Page 29 - Services Selection Board (SSB) Interviews

Alpha Series Completion 25

Concept Deviator

1. (c) AZ, GT, MN,___ , YB Now analyse the terms –
st
In this series letter contain AZ, second letter G. From 1 term = 1
this we can see that A to G increasing order but Z 2 term – It describes the 1 term that is has “one”
st
nd
rd
to T decreasing order. 3 letter also maintain same “one” or 11
nd
pattern respect to 2 letter. rd nd
Now, we are considering position of each letter 3 term It describes the 2 term that it has “two”
“one” or 21
according to the alphabetical order. th rd
A + 6 = G, G + 6 = M => M + 6 =S 4 term It describe the 3 term that it has “one””2”
and “one” “1” or 1211
Z – 6 = T, T – 6 = N =>N – 6 =H 5 term describes the 4 term that it has “one” “1”,
th
th
So, fourth term will be = SH “one” “2” and “two” “1” or 111221
2. (d) J2Z, K4X, I7V, ___, H16R, M22P Hence the 6 term would be – “Three” “1”, “two”
th
Here first letter contain 2 letters and one numeric “2” and “one” “1” or 312211
value. Solution from 7 to 9:
st
1 term of each letter : J K I __ H M
Here j to k increasing order but k to I decreasing A 26 J 17 S 8
order ,again h to k also increasing order. B 25 K 16 T 7
Each step + 1, – 2, + 3, – 4…… C 24 L 15 U 6
J + 1 = K, K – 2 = I, I + 3 = L, D 23 M 14 V 5
L – 4 = H, H + 5 = M, E 22 N 13 W 4
nd
2 term also increases by + 2, + 3, + 4, + 5 order. F 21 O 12 X 3
2 + 2 = 4, 4 + 3 = 7, 7 + 4 =11, 11 + 5 = 16, G 20 P 11 Y 2
16 + 6 = 22 H 19 Q 10 Z 1
3rd term decreases by 2 fixed order. I 18 R 9
Z – 2 = X, X – 2 = V, V – 2 = T 7. (b) Sum of total vowels in the sequence
Fourth letter will be L11T. = A (26) + E (22) + I (18) + O (12) + U (6)
3. (b) gfe ___ ig ___ eii ___ fei ___ gf ___ ii = 26 + 22 + 18 + 12 + 6 = 84
Above series cannot be solve directly because some 8. (c) 5 – 12 – 18 – 23 ⇒ V – O – I – D ⇒ VOID, is a
alphabet repeated. valid word.
2
2
2
So, take each option individually and put it . 9. (b) Z, W, R, K, – – ⇒ 1, 4, 9, 16, – – ⇒ 1 , 2 , 3 ,
2
Among all 4 given options only option 2 satisfied the 4 , – –
series : gfeii/gfeii/…. Observing the given sequence, we conclude that 52
4. (d) 18514 denotes AHEAD = 25 = B should be the next valid alphabet
Means 1 → A, 8 → H According to the alphabetical Solution From 10 to 12:
position. As per the given conditions –
Same as 31385, 3→A, 1→C, 3 → C, 8 → H, 5 → E Letter Number Letter Number
31385 ⇒ CACHE. A – 13 N 1
5. (c) Since E + E =G and maximum carry over is 1 B – 12 0 2
then C – 11 P 3
1 + D + D =H
D – 10 Q 4
From here we can not conclude that D must be E – 9 R 5
greater than 3
F – 8 S 6
6. (a) Given sequence is G – 7 T 7
1, 11, 21, 1211, 111221, ………