# HCL Placement Paper For Freshers part-2

**11. 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(x1)+56(x1) = 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 = (8580)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)