TCS Questions – 3
1.
For
climbing to the top of a hill 800 meters, Jack after climbing 16 meters in a minute
takes two minutes rest, or one minute rest after climbing 20 meters in a
minute. Paul takes one minute rest after climbing 10 meters in a minute and two
minutes rest after climbing 16 meters in a minute. If the time allowed to reach
the top of the hill is 2 hours (120 minutes) then what is the maximum number of
rests that can be taken?
Answer
: 40.
Let us take Jack first. He will take 50 minutes to climb the
hill at 16 meters per minute. In the process he will take 50 times rest of 2
minutes. Thus the total time will be 50 + 100 = 150 minutes. (But the time
allowed is only 120 minutes)
If
he climbs at 20 meters in 2 minutes then the time to reach the top of the hill
will be 80 minutes. He will 40 times
rest of one minute each. The total time to reach the top will be 80 + 40 =
120 minutes. (Exactly the time allowed)
Now
let us consider Paul. If he climbs
10 meters per minute then he will take 80 minutes to reach the top. In the
process he will take rest for 80 times of one minute each. Thus the total time
would be 80 + 80 = 160 minutes ( Time allowed is only 120 minutes). If he
climbs at 16 meters per minute then the time take to reach the top of the hill
will be 50 minutes. In the process he will take total rest time of 100 minutes.
Hence the total time to reach the top will be 50 + 100 = 150 minutes ( Time
allowed is only 120 minutes)
Thus the maximum number of rest that
one of them can take is 40
2. If the 20th term of an AP
is 560 and the 30th term of the AP is 840, then what is the sum of
the 5th term and 40th term of the series?
Answer is : 1260
The
30th term is 840 and the 20th term is 560. Hence the 10th
term is 840 – 560 = 280.
The
5th term is 280/2 = 140. The 40th term is 30th
term + 10th term ie 840 + 280 = 1120. Hence the sum of 5th term and 40th
term is 140 + 1120 = 1260.
3. Raj tossed three dices. If the results
are noted down, then what is the probability of Raj getting 10?
Answer is: 1/8
The
possibility of Raj getting 10 is in the following manner: 136/145/154/163.,
226/235/244/253/262; 316/325/334/343/352/361; 415/424/433/442/451/;
514/523/532/541/; 613/622/631/
27
possible manner as against total possibilities of 216 (6 x 6 x 6) ie 27/216 or
1/8
4.
The
diagonal of a square is twice the side of equilateral triangle. What is the
ratio of the area of the triangle to the area of the square?
Answer is: √3 : 8
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|
Let
the sides of the equilateral triangle be ‘ a ‘
Then the area of the equilateral triangle will be √3/4 a^2.
The
area of the square will be 2a^2/2 ie 4 a^2/2
Thus
the ratio will be √3/4 a^2 : 2 a^2
ie √3 : 8
5. Apple costs L rupees per kilogram for
first 30 kg Q rupees per kg for each additional kilogram. If the price of 33 kg
of apple is 11.67 and 36 kg of apple 12.48 then the cost of the first 10 kg of apples
is how much?
Answer: 3.62
36
kg costs 12.48 Cost of 33 kg 11.67
33
kg costs 11.67 Hence,
Cost of 3 kg 0.81
3 kg costs 0.81
Cost of 30 kg 10.86
Thus
cost of 10 kg 10.86/3 = 3.62
6.
In
the following series
1,2,2,3,3,3,4,4,4,4,1,1,2,2,2,2,3,3,3,3,3,3,4,4,4,4,4,4,4,4,…..
What
is the number in the 2320 position of the sequence?
Answer is: 1
The
first cycle in the sequence consists of 10 digits, the second 20 digits, the
third 30 digits and so on. So we have a
series 10 + 20 + 30 + 40 + ………….. +
x where x is the nearer value of 2320.
Thus we have 10 [ n (n+1)/2] nearer to 2320. N value 21 fill this condition and
we have x = 2310. This completes the cycle and the next cycle begins with the
digit 1. Hence the answer is 1.
7.
How
many 7’s are there between 0 and 400?
Answer: 79. From 0 to 100 there are 19 7’s and for each
100 numbers thereafter there are 20 7’s.
8.
The
letters in the word ABUSER are permuted in all possible ways and arranged in
alphabetical order. Then what would be the word in the 49th position
of this alphabetical order?
Answer: ARBESU
Arrange
the letters of ABUSER in the dictionary order; It will be ABERSU
Now
keep the first two letters AB. The remaining 4 letters can be arranged in 4! Ways.
ie 24 ways. Now take the next letter after B and keep AE aside. The remaining
four letters can again be arranged in 4! Ways ie 24 ways. Thus we have
completed 48 ways. The next is 49th way. Take A and R as first two
letters and the remaining four follow in the dictionary order. Thus we have the 49th word as
ARBESU
9.
15
students join a summer course. Every day 3 students are to stay back after
classes to clean the class room. After the course was over it was observed that
each pair of students were on duty only for one day. How many days the course
lasted?
Answer: 105 days.
The answer to this question can be arrived at
n two ways. Out of the 15 students it was observed that each pair had worked
only once. Hence the course lasted for 15C2 days. ie 105 days.
The
second way: Three students are to stay
back and clean the class rooms. If we number the students from 1 to 15 then we
have the following interesting thing.
Student
No 1 would have worked with the other students 14 days.
Student
No 2 would have worked with the other students 13 days
Student
No 3 …………………………………………………12 days. As you
go down
…….
……..
The
last Student would have worked with the other for 1 day only.
If you total the number of days you
will get 105 days which is the duration of the course.
10. Next number in the series: 1, 7, 8, 49, 50, 56, 57, 343, ?
Answer: 344
The
series goes in the following manner: 1*7
=7 7 + 1 = 8
7*7 = 49 + 1 =
50
7*8 = 56 + 1 =
57
7*49 = 343 + 1 = 344
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