# Infosys Latest Placement Paper 1

**Ans: x x**

**Sol:**It seems there is some problem with this question.

Let the distance between A and B is D km. As Paul is faster, take the speeds of Jake and Paul are s and 3s kmph.

As the speeds are in the ratio of 1 : 3, times taken by them should be 3 : 1. Take the times taken by them are 3x , x. But We know that 3x – x = 2 hour 15 min. So 2x = 9/4 hours, x = 9/8 hours. So time taken by the slower one (Jake) takes 3x time = 3 x 9/8 = 27/8 hours = 202.5 minutes.

(Or)

Take Jake speed = j and Paul = p kmph.

Now given that Dj−DpDj−Dp = 2 hr 15 min = 214214 hrs = 9/4 hrs

Also both of them together covered D distance in 2 hours. So Dj+Dp=2Dj+Dp=2

Adding these two equations will give us 2Dj=94+2=1742Dj=94+2=174 = 4 hours 15 minutes.

So in the above problem, some part is redundant.

2. A completes a work in 2 days, B in 4 days, C in 9 and D in 18 days. They form group of two such that difference is maximum between them to complete the work. What is difference 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+189×189+18 = 6 days.

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

Difference = 6 – 4/3 = 14/3 days.

**Ans: D**

**Ans: 252**

**Ans: 12**

**Ans: 121**

**Ans: 24**

**Ans: 373**

9. There are 1000 junior and 800 senior students in a class.And there are 60 sibling pairs where each pair has 1 junior and 1 senior. One student is chosen from senior and 1 from junior randomly.What is the probability that the two selected students are from a sibling pair?**Ans: 714 / 80000**

Sol:

Junior students = 1000

Senior students = 800

60 sibling pair = 2 x 60 = 120 student

One student chosen from senior = 800C1800C1=800

One student chosen from junior=1000C11000C1=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)=120C2120C2=7140

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

10. 161?85?65?89 = 100, then use + or – in place of ? and take + as m,- as n then find value of m-n.**Ans: – 1**

Sol:

161 – 85 – 65 + 89 = 100

so m’s =1, n’s = 2 => (m – n)= – 1

11. In a cycle race there are 5 persons named as J,K,L,M,N participated for 5 positions so 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.

12. Rahul took a part in cycling game where 1/5 ahead of him and 5/6 behind him excluding 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)15(x−1)+56(x−1) = x

31x – 31=30x

Total no. of participants x =31

13. If a refrigerator contains 12 cans such that 7 blue cans and 5 red cans. In how many ways can we remove 8 cans so that atleast 1 blue can and 1 red can remains in the refrigerator.

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

14. There are 16 people, they divide into four groups, now from those four groups select a team of three members,such that no two members in the team should belong to same group.**Ans: 256**

Sol:

We can select any three of the 4 groups in 4C34C3 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

15. How many five digit numbers are there such that two left most digits are even and remaining 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

16. 7 people have to be selected from 12 men and 3 women, Such that no two women can 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 12C6×3C112C6×3C1 = 2772

17. Tennis players take part in a tournament. Every player plays twice with each of his opponents. How many games are to be played?**Ans: 210**

Sol:

We can select two teams out of 15 in 15C215C2 ways. So each team plays with other team once. Now to play two games, we have to conduct 15C215C2 x 2 = 210 games.

18. 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 0. So the entire product has unit digit 0.

19. If [x^(1/3)] – [x^(1/9)] = 60 then find the value of x. **Ans: 4949**

Sol:

Let t = x1/9×1/9

So,

t3−t=60t3−t=60

Therefore, (t-1) x t x (t + 1) = 60 =3 x 4 x 5.

therefore, t = x1/9×1/9 =4.

hence, x = 4949

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