Placement Papers Infosys Paper - I

As we see the series 1,4,9,18......
we see that $ 2^{1}$-1 = 1 which is first term
$ 2^{2}$-0 = 4 which is second term
$2^{3}$+1 = 9 third term
$ 2^{4}$+2 = 18 fourth term
$2^{5}$+3 = 35 fifth term
$2^{6}$+4 = 68 sixth term
$2^{7}$+5 = 133 seventh term
$ 2^{8}$+6 = 262 eighth term
so answer is 262

Say M came first. The remaining 4 positions can be filled in 4! = 24 ways.

Now M came in second. N can finish the race in 3rd, 4th or 5th position. So total ways are 3 x 3! = 18.

M came in third. N can finish the race in 2 positions. 2 x 3! = 12.

M came in second. N can finish in only one way. 1 x 3! = 6

Total ways are 24 + 18 + 12 + 6 = 60.

Total ways of finishing the race = 5! = 120. Of which, M comes before N in half of the races, N comes before M in half of the races. So 120 / 2 = 60.

Junior student = 1000

Senior student = 800

60 sibling pair = 2 x 60 = 120 student

Probability that 1 student chosen from senior = 800

Probability that 1 student chosen from junior = 1000

Therefore,1 student chosen from senior and 1 student chosen from junior

n(s) = 800 x 1000 = 800000

Two selected student are from a sibling pair

n(E) = 120C2 = 7140

Therefore

P(E) = n(E)/n(S) = 7140⁄800000

50 p : 1 coin, 25 p : 2 coins, 10 p: 1 coin, 5 p : 1 coin, Total: 1.15 p

The given pattern is (Prime number - consecutive numbers starting with 1).

1 = 2 - 1

1 = 3 - 2

2 = 5 - 3

3 = 7 - 4

6 = 11 - 5

7 = 13 - 6

10 = 17 - 7

11 = 19 - 8

14 = 23 - 9

I. The Lorry reaches Mysore by 3 PM so it meets all the Lorries which starts after 6 a.m and before 3 p.m. So 9 lorries. Also the Lorry which starts at night 10 p.m on the previous day at Mysore reaches Bangalore in morning 7 a.m. So it also meets that Lorry. So the Lorry which starts at 6:00 am will cross 10 Lorries.

II. The lorry which has started at 6 p.m reaches destination by 3 a.m. Lorries which start at the opposite destination at 10 am reaches its destination at 7 pm. So all the lorries which starts at 10 am to 10 pm meets this lorry . So in total 13.

Coding = Sum of position of alphabets x Number of letters in the given word

GOOD = (7 + 15 + 15 + 4 ) x 4 = 164

BAD = (2 + 1 + 4) x 3 = 21

UGLY = (21 + 7 + 12 + 25) x 4 = 260

So, JUMP = (10 + 21 + 13 + 16) x 4 = 240

1. Number of ways that each team played once with other team = 16C216C2. To play with each team twice = 16 x 15 = 240

2. 16C216C2 = 120

3. Total 4 rounds will be played. Total number of matches required = 8 + 4 + 2 + 1 = 15

Formula: 15C215C2 x 2. So 15 x (15 - 1) = 15 x 14 = 210

We can understand it by writing in words

One

One time 1 that is = 11

Then two times 1 that is = 21

Then one time 2 and one time 1 that is = 1211

Then one time one, one time two and two time 1 that is = 111221

And last term is three time 1, two time 2, and one time 1 that is = 312211

So our next term will be one time 3 one time 1 two time 2 and two time 1

13112221 and so on.

N=(4*5-1)*5*5*5=2375
where
4 cases of first digit {2,4,6,8}
5 cases of second digit {0,2,4,6,8}
1 case of 44 for two leftmost digit
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}

13 - 46 - 8 - 180 + 210 + 75 = 64

m = 3

n = 4

m - n = - 1.

Possible ways of keeping atleast 1 blue and 1 red ball are drawing cans like (6,2) (5,3) (4,4)

(6,2) =>7C6×5C27C6×5C2 => 710 = 70

(5,3) =>7C5×5C37C5×5C3 => 21 x 10 = 210

(4,4) =>7C4×5C47C4×5C4 => 35 x 5 = 175

70 + 210 + 175 = 455.

$1^{2}+1$=2
$2^{2}-2$=2
$3^{2}+3$=12
$4^{2}-4$=12
$5^{2}+5$=30
$6^{2}-6$=30
So, 7th term is $7^{2}+7$=56
and the 8th term is=$8^{2}-8$=56
So, the answer is 56.

2121 - 1 = 1

2222 + 0 = 4

2323 + 1 = 9

2424 + 2 = 18

2525 + 3 = 35

So 2626 + 4 = 68, 2727 + 5 = 133, 2828 +6 = 262

Answer is 68, 133, 262.

Let x be the total number of participants including Rahul.

Excluding rahul = (x - 1)

15(x-1)+56(x-1)15(x-1)+56(x-1) = x

31x - 31 = 30x

Total number of participants x = 31.

Let the speeds be x and y

When moves in same direction the relative speed,

x - y = (85-80)37.5(85-80)37.5 = 0.13 - - - - - (I)

When moves in opposite direction the relative speed, x + y = 165/75 = 2.2 - - - - (II)

Now, equation I + equation II gives, 2x = 0.13 + 2.2 = 2.33 => x = 1.165

From equation l, x - y = 0.13 => y = 1.165 - 0.13 = 1.035

Therefore the speeds are 1.165 yards/sec and 1.035 yards/sec.

Let fathers age = 10x + y

Sons age = 10y + x (As, it is got by reversing digits of fathers age)

At that point

(10x + y) - 1 = 2{(10y + x) - 1}

=> x = (19y - 1)/8

Let y = 3 then x = 7.

For any other y value, x value combined with y value doesnt give a realistic age (like fathers age 120 etc)

So, this has to be solution.Hence fathers age = 73.

Sons age = 37.

Assume that height of the hill is 440 miles.

Let speed of Jack when going up = x miles/minute

and speed of Jill when going up = y miles/minute

Then speed of Jack when going down = 1.5x miles/minute

and speed of Jill wen going up = 1.5y miles/minute

Jack met jill 20 miles from the top. So Jill travelled 440 - 20 = 420 miles.

Time taken for Jack to travel 440 miles up and 20 miles down = Time taken for Jill to travel 420 miles up

440x+201.5x=420y440x+201.5x=420y

681.5x=420y681.5x=420y

68y = 63x

y = 63x6863x68 ---(1)

Time taken for Jack to travel 440 miles up and 440 miles down = Time taken for Jill to travel 440 miles up and 440 miles down - 1

440x+4401.5x=440y+4401.5y440x+4401.5x=440y+4401.5y - 1

440×53(1y−1x)=1440×53(1y−1x)=1-----(2)

Substitute (2) in (1) we get

x = 440×5×53×6340×5×53×63

t = 440×53(1x)440×53(1x)

t = 12.6min.

D because A is not first neither C and B is not the tallest person. The only person will be first is D.

So option (C). We can answer this question using both the statements together.