Infosys Quantitative Aptitude Questions With Answer, Here are the examples of questions that are asked in Infosys Placement Papers.
Questions Distributions:-
| Section | No. of Questions | Duration(minutes) |
|---|---|---|
| Quantitative Aptitude | 10 | 25 |
| Logical Reasoning | 15 | 35 |
| Verbal Ability | 40 | 35 |
| Total | 65 | 95 |
Infosys Quantitative Aptitude Question
1. Jayesh left point A for point B, 2 hours and 15 minutes later, Parul left A for B and arrived at B at the same time as Jayesh. if they both started simultaneously from A and B traveling towards each other, they would have met in 120 minutes. calculate the time required by slower one to reach A to B, if the ratio of speeds of the faster to slower is 3:1?
- a) 190 minutes
- b) 200 minutes
- c) 202.5 minutes
- d) 205 minutes
Show Answer
Answer= c)202.5 minutes
Let the distance between destination A and B is D km.
We know Parul is fast,
Let take the speeds of Jayesh and Parul are s and 3s respectively.
As the speeds are in the ratio of 1 : 3,
the time taken by them should be 3: 1.
Let the time taken by Jayesh and Parul be 3x, x.
But We know that 3x – x = 2 hours 15 min.
So 2x = 9/4 hours,
x = 9/8 hours.
So time taken by the slower one (Jayesh) takes 3x time = 3 x 9/8 = 27/8 hours = 202.5 minutes.
2. Alok can complete a piece of work in 2 days, Bushan in 4 days, Chetan in 9, and Dhiraj in 18 days. All the four of them formed groups of two such that difference is maximum between them to complete the work. What is the difference in the number of days they complete that work?
- a) 10 days.
- b) 14/3 says.
- c) 15 days.
- d) 16/3 days.
Show Answer
Answer: b) 14/3 days
If Chetan and Dhiraj form a pair and Alok and Bushan form a pair then the difference is maximum.
Now Chetan and Dhiraj together can complete the work = 9×189+18 = 6 days.
Alok and Bushan together can complete the work = 2×42+4 = 4/3 days.
Difference = 6 – 4/3 = 14/3 days.
3. 30 kgs of wheat costing Rs.18.40/-per/kg is mixed with how many kgs of wheat costing Rs.24/- per kg so that 15% profit can be obtained by selling the mixture at Rs.23/- per kg?
- a) 25 kg
- b) 17 kg
- c) 12 kg
- d) 9 kg
Show Answer
Answer: c) 12
Selling Price of 1 kg mixture = Rs.23.
Gain% = 15%.
Cost of 1 kg mixture = Rs.[(100/115) x 23] = Rs.20
Let the quantity of wheat costing Rs.24 is x kgs.
Using weighted average rule = x×24+30×18.4x+30=20
Solving we get x = 12
4. In a car race there are 5 persons named Joe, Kith, Larry, Monty, and Nathan who participated for 5 positions in how many ways can Monty finishes always before Nathan?
- a) 40
- b) 60
- c) 80
- d) 100
Show Answer
Answer b) 60
The total number of ways in which 5 persons can finish is 5! = 120 (there are no ties) Now in half of these ways Monty can finish before Nathan.
5. How many 4 digit numbers contain number ?
- a) 3170
- b) 3172
- c) 3178
- d) 3168
Show Answer
Answer d) 3168
Total number of 4 digit numbers are 9000 (between 1000 and 9999).
We find the numbers without any two in them. So total numbers are 8 x 9 x 9 x 9 = 5832.
So numbers with number two in them = 9000 – 5832 = 3168
6. Find out what will be the next number of the following sequence 7, 14, 55, 110, ?
- a) 112
- b) 118
- c) 121
- d) 127
Show Answer
Answer: c) 121
Next number = Previous number + Reverse of previous number So
7 ,7+7=14, 14+41 = 55, 55+55 = 110, 110+011 = 121
7. In Racing game rahul took part, where 1/5 ahead of him and 5/6 behind him excluding him. Find the total number of participants in the racing game?
- a) 21
- b) 23
- c) 27
- d) 31
Show Answer
Ans: 31
Let the total no of participants including Rahul = x
Excluding rahul=(x-1)
15(x−1)+56(x−1) = x
31x – 31=30x
8. Mehta’s family went for a vacation in Shimla. But it rained for 13 days when they were in Shimla. if it rains in the mornings, they had clear afternoons and vice versa. In all,t they are able to enjoy 11 mornings and 12 afternoons. How many days did Mehta family stay there totally?
- a) 15 Days.
- b) 18 Days.
- c) 21 Days.
- d) 24 Days.
Show Answer
Answer: b) 18
Total they enjoyed 11 mornings and 12 afternoons = 23 half days It rained for 13 days. So 13 half days.
So total days = (13 + 23) / 2 = 18)
9. A box consists of 12 tin cans such that 7 blue cans and 5 red cans. Find out in how many ways can we remove 8 tincans so that at least 1 blue can and 1 red can remain in the box?
- a) 275
- b) 300
- c) 325
- d) 455
Show Answer
Answer d) 455
Possible ways to draw 8 balls from the box which contains at least 1 blue and 1 red can after the drawing are (6,2) (5,3) (4,4).
For (6, 2) = ⇒7c6*5c2⇒7*10=70
For (5, 3) = ⇒7c5*5c3⇒21*10=210
(4, 4) = ⇒7c4*5c4⇒35*5=175
So Total ways = 70+210+175=455
10. If [x^(1/3)] – [x^(1/9)] = 60 then find the value of x.
- a) 49
- b) 51
- c) 59
- d) 78
Show Answer
Answer: a) 49
Let t = x1/9 So, t3−t=60
Therefore, (t-1) x t x (t + 1) = 60 =3 x 4 x 5.
therefore, t = x1/9 =4.
hence, x = 49
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