Q1. One year payment to an assistant is 120 and one gold coin. The assistant leaves after 8 months and receives 70 and a gold coin. Determine the value of the gold coin.
A) 20
B) 30
C) 40
D) 50
E) 60
Correct Option: B
Rationale: Let gold coin value = x.
Yearly payment = 120 + x.
8 months is 8/12 = 2/3 of a year.
Payment for 8 months = (2/3) * (120 + x)
Given: 70 + x = (2/3)(120 + x)
210 + 3x = 240 + 2x
x = 30.
Q2. Rohan purchased three mangoes, four apples, and five pears for 32. Had he purchased four mangoes, three apples, and two pears, it would have cost him only 22. How much would he pay for 7 mangoes, 7 apples, and 7 pears?
A) 42
B) 48
C) 54
D) 60
E) 64
Correct Option: C
Rationale: Let mango = M, apple = A, pear = P.
(1) 3M + 4A + 5P = 32
(2) 4M + 3A + 2P = 22
Adding (1) and (2):
(3 + 4)M + (4 + 3)A + (5 + 2)P = 32 + 22
7M + 7A + 7P = 54.
Q3. How many positive integer solutions exist for the equation 7x + 12y = 150?
A) 1
B) 2
C) 3
D) 4
E) 0
Correct Option: B
Rationale: 7x = 150 – 12y.
For x to be an integer, (150 – 12y) must be divisible by 7.
If y = 2: 150 – 24 = 126. 126/7 = 18. (18, 2) is a solution.
If y = 9: 150 – 108 = 42. 42/7 = 6. (6, 9) is a solution.
If y = 16: 150 – 192 is negative.
Only 2 positive integer solutions.
Q4. P, Q, R, S, and T play cards. P says to Q: “If you give me 5 cards, I’ll have as many as you.” S takes 10 from Q and equals T. P and R together = twice T. Q and S together = same as P and R. Total cards = 200. How many did Q have originally?
A) 45
B) 50
C) 55
D) 60
E) 65
Correct Option: B
Rationale: Equations:
(1) P+5 = Q-5 -> Q = P+10
(2) Q-10 = S+10? No, “S takes 10 from Q and equals T” -> Q-10 and S+10 is not stated, just S+10 = T.
(3) P+R = 2T
(4) Q+S = P+R -> Q+S = 2T
(5) P+Q+R+S+T = 200
Substitute (4) into (5): (P+R) + (Q+S) + T = 200 -> 2T + 2T + T = 200 -> 5T = 200 -> T = 40.
From (4): Q+S = 80. From (2): S+10 = 40 -> S = 30.
Then Q + 30 = 80 -> Q = 50.
Q5. When Amit’s age was the same as Sumit’s present age, the ratio of their ages was 5:3. When Sumit’s age becomes the same as Amit’s present age, find the ratio of ages of Amit & Sumit.
A) 5:7
B) 9:7
C) 4:5
D) 5:4
E) 3:5
Correct Option: B
Rationale: Let present ages be A and S. Amit is older.
Difference in age = (A-S).
When Amit was S, Sumit was S-(A-S) = 2S-A.
Ratio: S / (2S-A) = 5/3 -> 3S = 10S – 5A -> 5A = 7S -> A = 7/5 S.
When Sumit becomes A (after A-S years), Amit will be A+(A-S) = 2A-S.
New Ratio Amit:Sumit = (2A-S) / A = (2(7/5 S) – S) / (7/5 S)
= (14/5 S – 5/5 S) / (7/5 S) = (9/5 S) / (7/5 S) = 9:7.
Q6. If the sum of two numbers is 25 and their difference is 7, find the difference between their squares.
A) 150
B) 175
C) 200
D) 225
E) 250
Correct Option: B
Rationale: Let numbers be a and b.
a + b = 25, a – b = 7.
Difference of squares: a² – b² = (a+b)(a-b)
= 25 * 7 = 175.
Q7. A mother is now four times as old as her daughter. Six years ago, she was seven times as old as her daughter. What is the current age of the daughter?
A) 10
B) 12
C) 14
D) 16
E) 18
Correct Option: B
Rationale: Daughter = d, Mother = 4d.
6 years ago: (4d – 6) = 7(d – 6)
4d – 6 = 7d – 42
3d = 36 -> d = 12.
Q8. The sum of incomes of X and Y is more than Z and W. The sum of X and Z is the same as Y and W. X earns half as much as the sum of Y and W. If X’s income is 60,000 and the difference between Y and W is equal to X’s income, find Y’s income.
A) 70,000
B) 80,000
C) 90,000
D) 100,000
E) 110,000
Correct Option: C
Rationale: X = 60,000.
X = 1/2(Y+W) -> Y+W = 2X = 120,000.
Y-W = X = 60,000.
Adding equations: 2Y = 180,000 -> Y = 90,000.
Q9. A girl has enough money to buy 15 notebooks. If each notebook cost 40 paise less, she could buy three more notebooks and still have 30 paise left. How much money did she have originally?
A) 25.50
B) 30.00
C) 34.50
D) 37.50
E) 42.00
Correct Option: C
Rationale: Let cost be x paise. Total money = 15x.
New scenario: 18(x – 40) + 30 = 15x
18x – 720 + 30 = 15x
3x = 690 -> x = 230 paise (2.30 Rs).
Total money = 15 * 2.30 = 34.50.
Q10. The sum of ages of 6 children born at intervals of 2 years each is 66 years. What is the age of the oldest child?
Q11. If the length of a rectangle is decreased by 5 cm and the width increased by 2 cm, a square results. If the area of the rectangle and the square are equal, find the perimeter of the original rectangle.
Q12. M is twice as efficient as N. Working together they complete a task in 4 days. If working alone, N takes 6 days more than M. Find the time taken by N to complete the task alone.
A) 6 days
B) 8 days
C) 10 days
D) 12 days
E) 14 days
Correct Option: D
Rationale: Efficiency ratio M:N = 2:1.
Time ratio M:N = 1:2.
N – M = 6 days. Let M = x, N = 2x.
2x – x = 6 -> x = 6.
So M = 6 days, N = 12 days.
Combined: 1/6 + 1/12 = 3/12 = 1/4 (matches given 4 days).
Q13. Karan is 5 years older than Arjun, who is twice as old as Bheem. If the total of their ages is 55, how old is Arjun?
Equations and Algebra form the backbone of the Quantitative Aptitude section in MBA CET, and mastering them can significantly boost your overall score. These topics are not only concept-driven but also highly scoring when approached with the right strategies. Questions typically revolve around linear equations, quadratic equations, simultaneous equations, inequalities, and algebraic identities.
The key to excelling in algebra for MBA CET lies in speed and clarity of concepts. Instead of lengthy calculations, students should focus on smart techniques like substitution from options, elimination methods, and recognizing patterns. For example, many CET questions can be solved by directly putting values from the answer choices, saving valuable time during the exam. Similarly, understanding identities such as (a + b)², (a − b)², and a² − b² helps in solving questions quickly without heavy computation.
Another important aspect is translating word problems into equations. Questions involving ages, ratios, profit and loss, and mixtures often rely on forming correct equations. Practicing these regularly improves both accuracy and speed. Additionally, quadratic equations and inequalities are frequently tested, so being comfortable with factorization and sign analysis is crucial.
Time management plays a critical role in MBA CET, as the exam is speed-based. Students should aim to identify easy algebra questions and solve them quickly to maximize attempts. Regular practice with mock tests and previous year questions helps in recognizing common patterns and traps.
In conclusion, a strong grip on equations and algebra can act as a game-changer in MBA CET preparation. With consistent practice, concept clarity, and smart solving techniques, students can turn this section into their strongest area.
