TCS
Questions – 8
1.
A
person is taking an objective type exam consisting of 30 questions. His score
is found to be more than 80 using the formula S = 30 + 4C – W ( where S denotes
the score obtained, C the number of correct answers and W the number of wrong
answers). If the score is little less still more than 80 then it would be
difficult to answer. How many correct questions he answered?
Answer: 16 correct questions.
It
is trial and error method. Since the score was observed to be more than 80 the
only possible answer could be 4C – W should equal 50 or little more. The only
possible value for C is 16 in which case each wrong answer gets one negative
mark. Any other figure for C would not get you the proper answer.
2.
A
plane goes from Chicago to Columbus and returns. The time of travel either way
is the same. It leaves Chicago at 06.44 hours local time and arrives in
Columbus at 08.49 hours local time. The plane leaves Columbus at 16.25 hours
local time and arrives in Chicago at 16.38 hours local time. What is the time
taken by the plane to travel either way?
Answer: 69 minutes.
The
plane left Chicago at 06.44 hours in the morning and returns at 16.38 hours in
the evening. Hence the plane was away from Chicago for 9 Hours 54 minutes.
The
plane reaches Columbus at 08.49 hours in the morning and leaves at 16.25 hours
in the evening. Thus the plane was at Columbus for 7 hours and 36 minutes.
Hence
the plane was air borne for 9.54 hours minus 7.36 hours.
That
is 2 hours and 18 minutes or 138 minutes. This is the total air borne time for
both the ways. Hence the time taken by
the plane to travel either way is 139/2 = 69 minutes.
3.
If 3y + x > 2 (i) and x + 2y ≤ 3 (ii) then
what can be said of the value of y?
Answer: y > -1
Let
us try to eliminate one of the variables in the above two equations. Take the
second equation and multiply the same by -1. We now have
-x
-2y ≤ - 3. Adding this with equation (i) we
get y > -1
4.
What
is the sum of all even integers between 99 and 301?
Answer: 20200
The
first and last even digits between 99 and 301 are 100 and 300. We have to find
the sum of
(100
+ 102 + 104 + 106 + ……….. + 300) this
can be written as
2
( 50+51+52+53+54+……..+150) there are 101 terms in this series. Ie ‘n’ terms
Hence
the sum is given by n/2 ( a+l) where a and l are the first and the last terms.
Thus
the sum of (50+51+52+…….+150) will be 101/2 * ( 50+150).
Hence
the answer to the question is
2 * 101/2 * 200 = 20200
5.
There
are 20 red, blue or green balls. If there are 7 green balls and the sum of red
and green balls are less than 13, then at most how many red balls are there?
Answer: 5 balls. (simple logic)
R
+ B+ G = 20. G = 7. G + R < 13 7 + R < 13 Hence R < 6 (the maximum value) Hence
the maximum value of R could only be 5.
6.
If
‘n’ is the sum of two consecutive odd integers and is less than 100, then what
is the maximum value that ‘n’ could take?
Answer: 96
Let
us have the two integers as (2n + 1) and
(2n – 1).
Now
from the question, we have (2n+1) + (2n-1) < 100
So,
4n < 100 . Hence the maximum value for n could be 24 as 96 is the only
maximum integer divisible by 4 that is less than 100.
7.
M,
N, O and P are different individuals. M is the daughter of N. N is the son of
O. O is the father of P. Among the following statements which one is true?
(a) M is the daughter of P
(b) If B is the daughter of N, then M and
B are sisters.
(c) If C is the grand-daughter of O, then
C and M are sisters.
(d) P and N are brothers.
Answer: (b) Simple logic. Draw a family tree and then
answer.
8.
There
are 10 stepping stones numbered 1 to 10. A fly jumps every minute four steps
from its previous position. From 2 it would go to 6th and from there
to the 10th. Since there are only 10 steps, from the 10th
it would come down to 4th step and so on. If the fly starts from
step No 1, then after one hour in which step it would be?
Answer: Step No 1.
A
simple reasoning question. After every 5 minutes the fly will be in the same
step from which it started. After 60 minutes (one hour), it will again be in
step no 1.
9.
What is the sum of the base 7 numbers 1234 and
6543 n base 7?
Answer: 11110.
In base 7 there is no number 7 since it starts
from 0,1,2,3,…6. Hence to write 7 we start from 10, for 8, 11 and so forth.
A
B
C D
We
shall write the numbers as
1 2 3 4
6
5 4 3
Now
the sum of these …4+3=7 Under base 7 the value is 10. We take 0 (1 is carried)
3+4=7
plus carried forward 1 total 8. The base value for this is 11.Keep 1 and carry
1
5+2=7
plus 1 carry total 8. Base value again is 11. Keep 1 and carry forward 1.
6+1=7
plus carry 1, total 8. Base value for 8 is 11. Thus the answer is
11110.
10. A, B, C, D go on a picnic. A and B
weigh themselves on a scale and are 132 kg. B and C weigh 130 kg. C and D weigh
102 kg and B and D weigh 116 kg. What is the weight of D?
Answer: 44 kg.
B
+ C = 130 kg B + D = 116 kg Eliminating B from these two,
We
get C – D = 14 kg. C + D = 102 kg. Adding
these two we get 2C = 118 and C=58
Substituting
the value of C, We get the weight of D
as 44 kg.
--- x ---
No comments:
Post a Comment