latest placement papers of TCS
151 / 652
In a school, sixty percent of the students are girls and thirty five percent of the girls are poor. If a student is randomly selected, what is the probability of selecting a poor girl student?
A60%
B35%
C21%
DNone of these
Answer: Option C
Explanation:Lets assume 100 be the total strength
60% out of 100= 60 number of girls
now 35% of girls are poor i.e. (35/100)*60 =21
Therefore probability of selecting a poor girl out of 21 poor girls is 21 c 1=21
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152 / 652
Two beakers are on the table. The capacity of the first beaker is x liters and that of the second beaker is 2x liters. Two thirds of the first beaker and one fourth of the second beaker is filled with wine. The remaining space is filled with water. If the content in both the beakers are mixed in a large beaker of volume 3x liters, what is the proportion of wine in the beaker?
A12/11
B11/36
C7/6
D7/18
Answer: Option D
Explanation:capacity of beakers x,2x,3x
2/3(x)+1/4(2x)=7x/6
wine proportion is =7x/(6*3x)=7/18
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153 / 652
Three non negative numbers, X, Y and Z are such that the mean is M and the median is 5. If M is 10 more than the smallest number and 15 less than the biggest number, find the value of X+Y+Z.
A15
B5
C20
D30
Answer: Option D
Explanation:Given median=5,so y=5
x+y+z/3=10+z ---(1)
x+y+z/3=x-15 ---- (2)
By solving eq. (1) and (2) we get,
x=25,y=5,z=0
so x+y+z=30
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154 / 652
From 5 men and 11 women, in how many ways can a panel of 11 be formed such that the number of men is not more than 3?
A1650
B2255
C5522
DNone of these
Answer: Option D
Explanation:5C0*11C11+5C1*11C10+5C2*11C9+5C3*11C8=2256
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155 / 652
After 6 years Raju's father will be twice that of his age and two years ago, his mothers age was twice of that of Raju's age. What is the sum of Raju's parent's age?
A4 less than four times Raju's age
B2 more than four times Raju's age
C4 more than four times Raju's age
D2 less than four times Raju's age
Answer: Option C
Explanation:F+6=2(R+6)
F= 2R+6
M-2=2(R-2)
M= 2R-2
Therefore the sum of Raju's Parent's age is
F+M=2R+6+2R-2
F+M=4R+4
4 more than four times Raju's age
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156 / 652
John told Mark that if Mark gives 1/3rd of his money to him, he will have Rs 75. Mark told John that if john gives 1/2 his money to him, he will have Rs 75. How much money did they have totally?
A105
B125
C150
D75
Answer: Option A
Explanation:Here is no explanation for this answer
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157 / 652
A starts riding his bike at 10am with a speed of 20kmph and B also starts at 10am with a speed of 40kmph from the same point in the same direction. A turns south at 12 o'clock and B turns north at 11 am. What will be the distance between A and B at 2 pm?
A250 km
B120 km
C160 km
D145.6 km
Answer: Option C
Explanation:At 12 o clock A cover 40km and on the other side B at 11 o clock cover 40km, again they traveled towards each other(which is actually the distance between them), that is A has to traveled 2hr(From 12 to 2 at 20km/hr) i.e 2*20=40km and other side B has to traveled 3hr(From 11 to 2 at 40km/hr) i.e 3*40=120Km.
Then the total distance traveled by them is the Actual distance between them ie 40+120=160Km(Ans.)
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158 / 652
A circle has 29 points arranged in a clockwise manner numbered from 0 to 28. A bug moves clockwise around the circle according to the following rule. If it is at a point i on the circle, it moves clockwise in 1 second by ( 1 + r ) places, where r is the reminder ( possibly 0 ) when i is divided by 11. Thus if it is at position 5, it moves clockwise in one second by ( 1 + 5 ) places to point 11. Similarly if it is at position 28 it moves ( 1 + 6 ) or 7 places to point 6 in one second.
If it starts at point 28, at what point will it be after 9994 seconds?
A1
B5
C7
D3
Answer: Option D
Explanation:Here is no explanation for this answer
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159 / 652
The value of a scooter depreciates in such a way that its value of the end of each year is 3/4 of its value of the beginning of the same year. If the initial value of the scooter is Rs.40,000, what is the value at the end of 3 years ?
A22500
B30000
C16875
D12656
Answer: Option C
Explanation:40,000(3/4)^3=16875
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160 / 652
There is a pool of radius X and there is a pathway around the pool with a width of 4 feet. Find the radius of the pool if the path area/ pool area=11/25.
A12
B5
C25
D20
Answer: Option D
Explanation:(PI.(X+4)^2-PI.X^2)/PI.X^2=11/25
(X+4)^2/X^2-1=11/25
(X+4)^2/X^2=36/25
(X+4)/X=6/5
X=20
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