Ratio & Proportion
Asok Nadhani
1.1 Ratio
A Ratio is a comparison of the sizes of two
or more quantities of the same kind by division.
If x and y are two quantities of the same
kind (in same units), then the fraction x/y is called the ratio of x to y. It
is written as x : y. Thus, the ratio of x to y = x/y or x : y. The quantities x
and y are called the terms of the ratio, x is called the first term or
antecedent and y is called the second term or consequent.
For example, in the ratio 7 : 8, 7 & 8
are called terms of the ratio. 7 is called first term and 8 is called second
term.
Types
of Ratio
i.
Compounded Ratio: The ratio of the
product of antecedents to the product of the consequents of two or more ratios
is called compounded ratio.
For Example: For ratios x : y and u
: v the compounded ratio is x x u : y x v or xu : yv Thus
compounded ratio of 3 : 4 and 5 : 7 is 3 x 5 : 4 x 7 or 15 : 28 .
ii.
Duplicate Ratio: The ratio of squares
of its terms is called duplicate ratio.
For Example: For ratio x : y, x2 : y2
is the duplicate ratio.
Thus duplicate ratio
of 5 : 7 is 52 : 72 or 25: 49.
iii.
Sub Duplicate Ratio: The ratio of the
square roots of the terms of ratio is called as sub duplicate ratio.
For Example: For ratio x : y, √x : √y is the sub
duplicate ratio.
iv.
Triplicate Ratio: The ratio of cubes of
the terms of the ratio is called as triplicate ratio. For Example: For ratio x : y, x3 : y3 is the
triplicate ratio.
Thus triplicate ratio
of 7 : 9 is 73 : 93 or 343 : 729
v.
Inverse Ratio or
Reciprocal ratio: The
ratio of the reciprocals of the terms of the terms is called as inverse ratio
or reciprocal ratio.
For Example: For ratio x : y,
(1/x) :(1/y), or
y:x is the inverse or reciprocal ratio.
Thus
inverse or reciprocal ratio of 5 : 7 is (1/5) : (1/7) or 7:5
Note: The product of a ratio and its inverse is
always 1.
Thus 5 : 7 x 7 : 5 = 5
x 7 : 7 x 5 = 35 : 35 = 1.
vi.
Continued Ratio: The ratio between 3
or more quantities of same kind is called as continued ratio.
For Example: For quantities w, x, y, z, w: x: y : z is the
continued ratio
Thus continued ratio
of 3, 5, 7, 8 is 3 : 5 : 7 : 8.
Example
1 : Ratios
The ratio of the no. of Science students to
the no. of arts students in a college of 1440 students is 3 : 5. If 36 new arts
students are admitted in the college, find how many new Science students may be
admitted so that the ratio of the no. of Science students to the no. of arts
students may change to 2 : 3.
Solution: The ratio of the no.
of Science students to the no. of arts students
= 3 : 5.
Sum of the ratios = 3 + 5 = 8
So, the no. of Science students in the
college = (3 x 1440) / 8 = 540
And the no. of arts students in the college = (5 x 1440) / 8 = 900
Let the no. of new Science students admitted
be x, then the no. of Science students become
(540 + x).
After admitting 36 new arts students , the
no, of arts students become 900 + 36 =
936
According to given description of the
problem, (540 + x) / 936 = 2/3
Or, 3(540 + x) = 2 x 936
Or, 1620 + 3x = 1872 or, 3x = 252 or, x = 84.
Hence the no. of new Science students
admitted = 84.
Example
2 : Ratios
The monthly earning of two women are in the
ratio 4 : 5 and their monthly expenditures are in the ratio 7: 9. If each saves
Rs.150 per month, find their monthly earnings.
Solution:
Lets
the monthly earnings of two women be Rs.4x and Rs.5x so that the ratio is
Rs.4x : Rs.5x = 4 : 5. If each saves Rs.150
per months, then the expenditures of two women are Rs. (4x − 150) and Rs. (5x −
150).
(4x-150) / (5x−150) = 7/9,
or 36x − 1350 = 35x – 1050
or, 36x − 35x = 1350 − 1050, or, x = 30
Therefore, the monthly earnings of the two women
are Rs.4 x 300 and Rs.5 x 300 i.e. Rs.1200/-
and Rs.1500 /-.
Example
3: Ratios
The ratio of the prices of two flat was 16 : 25 . Two years later when the
price of the first has increased by 20% and that of the second by Rs.545, the
ratio of the prices becomes 11 : 20 .
Find the original prices of the two houses.
Solution:
Let
the original prices of two flat be Rs.16x and Rs.25x respectively. Then by the
given conditions
[{16x + (20% of 16x)}] / (25x + 545) = 11/20
Or, (16x + 3.2x) / (25x + 545) =
11/20
320x + 64x = 275x +
5995
or, 384x − 275x = 5995, or, 109x = 5995; or x
= 55
Therefore, the original prices of two houses
are Rs.16 x 55 and Rs. 25 x 55 i.e. Rs. 880 and
Rs. 1,375.
1.2
Proportion
An equality of two ratios is called a
proportion. Four quantities m, n, o, p are said to be in proportion if m: n = o
: p (also written as m : n : o: p) i.e. if m/n = o/p i.e. if mp = no.
The quantities m, n, o, p are called terms of
the proportion; m, n, o and p are called its first, second, third and fourth terms
respectively. First and fourth terms are called extremes (or extreme terms).
Second and third terms are called means (or middle terms).
If m : n = o : p then p is called fourth
proportional.
Cross
product rule.
If m : n = o : p are in proportion then m/n =
o/p i.e. mp = no
i.e. product of extremes = product of means.
Continuous
Proportion
Three quantities m, n, o of the same kind (in
same units) are said to be in continuous proportion if m : n = n : o i.e. m /n
= n / o i.e. n2 = mo
If m, n, o are in continuous proportion, then
the middle term n is called the mean proportional between m and o, m is the
first proportional and o is the third proportional.
Thus, if n is mean proportional, between m and
o, then n2 = mo i.e. n = Ömo.
When three or more numbers are so related
that the ratio of the first to the second, the ratio of the second to the
third, to the fourth etc, are all equal, the numbers are said to be in
continued proportion. We write it as
x / y = y / z = z /w = w / p = p /q . When x,
y, z, w, p and q are in continued proportion.
If a
ratio is equal to the reciprocal of the other, then either of them is in
inverse (or reciprocal) proportion of the other. For example 5 / 4 is in
inverse proportion of 4 / 5 and vice-versa.
Properties
of Proportion
- If
a : b = c : d, then ad = bc (By cross multiplication).
- If
a : b = c : d, then b : a = d : c (Invertendo)
- If
a : b = c : d, then a : c = b : d (Alternendo)
- If
a : b = c : d, then a + b : b = c + d : d (Componendo)
- If
a : b = c : d, then a – b : b = c – d : d (Dividendo)
- If
a : b = c : d, then a + b : a – b = c + d : c – d (Componendo and
Dividendo)
Example:
If (x/6) = (y/8) = (z/14), then prove that
{(x+y+z) / z} =2
We have, (x/6) = (y/8) =
(z/14) = (x + y + z) / (6 + 8 + 14) = (x + y + z) / 28
So, (x
+ y + z) / 28 = z/14. Or , (x + y +
z) / z = 28/14 =2
Example:
What quantity must be added to each of the
terms of the ratio a : b so that it may become equal to c : d ?
Let ‘x’ be added to each term of the ratio a :
b to make it c : d.
(a+x ) / (b+x) = c/d
So, d(a + x) = c(b + x), or da + dx = cb + cx, or dx-cx=cb-da, or (d – c)x = cb – da
Or x=( cb – da) / (d-c), or x= (ad – bc) /
(c-d)
Example:
An ornament weights 15 gm of which 5 gm is
pure silver and the rest alloy. Find the ratio of pure silver to alloy?
Weight of ornament = 15 gm. Weight of pure
silver = 5 gm. Weight of rest alloy =15 – 5 = 10 gm
(Weight of pure silver) / (Weight of alloy) =
5/10 = 1/2 = 1:2
Example:
What should be added to the terms of 3 : 13
so to get 1 : 3.
Let x be added to each terms of ratio of 3 : 13
to get 1 : 3
(3+x) / (13+x) = 1/3
or 3(3 + x) = (13 + x), or 9 + 3x = 13 + x, or
3x – x = 13 – 9, or 2x = 4, or x = 2.
Example:
What quantity must be added to each of the
terms of the ratio a : b so that it may become equal to c : d ?
Let ‘x’ be added to each term of the ratio a :
b to make it c : d.
(a+x ) / (b+x) = c/d
So, d(a + x) = c(b + x), or da + dx = cb + cx, or dx-cx=cb-da, or (d – c)x = cb – da
Or x=( cb – da) / (d-c), or x= (ad – bc) /
(c-d)
Example:
An ornament weights 15 gm of which 5 gm is
pure silver and the rest alloy. Find the ratio of pure silver to alloy?
Weight of ornament = 15 gm. Weight of pure
silver = 5 gm. Weight of rest alloy =15 – 5 = 10 gm
(Weight of pure silver) / (Weight of alloy) =
5/10 = 1/2 = 1:2
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