66. [3] + [[5] +11 [6]ன் மதிப்பு
A) [0]
B) [1]
C)[2]
D)3
67.a =13,6=5 மற்றும் a, 6 எனில் ax6ன் மதிப்பு
A) 45
B) 35
C) 25
D) 15)
Answers
Answer:
WBCS Exam Main Compulsory Question Paper on Arithmetic and Test of Reasoning 2016 solved paper
Profile picture for user arpita pramanik
Submitted by arpita pramanik on Thu, 02/01/2018 - 15:35
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WBCS Exam Main Compulsory Question Paper on Arithmetic and Test of Reasoning 2016 solved paper
1. How many cubes of 10 cm edge can be put in a cubic box of 1 m edge ?
(1) 10 (2) 100 (3) 1000 (4) 10000
Ans:- The volume of small cube of 10 cm edge is = 103=1000 cm3
The volume of 1 m = 100 cm cubic box is = 1003=1000000 cm3
Number of cubes are
1000000
1000
=1000
2. Next terms of the series 198, 194, 185, 169 .....
(1) 92 (2)112 (3)136 (4)144
Ans: The different between 1st and 2nd term is ( 198 - 194 ) = 4=22
The different between 2nd and 3rd term is ( 194 - 185 ) = 9 = 32
The different between 3rd and 4th term is ( 185 - 169 ) = 16 = 42
Then we can say the different between 4th and 5th term is 52=25
Then the next term is 169 - 25 = 144
3. Which fraction comes next in the sequence
1
2
,
3
4
,
5
8
,
7
16
(1)
9
32
(2)
10
17
(3)
11
34
(4)
12
35
Ans: The sequence is
1
2
,
3
4
,
5
8
,
7
16
Then
1
2
,
3
4
,
5
8
,
7
16
1
21
,
1+2
22
,
1+2+2
23
,
1+2+2+2
24
Now the next term will be
1+2+2+2+2
25
=
9
32
4. The last day of century cannot be
(1) Monday (2) Wednesday (3) Tuesday (4) Friday
Ans: 100 years contain 5 odd days. So last day of 1st century is Friday.
200 years contain (5×2)≡3 odd days. So last day of 2nd century is Wednesday .
300 years contain (5×3)=15≡1 odd day. So last day of 3rd century is Monday.
400 years contain 0 odd day. So the last day of 4th century is Sunday.
this cycle is repeated. So last day of century cannot be Tuesday, Thursday or Saturday.
5. The area of a square is equal to the area of a circle. The ratio between the side of a square and the radius of the circle is
(1)
√
π
:1 (2) 1:
√
π
(3) 1:π (4) π:1
Ans: Let the one side of the square is a. Then the area of the square is a2.
Let the radius of the circle is r. Then the area of the circle is πr2.
Now we can write
a2=πr2 ⇒a=
√
π
r ⇒
a
r
=
√
π
⇒a:r=
√
π
:1
Step-by-step explanation:
WBCS Exam Main Compulsory Question Paper on Arithmetic and Test of Reasoning 2016 solved paper
Profile picture for user arpita pramanik
Submitted by arpita pramanik on Thu, 02/01/2018 - 15:35
Share iconShare on Facebook Share on Twitter Share on LinkedIn Share via Messenger Share via Viber Share via WhatsApp
WBCS Exam Main Compulsory Question Paper on Arithmetic and Test of Reasoning 2016 solved paper
1. How many cubes of 10 cm edge can be put in a cubic box of 1 m edge ?
(1) 10 (2) 100 (3) 1000 (4) 10000
Ans:- The volume of small cube of 10 cm edge is = 103=1000 cm3
The volume of 1 m = 100 cm cubic box is = 1003=1000000 cm3
Number of cubes are
1000000
1000
=1000
2. Next terms of the series 198, 194, 185, 169 .....
(1) 92 (2)112 (3)136 (4)144
Ans: The different between 1st and 2nd term is ( 198 - 194 ) = 4=22
The different between 2nd and 3rd term is ( 194 - 185 ) = 9 = 32
The different between 3rd and 4th term is ( 185 - 169 ) = 16 = 42
Then we can say the different between 4th and 5th term is 52=25
Then the next term is 169 - 25 = 144
3. Which fraction comes next in the sequence
1
2
,
3
4
,
5
8
,
7
16
(1)
9
32
(2)
10
17
(3)
11
34
(4)
12
35
Ans: The sequence is
1
2
,
3
4
,
5
8
,
7
16
Then
1
2
,
3
4
,
5
8
,
7
16
1
21
,
1+2
22
,
1+2+2
23
,
1+2+2+2
24
Now the next term will be
1+2+2+2+2
25
=
9
32
4. The last day of century cannot be
(1) Monday (2) Wednesday (3) Tuesday (4) Friday
Ans: 100 years contain 5 odd days. So last day of 1st century is Friday.
200 years contain (5×2)≡3 odd days. So last day of 2nd century is Wednesday .
300 years contain (5×3)=15≡1 odd day. So last day of 3rd century is Monday.
400 years contain 0 odd day. So the last day of 4th century is Sunday.
this cycle is repeated. So last day of century cannot be Tuesday, Thursday or Saturday.
5. The area of a square is equal to the area of a circle. The ratio between the side of a square and the radius of the circle is
(1)
√
π
:1 (2) 1:
√
π
(3) 1:π (4) π:1
Ans: Let the one side of the square is a. Then the area of the square is a2.
Let the radius of the circle is r. Then the area of the circle is πr2.
Now we can write
a2=πr2 ⇒a=
√
π
r ⇒
a
r
=
√
π
⇒a:r=
√
π
:1