• Given, coded word : Z B Y X M N Q B – S T R A I G H T

GRAPE = 2 7 3 5 4. FOUR = 1 6 8 7.
This shows,
G = 2,
R = 7,
A = 3,
P = 5,
E = 4,
F = 1,
O = 6,
U = 8,
R = 7.
So, GROUP = 2 7 6 8 5.

From the given codes we have
G – 9,
L – 7,
O – 5,
S – 3,
R – 6,
Y – 2,
E – 1,
A – 8,
P – 4,
H – 0
So ‘geology’ is coded as GEOLOGY – 9157592.

From Question
M + 3 = P
O + 3 = R
U + 0 = U
S – 2 = Q
E – 2 = C
In the same way Same way,
S + 3 = V
H + 3 = K
I + 0 = I
F – 2 = D
T – 2 = R

A = 1,
ACT = 1 + 3 + 20 = 24 (Letters represented by their position in the alphabet)
FAT = 6 + 1 + 20 = 27

The given logic is as follows, Sum of the digits + second digit = Answer.
i.e., 1 + 4 = 5 + 4 = 9.
2 + 8 = 10 + 8 = 18.
Similarly, 7 + 8 = 15 + 8 = 23.

Code for Consonant is it’s place value, and code for vowels are given below:
A ⇒ 1,
E ⇒ 2,
I ⇒ 3,
O ⇒ 4,
U ⇒ 5
Therefore M ⇒ 13,
A ⇒ 1,
N ⇒ 14,
G ⇒ 7,
O ⇒ 4

• Move +2, +3, +4 forward for first 3 letters
• Move -2, -3, -4 backward for last 3 letters.

C + 2 = E
A + 3 = D
S + 4 = W
K – 2 = I
E – 3 = B
T – 4 = P

• If we move two letter back after H it comes F.
• If we move two letter back after K it comes I.
• If we move two letter back after I it comes R.
• If we move two letter back after G it comes E.

1. If only conclusion I follows
2. If only conclusion II follows
3. If either I or II follows
4. If neither I nor II follows and
5. If both I and II follow.

1. If only conclusion I follows
2. If only conclusion II follows
3. If either I or II follows
4. If neither I nor II follows and
5. If both I and II follow.

1. If only conclusion I follows
2. If only conclusion II follows
3. If either I or II follows
4. If neither I nor II follows and
5. If both I and II follow.

Assume that in x days, one cow will eat one bag of husk.
More cows, less days (Indirect proportion)
More bags, more days (Direct proportion)
Hence we can write as
1 * 20 * x = 20 * 1 * 20

x = 20.

Let the required length be x meters
More men, More length built (Direct proportion)
Less days, Less length built (Direct Proportion)
(30 x 5 x X) = (40 x 3 x 56)
=> x = 44.8
Hence, the required length is 44.8 m.

This is a situation of direct variation.

Less number of weeks —-> amount raised will be less .
Given :
Miguel raised $2,372.50 in 36.5 weeks
Amount raised in one week is
= 2372.5 / 36.5
= $65
Amount raised in 20 weeks is
= 65 x 20
= $1300
Hence, the money raised in 20 weeks is $1300

This is a situation of direct variation.

Because, less number of basket balls —–> cost will be less.
Given :
75 basketballs cost = $1,143.75.
Cost of one basket ball = 1143.75 / 75 = 15.25
Then, cost of 26 basket balls = 15.25 x 26 = 396.50.
Hence, the cost of 26 basket balls is $ 396.50.

Given that fort had provision for 500 persons for 27 days Hence, after 3 days, the remaining food is sufficient for 500 persons for 24 days.
Remaining persons after 3 days = 500 + 300 = 800
Assume that after 10 days,the remaining food is sufficient for 800 persons for x days.

More men, Less days (Indirect Proportion)
⇒ Men 500 : 800 :: x : 24
⇒ 500 × 24 = 800x
⇒ 5 × 24 = 8x
⇒ x = 5 × 3 = 15

Let the required number days be x.
Less spiders, More days (Indirect Proportion)
Less webs, Less days (Direct Proportion)
Hence we can write as,

• 1 x 7 x x = 7 x 1 x 7

x = 7.

Assume that in x days, one cow will eat one bag of husk.
More cows, less days (Indirect proportion)
More bags, more days (Direct proportion)
Hence we can write as

• 1 x 40 x x = 40 x 1 x 40

x = 40.