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Solve the following question by using the matrix given below:-
9 10 11 12
5 13 17 19
4 8 12 16
7 2 3 5
Q.No. : 1
Question : Change row(3) by col(3) and then column(3) by column(1) then the element at (4 x 3) is
A:
7
B: 11
C (Your Answer): 12
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D: 3
Answer: A
Solution
Given Matrix:
Change row(3) by col(3)
9 10 11 12
5 13 17 19
4 8 12 16
7 2 3 5
Step 1: Change row(3) by col(3)
9 10 4 12
5 13 8 19
11 17 12 3
7 2 16 5
Step 3: Change col(3) by col(1)
4 10 9 12
8 13 5 19
12 17 11 3
16 2 7 5
the element at (4 x 3) is 7
Q.No. : 2
Question : Change column(1) by column(3) after that the row(2) by row(4) then which row contains only prime numbers
A: 2
B: 1
C (Your Answer): 4
D:
2 and 4 both
Answer: D
Q.No. : 3
Question : Change row(2) by row(3) then row(3) by column(3), then the element at the
position (2 x 3 ) is
A: 17
B (Your Answer):
13
C: 8
D: 5
Answer: B
Solution
After changing row(2) by row(3) matrix will convert to,
9 10 11 12
4 8 12 16
5 13 17 19
7 2 3 5
then changing row(3) by column(3) we get
9 10 5 12
4 8 13 16
11 12 17 3
7 2 19 5
So at the position (2 x 3 ) element is 13.
Q.No. : 4
Question : Change all the elements diagonally by keeping 7 fixed then change the row(3) by column(3), find the element at position (4 x 3)
A: 3
B: 16
C:
10
D (Your Answer): 8
Answer: C
Solution
The given Matrix is:
9 10 11 12
5 13 17 19
4 8 12 16
7 2 3 5
Step 1: Change all the elements diagonally by keeping 7 fixed
Take mirror image about the diagonal containing 7:
5 16 19 12
3 12 17 11
2 8 13 10
7 4 5 9
Step 2: change the row(3) by column(3)
5 16 2 12
3 12 8 11
19 17 13 5
7 4 10 9
Element at position (4 x 3) is 10
Q.No. : 5
Question : Change the row(2) by row(4) then row(4) by row(1), find the column whose sum of element is least
A:
1
B: 2
C: 3
D (Your Answer): 4
Answer: A
Solution
Final matrix after rotation
5 13 17 19
7 2 3 5
4 8 12 16
9 10 11 12
So least sum column is first column
Solve the following question using the matrix shown below:-
{a b c d
e f g h
i j k l
m n o p}
Q.No. : 6
Question : Exchange the row(2) by col(2) then the row(2) by row(4) then row(4) by col(4), the element at the position (4 x 3) is
A: k
B: f
C:
l
D: j
Answer: C
Solution
Step 1. Exchange the row(2) by col(2)
a e c d
b f j n
i g k l
m h o p
Step 2. Exchange the row(4) by col(4)
a e c m
b f j h
i g k o
d n l p
element at the position (4 x 3) is l
Q.No. : 7
Question : In the above question change the row(3) by column(3), then the element at (3 x 2) is
A: l
B:
g
C: j
D: k
Answer: B
Solution
Given Matrix:
a b c d
e f g h
i j k l
m n o p
Change the row(3) by column(3) :
a b i d
e f j h
c g k o
m n l p
Element at (3x2) is g
Q.No. : 8
Question : Change all the rows by the column of same numbers, and doing the same by keeping the element d fixed for the next, then the element at position (3 x 3) is
A:
f
B: h
C: k
D: n
Answer: A
Solution
Change all the rows by the column of same numbers = Transposing the matrix by
keeping 'a' fixed=
a e i m
b f j n
c g k o
d h l p
Now we will transpose by keeping 'd' fixed
p o n m
l k j i
h g f e
d c b a
so element at 3x3=f
Q.No. : 9
Question : Change the column(2) by row(2) and then row(3) by row(2) then what is the element at (3 x 2).
A: e
B: b
C:
f
D: j
Answer: C
Q.No. : 10
Question : Change the row(3) by column(3) then what is the element at position (3 x 3)
A: g
B: c
C: j
D:
k
Answer: D
Solve the following question using the matrix given below:-
a b c d
p q r s
u v w x
1 2 3 4
Q.No. : 11
Question : What is the value of trace(M), if a=1 and q+w=0.
A:
5
B: 0
C: 3
D: 1
Answer: A
Solution
The trace of an n-by-n square matrix is defined to be the sum of the elements on the main
diagonal
Trace = a + q + w + 4
Given, a = 1, q + w = 0
So Trace = 1 + 0 + 4 = 5
Q.No. : 12
Question : If we diagonally change the elements by keeping a fixed then the element above r is?
A: c
B:
q
C: w
D: 3
Answer: B
Solution
By keeping a fix change the element diagonally
means keep the diagonal containing a intact and treat it as a mirror
a b c d
p q r s
u v w x
1 2 3 4
will change to
a p u 1
b q v 2
c r w 3
d s x 4
In other word "Transpose"
So element above "r" will be "q"
Q.No. : 13
Question : Change the row(2) by row(4) then the column(3) by column(1) then what is the element between the vowel..
A: 2
B:
1
C: p
D: s
Answer: B
Q.No. : 15
Question : If we change the column(1) by column(4) and then row(1) by row(3) then the element above 2 is
A: v
b
Q.No. : 14
Question : If we change the numbers by the alphabets placed at the same position of alphabets then the value of det(M) is
A: (pqv-wqb)cu
B: 2(pq-rs)(ab-cd)
C: (pq-rs)/(ab-cd)
D:
0
Answer: D
Solution
det(M) is determinant value of the matrix M.
Det(M) =
12 3 4
16 17 18 19
21 22 23 24
1 2 3 4
= 0
B:
C: q
D: none of these
Answer: B
Q.No. : 16
Question : 1,2,3,2,3,4,3,4,_____
A: 3
B: 4
C:
5
D: 6
Answer: C
Q.No. : 17
Question : 3,15,35,63,_____
A: 71
B: 73
C: 95
D:
99
Answer: D
Solution
1 * 3 = 3
3 * 5 = 15
5 * 7 = 35
7 * 9 = 63
9 * 11 = 99
Q.No. : 18
Question : 1,3,7,13,_____
A: 20
B: 17
C: 19
D:
21
Answer: D
Q.No. : 19
Question : 2,6,14,26,_____
A: 38
B: 36
C:
42
D: 34
Answer: C
Q.No. : 20
Question : 1,3,4,5,7,8,11,13,_____
A:
14
B: 10
C: 15
D: 16
Answer: A
Solution
This can be considered of 3 number series,
1,3,4,5,7,8,11,13, ?
|_____|____|
|____|______|
|____|_______|
1 + 4 = 5, 5 + 6 = 11
3 + 4 = 7, 7 + 6 = 13
4 + 4 = 8, 8 + 6 = 14
Hence 14 is answer
Q.No. : 21
Question : 101,122,145,_____
A:
170
B (Your Answer): 163
C: 197
D: 168
Answer: A
Q.No. : 22
Question : 1,4,2,5,3,_____
A: 1
B:
6
C (Your Answer): 2
D: 0
Answer: B
Q.No. : 23
Question : 25,26,24,25,23,24,_____
A (Your Answer): 25
B: 21
C: 23
D:
22
Answer: D
Q.No. : 24
Question : 25,26,25,26,27,26,_____
A: 26
B (Your Answer): 28
C:
27
D: 25
Answer: C
Q.No. : 25
Question : 3,8,15,24,_____
A:
35
B (Your Answer): 30
C: 32
D: 42
Answer: A
Q.No. : 26
Question : 3,14,39,_____
A (Your Answer): 74
B: 64
C: 72
D:
84
Answer: D
Solution
The series goes like
3 * 1 = 3
7 * 2 = 14
13* 3 = 39
21 * 4 = 84
Q.No. : 27
Question : 2,6,12,20,_____
A (Your Answer):
30
B: 28
C: 32
D: 25
Answer: A
Q.No. : 28
Question : 20,21,21,22,22,_____
A (Your Answer): 23
B:
22
C: 24
D: 21
Answer: B
Solution
20 - 1 time
21 - 2 times
22 - 3 times
Hence 22
Q.No. : 29
Question : 6,24,60,120,_____
A: 220
B (Your Answer):
210
C: 240
D: 260
Answer: B
Q.No. : 30
Question : 6,24,60,120,_____
A (Your Answer): 220
B:
210
C: 240
D: 260
Answer: B
Solution
6*1 = 6
6*4 = 24
6*10=60
6*20=120
6*? = ?
1,4,10,20,?
(4-1)=3 (10-4)=6 (20-10)=10
3,6,10,?
Next will be 15 offcourse ( 3+3=6 6+4=10 10+5=15)
So After 1,4,10,20 it will be 20+15 = 35
So Next number will be 6*35=210
Q.No. : 31
Question : 1,2,3,2,4,6,4,8,_____
A: 8
B (Your Answer): 6
C: 16
D:
12
Answer: D
Q.No. : 32
Question : 7,26,63,124,_____
A (Your Answer): 243
B: 247
C:
215
D: 245
Answer: C
Q.No. : 33
Question : 1,1,2,3,5,8,13,___
A: 17
B: 18
C: 19
D:
21
Answer: D
Q.No. : 34
Question : 21,22,21,22,23,22,23,24,23,24,_____
A: 23
B: 24
C:
25
D: 26
Answer: C
Q.No. : 35
Question : 1,1,2,6,24,____
A: 96
B: 48
C: 72
D:
120
Answer: D
Q.No. : 36
Question : The regular hexagon of side 2m, inside which a circle is drawn by touching all its side. Find the area of the circle.
A: pie
B: 2*pie
C:
3*pie
D: pie/2
Answer: C
Q.No. : 37
Question : 7 spider makes 7 webs in 7 days then 1 spider makes 1 web in how many days?
A: 1
B:
7
C: 1/7
D: 7/2
Answer: B
Solution
7 spider makes 7 webs in 7 days
1 spider makes 7 webs in 7*7 days
1 spider makes 1 webs in 7*7/7 days = 7 days
Q.No. : 38
Question : In a regular week, there are 5 working days and for each day, the working hours are 8. A man gets Rs. 2.40 per hour for regular work and Rs. 3.20 per hours for overtime. If he earns Rs. 432 in 4 weeks, then how many hours does he work for ?
A: 160
B:
175
C: 180
D: 195
Answer: B
Solution
In 4 weeks, working days = 4*5= 20
So total working hours = 20 * 8 = 160
Earnings from working hours = 160 * 2.4 = 384 Rs
So earning from overtime = 432 - 384 = 48 Rs.
Total overtime hours = 48/3.2 = 15 Hrs.
So Total hours worked = 160 + 15 = 175
Q.No. : 39
Question : The banker's gain of a certain sum due 2 years hence at 10% per annum is Rs. 24. The present worth is:
A: 480
B: 520
C:
600
D: 960
Answer: C
Solution
Let x be the T.D
x*10*2/100=24
x=120
Present worth*10*2/100=120
Present worth=600
Q.No. : 40
Question : In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together?
A: 120
B:
720
C: 4320
D: 2160
Answer: B
Solution
We divide OPTICAL into two groups : consonants and vowels
Consonants : PTCL
Vowel : OIA
Now vowels has to be together so we consider it as one group
Different ways in which letters of the word 'OPTICAL' be arranged = 5!*3! =720
Q.No. : 41
Question : Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?
A: 210
B: 1050
C:
25200
D: 21400
Answer: C
Solution
Ways to choose 3 consonants out of 7 consonants =7C3Ways to choose 2 consonants out of 4
consonants =4C2
Ways to arrange 5 letters = 5!
So total ways = 7C3 *4C2*5!
= 25200
Q.No. : 42
Question : 4 men and 6 women can complete a work in 8 days, while 3 men and 7 women can complete it in 10 days. In how many days will 10 women complete it?
A: 35
B:
40
C: 45
D: 50
Answer: B
Solution
Let one men completes work in x days and one women in y days
(4*8)/x+(6*8)/y=1 =>4/x+6/y=1/8
and
(3*10)/x+(7*10)/y=1 => 3/x+7/y=1/10
On solving y=400
One women completes work on 400 days
therefore 10 women completes work in = 400/10 =40 days
Q.No. : 43
Question : A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the first pipe is:
A: 6 hour
B: 10 hour
C:
15 hour
D: 20 hour
Answer: C
Solution
Let time required by the first pipe = x hrs
So time taken by second pipe = (x-5) hrs
And time taken by third pipe = (x-5-4) = (x-9) hrs
Now, The first two pipes operating simultaneously fill the tank in the same time during which
the tank is filled by the third pipe alone
So 1/x + 1/(x-5) = 1/(x-9)
or x = 15 hrs
Q.No. : 44
Question : A watch which gains 5 seconds in 3 minutes was set right at 7 a.m. In the afternoon of the same day, when the watch indicated quarter past 4 o'clock, the true time is:
A: 59(7/12) min past3
B:
4pm
C: 58(7/11) min past 3
D: 2(3/11) min past 4
Answer: B
Solution
Short Method:
Since clock gains 5sec in 3 mins.
So it will gain 100 sec in 60mins (1 hour)
So it will gain 900 sec in 9 hours.
i.e clock will be 900 sec (15 mins ) fast when actual time is 4 o'clock
i.e. clock will show quarter past 4 when actual time is 4.
Actual Method:
Suppose clock gains t sec when it shows 4 mins 15 sec.
So the actual time should be (75-t/60) minutes past 3 o'clock.
So the duration of time in which clock shows quarter past 4 = 8 hrs + (75 - t/60) mins
= (550-t/60) minutes
In (550-t/60) minutes clock will gain = 5/3 * (550-t/60)
So 5/3 * (550-t/60) = t
or t = 900 sec
i.e. 15 minutes
So actual time will be quarter past 4'oclock- 15 minutes = 4'oclock
Q.No. : 45
Question : In Arun opinion, his weight is greater than 65 kg but less than 72 kg. His brother doesnt not agree with Arun and he thinks that Aruns weight is greater than 60 kg but less than 70 kg. His mother's view is that his weight cannot be greater than 68 kg. If all are them are correct in their estimation, what is the average of different probable weights of Arun?
A:
67 kg
B: 68 kg
C: 69 kg
D: Data inadequate
Answer: A
Solution
According to Arun his weight can be 66, 67, 68, 69, 70, 71
According to his brother: 61, 62, 63, 64, 65, 66, 67, 68, 69
According to his mother: 68, 67, 66, 65, ....
All are correct, so possible weight: 66, 67, 68
Average = 67
Hence A
Q.No. : 46
Question : A metallic sheet is of rectangular shape with dimensions 48 m x 36 m. From each of its corners, a square is cut off so as to make an open box. If the length of the square is 8 m, the volume of the box (in m3) is:
A: 4830
B:
5120
C: 6420
D: 8960
Answer: B
Solution
Clearly, l = (48 - 16)m = 32 m,
b = (36 -16)m = 20 m,
h = 8 m.
Volume of the box = (32 x 20 x 8) m3 = 5120 m3.
Q.No. : 47
Question : 50 men took a dip in a water tank 40 m long and 20 m broad on a religious day. If the average displacement of water by a man is 4 m3, then the rise in the water level in the tank will be( in cm):
A: 20
B:
25
C: 35
D: 50
Answer: B
Solution
Total volume of water displaced = (4 x 50) m3 = 200 m3.
Let in water be x
40*20*x=50*4
x=0.25m= 25cm