Hexaware Quantitative Aptitude Questions And Answers
Hexaware Quantitative Aptitude MCQs : This section focuses on "Quantitative Aptitude" for Hexaware Exam. These Quantitative Aptitude MCQs are asked in previous Hexaware placements/recruitment exams and will help you to prepare for upcoming Hexaware drives.
1. I make a profit of 20% by selling an article. What would be the profit percent if it were calculated on the selling price instead of the cost price?
Explanation: Let us assume
Cost Price = 100
Then 20% of profit means
Selling Price is = 120.
Now, Profit %, if it was calculated on SP, will be
20/120 × = 16.67%.
2. If log10 (87.5) = 1.9421, then the number of digits in (875)^10 is?
Explanation: X = (875)^10 = (87.5 x 10)^10
Therefore, log10 (X) = 10(log10 (87.5) + 1)
= 10(1.9421 + 1)
= 10(2.9421) = 29.421
X = antilog(29.421)
Therefore, number of digits in X = 30.
3. How many parallelograms will be formed if 7 parallel horizontal lines intersect 6 parallel vertical lines?
Explanation: Parallelograms are formed when any two pairs of parallel lines (where each pair is not parallel to the other pair) intersect. Hence, the given problem can be considered as selecting pairs of lines from the given 2 sets of parallel lines. Therefore, the total number of parallelograms formed = 7C2 x 6C2 = 315
4. Prasanth bought 20kg of rice at rupees 30 per kg and 40 kg of rice at 35 rupees per kg. Now he sold the entire lot at 45 rupees per kg. Find the amount of loss and profit made by Prasanth.
Explanation: Total cost price = 20*30 + 40*35 = 2000
Total selling price = 45*60 = 2700
Hence, Profit made by Prasanth = 700 Rupees.
5. Without any stoppage, a person travels a certain distance at an average speed of 42 km/h, and with stoppages he covers the same distance at an average speed of 28 km/h. How many minutes per hour does he stop?
Explanation: Let the total distance to be covered is 84 kms.
Time taken to cover the distance without stoppage = 84 / 42 hrs = 2 hrs
Time taken to cover the distance with stoppage = 84 / 28 = 3 hrs.
Thus, he takes 60 minutes to cover the same distance with stoppage.
Therefore, in 1 hour he stops for 20 minutes.
6. If 12% of x is equal to 6% of y, then 18% of x will be equal to how much % of y ?
Explanation: We have ,
12% of X = 6% of Y
=> 2% of X = 1% of Y
=>(2 x 9)% of X = ( 1 x 9)% of Y
Thus, 18% of X = 9% of Y.
7. Walking at the rate of 4 kmph a man cover a certain distance in 2 hr 45 min. Running at a speed of 16.5 kmph the man will cover the same distance in?
Explanation: Distance = Speed × time
Here time = 2hr 45 min = 11/4 hr
Distance = 4×11/4=11 km
New Speed =16.5 kmph
Therefore time = Distance/Speed=11/16.5=40 min.
8. A man walks 6 km at a speed of 1 1/2 kmph, runs 8 km at a speed of 2 kmph and goes by bus another 32 km. Speed of the bus is 8 kmph. Find the average speed of the man.
Explanation: Man walked 6 km at 1.5 kmph, again he walked 8 km at speed of 2 kmph and 32 km at a speed of 8kmph time taken indivisually:
=> 6/1.5 = 4 m
=> 8/2 = 4 m
=> 32/8 = 4 m
Average speed of man= total distance/ total time
=> 46/12 = 3 (5/6)
9. A teacher has to choose the maximum different groups of three students from a total of six students. Of these groups, in how many groups there will be included in a particular student?
Explanation: If students are A, B, C, D, E and F; we can have 6C3 groups in all. However, if we have to count groups in which a particular student (say A) is always selected we would get 5C2 = 10 ways of doing it.
10. Two trains starting at the same time from 2 stations 200 km apart and going in the opposite direction cross each other at a distance of 110 km from one of the stations. What is the ratio of their speeds?
Explanation: By Analyzing the given information
In the same time, they cover 110 km and 90 km respectively.
For the same time, speed and distance are inversely proportional.
So ratio of their speed = 110: 90 = 11: 9.