Quantitative aptitude questions play a crucial role in competitive exams and campus placements. Here, we provide a set of frequently asked quantitative questions, along with their answers and explanations.
How many liters of a 90% concentrated acid need to be mixed with a 75% solution of concentrated acid to get a 30-liter solution of 78% concentrated acid?
Options: a) 3 b) 4 c) 6 d) 10
Answer: 6 liters
Solution: Using the alligation method:
xx+y×90+yx+y×75=78\frac{x}{x + y} \times 90 + \frac{y}{x + y} \times 75 = 78
Let x be the quantity of 90% acid and y be the quantity of 75% acid. Given total solution is 30 liters:
x+y=30x + y = 30
Solving, x = 6 liters.
If f(x)=ax4−bx2+x+5f(x) = ax^4 – bx^2 + x + 5 and f(−3)=2f(-3) = 2, then f(3)f(3) = ?
Options: a) 1 b) -2 c) 3 d) 8
Answer: 8
Solution: Since f(x) is symmetric (even function except for x), f(−3)=f(3)f(-3) = f(3). Hence, f(3)=2f(3) = 2.
Of a set of 30 numbers, the average of the first 10 numbers is equal to the average of the last 20 numbers. Then, the sum of the last 20 numbers is?
Options: a) Cannot be determined. b) 2 ×\times sum of last ten numbers c) 2 ×\times sum of first ten numbers d) sum of first ten numbers
Answer: 2 ×\times sum of first ten numbers
Solution: Since averages are the same: S1010=S2020\frac{S_{10}}{10} = \frac{S_{20}}{20} S20=2S10S_{20} = 2S_{10}
A play school has chocolates sufficient for 50 students for 30 days. For the first ten days, only 20 students were present. How many more students can be accommodated so that chocolates last for exactly 30 days?
Options: a) 45 b) 60 c) 55 d) 70
Answer: 45
Solution: Using the formula: (50×30)−(20×10)=(50+x)×20(50 \times 30) – (20 \times 10) = (50 + x) \times 20 Solving for x gives 45 students.
The circumference of the front wheel of a cart is 133 uneves and that of the back wheel is 190 uneves. What is the distance traveled when the front wheel has done nine more revolutions than the rear wheel?
Options: a) 570 b) 1330 c) 3990 d) 399
Answer: 3990 uneves
Solution: Using relative revolutions and LCM approach, distance covered is 3990 uneves.
There are 20 people sitting in a circle, including 18 men and 2 sisters. How many arrangements are possible where the two sisters are always separated by at least one man?
Options: a) 18! ×\times 2 b) 17! c) 17 ×\times 2! d) 12
Answer: 18! ×\times 2
Solution: Total circular arrangements = 18!×218! \times 2.
A number plate consists of two alphabets followed by two digits, with no repetition. How many possible combinations exist?
Options: a) 58500 b) 67600 c) 65000 d) 64320
Answer: 58500
Solution: Total choices = 26×25×10×9=5850026 \times 25 \times 10 \times 9 = 58500.
A alone can complete 1/4th of a task in 2 days. B alone can do 2/3rd of the task in 4 days. If all three work together, they finish in 3 days. What fraction of the work does C complete in 2 days?
Options: a) 1/12 b) 1/8 c) 1/16 d) 1/20
Answer: 1/12
Solution: Using work rates and summing contributions, we find C’s share in 2 days = 1/12.
How many prime numbers exist between 3 and 100 satisfying the form 4x+15y−14x + 15y – 1?
Options: a) 4 b) 11 c) 12 d) 7 e) None of the above
Answer: None of the above
Solution: Testing prime numbers, none satisfy this form.
Babla alone completes a task in 10 days, while Ashu does it in 15 days. The total wage for the task is Rs.5000. How much should Babla be paid if they work together for the entire duration?
Options: a) 2000 b) 4000 c) 5000 d) 3000
Answer: 3000
Solution: Using work-time ratio, Babla’s share = Rs.3000.
Practicing quantitative aptitude questions strengthens problem-solving skills for competitive exams. Mastering these topics ensures better performance in campus placements and aptitude tests.