# Reasoning: Machine Input-Output Questions Set 26

Directions (1-5): A string of numbers is given as input. The further steps given are obtained by applying certain logic. Numbers of step II have been obtained by using at least 1 digit of each number in step I. Each step is a resultant of previous step only. None of the exact logic is repeated in any step. Input: 1. What is the average of numbers obtained in step II and step IV?
A) 33.4
B) 16.3
C) 22.3
D) 23.6
E) Other than options given
Option C
Solution:
For given example:

Step I: See arrows – multiplication is done for digits in blocks. 1st digit with 1st digit, 2nd with 2nd. 2*3 = 6, 3*3 = 9 so 69 in 1st place. 1*5 = 5, 7*1 = 7, so 57, 4*2 = 8, 2*3 = 6
Step II: Instructions say Numbers of step II have been obtained by using at least 1 digit of each number in step I.
So 6*5 – 8 = 22, 9*7 – 6 = 57
Step III: (2+2)/2 = 2, (5+7)/2 = 6
Step IV: 6-2 = 4, it could have been (2+6)/2 = 4 but given that (None of the exact logic is repeated in any step.)
For given input:::
Step I: 4*1, 3*2 = 46…………8*1, 2*4 = 88……..1*6, 3*3 = 69
Step II:       4*8- 6= 26………………6*8-9 = 39
Step III:      (2+6)/2 = 4………………(3+9)/2 = 6
Step IV:                         6-4 = 2
So
Step I: 46……88…..69
Step II:     26……39
Step III:      4…….6
Step IV:         2
2. What is the second smallest number obtained in any step of given input?
A) 2
B) 4
C) 6
D) 5
E) 7
Option B
3. Find the difference between sum of numbers obtained in 1st step and sum of numbers obtained in all other steps.
A) 118
B) 102
C) 112
D) 173
E) 126
Option E
Solution:
Required difference = (46+88+69) – (26+39+4+6+2)
4. What is the difference between the second largest and the third smallest numbers obtained in any steps?
A) 63
B) 40
C) 74
D) 61
E) 55
Option A
Solution:
Required difference = 69-6 = 63
5. What is the average of numbers obtained in last 2 steps?
A) 6
B) 2
C) 5
D) 4
E) None of these
Option D
Solution:
Required average =(4+6+2)/3 = 4

Directions (1-5): A string of alphabets is given as input. The further steps given are obtained by applying certain logic. Alphabets of step II have been obtained by using at least 1 digit of each number in step I. Each step is a resultant of previous step only. None of the exact logic is repeated in any step. Input: 1. Which alphabet occurs exactly 3 times in any steps?
A) L
B) F
C) R
D) U
E) Other than options given
Option B
Solution:
For given example:

Step I: See arrows, takes first alphabet with 1st and 2nd with 2nd.
C = 3, H = 8, 3*8 = 24=X,
S = 19, A = 1, 19*1 = 19 = S so XS
Next B= 2, K = 11, 2*11 = 22 = V, ….and so on VZ and NT
Step II: X = 24, V = 22, N = 14 — 24+22 – 14 = 32, 3 = C, 2=B
S= 19, Z = 26, T = 20 — 19+26-20 = 25, 2 = B, 5 = E
Step III: CB = 3+2 = 5 = E
BE = 2+5 = 7 = G
Step IV: G= 7,E = 5 — 7-5 = 2 = B
For given input:::
Step I: See arrows, takes first alphabet with 1st and 2nd with 2nd.
D = 4, C = 3, 4*3 = 12=L,
R = 18, A = 1, 18*1 = 18 = R so LR
Next E= 5, D = 4, 5*4 = 20 = T, ….and so on TU and PM
Step II: L = 12, T = 20, P = 16 — 12+20 – 16 = 16, 1 = A, 6=F
R= 18, U = 21, M = 24 — 18+21-13 = 26, 2 = B, 6 = F
Step III: AF = 1+6 = 7 = G
BF = 2+6 = 8 = H
Step IV: H= 8, G = 7 — 8-7 = 1 = A
So
Step I: LR….…TU……..PM

Step II:     AF……BF
Step III:    G…….H
Step IV:        A
2. What is the sum of numbers corresponding to each alphabet in step II? (Taking A = 1, B = 2, …..Z = 26)
A) 18
B) 19
C) 15
D) 13
E) 11
Option C
Solution:
Step II: A+F+B+F = 1+6+2+6 = 15
3. Let @ is the sum of numbers corresponding to each alphabet in step III. What is the alphabet corresponding to @? (Taking A = 1, B = 2, …..Z = 26)
A) L
B) M
C) P
D) O
E) Q
Option D
Solution:
Step III: G+H = 7+8 = 15 = O
4. How many alphabets appear more than once in any step?
A) 3
B) 2
C) 4
D) 1
E) 5
Option B
Solution:
A and F
5. In step I, what is the number corresponding to the alphabet which is also present in ‘PACT’?
A) 16
B) 3
C) 1
D) 20
E) None of P, A, C, T is present
Option D
Solution:
Out of P, A, C, T, — T is present in step I, T = 20

## 23 Thoughts to “Reasoning: Machine Input-Output Questions Set 26”

1. Jellyfish

Thx az

2. yuvi

thanku mam..2nd wala sach me kafi aacha tha.:)

3. yuvi

@subhra mam..2nd me error hai..ans should be..

LR TU FX

BF AE

H F

B

YE HONA CHHAIYE….MAM BECOZ WANHA PE 20+12-6=26 HOGA…GALTI SE 16 HO GAYA HAI…ND18+21-24=15 HOGA WNHA GALTI SE..26 HO GAYA HAI..

PLZZ CHECK..

1. .-•~¹°” Ambika : ˆ”°¹~•

hmm error ho gya ..mera bhi BF and AE aa rha

1. yuvi

hmm..is hisab se phir syad sabke ans change ho jayenge…wase how r u mam?

1. .-•~¹°” Ambika : ˆ”°¹~•

yaa i am good :)) hows u

1. yuvi

yeah now i m also good mam..

2. yuvi

or BF nd AE aayega na mam..AF nhi na…

1. .-•~¹°” Ambika : ˆ”°¹~•

haan AE hi

2. Shubhra

yes numbers got exchanged

now check the input

1. yuvi

ohh ok..so input se prob tha..thanx mam..

4. ครђ

😉 Thanks Mam

5. jaga

mam ..q2 kese karna hai>

1. Shubhra

in input question

CS….. HA

C = 3, H = 8 3*8 = 24, and 24 = X
S = 19, A = 1, 19*1 = 19, 19 = S
So

CS….. HA points to XS
this is for 1st step

further see other steps – different logic as explained , if doubts ask again

1. jaga

1st wala ka no 2 mam

1. Shubhra

Second smallest number pucha hai na

1. jaga

samjh gya mam thank u,..

6. _/_

ty 2nd qs is so good
🙂

7. hi

2nd is very good AZ is rockstar ….☺☺

8. Jahnavi

mam plz provide more new pattern i/o

9. dream girl ;;;;/@ :)

mam 2nd questn mai STEP 1 mai PM hoga na instead of FX?

1. Shubhra

yes PM as explained above

1. dream girl ;;;;/@ :)

ok thank u mam