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Practice Questions on Ratio and Proportion | Quantitative Techniques for CLAT PDF Download

Q1: A car travels 240 miles in 4 hours. What is its average speed in miles per hour (mph)?
(a) 60      
(b) 40      
(c) 50    
(d) 70
Ans:
(a)
Average Speed = Total Distance / Total Time
Average Speed = 240 miles / 4 hours
Average Speed = 60 mph

Q2: Three friends, A, B, and C, decide to share a sum of money. A gets 2/5 of the money, B gets 1/4 of the money, and C gets the rest. If C receives $300 less than A, and the total sum of money they share is $4,000, how much does each friend receive?
(a) 1000    
(b) 1100    
(c) 1200  
(d) 1300
Ans:
(d)
A’s share = (2/5) * $4,000 = $1,600
B’s share = (1/4) * $4,000 = $1,000
Now, we need to find C’s share, knowing that C gets $300 less than A.
C’s share = A’s share – $300 = $1,600 – $300 = $1,300
So, C’s share is $1,300.
To verify: A’s share ($1,600) + B’s share ($1,000) + C’s share ($1,300) = Total ($4,000)

Q3: A mixture contains sugar solution and colored water in the ratio of 4 : 3. If 10 liters of colored water is added to the mixture, the ratio becomes 4: 5. Find the initial quantity of sugar solution in the given mixture.
(a) 10    
(b) 20    
(c) 30    
(d) 40
Ans:
(b)
The initial ratio is 4 : 3.
Let ‘k’ be the common ratio.
⇒ Initial quantity of sugar solution = 4 k
⇒ Initial quantity of colored water = 3 k
⇒ Final quantity of sugar solution = 4 k
⇒ Final quantity of colored water = 3 k + 10
Final ratio = 4 k : 3 k + 10 = 4 : 5
⇒ k = 5
Therefore, the initial quantity of sugar solution in the given mixture = 4 k = 20 liters

Q4: A invested Rs. 70,000 in a business. After a few months, B joined him with Rs. 60,000. At the end of the year, the total profit was divided between them in the ratio of 2: 1. After how many months did B join?                                                      
(a) 5    
(b) 10    
(c) 15  
(d) 20
Ans:
(a)
Let A work alone for ‘n’ months.
⇒ A’s input = 70,000 x 12
⇒ B’s input = 60,000 x (12 – n)
So, (70,000 x 12) / [60,000 x (12 – n)] = 2 / 1
⇒ (7 x 12) / [6 x (12 – n)] = 2 / 1
⇒ 12 – n = 7
⇒ n = 5
Therefore, B joined after 5 months.

Q5:  5 notebooks cost as much as 12 pens, 7 pens as much as 2 geometry, 3 geometry as much as 2 bottles. If the cost of 5 bottles is Rs 875, then find the cost of a notebook.
(a) 80  
(b) 85    
(c) 90  
(d) 95
Ans: 
(a)
5 bottles = 875
2 bottles = 875 * 2/5 = 3 geometry
2 geometry cost= 875 * 2/5 * 2/3 = 7 pens cost
12 pens cost = 875 * 2/5 * 2/3 * 12/7 = 5 notebook
1 notebook= 875 * 2/5 * 2/3 * 12/7 * 1/5 = Rs 80

Q6: What number must be subtracted from the numerator and denominator 43/37 that it becomes 3 : 2?
(a) 30  
(b) 25  
(c) 35  
(d) 20
Ans:
(b)
Let x be the number subtracted from the fraction.
(43 – x) / (37 – x) = 3/2
86 – 2x = 111 – 3x
x = 25

Q7: The ratio of the ages of a father and his son is 5:2. If the father is 40 years older than the son, what are their ages?
(a) 66  
(b) 76  
(c) 86  
(d) 56
Ans:
(a)
Let the son’s age be “2x” years. Then, the father’s age is “5x” years.
According to the given information, the father is 40 years older than the son, so we can write the equation:
5x – 2x = 40
3x = 40
To find the value of “x,” divide both sides by 3:
x = 40 / 3 x = 13.33 (rounded to two decimal places)
So, the son’s age is approximately 13.33 years,
and the father’s age is: 5x ≈ 5 * 13.33 ≈ 66.67 years

Q8: In a mixture of milk and water, the ratio of milk to water is 5:2. If 8 liters of water are added to the mixture, the ratio becomes 5:3. How many liters of milk were in the original mixture?
(a) 10  
(b) 20  
(c) 30  
(d) 40
Ans:
(d)
Milk = 5x liters Water = 2x liters
Now, when 8 liters of water are added, the ratio becomes 5:3. So, we have:
Milk = 5x liters Water = (2x + 8) liters
According to the new ratio, we can write:
5x / (2x + 8) = 5 / 3
=3 * 5x = 5 * (2x + 8)
=15x = 10x + 40
=5x = 40 = x = 8
Milk = 5x = 5 * 8 = 40 litres.

Q9: The ages of A, B, and C are in the ratio 5:7:9, and the sum of their ages is 63 years. How old is C?
(a) 25  
(b) 26  
(c) 27  
(d) 28
Ans:
(c)
Let the ages of A, B, and C be 5x, 7x, and 9x, respectively, where x is a positive integer representing the common ratio.
According to the problem, the sum of their ages is 63 years: 5x + 7x + 9x = 63
21x = 63
x = 63 / 21
x = 3
Now that we know the value of x, we can find the age of C:
C’s age = 9x C’s age = 9 * 3
C’s age = 27 years

Q10: Two numbers A and B are such that the sum of 5% of A and 4% of B is two-third of the sum of 6% of A and 8% of B. Find the ratio of A : B is-                        
(a) 4:3  
(b) 3:4  
(c) 2:4  
(d) 4:2
Ans:
(a)
From the equation, we get
5% of A + 4% of B = 2/3 (6% of A + 8% of B)
⇒ 15% of A + 12% of B = 12% of A + 16% of B
⇒ [15 – 12] % of A = [16 – 12] % of B
⇒ 3 % of A = 4% of B
Therefore, A : B = 4 : 3.

The document Practice Questions on Ratio and Proportion | Quantitative Techniques for CLAT is a part of the CLAT Course Quantitative Techniques for CLAT.
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FAQs on Practice Questions on Ratio and Proportion - Quantitative Techniques for CLAT

1. What is ratio and proportion?
Ans. Ratio is a comparison between two numbers or quantities, while proportion is an equation stating that two ratios are equal. In other words, proportion is a statement that shows how one ratio is equivalent to another.
2. How do you simplify a ratio?
Ans. To simplify a ratio, you need to find the greatest common divisor (GCD) of the two numbers in the ratio and divide both numbers by the GCD. This will give you the simplified form of the ratio.
3. How can ratios be used in real-life situations?
Ans. Ratios are commonly used in various real-life situations such as cooking recipes, financial analysis, and scale models. For example, in a recipe, the ratio of ingredients determines the proportions needed to make the dish.
4. What is the difference between direct and inverse proportion?
Ans. In direct proportion, as one quantity increases, the other quantity also increases at the same rate. Inverse proportion, on the other hand, occurs when one quantity increases while the other decreases, or vice versa, in such a way that their product remains constant.
5. How can proportion be used to solve problems?
Ans. Proportion can be used to solve various types of problems, such as finding a missing value in a ratio, determining an unknown quantity based on a given proportion, or solving problems involving direct or inverse proportionality. By setting up and solving equations using proportions, you can find the desired solution.
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