Ratios and Proportions Questions

FACTS  AND  FORMULAE  FOR  RATIO  AND  PROPORTION  QUESTIONS

 

 

1. RATIO: The ratio of two quantities a and b in the same units, is the fraction a/b and we write it as a:b.

In the ratio a:b, we call a as the first term or antecedent and b, the second term or consequent.

Ex. The ratio 5: 9 represents 5/9 with antecedent = 5, consequent = 9.

Rule: The multiplication or division of each term of a ratio by the same non-zero number does not affect the ratio. Ex. 4: 5 = 8: 10 = 12: 15 etc. Also, 4: 6 = 2: 3.

 

2. PROPORTION : The equality of two ratios is called proportion

If a: b = c: d, we write, a: b :: c : d and we say that a, b, c, d are in proportion . Here a and d are called extremes, while b and c are called mean terms.

Product of means = Product of extremes.

Thus, a: b :: c : d <=> (b x c) = (a x d).

 

 3(i) Fourth Proportional : If a : b = c: d, then d is called the fourth proportional to a, b, c.

     (ii) Third Proportional : If a: b = b: c, then c is called the third proportional to a and b.

     (iii) Mean Proportional : Mean proportional between a and b is ab

 

 4. (i) COMPARISON OF RATIOS : We say that (a: b) > (c: d)  (a/b)>(c /d).

     (ii) COMPOUNDED RATIO : The compounded ratio of the ratios (a: b), (c: d), (e : f)  is    (ace: bdf)

 

5. (i) Duplicate ratio of (a : b) is a2:b2.

    (ii) Sub-duplicate ratio of (a : b) is (√a : √b).

    (iii)Triplicate ratio of (a : b) is a3:b3.

    (iv) Sub-triplicate ratio of (a : b) is a13: b13.

    (V) If ab=cd, thena+ba-b=c+dc-d (Componendo and dividendo)

 

 6. VARIATION:

(i) We say that x is directly proportional to y, if x = ky for some constant k and we write, xy

(ii) We say that x is inversely proportional to y, if xy = k for some constant k and we write, x1y

Q:

The number of oranges in three baskets are in the ratio of 3 : 4 : 5. In which ratio the no. of oranges in first two baskets must be increased so that the new ratio   becomes 5 : 4 : 3?

A) 1:3 B) 2:1
C) 3:4 D) 2:3
 
Answer & Explanation Answer: B) 2:1

Explanation:

Let,   B1 : B2 : B3 = 3x : 4x : 5x 

 

again  B1 : B2 : B3 = 5y : 4y : 3y

 

Since there is increase in no.of oranges in first two baskets only, it means the no. of oranges remains constant in the third basket

 

Therefore,   5x = 3y

 

Hence    3x : 4x : 5x   =>   9y5:12y5:15y5 = 9y:12y:15y

 

and       5y : 4y : 3y   =>   25x : 20x : 15x

 

Therfore, increment in first basket = 16

 

Increment in second basket = 8

 

Thus, required ratio = 16/8 = 2:1

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Q:

The concentration of petrol in three different mixtures (petrol and kerosene) is 1/2 , 3/5 and 4/5 respectively. If 2 litres, 3 litres and 1 litre are taken from these three different vessels and mixed. what is the ratio of petrol and kerosene in the new mixture?

A) 4:5 B) 3:2
C) 3:5 D) 2:3
 
Answer & Explanation Answer: B) 3:2

Explanation:

Concentration of petrol in    A        B        C

                                                1/2      3/5     4/5

Quantity of petrol taken from A = 1 litre out of 2 litre

Quantity of petrol taken from B = 1.8litre out of 3 litre

Quantity of petrol taken from C = 0.8 litre out of 1 litre

Therefore, total petrol taken out from A, B and C = 1+1.8+0.8 =3.6 litres

So, the quantity of kerosen =(2+3+1) - 3.6 =2.4 litre

Thus, the ratio of petrol to kerosene = 3.6/2.4 = 3/2

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96 22161
Q:

If X : Y : Z = 3 : 4 : 5, then what will be the ratio of (X/Y) : (Y/Z) : (Z/X)?

 

A) 5 : 4 : 3 B) 30 : 40 : 27
C) 45 : 48 : 100 D) 20 : 15 : 12
 
Answer & Explanation Answer: C) 45 : 48 : 100

Explanation:
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4 21097
Q:

In a competitive exam, the number of passed students was four times the number of failed students. If there had been 35 fewer appeared students and 9 more had failed, the ratio of passed and failed students would have been 2 : 1, then the total number of students appered for the exam ?

A) 175 B) 165
C) 155 D) 145
 
Answer & Explanation Answer: C) 155

Explanation:

Let the number of failed students be x

=> Number of passed students = 4x 

So total number of students was 5x

From the given data,

If total number of students be 5x – 35

=> 4x-35-9/x+9 = 2/1

=> 4x – 44 = 2(x + 9)

=> 4x – 2x = 18 + 44

=> x = 31

Total number = 31×5

= 155

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43 20631
Q:

The incomes of two persons A and B are in the ratio 3 : 4. If each saves Rs.100 per month, the ratio of their expenditures is Rs. 1 : 2. Find their incomes.

A) Rs. 100 and Rs.150 B) Rs. 150 and Rs.200
C) Rs.200 and Rs.250 D) Rs.250 and Rs.300
 
Answer & Explanation Answer: B) Rs. 150 and Rs.200

Explanation:

Let the incomes of A and B be 3P and 4P.

If each saves Rs. 100 per month, then their expenditures = Income - savings = (3P - 100) and (4P - 100).

 

The ratio of their expenditures is given as 1 : 2.

Therefor, (3P - 100) : (4P - 100) = 1 : 2

Solving, We get P = 50. Substitute this value of P in 3P and 4P.

Thus, their incomes are : Rs.150 and Rs.200

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43 20194
Q:

Select the missing number based on the given related pair of numbers.

 

158 : 384 :: 140 : ___

A) 347 B) 346
C) 348 D) 349
 
Answer & Explanation Answer: C) 348

Explanation:
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9 19577
Q:

The ratio of male and female in a city is 7 : 8 respectively and percentage of children among male and female is 25 and 20 respectively. If number of adult females is 156800, what is the total population of the city?

A) 4,12,480 B) 3,67,500
C) 5,44,700 D) 2,98,948
 
Answer & Explanation Answer: B) 3,67,500

Explanation:

Let the total population be 'p'

Given ratio of male and female in a city is 7 : 8

In that percentage of children among male and female is 25% and 20%

=> Adults male and female % = 75% & 80%

But given adult females is = 156800

=> 80%(8p/15) = 156800

=> 80 x 8p/15 x 100 = 156800

=> p = 156800 x 15 x 100/80 x 8

=> p = 367500

Therefore, the total population of the city = p = 367500

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75 19370
Q:

Vinod have 20 rupees. He bought 1, 2, 5 rupee stamps. They are different in numbers by the reason of no change, the shop keeper gives 3 one rupee stamps. So how many stamps Vinod have ?

A) 10 B) 18
C) 12 D) 15
 
Answer & Explanation Answer: A) 10

Explanation:

Given total rupees = 20 Rs
No. of one rupee stamps = 3
Now, remaining money = Rs. 17
With that he buys only 2 and 5 rupee stamps
Let number of Rs. 5 stamps = K
Let number of Rs. 2 stamps = L
5K + 2L = 17
K = 3, L = 1 (possible)
L = 6, K = 1 (possible)
=> But given that they are different in number so, K is not equal to 3
one rupee stamps = 3
2 two stamps = 6
5 rupee stamps = 1
Total number of stamps = 10.

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45 19254