Multiple choice questions on ratio and proportion play a major role quantitative aptitude test. There is no competitive exam without the questions from this topic. We have already learned this topic in our lower classes.Even though we have been already taught this topic in our lower classes, we need to learn some more short cuts which are being used to solve the problems in the above topic.

The only thing we have to do is, we need to apply the appropriate short cut and solve the problems in a limited time. This limited time will be one minute or less than one minute in most of the competitive exams.

1. a/b is the ratio of a to b. that is a:b. |

2. When two ratios are equal, they are said to be in proportion. If a/b = c/d, then a,b,c,d are in proportion. It can be written as a:b=c:d. |

3. Cross product rule in proportion. Product of the extremes = product of the means. That is ad= bc. |

4. If two ratios are inverse to each other, then the product of those two ratios will be 1 or 1:1. |

5. If “a” is directly proportional to “b”, then it can be written as a=kb.If “a” is inversely proportional to “b”, then it can be written as a=k/b. |

6. If “m” kg of one kind costing “a” rupees/kg is mixed with “n” kg of another kind costing “b” rupees/kg.Then the price of the mixture is = (ma+nb)/(m+n). |

7. If one quantity increases or decreases in the ratio a:b, then the new quantity is = b of the original quantity/a.That is new quantity = (“b” x original quantity) /a. |

Students who are preparing to improve their aptitude skills and those who are preparing for this type of competitive test must prepare this topic in order to have better score. Because, today there is no competitive exam without questions from the topic ratio and proportion . Whether a person is going to write placement exam to get placed or a students is going to write a competitive exam in order to get admission in university, they must be prepared to solve time and work problems. This is the reason for why people must study this topic.

As we mentioned in the above paragraph, a person who wants to get placed in a company and a students who wants to get admission in university for higher studies must write competitive exams like placement test and entrance exam. To meet the above requirement, it is very important to score more marks in the above mentioned competitive exams. To score more marks, they have to prepare this topic. Preparing this topic would definitely improve their marks in the above exams. Preparing this topic is not difficult task. We are just going to remember the stuff that we have already learned in our lower classes

Students have to learn few basic operations in this topic time and work and some additional tricks. Already we are much clear with the four basic operations which we often use in math. They are addition, subtraction, multiplication and division. Even though we are much clear with these four basic operations, we have to be knowing some more stuff to do the problems which are being asked from this topic in competitive exams. The stuff which I have mentioned above is nothing but the tricks and shortcuts which need to solve the problems in a very short time.

Short cut is nothing but the easiest way to solve Multiple choice questions on ratio and proportion. In competitive exams, we will have very limited time to solve each problem. Then only we will be able to attend all the questions. If we do problems in competitive exams in perfect manner with all the steps, it will definitely take much time and we may not able to attend the other questions. So we need some other way in which the problems can be solved in a very short time. The way we need to solve the problem quickly is called as shortcut.

**Here, we are going to have some problems on ratio and proportion. You can check your answer online and see step by step solution.**

1. The average age of three boys is 25 years and their ages are in the proportion 3:5:7. The age of the youngest boy is

From the given ratio, ages of the three boys are 3x, 5x and 7x.

Average of the ages = 5x

Average of the ages = 25 (given)

Then we have 5x=25 ===> x=5

Hence, the age of the youngest boy = 3x5 = 15 years.

Average of the ages = 5x

Average of the ages = 25 (given)

Then we have 5x=25 ===> x=5

Hence, the age of the youngest boy = 3x5 = 15 years.

2. John weighs 56.7 kilograms. If he is going to reduce his weight in the ratio 7:6, find his new weight.

Original weight of John = 56.7 kg (given)

He is going to reduce his weight in the ratio 7:6

[Hint:If a quantity increases or decreases in the ratio a:b, then new quantity = 'b' of the original quantity divided by 'a']

His new weight = (6x56.7)/7 = 6x8.1 = 48.6 kg.

He is going to reduce his weight in the ratio 7:6

[Hint:If a quantity increases or decreases in the ratio a:b, then new quantity = 'b' of the original quantity divided by 'a']

His new weight = (6x56.7)/7 = 6x8.1 = 48.6 kg.

3. The ratio of the no. of boys to the no. of girls in a school of 720 students is 3:5. If 18 new girls are admitted in the school, find how many new boys may be admitted so that the ratio of the no. of boys to the no. of girls may change to 2:3.

Sum of the terms in the given ratio = 3+5 = 8

for the given ratio,

No. of boys in the school = 720x(3/8)= 270

No. of girls in the school = 720x(5/8)= 450

Let "x" be the no. of new boys admitted in the school.

18 new girls are admitted (given)

After the above new admissions,

no. of boys in the school = 270+x

no. of girls in the school = 450+18 = 468

From the given information,

(270+x):468 = 2:3

3(270+x)=468x2

810+3x = 936

3x=126

x=42

Hence the no. of new boys admitted in the school is 42

for the given ratio,

No. of boys in the school = 720x(3/8)= 270

No. of girls in the school = 720x(5/8)= 450

Let "x" be the no. of new boys admitted in the school.

18 new girls are admitted (given)

After the above new admissions,

no. of boys in the school = 270+x

no. of girls in the school = 450+18 = 468

From the given information,

(270+x):468 = 2:3

3(270+x)=468x2

810+3x = 936

3x=126

x=42

Hence the no. of new boys admitted in the school is 42

4. The monthly incomes of two persons are in the ratio 4:5 and their monthly expenditures are in the ratio 7:9. If each saves $50 per month, find the monthly income of the second person.

From the given information,

Income of the 1st person = 4x

Income of the 2nd person = 5x

Expenditure of the 1st person = 4x-50

Expenditure of the 2nd person = 5x-50

(Because each saves $50 per month and Expenditure = Income - Savings)

Expenditure ratio = 7:9

(4x-50):(5x-50) = 7:9

9(4x-50) = 7(5x-50)

36x-450 = 35x-350

x = 100

Hence income of the second person = 5X100 = $500 .

Income of the 1st person = 4x

Income of the 2nd person = 5x

Expenditure of the 1st person = 4x-50

Expenditure of the 2nd person = 5x-50

(Because each saves $50 per month and Expenditure = Income - Savings)

Expenditure ratio = 7:9

(4x-50):(5x-50) = 7:9

9(4x-50) = 7(5x-50)

36x-450 = 35x-350

x = 100

Hence income of the second person = 5X100 = $500 .

5. The ratio of the prices of two houses was 16:23. Two years later when the price of the first has increased by 10% and that of the second by $477, the ratio of the prices becomes 11:20. Find the original price of the first house.

From the given information,

original of the 1st house = 16x

original of the 2nd house = 23x

After increment in prices,

price of the 1st house = 16x + 10%of16x=16x+1.6x = 17.6x

price of the 2nd house = 23x+477

After increment in prices, the ratio of the prices becomes 11:20

Then we have, 17.6x:(23x+477) = 11:20

20(17.6x) = 11(23x+477)

352x = 253x + 5247

99x = 5247

x = 53

Hence original price of the first house = 16X53 = $848 .

original of the 1st house = 16x

original of the 2nd house = 23x

After increment in prices,

price of the 1st house = 16x + 10%of16x=16x+1.6x = 17.6x

price of the 2nd house = 23x+477

After increment in prices, the ratio of the prices becomes 11:20

Then we have, 17.6x:(23x+477) = 11:20

20(17.6x) = 11(23x+477)

352x = 253x + 5247

99x = 5247

x = 53

Hence original price of the first house = 16X53 = $848 .

6. Find in what ratio will the total wages of the workers of a factory be increased or decreased, if there be a reduction in the number of workers in the ratio 15:11 and an increment in their wages in the ratio 22:25.

Let x and y be the no. of workers and wages per worker respectively.

The, total wages of the workers = xy

After reduction, no. of workers in the factory = (11x)/15

After increment, wages per worker = (25y)/22

[Hint:If a quantity increases or decreases in the ratio a:b, then new quantity = 'b' of the original quantity divided by 'a']

After the above two changes, total wages of the workers = (11x/15)(25y/22)

By simplification, total wages of the workers = 5xy/6 = (0.83)xy

Before changes, total wages = xy = 1xy

After changes, tital wages = 5xy/6 = (0.83)xy

Comparing the above two, it is very clear that the total wages of the workers has been reduced after reduction in the number of workers and increment in wages per worker.

Now we need the ratio. That can be taken from the term "5xy/6". That is 6:5. Since the total wages has been decreased, it has to be decrement ratio. So first term to be greater than the second term.

Hence the total wages of the workers will be decreased in the ratio 6:5. .

The, total wages of the workers = xy

After reduction, no. of workers in the factory = (11x)/15

After increment, wages per worker = (25y)/22

[Hint:If a quantity increases or decreases in the ratio a:b, then new quantity = 'b' of the original quantity divided by 'a']

After the above two changes, total wages of the workers = (11x/15)(25y/22)

By simplification, total wages of the workers = 5xy/6 = (0.83)xy

Before changes, total wages = xy = 1xy

After changes, tital wages = 5xy/6 = (0.83)xy

Comparing the above two, it is very clear that the total wages of the workers has been reduced after reduction in the number of workers and increment in wages per worker.

Now we need the ratio. That can be taken from the term "5xy/6". That is 6:5. Since the total wages has been decreased, it has to be decrement ratio. So first term to be greater than the second term.

Hence the total wages of the workers will be decreased in the ratio 6:5. .

7. The angles of a triangle are in the ratio 2:7:11. The angles are

From the given information, the three angles of the triangle are 2x, 7x and 11x

In any triangle, sum of the three angles = 180 degrees

Then we have, 2x+7x+11x = 180

20x = 180

x = 9

The first angle = 2(9)= 18

The second angle = 7(9)= 63

The third angle = 11(9) =99

Hence the three angle of the triangle are (18, 63, 99) .

In any triangle, sum of the three angles = 180 degrees

Then we have, 2x+7x+11x = 180

20x = 180

x = 9

The first angle = 2(9)= 18

The second angle = 7(9)= 63

The third angle = 11(9) =99

Hence the three angle of the triangle are (18, 63, 99) .

8. The ratio of two numbers is 7:10. Their difference is 105. The numbers are

From the given information, the two numbers are 7x and 10x

Their difference = 105 (given)

Then we have, 10x-7x = 105

3x = 105

x = 35

The first number = 7(35)= 245

The second number = 10(35)= 350

Hence the two numbers are (245, 350) .

Their difference = 105 (given)

Then we have, 10x-7x = 105

3x = 105

x = 35

The first number = 7(35)= 245

The second number = 10(35)= 350

Hence the two numbers are (245, 350) .

9. A , B and C are three cities. The ratio of average temperature between A and B is 11:12 and that between A and C is 9:8. The ratio between the average temperature of B and C is

From the given two ratios A:B = 11:12 and A:C = 9:8, we find the common element A. The value corresponding to A in the first ratio is 11 and in the second ratio is 9.

Now, we have to find L.C.M of 11,9. That is 99.

Then we have to make the value corresponding to A in both ratios as 99 by multiplication.

In the 1st ratio,A:B = 11X9:12X9 = 99:108-----(1)

In the 2nd ratio,A:C = 9X11:8X11 = 99:88------(2)

From (1) and (2), the value corresponding to B is 108 and C is 72.

So, B:C = 108:72

By simplification, B:C = 27:22

The ratio between the average temperature of B and C is 27:22.

Now, we have to find L.C.M of 11,9. That is 99.

Then we have to make the value corresponding to A in both ratios as 99 by multiplication.

In the 1st ratio,A:B = 11X9:12X9 = 99:108-----(1)

In the 2nd ratio,A:C = 9X11:8X11 = 99:88------(2)

From (1) and (2), the value corresponding to B is 108 and C is 72.

So, B:C = 108:72

By simplification, B:C = 27:22

The ratio between the average temperature of B and C is 27:22.

10. The ratio between the speeds of two trains is 7:8. If the second train runs 400 kms. in 5 hours, the speed of the first train is

From the given information,

Speed of the 1st train = 7x

Speed of the 2nd train = 8x -------(1)

[Hint:Speed = Distance/Time]

Speed of the 2nd train = 400/5 = 80km/hr-------(2)

From (1) and (2), we get, 8x = 80

x = 10

Hence speed of the 1st train = 7X10 = 70 km/hr.

Speed of the 1st train = 7x

Speed of the 2nd train = 8x -------(1)

[Hint:Speed = Distance/Time]

Speed of the 2nd train = 400/5 = 80km/hr-------(2)

From (1) and (2), we get, 8x = 80

x = 10

Hence speed of the 1st train = 7X10 = 70 km/hr.

HTML Comment Box is loading comments...