Ratios are numerical way/method used to compare two quantities of the same type, the quantities can be weight, money, ages, marks, length, volume, Area etc.

When two quantities are being compared, they can be in fraction or separated by column in-between them, example x/y or x:y respectively.

If a/b and x/y are two different ratios given, and the two ratios are equal to each other.

**There are three Rules that should be known when solving a problem related to the one above. Which are:**

## First Rule (**Ratios**): If a/b = x/y

The product of the numerator of the first fraction(a) and the denominator of the second fraction(y) is equal to the product of the numerator of the second fraction(x) and the denominator of the first fraction(b). That is ay = bx.

## Second Rule (**Ratios**): If a/b = x/y

The inverse of the two quantities are still equal (that is when the numerator and denominator of each quantities interchange their positions). b/a = y/x.

## Third Rule (**Ratios**): If a/b = x/y

The two quantities are still equal even when both the numerator and the denominator of the first quantity (a and b) are used as new numerators and their previous corresponding terms (x and y) are now made to be their denominators respectively. a/x = b/y.

Once these three Rules above are mastered by the students, any problems related to Ratios will be resolved with ease.

## QUESTIONS AND SOLUTIONS:

### Q1: The ratio of the circumference of a circle to its diameter is 22:7. What is the circumference of a circle of diameter 15.6m?

**Solution:**

Circumference of the first circle = 22

Diameter = 7

Let the circumference of the second circle be y

Diameter = 15.6m

22/7 = y/15.6

Using rule 1, we have to cross multiply.

y x 7 = 22 x 15.6

7y = 343.2

Divide both sides by the coefficient of y.

y = 343.2/7

y = 49.03m (2d.p)

### Q2: In preparing a recipe for cake, the ratio of flour to sugar is 40:3. Find the required quantity of flour with 18kg of sugar.

**Solution**:

First flour = 40

First sugar = 3

Second flour = ?

Second sugar = 18kg

Let the second quantity of flour be “f”

Then, 40/3 = f/18

Cross multiply

f x 3 = 40 x 18

3f = 720

Divide both sides by the coefficient of f.

f = 720/3

f = 240kg

### Q3: Ijezie has twice as many apples as Muhammed and Mohammed has three times as many apples as Bamidele. If the three men have a total of 360 apples. Find the number owned by Mohammed.

**Solution**:

Let the number of apples owned by Mohammed = M.

Ijezie = 2 x M = 2M

Bamidele = M/3

Total number of Apples = 360

Then:

M + 2M + M/3 = 360

3M + 6M + M = 1080

10M = 1080

Divide both sides by the coefficient of M

10M = 1080

M = 1080/10

M = 108 apples

### Q4: The ages of a mother and her daughter are in the ratio 8:5. If the daughter’s age now is 15, what is the mother’s age 6 years ago?

**Solution**:

Mother’s age = 8

Daughter’s age = 5

Mother’s age now = ?

Daughter’s age now = 15

Let the mother’s age now be x.

8/5 = x/15

Cross multiply

5x = 8 x 15

5x = 120

Divide both sides by the coefficient of x.

x = 120/5

x = 24

The mother is now 24 years, but her mother’s age 6 years ago = 24 – 6 = 18 years.

### Q5: A workers income is increased in the ratio 36:30. Find the increase percent.

**Solution**:

Let the increase percent be y

The worker’s increase income = 36 and its total percentage is (100 + y)%

The worker’s previous income = 30 and its percentage is 100%.

36/30 = (100 + y)/ 100

Cross multiply

30(100 + y) = 36 x 100

3000 + 30y = 3600

30y = 3600 – 3000

30y = 600

Divide both sides by the coefficient of y.

30y = 600

y = 600/30

y = 20%

### Q6: The cost price #450 of an article is reduced by #150. Find the ratio by which the price is reduced.

**Solution**:

Cost price was #450

It is reduced to (#450 – #150) = #300

The ratio by which the price is reduced = 450/300

= 3/2 = 3:2

### Q7: In each case, find which of the two ratios is greater.

#### (a) 16:7 or 17:6

#### (b) 2.5g : 2kg or 0.4kg : 300kg

#### (c) #1.60 : #4 or #6 : #11

**Solution**:

(a) 16:7 or 17:6

16/7 or 17/6

First find the LCM of their denominators.

LCM = 42

Multiply each ratio by the LCM.

16/7 x 42 = 96

17/6 x 42 = 119

Therefore 17:6 is greater than 16:7

(b) 2.5g : 2kg or 0.4kg : 300kg

[(2.5/2) x 1/1000]kg or (0.4/300)kg

2.5/2000 or 0.4/300

You can also divide the Numerator by its denominator.

2.5/2000 = 0.00125

0.4/300 = 0.001333

Therefore 0.4kg : 300kg is greater.

c) #1.60 : #4 or #6 : #11

1.6/4 or 6/11

Divide.

0.4 or 0.55

So, #6 : #11 is greater

### Q8: Divide 490 in the ratio of 2:5:7

**Solution**:

First sum up the ratios

2 + 5 + 7 = 14

14 is equivalent to 490

Let x be equivalent to 2

y be equivalent to 5

And z be equivalent to 7

2/14 = x/490

Cross multiply.

14x = 2 x 490

14x = 980

Divide both sides by the coefficient of x.

x = 980/14

x = 70

5/14 = y/490

Cross multiply

14y = 490 x 5

14y = 2450

y = 2450/14

y = 175

7/14 = z/490

1/2 = z/490

Cross multiply

2z = 490

Divide both sides by the coefficient of z.

z = 490/2

z = 245

### Q9: The areas of two squares are in the ratio 9:16. The smaller square has sides of length 15cm. Find the length of the sides of the lager square.

**Solution**:

Area of smaller square = 9

Area of bigger square = 16

Sides of the smaller square= 15cm

area = L² = 15 x 15 = 225cm²

Let area of the bigger square be A.

So,

9/16 = 225/A

Cross multiply,

9A = 16 x 225

9A = 3600

A = 3600/9

A = 400cm²

But the length of the lager square is the square root of its area.

L² = A

L = ✓A = ✓400

L = 20cm

### Q10: What number must be added to each term of the ratio 11:15, so that it becomes the ratio 7:8?

Solution:

Let the number be y

(11 + y)/(15 + y) = 7/8

Cross multiply.

8(11 + y) = 7(15 + y)

88 + 8y = 105 + 7y

Collection of like terms.

8y – 7y = 105 – 88

y = 17

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