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HCL Placement Paper For Freshers part-2

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by February 26, 2016 HCL, Placement Papers with Answers

hcl 111. If Ever + Since = Darwin then D + a + r + w + i + n is ?

Sol: Tough one as it has 10 variables in total. 4 digit number + 5 digit number = 6 digit number. So left most digit in the answer be 1. and S = 9, a = 0. Now we have to use trial and error method.

Here E appeared 3 times, I, R, N two times each. Now E + I or E + I + 1 is a two digit number with carry over. What could be the value of E and I here. 8 and 7 are possible. But from the second column, 8 + C = 7 or 17 not possible. Similarly with 7 and 6. If E = 5, then the remaining value can be filled like above.
5653 + 97825 = 103478
Answer is 23


12. There are 16 hockey teams. find :
(1) Number of matches played when each team plays with each other twice.
(2) Number of matches played when each team plays each other once.
(3) Number of matches when knockout of 16 team is to be played

Sol:
1. Number of ways that each team played once with other team = 16C2. To play with each team twice = 16 x 15 = 240
2. 16C2 = 120
3. Total 4 rounds will be played. Total number of matches required = 8 + 4 + 2 + 1 = 15

13. 15 tennis players take part in a tournament. Every player plays twice with each of his opponents. How many games are to be played?
A. 190
B. 200
C. 210
D. 220
E. 225

Sol:
Formula: 15C2 x 2. So 15 x (15 – 1) = 15 x 14 = 210

14. 1, 11, 21, 1211, 111221, 312211, . . . . . what is the next term in the series?

Sol:
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

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

Sol:
N = 4 x 5 x 5 x 5 x 5 = 2375
Where
4 cases of first digit {2,4,6,8}
5 cases of second digit {0,2,4,6,8}
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}

16. 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.

Sol:
Possible ways of keeping atleast 1 blue and 1 red ball are drawing cans like (6,2) (5,3) (4,4)
(6,2) ?7C6×5C2 ? 710 = 70
(5,3) ?7C5×5C3 ? 21 x 10 = 210
(4,4) ?7C4×5C4 ? 35 x 5 = 175
70 + 210 + 175 = 455

17. Find the 8th term in series?
2, 2, 12, 12, 30, 30, – – – – –

Sol:
11 + 1 = 2
22 – 2 = 2
32 + 3 = 12
42 – 4 = 12
52 + 5 = 30
62 – 6 = 30
So 7th term = (72 + 7) = 56 and 8th term = ({82} – 8) = 56
Answer is 56

18. Rahul took a part in cycling game where 1/5 ahead of him and 5/6 behind him then total number of participants =

Sol:
Let x be the total number of participants including Rahul.
Excluding rahul = (x – 1)
15(x–1)+56(x–1) = x
31x – 31 = 30x
Total number of participants x = 31

19. Data sufficiency question:
What are the speeds two trains travels with 80 yards and 85 yards long respectively? (Assume that former is faster than later)
a) they take 75 seconds to pass each other in opposite direction.
b) they take 37.5 seconds to pass each other in same direction

Sol:
Let the speeds be x and y
When moves in same direction the relative speed,
x – y = (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.

20. Reversing the digits of father’s age we get son’s age. One year ago father was twice in age of that of his son? find their current ages?

Sol:
Let father’s age = 10x + y
Son’s 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 doesn’t give a realistic age (like father’s age 120 etc)
So, this has to be solution.Hence father’s age = 73.
Son’s age = 37.

(94)

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