ratio of the number of puppies to kittens is 4:7. When 7 more puppies more puppies are received, the ratio of the number of puppies to the number of the kittens charges to 5:7. How many pets does the pet store now have?

Sol. Let the number of puppies be \(4x\)  & number of kitters be \(7x\).

\(\frac { 4x\quad +\quad 7 }{ 7x } =\quad \frac { 5 }{ 7 }\)

 

\((4x\quad +\quad 7)(7)\quad =\quad (7x)(5)\)

 

\(28x\quad +\quad 49\quad =\quad 35x\)

 

\(7x\) = 49

 

\(x\) = 7

 

No. of puppies = \(4\times 7+7\) = 35

No. of kittens = \(7\times 7\) = 49

No. of pets = 35 + 49 = 84

 

Problem 2) Liquids A & B are in the ratio 2:1 in the 1st container, and 1:2 in the 2nd container. In what ratio should the contents of the two containers be mixed to obtain a mixture of A & B in the ratio 1:1?

Sol.  Let 1st container contains 1 litre of liquid. 1st container contains \(\frac { 2 }{ 3 }\) litre   of liquid A. 1st container contains \(\frac { 1 }{ 3 }\) litre  of liquid B. Let 2nd container contains 1 litre of liquid. second container contains \(\frac { 1 }{ 3 }\) litre  of liquid A. 2nd container contains \(\frac { 2 }{ 3 }\) litre   of liquid B.

Let \(x\) of liquid is taken from first container & \(1-x\) of liquid is taken from 2nd container.

New mixture contains \(\frac { 2 }{ 3 } x\quad +\quad \frac { 1 }{ 3 } (1-x)\) litre of liquid A.

New mixture contains \(\frac { 1 }{ 3 } x\quad +\quad \frac { 2 }{ 3 } (1-x)\) litre  of liquid B.

\(\frac { 2 }{ 3 } x\quad +\quad \frac { 1 }{ 3 } (1-x)\quad =\quad \frac { 1 }{ 3 } x\quad +\quad \frac { 2 }{ 3 } (1-x)\) \(x=\frac { 1 }{ 2 }\) Litre

\(\frac { 1 }{ 2 }\) of liquid is taken from first container & \(\frac { 1 }{ 2 }\) of liquid is taken from 2nd container.

Ans=1:1

 

Problem 3) If \(\frac { 1 }{ 2 }\)  of the number of white roses in a garden is  \(\frac { 1 }{ 8 }\)  of the total number of roses, and  \(\frac { 1 }{ 3 }\)  of the number of red roses is  \(\frac { 1 }{ 9 }\)  of the total number of roses, then what is the ratio of white roses to red roses?

Sol.  Let number of white roses be \(x\).

No. of red roses be \(y\).

Total roses be \(x + y\).

\(\frac { 1 }{ 2 } x\quad =\quad \frac { 1 }{ 8 } (x\quad +\quad y)\)

 

\(\frac { 1 }{ 3 } y\quad =\quad \frac { 1 }{ 9} (x\quad +\quad y)\)

 

\(4x\quad =\quad (x\quad +\quad y)\)

 

\(3y\quad =\quad (x\quad +\quad y)\)

 

\(4x\quad =\quad 3y\)

 

\(\frac { x }{ y } =\frac { 3 }{ 4 }\)

Ratio=3:4

 

Problem 4)   The ratio between two numbers is 3:4 & and their L.C.M. is 180. Find the first number?

Sol. Let 1st number be \(3x\)  & 2nd number be \(4x\).

L.C.M. of \(3x\) and \(4x\) = \(3\times 4\times x \) = \(12x\)

 

\(12x\) = 180

 

1st number =  \(\frac { 180 }{ 12 } \times 3\) = \(\frac { 180 }{ 4 }\) = 45 Ans

 

Problem 5)   If  various inversely as \({ y }^{ 2 }-1\)  & is equal to 24 when y=10. Find x when y=5?

Sol.  \(x\quad =\quad \frac { k }{ { y }^{ 2 }-1 }\)                                 ;        (x=10, when y=10)

             \( 24\quad =\quad \frac { k }{ 100-1 }\)   

                              k = (99)(24)              ;

\(x\quad =\quad \frac { k }{ 25\quad -\quad 1 }\)

 

\(x\quad =\quad \frac { 99\times 24 }{ 24 }\)

 

x = 99

 

Problem 6)   The total number of pupils in three classes of a school is 333. The number of pupils in class I & II are in the ratio 3:5 & those in classes II & III are in the ratio 7:11. Find the number of pupils in the class that had the highest number of pupils?

a) 63

b) 105

c) 165

d) 180

Sol.  Let number of pupils in classes I & II are 3x & 5x.

          Let number of students in class II & III are 7y & 11y.

\(5x\quad =\quad 7y\)

 

\(y\quad =\quad \frac { 5x }{ 7 }\)

 

\(3x\quad +\quad 7y\quad +\quad 11y\quad =\quad 333\)

 

\(\frac { (3)(7y) }{ 5 } +7y+11y\quad =\quad 333\)

 

\(y\quad =\quad \frac { 333(5) }{ 111 }\)

 

\(y\quad =\quad 21\)

 

No. of student in class I = 21(3) = 63

No. of student in class II = 21(5) = 105

No. of student in class III = 33(5) = 165

 

Problem 7)   If a:b=c:d, e:f=g:h, then (ae+bf)(ae-bf) = ?

a) \(\frac { e+f }{ e-f } \)

b) \(\frac { cg\quad +\quad dh }{ cg\quad -\quad dh }\)

c) \(\frac { cg\quad -\quad dh }{ cg\quad +\quad dh }\)

d) \(\frac { e\quad -\quad f }{ e\quad +\quad f }\)

Sol. \(\frac { a }{ b } \quad =\quad \frac { c }{ d } ……..(1)\)

 

\(\frac { e }{ f } =\frac { g }{ h } ………….(2)\)

          Multiply equation (1) & (2)

 

\(\frac { a\quad \times \quad e }{ b\quad \times \quad f } \quad =\quad \frac { c\quad \times \quad g }{ d\quad \times \quad h }\)

 

\(\frac { ae }{ bf } \quad =\quad \frac { cg }{ dh }\)

By using componendo and dividendo

\(\frac { ae\quad +\quad bf }{ ae\quad -\quad bf } \quad =\quad \frac { cg\quad +\quad dh }{ cg\quad -\quad dh }\)

 

Problem 8) A factory employs skilled workers, unskilled workers & clerks in the proportion 8:5:1 & the wages of a skilled worker, an unskilled workers & clerks are in the ratio 5:3:2. Total wages of all workers amount to Rs. 31,800. The wages paid to each category of workers are

a) 24000, 6000, 1800

b) 15000, 10800, 6000

c) 20000, 9000, 2800

d) 25000, 5000, 1800

Sol.  Let number of skilled workers, unskilled workers & clerks be

\(8x,\quad 5x\quad \quad and\quad x\).

Let wages of a skilled workers, unskilled workers & clerks be

\(5y,\quad 2y\quad \quad and\quad 3y\).

Total wages of skilled workers = \(8x\quad \times \quad 5y\quad =\quad 40xy\)

Total wages of unskilled workers = \(5x\quad \times \quad 2y\quad =\quad 10xy\)

Total wages of clerks = \(x(3y)\quad =\quad 3xy\quad\)

 

\(53xy\quad =\quad 31800\quad\)

 

\(xy\quad =\quad 600\quad\)

 

Total wages of skilled workers = 40(600) = 24000/-

Total wages of unskilled workers = 10(600) = 6000/-

Total wages of clerks = 3(xy) = 3(600) = 1800/-

 

Problem 9) On his deathbed, Mr. Kalu called upon his three sons & told them to distribute all his assets worth Rs. 525,000 in the ratio of  amongst themselves. Find the biggest share amongst the three portions.

Sol.  Let the share of three sons be \(\frac { x }{ 15 } ,\quad \frac { x }{ 21 } \quad and\quad \frac { x }{ 35 }\).

\(\frac { x }{ 15 } +\frac { x }{ 21 } +\frac { x }{ 35 } =525000\\ 7x\quad +\quad 5x\quad +\quad 3x\quad =\quad 525000\times 3\times 5\times 7\\ 15x\quad =\quad (525000)\quad \times \quad (15)\quad \times \quad (7)\\ x\quad =\quad (525000)\times (7)\\ Biggest\quad share\quad =\quad \frac { 525000\times 7 }{ 15 } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad (7000)\times (35)\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad 245,000/-\)

 

Problem 10) The ratio of 16 to g is equal to the ratio of g to 49.

Quantity A                  Quantity B

g                                      28

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. \(\frac { 16 }{ g } =\frac { g }{ 49 } \\ { g }^{ 2\quad }=\quad 49\times (16)\\ g\quad =\quad \pm 28\)

“The relationship cannot be determined from the information given”.

\({ g }^{ 2\quad }=\quad 16\quad \times \quad 49\)

Remember that when “unsquaring” an equation with a squared variable, you must account for the negative possibility. The value of g could be either positive 28 or negative 28.

 

Problem 11) In a group of adults, the ratio of women to men is 5 to 6, while the ratio-handed people is 7 to 9. Everyone is either left or right-handed; no is both.

Quantity A    The number of women in the group.        :

Quantity B :The number of left-handed people in the group.

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 number of woman be \(5x\) & number of men be \(6x\).

Let number of left-handed people be \(7y\)  & number of right-handed people be \(9y\).

Total number of people =

\(5x\quad +\quad 6x\quad =\quad 9y\quad +\quad 7y\\ 11x\quad =\quad 16y\\ x\quad =\quad \frac { 16y }{ 11 } \\ y\quad =\quad \frac { 11x }{ 16 }\)

Number of woman in the group =

\(5x\)

Number of left-handed people in the group = \(7\times \frac { 11x }{ 16 } \quad =\quad \frac { 77x }{ 16 } \\ 5\quad >\quad \frac { 77 }{ 16 }\)

Number of woman is the group are more than the number of left-handed people in the group.

(a) is correct.

 

Problem 12) A pantry holds x cars of bears, twice as many cars of soup & half as many cars of tomato paste as there are cars of bears. If there are no other cars in the pantry, which of the following could be the total number of cars in the pantry? Indicate two such numbers.

a) 6

b) 7

c) 36

d) 45

e) 63

Sol. Let number of cans of soup be y..

Let number of cans of bears be x.

\(y\quad =\quad 2x\)

Cans of tomato = \(\frac { x }{ 2 }\)

Cans of soup = \(2x\)

Total number of cars = \(x\quad +\quad 2x\quad +\quad \frac { x }{ 2 } \quad =\quad 3x\quad +\quad \frac { x }{ 2 } \quad =\quad \frac { 7x }{ 2 }\)

x must be an even integer. Total number of cars could be 7, 14, 21, 28, 35……..etc.

7 & 63 are correct.

 

Problem 13) Party Cranberry is 3 parts cranberry juice & 1 party seltzer. Fancy Lemonade is 1 part lemon juice & 2 parts seltzer. An amount of Party Cranberry is mixed with an equal amount of Fancy Lemonade.

Quantity A                                                 Quantity B:

The fraction of the resulting                   The fraction of the

mix that is cranberry juice.                     resulting mix that is seltzer.

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. Party Cranberry = 3(Cranberry Juice)+(seltzer)

Fancy Lemonade = (Lemon Juice)+2(seltzer)

For Party Cranberry,

Cranberry: seltzer: whole = 3:1:4

For fancy Lemonade,

Lemon: Seltzer: Whole = 1:2:3

Let 1 litre of party Cranberry & 1 litre of fancy lemonade was taken.

Cranberry juice in final mixture = \(\frac { 3 }{ 4 } \quad =\quad \frac { 9 }{ 12 }\)

Seltzer in final mixture = \(\frac { 1 }{ 4 } +\frac { 2 }{ 3 } =\frac { 11 }{ 12 }\)

 

\( \frac { 11 }{ 12 } >\frac { 9 }{ 12 }\)

 

(Seltzer in final mixture) > (Cranberry juice in final mixture)

 

Problem 14) In a parking lot, \(\frac { 1 }{ 3 }\)   of the vehicles are black & \(\frac { 1 }{ 5 }\) of the remainder are white. How many vehicles could be parked on the lot?

a) 8

b) 12

c) 20

d) 30

e) 35

Sol.  Black vehicle = \(\frac { 1 }{ 3 }\) (Total Vehicle)

White vehicle = \(\frac { 2 }{ 3 } \times \frac { 1 }{ 5 }\)(Total Vehicle) = \(\frac { 2 }{ 15 }\)(Total Vehicle)

Number of white cars must be countable with whole numbers. Number of total vehicles must be divisible by 15 of the answer choices, only 30 are divisible by 15.

 

Problem 15) In model town, the ratio of school going children to non-school going children is 5:4, if in the next year 20% of school going children turned into non-school going children making it to 35,400. What is the new ratio of school going children to non-school going children?

a) \(\frac { 4 }{ 5 } \)

b) \(\frac { 5 }{ 4 } \)

c) \(\frac { 3 }{ 4 } \)

d) \(\frac { 5 }{ 3 } \)

Sol. Number of school going children = \(5x\)

Number of non-school going children = \(4x\)

New number of school going children = \(5x(\frac { 80 }{ 100 } )\quad =\quad \frac { 400x }{ 100 }\)

New number of school going children = \(4x\quad +\quad 5x(\frac { 20 }{ 100 } )\quad =\quad x(\frac { 500 }{ 100 } )\)

New ratio = \(\frac { 4 }{ 5 } \)

 

Problem 16) In the famous island of Italy, there are four men for every three women & four children for every three me. How many children are there in the island if it has 531 women?

Sol. \(\frac { Men }{ Women } =\frac { 4 }{ 3 }\)

 

\(\frac { Children }{ Men } =\frac { 5 }{ 3 }\)

 

\(\frac { Men }{ 531 } =\frac { 4 }{ 3 }\)

 

Men = \(\frac { 4 }{ 3 } \quad (531)\)

                 

Children = \(\frac { 5 }{ 3 } (\frac { 4 }{ 3 } )(531)\quad =\quad 1180\)  Ans.

 

 

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