Synexus clinical research ltd Interview Questions, Process, and Tips

speed = distence / time
60km/hr=distance / 9 sec
km / hr to m / s==> 60*1000 / 1* 3600 = 16.667
16.667*9 = distance
150 m is the answer

let present age of ANAND AND BALA BE( A AND B ) RESPECTIVELY

A.T.Q

(A-10) = 1/3[B-10]. ……… – {1}

given

B = A+ 12 PUTTING IN 1

THEREFORE A-10 = 1/3 (A+2) = A+2 = 3A – 30

2A =32

A=16

Rs. 1500

(3/4)*(L/Sp)=30
L/Sq=75
(3/4)*(Sq/Sp)=30/75
Sq/Sp=8/15
Sp/Sq=15/8

47

343 : 729 :: 125 : ?
7^3 : 9^3 :: 5^3:7^3

2:3

34

(x-2)^3==x^3-6x^2+12x-8
(x-2)^3==(2^(2/3)+2^(1/3))^3
therefore,by solving
Ans: x^3-6x^2+x== -8-5*2^(2/3)-5*2^(1/3)

CPNCBZ

Ans: 60 kph

Suppose Person meets the train everyday at 3 PM at Station A.

His speed is 12kph.

So normally he reaches 5 km before the meeting point (pt B) at (5/12 hr = 25 min before) 2:35PM.

But if he is late by 30 min, then he will reach that point (pt B) by 3:05 PM.

Train is traveling at its normal speed so it covers the distance of 5 Km in 5 min starting from Station A and reaches the meeting point (pt B) at 3:05 PM.

So speed of the train is 5KM/5min = 60 kph.

Thanq saddha