Aptitude and Reasoning Questions

1 man's 1 day work =1/96 ; 1 woman's 1 day work = 1/192

Work done in 6 days=6896+8192=6×18 =34 

Remaining work = 1/4

(8 men +8 women)'s 1 day work = 1896+8192 =1/8

Remaining work =1/4 -  1/8 = 1/8

 1/96 work is done in 1 day by 1 man

Therefore, 1/8 work will be done in 1 day by 96 x (1/8) =12 men

Total weight increased = (8 x 2.5) kg = 20 kg.

Weight of new person = (65 + 20) kg = 85 kg.

Let the sum be Rs.P.then
P(1+R/100)^3=6690…(i) and P(1+R/100)^6=10035…(ii)
On dividing,we get (1+R/100)^3=10025/6690=3/2.
Substituting this value in (i),we get:
P*(3/2)=6690 or P=(6690*2/3)=4460
Hence,the sum is rs.4460.

Given Number:     4  2  1  5  7  9  3  6  8

Ascending order:  1  2  3  4  5  6  7  8  9

Hence the required pairs are  12, 49, 16

Let Rs. x be the fare of city B from city A and Rs. y be the fare of city C from city A.

Then, 2x + 3y = 77 ...(i) and

3x + 2y = 73 ...(ii)

Multiplying (i) by 3 and (ii) by 2 and subtracting, we get: 5y = 85 or y = 17.

Putting y = 17 in (i), we get: x = 13.

From the given data,

$$\begin{gathered} 40100 x 75 + 80100 x 25 = K100 x 250 \\ 30 + 20 = 5K2 \\ 50 x 2 = 5K \\ K = 1005 = 20 \end{gathered}$$

Relative speed of the thief and policeman  =  (11 – 10) km/hr = 1 km/hr 

Distance covered in 6 minutes  = (1/60) x 6 km   = 1/10 km = 100 m

Therefore, Distance between the thief and policeman = (200 – 100) m = 100 m.

 The alphabets are coded as shown :

$$\begin{gathered} R O S E C H A I P \\ 6 8 2 1 7 3 4 5 9 \end{gathered}$$

So, in SEARCH,

S ia coded as 2,
E as 1,
A as 4,
R as 6,
C as 7,
H as 3.
Thus, the code for SEARCH is 214673