Download Ratio and Proportion

RATIO AND PROPORTION
The ratio of two quantities of the same kind and in the same unit is the fraction that one quantity is of the
a
other. The ratio of a to b is the fraction , written as a : b.
b
In the ratio a : b, a is called the first term or antecedent and b is called the second term or consequent.
Eg:
Ratio of a week to a leap year is 7 : 366.
Property: The value of a ratio remains unchanged if both its terms are multiplied or divided by the same
non zero number.
a ma

[b  0]
b mb
a a/m
=
[m  0] [b  0]
b b/m
The ratio a : b is always written in its lowest terms.
Eg:
27 : 81 = 1 : 3
Commensurable quantities: Two quantities are said to be commensurable if their ratio can be
expressed as the ratio of two integers. Otherwise they are incommensurable.
Some results:
Ratio of greater inequality, less inequality and Equality:
A ratio a : b is called:
(i) a ratio of greater inequality, if a > b, e.g. (5 : 2);
(ii) a ratio of less inequality, if a < b e.g. (3 : 8)
(iii) a ratio of equality, if a = , e.g. (3 :3)
Rule1: If the same positive number is added to both the terms of a ratio of greater inequality, then the ratio is
diminished.
For example, (7 : 4) is a ratio of greater inequality and (7 + 3 : 4 + 3) < (7 : 4). as 10 x 4 < 7 x 7.
Rule 2: If the same positive number is added to both the terms of a ratio of less inequality, then the ratio is
increased.
For example, (2 : 5) is a ratio of greater inequality and (2 + 4 : 5 + 4) < (2 : 5), as 6 x 5 > 9 x 2
Composition of Ratios:
(i) The Compounded ratio of (a : b) and (c : d) is (ac : bd)
(ii) The Duplicate ratio of (a : b) is (a2 : b2)
(iii) The Triplicate ratio of (a : b) is (a3 : b3)
(iv) The Sub-duplicate ratio of (a : b) is ( a : b )
(v) The Sub-triplicate ratio of (a : b) is (a1/3 : b1/3)
 1 1
(vi) The Reciprocal ratio of (a : b) is  :  = (b : a)
a b
Illustration 1:
Find the compounded ratio of:
(i) (6 : 5) and (8 : 12)
Solution:
(ii) (3 : 7), (6 : 11) and (14 : 27)
(i) Compounded ratio of (6 : 5) and (8 : 12) = (6 x 8:5 x 12) = (48 : 60) = (4 : 5)
(ii) Compounded ratio of (3 : 7). (6 : 11) and (14 : 27) = (3 x 6 x 14 : 7 x 11 x 27)
3 x 6 x14
4
=
=
= 4 : 33.
7 x11x 27 33
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Illustration 2:
Find
(i) The triplicate ratio of 4 : 3;
(ii) The duplicate ratio of 2 3 : 3 2 ;
(iii) The sub-duplicate ratio of 36 : 25;
(iv) The sub-triplicate ratio of 8x3 : 27y3
(v) The reciprocal ratio of 6 : 11.
Solution:
(i) The triplicate ratio of 4 : 3 = 43 : 33 = 64 : 27
(ii) The duplicate ratio of 2 3 :3 2  (2 3)2 :(3 2)2 12:18  2:3.
(iii) The sub-duplicate ratio of 36 : 25 =
3
36: 25  6:5
3
(iv) The sub-triplicate ratio of 8x : 27y = (8x3)1/3: (27y3)1/3 = 2x : 3y.
1 1
(v) The reciprocal ratio is 6 : 11 = : 11:6.
6 11
Illustration 3:
If a : b = 4 : 9 and b : c = 6 : 5, find (i) a : c (ii) a : b : c.
Solution:
(i) a : b = 4 : 9 and b : c 
a 4
b 6
 and  .
b 9
c 5
a a b 4 6 8
  x  =  x  =
.
c  b c   9 5  15
 a : c = 8 : 15
(ii) a : b : c = 4 : 9 :
Illustration 4:
Solution:
15
= 8 : 18 : 15.
2
If (3x + 4y) : (6x + 5y) = 11 : 19, find ( x : y)
(3x + 4y) : (6x + 5y) =11 : 19
3x  4y 11

=
6x  5y 19
 19 (3x + 4y) = 11 (6x +5y) [By cross multiplication]
 57x + 76y = 66x + 55y
 9x = 21y
x
21 7

=

9 3
y
 x : y = 7 : 3.
Exercise 1:
Exercise 2:
Find compound ratio of
(a)
3 : 4; 5 : 6
(b)
a + b : a – b, a2 + b2 : ( a + b )2 and (a2 – b2)2 : (a4 – b4)
Calculate the duplicate ratio of the following
(a) 2 2 : 3 3
(b)
Exercise 3:
x y
:
3 7
Calculate the triplicate ratio of the following
(a) a : b
(b) 2 : 7
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Exercise 4:
Calculate the sub duplicate ratio of the following 64 b2 :
Exercise 5:
Calculate the sub triplicate ratio of 125x3 : 8y3
Illustration 5:
If x : y = 4 : 5. Find the ratio 10x + 3 y : 5 x + 2 y
Solution:
(10 x + 3 y) : ( 5 x + 2 y ) =
a2
49
10x  3y
5x  2y
x
10    3 10  4  3
8  3 11
y
5
=
=

42 6
4
x
5

2
5   2
5
 
y
Illustration 6:
If 3x + 7y : 2x + 5y = 27 : 19 find x : y
Solution:
3x  7y
27
=
19
2x  5y
57x + 133y = 54x + 135y
3x = 2y
x
2
=
3
y
Comparison of two or more ratios
Illustration 7:
Which is greater 8 : 9 or 10 : 11
Solution:
Let
10
a 8
c
=
 and
b 9
d
11
ad 88
=
 ad < bc
bc 90
a c
 < .
b d
To divide a given quantity in a given ratio:
Illustration 8:
Solution:
Divide 105 in the ratio 17 : 4
Exercise 6:
Divide the sum of Rs. 1216 between A and B in the ratio
Exercise 7:
At present the ages of Ajay and Basu are in the ratio 7 : 5. Two years back, the ratio of their
Ratio 17 : 4
Sum of ratio = 21
17
1st term
 105 = 85
21
4
2nd term
 105 = 20.
21
1 1
: .
3 5
ages was 3 : 2. Find their ages.
Exercise 8:
If 3x + 4y : 8x + 5y = 29 : 49, find x : y
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Exercise 9:
If x : y = 8 : 9, find the ratio 7x – 4y : 3x + 3y
Exercise 10:
If 6a2 – ab : 2 ab – b2 = 6 : 1, find a : b
Exercise 11:
If
3x  4y
5x  6y
=
find x : y
2x  3y
4x  5y
PROPORTION:
Four quantities a, b, c, d are said to be in proportion if a : b = c : d, and we write a : b : : c : d.
If four quantities a, b, c, d, are in proportion, then we also say that they are proportional. The terms ‘a and d’
are called extremes and ‘b and c’ called means. The term d is called the fourth proportional to a, b, c.
Eg:
2, 6, 18, 54 are in proportion because 2 : 6 = 18 : 54. Clearly, 2 and 54 are extremes while 6 and 18
are means. Also, the fourth proportional to 2, 6, 18, is 54
Theorem:
The four quantities a, b, c, d are in proportion if and only if the product of extremes is equal
to the product of means.
Proof: a, b, c, d are in proportion  a : b : : c : d
a:b=c:d
a c

b d
 ad = bc
 Product of extremes = product of means.
Conversely, let a, b, c, d be four quantities such that ad = bc. Then,
ad
bc
ad = b c 
=
[ Dividing both sides by bd]
bd
bd
a c
 
b d
 a : b = c :d  a : b : : c : d
 a, b, c, d are in proportion
Hence, a : b : : c : d  ad = bc.
Continued proportion: Three quantities are said to be in continued proportion if the ratio of the first and the
second is equal to the ratio of the second and third.
Thus, a, b, c are in continued proportion if
a b
a : : b : c i. e.   b2 = ac.
b c
Here, b is called the mean proportional between a and c is called the third proportional to a and b.
Four or more quantities are said to be continued proportion if the ratio of the first and the second is equal to
the ratio of the second and the third and so on.
Thus, a, b, c, d, e, f ….. are in continued proportion if,
a b c d e
     ......
b c d e f
Eg 1:
4, 12, 36 are in continued proportion as 4 : 12 : : 12 : 36.
Eg 2:
4, 8, 16, 32 are in continued proportion the product of the extremes is equal to the square of
the mean.
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Theorem:
Proof:
If three quantities are in continued proportion the product of the extremes is equal to the
square of the mean.
Let a, b, c be in continued proportion. Then,
a
b
a:b=b:c
=
 b2 = ac
b
c
Illustration 9:
Find the value of x if 3 : x : : 12 : 20.
Solution:
We have, 3 : x : : 12 : 20 
Illustration 10:
Find the value of x if x : 6 : : 5 : 3.
Solution:
We have, x : 6 : : 5 : 3 
Illustration 11:
What number must be added to each of the numbers 7, 16, 43, 79 to make the number in pro
3
3  20
12
=
x=
=5
20
x
12
65
x 5
=
x=
= 10
6 3
3
portion?
Solution:
Let x be the required number: Then,
(7 + x) : (16 + x) : : ( 43 + x) : (79 + x)
7x
43  x

=
 (7 + x) (79 + x) = (16 + x) (43 + x)
16  x
79  x
 553 + 86x + x2 = 688 + 59x + x2  27x = 135  x = 5.
Illustration 12:
Solution:
Find the fourth proportional to : 8, 14, 16
Illustration 13:
Solution:
Find the third proportional to 5, 10
Illustration 14:
Solution:
Find the mean proportional between 6, 54
Illustration 15:
Find the number such that the mean proportional between them is 14 and third proportional to
Let the fourth proportional be x. Then,
8
16
16  14
8 : 14 : : 16 : x 
=
x=
= 28.
8
14
x
Let the third proportional to 5, 10 be x. Then,
10
5
5 : 10 :: 10 : x 
=
 5x = 100  x = 20.
10
x
Let the mean proportional to 6, 54 be x. Then,
6
x
6 : x :: x : 54 
=
 x2 = 6  54  x2 = 324  x = 18.
54
x
them is 112.
Solution:
Let x and y be the required numbers. Then,
Mean proportional between x and y is 14
x
14
 x : 14 :: 14:y 
=
 xy = 196…………(i)
14
y
Third proportional to x and y = 112
x
y
 x : y :: y : 112 
=
 y2 = 112x………(ii)
y
112
Now, xy = 196  x =
196
y
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On putting x =
y2 = 112 
196
in (ii), we get
y
196
 y3 = 112  196 = (4  4  7)  (7  7  4)
y
 y3 = 43  73  y = 4  7 = 28
196
= 7.
28
Hence, the required numbers are 7, 28.
Putting the value of y in (i), we get x =
a2  b2 + c 2
= b4
a  b 2 + c 2
Illustration 16:
If b is the mean proportional between a and c, prove that
Solution:
b is the mean proportional between a and c
a
b
 a : b :: b : c 
=
 b2 = ac
b
c
a2  b2  c 2
a2  ac  c 2
a2  b2  c 2
LHS = 2
=
=
[ b2 = ac]
1
1
1
1
1
1
a  b2  c 2




a2 b2 c 2
a2 ac c 2
a2c 2 (a2  ac  c 2 )
a2  ac  c 2
= 2
=
= a2c2 = (ac)2 = b4 = RHS. [ b2 = ac]
c  ac  a2
a2  ac  c 2
a2 c 2


1 1 2
+ =
then prove that p : q = r : s
q s r
Illustration 17:
If p + r = 2q and
Solution:
We have : p + r = 2q and

2
1 1 2
 
q s r
2
1 2
 
pr s r
[Substituting q =
pr
]
2
r  p  r 
 1
1
2
2 1
1
1
– +
= 0  2
=02 
  +
 + =0
s
s
s
pr
r
 r  p  r  
p  r r 
2p
2p 1
1
–
+ =0–
+ =0
r p  r  s
2qr s


p 1
p r
   p:q=r:s
qr s
q s
Exercise 12:
Find the value of x, If 23 : 33 : : 3 : x
Exercise 13:
What number must be added to each of the four numbers 10, 18, 22, 38 to make then in
proportion?
Exercise 14:
Find the 4th proportional to x2 – 5x + 4, x2 + x – 2, x2 - 16
Exercise 15:
Find the 3rd proportional to ab, ab2.
Exercise 16:
Find two numbers such that the mean proportional between them is 24 and the third
proportional to them is 192.
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K – method
In some problems we equate each of the equal ratios to K.
a b
eg.
for three quantities a, b, c in continued proportion.
 = K;
b c
a c
Or
for four quantities a, b, c, d in continued proportion.
 = K;
b d
a b c
Or
  = K; for four quantities in continued proportion.
b c d
Illustration 18:
If a, b, c are in continued proportion, prove that: abc (a + b + c)3 = (ab + bc + ca)3
Solution:
Let
LHS =
=
=
=
=
RHS =
=
=
=
=
Illustration 19:
a b
 k
b c
a = bk
b = ck
a = ck 2
abc (a + b + c)3
c3 k 3 [c (1 + k + k 2)]3
c3 k 3 c3 (1 + k + k 2)]3
c6 k 3 (1 + k + k2)3
(ab + bc + ca)3
(ck2. ck + ck . c + c . ck2)3
(c2k (1 + k + k2))3
c6 k3 (1 + k + k2)3
= LHS.
If a, b, c, d are in proportion, prove that
a4 + c4
b4 + d 4
=
m a 2 + nc 2
m b 2 + nd 2
.
a c
 k
b d
 a = bk, c = dk
Substituting in LHS,
Solution:
Let
a4  c 4
=
b d
4
Illustration 20:
Solution:
b d
4
4
= k2
mb k  nd2k 2
ma  nc
=
= k2 
mb2  nd2
mb2  nd2
2
RHS =
4
b4k 4  d4k 4
2
2 2
LHS = RHS.
If a, b, c, d are in continued proportion, prove that
a3 + b3 + c 3
3
3
b +c +d
a b c
Let    K
b c d
a = bk,
b = ck,
c = dk

b = dk2

a = ck2  a = dk3
a3  b 3  c 3
d3k 9  d3k 6  d3k 3
a
LHS = 3
=
= k3 =
= RHS
3
3
3 6
3 3
3
d
b c d
d k d k d
3
=
a
d
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Exercise 17:
(i)
(ii)
(iii)
(iv)
If a, b, c, d are in proportion, prove that:
(9a + 13b) : (9c + 13d) = (9a – 13b) :(9c – 13d)
(a2 + ab) : (c2 + cd) = (b2 – 2ab) : (d2 – 2cd)
(a + c)3 : (b + d)3 = a (a – c)2 : b (b – d)2
(ma + nb) : b = (mc + nd) : d
 a  c  b2 =  a  b  ab 
 b  d  cd c 2  d 2  cd 
2
(v)
Exercise 18:
If
a
b
=
c
=
d
e
f
2
, prove that:
(i)
 a 2 c 2 e 2   ac ce ae 
+
+
 2+ 2+ 2 =
.
d
f   bd df bf 
b
(ii)
a+b c+d e+f 
(bdf) 
+
+
 = 27 (a + b) (c + d) (e + f).
d
f 
 b
3
(iii)
Exercise 19:
a 3 ce
b 3 df
If
x
a
=
a5 c 3 e 2
=
b5 d 3 f 2
y
b
=
z
c
,
ax  by
prove that
 a + b  x  y 
Exercise 20:
.
+
 by  cz 
 b + c  (y  z)
+
= 3.
If a, b, c, d are in continued proportion, prove that
2
2
(i)
Exercise 21:
cz  ax 
c + a  z  x 
If
a b a c 
d b d c 
2
 c + b  –  c + b  = (a – d) .




x
b+c a
=
y
c+ab
=
z
a+ b c
1 
 1
 2  2
c
b


, prove that each ratio is equal to
x+y+z
 a+b+c .


Properties of proportion:
1.
Invertendo : If a : b = c : d, then b : a = d : c.
Proof.
a:b=c:d
2.
Alternendo : If a : b = c : d, then a : c = b : d.
Proof.
a:b=c:d
a c

b d
b d
 
[Taking reciprocal of both sides]
a c
 b : a = d : c.
Hence, a : b = c : d  b : a = d : c.

a c

b d
 ad = bc
[By cross multiplication]
a b
 
c d
Hence, a : b = c : d  a : c = b : d.

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3.
Componendo : If a : b = c : d, then a + b : b = c + d : d.
Proof.
a:b=c:d
4.
Dividendo : If a : b = c : d, then (a – b) : b = (c – d) : d.
Proof.
a:b=c:d
5.
Componendo and Dividendo : If a : b = c : d, then (a + b) : (a – b) = (c – d) : (c – d).
Proof.
a:b=c:d
6.
Convertendo : If a : b = c : d, then a : (a – b) = c : (c – d).
Proof:
a:b=c:d
a c

b d
a
c
  1   1 [Adding 1 on both sides]
b
d
ab
cd

=
b
d
 (a + b) : b = ( c + d) : d
Hence, a : b = c : d  (a + b) : b = (c + d) : d.

a c

b d
a
c
 1 = 1
[Subtracting 1 from both sides]
b
d
ab
cd

=
b
d
 (a – b) : b = (c – d) : d.
Hence, a : b = c : d  (a – b) : b = (c – d) : d.

a c

b d
ab c d

…(i)
[By componendo]

b
d
ab c d
and
…(ii) [By dividendo]

b
d
ab c d

[On dividing (i) by (ii)]

ab c d
 (a + b) : (a – b) = (c + d) : (c – d).
Hence, a : b = c : d  (a + b) : (a – b) = (c + d) : (c – d).

a c

b d
b d
 
[By invertendo]
a c
ab c d

[Multiplying both sides by -1]

a
c
a
c

[By dividendo]

ab c d
Hence, a : b = c : d  a : (a – b) = c : (c – d).

Summary of the Results :
The above results may be expressed as given below:
a
=
b
a
(ii) Alternendo : 
b
(i) Invertendo :
b d
c
  .
d
a c
c
a b
  .
d
c d
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a c
ab c d
.
 

b d
b
d
a c
ab
cd
(iv) Dividendo :
=
.
 
c d
b
d
a c
ab c d
(v) Componendo and Devidendo :  
.

b d
ab c d
a c
a
c
(vi) Convertendo :
.
 

b d
ab c d
(iii) Componendo :
Illustration 21:
Solution :
If p =
4xy
, find the value of
x+y
We have:
p + 2x
p  2x
p=
4xy
p
2y


xy
2x x  y

2y  x  y
p  2x
=
2y   x  y 
p  2x

p  2x 3y  x

p  2x
yx
p + 2y
p  2y
[By componendo and dividendo]

p
2x

2y x  y
p  2y
2x  x  y

p  2y 2x   x  y 

.
….(i)
4xy
xy
Again, p =
+
[By componendo and dividendo]
p  2y 3x  y
…..(ii)

p  2y
xy
Adding (i) and (ii), we get :
p  2x p  2y 3y  x 3x  y
=


p  2x p  2y
yx
xy

=
3y  x 3x  y 3y  x  3x  y 2y  2x
=
=

yx
yx
yx
yx
=

Illustration 22:
Solution:
2y  x
y  x
p  2x p  2y

2
p  2x p  2y
Solve for x :
a+ x + a  x
a+ x  a  x
ax  ax
ax  ax


2

=
c
d
.
c
d
[ a  x  a  x]  [ a  x  a  x]
[ a  x  a  x]  [ a  x  a  x]
2 ax
2 ax

cd
, i.e.
cd
ax
ax
=
=
cd
cd
[By componendo and dividendo]
 c  d
(c  d)
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
a  x  c  d

a  x  c  d 2

 a  x    a  x   c  d   c  d

 a  x    a  x   c  d 2   c  d 2

2a c 2  d2

2x
2cd
2
[On squaring both sides]
2
Hence, x =
Illustration 23:
3x  9x  5
2

=5.
3x  9x 2  5
3x  9x 2  5
Solution :

5
1
[3x  9x 2  5]  [3x  9x 2  5]
=
[3x  9x  5]  [3x  9x  5]
2

[By componendo and dividendo]
2acd
.
c 2  d2
3x + 9x 2  5
Solve :
2
2
6x
2 9x  5
2

6

4
x
9x  5
2

5 1
[By componendo and dividendo]
5 1
1
2
2
x
1
[On squaring both sides]

9x 2  5 4
 4x2 = 9x2 – 5, i.e. 5x2 = 5
 x2 = 1
 x = 1 or x = –1
But, x = –1 does not satisfy the given equation.
Hence, x = 1.

Illustration 24:
If
Solution:


a 3 + 3ab2
2
3a b + b
a
a
3
=
x 3 + 3xy 2
2
3
(3x + y )
, prove that :
 
 = x
  3a b  b   x
x
a
=
y
.
b
 

  3x y  y 
3
 3ab2  3a2b  b3
3
 3xy 2  3x 2 y  y3
3
 3ab
3
 3xy
2
2
3
a3  b3  3ab  a  b 
a3  b3  3ab  a  b 

a  b
3
a  b

ab x  y

ab x  y

a  b  a  b
a  b  a  b

2a 2x
a x
, i.e. 

2b 2y
b y

x y
 .
a b
3
=
2
2
3
[By componendo and dividendo]
x3  y3  3xy  x  y 
x3  y3  3xy(x  y)
x  y

3
x  y
3
[Taking cube root of each side]
=
 x  y   x  y
 x  y   x  y
[By componendo and dividendo]
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Illustration 25:
Solution:
If (4a + 9b) x (4c – 9d) = (4a – 9b) (4c + 9d), prove that a : b :: c : d.
(4a + 9b) x (4c – 9d) = (4a – 9b) (4c + 9d)
4a  9b 4c  9d


4a  9b 4c  9d
 4a  9b    4a  9b   4c  9d   4c  9d


[By componendo and dividendo]
 4a  9b    4a  9b   4c  9d   4c  9d
8a
8c

18b 18d
a c
  , i.e. a : b :: c : d.
b d
Hence, a : b :: c : d.

Illustration 26:
Solution:
If
b+c a
=
c+ab
=
a+bc
, then prove that each ratio is equal to
y+zx
z+ x y
x+y z
We have
Sum of antecedents
bc a
c ab
abc
=
=
=
Sum of consequents
yzx
zxy
xyz
bc ac ababc
abc
=
yzxzxyxyz
xyz
=
bc a abc
bc a y zx



yzx xyz
abc x yz
Now,

a b c
= = .
x y z
b  c  a    a  b  c 
abc

2x
2a
=
abc
xyz

a
x

abc x yz

a abc
.

x xyz
=
 y  z  x    x  y  z
xyz
[By alternenedo]
[By dividendo]
Similarly,
c ab abc
b abc
 

zxy xyz
y xyz
and,
c
abc abc
abc

=

z
xyz xyz
xyz
Hence, each of the given ratio =
Exercise 22:
Solve
Exercise 23:
Solve
x+2 +
x 3
x+2 
x 3
a + x + a2  x 2
a+ x  a  x
2
a b c
 
x y z
=5
=
2
b
x
3
Exercise 24:
Solve
a+ x
a x 
16 
=

ax
a+ x 
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Exercise 25:
Solve
Exercise 26:
If x =
Exercise 27:
If x =
Exercise 28:
If
2x 2  7x + 9
7x  9
=
2a + 1 + 2a  1
2a + 1  2a  1
3x 2  8x + 5
8x  5
, prove that : x2 – 4ax + 1 = 0.
2a + 3b + 2a  3b
2a + 3b  2a  3b
x+y
ax + by
=
y+z
ay + bz
=
, prove that : 3bx2 – 4 ax + 3b = 0
z+ x
az + bx
, prove that each of these ratios is equal to
2
a+b
, where
x + y + z  0.
ANSWERS TO EXRECISES
Exercise 1: (a) 5 : 8 ,(b)1 : 1
Exercise 2: (a) 8 : 27 ,(b) 49x2 : 9y2
Exercise 3: (a) a3 : b3 ,(b)8 : 343
Exercise 4: 56b : a
Exercise 5: 5x : 2y
Exercise 6: Rs.760, Rs. 456
Exercise 7: 14, 15
Exercise 8: 3 : 5
Exercise 9: 20 : 51
Exercise 10: 3 : 2 or 2 3
Exercise 11: 1:1
Exercise 12: 81/8
Exercise 13: 2
Exercise 14: X2 + 6X + 8
Exercise 15: ab3
Exercise 16:
12, 48
Exercise 22:
x=7
Exercise 23:
x=b
Exercise 24:
x = a/3,3a
Exercise 25:
x = 0, x = 17
5
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ASSIGNMENTS
SUBJECTIVE
LEVEL – I
1.
Show that the side of a square and its diagonal are incommensurable quantities.
2.
Compare the ratios (4 : 9) and (2 : 3).
3.
If (4x + 3) : (9x + 10) is the triplicate ratio of 3 : 4, find the value of x.
4.
The sides of a triangle are in the ratio
1 1 1
and its perimeter is 91 cm. Find the lengths of the
: :
2 3 4
sides of the triangle.
5.
Two numbers are in the ratio of 1/3 : 1/2. The difference of their squares is 405. Find the numbers.
6.
Find the fourth proportional to 1.4, 3.2 and 7.
7.
a b
Find the third proportional to    and
b a
8.
Find the mean proportional between (a – b) and (a3 – a2 b)
9.
What must be added to each of the numbers 7, 15, 19 and 35. So that the resulting numbers are in
proportion
10.
If b is the mean proportional between a and c, prove that (ab + bc) is the mean proportional between
(a2 + b2) and (b2 + c2).
a2  b2
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LEVEL – II
1.
A and B have Rs. 40 and Rs. 50 respectively, A spends Rs. 18 and B spends Rs. 21 whose saving
is more.
2.
If (3x + 3) : ( 9x + 7) is the duplicate ratio of 3 : 5, find the value of x
3.
If 24(3x2 – y2) = 37xy, find x : y
4.
(3x + 5y) : ( 2x – y) = (12 x – y) : ( 5x – 3y) find x : y
5.
Divide Rs 1870 into three parts in such a way that half of the first part, one third of the second part
and one sixth of the third part are equal.
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OBJECTIVES
LEVEL – I
1.
2.
a 2
5a  9b
is
 then the value of
b 9
2a  b
101
71
(A)
(B)
13
13
If
(C)
71
9
if a : (b+c) = b : (c + a) = c : (a + b) then each of these ratio is
1
1
1
(A) either
or –1
(B) either
or 1
(C) either –1 or
2
2
2
(D) none of these
(D) none of these
3.
A bag contains Rs 87 in the form of a rupee, 50 paise, 10 paise and 5 paise coins in the ratio 3 : 4 ::
5 : 6. the number of 10 paise coins is
(A) 15
(B) 75
(C) 45
(D) none of these
4.
The number that should be subtracted from each of the numbers 37, 52, 55, 79 in order that the
remainder may be in proportion is
(A) 5
(B) 6
(C) 7
(D) 1
5.
Two number are in the ratio 6 : 11. The greater exceeds the smaller by 35. The numbers are
(A) 18 : 33
(B) 30 : 55
(C) 36 : 66
(D) 42 : 77
6.
If
7.
If a varies jointly as b and c, and if a is 6 when b is 4 and c is 3, then
1
1
1
of A,
of B and
of C are equal then the ratio between A, B, C is
5
4
7
1 1 1
1 1 1
(A) : :
(B) 4 : 5 : 7
(C) 7 : 5 : 4
(D) : :
4 5 7
7 5 4
(A)
8.
3
2
(B) 2
A fourth proportional to ab, bc, cd is
c 2d
(A)
(B) ad
a
a
is equal to.
bc
(C)
1
2
(D) 12
(C)
c
a
(D)
a
d
9.
In the journey of 96 km performed by taxi, train and bus, the distance traveled by three ways in that
order are in the ratio 5 : 8 : 3 and the charges per km are in the ratio 4 : 3 : 2. If the taxi charges per
km was Rs 2. The total charge for the journey is Rs____.
(A) 150
(B) 300
(C) 180
(D) 200
10.
Find x : y given that 3x – 2y : x – 5y = 1 : 3
(A) 1 : 6
(B) 2 : 5
(C) 1 : 8
(D) 2 : 3
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LEVEL – II
1.
2.
If 3x2 – 10xy + 3y2 = 0, then x : y =
(A) 1 : 1
(B) 1 : 3
Find x : y given
(A)
3.
4.
4
7
(C) 1 : 3 or 3 : 1
(D) none of these
3x 2  33y 2
= x + 7y
x  4y
(B)
4
21
5
a2  b2
ab
= , find the value of 2
.
ab
4
a  b2
(A) 41/40
(B) 9
(C)
5
2
(D)
1.21
12
If
(C) 5/4
If x + 24 : 6(x – 1) is the duplicate ratio of 7 : 12, x =
(A) 21
(B) 25
(C) 37/15
(D) none of these
(D) none of these
x
y
z
=
=
then (b-c) x + (c – a) y + (a – b) z =
bc a
c ab
abc
(A) 1
(B) n
(C) a
(D) 0
5.
If
6.
An employer reduces the number of his employees in the ratio 16 : 15 and increases wages in the
ratio 21 : 22. The wages paid will decrease in the ratio
(A) 55 : 56
(B) 56 : 55
(C) 315 : 352
(D) 352 : 315
7.
A sum of money is divided into two parts in the ratio p : q. The first part is then divided between two
persons in the ratio a : b and the second between the same persons in the ratio c : d. If the total
amounts received by each be equal, then
(A) p : q = (a + b) (c + d) : (a – b) (c – d)
(B) p : q = (a + b) (d – c) : (a – b) (c + d)
(C) p : q = ( a + c) (b + d) : (a – c) (b – d)
(D) None of these
8.
Rs 600 was divided among P, Q, R and S so that P and Q got four times as much as R and S
1
together, Q got 3 times of what R got and R got 1 times as much as S. P got Rs
2
(A) 216
(B) 264
(C) 72
(D) None of these
9.
If the ratio of the work done by (x – 2) men in (x + 2) days to the work done by (x + 3) men in (x – 2)
days 15 : 16 then x is
(A) 2
(B) 13
(C) 15
(D) None of these
10.
The expenses of a lunch are partly constant and partly proportional to the number of guests. The
expenses amount to Rs.65 for 7 guests and Rs. 97 for 11 guests. Then the expenses for 18 guests
will amount to
(A) Rs. 162
(B) 153
(C) Rs. 176
(D) None of these
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ANSWERS
SUBJECTIVE
LEVEL – I
2.
2:3>4:9
3.
6
4.
42 cm, 28 cm, 21 cm
5.
18, 27
6.
16
7.
ab
8.
a2 – ab
9.
5
LEVEL – II
1.
B having more saving
3.
8:9
5.
Rs. 340, Rs. 510, Rs. 1020
4.
2.
27
2 : 3 or 8 : 3
OBJECTIVE
LEVEL – I
1.
B
2.
A
3.
B
4.
C
5.
D
6.
B
7.
C
8.
A
9.
A
10.
C
LEVEL – II
1.
C
2.
C
3.
A
4.
B
5.
D
6.
B
7.
B
8.
B
9.
B
10.
B
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