Dear Readers, Important Practice Reasoning Questions for NICL AO & RBI Grade B 2017 was given here with Solutions. Aspirants those who are preparing for Bank exams & all other Competitive examination can use this material.
Directions (Q.1-5): These questions are based on the following letter / number/ symbol arrangement. Study it carefully and answer the questions that follow
B # A R 5 8 E % M F 4 J 1 U @ H 2 © 9 T I 6 * W 3 P # K 7 $ Y S
Directions (Q.1-5): These questions are based on the following letter / number/ symbol arrangement. Study it carefully and answer the questions that follow
B # A R 5 8 E % M F 4 J 1 U @ H 2 © 9 T I 6 * W 3 P # K 7 $ Y S
1.Which of the following is the twelfth to the left of the twentieth from the left end of the above arrangement?
1). Twelfth to the left of the twentieth from the left end = 20-12 = 8th from left end = %
Answer: A
2.If all the numbers in the above arrangement are dropped, which of the following will be the eleventh from the right end?Answer: A
2). B # A R E % M F J U @ H © T I * W P # K $ Y S
11th from right end is ©
Answer is: e)
11th from right end is ©
Answer is: e)
3. Which of the following is the fifteenth to the left of the seventh from the right end of the above arrangement?
3). Fifteenth to the left of the seventh from the right end = 15 + 7 = 22nd from right end = 4
Answer is: b)
Answer is: b)
4.How many such numbers are there in the above arrangement each of which is immediately preceded by a consonant and also immediately followed by a symbol?
4). H 2 © and K 7 $
Answer is: c)
5. Four of the following are alike in a certain way based on their positions in the above arrangement and so form a group. Which is the one that does not belong to that group?Answer is: c)
5). All values are
1 + 2 = @ ; 1 – 2 = 4
© + 2 = T ; © - 2 = H
W + 2 = P ; W – 2 = 6
9 – 2 = 2 ; 9 + 2 = I
# + 2 = 7 ; # - 2 = 3
So 9 2 I is not belong to that group.
Answer is: d)
Directions (Q.6-10): In a certain instruction system the different computation processes are written as follows:1 + 2 = @ ; 1 – 2 = 4
© + 2 = T ; © - 2 = H
W + 2 = P ; W – 2 = 6
9 – 2 = 2 ; 9 + 2 = I
# + 2 = 7 ; # - 2 = 3
So 9 2 I is not belong to that group.
Answer is: d)
(i) a $ b «c means c is subtracted from the product of a and b.
(ii) a @ b # c means a is multiplied by the sum of b and c.
(iii) a * b & c means c is subtracted from b and the resultant is added to a.
(iv) a © b % c means b is divided by c and the resultant is added to square of a.
In each of the following questions, a set of instruction sequence is given. You are required to find out the outcome which should come in place of the question mark (?) in each of the given sets of sequence.
6.m * 78 & 56 = 50
m @ 7 # 13 = ?
6). . m * 78 & 56 = 50
(78 - 56) + m = 50 [Use instruction (iv)]
m = 50 - 22 = 28
Now, we have
28 @ 7 # 13
28 × (7 + 13) [Use instruction (i)]
= 560
? = 560
Answer is: a)
(78 - 56) + m = 50 [Use instruction (iv)]
m = 50 - 22 = 28
Now, we have
28 @ 7 # 13
28 × (7 + 13) [Use instruction (i)]
= 560
? = 560
Answer is: a)
7. 17 $ 4 « 8 = t
7 © t % 15 = ?
7). 17 $ 4 « 8
(17 × 4) - 8 [Use instruction (iii)]
= 60
Hence t = 60
Now, we have
7 © 60 % 15
(60 ÷ 15) + 72 [Use instruction (ii)]
= 53
? = 53
Answer is: b)
8. 5 © 49 % 7 = a(17 × 4) - 8 [Use instruction (iii)]
= 60
Hence t = 60
Now, we have
7 © 60 % 15
(60 ÷ 15) + 72 [Use instruction (ii)]
= 53
? = 53
Answer is: b)
a * 87 & 29 = ?
8). 5 © 49 % 7
(49 ÷ 7) + 52 [Use instruction (ii)]
= 32
Hence a = 32
Now, we have
32 * 87 & 29
(87 - 29) + 32 [Use instruction (iv)]
= 58 + 32 = 90
?= 90
Answer is: e)
9. 13 @ 4 # 3 = p(49 ÷ 7) + 52 [Use instruction (ii)]
= 32
Hence a = 32
Now, we have
32 * 87 & 29
(87 - 29) + 32 [Use instruction (iv)]
= 58 + 32 = 90
?= 90
Answer is: e)
p $ 5 « 55 = ?
9). 13 @ 4 # 3
13 × (4 + 3) [Use instruction (i)]
= 91
Hence p = 91
Now, we have
91 $ 5 « 55
(91 × 5) - 55 [Use instruction (iii)]
= 400
?= 400
Answer is: d)
13 × (4 + 3) [Use instruction (i)]
= 91
Hence p = 91
Now, we have
91 $ 5 « 55
(91 × 5) - 55 [Use instruction (iii)]
= 400
?= 400
Answer is: d)
10.b $ 15 « 18 = 42
b © 36 % 9 = ?
10). b $ 15 « 18 = 42
(b × 15) - 18 = 42 [Use instruction (iii)]
(b × 15) = 42 + 18
b = 4
Now, we have
4 © 36 % 9
(36 ÷9) +16 [Use instruction (ii)]
= 20
? = 20
Answer is: c
(b × 15) - 18 = 42 [Use instruction (iii)]
(b × 15) = 42 + 18
b = 4
Now, we have
4 © 36 % 9
(36 ÷9) +16 [Use instruction (ii)]
= 20
? = 20
Answer is: c
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