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June 17th, 2016, 08:53 AM
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Join Date: Mar 2012
Re: HCL Aptitude Questions And Answers

The HCL Placement Drive is Consist of the Three Rounds that are as follow :

Aptitude test
Group Discussion
Interview

The Aptitude question paper of HCL is given below :
1. What is the 8th term in the series 1,4, 9, 25, 35, 63, . . .

Sol:
1, 4, 9, 18, 35, 68, . . .
The pattern is
1 = 21 – 1
4 = 22 – 0
9 = 23 + 1
18 = 24 + 2
35 = 25 + 3
68 = 26 + 4
So 8th term is 28 + 6 = 262

2. USA + USSR = PEACE ; P + E + A + C + E = ?

Sol:
3 Digit number + 4 digit number = 5 digit number. So P is 1 and U is 9, E is 0.
Now S repeated three times, A repeated 2 times. Just give values for S. We can easily get the following table.

USA = 932
USSR = 9338
PEACE = 10270
P + E + A + C + E = 1 + 0 + 2 + 7 + 0 = 10

3. 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 make always before N?

Sol:
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.

Shortcut:
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.

4. If POINT + ZERO = ENERGY, then E + N + E + R + G + Y = ?

Sol:
4 digit number + 5 digit number = 6 digit number. So E = 1, P = 9, N = 0
Observe R + 0 = G. But R = G not possible. 1 + R = G possible. So R and G are consecutive. G > R.
1 + I = R, So I and R are consecutive. R > I. i.e., G > R > I. and G, R, I are consecutive. Now O + T should give carry over and O + Z also give carry over. So O is bigger number. Now take values for G, R, I as 8, 7, 6 or 7, 6, 5 etc. and do trial and error.

POINT = 98504, ZERO = 3168 and ENERGY = 101672.
So E + N + E + R + G + Y = 1 + 0 + 1 + 6 + 7 +2 = 17

5. 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.1 student is chosen from senior and 1 from junior randomly.What is the probability that the two selected students are from a sibling pair?

Sol:
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

6. SEND + MORE = MONEY. Then what is the value of M + O + N + E + Y ?

Sol:
Observe the diagram. M = 1. S + 1 = a two digit number. So S = 1 and O cannot be 1 but 0. Also E and N are consecutive. Do trial and error.

SEND = 9567, MORE = 1085, MONEY = 10652
SO M + O + N + E + Y = 1 + 0 + 6 + 5 + 2 = 14

7. A person went to shop and asked for change for 1.15 paise. But he said that he could not only give change for one rupee but also for 50p, 25p, 10p and 5p. What were the coins he had ?

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

8. 1, 1, 2, 3, 6, 7, 10, 11, ?

Sol:
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

9. A Lorry starts from Banglore to Mysore At 6.00 a.m, 7.00 a.m, 8.00 a.m.....10 p.m. Similarly another Lorry on another side starts from Mysore to Banglore at 6.00 a.m, 7.00 a.m, 8.00 a.m.....10.00 p.m. A Lorry takes 9 hours to travel from Banglore to Mysore and vice versa.
(I) A Lorry which has started At 6.00 a.m will cross how many Lorries.
(II) A Lorry which has started At 6.00 p.m will cross how many Lorries.

Sol:
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.

10. GOOD is coded as 164 then BAD coded as 21.if ugly coded as 260 then JUMP?

Sol:
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

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}


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