HCL Aptitude Questions

                                                       HCL Aptitude Questions

HCL solved aptitude questions for preparation :

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}

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.

21. The hour hand lies between 3 and 4. Tthe difference between hour and minute hand is 50 degree.What are the two possible timings?

Sol:
The angle between the hour hand and minute hand at a given time H:MM is given by
θ = 30×H – 211×MM
The time after H hours, hour hand and minute hand are at
MM = | 211×((30×H)±θ) |
given H = 3, MM = 50
Substituting the above values in the formula
θ = 8011, 28011

22. Jack and Jill went up and down a hill. They started from the bottom and Jack met Jill again 20 miles from the top while returning. Jack completed the race 1 min a head of Jill. If the hill is 440 miles high and their speed while down journey is 1.5 times the up journey. How long it took for the Jack to complete the race ?

Sol:
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

Case 1 :
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=420y
681.5x=420y
68y = 63x
y = 63x68 ---(1)

Case 2 : 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+4401.5y – 1
440×53(1y−1x)=1-----(2)

Substitute (2) in (1) we get
x = 440×5×53×63
t = 440×53(1x)
t = 12.6min

23. Data Sufficiency question:
A, B, C, D have to stand in a queue in descending order of their heights. Who stands first?
I. D was not the last, A was not the first.
II. The first is not C and B was not the tallest.

Sol:
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.

24. One of the longest sides of the triangle is 20 m. The other side is 10 m. Area of the triangle is 80 m2. What is the another side of the triangle?

Sol:
If a,b,c are the three sides of the triangle.
Then formula for Area = (s(s–a)×(s–b)×(s–c))−−−−−−−−−−−−−−−−−−−−√
Where s = (a+b+c)2=12×(30+c)
[Assume a = 20 ,b = 10]
Now,
Check the options.

25. Data Sufficiency Question:
a and b are two positive numbers. How many of them are odd?
I. Multiplication of b with an odd number gives an even number.
II.a2 – b is even.

Sol:
From the 1st statement b is even, as when multiplied by odd it gives even
a2 – b = even
⇒ a is even
Here none of a and b are odd

26. Mr. T has a wrong weighing pan. One arm is lengthier than other. 1 kilogram on left balances 8 melons on right, 1 kilogram on right balances 2 melons on left. If all melons are equal in weight, what is the weight of a single melon.

Sol:
Let additional weight on left arm be x.
Weight of melon be m
x + 1 = 8 x m - - - - - - (1)
x + 2 x m = 1 - - - - - - (2)
Solving 1 & 2 we get.
Weight of a single Melon = 200 gm.

27. a, b, b, c, c, c, d, d, d, d, . . . . . . Find the 288th letter of this series. 

Sol:
Observe that each letter appeared once, twice, thrice ....

They form an arithmetic progression. 1+2+3......
We know that sum of first n natural numbers = n(n+1)²
So n(n+1)² ≤ 288
For n = 23, we get 276. So for n = 24, the given series crosses 288.
Ans is X

28. There are three trucks A, B, C. A loads 10 kg/min. B loads 13 1/3 kg/min. C unloads 5 kg/min. If three simultaneously works then what is the time taken to load 2.4 tones?

Sol:
Work done in 1 min =10 + 403 – 5= 553 kg/min
For 1 kg = 3/55 min
For 2.4 tonnes = 3/55 x 2.4 x 1000 = 130 mins = 2hrs 10min

29. A person is 80 years old in 490 and only 70 years old in 500 in which year is he born?
a) 400
b) 550
c) 570
d) 440

Sol:
He must have born in BC 570
Hence in BC 500 he will be 70 years
And in BC 490 he will be 80 years

30. Lucia is a wonderful grandmother and her age is between 50 and 70. Each of her sons have as many sons as they have brothers. Their combined ages give Lucia's present age.what is the age?

Sol:
The question basically states that if Lucia were to have say 10 sons, then each son would have 9 sons (Lucia's grandsons – since each son has 9 brothers). So the total in this case would be 9×10 grandsons + 10 sons = 100.
Let us assume Lucia has got x sons. Now each son has (x - 1) sons. So total = x + (x - 1) x. For x = 8 we get 64 which is in between 50 and 60. ( 7 x 8 grandsons + 8 sons = 64 )

31. 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?

Sol:
Clearly 11 mornings and 12 afternoons = 23 half days
since 13 days raining means 13 half days.
so 23 – 13 =10 half days ( not affected by rain )
so 10 half days = 5 full days
Total no. of days = 13 + 5 = 18 days.

32. Find the unit digit of product of the prime number up to 50 .

Sol:
Prime number up to 50 are
2,3,5,7,11,...,43,47
Product = 2×3×5×7×11×−−−×43×47
There's a term 2×5=10
So unit digit of product = 0

33. Complete the series..
2 2 12 12 30 30 ?

Sol:
Answer is 56.
It follows the series as:
1 x 2 = 2
2 x 1 = 2
3 x 4 = 12
4 x 3 = 12
5 x 6 = 30
6 x 5 = 30
7 x 8 = 56
This is the required number for the series.

34. An escalator is descending at constant speed. A walks down and takes 50 steps to reach the bottom. B runs down and takes 90 steps in the same time as A takes 10 steps. How many steps are visible when the escalator is not operating?

Sol:
Lets suppose that A walks down 1 step / min and
escalator moves n steps/ min
It is given that A takes 50 steps to reach the bottom
In the same time escalator would have covered 50n steps
So total steps on escalator is 50 + 50n.
Again it is given that B takes 90 steps to reach the bottom and time 
taken by him for this is equal to time taken by A to cover 10 steps i.e 
10 minutes. So in this 10 min escalator would have covered 10n steps.
So total steps on escalator is 90 + 10n

Again equating 50 + 50n = 90 + 10n we get n = 1
Hence total number of steps on escalator is 100.

35. How many ways can one arrange the word EDUCATION such that relative positions of vowels and consonants remains same?

Sol:
The word EDUCATION is a 9 letter word with none of letters repeating
The vowels occupy 3,5,7th & 8th position in the word & remaining five positions are occupied by consonants.
As the relative position of the vowels & consonants in any arrangement should remain the same as in the word EDUCATION 
The four vowels can be arranged in 3rd,5th,7th & 8th position in 4! ways.
similarly the five consonants can be arranged in 1st ,2nd ,4th, 6th & 9th position in 5! ways
Hence the total number of ways = 5!×4!=120×24=2880