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**DIRECTIONS: Read the following given below and answer the questions that follow.**

i) Eleven students Anwar, Baham, Chetan, Dayal, Eshwar, Farooq, Gajendra, Hariom, Inayat,

Jatin and Kishore are sitting on a bench in a lecture room facing towards the teacher.

ii) Dayal is towards the left of Farooq and second to the right of Chetan.

iii) Eshwar is second to the left of Anwar and sitting on one end.

iv) Jatin is the neighbourer of Anwar and Baham and is third to the left of

Gajendra.

v) Hariom is next to left of Dayal and is third to the right of Inayat.

**1) Which two students are sitting in two ends?**a) Kishore and Dayal b) Eshwar and Dayal

c) Eshwar and Farooq d) Kishore and Farooq

**2) Which group of students are sittting just next to the right of Gajendra?**a) Chetan, Hariom, Dayal, Eshwar b) Chetan, Hariom, Inayat, Baham

c) Chetan, Hariom, Inayat, Dayal d) none of the above

**3) Who is sitting just in the middle?**a) Inayat b) Chetan c) Baham d) Jatin

**4) Which of the five given statements is unnecessary?**a) (i) and (ii) b) (i) c) (iii) d) All four are necesssary

**5) Which of the following is correct?**a) Eshwar and Anwar are the closest neighbourers of Jatin

b) Gajendar, Inayat and Baham are sitting to the left of Chetan

c) Hariom is in the middle of the line

d) Anwar, Kishore and Eshwar are to the right of Jatin

**Ques. A man sold two cows for Rs. 210 at a total profit of 5 %. He sold one cow at a lossof 10% and another at a profit of 10%. What is the price of each cow?**

- 50, 100
- 100, 50
- 150, 100
- 50, 150
**Ans**:50 and 150

**Solution**: after profit, price is 210

so cost price of 2 cows = 200

x+y=200

selling price of 2 cows =210

0.9x + 1.1y =210 //1 cow sold at 10 % loss and another is at 10% profit

solving both equations x=50 and y=150

so one cow is of x = Rs. 50

and Other is of 200-x = (200-50) = Rs. 150

**Ques. In a class there are less than 500 students . when it is divided by 3 it gives awhole number. Similarly when it is divided by 4,5 or 7 gives a whole number. find theno. of students in the class**

- 430
- 420
- 410
- 440

**A completes a work in 2 days, B in 4 days, C in 9 and D in 18 days. They form groupof two such that difference is maximum between them to complete the work. What isdifference in the number of days they complete that work?**

**Ans:** 14/3 days.**Sol:** If C and D form a pair and A and B form a pair the difference is maximum. Now C

and D together can complete the work = 9×189+18 = 6 days.

A and B together can complete the work = 2×42+4 = 4/3 days.

Difference = 6 – 4/3 = 14/3 days.

**How many 4 digit numbers contain number 2.**

a. 3170

b. 3172

c. 3174

d. 3168**Ans: **D**Sol:**

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

**How many three digit numbers abc are formed where at least two of the three digitsare same.**

**Ans: 252**

Sol:

Total 3 digit numbers = 9 x 10 x 10 = 900

Sol:

Total number of 3 digit numbers without repetition = 9 x 9 x 8 = 648

So number of three digit numbers with at least one digit repeats = 900 – 648=252

**How many kgs of wheat costing Rs.24/- per kg must be mixed with 30 kgs of wheatcosting Rs.18.40/- per kg so that 15% profit can be obtained by selling the mixture atRs.23/- per kg? Ans: 12**

Sol:

S.P. of 1 kg mixture = Rs.23. Gain = 15%.

C.P. 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 = 12What is the next number of the following sequence 7, 14, 55, 110, ….?Ans: 121**

Sol:

Next number = Previous number + Reverse of previous number So

7 ,7+7=14, 14+41 = 55, 55+55 = 110, 110+011 = 121

**How many numbers are divisible by 4 between 1 to 100**Ans: 24

Sol: There are 25 numbers which are divisible by 4 till 100. (100/4 = 25). But we should

not consider 100 as we are asked to find the numbers between 1 to 100 which are

divisible by 4. So answer is 24.

**(11111011)2= ()8**

Ans: 373

Sol: 11111011)2=(251)10=(373)8 or

You can group 3 binary digits from right hand side and write their equivalent octal form.

**There are 1000 junior and 800 senior students in a class.And there are 60 siblingpairs where each pair has 1 junior and 1 senior. One student is chosen from senior and1 from junior randomly.What is the probability that the two selected students are from asibling pair?**

Ans: 714 / 80000

Sol:

Junior students = 1000

Senior students = 800

60 sibling pair = 2 x 60 = 120 student

One student chosen from senior = 800C1

=800

One student chosen from junior=1000C1=1000

Therefore, one student chosen from senior and one student chosen from junior n(s) =

800 x 1000=800000

Two selected students are from a sibling pair n(E)=120C2=7140

therefore,P(E) = n(E) / n(S)=7140/800000 = 714/80000

**161?85?65?89 = 100, then use + or – in place of ? and take + as m,- as n then findvalue of m-n.**

Ans: -1

Sol:

161 – 85 – 65 + 89 = 100

so m’s =1, n’s = 2 => (m – n)= – 1

**In a cycle race there are 5 persons named as J,K,L,M,N participated for 5 positionsso that in how many number of ways can M finishes always before N?**

Ans: 60

Sol: Total number of ways in which 5 persons can finish is 5! = 120 (there are no ties)

Now in half of these ways M can finish before N.

**Rahul took a part in cycling game where 1/5 ahead of him and 5/6 behind himexcluding him. Then total number of participants are**

Ans: 31

Sol:

Let the total no of participants including Rahul = x Excluding rahul=(x-1)

15(x−1)+56(x−1) = x

31x – 31=30x

Total no. of participants x =31

**If a refrigerator contains 12 cans such that 7 blue cans and 5 red cans. In how manyways can we remove 8 cans so that atleast 1 blue can and 1 red can remains in therefrigerator.**

Ans:

Sol:

Possible ways to draw 8 balls from the refrigerator which contains atleast 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

For (4, 4) = ⇒7c4

*5c4⇒35*5=175 So Total ways = 70+210+175=455

**There are 16 people, they divide into four groups, now from those four groups selecta team of three members,such that no two members in the team should belong to samegroup.**

Ans: 256

Sol:

We can select any three of the 4 groups in 4C3

ways. Now from each of these groups we can select 1 person in 4 ways. So total ways

= 4 x 4 x 4 x 4 = 256

**How many five digit numbers are there such that two left most digits are even andremaining are odd and digit 4 should not be repeated.**

Ans: 2375

Sol:

We have

4 cases of first digit {2,4,6,8}

5 cases of second digit {0,2,4,6,8}

But 44 is one case we have to omit. So total ways for leftmost two digits are 4 x 5 – 1 =

19 5 cases of third digit {1,3,5,7,9}

5 cases of fourth digit {1,3,5,7,9}

5 cases of fifth digit {1,3,5,7,9}

So total ways = 19 x 5 x 5 x 5 = 2375

**7 people have to be selected from 12 men and 3 women, Such that no two womencan come together. In how many ways we can select them?**

Ans: 2772

Sol:

We can select only one woman, and remaining 6 from men. So 12C6×3C1

= 2772

**Tennis players take part in a tournament. Every player plays twice with each of hisopponents. How many games are to be played?**

Ans: 210

Sol:

We can select two teams out of 15 in 15C2 ways. So each team plays with other team

once. Now to play two games, we have to conduct 15C2 x 2 = 210 games.

**Find the unit digit of product of the prime number up to 50 .**Ans: 0

Sol: No need to write all the primes upto 50. There are two primes 2, 5 gives unit digit of

So the entire product has unit digit 0.

If [x^(1/3)] – [x^(1/9)] = 60 then find the value of x.

Ans: 49

Sol:

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

**A family X went for a vacation. Unfortunately it rained for 13 days when they werethere.But whenever it rained in the mornings, they had clear afternoons and vice versa. In allthey enjoyed 11 mornings and 12 afternoons. How many days did they stay theretotally?**

Ans: 18

Sol:

Total they enjoyed on 11 mornings and 12 afternoons = 23 half days It rained for 13

days. So 13 half days.

So total days = (13 + 23) / 2 = 18

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