TCS Questions - 13

TCS Questions - 13

1.       The savings of employee equals income minus expenditure. If the income of A,B,C are in the ratio 1:2:3 and their expense ratio 3:2:1 then what is the order of the employees in increasing order of their size of their savings?

Answer: A, B, C

The ratio of income of A, B, and C are 1:2:3. Let us assume their Incomes as 10000, 20000 and 30000

The ratio of their expenditure is 3:2:1. (C spends the least  while A spends the maximum). Assuming the expenditure of A, B and C as 9000, 6000 and 3000, then the

Savings of A, B and C works out to 1000, 14000 and 27000 the ratio being  1:14:27 The increasing order of employees’ savings is A, B and C.

2.       Two full tanks one shaped like the cylinder and the other like a cone contain liquid fuel. The cylindrical tank held 500 litres more than the conical tank. After 200 litres of fuel is pumped out from each tank the cylindrical tank now contains twice the amount of fuel in the conical tank.  How many litres of fuel did the cylindrical tank had when it was full?

Answer: 1200 litres

Let the initial quantity of fuel in the conical tank be ‘x’.  Then the quantity of fuel in the cylindrical tank is (x+500). Now 200 litres of fuel are removed from each tank.

The fuel in the conical tank now is (x-200). The fuel in the cylindrical tank will be

(x+500)-200=(x+300). From the question we know (x+300) = 2*(x-200). Solving this we get the value of ‘x’ as 700 litres. Hence the initial quantity of fuel in the cylindrical tank is (700+500)= 1200 litres.

3.       Eesha bought two varieties of rice costing 50Rs per kg and 60 Rs per kg and mixed them in some ratio. Then she sold that mixture at 70 Rs per kg making a profit of 20 %. What was the ratio of the mixture?

Answer: 1:4

The sale price of Rs 70/kg includes 20% profit. If the profit element is eliminated, we get the sale price as 70 * 100/120 = 58.33. (This can be rounded off to Rs 58.) Now applying Rule of alligation we have

                                                                                          A   50                           60 B

                                                                                                                58 X

                                                                                    (B-X)      2                         8 (X-A)

Thus the ratio of mixture of the two varieties Rs 50 and Rs 60 is 2:8 or 1 : 4.

4.       In a horse racing competition there were 18 horses numbered 1 to 18.The organizers assigned a probability of winning the race to each horse based on horses health and training. The probability that horse one would win is 1/7, that horse two would win is 1/8, and that horse three would win is 1/7. Assuming that tie is impossible, find the chance that one of these three horses would win the race?

Answer: 15/49

The chances of Horse No 1 winning    1/7        Probability of horse No 1 not winning     6/7

The chances of horse No 2 winning     1/8        Probability of horse No 2 not winning    7/8

The chances of horse No 3 winning     1/7        Probability of horse No 3 not winning     6/7.

The chances of any one of these three horses winning the race will be:

(A winning, B and C not winning) + (B winning and C and A not winning) + (C winning and A and B not winning)

Now substituting the relevant information in the above we get the following.

Probability of any one of the three horses winning is arrived at as

(1/7 + 7/8 + 6/7) + ( 1/8 + 6/7 + 6/7) + ( 1/7 + 6/7 + 7/8)   Reducing this we get the value

Of probability of any one of these three horses winning as 15/49.

5.       The missing number in the sequence
5, 6, 15, ?, 89, 170, 291, is

(a) 50                     (b) 40                    (c) 42                     (d) 32

Answer:  (b) 40

6 – 5 = 1  (1^2)       15 – 6 = 9 (3^2)

170 – 89 = 81 (9^2)     291- 170 = 121 (11^2) Going by this we observe that between 89 and 15 we should have two digits namely 5^2 and 7^2. Hence the missing number is

15      5^2 = 40.

6.       Two trains starting at same time, one from Bangalore to Mysore and other in opposite direction arrive at their destination 1 hour and 4 hours respectively after passing each other. How much faster is one train from other?

Answer: One train is four times faster than the other train.

The distances between the two stations remain the same.
The ratio of speeds will be inversely proportional to their respective times.
Let speed of train 1 be S1 and that of train 2 be S2.
Let time taken by train 1 be T1=1hr
and time taken by train 2 be T2=4hr
We now have a ratio

S1:S2 = 1/T1:1/T2 = T2:T1 = 4:1
So train 1 is 4 times faster than train 2.

7.       If f(f(n))+f(n)=2n+3 and f(0)=1, what is the value of f(2012)?

a.2011   b.2012   c.2013   d.4095

Answer: C

                We know f(0) = 1 and f{f(n)} + f(n) = 2n + 3

                Now substituting 0 for n in the above equation, we have f(f(0) + f(0) = 2(0) + 3.

                Since f(0) = 1 substituting this we have f(1) + 1 = 3. Hence f(1) = 2

                Similarly if we put the value of n = 1, then we will get f(2) = 3 and so on .

                The value of f(n) will be n+1

                On this basis f(2012) will be 2013. Hence the answer is ‘c’.

8.       A wooden rod AB is 1 meter long. Raju makes a mark every 3 cm along the rod starting from the end A. Vikram makes a mark every 7 cm and Aditya every 9 cm both starting from the end B. If the rod is cut at each of the points where there is a mark, then how many pieces would it be cut into?

Answer: 51 pieces

Raju will be marking at 33 places starting from the end A. Thus it will give 33 pieces.

Vikram will be marking at 14 places starting from the end B. Of these markings five markings at 93 cm, 72 cm, 51, cm, 30 cm, and 9 cm will coincide with the markings of Raju. Also the marking at 37 cm will coincide with that of Aditya. Thus six markings are common with either Raju or Aditya. Leaving aside this, Vikrams markings will give 8 pieces.

Aditya’s  cutting starting from the end B will give 11 pieces of which the marking at one place 37 cm is common with that of Vikram. Thus the number of pieces will be 10.

Hence the total number of pieces are 33 + 8 + 10 = 51 pieces.

9.       The crew of a rowing team of 8 members is to be chosen from 12 men  (M1, M2, …., M12) and 8 women (W1, W2,…., W8), such that there are two rows, each row occupying one of the 2 sides of the boat and that each side must have 4 members including at least one women. Further it is also known W1 and M7 must be selected for one of its sides while M2, M3 and M10 must be selected for other side. What is the number of ways in which rowing team can be arranged?

Answer:   70 ways.

Side One: Total Number 4.   M2, M3 and M10 are must. This leaves one seat for woman. Out of the eight women, W1 should be on side two. That leaves remaining 7 women from whom one woman can be selected in 7 ways.

Side Two: Total Number 4. W1 and M7 are must. That leaves two remaining seats that can be filled exclusively by either women or one woman and one man. We have 6 remaining women and the two seats can be filled in 6C2 ways ie 15 ways. Suppose we fill one man and one woman this can be done in the following manner. Since there are 6 women and 8 men available the filling can be done in (6C1)*(8C1) = 48 ways.

Thus the remaining seats in the boat can be filled in 7 + 15 + 48 = 70 ways.

10. At the end of 1994, Rohit, was half as old as his grandmother. The sum of the years in which they were born is 3844. How old Rohit was at the end of 1999?

A) 48                                      B) 55                                      C) 49                                      D) 53

Answer: (d)  53 years

Let x and y be the years when the Rohit and grandmother were born.

Thus we have x + y = 3844   …….  (I)

                In 1994 Rohit is half the age of his grandmother. We now have another equation

                (1994 – y) = 2(1994 – x) this will give a value of 2x – y = 1994………..(II)

                Adding (I) and (II) we get 3x = 5838 and x = 1946 Hence Y = 3844 – 1946 = 1898

                In 1994 Rohit’s age was 1994 – 1946 = 48 years.

                In 1994 grand-mother’s age was  1994 – 1898 = 96 years (Twice the age of Rohit)

                So, Rohit’s age in 1999 will be 48 + 5 = 53 years.

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TCS Questions - 12

TCS – Questions -12

1.       Raja and Rakesh started a business and invested Rs 20,000 and Rs 25,000 respectively. After 4  months  Rakesh left and Mukesh joined by investing Rs 15,000. At the end of the year there was a profit of Rs 4600. what is the share of profit of Mukesh?

Answer: Rs 1,200

Raja’s investment of Rs 20,000 was in the business for 12 months. Total amount Rs 2,40,000

Rakesh’s investment of Rs 25,000 was there for 4 months. Total amount                Rs 1,00,000

Mukesh’s investment of Rs 15,000 was there for 8 months. Total amount              Rs  1,20,000

Thus total turnover of business is Rs 4,60,000. On this turnover the profit earned is Rs 4,600

Mukesh share of profit is 4,600 * 1,20,000/4,60,000 = Rs 1,200 

2.       16 men can complete a work in 24 days , while 48 children can do it in 16 days. 12 men started the work , after 14 days 12 children joined the work. In how many days will all of them do the remaining work ?

Answer: 12 days

16 Men can complete the work in 24 days. 16 Men in one day can do 1/24 work.

1 man in one day can do 1/24*16 = 1/384 work.

48 children can do the work in 16 days. 48 Children in one day can do 1/48 work.

1 child in one day can do 1/48*16 = 1/768 work.

Now, 12 men work for 14 days. The completed work is 12*14/384 = 168/384 work.

The remaining work is 1 – 168/384 = 216/384.

Now, 12 men and 12 children in one day can do 12/384 + 12/768 = 36/768 work.

The remaining work is 216/384 and the 12 men and 12 children will take

(216/384) / (36/768) days. Reducing this you will get a value 12 days.

3.       How many motor vehicle registration number plates can be formed with the digits 1,2,3,4,5(no digits being repeated ) if given that registration number can have 1 to 5 digits? 

Answer: 35

Number plates with one digit    =5c1ways  =   5
Number plates with two digits   =5c2ways  = 10
Number plates with three digits =5c3 ways = 10
Number plates with four digits   =5c4 ways =   5
Number plates with five digits    =5c5 ways =   5
Total=35ways

4.       5000 voted in an election between two candidates. 14% of the votes were invalid. The winner won by a margin approximately closer to 15%. Find the number of votes secured by the winner.

a)2650                   b)2564                  c)2473                   d)2360

             Answer: C.

Total votes polled 5000. 14% was invalid. Hence valid votes were 5000 – 700 = 4300

The winner secured a margin of 15% over the rival. Thus the winner’s votes were

4300*15/100 = 645 votes more than the rival. Excluding this the remaining votes were equally shared by both the candidates. Each candidate got (4300 – 645) / 2 = 1827.5.

Adding the winner’s margin of 645 votes to this figure, we get the total number of votes secured by the winner approximately as 1827.5 + 645 = 2473.

5.       In 2003 there are 28 days in February and 365 days in the year. In 2004 there are 29 days in February and 366 days in the year. If the date March 11 2003 is Tuesday, then which one of the following would be the day for date March 11 2004? 

a)    Monday       b)  Wednesday      c)   Thursday                  d)    Friday

Answer: C.

Simple. If the year 2004 was not a leap year then the date March 11, would be a day ahead, ie Wednesday. Since 2004 is a leap year and March comes after February that has one extra day the date March 11, 2004 would be Thursday.

6.       A child was looking for his father. He went 90 mts in the east before turning to his right. He went 20 mts before turning to his right again to look for his father to his uncle’s place 30 mts from this point. His father was not there. From here on he went 100 mts to the north before meeting his father in the street. How far away from the starting point did the son meet his father?

Answer: 100 meters. 

The route the boy takes would appear as under. We have to find the distance from the starting point to end point.

Distance between A and B 90 mtrs.    B and C  20 meters .  Distance between C and D 30 meters.  Distance between D and F 100 meters.  From this information we now arrive at the following:   Distance between A and E = 90 – 30 = 60 mtrs.

Likewise distance between E and F = 100 – 20 = 80 mtrs.  Now applying the Pythagoras theorem we arrive the distance of A and F as 100 meters.

7.       How many positive integers less than 500 can be formed using the numbers 1,2,3,and 5 for digits, each digit being used only once.

Answer: 74

Only four numbers are available and repetition is not allowed.

Number of single digit numbers 4C1  =                                                        4

Number of double digit numbers 4C2  =        6 * 9                               54

Number of three digit numbers    4C3  =        4 * 4                               16

Total number of positive digits                                                   74

8.       A circular swimming pool is surrounded by a concrete wall 4 feet wide.if the area of the wall is 11/25 of the area of the pool, then the radius of the pool in feet is?

Answer: 20 feet.

Let the radius of the pool be ‘r’.  The thickness of the wall is 4 ft.

Area of the pool +area of the wall = pi(r+4)²

Area of the pool = pi*r^2.

Area of the wall alone =   pi(r+4)² – pi*r² = 11/25 (pi*r²)

Solving this we get the value of ‘r’ as 20 ft.  

9.       A survey of n people in the town of Badaville found that 50% of them prefer brand A.

Another survey of 100 people in the town Chottaville found that 60% prefer brand A. In total 55% of all the people surveyed together prefer Brand A. What is the total number of people surveyed? 

Answer: 200 persons (a very simple question)

Assume the number of people surveyed Badaville as 100. Then those preferring Brand A is 50

In Chotaville those preferring brand A is 60% of 100 surveyed. ie 60.

Total preference for brand A in both the places is 110 which is 55% of the number surveyed. Hence, the total number of people surveyed is 200.

10.   One day Eesha started 30 min late from home and reached her office 50 min late while driving 25% slower than her usual speed. How much time in min does Eesha usually take to reach her office from home?

Answer: 60 minutes

Let the distance be D and the usual speed S. Then the time taken to travel is D/S=T     (1)

The new speed is (S- 1/4S) = ¾ S. The new time taken is  D/3/4 S = T + 20 (50 – 30)  (2)

Equating (1) and (2) we get the value of T as 60 minutes.            

 

 

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TCS Questions - 11

TCS Questions - 11

1.      
From the top of a 9 mtr high building AB, the angle of elevation to the top of a tower CD is 30⁰ and the angle of depression of the foot of the tower is 60⁰. What is the height of the tower?                                                                                             C                                        

 A:  12,   B:   18      C:   9     D:    15                                                                           

Answer:    B      

Angle of elevation from A to C is 30⁰ and angle of depression            E                                 A

from A to D is 60^0.                                                                                                                                           

We now know ∟BAD is 30⁰ and ∟EAC is 30^{0                                                                                                                                                                                                                                                                                                                9 mtrs}

Let the distance BD be x. Then the distance EA also is x                       D                            B

Let EC be Y

                                                                                                                                       

Tan BAD  = BD/BA = x/9 = √3.  Hence x = 9√3

Tan EAC  =  EC/ EA = Y/9√3 = 1/√3 Hence Y = 9

AB = 9 mtrs = ED. Hence the height of CD the tower is 9 + 9 = 18 mtrs.

               

2.       In a mixture of A, B, and C if A and B are mixed in the ratio 3:5 and B and C are mixed in the ratio 8:5 and if the final mixture is 35 ltrs, then what is the quantity of B?

A: 13.34       B:  15.73     C:    16.73       D:   9.45

Answer: B

We have two different values for B in the above ratio.

A : B :: 3 :5    and B : C :: 8 : 5   We shall now make the value of B the same in both the ratios.

Multiplying A : B by 8 and B : C by 5 we get new ratios where the value of B is the same.

A : B :: 24 : 40   and B : C :: 40 : 25

Combining all the three we have the ratio A :B : C :: 24 : 40 : 25   The total quantity of the final mixture is 35 ltrs.

Quantity of B in the mixture is  (35 x 40) / 89 = 15.73

3.       My flight takes off at 2.00 A.M. from a place 18N 10E and after 10 hours flight lands at a place with coordinates 36N 70W. What is the local time when the plane lands at the destination?

A:  12 noon     B:   6.40 a.m.     C:    5.20 p.m.      D:   6.50 a.m.

Answer: B

The longitudinal difference between the two stations is 70W + 10E = 80⁰

For every one degree variance in the longitude the time difference is 4 mts. As you go towards East the time increases, and as you go west the time decreases. In this case the flight is towards west and the destination time will be less than the originating station.

The time variance between the two stations is 4 x 80 = 320 minutes or 5 hours and 20 minutes. 

The flying time was 10 hours. The time of departure was 2.00 a.m. Hence after 10 hours the time at the originating station was 12.00 noon. ( 2 + 10 ).  The time difference between the two stations is 5 hrs 20 mts. Hence the time of landing at the destination was

12.00 – 5.20 = 6.40 a.m.

4.       In a100 mtrs race, A beats B by 15 mtrs and B defeats C by 10 mtrs. If A beats C by 5 seconds then find the speed of C.

A:  5 mtrs/sec      B:   4.7 mtrs/sec              C:  4 mtrs/sec         D:  4.5 mtrs/Sec

Answer: B

When A covers 100 meters, B was at a distance of 85 meters. When B covers 100 meters C was at a distance of 90 meters.

Now, When A covers 100 meters C was at a distance equal to (85 x 90)/100 = 76.50 meters.

A defeats C by 5 seconds. In other words to run 100 – 76.50 = 23.50 meters C takes 5 seconds. Hence the speed of C = 23.50/5 = 4.70meters per sec.

5.       Oshkosh used a study of colours in African National Flags.  He found 38 flags have Red, 20 have Blue and 13 have both Red and Blue and 8 have neither Red nor Blue. How many flags have:

a)      Have Red but not Blue?

b)      Have Blue but not Red?

c)       Were included in the study?

A:   25, 7, 53        B:   38, 20, 79        C:    25, 7, 66         D:    30, 20, 60

Answer: A

38 flags have Red less 13 flags having Red and Blue. Hence flags having only Red are 25

                20 flags have Blue less 13 flags having Red and Blue. Hence flags having only Blue are 7

Total number of flags studied:  Only Red 25. Only Blue 7. Both Red and Blue 13. Neither Red nor Blue 8. Total flags 25+7+13+8 = 53.

6.       How much per cent must be added to the cost price of goods so that a profit of 20% must be made after giving a discount of 10% from the market price.

A) 204
B) 30
C) 33 3/3
D) 25
E) 10

Answer : C  

This question can be answered from the choices given. The marked price will be 133.33% .10% discount on this will be 13.33%. Reducing this we get the sale price as 120% which gives 20% profit.

7.       In a class of boys and girls Vikas's rank is 9th and Tanvi's Rank is 17th . Vikas's Rank among the boys in that class is 4th from the top and 18th from the bottom and Tanvi's rank among the girls is 8th from top and 21st from bottom. In the order of rank, how many girls are there between Tanvi and Vikas ?

Answer: 2 Girls.

                                                                Vikas                                                                                     Tanvi

Vikas rank                                              9                                                                              17

Vikas rank among the boys             4 (from top)     Tanvi’s rank among the Girls           8 (from top)

Vikas rank among boys                   18 (from last)     Tanvi’s rank among the girls           21(from last)

Total number of Boys    (4+18-1)  21                         Total number of Girls (8+21-1)       28

Total Students in the class  21 + 28 = 49

There are 5 Girls and 3 boys above Vikas rank which is 9.

There are 7 Girls and 9 boys above Tanvis rank which is 17. Hence the number of Girls between Vikas and Tanvi are 2

8.       Two men together start a journey in the same direction. They Travel 9km/hr and 15km/hr respectively. After travelling for 6 hours the man travelling at 9km/hr doubles his speed and both of them finish the distance in the same time. How many hours will they take to reach their destination?

Answer: 18 hours.

Let the two men be A and B. A travels at 9kmph and B travels at 15kmph.

After 6 hours A would have covered  9 x 6 = 54 kms

After 6 hours B would have covered  15 x 6 = 90 kms.    B is ahead of A by 36 kms.

A now increases his speed to 18kmph and B continues at 15 kmph. The relative speed between A and B now is 18 – 15 = 3kmph. To cover the difference in distance of 36 kms A would take 36/3 = 12 hours.

Thus A would take totally 18 hours to cover a distance of 270kms.  (6*9 + 12*18)

B also would take 18 hours to cover the distance of 270 kms ( 15*18).

9.       A train covers a distance between stations A and B in 45 Minutes. If the speed is reduced by 5km/hour. It will cover the same distance in 48 Minutes, What is the distance between station A & B in kms?

Answer: 60 km

Speed * time would give the distance.  Let the initial speed be x kmph. The time taken is 45 minutes or ¾ hrs.

Hence Distance = 3x/4.     Now the speed is reduced by 5 kms and the time taken is 48 minutes or 4/5 hrs.

Hence Distance = (x-5)*4/5 . Both these are equal. So we have an eqution

3x/4 = (4x-20)/5.    Solving we get the value of x as 80kmph. At 80 kmph the time taken to cover the distance

Was ¾ hours. Hence the distance between A and B = 80 * ¾ = 60 km.

10.   The average set of numbers is 46, if 4 numbers whose average is 52 are subtracted from this set , the

average becomes 44.5...find the original number of numbers in the set ?..

Answer: 20.

Let the set of numbers be X. Then the total of the numbers are 46 * X = 46X.

4 numbers with average of 52 are removed. Total of these 4 numbers is 4*52 = 208.

The new average for (X-4) numbers is 44.5. Thus we have an equation

46X = 44.5(X-4) + 208.    Solving we get the value for X as 20.

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