HCL Time & Work Problems

Solved Time & Work Problems For HCL Aptitude Tests

Dear Reader,

Below are three problems on time and work which you can expect in HCL and other companies placement tests.

Question 1

In a factory, there are two machines A and B. A machine can produce 1000 watts power in 5 hours and by running both the machines together can produce same watts power in 3 hours. Find the time taken by B machine alone to produce same watts of power.

a) 5 hrs b) 7 hrs & 30 mins c) 6 hrs d) 6 hrs & 30 mins

Answer : b) 7 hrs & 30 mins

Solution :

A type machine can produce 1000 watts power in 5 hours.
Then A's 1 hour work = 1/5.
Both A and B can produce in 3 hours.
And 1 hour work of (A + B)'s = 1/3.
B's 1 hour work = 1/3 = 1/5 = 2/15.
Therefore, machine B can produce 1000 watts power in 15/2 hours.
i.e., 15/2 hours = 7.5 hours = 7 hours and 30 minutes.

Question 2

There is a building with 4 floors and each floor has equal area. There are three sweepers P, Q and R. P can clean 1 floor in 10 hours, Q can clean 8/5 of the floors in 40 hours and R can clean 4/3 of the floors in 13 hours. Then who will clean all the 4 floors at first?

a) R b) R & P c) Q d) Q & P

Answer : a) R

Solution :

P can sweep 1 floor in 10 hours.
Then 4 floors will be cleaned by P in = 10 x 4 = 40 hours.

Q can sweep 8/5 part in 40 hours.
Then 1 floor will be done by Q in 40/(8/5) = 25 hours.
And 4 floors will be done by Q in 25 x 4 = 100 hours.

R can clean 4/3 part of the floors in 13 hours.
Then 1 floor will be cleaned by R in 13/(4/3) = 13 x 3/4 hours.
And 4 floors will be done by R in 13 x 3/4 x 4 = 13 x 3 = 39 hours.

Thus P, Q and R takes 40, 100 and 39 hours respectively.
Hence the sweeper R will complete the work at first.

Question 3

A man and his son planned to paint their own house. The man alone can do in 20 days. He paints for 4 days and then his son completed in 8 days. Find how many days will they take working together?

a) 12 days b) 11 days c) 7 days d) 8 days

Answer : c) 7 days

Solution :

Man's 1 day work = 1/20
Then 4 days work = 4/20 = 1/5
Remaining work = 1 - 1/5 = 4/5

Now, 4/5 work is done by the son in 8 days.
Then, whole work will be done by him in (8 x 5/4) days = 10 days.
Then his 1 day work = 1/10

The man and his son's 1 day work = 1/10 + 1/10 = 3/10
They will complete the work together in 20/3 days = 6 2/3 days = 7 days(nearly).
Hence the answer is 7 days.

Infosys Sample Clock Problems

Clock Problems for infosys placement

Dear Reader,

Below are three problems based on clocks, which you can expect in infosys and other companies placement test.

Question 1

Three analog clocks were set to correct time. First one run with the exact time. Second one loss one hour per day. Third one gains one hour per day. After how many days will they show the same time?

a)180 days & 10 hours b)287 days & 12 hours c)212 days d)199 days

Answer : b) 287 days & 12 hours

Solution :

Let a day be a cycle of 24 hours.
First one completes a cycle within a day
Second one completes a cycle in 25 hours
And third one completes a cycle in 23 hours.
Now, the LCM of 23,24,25 is 23 x 24 x 25 = 13800.

Therefore, after 13,800 hours (least period) they will show the correct time together. (for 24 hours cycle)
i.e., 13,800/24 = 575
On 575th day they will show the correct time.
Since the clocks are analog (12 hours cycle), after 287 days and 12 hours (575/2 = 287.5)they will show the same time.

Question 2

Three clocks were set to true time and starts to run together, makes a beep sound at intervals of 8,10 and 12 minutes respectively. In 2 days, how many times do they will make the sound together?

a)12 b)15 c)25 d)30

Answer : c)25

Solution:

The clocks make sound at intervals of 8, 10 and 12 minutes respectively.
The LCM of 8,10 and 12 is 120.
So the clocks will make sound together after every 120 minutes.
i.e., after every 2 hours.

In 2 days(48 hours), they will make the sound together (48/2) = 24 times.
At the end of 2 day,they will make one more beep together.
Therefore, totally (24+1) 25 times beep sound is made together.
Hence the answer is 25 times.

Question 3

Three analog clocks were set to the exact time at 12 am and they run together and gain 3, 6 and 9 minutes per hour respectively. At what time, they will show the same time again?

a)2.42am b)12.02am c)1.58pm d)12.02pm

Answer : a)2.42am

Solution :

Let an hour(60 minutes) be a cycle.
The clocks complete a cycle early by 3, 6 and 9 minutes respectively.
i.e., a cycle is completed in 57, 54 and 51 minutes respectively.
LCM of 57,54 and 51 is 17442 minutes.
The clocks show the same time after 17442 minutes

To convert 17442 minutes, we have to divide by 60 :

17442/60 = 17400/60 + 42/60 = 290 hours + 42/60 hours = 290 hours and 42 minutes
(42/60 hours = 42/60 x 60 minutes = 42 minutes)

After 290 hours and 42 minutes from 12 am, an analog clock shows 2.42am. (since 290 = 24 x 12 + 2).

Hence, they show the exact time again at 2.42 am .

HCL Distance Puzzles

Solved Distance Puzzles For HCL Placement Tests

Dear Reader,

Below are three problems based on distance and time calculations.

Question 1

A boy drops a bouncing toy from 1st floor to ground floor. Each time, it bounces vertically to a height of 1/2 of the preceding bounce's height. Find the distance the toy will cross if it totally takes 6 bounces and the distance between each floor is 10m.

a)39 m b)30 m c)40 m d)15 m

Answer : b)30 m

Solution :

The distance between two floors is 10m.
The boy dropped the bouncing toy from a height of 10 m.
As the toy is initially dropped, the toy falls 10 meters. Then, the toy rebounds to 10 x 1/2 = 5 meters and falls back down 5 meters. This process will repeat for 5 more times.

Then:

Distance that the toy travels down = 10 + 10x1/2 + 10x 1/2 x 1/2 + ...+ 10x(1/2)⁶
= 10 + 10x(1/2)¹ + 10(1/2)² + 10(1/2)³ + 10x(1/2)⁴ + 10(1/2)⁵ + 10(1/2)⁶ which is a g.p series with 7 terms and first term = 10 common term = 1/2.

we know that, sum of n terms in a g.p series is
sum = a(1 - rⁿ)/1 - r when r is less than one
Here, distance = 10 x [1 - (1/2)⁷]/(1-1/2) = 10 x 127/128 / 1/2
= 20x127/128 = 19.84 meters.
Distance that the toy travels up = 10x(1/2)¹ + 10(1/2)² + 10(1/2)³ + 10x(1/2)⁴ + 10(1/2)⁵ + 10(1/2)⁶
= 19.84 - 10 = 9.84 meters.

Thus, the total distance is 19.84 + 9.84 = 29.68 meters = 30 meters (approximately)

Question 2

A spider starts jumping up from the bottom of a ladder with 32 steps. Each minute, it jumps 4 steps and slips back 2 steps. How much time would the spider take to reach the top of the ladder?

a)15 minutes b)14 minutes c)16 minutes d)18 minutes

Answer : a)15minutes.

Solution :

Think of it like a number line: In the first minute, the spider goes 4 steps up, and slips back 2 steps.

0...1...2...3...4...5...6
|---|---|---|---|---|---|
---------------->
........<-------

(Imagine that's going up instead of horizontal.)

The spider effectively climbs 4 steps - 2 steps = 2 steps higher after the first minute.
In the second minute, the spider goes up 4 more steps, and slips back 2 steps. So he's at 2 steps + 4 steps - 2 steps = 4 steps.
Every minute, it gains 4 steps and loses 2 steps, for a total gain of 2 steps. So after the 14th minute, it is at 28 steps height.
Now, here's the twist: in the 15th minute, it jumps up 4 steps. Since 28 steps + 4 steps = 32 steps, it is at the top of the ladder.
Hence the answer is 15 minutes.

Question 3

A man throws a ball from certain height to the ground and it covers a distance of X meters on the ground. Each time it's bounce covers 2/3 of previous distance. The ball stops bouncing after 4 bounces. Then the total distance it covers on the ground is:

a)191X/43 meters b)250X/21 meters c)172X/75 meters d)211X/81 meters

Answer : d)211X/81 meters

Solution :

This problem is slightly different from the 1st one.
Here, we have to find the distance which covered by the ball on the ground.
Given that, it covers X meters by the through.
And it takes four bounce, each bounce covers 2/3 of previous distance.
Then the distance covered in 1st bounce = (2/3)X

The distance covered in 2nd bounce = (2/3) of (2/3)X = X(2/3)²
The distance covered in 3rd bounce = (2/3) of X(2/3)² = X(2/3)³
And the distance covered in 4th bounce = (2/3) of X(2/3)³ = X(2/3)⁴

Required distance = X + (2/3)X + X(2/3)² + X(2/3)³ + X(2/3)⁴
The above series is a G.P series of 5 terms with a = X and r = 2/3
Then sum = a(1 - rⁿ)/(1 - r)
= X[1 - (2/3)⁵]/(1 - 2/3)
= X[1 - 32/243]/(1/3)
= 3X(211/243) = 211X/81
Hence the required answer is 211X/81 meters.

TCS Measurement Problems

Solved Measurement Problems TCS Placement Tests

Dear Reader,

Below are three problems dealing with parameters of circles and squares.

Question 1

Let ABCD be a square. Alex wants to draw 5 circles of equal radius 'r' with their centres on BD such that the two extreme circles touch two sides AB, BC and AD, CD of the square respectively and each middle circle touches two circles on either side. Find the ratio of r to that of BD.

a) 2 : 11.828 b) 5 : 19.214 c) 1 : 10.828 d) 1 : 12.515

Answer : c) 1 : 10.828

Solution :

As per the statements in question, Alex would have drawn a diagram as shown below.

Let the radius of each circle be r.
Let the side of the square ABCD be a.
Then Diagonal BD = AC = a x sqrt 2 (You will get this formula by applying Pythagoras theorem)
We have to find this diagonal length as follows:
Given, he had drawn 5 circles on diagonal BD.
After sketching 5 circles according to the question, we see that some gap between corner of square and extreme circles. Now, let us calculate that space:

Draw perpendicular lines from sides AB and BC to extreme circle. Now you get a small square of side r and of diagonal r x sqrt2.
In the below diagram, Here, Y is the centre of left extreme circle, BXYZ is a small square with side r and diagonal r sqrt2.
Similar observation can be made on the other extreme circle as well.

Summing up our observations in the main diagram, we get,

Hence the diagonal of big square = r sqrt2 + r + 6r + r + r sqrt2 = 8r + 2r sqrt2
Then the required ratio = r : 8r + 2r sqrt2 = r : r(8 + 2sqrt2)
We can remove r on both sides as it is common. Therefore, we get,
= 1 : 8+2 sqrt2
= 1 : 8+2(1.414) = 1:10.828
Hence the required ratio is 1 : 10.828.

Question 2

In an exam, students were asked to find the perimeter of a square which contains three circles such that their centres on the diagonal of the square, the middle circle touches two extreme circles on either side and two extreme circles touch two sides of the square and the radius of the three are equal. And at what times the radius equals the perimeter?

a) 19.312 b) 12.312 c) 21.312 d) 17.312

Answer : a) 19.312

Solution:

Let the radius of circle be r.
Let the side of the square be a.
Then Diagonal = a(sqrt2)
By applying a similar logic as that of problem 1, we will get a diagram as follows.

a(sqrt2) = diagonal = r sqrt2 + r + 2r + r + r sqrt2 = 4r + 2r sqrt2
a = (4r + 2r sqrt2) / sqrt2
a = 4r/sqrt2 + 2r sqrt2 / sqrt2
a = (2 x 2r) / sqrt2 + 2r

But we know, 2/sqrt2 = sqrt2. Therefore, above equation becomes,
a = 2r sqrt2 + 2r = 2r(sqrt2+1)
Then perimeter of the square = 4a = 4 x 2r(sqrt2+1) = 8r(sqrt2+1)
= 8r(1.414+1) = 8r(2.414) = r(19.312)
Hence 19.312 time radius equals the perimeter of the square.

Question 3

Find the length of diagonal BD of a square ABCD when 7 circles of radius 2 cm are located in ABCD such that their centres on BD, two extreme circles touch two sides of ABCD and each middle circle touches two circles on either side.

a)32.98cm b)29.65cm c)31.12cm d)28.56cm

Answer : b)29.65cm

Solution :

Given that, ABCD is a square where AB, BC, CD and AD are sides of square and BD, AC are diagonals of the square.
Applying similar logic to that of first and second questions, we will get the below diagram corresponding to this question :

The length of the diagonal BD = r sqrt2 + r + 10r + r + r sqrt2
= 2 rsqrt2 + 12r
Given that r = 2 cm
Then BD = 2(2)sqrt2 + 12(2) = 4 sqrt2 + 24 = 4(1.414) + 24 = 29.65 cm.
Hence the required answer is 29.65cm

Simplification Problems HCL Aptitude Tests

Simplification Problems To Prepare For HCL Aptitude Tests

Dear Reader,

Below are three fraction based simplification problems which you can expect in HCL and all other companies aptitude tests.

Question 1

A man spent 5/16 of his age plus two as student, 1/40 plus 1 as a husband, 1/4 as a good politician and 3/40 as a father. And the remaining 6 years as a good grand father. Then the living days of the man is:

a) 72 b) 89 c) 80 d) 69

Answer : c)80

Solution :

Lets say his age is X.

So years spent as student = 5X / 16 + 2,
as Husband = X/4 + 1,
as Politician = X/4,
and as Father = 3X/40.

Remaining 6 years (as grandfather) = X - { 5X/16 + 2 + X/4 + 1 + X/4 + 3X/40}
i.e., 6 = X - { 5X / 16 + 2 + X / 4 + 1 + X / 4 + 3X / 40}
6 = X - [13X / 16 + 3 + 3X / 40]
6 = X - [71X / 80 + 3]
9X/80 = 9
X = 80.
Hence the age of the man is 80 years.

Question 2

In a village, every weekend, 1/4 th of men and 1/6 th of women participate in the social activity. If the total number of participants is 75, out of them 25 are men then the total number of men and women in the village is:

a) 100 b) 200 c) 300 d) 400

Answer : d)400

Solution :

Number of men who participated in social activities = 25
Therefore, 1/4 th of the men = 25
Or, total number of men in the village = 100

Number of women who participated in social activities = 75 - 25 = 50
Therefore, 1/6 th of the women = 50
Or, total number of women in the village = 300
Hence the required answer is 300 + 100 = 400.

Question 3

A father has divided his properties in such a way that one-half of total properties goes to A, two-third of the remaining shared equally to B, C and D and the rest to E, F, G and H such they equally gets Rs.30,000 by their sharing property. Then the amount will D get is:

a) Rs.80,000 b) Rs.68,000 c) Rs.24,000 d) Rs.72,000

Answer : a) Rs.80,000

Solution :

A's share = 1/2
Then remaining = 1 - 1/2 = 1/2 ...(A)

2/3 of this remaining 1/2 (as in eq A) goes to B,C and D.
Total share of B, C and D = 2/3 of 1/2 = 1/3

Therefore, now remaining property = Total share of E,F,G and H
= value of (A) - total share of B,C and D
= 1/2 - 1/3 = 1/6

Then, E, F, G and H 's individual shares = (1/6) / 4 = 1/6 x 1/4 = 1/24
Given 1/24 = Rs.30,000
Then, total property amount = Rs.24 x 30,000 = Rs.7,20,000
Total share of B, C and D = 1/3 x 7,20,000
Then their individual share amount = (1/3 x 7,20,000) / 3 = Rs.80,000.
Hence D's share is Rs.80000.

Infosys Aptitude Questions

3 Solved Sibling Problems To Prepare For Infosys Aptitude Tests

Dear Reader,

Below are three problems based on the concept of number of brothers and sisters, you have to find the number of siblings according to given statements.

Question 1 :

If Mr.X, Mr.Y and Mr.Z are three siblings and following three statements are true
1) Mr.X has one older brother and three younger sisters.
2) Mr.Y has two older brothers and two younger sisters
3) Mr.Z has three older brothers and one younger sister.

Then the least possible number of siblings in their family is:

a) 5 b) 6 d) 7 d) 8

Answer : b)6

Solution :

First, we need to think about X, Y and Z as who is the oldest among the three and who is the youngest among the three.
X has one older brother while Y has two. Therefore, X is older than Y. Similarly Y is older than Z. Therefore, among X,Y and Z
X is the oldest, Y is next and Z is the youngest.
X has an older brother and Z has a younger sister.

Try making a chart laying out all the three conditions given in question, and mark the gender of each known sibling.
Then, it is clear that X is male and Y is female.

 M M F M F F
 1st 2nd 3rd 4th 5th 6th
  X Y  Z

Based on the above chart, there should be at least 6 siblings.

Question 2:

If A, B and C are triblet. A,C are males and B is female,
1) A & C say " We have 2 more brothers than the number of sisters"
2) B says " I have 2 more brothers than twice the sisters "

If the above two statements are true then the number of siblings in their family is:

a) 6 b) 12 c) 8 d) 9

Answer : b) 12

Solution :

Let g be the number of girls in the family.
Let b be the number of boys in the family.

Consider statement 1, A and C would have b-2 brothers (excluding them). This b-2 is equal to two more than the number of sisters they have.
Therefore, b-2 = g+2
Or, b-g = 4 -----(1)

Now, from statement 2, B would have g-1 sisters (excluding herself) and b brothers. She has 2 more brothers than twice the number of sisters.
Therefore, Twice the number of sisters = 2(g-1) which is equal to 2 + number of brothers.
Then we have, 2(g-1) + 2 = b
2g - 2 + 2 = b
b-2g = 0 -------(2)
Solving (1) and (2), we have g = 4 and b = 8.
Hence the number of siblings = 4+8 = 12.

Question 3 :

Aravind and Bavya are friends.
1)Aravind's sister has thrice as many sisters of Bavya as brothers.
2)Aravind has 1 more brother than than that of Bavya
3)Twice Bavya's brother's sisters equals the number of boys in Aravind's family.

If the above all 3 statements are true then the number of brothers what Aravind has more than Bavya is:

a) 1 b) 2 c) 3 d) none of these

Answer : a) 1

Solution :

Let the number of boys in Aravind's family be a
Let the number of girls in Aravind's family be b
Let the number of boys in Bavya's family be c
Let the number of girls in Bavya's family be d.

consider statement 1, "Aravind's sister has thrice as many sisters of Bavya as brothers" :

Number of Bavya's sisters = d-1 (excluding her)
Thrice the Bavya's sisters = 3(d-1)
Aravind's sister has 'a' number of brothers.
Therefore, 3(d-1) = a
d-1 = a/3....(1)

Consider statement2,"Aravind has 1 more brother than that of Bavya" :

Number of Aravind's brothers = a-1
Number of Bavya's brother = c
Since, Aravind has 1 more brother than Bavya then a-1 = c+1
a-c = 2...(2)

Consider statement3,"Twice Bavya's brother's sisters equals the number of boys in Aravind's family"

The number of sisters of Bavya's brother = d
Number of boys in Aravind's family = a
Therefore 2d = a
Or d = a/2
Then, d-1 = a/2 - 1 ...(3)

Now, let us solve equations we obtained in previous steps :

From (1) and (3),
a/3 = a/2 - 1
a = 6
Put a value in (2), we have c = 4
Put a value in (3), we have d = 3

Therefore Aravind has a-1 = 6-1 = 5 brothers
And Bavya has c = 4 brothers.
Hence, Aravind has 1 more brother than Bavya.