TCS Apititude Questions

 1) The water from one outlet, flowing at a constant rate, can fill the swimming pool in 9 hours. The water from second outlet, flowing at a constant rate can fill up the same pool in approximately in 5 hours. If both the outlets are used at the same time, approximately what is the number of hours required to fill the pool? 

Ans: Assume tank capacity is 45 Liters. 

Given that the first pipe fills the tank in 9 hours. 

So its capacity is 45 / 9 = 5 Liters/ Hour. 

Second pipe fills the tank in 5 hours. So its capacity is 45 / 5 = 9 Liters/Hour.

 If both pipes are opened together, then combined capacity is 14 liters/hour. 

To fill a tank of capacity 45 liters, Both pipes takes 45 / 14 = 3.21 Hours.

2) If 75 % of a class answered the first question on a certain test correctly, 55 percent answered the second question on the test correctly, and 20 percent answered neither of the questions correctly, what percentage answered both correctly? 

Ans : It is a problem belongs to sets. 

We use the following formula n(A∪∪B) = n(A) + n(B) - n(A∩∩B) Here n(A∪∪B) is the people who answered atleast one of the questions.

 It was given that 20% answered neither question then the students who answered atleast one

 question is 100% - 20% = 80% 

Now substituting in the formula we get 80% = 75% + 55% - n(A∩∩B) ⇒⇒ n(A∩∩B) = 50%  .

3) A student's average ( arithmetic mean) test score on 4 tests is 78. What must be the students score on a 5th test for the students average score on the 5th test to be 80?

 Ans: We know that Average =Sum of the observations

 No of observations=Sum of the observations 

No of observations So Sum of 4 test scores = 78××4=312 Sum of 5 tests scores = 80××5=400 ⇒⇒ 5th test score=400-312=88 

Alternative method: If the student scores 78 in the fifth test also, what could be his average? No change. Is it not? 

But to bring the average to 80, he must have scored enough marks extra so that each of the five subject scores increase upto 80. i.e., he should have scored 2 x 5 = 10 runs extra in the fifth subject. So 5th subject score is 78 + 10 = 88.

4) Rural households have more purchasing power than do urban households at the same income level, since some of the income urban and suburban households use for food and shelter can be used by the rural households for other needs. Which of the following inferences is best supported by the statement made above? 

(A) The average rural household includes more people than does the average urban or suburban household

(B) Rural households have lower food and housing costs than do either urban or suburban households.

(C) Suburban households generally have more purchasing power than do either rural or urban households. 

(D) The median income of urban and suburban households is generally higher than that of rural households.

 (E) All three types of households spend more of their income on housing than on all other purchases combined.   

Ans: If average rural household includes more people, then how come they have more purchasing power? Infact, they have less purchasing power as they have to feed more people. 

Option A ruled out. 

Option C does not explain why rural households have more purchasing power than urban. Ruled out. If median income of urban and suburban households is generally higher than rural households they are likely to have more purchasing power, assuming other parameters constant. But this does not explain why rural households have more purchasing power.

 Options D ruled out.

 Option E does not provide any explanation why rural households have more purchasing power. Ruled out.

 Option B is correct as, If rural households spend less income on food and shelter due to less prices they definitely have more disposable income to spend.  

5) Jose is a student of horticulture in the University of Hose. In a horticultural experiment in his final year, 200 seeds were planted in plot I and 300 were planted in plot II. If 57% of the seeds in plot I germinated and 42% of the seeds in plot II germinated, what percent of the total number of planted seeds germinated?

 Ans: Total seeds germinated in Plot I = 57% of 200 = 114

 Total seeds germinated in Plot II = 42% of 300 = 126 

Total germinated seeds = 114 + 126 = 240

 The percentage of germinated seeds of the total seeds = 240500×100240500×100 = 48%

6) A closed cylindrical tank contains 36ππ cubic feet of water and its filled to half its capacity. When the tank is placed upright on its circular base on level ground, the height of water in the tank is 4 feet. When the tank is placed on its side on level ground, what is the height, in feet, of the surface of the water above the ground? 

Ans: We know that the volume of cylinder = πr2hπr2h 

Given tank hight = 4ft. 

⇒⇒ πr24πr24 = 36ππ 

⇒⇒ r = 3

 So the radius is 3 which means the diameter is 6. 

As the cylinder is filled to initially exactly half of the capacity, When this cylinder is placed on its side, Water comes upto the height of the radius. 

So water comes upto 3 ft. 

7) The present ratio of students to teachers at a certain school is 30 to 1. If the student enrollment were to increase by 50 students and the number of teachers were to increase by 5, the ratio of the teachers would then be 25 to 1 What is the present number of teachers? 

Ans : Assume the present students and teachers are 30K, K After new recruitments of students and teachers the strength becomes 30K + 50, K + 5 respectively.

 But given that this ratio = 25 : 1

 ⇒30K+50K+5=251⇒30K+50K+5=251 

Solving we get K = 15 

So present teachers are 15.  

8) College T has 1000 students. Of the 200 students majoring in one or more of the sciences,130 are majoring in Chemistry and 150 are majoring in Biology. If at least 30 of the students are not majoring in either Chemistry or Biology, then the number of students majoring in both Chemistry and Biology could be any number from .

Ans: If we assume exactly 30 students are not majoring in any subject then the students who take atleast one subject = 200 - 30 = 170 

We know that n(A∪∪B) = n(A) + n(B) - n(A∩∩B) ⇒⇒ 170 = 130 + 150 - n(A∩∩B) Solving we get n(A∩∩B) = 110. 

i.e., Students who can take both subjects are 110 But If more than 30 students are not taking any subject, what can be the maximum number of students who can take both the subjects?

 As there are 130 students are majoring in chemistry, assume these students are taking biology also.

So maximum students who can take both the subjects is 130 .

So the number of students who can take both subjects can be any number from 110 to 130.

9. If f(x) = (1+x+x2+x3+.......x2012)2−x2012

(1+x+x2+x3+.......x2012)2−x2012

g(x) = 1+x+x2+x3+.......x20111+x+x2+x3+.......x2011 

Then what is the remainder when f(x) is divided by g(x) 

Ans : Let us multiply g(x) with x on the both sides

 x.g(x) = x+x2+x3+.......x2012x+x2+x3+.......x2012 

add 1 on both sides

 x. g(x) + 1 = 1+x+x2+x3+.......x20121+x+x2+x3+.......x2012 

Substitute this value in f(x)

 then f(x) = (x.g(x)+1)2−x2012(x.g(x)+1)2−x2012 

f(x) = x2.g(x)2+2.g(x)+1−x2012x2.g(x)2+2.g(x)+1−x2012 

Now f(x) is divisible by g(x) first two terms are exactly divisible by g(x) and we get 1 - x2012x2012 But 1 - x2012x2012 = (1 - x)(1+x+x2+x3+.......x20111+x+x2+x3+.......x2011)

 if this expression is divisible by g(x) we get 0 as remainder.

10.In a class there are 60% of girls of which 25% poor. What is the probability that a poor girl is selected is leader?

Ans: 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%.

11) letters in the word ABUSER are permuted in all possible ways and arranged in alphabetical order then find the word at position 49 in the permuted alphabetical order?

 a) ARBSEU 

b) ARBESU 

c) ARBSUE

 d) ARBEUS 

Ans: The best way to solve this problems is Just ask how many words starts with A. 

If we fix A, then the remaining letters can be arranged in 5! ways = 120. 

So the asked word must start with A. Arrange all the given letters in alphabetical order. 

ABERSU Let us find all the words start with AB. AB**** = 4!= 24 ways Now we find all the words start wit AE. 

AE****= 4!= 24 ways So next word start with AR and remaining letters are BESU

 So option B.

12) From the top of a 9 metres high building AB, the angle of elevation of the top of a tower CD is 30º and the angle of depression of the foot of the tower is 60º. What is the height of the tower?

 

Ans: We have to find the value of CD. We use Sine rule to find the answer easily. 

Sine rule is aSinA=bSinB=cSinCaSinA=bSinB=cSinC

 In triangle BDE, 9Sin60=xSin309Sin60=xSin30 

So 9√ 3 2=x12⇒x=9√ 3 932=x12⇒x=93

 In triangle BCD, CDSin30=9√ 3 Sin60CDSin30=93Sin60 

CD12=9√ 3 √ 3 2⇒CD=3CD12=9332⇒CD=3

 So height of the tower = 9 + 3 = 12.

13) If there are 30 cans out of them one is poisoned if a person tastes very little he will die within 14 hours so if there are mice to test and 24 hours to test, what is the minimum no. of mice’s required to find poisoned can? 

Ans: If only 3 person are used, by giving wine drops suggested by the diagram, we can find the poisoned casks upto 8.

 for example, If the 2nd and 3rd persons die, then 7th cask is poisoned. 

As a rule of thumb, If we have n mice, we can easily find the poison casks upto 2n2n. 

As the number of casks are less than 32 we can use only 5 mice.  

14). A hare and a tortoise have a race along a circle of 100 yards diameter. The tortoise goes in one direction and the hare in the other. The hare starts after the tortoise has covered 1/5 of its distance and that too leisurely. The hare and tortoise meet when the hare has covered only 1/8 of the distance. By what factor should the hare increase its speed so as to tie the race? 

Ans :Assume the circumference of the circle is 200 meters. 

Hare and tortoise started at the same point but moves in the opposite direction.

 It is given that by that time tortoise covered 40 m (1/5th of the distance), Hare started and both met after hare has covered 25. 

This implies, in the time hare has covered 25m, hare has covered 200 - 40 - 25 = 135 meters.

 So Hare : tortoise speeds = 25 : 135 = 5 : 27 Now Hare and tortoise has to reach the starting point means, Hare has to cover 175 meters and Tortoise has to cover only 25 meters in the same time.

 As time =DistanceSpeed=2527=1755×KDistanceSpeed=2527=1755×K 

Ie., Hare has to increase its speed by a factor K. 

Solving we get K = 37.8 .

15) A cow and horse are bought for Rs.2,00,000. The cow is sold at a profit of 20% and the horse is sold a t a loss of 10%. The overall gain is Rs.4000, the Cost price of cow? 

a) 130000

 b) 80000

 c) 70000 

d) 120000

 Ans: Overall profit = 4000200000×100=2%4000200000×100=2%

 By applying alligation rule,

 we get So cost price of the cow = 2/5 x 200000 = 80,000

16) A circle has 29 points arranged in a clock wise manner from o to 28. A bug moves clockwise manner from 0 to 28. A bug moves clockwise on the circle according to following rule. If it is at a point i on the circle, it moves clockwise in 1 sec by (1 + r) places, where r is the remainder (possibly 0) when i is divided by 11. If it starts in 23rd position, at what position will it be after 2012 sec.  

Ans: After 1st second, it moves 1 + (23/11)r = 1 + 1 = 2, So 25th position 

After 2nd second, it moves 1 + 25/11 = 1 + 3 = 4, So 29th position = 0 

After 3rd second, it moves 1 + 0/11 = 1 + 0 = 1, So 1st position 

After 4th second, it moves 1 + 1 = 3rd position 

after 5th, 1 + 3/11 = 4 So 7th 

After 6th, 1 + 7/11 = 8 so 15th After 7th, 1 + 15/11 = 5 so 20th 

After 8th, 1 + 20/11 = 10th, So 30th = 1st So it is on 1st

 after every 3 + 5n seconds. So it is on 1st position after 2008 seconds (3 + 5 x401) So on 20th after 2012 position.

17) Sum of two number is 50 & sum of three reciprocal is 1/12 so find these two numbers

Sol :  x+y = 50 .....(1) x=50-y ....(2) 

1x+1y=1121x+1y=112 ⇒y+xxy=112⇒12(y+x)=xy⇒y+xxy=112⇒12(y+x)=xy ...(3) 

 put (2) in (4)

 ⇒⇒ 12(y+50-y)=(50-y)y 

⇒⇒ 12y+600-12y=50y-y2y2

 ⇒⇒ y2y2-50y+600=0 

⇒⇒ y2y2-30y-20y+600=0

 ⇒⇒ y(y-30)-20(y-30)=0 

⇒⇒ (y-20) (y-30)=0 

 y=20 or y=30

 if y=20 then x = 30 

 or y=30 then x = 20 

 two numbers are 30 & 20

18) Dinalal divides his property among his four sons after donating Rs.20,000 and 10% of his remaining property. The amounts received by the last three sons are in arithmetic progression and the amount received by the fourth son is equal to the total amount donated. The first son receives as his share RS.20,000 more than the share of the second son. The last son received RS.1 lakh less than the eldest son. 10. Find the share of the third son. 

a) Rs.80,000 

b) Rs.1,00,000 

c) Rs.1,20,000

 d) Rs.1,50,000

Ans: Assume the amounts received by the 2nd, 3rd, and 4th sons are a+d, a, a-d (as they are in AP)

 Now Eldest son received Rs.20,000 more than the 2nd son. 

So He gets a+d+20,000 

Last son received 1 lakh less than the eldest son.

 So (a+d+20,000) - (a-d) = 1,00,000 ⇒⇒ 2d = 80,000 ⇒⇒ d = 40,000 

So Amounts received by the 4 sons are a + 60,000, a+40,000, a, a - 40,000. 

Assume His property = K rupees.

 It was given that the youngest son's share is equal to 20,000 + 110(K−20,000)110(K−20,000)

 Then 20,000 + 110(K−20,000)110(K−20,000) = a - 40,000 ...........(1) and

 the Remaining property = Sum of the properties received by all the four son's together.

 Remaining property = 910910(K-20,000)

 ⇒910⇒910(K-20,000) = ( a + 60,000 ) + (a+40,000) + a +( a - 40,000) ..(2)

 Solving We get K = 40,000 and a = 1,20,000 So third son got Rs.1,20,000 .

19)A man sold 12 candies in $10 had loss of b% then again sold 12 candies at \$12 had profit of b% find the value of b.

Ans: Here 12 candies is immaterial. 

Loss % = CP−SPCP×100CP−SPCP×100

 So Here SP = 10 and loss% = b% CP−10CP×100=b⇒CP−10CP=b100CP−10CP×100=b⇒CP−10CP=b100

 In the second case he got a profit of b% So Profit % = SP−CPCP×100SP−CPCP×100

 So Here SP = 12 and profit% = b% 12−CPCP×100=b⇒12−CPCP=b10012−CPCP×100=b⇒12−CPCP=b100 

Solving 1 and 2 we get b = 1/11 or 9.09%  

20)15. A rectangular park 60 m long and 40 m wide has concrete crossroads running in the middle of the park and rest of the park has been used as a lawn.if the area of the lawn is 2109 sq.m,then what is the width of the road.

 a. 2.91 m 

b. 3m 

c. 5.82 m 

d. none 

Ans : Option : B Let us shift the path to the left hand side and top.

 This does not change the area of the lawn. 

Now lawn area = (60 - x) (40 - x) 

for x = 3, we get lawn area = 2109.