Proportion
A
statement expressing the equality of two ratios is called a proportion,
i.e., if a : b = c : d or a/b =c/d
then a,b,
c & d are said to be in proportion. Here
a and d are called the extremes
and b
and c are called the
means.
means.
Also
d is called the fourth proportional to
a, b and c. Thus, we can write a : b = c : d,
a is the first proportional,
b is the second proportional,
c is the third proportional,
d is the fourth proportional.
For example, 5 : 10 = 22 : 44 is in proportion.
Each quantity in a proportion is called a term. The first and the last terms are known as the extremes while the second and the third term are called the means. For the four quantities to be in proportion,
Product of means = bc
a:b::c:d
Product of extremes = ad
Product of extremes = ad
Product of means = Product of extremes.
Continued Proportion
Three or more quantities are said to be in continued
proportion, when the ratio of the first and the second is equal to ratio of the second and the third and so on. Thus a, b, c are
in continued proportion if a : b :: b : c
i.e., a/b = b/c
Similarly, a, b, c, d, ... are in continued proportion if,a : b :: b
: c :: c : d ... , i.e. a = b = c = ……
b c d
Proportions are equations and can be transformed using procedures for equations. Some of the transformed
equations are used frequently and are called the laws of proportion.
a : b = c : d, then
ad = bc {Product of extremes = Product of means}
|
|
b/a = d/c
|
(Invertendo)
|
a/c = b/d
|
(Alternendo)
|
In the proportion a : b = b : c, c is
called the third proportional to a and b, and b is called a mean
proportional between
a and c. Thus, b2
= ac. This is known as continued proportion.
E20.
Find the
fourth proportional to 12a2, 9a2b and 6ab2.
Sol. Let x be the fourth
proportional, then
12a2 = 6ab2 x =
6ab2•9a2b = 9 ab3
9a2b x 12a2 2
E21.
Find the
third proportional to 12r2, 4rs.
Sol. Let x be the third proportional 12r2/ 4rs = 4 rs/x , x = 4/3 s2
E22.
Find
the mean proportional between 9a2b and 25b3.
Sol. Let x be the mean proportional.
9a2b = x
x 25b3 , x2 =
9a2b.25b3 = x = ± 15ab2
E23.If a/b = c/d = e/f , then show that a2 – ac + e2 = ae
b2-bd+f2 bf
Sol. Let a/b =c/d =e/ f = k (where k is a constant). Then , a = bk,
c = dk,
e = fk.
LHS = a2 - ac+e2 = b2k2 - bdk2 + f2k2
b2-bd+f2 b2-bd+f2
k2(b2 - bd + f2) = k2
b2 -bd+f2
RHS = ae/bf = bfk2
/ bf = k2 .
LHS = RHS.
E24.If a/b = b/c , then
prove that a3 +b3
= ab
b3 +c3
c2
Short-cut: Let a/b = b/c = k. Then, b =
ck, a = bk = ck2.
a3 + b3 = c3k6 + c3k3 = c3k3 (k3 + 1) = k3
LHS = b3
+ c3 c3k3
+c3 c3(k3
+ 1)
RHS = ab / c2 = ck2 . ck/ c2
= k3
LHS = RHS.
Variation
Most of us
would still remember statements like "The pressure of an enclosed gas varies directly as the
temperature." This and similar
statements have precise mathematical meanings and they represent a
specific type of function called variation functions.
The three
general types of variation functions are direct variation,
inverse variation and joint variation.
Direct Variation
If two quantities X & Y are
related such that any increase or decrease
in 'Y' produces a proportionate increase or decrease in 'X' or vice
versa, then the two quantities are said to be in direct proportion.
X is directly proportional to Y is written as X - Y or X =
KY.
In other
words X : Y = X/Y = K. Here K is a constant whose value for a particular variation is same.
Consider X1= KY1
and X2 = KY2, dividing the two we get X1 / X2 = Y1 / Y2
Thus, the chances of your success in the test are directly proportional to the number
of hours of sincere work devoted every day.
E25.If x = y and x = 9 when y = 30, then find the
relation between x and y. Find x when y = 7 ½and y when x = 6.
Sol.Letx=ky,then9=k(30), k= 3/10 ,i.e.,x= 3/10 y
When y=7½,x=13/10(7½) =2¼
When x = 6, x = 3/10y
.•.y=20.
E26. Different sizes of the car have different models. The
weight of a car model varies directly as the cube
of its length. The weight of a car model of length 3 cm is 10 gm. What is the
weight of a car model of length 12 cm?
Sol.
Let W gm be the
weight of a car model and L cm be its length.
.•. W → L3 or W = kL3 (where k is a
constant)
10 = k(3)3. = k=10 i.e.
W =10/ 27L3
When L = 12, W = 10/ 27(12)3 =640 gm.The required weight is 640 gm.
Inverse Variation
Here two quantities X & Y are related such that, any increase
in X
would lead to a decrease in Y or any decrease in X would lead to an
increase in Y. Thus the quantities X & Y are said to be inversely related and X is inversely proportional
to Y is written as
X = 1/Y or X = k/Y or XY = k (constant)
Therefore X1Y1 = X2Y2
Or the product of two quantities
remains constant.
Thus, the chances that you will be able to cheat in a test
are inversely proportional to the smartness of the invigilator.
E27.If
y varies inversely as x, and y = 3 when x = 2, then find x when y = 21.
Sol.
y→ 1/x or y = k/x
(where k is a constant)
then 3= k/2, k=
6 i.e. y= 6/x
When y = 21, 21 =6/x , x=2/7
E28. The pressure P of a .given mass of ideal gas varies
inversely as the volume V and directly as the temperature T. To what pressure must 100 cubic feet of helium at
1 atmospheric pressure and 253°C temperature
be subjected to be compressed to 50
cubic feet when the temperature is 313°C? Given : Apply Mathematical
concepts only.
Sol. P = kT or k = PV = 1(100) _ = 100
V
T 253
253
Thus the required pressure is
P=100T
=100(313)
253V 253
(50)
=2.47 atmospheric
pressure.
Joint Variation
(a)
A varies jointly as B and C and
is denoted by A : BC Or A = kBC (where k is a
constant).
(b)
A varies directly as B and inversely as C and is denoted by A: B/C or
A= kB/C (where k is a constant).
(c)
If A varies
as B when C is constant, and if A varies as C when B is constant then A varies as BC when B
and C both vary. So, A : BC. Or A = kBC (where k is a
constant).
E29. Given, a varies as b when c is constant, and as c2 when
b is constant. If a = 770, then b = 15 &
c = 7, and when c = 3 & a = 132, find b.
Sol. As a : bc2 or a = kbc2 (where k
is a constant)
.•.770
= k (15) (7)2.
= k=22/21 i.e. a=2/21bC2.
When c = 3,a= 132
132 = 22/21 (b)(3)2
.•. b = 14.
Variation by Parts
(a) If x is partly constant and partly varies as y, then x =
k + kly (where k and kl are constants).
(b) If x is partly
constant and partly varies as the inverse of y,
then x = k + kl (where k and kl are constants).
y
(c) If x partly varies as y and partly varies as z, then x =
ky + kiz (where k and kl are constants).
(d) If x partly varies as y and partly varies as the inverse of z, then x = ky+ kI /z (where k and kl are constants).
E30.Total expenses of a boarding house are partly fixed and partly varying linearly with
the number of boarders. The average expense per boarder is Rs.700 when there are 25 boarders and
Rs.600 when there are 50 boarders. What is the average expense per boarder when there are 100 boarders?
Sol. Let
F is the fixed cost & K is
a constant.
25 boarders : Average
cost = Rs.700
Total cost = 25 x 700 = 17500 = F +
25K ...(1)
50
boarders : Average cost = Rs.600
= Total cost = 50 x
600 = 30000 = F + 50K ...(2)
Solving (1) and (2), we get K = 500, F = 5000. The
average expense per boarder = (F + 100K)/100.
= (5000 + 100 x 500)/100 = 550.
Also, with an increment of 25 boarders, the cost
increases by Rs.12500.
Important
IfA : Band B:C,thenA:C.
If A : C and B : C, then (A ± B) : C and √(AB) :
C.
IfA:BC,then
A/C:B and
A/B:C.
If A:B and C:D, then AC:BD.
If A: then An:Bn.
If A:B and A:C then A:(B—C)andA:(B+C).
If A : B, then AP : BP where P is any quantity, constant or variable
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