Whether you are using units from the Metric system (as we do in this post) or US measurement system (the GMAT being an American test), the concepts don’t change.

Ratio is the quantitative relation between two amounts showing the number of times one value contains or is contained within the other.

(*Reference : Oxford dictionary*)

**Notation**: Ratio of two values a and b is written as a:b or a/b or a *to* b.

For instance, the ratio of number of boys in a class to the number of girls is 2:3. Here, 2 and 3 are not taken as the exact count of the students but a multiple of them, which means the number of boys can be 2 or 4 or 6…etc and the number of girls is 3 or 6 or 9… etc. It also means that in every five students, there are two boys and three girls.

**Question**: In a certain room, there are 28 women and 21 men. What is the ratio of men to women? What is the ratio of women to the total number of people?

**Solution**:

Men : women = 21 : 28 = 3:4

Women : total number of people = 28 : 49 = 4 : 7

**Question**: In a group, the ratio of doctors to lawyers is 5:4. If the total number of people in the group is 72, what is the number of lawyers in the group?

**Solution**:

Let the number of doctors be 5x and the number of lawyers be 4x.

Then 5x+4x = 72 → x=8.

So the number of lawyers in the group is 4*8 = 32.

**Question**: In a bag, there are a certain number of toy-blocks with alphabets A, B, C and D written on them. The ratio of blocks A:B:C:D is in the ratio 4:7:3:1. If the number of ‘A’ blocks is 50 more than the number of ‘C’ blocks, what is the number of ‘B’ blocks?

**Solution**:

Let the number of the blocks A,B,C,D be 4x, 7x, 3x and 1x respectively

4x = 3x + 50 → x = 50.

So the number of ‘B’ blocks is 7*50 = 350.

**Question**:

If the ratio of chocolates to ice-cream cones in a box is 5:8 and the number of chocolates is 30, find the number of ice-cream cones.

**Solution**:

Let the number of chocolates be 5x and the number of ice-cream cones be 8x.

5x = 30 → x = 6.

Therefore, number of ice-cream cones in the box = 8*6 = 48.

A lot of questions on ratio are solved by using proportion.

A proportion is a comparison of two ratios. If a : b = c : d, then a, b, c, d are said to be in proportion and written as a:b :: c:d or a/b = c/d.

a, d are called the extremes and b, c are called the means.

For a proportion a:b = c:d, product of means = product of extremes → b*c = a*d.

Let us take a look at some examples:

**Question**:

In a mixture of 45 litres, the ratio of sugar solution to salt solution is 1:2. What is the amount of sugar solution to be added if the ratio has to be 2:1?

**Answer**:

Number of litres of sugar solution in the mixture = (1/(1+2)) *45 = 15 litres.

So, 45-15 = 30 litres of salt solution is present in it.

Let the quantity of sugar solution to be added be x litres.

Setting up the proportion,

sugar solution / salt solution = (15+x)/30 = 2/1 → x = 45.

Therefore, 45 litres of sugar solution has to be added to bring it to the ratio 2:1.

**Question**:

A certain recipe calls for 3kgs of sugar for every 6 kgs of flour. If 60kgs of this sweet has to be prepared, how much sugar is required?

**Solution**:

Let the quantity of sugar required be x kgs.

3 kgs of sugar added to 6 kgs of flour constitutes a total of 9 kgs of sweet.

3 kgs of sugar is present in 9 kgs of sweet. We need to find the quantity of sugar required for 60 kgs of sweet. So the proportion looks like this.

3/9 = x/60 → x=20.

Therefore, 20 kgs of sugar is required for 60 kgs of sweet.

**Question**:

If a 60 ml of water contains 12% of chlorine, how much water must be added in order to create a 8% chlorine solution?

**Solution**:

Let x ml of chlorine be present in water.

Then, 12/100 = x/60 → x = 7.2 ml

Therefore, 7.2 ml is present in 60 ml of water.

In order for this 7.2 ml to constitute 8% of the solution, we need to add extra water. Let this be y ml.

Then, 8/100 = 7.2/y → y = 90 ml.

So in order to get a 8% chlorine solution, we need to add 90-60 = 30 ml of water.

**Question**:

There is a 20 litres of a solution which has 20% of bleach. Extra bleach is added to it to make it to 50% bleach solution. How much water has to be added further to bring it back to 20% bleach solution?

**Answer**:

This question has 3 parts.

In the first part, there is 20% of bleach in 20 L of solution → 4 L of bleach in 16 L of water = 20 L of solution. Let’s note the details in a table for better clarity and understanding.

Bleach | Water | Total | ||

% | Quantity in L | % | Quantity in L | |

20% | 4 L | 80% | 16 L | 20 L |

In the second part, Extra bleach is added to bring it to 50% of total solution. Let the amount of bleach added be x litres.

Bleach | Water | Total | ||

% | Quantity in L | % | Quantity in L | |

20% | 4 L | 80% | 16 L | 20 L |

50% | 4+x | 50% | 16 L | 20+x |

Then, (4+x)/(20+x) = 50/100 → x = 12 L of bleach is added.

Now, there is 4+12 = 16 L of bleach in 16 L of water in a total of 32 L of solution.

Now, to bring the bleach percentage back to 20%, extra water is added and the amount of bleach remains the same. Let this extra amount of water be y litres.

Bleach | Water | Total | ||

% | Quantity in L | % | Quantity in L | |

20% | 4 L | 80% | 16 L | 20 L |

50% | 16 L | 50% | 16 L | 32 L |

20% | 16 L | 80% | 16+y | 32+y |

16 L of bleach constitutes 20% of the solution →

16/(32+y) = 20/100 → y = 48.

Therefore, 48 litres of water has to be added to the solution if bleach has to be 20% of the whole solution.

**Question**:

1 kg of cashews costs Rs. 100 and 1 kg of walnuts costs Rs. 120. If a mixture of cashews and walnuts is sold at Rs. 105 per kg,then what fraction of the total mixture are walnuts?

**Solution**:

For this type of problems, first step is to determine how much each of the items is above or below the target.

Our target price is Rs. 105. Cashews price is Rs. 5 below the target price and walnuts price is Rs. 15 above the target price.

So, for each kg of cashews added, let’s consider it as ‘-5’ and for each kg of walnuts added, let’s consider it as ‘+15’. These two have to be added in such a way that they cancel out each other. Adding ‘-5’ thrice gives a ‘15’ and adding ‘+15’ once results in cancellation of the terms.

This means that adding 3 kgs of cashews and 1 kg of walnuts gives a mixture that can be sold at Rs. 105 per kg.

So, 3 kgs of cashews present for every 1 kg of walnuts. The ratio of cashews to walnuts is 3:1. Fraction of walnuts in the mixture = 1/(3+1) = 1/4 of the total mixture

Problem 1

On a certain map, 1 cm = 12 km actual distance. If two places are 96 km apart, what is their distance on map?

A. 10 cm

B. 12 cm

C. 96 cm

D. 8 cm

Answer 1

D.

**Explanation**:

1cm/12 km = x cm/100 km → x = 8 cm

Problem 2

A person types 360 words in 4 minutes. How much time does he take to type 900 words?

A. 15

B. 90

C. 10

D. 9

Answer 2

C.

**Explanation**

4/360 = x/900 → x=10

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## 1 Comment

Hey can you please emphasize on the working of the cashews and walnuts question?

I’m stuck on how we’re assuming this:

So, for each kg of cashews added, let’s consider it as ‘-5’ and for each kg of walnuts added, let’s consider it as ‘+15’. These two have to be added in such a way that they cancel out each other. Adding ‘-5’ thrice gives a ‘15’ and adding ‘+15’ once results in cancellation of the terms.

Why do they have to be added to cancel out eaach other?

Would appreciate the help.

Thanks!