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CAC18301
QUANTITATIVE APTITUDE
&
LOGICAL REASONING - II
Syllabus
Unit - 2
 RATIO & PROPORTION
Ratio:
The ratio of two quantities a and b in the same
units, is the fraction and we write it as a : b
where the first term a is called antecedent and the
second term b is called consequent.
For example, in 3 : 7, 3 is the antecedent and 7 is
the consequent.
Proportion:
The equality of two ratios is called proportion.
b
a
Also if a : b = c : d, then this equality of ratios
can also be written as a : b : : c : d and we say
that a , b, c and d are in proportion.
Here a and d are called extremes, while b and
c are called mean terms.
Or equivalently,
i.e., Product of means = Product of extremes
d
c
b
a
 )()( dacb 
Some Important Facts and Formulae
 If a : b = c : d, then d is called the fourth
proportional to a, b, c.
 If a : b = b : c, then c is called the third
proportional to a and b.
 Mean Proportional
Mean proportional between a and b is
 Comparison of Ratios
(a : b) > (c : d)
ab
d
c
b
a

 COMPOUNDED RATIO
The compounded ratio of the ratios (a : b), (c : d),
(e : f) is (ace : bdf )
 Duplicate Ratio of (a : b) is (a2 : b2)
 Sub-duplicate ratio (a : b) is
 Triplicate ratio of (a : b) is (a3 : b3)
 Sub-triplicate ratio of (a : b) is (a1/3 : b1/3)
 If , then
which is called Componendo and dividendo
 ba :
d
c
b
a

dc
dc
ba
ba





Finally,
 VARIATION: (Proportionality)
(i) x is directly proportional to y, if x = ky, for
some constant k, and it is expressed in
symbol as
(ii) x is inversely proportional to y, if
for some constant k, and in symbol,
we represent it by
yx 
y
k
x 
y
x
1

Solved Examples
1. If a : b = 5 : 9 and b : c = 4 : 7, find a : b : c.
Solution:
Given that a : b = 5 : 9 and b : c = 4 : 7
where the values at the places of b must be
same for the required ratio a : b : c.
So, consider b : c = 4 : 7













4
9
7:
4
9
4
4
63
:9
Therefore,
a : b = 5 : 9 and b : c
Now combining the ratios, we get
a : b : c
4
63
:9
4
63
:9:5
63:36:20
2. Find the (i) fourth proportional to 4, 9, 12
(ii) the third proportional to 16 and 36
(iii) the mean proportional between 0.08 and
0.18
Solution:
(i) Let the fourth proportional 4, 9, 12 be x.
i.e., 4 : 9 : : 12 : x
Therefore, the fourth proportional to 4, 9, 12 is 27.
27
4
129
1294




x
x
(ii) Let the third proportional to 16 and 36 be x.
Then we have 16 : 36 : : 36 : x
Therefore, the third proportional to 16 and
36 is 81.
(iii) Mean proportional between 0.08 and 0.18 is
81
16
3636
363616




x
x
12.0
100
12
10000
144
100
18
100
8
18.008.0 
3. If x : y = 3 : 4, find (4x + 5y) : (5x – 2y).
Solution:
Given that x : y = 3 : 4.
(taking out y from
both the nr and dr
and then cancelling
both the y’s outside)
7
32
4
7
8
2
4
3
5
5
4
3
4
25
54
25
54



















y
x
y
x
yx
yx
4. Divide Rs. 672 in the ratio 5 : 3.
Solution:
The given ratio is 5 : 3.
Therefore the sum is to be divided by the total
number, 5+3 = 8 of equal parts.
Then the required split is given by
5 × 84 : 3 × 84 = 420 : 252
Hence the required parts of the sum Rs. 672 are
Rs. 420 and Rs. 252.
84
8
672

5. Divide Rs. 1162 among A, B, C in the ratio
35 : 28 : 20.
Solution:
The total no. of equal splits = 35+28+20 = 83.
Therefore, the required split in the given ratio is
= 35 × 14 : 28 × 14 : 20 × 14
= 490 : 392 : 280
Hence the required split of the Amount Rs. 1162
is Rs. 490, Rs. 392 and Rs. 280.
14
83
1162

6. A bag contains 50 p, 25 p and 10 p coins in
the ratio 5 : 9 : 4, amounting to Rs. 206. Find
the number of coins of each type.
Solution:
Let the no. of 50 p, 25 p and 10 p coins be 5x,
9x and 4x respectively.
Then representing the counts in the given
paises in rupees,
206
10
4
4
9
2
5

xxx
Taking LCM for the denominators, we obtain that
Hence the required number of coins of each type
are 5×40 = 200 of 50 p,
9×40 = 360 of 25 p
and 4×40 = 160 of 10 p
40
4120103
412084550
206
20
84550





x
x
xxx
xxx
Problems on mixtures and solutions
7. A mixture contains alcohol and water in the ratio 4 : 3. If 5 litres
of water is added to the mixture, the ratio becomes 4 : 5. Find the
quantity of alcohol in the given mixture.
Solution:
Let the quantity of alcohol and water be 4x litres and 3x litres
respectively.
Then by given, we have
20x = 4(3x + 5)
20x – 12x = 20
8x = 20 x = 2.5
Therefore, the quantity of alcohol = 4 × 2.5 litres = 10 litres
5
4
53
4

x
x
8. In a mixture of 60 litres, the ratio of milk and water
is 2 : 1. If this ratio is to be 1 : 2, then the quantity
of water to be further added is
(a) 20 liters (b) 30 litres (c) 40 litres (d) 60 litres
(Ans : (d))
9. The sides of a triangle are in the ratio
and its perimeter is 104 cm. The length of the
longest side is :
(a) 52 cm (b) 48 cm (c) 32 cm (d) 26 cm
(Ans : (b))
(Hint: taking LCM of 2, 3, 4 , we obtain the given ratio as
6 : 4 : 3. Hence the longest side is )
4
1
:
3
1
:
2
1
cm48
13
6
104 
10. The speeds of three cars are in the ratio 5 : 4 : 6. The
ratio between the time taken by them to travel the same
distance is
(a) 5 : 4 : 6 (b) 6 : 4 : 5 (c) 10 : 12 : 15 (d) 12 : 15 : 10
(Ans: (d))
(Hint: the required ratio is . Hence by taking
LCM of the denominators, 60, we simplify this ratio as
12 : 15 : 10)
11. Which of the following ratios is greatest?
(a) 7 : 15 (b) 15: 23 (c) 17 :25 (d) 21 :29
(Ans: (d))
(Hint: By getting the fraction values of the ratios and
comparing the values, we have the highest fraction obtained
from (d))
6
1
:
4
1
:
5
1
TRY YOURSELF & CHECK YOUR ANSWERS :
12. In a ratio, which is equal to 3 : 4, if the
antecedent is 12, then the consequent is :
(a) 9 (b) 16 (c) 20 (d) 24
(Ans: (b))
13.An alloy is to contain copper and zinc in the
ratio 9 : 4. The zinc required to be melted with
24 kg of copper is :
(a) (b) (c) (d)
(Ans: (a))
kg
3
2
10 kg
3
1
10 kg
3
2
9 kg9
14. The compounded ratio of (2 : 3), (6 : 11) and
(11 : 2) is
(a) 1 : 2 (b) 2 : 1 (c) 11 : 24 (d) 36 : 121
(Ans: (b))
15. What is the ratio whose terms differ by 40
and the measure of which is ?
(a) 16 : 56 (b) 14 : 56 (c) 15 : 56 (d) 16 : 72
(Ans: (a))
16. If 10% of x = 20% of y, then x : y is equal to :
(a) 1 : 2 (b) 2 : 1 (c) 5 : 1 (d) 10 : 1
(Ans: (b))
7
2
17. If A : B = 3 : 4 and B : C = 8 : 9, then A : C is
(a) 1 : 3 (b) 3 : 2 (c) 2 : 3 (d) 1 : 2
(Ans: (c))
( Hint: By given, and
)
4
3

B
A
9
8

C
B
3:2:
3
2
3
2
9
8
4
3


CA
C
A
C
B
B
A

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CAC18301 unit 2 ppt along with syllabus and marks splits

  • 3.
  • 4.
  • 5. Unit - 2  RATIO & PROPORTION Ratio: The ratio of two quantities a and b in the same units, is the fraction and we write it as a : b where the first term a is called antecedent and the second term b is called consequent. For example, in 3 : 7, 3 is the antecedent and 7 is the consequent. Proportion: The equality of two ratios is called proportion. b a
  • 6. Also if a : b = c : d, then this equality of ratios can also be written as a : b : : c : d and we say that a , b, c and d are in proportion. Here a and d are called extremes, while b and c are called mean terms. Or equivalently, i.e., Product of means = Product of extremes d c b a  )()( dacb 
  • 7. Some Important Facts and Formulae  If a : b = c : d, then d is called the fourth proportional to a, b, c.  If a : b = b : c, then c is called the third proportional to a and b.  Mean Proportional Mean proportional between a and b is  Comparison of Ratios (a : b) > (c : d) ab d c b a 
  • 8.  COMPOUNDED RATIO The compounded ratio of the ratios (a : b), (c : d), (e : f) is (ace : bdf )  Duplicate Ratio of (a : b) is (a2 : b2)  Sub-duplicate ratio (a : b) is  Triplicate ratio of (a : b) is (a3 : b3)  Sub-triplicate ratio of (a : b) is (a1/3 : b1/3)  If , then which is called Componendo and dividendo  ba : d c b a  dc dc ba ba     
  • 9. Finally,  VARIATION: (Proportionality) (i) x is directly proportional to y, if x = ky, for some constant k, and it is expressed in symbol as (ii) x is inversely proportional to y, if for some constant k, and in symbol, we represent it by yx  y k x  y x 1 
  • 10. Solved Examples 1. If a : b = 5 : 9 and b : c = 4 : 7, find a : b : c. Solution: Given that a : b = 5 : 9 and b : c = 4 : 7 where the values at the places of b must be same for the required ratio a : b : c. So, consider b : c = 4 : 7              4 9 7: 4 9 4 4 63 :9
  • 11. Therefore, a : b = 5 : 9 and b : c Now combining the ratios, we get a : b : c 4 63 :9 4 63 :9:5 63:36:20
  • 12. 2. Find the (i) fourth proportional to 4, 9, 12 (ii) the third proportional to 16 and 36 (iii) the mean proportional between 0.08 and 0.18 Solution: (i) Let the fourth proportional 4, 9, 12 be x. i.e., 4 : 9 : : 12 : x Therefore, the fourth proportional to 4, 9, 12 is 27. 27 4 129 1294     x x
  • 13. (ii) Let the third proportional to 16 and 36 be x. Then we have 16 : 36 : : 36 : x Therefore, the third proportional to 16 and 36 is 81. (iii) Mean proportional between 0.08 and 0.18 is 81 16 3636 363616     x x 12.0 100 12 10000 144 100 18 100 8 18.008.0 
  • 14. 3. If x : y = 3 : 4, find (4x + 5y) : (5x – 2y). Solution: Given that x : y = 3 : 4. (taking out y from both the nr and dr and then cancelling both the y’s outside) 7 32 4 7 8 2 4 3 5 5 4 3 4 25 54 25 54                    y x y x yx yx
  • 15. 4. Divide Rs. 672 in the ratio 5 : 3. Solution: The given ratio is 5 : 3. Therefore the sum is to be divided by the total number, 5+3 = 8 of equal parts. Then the required split is given by 5 × 84 : 3 × 84 = 420 : 252 Hence the required parts of the sum Rs. 672 are Rs. 420 and Rs. 252. 84 8 672 
  • 16. 5. Divide Rs. 1162 among A, B, C in the ratio 35 : 28 : 20. Solution: The total no. of equal splits = 35+28+20 = 83. Therefore, the required split in the given ratio is = 35 × 14 : 28 × 14 : 20 × 14 = 490 : 392 : 280 Hence the required split of the Amount Rs. 1162 is Rs. 490, Rs. 392 and Rs. 280. 14 83 1162 
  • 17. 6. A bag contains 50 p, 25 p and 10 p coins in the ratio 5 : 9 : 4, amounting to Rs. 206. Find the number of coins of each type. Solution: Let the no. of 50 p, 25 p and 10 p coins be 5x, 9x and 4x respectively. Then representing the counts in the given paises in rupees, 206 10 4 4 9 2 5  xxx
  • 18. Taking LCM for the denominators, we obtain that Hence the required number of coins of each type are 5×40 = 200 of 50 p, 9×40 = 360 of 25 p and 4×40 = 160 of 10 p 40 4120103 412084550 206 20 84550      x x xxx xxx
  • 19. Problems on mixtures and solutions 7. A mixture contains alcohol and water in the ratio 4 : 3. If 5 litres of water is added to the mixture, the ratio becomes 4 : 5. Find the quantity of alcohol in the given mixture. Solution: Let the quantity of alcohol and water be 4x litres and 3x litres respectively. Then by given, we have 20x = 4(3x + 5) 20x – 12x = 20 8x = 20 x = 2.5 Therefore, the quantity of alcohol = 4 × 2.5 litres = 10 litres 5 4 53 4  x x
  • 20. 8. In a mixture of 60 litres, the ratio of milk and water is 2 : 1. If this ratio is to be 1 : 2, then the quantity of water to be further added is (a) 20 liters (b) 30 litres (c) 40 litres (d) 60 litres (Ans : (d)) 9. The sides of a triangle are in the ratio and its perimeter is 104 cm. The length of the longest side is : (a) 52 cm (b) 48 cm (c) 32 cm (d) 26 cm (Ans : (b)) (Hint: taking LCM of 2, 3, 4 , we obtain the given ratio as 6 : 4 : 3. Hence the longest side is ) 4 1 : 3 1 : 2 1 cm48 13 6 104 
  • 21. 10. The speeds of three cars are in the ratio 5 : 4 : 6. The ratio between the time taken by them to travel the same distance is (a) 5 : 4 : 6 (b) 6 : 4 : 5 (c) 10 : 12 : 15 (d) 12 : 15 : 10 (Ans: (d)) (Hint: the required ratio is . Hence by taking LCM of the denominators, 60, we simplify this ratio as 12 : 15 : 10) 11. Which of the following ratios is greatest? (a) 7 : 15 (b) 15: 23 (c) 17 :25 (d) 21 :29 (Ans: (d)) (Hint: By getting the fraction values of the ratios and comparing the values, we have the highest fraction obtained from (d)) 6 1 : 4 1 : 5 1
  • 22. TRY YOURSELF & CHECK YOUR ANSWERS : 12. In a ratio, which is equal to 3 : 4, if the antecedent is 12, then the consequent is : (a) 9 (b) 16 (c) 20 (d) 24 (Ans: (b)) 13.An alloy is to contain copper and zinc in the ratio 9 : 4. The zinc required to be melted with 24 kg of copper is : (a) (b) (c) (d) (Ans: (a)) kg 3 2 10 kg 3 1 10 kg 3 2 9 kg9
  • 23. 14. The compounded ratio of (2 : 3), (6 : 11) and (11 : 2) is (a) 1 : 2 (b) 2 : 1 (c) 11 : 24 (d) 36 : 121 (Ans: (b)) 15. What is the ratio whose terms differ by 40 and the measure of which is ? (a) 16 : 56 (b) 14 : 56 (c) 15 : 56 (d) 16 : 72 (Ans: (a)) 16. If 10% of x = 20% of y, then x : y is equal to : (a) 1 : 2 (b) 2 : 1 (c) 5 : 1 (d) 10 : 1 (Ans: (b)) 7 2
  • 24. 17. If A : B = 3 : 4 and B : C = 8 : 9, then A : C is (a) 1 : 3 (b) 3 : 2 (c) 2 : 3 (d) 1 : 2 (Ans: (c)) ( Hint: By given, and ) 4 3  B A 9 8  C B 3:2: 3 2 3 2 9 8 4 3   CA C A C B B A