Reasoning Aptitude
PUZZLES MCQs
Total Questions : 244
| Page 23 of 25 pages
Answer: Option E. -> There is no such teacher
English
Hindi
MathematicsGeography
History
French
A
y
y
y
B
y
y
y
y
y
C
y
y
D
y
y
E
y
y
None of the teachers was teaching less than two subjects
Answer: Option B. -> Hindi
Three teachers were teaching Hindi -- A, B and D
English
Hindi
MathematicsGeography
History
French
A
y
y
y
B
y
y
y
y
y
C
y
y
D
y
y
E
y
y
Three teachers were teaching Hindi -- A, B and D
Question 223. In a lake, there are 10 steps labelled using alphabets from A to J. Starting from step A, every minute a frog jumps to the 4th step from where it started - that is from the step A it would go to the step E and from E it would go to the step I and from I it would go to C etc. Where would the frog be at the 60th minute if it starts at the step A ?
Answer: Option B. -> A
The steps are labelled using alphabets from A to J.
TO MAKE IT EASY, WE HAVE ASSIGNED NUMBERS TO THE STEPS AS FOLLOWS:
Labelled with alphabets : A B C D E F G H I J
Labelled with numbers : 1 2 3 4 5 6 7 8 9 10
The frog takes total 60 minutes and takes 4 step length jumps every time.
Thus,
1st minute : 1 + 4 = 5th step (E)
2nd minute : 5 + 4 = 9th step (I)
3rd minute : 9 + 4 = 3rd step (C)
4th minute : 3 + 4 = 7th step (G)
5th minute : 7 + 4 = 11 = 10+1 = 1st step (A)
The same process is repeated 12 x 5 times.
Then, the jumping positions are
1 5 9 3 7
1 5 9 3 7
1 5 9 3 7 and so on.
After 15 cycles, frog will be in the 1st position
i.e., at 5th minute, 10th minute, 15minutes....60th minute frog will be in the 1st position.
i.e., at 60t
The steps are labelled using alphabets from A to J.
TO MAKE IT EASY, WE HAVE ASSIGNED NUMBERS TO THE STEPS AS FOLLOWS:
Labelled with alphabets : A B C D E F G H I J
Labelled with numbers : 1 2 3 4 5 6 7 8 9 10
The frog takes total 60 minutes and takes 4 step length jumps every time.
Thus,
1st minute : 1 + 4 = 5th step (E)
2nd minute : 5 + 4 = 9th step (I)
3rd minute : 9 + 4 = 3rd step (C)
4th minute : 3 + 4 = 7th step (G)
5th minute : 7 + 4 = 11 = 10+1 = 1st step (A)
The same process is repeated 12 x 5 times.
Then, the jumping positions are
1 5 9 3 7
1 5 9 3 7
1 5 9 3 7 and so on.
After 15 cycles, frog will be in the 1st position
i.e., at 5th minute, 10th minute, 15minutes....60th minute frog will be in the 1st position.
i.e., at 60t
Answer: Option B. -> 14
1
2
3
4
5
6
7
8
9
10
11
12
Ramesh
14
...
...
...
...
...
...
...
...
...
...
Suresh
5
4
3
2
1
Total Number of Student = 33
Number of student in between Ramesh and Suresh is,= 33 - (13+6)
= 14.
1
2
3
4
5
6
7
8
9
10
11
12
Ramesh
14
...
...
...
...
...
...
...
...
...
...
Suresh
5
4
3
2
1
Total Number of Student = 33
Number of student in between Ramesh and Suresh is,= 33 - (13+6)
= 14.
Answer: Option A. -> 8 , 6
Let B be the number of brothers and S be the number of sisters in the family.
Consider any two boys. They would be having (B - 2) brothers (excluding the two). But this number is equal to the number of sisters they have.
Therefore,
B - 2 = S
or , B - S = 2 ............(1)
Each girl will have (S - 1) sisters. Twice the number of sisters = 2(S - 1).
Since, each girl has twice as many brothers as sisters, we have, 2(S-1)-2 = B
2S - 4 = B ........... (2)
Substituting, eqn (2) in Eqn (1), we get
2S - 4 - S = 2
S = 6
On substituting S = 6 in eqn (1) , we get
B - 6 = 2
B = 8.
Let B be the number of brothers and S be the number of sisters in the family.
Consider any two boys. They would be having (B - 2) brothers (excluding the two). But this number is equal to the number of sisters they have.
Therefore,
B - 2 = S
or , B - S = 2 ............(1)
Each girl will have (S - 1) sisters. Twice the number of sisters = 2(S - 1).
Since, each girl has twice as many brothers as sisters, we have, 2(S-1)-2 = B
2S - 4 = B ........... (2)
Substituting, eqn (2) in Eqn (1), we get
2S - 4 - S = 2
S = 6
On substituting S = 6 in eqn (1) , we get
B - 6 = 2
B = 8.
Answer: Option C. -> 20
Total number of trees,
= 7+14-1
= 20
Total number of trees,
= 7+14-1
= 20
Answer: Option A. -> B
Let, A = x
Then, B = 2x
and, F = 4x
C = x/2
and D = x/4
Thus, The second oldest is B.
Let, A = x
Then, B = 2x
and, F = 4x
C = x/2
and D = x/4
Thus, The second oldest is B.
Answer: Option B. -> 18 years
Age of Shan = 55 years
Age of Sathian = 55-5 = 50 years
Age of Balan = 50-6 = 44 years
Age of Deven = 44-7 = 37 years
Thus, difference between Devan and Shan, = 55 -37
= 18.
Age of Shan = 55 years
Age of Sathian = 55-5 = 50 years
Age of Balan = 50-6 = 44 years
Age of Deven = 44-7 = 37 years
Thus, difference between Devan and Shan, = 55 -37
= 18.
Answer: Option C. -> 3
The ascending order of the number is 123456789.
Required difference,
6-3 = 3.
The ascending order of the number is 123456789.
Required difference,
6-3 = 3.
Question 230. Rohit was walking on the street, one boy requested him to donate for cancer patients welfare fund. He gave him a rupee more than half the money he had. He walked a few more steps. Then came a girl who requested him to donate for poor people's fund for which he gave two rupees more than half the money he had then. After that, again a boy approached him for an orphanage fund. He gave three rupees more than half of what he had. At last he had just one rupee remaining in his hand.
How much amount did Ram have in his pocket when he started?
How much amount did Ram have in his pocket when he started?
Answer: Option C. -> Rs.42
Let X be the rupees he initially had.
He gave for the cancer fund one rupee more than half of what he had.
i.e.,[1 +(X/2)].
Remaining money = X-(1+X/2) = [(X/2) - 1.
he gave for poor people's, rupee 2 more than half what he remain with,
= [2+{1/2*(X/2-1)}]
= [2+{(X-2)/4}]
= (6+X)/4
Now, remaining money = ((X/2)-1) - ((6+X)/4)
= (X-10)/4.
Again he gave 3 rupees more than half of what he had for orphanage,
[3+(1/2*((X-10)/4))]
= 3+[(X-10)/8]
= (14+X)/8
now left money,[{(X-10)/4]-[(14+X)/8]}
= [(2X-X-20-14)/8]
= (X-34)/8
As given, finally he had one rupee remaining so (X-34)/8 = 1
So,X-34 = 8
X = 8+34 = 42
Hence, Rohit had Rs. 42 initially in his pocket.
Let X be the rupees he initially had.
He gave for the cancer fund one rupee more than half of what he had.
i.e.,[1 +(X/2)].
Remaining money = X-(1+X/2) = [(X/2) - 1.
he gave for poor people's, rupee 2 more than half what he remain with,
= [2+{1/2*(X/2-1)}]
= [2+{(X-2)/4}]
= (6+X)/4
Now, remaining money = ((X/2)-1) - ((6+X)/4)
= (X-10)/4.
Again he gave 3 rupees more than half of what he had for orphanage,
[3+(1/2*((X-10)/4))]
= 3+[(X-10)/8]
= (14+X)/8
now left money,[{(X-10)/4]-[(14+X)/8]}
= [(2X-X-20-14)/8]
= (X-34)/8
As given, finally he had one rupee remaining so (X-34)/8 = 1
So,X-34 = 8
X = 8+34 = 42
Hence, Rohit had Rs. 42 initially in his pocket.
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