Properties of Ratios: CAT Quantitative Aptitude – Part 2

Properties of Ratios (Continued)

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8. A ratio of greater inequality is diminished, and a ratio of less inequality is increased, by adding the same quantity to both its terms.

Mathematically, let a/b be the ratio, and let be the ratio, and let   be the new ratio formed by adding x to both its terms.

Now,

and (a – b) is positive or negative accordingly as a is greater or less than b.

Hence, if ,

 and if 

9. A ratio of greater inequality is increased, and a ratio of less inequality is diminished by taking the same quantity from both its terms.

10. When two or more ratios or a series of fractions are equal, then

where p, q, r, s, n...are any quantities what so ever.

11. When a series of fractions are equal, each of them is equal to the sum of all the numerators divided by the sum of all the denominators.

12. If an equation is homogeneous with respect to certain quantities, we may for these quantities, substitute in the equation any other proportional to them.

(For our understanding, a homogeneous equation is one where the sums of the powers of every term are the same. Therefore,

yx²+xy²+y³=0 is a homogeneous equation of degree 3

but, yx+xy²+Y² =0 is not a homogeneous equation as all the terms do not have an equal degree.

For instance, the equation yx² +mxy² + ny³= 0 is homogeneous in x and y.
Let α and β be the quantities proportional to x and y respectively.

Put k= x/α = y/β, so that  = αk; y= βk

Then, βk.(αk)²+ m.α.k.(βk)² + (βk)³ = 0

βα² + m αβ² + nβ³ =0                   ... dividing both sides by k³

(an equation of the same form as the original one, yx² + mxy² + ny³ = 0 but with α and β in the places of x and y respectively).

13.  If   be unequal fractions with all denominators of the same sign (either positive or negative), then the fraction    lies in magnitude between the greatest and least of them.

that is ,

Solved Examples:

Problem 1: What number must be added to each term of the ratio 5:37 to make it equal to 1:3?

Solution:

Let x be added to each term of the ratio 5:37

Then,

Therefore 3x+15 = x + 37

Therefore, x = 11

Therefor 11+5/11+37 = 16/48 = 1/3

Problem 2: If the ratios     then the expression  is equal to:

[1] a/b

[2] 1

[3] a²/b²

[4] a⁴/b⁴

Solutions:

Given that a/b = c/d = e/f, we can get a f = eb

or e^₂ b² = a² f²

Putting these values in the given expression, we have

Therefore, the correct answer is choice [4].
       
Problem 3: I used 6 litres of oil paint to paint a map of India 6 metres high. How many litres of paint would I need for painting a proportionately scaled map that is 18 metres high?

[1] 18

[2] 30

[3] 54

[4] Cannot be determined

Solution:

This is a problem taken from CAT 1993 Admission Bulletin.
If the heights of maps are in the ratio of 6:18 or 1:3, the areas of the maps will be in the ratio of 12:32 (as area is a square of height-dimension). Therefore, the quantity (of paint) required to paint the map that is 18 metres high = (6 × 9) litres or 54 litres.