Wipro previous placement questions - 2

Wipro previous placement questions - 2

1.  At how many points between 10 O'clock and 11 O'clock are the minute hand and hour hand of a clock at an angle of 30 degrees to each other?

Sol:
Between 10 and 11, the minute hand and hour hand are at an angle of 30o to each at (12/11) x  45 minutes past 10 = 49 1/11 minutes past 10. The next time they will be at angle of 30o to each other will be at 11.

2. The egg vendor calls on his first customer & sells half his eggs & half an egg. To the 2nd customer he sells half of what he sells half of what he had left & half an egg. & to the 3rd customer he sells half what he had then left & half an egg. By the way he did not break any eggs. In the end three eggs were remaining . How many total eggs he was having ?
Sol:
31 eggs.
After selling to 3 persons , he was left with 3 eggs.
After selling to 2 persons , he was left with 3 x 2 + 1 = 7 eggs.
After selling to 1 person , he was left with 7 x 2 + 1 = 15 eggs.
Before selling to 1 st person , he was having 15 x 2 + 1 = 31 eggs.

3. There are some people in party, 1/3rd left the party . Then 2/5th of the remaining left the party , then 2/3rd of the remaining left the party . At last 6 were remaining . How many people were in total ?
Sol:
45
If x persons were there in total , then
x × (1 – 1/3)× (1 – 2/5) ×(1 – 2/3) = 6
x×2/3 × 3/5 × 1/3 = 6
x = 6 × 5 × 3/2 = 45

4. Two trains are traveling from point A to point B such that the speed of first train is 65 kmph and the speed of 2 train is 29 kmph. Where is the distance b/w A and B such that the slower train reached 5 hrs late compared to the faster?
Sol:
If x is the distance, then
x/29 – x/65 = 5
Then x = 5×29×6565−29$5\times {\displaystyle \frac{29\times 65}{65-29}}$ = 261.8055 kms

5. A person was fined for exceeding the speed limit by 10 km/hr.Another person was also fined for exceeding the same speed limit by twice the same.If the second person was traveling at a speed of 35 km/hr,find the speed limit.
a) 19 km/hr
b) 27 km/hr
c) 30 km/hr
d) 15 km/hr
Sol:
If x is speed limit,
Speed of first person = x + 10
Speed of 2nd person = x + 20
But speed of 2nd person = 35 kmph
x + 20 = 35
x = 15 kmph.
so speed limit is 15 kmph option D

6. The average of ten numbers is 7. If each number is multiplied by 12 ,then the average of new set of numbers is :
a) 7
b) 19
c) 82
d) 84
Sol:
The avg will be = 12×7= 84

7. The average of eight numbers is 14. The average of six of these numbers is 16.The average of the remaining two numbers is :
a) 4
b) 8
c) 16
d) none
Sol:
Average of eight numbers = 14
Average of six numbers = 16
Average will be = (14×8 – 16×6)/2

8. The average age of a class of 39 students is 15 years.  If the age of the teacher be included, then the average increases by 3 months .Find the age of the teacher.
a) 25 years
b) 27 years
c) 35 years
d) 28 years
Sol:
Sum of the ages of the students = 39×15 = 585
New average = 15 years 3 months = 15 + 14${\displaystyle \frac{1}{4}}$ year
Sum of the ages of students and teacher = 40×1514$15{\displaystyle \frac{1}{4}}$ = 40×614${\displaystyle \frac{61}{4}}$ = 610
Teacher age = 610  – 585 = 25 years.

9. Two trains start from stations A and B spaced 50 kms apart at the same time and speed. As the trains start, a bird flies from one train towards the other and on reaching the second train, it flies back to the first train. This is repeated till the trains collide. If the speed of the trains is 25 km/h and that of the bird is 100 km/h. How much did the bird travel till the collision.
Sol:
Since the trains is travelling at 25 kmph, at each other, the relative speed is 50 kmph.
Speed = 50 kmph
Distance = 50 km
Time to collision = distance / speed = 1 hr
Speed of bird = 100 kmph
Time flying = 1 hr (the bird is flying till the trains collide)
Distance travelled = speed × time = 100 km

10. There are 20 poles with a constant distance between each pole. A car takes 24 second to reach the 12th pole.  How much will it take to reach the last pole.
Sol:
Assuming the car starts at the first pole.
To reach the 12th pole, the car need to travel 11 poles (the first pole doesn't count, as the car is already there).
11 poles 24 seconds
1 pole (24/11) seconds
To reach the last (20th) pole, the car needs to travel 19 poles.
19 pole 19 x (24/11) seconds
= 41.4545 seconds

11. Father's age is three years more than three times the son's age. After three years, father's age will be ten years more than twice the son's age. What is the father's present age?
Sol:
Let the son's present age be x years.then father's present age will be 3x + 3 years.
After 3 years,3x + 3 + 3 = 2 (x + 3) + 10
Solving we get, x = 10.
Substituting x =10 in 3x + 3,
Hence father's present age will be x = 33 years.

12. In a railway station, there are two trains going. One in the harbor line and one in the main line, each having a frequency of 10 minutes. The main line service starts at 5 o'clock and the harbor line starts at 5.02 A.M. A man goes to the station every day to catch the first train that comes. What is the probability of the man catching the first train?
Sol:
For each 10 min interval, if man comes in first 2 min, he'll catch the 1st train, if he comes in next 8 min, he'll catch the 2nd train.
Hence for harbor line = (2/10) = 0.2 and for main line 0.8.

13. A ship went on a voyage. After it had traveled 180 miles a plane started with 10 times the speed of the ship.  Find the distance when they meet from starting point.
Sol:
Let the speed of the ship = m miles/hr. and plane took 't' hours to meet the ship
Then, m×t is the distance ship traveled after plane started
So we have, mt + 180 = 10mt
⇒ 9mt = 180
⇒ mt = 20
Hence distance = 180 + 20 = 200 miles

14. On 8th Feb, 2005 it was Tuesday. What was the day of the week on 8th Feb, 2004?
a. Tuesday 
b. Monday
c. Sunday 
d. Wednesday
Sol:
Sunday
The year 2004 is a leap year and therefore, two days will be preceded from Tuesday

15. At what time between 2 and 3 o'clock will the hands of a clock be together?
a. 10×10/11 
b. 10×11/10
c. 11×10/11 
d. 12×10/11
Answer : d
Sol:
The hands of a clock would be together when the angle between The hour hand and minute hand is Zero.  Now apply the formula: θ=∣∣∣30h−112m∣∣∣$\theta =\left|30h-{\displaystyle \frac{11}{2}}m\right|$
Here θ$\theta$ = 0
⇒11/2m – 30h = 0
⇒11/2m – 30×2 = 0
⇒ m = 120/11

16. At what angle the hands of a clock are inclined at 15 minutes past 5?
a. 117/2 ° 
b. 64 °
c. 135/2 ° 
d. 145/2 °
Sol:
Apply the formula:
θ=∣∣∣30h−112m∣∣∣$\theta =\left|30h-{\displaystyle \frac{11}{2}}m\right|$
⇒ Angle = 30 × 5 –11/2 × 15 = 150 – 165/2 = 135/2

17. At 3.40, the hour hand and the minute hand of a clock form an angle of
a. 120° 
b. 125°
c. 130° 
d. 135°
Answer: C
Sol:
Use formula  θ=∣∣∣30h−112m∣∣∣$\theta =\left|30h-{\displaystyle \frac{11}{2}}m\right|$
Angle = 30×3 – 11/2 × 40 = 90 – 220 = 130°

18. How many times in a day, the hands of a clock are straight?
a. 22
b. 24
c. 44 
d. 48
Sol:
The hands of a clock point in opposite directions (in the same straight line) 11 times in every 12 hours. (Because between 5 and 7 they point in opposite directions at 6 o'clock only).
So, in a day, the hands point in the opposite directions 22 times.

19. Find the angle between the hour and the minute hand of a clock when the time is 3.25.
a. 47 ½ 
b. 49 ½
c. 55 ½ 
d. 57 ½
Sol:
Formula : θ=∣∣∣30h−112m∣∣∣$\theta =\left|30h-{\displaystyle \frac{11}{2}}m\right|$
Angle = 11/2 × 25 – 30×3 = 95/2 = 47.5

20. At what time, in minutes, between 3 o'clock and 4 o'clock, both the needles will coincide each other?
A. 5 1/11 ° 
B. 12 4/11 °
C. 13 4/11° 
D. 16 4/11°
sol:
Formula : θ=∣∣∣30h−112m∣∣∣$\theta =\left|30h-{\displaystyle \frac{11}{2}}m\right|$
Here angle is 0. So
11/2 m – 30 h = 0
11/2 m – 30 × 3 = 0
m = 180/11
= 16 4/11
Ans:: D

TCS placement questions - 2

TCS placement questions - 2

1. If f(x) = (1+x+x2+x3+.......x2012)2−x2012$(1+\mathrm{x}+{\mathrm{x}}^2+x^3+.......x^{2012}{)}^2-x^{2012}$
g(x) = 1+x+x2+x3+.......x2011$1+\mathrm{x}+{\mathrm{x}}^2+x^3+.......x^{2011}$ 
Then what is the remainder when f(x) is divided by g(x)
Let us multiply g(x) with x on the both sides
x.g(x) = x+x2+x3+.......x2012$\mathrm{x}+{\mathrm{x}}^2+x^3+.......x^{2012}$
add 1 on both sides
x. g(x) + 1 = 1+x+x2+x3+.......x2012$1+\mathrm{x}+{\mathrm{x}}^2+x^3+.......x^{2012}$
Substitute this value in f(x)
then f(x) = (x.g(x)+1)2−x2012${\left(x.g(x)+1\right)}^2-x^{2012}$
f(x) = x2.g(x)2+2.g(x)+1−x2012$x^2.g(x{)}^2+2.g(x)+1-x^{2012}$
Now f(x) is divisible by g(x) first two terms are exactly divisible by g(x) and we get 1 - x2012$x^{2012}$
But 1 - x2012$x^{2012}$ = (1 - x)(1+x+x2+x3+.......x2011$1+\mathrm{x}+{\mathrm{x}}^2+x^3+.......x^{2011}$)
if this expression is divisible by g(x) we get 0 as remainder.

2. A number has exactly 3 prime factors, 125 factors of this number are perfect squares and 27 factors of this number are perfect cubes. overall how many factors does the number have?
We know that the total factors of a number N = ap.bq.cr$a^p.b^q.c^r$ ....
Now the total factors which are perfect squares of a number N = ([p2]+1).([q2]+1).([r2]+1)....$\left(\left[{\displaystyle \frac{p}{2}}\right]+1\right).\left(\left[{\displaystyle \frac{q}{2}}\right]+1\right).\left(\left[{\displaystyle \frac{r}{2}}\right]+1\right)....$
where [x] is greatest intezer less than that of x.
Given ([p2]+1).([q2]+1).([r2]+1)....$\left(\left[{\displaystyle \frac{p}{2}}\right]+1\right).\left(\left[\frac{q}{2}\right]+1\right).\left(\left[\frac{r}{2}\right]+1\right)....$ = 125
So [p2]+1$\left[{\displaystyle \frac{p}{2}}\right]+1$ = 5;  [q2]+1$\left[{\displaystyle \frac{q}{2}}\right]+1$ = 5; [p2]+1$\left[{\displaystyle \frac{p}{2}}\right]+1$ = 5
[p2]$\left[{\displaystyle \frac{p}{2}}\right]$ = 4 ⇒$\Rightarrow$ p = 8 or 9, similarly q = 8 or 9, r = 8 or 9
Given that 27 factors of this number are perfect cubes
so ([p3]+1).([q3]+1).([r3]+1)....$\left(\left[{\displaystyle \frac{p}{3}}\right]+1\right).\left(\left[{\displaystyle \frac{q}{3}}\right]+1\right).\left(\left[{\displaystyle \frac{r}{3}}\right]+1\right)....$ = 27
So [p3]+1$\left[{\displaystyle \frac{p}{3}}\right]+1$ = 3 ⇒$\Rightarrow$ = [p3]$\left[{\displaystyle \frac{p}{3}}\right]$ = 2
⇒$\Rightarrow$ p = 6, 7, 8
By combining we know that p = q = r = 8
So the given number should be in the format = a8.b8.c8$a^8.b^8.c^8$ ....
Number of factors of this number = (8+1).(8+1).(8+1) = 729

3. In a class there are 60% of girls of which 25% poor.  What is the probability that a poor girl is selected is leader?
Assume total students in the class = 100
Then Girls = 60% (100) = 60
Poor girls = 25% (60) = 15
So probability that a poor girls is selected leader = Poor girls / Total students = 15/100 = 15%

4. A and B are running around a circular track of length 120 meters with speeds 12 m/s and 6 m/s in the same direction.  When will they meet for the first time?
A meets B when A covers one round more than B.
A's relative speed = (12 - 6) m/s.  So he takes 120 / 6 seconds to gain one extra round.
So after 20 seconds A meets B.

5. A completes a work in 20 days B in 60 days C in 45 days.  All three persons working together on a project got a profit of Rs.26000 what is the profit of B?
We know that profits must be shared as the ratio of their efficiencies.  But efficiencies are inversely proportional to the days.  So efficiencies of A : B : C = 1/20 : 1/60 : 1/45 = 9 : 3 : 4
So B share in the total profit = 316×26000${\displaystyle \frac{3}{16}}\times 26000$ = Rs.4875

6. A completes a piece of work in 3/4 of the time in B does, B takes 4/5 of the time in C does.  They got a profit of Rs. 40000  how much B gets?
Assume C takes 20 Days.  Now B takes 4/5 (20) = 16 days.  A takes 3/4(16) = 12
Now their efficiencies ratio = 1/20 : 1/16 : 1/12 = 12 : 15 : 20
B's share in the profit of Rs.40000 = 15/47 (40000) = Rs.12765

7. An empty tank be filled with an inlet pipe ‘A’ in 42 minutes. After 12 minutes an outlet pipe ‘B’ is opened which can empty the tank in 30 minutes. After 6 minutes another inlet pipe ‘C’ opened into the same tank, which can fill the tank in 35 minutes and the tank is filled. Find the time taken to fill the tank?
Assume total tank capacity = 210 Liters
Now capacity of  pipe A = 210/42 = 5 Liters
Capacity of B =  210 / 30 = - 7 Liters
Capacity of C =  210 / 35 = 6 min
Assume tank gets filled in x min after the third pipe got opened.
So x×5+6×(−2)+4x=210$x\times 5+6\times (-2)+4x=210$
⇒48+4x=210⇒4x=162⇒x=40.5$\Rightarrow 48+4x=210\Rightarrow 4x=162\Rightarrow x=40.5$
Total time taken to fill the tank = 40.5 + 12 + 6 = 51.5

8. Mother, daughter and an infant combined age is 74, and mother's age is 46 more than daughter and infant.  If infant age is 0.4 times of daughter age, then find daughters age.
Assume M + D + I = 74; .................(1)
Also given M - D - I = 46 ⇒$\Rightarrow$ M = D + I + 46
Also I = 0.4 D ⇒$\Rightarrow$ I = 2/5 D
Substituting M and I values in the first equation we get D - 25${\displaystyle \frac{2}{5}}$D - 46 + D + 25${\displaystyle \frac{2}{5}}$D = 74
Solving D = 10

9. A Grocer bought 24 kg coffee beans at price X per kg. After a while one third of stock got spoiled so he sold the rest for $200 per kg and made a total profit of twice the cost. What must be the price of X?
Total Cost price = 24×$\times$X
As 1/3rd of the beans spoiled, remaining beans are 2/3 (24) = 16 kgs
Selling price = 200 ×$\times$ 16 = 3200
Profit = Selling price - Cost price = 3200 - 24×$\times$X
Given Profit = 2 ×$\times$ Cost price
3200 - 24×$\times$X = 2 ×$\times$ (24×$\times$X)
Solving X = 44.44

10. Bhanu spends 30% of his income on petrol on scooter 20% of the remaining on house rent and the balance on food. If he spends Rs.300 on petrol then what is the expenditure on house rent?
Given 30% (Income ) = 300 ⇒$\Rightarrow$ Income = 1000
After having spent Rs.300 on petrol, he left with Rs.700.
His spending on house rent = 20% (700) = Rs.140

11. Let exp(m,n) = m to the power n. If exp(10, m) = n exp(2, 2) where to and n are integers then n =
Given 10m=n.22${10}^m=n{.2}^2$
⇒$\Rightarrow$ 2m×5m=n.22⇒2m−2×5m=n$2^m\times 5^m=n{.2}^2\Rightarrow 2^{m-2}\times 5^m=n$
For m = 2 we get least value of n = 25, and for m > 2 we get infinite values are possible for n.

12. How many kgs. of wheat costing Rs. 5 per kg must be mixed with 45 kg of rice costing Rs. 6.40 per kg so that 20% gain may be obtained by  selling the mixture at Rs. 7.20 per kg ?
If the selling price of the mixture is Rs.7.2 when sold at 20% profit then
CP ×120100$\times {\displaystyle \frac{120}{100}}$ = 7.2 ⇒$\Rightarrow$ CP = Rs.6
Now by applying weighted average formula = K×5+45×6.4K+45=6${\displaystyle \frac{K\times 5+45\times 6.4}{K+45}=6}$
⇒$\Rightarrow$ K = 18 kgs

13. The diagonal of a square is twice the side of equilateral triangle then the ratio of Area of the Triangle to the Area of Square is?
Let the side of equilateral triangle = 1 unit.
We know that area of an equilateral triangle = 3√4a2${\displaystyle \frac{\sqrt{3}}{4}{a}^2}$
As side = 1 unit area of the equilateral triangle = 3√4${\displaystyle \frac{\sqrt{3}}{4}}$
Now Diagonal of the square = 2 (side of the equilateral triangle) = 2
We know that area of the square = 12D2${\displaystyle \frac{1}{2}{D}^2}$ where D = diagonal
So area of the square = 12(22)=2${\displaystyle \frac{1}{2}(2^2)=2}$
Ratio of the areas of equilateral triangle and square = 3√4${\displaystyle \frac{\sqrt{3}}{4}}$ : 2 ⇒$\Rightarrow$ 3√:8$\sqrt{3}:8$

14. Raj tossed 3 dices and there results are noted down then what is the probability that Raj gets 10?
Always remember when 3 dice are rolled the number of ways of getting n ( where n is the sum of faces on dice)
= (n−1)C2$(n-1)C_2$ where n = 3 to 8
= 25 where n = 9, 12
= 27 where n = 10, 11
= (20−n)C2$(20-n)C_2$ where n = 13 to 18
The required probability = 2763${\displaystyle \frac{27}{6^3}}$ = 27216