HCL Paper 1

1. In a cycle race there are 5 persons named as J, K, L, M, N participated for 5 positions so that in how many no. of ways can M make always before N?

ans: 60.

2. 15.002 × ? × 25.0210 = 7113.018
a. 19
b. 26
c. 11
d. 31
e. 35

3. Statement: K > I > T > E; O < R < K
Conclusion: I. R < E
II. O < T
a. Neither conclusion I nor II follows
b. Both conclusions I and II follows
c. Only conclusion II follows
d. Either conclusion I or II follows
e. Only conclusion I follows

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

Ans:
3 Digit no. + 4 digit no. = 5 digit no.. 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

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

Ans:
4 digit no. + 5 digit no. = 6 digit no.. 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 no.. 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

6. 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?

Ans:
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

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

Ans:
Observe the diagram. M = 1. S + 1 = a two digit no.. 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

8. 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 ?

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

9. 1, 1, 2, 3, 6, 7, 10, 11, ?

10. 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.

Ans:
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.

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

Ans:
Coding = Sum of position of alphabets x no. 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

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

Ans: Tough one as it has 10 variables in total. 4 digit no. + 5 digit no. = 6 digit no.. 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 no. 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

13. There are 16 hockey teams. find :
(1) no. of matches played when each team plays with each other twice.
(2) no. of matches played when each team plays each other once.
(3) no. of matches when knockout of 16 team is to be played

Ans:
1. no. 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 no. of matches required = 8 + 4 + 2 + 1 = 15

14. 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

Ans:
c) 210

15. 1, 11, 21, 1211, 111221, 312211, . . . . . what is the next term in the series?

Ans:
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

16. How many five digit no.s are there such that two left most digits are even and remaining are odd.

Ans:
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}

17. 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.

Ans:
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

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

Ans:56

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

Ans:
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?

Ans:
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?

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

Ans:
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 = 63×68 —(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.

Ans:
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?

Ans:
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,

25. Data Sufficiency Question:
a and b are two positive no.s. How many of them are odd?
I. Multiplication of b with an odd no. gives an even no..
II.a2 – b is even.

Ans:
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.

Ans:
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.

Ans:
Observe that each letter appeared once, twice, thrice ….

They form an arithmetic progression. 1+2+3……
We know that sum of first n natural no.s = 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?

Ans:
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

Ans:
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. Statements: All alphabets are no.s, some alphabets are digits
Conclusions: I. At least some digits are no.s
II. No digit is a no.
a. Either conclusion I or II follows
b. Neither conclusion I nor II follows
c. Only conclusion II follows
d. Only conclusion I follow
e. Both conclusions I and II follow

32. 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?

33. Find the unit digit of product of the prime no. up to 50 .

Ans:
Prime no. 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

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

Ans:
Answer 56.
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

35. A basket contains 3 blue, 5 black and 3 red balls. If 3 balls are drawn at random what is the probability that all are black?
a. 2/11
b. 1/11
c. 3/11
d. 8/33
e. None of these