# Basic Concepts - Ratio

The ratio of two quantities in the same units is a fraction that one quantity is of the other. Thus, a to b is a ratio (a/b), written as a : b. The first term of a ratio is called antecedent, while the second term is known as consequent. Thus, the ratio 4:7 represents 4/7 with antecedent 4 and consequent 7.

The multiplication or division of each term of a ratio by a same non-zero number does not effect the ratio. Thus, 3 : 5 is the same as 6 : 10 or 9 : 15 or 12 : 20 etc.

**Proportion:**

The equality of two ratios is called proportion.

Thus, 2 : 3 = 8 : 12 is written as 2 : 3 : : 8 : 12 and we say that 2, 3, 8 and 12 are in proportion.

In a proportion, the first and fourth terms are known as extremes, while second and third terms are known as means.

In a proportion, we always have:

**Product of Means = Product of Extremes.**

# Concept 1: Fourth proportional

**►***Fourth proportional ***means: the fourth number which has the same ratio with the third number as the second number has with the** **first number.**

**Illustration 1:**

Find the fourth proportional to 8, 12 and 16.

You are supposed to calculate a number, which is having the same ratio with 16 as 12 is having with 8 i.e.

In this case, the product of extremes, would be equal to product of middle values. 8×x=12×16⇒x =24

# Concept 2: Third Proportional

**►**** Third proportional means: the third number, which has the same ratio with the second number as the second number is having with**

**the first number.**

**Illustration 2 :**

Find the third proportional to 9 and 12.

In this case, the second number is repeated once as shown below Again, the product of extremes, would be equal to the product of middle values.

9 × x = 12 × 12 ⇒ x = 16

# Concept 3: Second, or Mean Proportional

**►***Second proportional ***means: the second number which has the same ratio with the first number as the third number has with it. This** **is also known as mean proportional.**

If a:b = b:c, then b is called the mean proportional and is equal to the square root of the product of a and c b2= a.c ⇒ b= *ac*

**This is also known as geometric mean.**

**Illustration 3**:

Find the mean proportional between 4 and 16.

Mean proportional between 4 & 16 = √4×16 = √64 = 8

**Illustration 4:**

** **If (23-x) is the geometric mean between (28-x) and (19-x) find the value of x.

By the formula given above, (28 - x)(19 - x) = (23 – x)2. Solving this equation, we get x = 3.

If a/b=b/c=c/d etc. then a,b,c,d are in geometric progression.

# Concept 4: Dividing numbers into ratios

**If any number/amount is to be divided in the ratio of a : b. Then their respective shares can be calculated as **⇒ **number × a/(a+b) and** **number × b/(a+b).**

**Illustration 5:**

Divide Rs 455 in the ratio 4 : 3.

To divide a sum in any ratio, we divide by the sum of the ratio and multiply by the parts individually. Hence one part = 455 × 4/7 = 260 Second part = 455 × 3/7 = 195.

**Illustration 6:**

A certain amount was divided between Kavita and Sheena in the ratio 4 : 3. If Sheena’s share was Rs 2400, the amount is?

We can make a simple equation: Sheena’s share is 3/7 x = 2400, then x = 2400 × 7/3 = 5600.

**Illustration 7:**

A stick 1.4 m long casts a shadow 1.3 m long and the same time when a pole casts a shadow 5.2 m long. Find the length of the pole.

We can make the ratio as 1.4 : 1.3 : : x : 5.2.

Notice the order of the terms. Solving this, we get 1.3x = 1.4 × 5.2, hence x = 5.6.

# Concept 5: Resolving ratios with 3 numbers

**When three terms are given, we can resolve them by finding the LCM of the common term.**

**Illustration 8:**

If a : b = 2 : 3 and b : c = 5 : 7, find a : c and a : b : c.

The LCM of the common term b (3, 5) is 15.

Then a : b = 10 : 15 and b : c = 15 : 21.

Hence we can resolve the ratio as: a : b : c = 10 : 15 : 21.

Similarly, we can resolve more than three ratios also.

# Concept 6: Using proportion in sums

**When any ratio is given, we can assume the terms as multiples of ratio.**

**Illustration 9:**

Find three numbers in the ratio 2 : 3 : 5, the sum of whose squares is 608.

Take the terms as 2x: 3x: 5x. Then sum of the squares = 4x2 +9x2 +25x2 = 608. On solving we get 38x2 = 608 and x = 4.

**Illustration10:**

If 5m-n = m+2n, find the value of (4m+n) : (4m-n).

5m – n = m + 2n, hence 4m = 3n. Substitute in the given expression to get: Required ratio = (3n + n : 3n – n) = 4n : 2n, hence the ratio is 2 : 1.

**Illustration 11:**

A girl got 257 marks in English, Mathematics and History together. The ratio of the marks obtained by her in English and Mathematics is 7 : 8, while that in English and History is 12 : 11. Find the number of marks obtained by her in each subject.

E + M + H = 257.

E : M = 7 : 8.

E : H = 12 : 11.

E : M : H = 84 : 96 : 77 (taking LCM).

Hence marks in English = 84, Marks in Maths = 96 and Marks in History = 77.

**Illustration 12:**

The ratio of the first and second-class fares between the two stations is 4 : 1 and the number of passengers traveling by first and second-class is 1 : 40. If Rs. 1100 is collected as fare, what is the amount collected from first class passengers?

Ratio of the amounts collected from 1st and 2nd class = (4 × 1) : (1 × 40) =1:10.

Hence Amount collected from 1st class passengers = 1/11 × 1100 = 100.

# Concept 7: Sums using mixtures and liquids

**To solve sums on liquids, make a fraction and add the liquid to the relevant term.**

**Illustration 13:**

A mixture contains alcohol and water in the ratio 4 : 3. If 7 litres of water is added to the mixture, the ratio of alcohol and water becomes 3 : 4. Find the quantity of alcohol in the mixture.

Let the alcohol : water be 4x : 3x.

Adding 7 litres of water, the fraction becomes 4x/(3x + 7) = 3⁄4. On solving, we get x = 3 and alcohol = 4x = 12.

**Illustration 14:**

729 ml of a mixture contains milk and water in the ratio 7 : 2. How much more water is to be added to get a new mixture containing milk and water in the ratio of 7 : 3?

1. 60 ml 2. 70 ml 3. 81 ml 4. 90 ml

Milk and water in the original liquid = 7/9 × 729 = 567 and water = 2/9 × 729 = 162.

Let water to be added = x.

Then, 567/(162 + x) = 7/3

hence we get 1701 = 1134 + 7x; or 7x = 567; or x = 81

In case of taking out of a liquid and mixing some other liquid repeatedly, use the formula:

Formula: If from a vessel containing 'a' litres of wine, 'b' litres are drawn and replaced with water, and this operation in repeated 'n' times, then

Wine left after the operation/ Total quantity of mixture =

** **

**Illustration 15:**

A 36 litre vessel contains pure spirit to the brim. 6 litres are drawn off and substituted with water. The mixture is stirred thoroughly and the procedure repeated twice. What is the final percentage purity?

In this case, the final percentage will be [(36 – 6)/36]3, applying the above formula. Since percentage is asked for, the answer would be: (5/6)3 × 100%.

** ****☺****Time Saver**

Remember that in case there are two vessels filled with different liquids (milk and water for instance) and we take 10% of one and mix it with the other and then take 10% of the second vessel and mix it with the first, the ratio of milk and water in the first, to the water and milk in the second, after mixing a number of times will always be equal.

# Concept 8: Variation

*Variation*

**►**** If one object ‘m’ is directly related with the other object ‘n’, then this is written as m = kn, where ‘k’ is a constant. ****►****Similarly if the object is inversely related with the other object, then m = k/n.**

A ∝ B (A varies as B)

Then, A = kB, where k is a constant.

A quantity A is said to vary inversely as quantity B when A varies directly as the reciprocal of B.

A = k/B where k is constant

A quantity is said to vary jointly as number of others when it varies directly as their product. A = k(B × C)

A is said to vary directly as B and inversely as C if A varies as B/C.

**Direct Variation: **If two variables a and b are related in such a way that for any increase or decrease in “a”, “b” will also increase or decrease and vice versa, then the two magnitudes are in direct variation or direct proportion to each other.

**Inverse Variation: **If two variables “a” and “b” are related in such a way that for any increase (or decrease) in “a”, quantity “b” decreases (or increases) in the same ratio, then these two magnitudes are said to vary inversely to each other and the proportion in that case is called inverse proportion or inverse variation

**Illustration 16:**

The marks of a person are directly related with the number of hours a person spends in the studies everyday. Ramesh scored 80%, after studying 4 hrs a day. What would be the marks of Sudesh, who spends 3 hrs a day on studies?

Let the marks scored by a person be = m, and the hrs he spends = h.

As marks are directly related with hrs, we can write m = kh. When‘m’=80, then‘h’is4 ⇒80=4k.Sovalueofk =20. Now for Sudesh, when h is given to be 3 then m = 20h ⇒ 20×3 = 60%

**Illustration 17:**

A = 4 when B = 24, find B when A = 7, where B varies directly as A.

Since B varies directly as A, we can write B = kA where k is a constant.
Hence 24 = k × 4, on solving we get k = 4.

The equation is thus: B = 4A. If A = 7, B = 28.

** **

**Illustration 18:**

Square of x varies as cube of y. Given that x = 4 when y = 2, find x when y is 4.

Since square of x varies as cube of y, we can write x2 = ky3.
Substituting the values of x and y, we get 16 = k × 4, hence k = 4.

When y=4, x2 =ky3 = 4×43.Hence x=16

** **

**Illustration 19:**

Area of a triangle varies as its base when height is constant and as height when base is constant. If the base is 4 cm and height is 10 cm, the area of the triangle is 20 cm2. Find the length of the base if the area of the triangle is 25 cm2 and its height is 5 cm.

According to the above, Area = k × b × h.

Substituting the values, we get 20 = k × 10 × 4; hence k = 1⁄2.

So area = 1⁄2 × b × h
Substituting the second set of values, we get 25 = 1⁄2 × 5 × base.

Hence base = 10.

** **

**Illustration 20:**

If y is proportional to x and x = 6 when y = 2; Find

1) The value of y when x = 15.

2) The value of x when y = 6.

We know that y = kx. Substituting values, we get 2 = 6k; or k = 2/6. Equation becomes y = x/3.

1) When x = 15, y = 15/3 = 5.

2) When y = 6, then 6 = x/3, or x = 18.

**
**

**Illustration 21:**

The number of days required to finish a job is inversely proportional to the number of labourers employed. 15 women finish the work of harvesting a groundnut crop in 8 days. Find the number of women to be employed if the same job is to be completed in 6 days.