HCL Aptitude Questions | 103572

** HCL Aptitude Questions**

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

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

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.

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

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) = 120C

Therefore

P(E) = n(E)/n(S) = 7140?800000

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

Sol:

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

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

(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.

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

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

(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 = 16C

2. 16C

3. Total 4 rounds will be played. Total number of matches required = 8 + 4 + 2 + 1 = 15

A. 190

B. 200

C. 210

D. 220

E. 225

Sol:

Formula: 15C

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

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}

Sol:

Possible ways of keeping atleast 1 blue and 1 red ball are drawing cans like (6,2) (5,3) (4,4)

(6,2) ?7C

(5,3) ?7C

(4,4) ?7C

70 + 210 + 175 = 455

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

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

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.

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.

Sol:

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

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

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.

Sol:

If a,b,c are the three sides of the triangle.

Then formula for Area = (s(sa)×(sb)×(sc))?????????????????????

Where s = (a+b+c)2=12×(30+c)

[Assume a = 20 ,b = 10]

Now,

Check the options.

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

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.

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