Ratio, Proportion, Averages and Arithmetic Word Problems Previous Year Questions (PYQs) for XAT: 7+ Solved Questions with Step-by-Step Solutions

Solve 7+ Ratio, Proportion, Averages and Arithmetic Word Problems previous year questions for XAT with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Arithmetic & Word Problems

Chapter Roadmap: Ratio, Averages & Word Problems

Your Journey Through This Chapter

Step 1: Ratio, Proportion & Mixtures (Current Topic)
Focus: The foundation of comparative quantities and blending solutions.
Mastery: You will learn to handle fractional parts, integer constraints, and the alligation method.

Step 2: Averages & Constraints
Focus: Central tendency and optimizing values within strict boundaries.
Mastery: You will learn to find maximums and minimums using the concept of deviations and weighted averages.

Step 3: Arithmetic Word Problems & Sets
Focus: Real-world scenarios involving sets, Venn diagrams, and complex logical conditions.
Mastery: You will learn to translate dense paragraphs into clean equations and visual models.

By the end of this chapter, you will have mastered the core arithmetic tools required to decode almost any quantitative word problem in the exam.

Ratio, Proportion & Mixtures

The Hook: Ratios are just fractions in disguise, and mixtures are just weighted averages.

What you will learn here

The Multiplying Factor: How to translate ratios into actual quantities using a common variable.
Integer Constraints: Solving linear equations when prices and quantities must be whole numbers.
The Alligation Method: A visual shortcut to find mixing ratios without heavy algebra.
Mixture Replacements: Handling scenarios where a portion of a mixture is repeatedly removed and replaced.

Context: This is the foundation of the chapter. Mastering ratios here will make the averages and word problems in the next topics significantly easier.

Ratio, Proportion, Averages and Arithmetic Word Problems: Solved Questions with Step-by-Step Explanations (5 Problems)

Question 1 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

Common Description:

Brijbhushan, a micro financier, lends money at the rate of Rs.10 per square meter to small farmers at a village.

He charges an annual interest rate of 10%. All the farming plots in that village are rectangular, with areas varying between a minimum of 1000 square meters and a maximum of 10,000 square meters.

This year, Brijbhushan has lent money only to five farmers: Aditya, Binod, Chhuttan, Dabloo and Govind. The perimeter of Chhuttan’s plot is 250 meters, with the length and width being at a ratio of 4:1. Aditya’s plot has an area three times the area of Govind’s plot. The area of Aditya’s plot is also the average of the areas of Govind’s plot and Dabloo’s plot. The plots belonging to Aditya, Binod and Dabloo are of the same width, but of different lengths. Moreover, the length of Binod’s plot is the sum of the lengths of Aditya’s plot and Dabloo’s plot.

If the width of the Aditya’s plot is 25 meters, what is the MINIMUM possible length of Binod’s plot?

A.
360 meters
B.
320 meters
C.
250 meters
D.
400 meters
E.
280 meters

Correct Answer:

Step-by-Step Solution

Key idea: This is an arithmetic word problem with geometric constraints, recognizable by the relationships between areas, lengths, and widths of rectangular plots.

Step 1: Analyze Chhuttan's plot: Perimeter = 250m, Length:Width = 4:1.

$2 (L + W) = 250 \Rightarrow L + W = 125$ .

$4 x + x = 125 \Rightarrow 5 x = 125 \Rightarrow x = 25$ .

So, Chhuttan's width = 25m, length = 100m, Area = 2500 sq m. (This is consistent with the 1000-10000 range).

Step 2: Analyze area relationships: Let Govind's area be $G$ .

Aditya's area $A = 3 G$ .

Aditya's area is the average of Govind's and Dabloo's: $A = (G + D) /2$ .

Substitute $A = 3 G$ : $3 G = (G + D) /2 \Rightarrow 6 G = G + D \Rightarrow D = 5 G$ .

Step 3: Apply area constraints: All areas are between 1000 and 10000.

$1000 \leq G \leq 10000$

$1000 \leq 3 G \leq 10000 \Rightarrow G \geq 333.3$

$1000 \leq 5 G \leq 10000 \Rightarrow G \leq 2000$ .

So, $1000 \leq G \leq 2000$ .

Step 4: Analyze Binod's plot: Aditya, Binod, and Dabloo have the same width, $W_{A B D} = 25$ m.

Length of Binod $L_{B} = L_{A} + L_{D}$ .

Area of Binod $B = W_{A B D} \times L_{B} = W_{A B D} \times (L_{A} + L_{D}) = (W_{A B D} \times L_{A}) + (W_{A B D} \times L_{D}) = A + D$ .

So, $B = 3 G + 5 G = 8 G$ .

Step 5: Find minimum length of Binod's plot:

$L_{B} = B / W_{A B D} = 8 G /25$ .

To minimize $L_{B}$ , we must minimize $G$ . The minimum valid $G$ is 1000.

Step 6: Calculate minimum $L_{B} = 8 (1000) /25 = 8000/25 = 320$ meters.

Answer: 320 meters

Question 2 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

Rajan is a fruit seller. On any day, he sells only one kind of fruit. On the first day, he buys 9 kg of blueberries.

On the second day, he buys 22 kg of kiwis. On the third day, he buys 50 kg of peaches. The per kg purchase price of each fruit is an integer. Further, on each of these three days, he spends the same amount to purchase fruits. On the fourth day, he buys mangoes at Rs. 35/kg and spends Rs. 15 less than any of the previous three days.

If he then sells all the mangoes at Rs. 50/kg, what is his MINIMUM possible profit on the fourth day?

A.
Rs. 2130
B.
Rs. 2115
C.
Rs. 2085
D.
Rs. 2265
E.
None of the other options is correct

Correct Answer:

Step-by-Step Solution

Key idea: This is an integer constraint and LCM problem, recognizable because the same total amount is spent on different quantities, implying the total is a common multiple.

Step 1: Let the amount spent on each of the first three days be $X$ .

Step 2: Since he buys 9 kg, 22 kg, and 50 kg, and the per kg price is an integer, $X$ must be divisible by 9, 22, and 50.

Step 3: Find the LCM of 9, 22, and 50.

$9 = 3^{2}$

$22 = 2 \times 11$

$50 = 2 \times 5^{2}$

LCM = $2 \times 3^{2} \times 5^{2} \times 11 = 2 \times 9 \times 25 \times 11 = 4950$ .

So, $X = 4950 m$ for some positive integer $m$ .

Step 4: On the fourth day, he spends Rs. 15 less, so the amount is $X - 15 = 4950 m - 15$ .

Step 5: He buys mangoes at Rs. 35/kg, so $4950 m - 15$ must be divisible by 35.

Step 6: Check divisibility by 35 (which means divisible by 5 and 7).

$4950 m - 15$ is clearly divisible by 5.

For divisibility by 7: $4950 \equiv 1 (mod 7)$ (since $4950 = 7 \times 707 + 1$ ).

Also, $15 \equiv 1 (mod 7)$ .

So, $4950 m - 15 \equiv 1 m - 1 (mod 7)$ .

We need $m - 1 \equiv 0 (mod 7) \Rightarrow m \equiv 1 (mod 7)$ .

Step 7: The minimum positive integer $m$ is 1.

Step 8: If $m = 1$ , $X = 4950$ . Amount spent on Day 4 = $4950 - 15 = 4935$ .

Step 9: Quantity of mangoes = $4935/35 = 141$ kg.

Step 10: Revenue from selling at Rs. 50/kg = $141 \times 50 = 7050$ .

Step 11: Profit = Revenue - Cost = $7050 - 4935 = 2115$ .

Answer: Rs. 2115

Question 3 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

Eight employees of an organization have been rated on a scale of 1 to 50 for their performance. All ratings are integers. The overall average rating of the eight employees is 30. While the five employees with the highest ratings average 38, the five employees with the lowest ratings average 25.

Which of the following, about the ratings obtained by the eight employees, is DEFINITELY FALSE?

A.
The second highest rating obtained is 38.
B.
The median of the eight ratings is 37.5.
C.
The lowest rating obtained is 1.
D.
The highest rating obtained is 40.
E.
The third lowest rating obtained is 37.

Correct Answer:

Step-by-Step Solution

Key idea: This is an overlapping averages (sliding window) problem, recognizable by the overlapping groups of "highest" and "lowest" ratings.

Step 1: Total sum of 8 ratings = $8 \times 30 = 240$ .

Step 2: Sum of top 5 ratings = $5 \times 38 = 190$ .

Step 3: Sum of bottom 5 ratings = $5 \times 25 = 125$ .

Step 4: The top 5 and bottom 5 overlap at the 4th and 5th ratings (let's call them $x_{4}$ and $x_{5}$ ).

Step 5: Sum of all 8 = (Sum of top 5) + (Sum of bottom 5) - ( $x_{4} + x_{5}$ ).

Step 6: $240 = 190 + 125 - (x_{4} + x_{5}) \Rightarrow x_{4} + x_{5} = 315 - 240 = 75$ .

Step 7: The median of 8 ratings is the average of the 4th and 5th ratings: $(x_{4} + x_{5}) /2 = 75/2 = 37.5$ . (Option B is definitely true).

Step 8: We need to find which option is DEFINITELY FALSE. Let's test the maximum possible rating ( $x_{8}$ ).

Step 9: Since $x_{4} + x_{5} = 75$ and $x_{4} \leq x_{5}$ , we must have $x_{5} \geq 37.5 \Rightarrow x_{5} \geq 38$ (since ratings are integers).

Step 10: The top 5 sum is $x_{4} + x_{5} + x_{6} + x_{7} + x_{8} = 190$ .

Step 11: Substitute $x_{4} + x_{5} = 75$ : $75 + x_{6} + x_{7} + x_{8} = 190 \Rightarrow x_{6} + x_{7} + x_{8} = 115$ .

Step 12: Since $x_{5} \leq x_{6} \leq x_{7} \leq x_{8}$ , all three must be $\geq 38$ .

Step 13: If $x_{8} = 40$ , then $x_{6} + x_{7} = 75$ . Since $x_{6} \geq 38$ , the maximum $x_{7}$ can be is $75 - 38 = 37$ . But $x_{6} \leq x_{7}$ , so $38 \leq 37$ , which is a contradiction.

Step 14: Thus, $x_{8}$ cannot be 40. The maximum possible rating is 39.

Answer: The highest rating obtained is 40.

Question 4 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

In a small college, students are allowed to take only one specialization. Traditionally, only two specializations are offered: Science and Arts. Students enrolled to specialize in Science must take Physics and Mathematics subjects, while students enrolled to specialize in Arts must take Economics and Political Science subjects.
Students enrolled in Science are not allowed to take either Economics or Political Science, while students enrolled in Arts are not allowed to take either Physics or Mathematics.
Recently, the college has started a third specialization called MatEco that requires students to take Economics and Mathematics. However, MatEco students would not be allowed to take either Physics or Political Science. When the college opens this new specialization for enrolment, it allows students, originally enrolled in Science or Arts, to switch to MatEco.
From among the students originally enrolled in Arts, 20 students switch to MatEco. This makes the number of Science students twice the number of Arts students. After this, from among the students who originally enrolled in Science, 45 students switch to MatEco. This makes the number of Arts students twice the number of Science students.
In total, how many students, from among those originally enrolled in Science or Arts, are now taking Economics?

A.
45
B.
65
C.
80
D.
95
E.
None of the remaining options is correct.

Correct Answer:

Step-by-Step Solution

Key idea: This is a set theory and algebraic translation problem, recognizable by the shifting of students between distinct groups and the resulting ratio changes.

Step 1: Let the initial number of Science students be $s$ and Arts students be $a$ .

Step 2: 20 students switch from Arts to MatEco. The remaining Arts students are $a - 20$ . Science remains $s$ .

Step 3: The problem states Science is now twice Arts: $s = 2 (a - 20) \Rightarrow s = 2 a - 40$ .

Step 4: Next, 45 students switch from Science to MatEco. The remaining Science students are $s - 45$ . Arts remains $a - 20$ .

Step 5: The problem states Arts is now twice Science: $a - 20 = 2 (s - 45) \Rightarrow a - 20 = 2 s - 90 \Rightarrow a = 2 s - 70$ .

Step 6: Substitute $s$ from Step 3 into Step 5: $a = 2 (2 a - 40) - 70 \Rightarrow a = 4 a - 80 - 70 \Rightarrow 3 a = 150 \Rightarrow a = 50$ .

Step 7: Find $s$ : $s = 2 (50) - 40 = 60$ .

Step 8: We need the total number of students originally in Science or Arts who are NOW taking Economics.

Step 9: Originally, only Arts students took Economics. Science students did not.

Step 10: After the switches, the students taking Economics are the remaining Arts students and ALL MatEco students (since MatEco requires Economics).

Step 11: Remaining Arts = $50 - 20 = 30$ . Total MatEco = $20 (from Arts) + 45 (from Science) = 65$ .

Step 12: Total taking Economics = $30 + 65 = 95$ .

Answer: 95

Question 5 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

Common Description: Aman has come to the market with Rs. 100. If he buys 5 kilograms of cabbage and 4 kilograms of potato, he will have Rs. 20 left; or else, if he buys 4 kilograms of cabbage and 5 kilograms of onion, he will have Rs. 7 left.
The per kilogram prices of cabbage, onion and potato are positive integers(in rupees), and any type of these vegetables can only be purchased in positive integer kilogram, or none at all. Aman decides to buy only onion, in whatever maximum quantity possible(in positive integer kilogram), with the money he has come to the market with. How much money will he be left with after the purchase?

A.
Rs. 12
B.
Rs. 9
C.
Rs. 7
D.
Rs. 5
E.
Rs. 1

Correct Answer:

Step-by-Step Solution

Key idea: This is a linear Diophantine equation problem with integer constraints, recognizable by the requirement that prices are positive integers and we have two spending scenarios.

Step 1: Let the prices per kg of cabbage, potato, and onion be $C$ , $P$ , and $O$ respectively.

Step 2: From the first scenario, $5 C + 4 P = 100 - 20 = 80$ .

Step 3: From the second scenario, $4 C + 5 O = 100 - 7 = 93$ .

Step 4: Analyze $5 C + 4 P = 80$ . Since $4 P$ and $80$ are multiples of 4, $5 C$ must also be a multiple of 4. Thus, $C$ must be a multiple of 4.

Step 5: Test possible positive integer values for $C$ :

If $C = 4$ , $5 (4) + 4 P = 80 \Rightarrow 4 P = 60 \Rightarrow P = 15$ .
If $C = 8$ , $5 (8) + 4 P = 80 \Rightarrow 4 P = 40 \Rightarrow P = 10$ .
If $C = 12$ , $5 (12) + 4 P = 80 \Rightarrow 4 P = 20 \Rightarrow P = 5$ .
If $C = 16$ , $P = 0$ , but prices must be positive integers.

Step 6: Now use the second equation $4 C + 5 O = 93$ to find which $C$ gives an integer $O$ :

If $C = 4$ , $16 + 5 O = 93 \Rightarrow 5 O = 77$ (not an integer).
If $C = 8$ , $32 + 5 O = 93 \Rightarrow 5 O = 61$ (not an integer).
If $C = 12$ , $48 + 5 O = 93 \Rightarrow 5 O = 45 \Rightarrow O = 9$ .

Step 7: The only valid price for onion is $O = 9$ Rs/kg.

Step 8: Aman buys only onions with Rs. 100. Maximum quantity = $⌊ 100/9 ⌋ = 11$ kg.

Step 9: Cost for 11 kg = $11 \times 9 = 99$ Rs. Money left = $100 - 99 = 1$ Rs.

Answer: Rs. 1

Ratio, Proportion, Averages and Arithmetic Word Problems Previous Year Questions (PYQs) for XAT: 7+ Solved Questions with Step-by-Step Solutions

Chapter Roadmap: Arithmetic & Word Problems

Chapter Roadmap: Ratio, Averages & Word Problems

Your Journey Through This Chapter

Ratio, Proportion & Mixtures

Ratio, Proportion & Mixtures

What you will learn here

Ratio, Proportion, Averages and Arithmetic Word Problems: Solved Questions with Step-by-Step Explanations (5 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

More previous year questions (pyqs) in this unit

A question from this chapter

Login to view the complete previous-year questions and solutions

Personalised first. High quality throughout.

Built around you, not around a syllabus PDF

Revision that hits your weak spots

Questions calibrated to the real exam

Notes written for recall, not for volume

One place for everything

Honest progress

Unlock the whole course