Basic Concept and Understanding
| Quants – Ravi Prakash | ||||||||
| Arithmetic | Playlist | |||||||
| Speed Math | SM 1 | SM 2 | SM 3 | SM 4 | SM 5 | SM 6 | ||
| Averages | AVG 1 | AVG 2 | AVG 3 | AVG 4 | ||||
| Alligation & Mixtures | AM 1 | AM 2 | AM 3 | AM 4 | AM 5 | AM 6 | ||
| Ratio | R1 | R2 | R3 | R4 | R5 | R6 | ||
| Proportion, Variation | PV 1 | PV 2 | ||||||
| Percentages, Profit & Loss | PPL 1 | PPL 2 | PPL 3 | PPL 4 | PPL 5 | PPL 6 | PPL 7 | PPL 8 |
| Simple & Compund Interest | SCI 1 | SCI 2 | SCI 3 | SCI 4 | SCI 5 | SCI 6 | ||
| Time & Work | TW 1 | TW 2 | TW 3 | |||||
| Time, Speed & Distance | TSD 1 | TSD 2 | TSD 3 | TSD 4 | TSD 5 | TSD 6 | TSD 7 | TSD 8 |
| Boats & Streams | BS 1 | BS 2 | ||||||
| Relative Speed | RS 1 | RS 2 | RS 3 | RS 4 | RS 5 | |||
| Linear Tracks | LT 1 | LT 2 | LT 3 | |||||
| Circular Tracks | CT 1 | CT 2 | CT 3 | |||||
| Linear Races | LR 1 | LR 2 | ||||||
| Clocks | C1 |
CAT PYQ’S WITH SOLUTION
CAT Arithmetic Questions with Solutions (2024)
CAT 2024 – Question 1: Precious Stone Price
Question:
The price of a precious stone is directly proportional to the square of its weight. Sita has a precious stone weighing 18 units. If she breaks it into four pieces with each piece having a distinct integer weight, then the difference between the highest and lowest possible values of the total price of the four pieces will be 288000. What is the price of the original precious stone?
Options:
(a) 1,944,000
(b) 972,000
(c) 1,620,000
(d) 1,296,000
Solution:
Let price ∝ square of weight ⇒ C = k × w²
Original weight = 18 ⇒ C = k × 18² = 324k
Maximum price: weights = 1, 2, 3, 12 ⇒ C = k(1² + 2² + 3² + 12²) = k(158)
Minimum price: weights = 3, 4, 5, 6 ⇒ C = k(9 + 16 + 25 + 36) = k(86)
Difference = 72k = 288000 ⇒ k = 4000
Original Price = 324k = 324 × 4000 = 1,296,000
Answer: (d) 1,296,000
CAT 2024 – Question 2: Work Completion Time
Question:
Working together, Rahul, Rakshita, and Gurmeet would have taken more than 7 days to finish a job. Rahul and Gurmeet, working together, would have taken less than 15 days. However, all three worked together for 6 days, then Rakshita worked alone for 3 days to finish the job. If Rakshita had worked alone on the job, the number of days she would have taken cannot be:
Options:
(a) 20
(b) 17
(c) 16
(d) 21
Solution:
Let total work = 1 unit
Let R, K, G be work/day by Rahul, Rakshita, Gurmeet respectively
Given:
6(R + K + G) + 3K = 1
R + G alone would take >15 days ⇒ R + G < 1/15
Try possible values:
If K = 1/16, it contradicts the total work equation and constraints.
Answer: (c) 16
CAT 2024 – Question 3: Mixture Percentage
Question:
Anil mixes cocoa and sugar in 3:2 ratio (Mixture A), and coffee and sugar in 7:3 ratio (Mixture B). He mixes A and B in 2:3 ratio to form Mixture C. Then he adds an equal amount of milk to C. What is the percentage of sugar in the final drink?
Options:
(a) 17%
(b) 16%
(c) 21%
(d) 24%
Solution:
Assume:
Mixture A = 200 ml ⇒ cocoa = 120, sugar = 80
Mixture B = 300 ml ⇒ coffee = 210, sugar = 90
Mixture C = 200 + 300 = 500 ml
Sugar = 80 + 90 = 170 ml
Add 500 ml milk ⇒ final volume = 1000 ml
Sugar % = (170 / 1000) × 100 = 17%
Answer: (a) 17%
CAT 2024 – Question 4: Profit and Discount
Question:
A merchant buys cloth at Rs.100/meter and gets 5 cm free on every 100 cm. He sells at Rs.110/meter but gives only 95 cm per 100 cm sold. If he offers a 5% discount, what is his profit percentage?
Options:
(a) 4.2%
(b) 9.7%
(c) 15.5%
(d) 16%
Solution:
Cost price:
105 cm for Rs.100 ⇒ CP per cm = 100 / 105 ≈ 0.952
Selling price:
1 meter = Rs.110 ⇒ 110 / 100 = 1.1 Rs/cm
5% discount ⇒ 1.1 × 0.95 = 1.045 Rs/cm
But customer gets only 95 cm for 1 meter ⇒ effective SP = 1.045 × (100 / 95) = 1.1
Profit = 1.1 – 0.952 = 0.148
Profit % = (0.148 / 0.952) × 100 ≈ 15.5%
Answer: (c) 15.5%
CAT 2023 – Question 1: Average Weight Increase
Question:
The average weight of students in a class increases by 600 grams when some new students join. If the average weight of the new students is 3 kg more than the average weight of the original students, what is the ratio of the number of original students to the number of new students?
Options:
(a) 1:4
(b) 1:2
(c) 4:1
(d) 3:1
Solution:
Let the average weight of original students be x kg.
Average weight of new students = x + 3 kg.
Combined average increases by 0.6 kg.
Let the number of original students be O and new students be N.
Using the formula for combined average: Ox+N(x+3)O+N=x+0.6\frac{Ox + N(x + 3)}{O + N} = x + 0.6
Solving: Ox+Nx+3NO+N=x+0.6\frac{Ox + Nx + 3N}{O + N} = x + 0.6 x(O+N)+3NO+N=x+0.6\frac{x(O + N) + 3N}{O + N} = x + 0.6 x+3NO+N=x+0.6x + \frac{3N}{O + N} = x + 0.6 3NO+N=0.6\frac{3N}{O + N} = 0.6 3N=0.6(O+N)3N = 0.6(O + N) 3N=0.6O+0.6N3N = 0.6O + 0.6N 3N−0.6N=0.6O3N – 0.6N = 0.6O 2.4N=0.6O2.4N = 0.6O O/N=2.4/0.6=4O/N = 2.4 / 0.6 = 4
Answer: (c) 4:1
CAT 2023 – Question 2: Train Meeting Problem
Question:
Trains A and B start traveling at the same time towards each other with constant speeds from stations X and Y, respectively. Train A reaches station Y in 10 minutes while train B takes 9 minutes to reach station X after meeting train A. What is the total time taken, in minutes, by train B to travel from station Y to station X?
Options:
(a) 6
(b) 15
(c) 10
(d) 12
Solution:
Let the meeting point be M.
Let the distance from X to M be D1 and from Y to M be D2.
Speed of A = D1 / t
Speed of B = D2 / t
Given:
- Train A takes 10 minutes to reach Y ⇒ total distance = D1 + D2 ⇒ D1 + D2 = speed of A × 10
- After meeting, Train B takes 9 minutes to reach X ⇒ D1 = speed of B × 9
From above:
- D1 = speed of B × 9
- D2 = speed of B × t
So, D1 + D2 = speed of B × (t + 9) = speed of A × 10
But speed of A = D1 / t = (speed of B × 9) / t
Therefore:
speed of B × (t + 9) = ((speed of B × 9) / t) × 10
Solving:
(t + 9) = (9 × 10) / t
t + 9 = 90 / t
t² + 9t – 90 = 0
Solving the quadratic equation: t = 6 (positive root)
Total time taken by Train B = t + 9 = 6 + 9 = 15 minutes
Answer: (b) 15
CAT 2023 – Question 3: Nut Mixture Profit
Question:
Ankita buys 4 kg cashews, 14 kg peanuts, and 6 kg almonds when the cost of 7 kg cashews is the same as that of 30 kg peanuts or 9 kg almonds. She mixes all the three nuts and marks a price for the mixture to make a profit of ₹1752. She sells 4 kg of the mixture at this marked price and the remaining at a 20% discount on the marked price, thus making a total profit of ₹744. What amount, in rupees, did she spend in buying almonds?
Options:
(a) ₹1680
(b) ₹1176
(c) ₹2520
(d) ₹1440
Solution:
Let the cost per kg of cashews be C.
Then, cost per kg of peanuts = (7C) / 30
Cost per kg of almonds = (7C) / 9
Cost of 4 kg cashews = 4C
Cost of 14 kg peanuts = 14 × (7C / 30) = (98C) / 15
Cost of 6 kg almonds = 6 × (7C / 9) = (14C) / 3
Total cost = 4C + (98C / 15) + (14C / 3)
Finding a common denominator and summing:
Total cost = (60C + 98C + 70C) / 15 = (228C) / 15
Total weight = 4 + 14 + 6 = 24 kg
Marked price per kg = (Total cost + 1752) / 24
She sells 4 kg at marked price and 20 kg at 20% discount:
Total revenue = 4 × MP + 20 × 0.8 × MP = (4 + 16) × MP = 20 × MP
Total revenue = Total cost + 744
So, 20 × MP = (228C / 15) + 744
But MP = (228C / 15 + 1752) / 24
Now, solve for C:
20 × ((228C / 15 + 1752) / 24) = (228C / 15) + 744
Multiply both sides by 24:
20 × (228C / 15 + 1752) = 24 × (228C / 15 + 744)
Compute both sides:
Left: (4560C / 15) + 35040
Right: (5472C / 15) + 17856
Now, equate both sides:
(4560C / 15) + 35040 = (5472C / 15) + 17856
Subtract (4560C / 15) and 17856 from both sides:
35040 – 17856 = (5472C / 15) – (4560C / 15)
17184 = (912C) / 15
C = (17184 × 15) / 912 = 282
Cost per kg of almonds = (7C) / 9 = (7 × 282) / 9 = 219.33
Total cost for 6 kg almonds = 6 × 219.33 ≈ ₹1316
Answer: Approximately ₹1316 (Note: This doesn’t match any of the provided options, suggesting a possible discrepancy in the question or options.)
CAT 2023 – Question 4: Average of Integers
Question:
The average of three integers is 13. When a natural number n is included, the average of these four integers remains an odd integer. What is the minimum possible value of n?
Options:
(a) 3
(b) 4
(c) 5
(d) 1
Solution:
Sum of three integers = 3 × 13 = 39
Let the new average be an odd integer A. Then:
(39 + n) / 4 = A ⇒ 39 + n = 4A ⇒ n = 4A – 39
We need the smallest natural number n such that A is an integer and n is natural.
Try A = 10 ⇒ n = 4×10 – 39 = 1
Try A = 11 ⇒ n = 4×11 – 39 = 5
Try A = 12 ⇒ n = 4×12 – 39 = 9
The smallest natural number n is 1 when A = 10 (which is even), but the average must be an odd integer. So, A must be odd.
Try A = 11 ⇒ n = 5
Try A = 13 ⇒ n = 13
Minimum possible n is 5
Answer: (c) 5
CAT 2023 – Question 5: Mixture Ratio
Question:
A mixture contains lemon juice and sugar syrup in equal proportion. If a new mixture is created by adding this mixture and sugar syrup in the ratio 1:3, what is the ratio of lemon juice to sugar syrup in the new mixture?
Options:
(a) 1:4
(b) 1:5
(c) 1:6
(d) 1:7
Solution:
1:7
