East renfrewshire council Interview Questions, Process, and Tips

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.

2.01

D=5, G=1

1km is equal to 1000 meter
Therefore 225 meter is how many km
Then we have to cross multiply
1km – 1000 meter
? – 225 meter
0.225 km
Then we have to apply formula
Speed = Distance/Time
Speed = 0.225/12 sec
= 0.01875×3600sec
= 67.5
Therefore speed of train is = 67.5

427 is the answer
Because
385 means (3+5=8) or (8-3=5) ,
462 means (4+2=6) or (6-2=4) ,
572 means (5+2=7) or (7-5=2) ,
396 means (6+3=9) or (9-3=6) ,
671 means (6+1=7) or (7-6=1) ,
264 means (4+2=6) or (6-2=4) ,
but only 427 is the number in which if we do addition of any of its two digits (e.g 4+2 or 4+7) then its answer doesn’t come 3rd digit
like if we do 4+2 then answer will be 6 so the number should be 426 but it is not
or if we do 7-4 then answer will be 3 (not 2)
so the right answer is 427

600

360

zero – as one letter cant be wrong – atleast two are to be wrong

26

(x**2 – 6* x + 5) = (x-1)*(x-5)
(x**2 + 2 * x + 1) = (x + 1) * (x+1) = (x+1)**2
For what x is (x-1)*(x-5)/( (x+1)**2) a minimum?
One way to answer this question is by using calculus.
Take the derivative, and set to zero.
Since this is a fraction of polynomials, and a fraction is
zero only if it’s numerator is zero, we need calculate only
the numerator of the derivative and set it to zero.
The numerator of the
Derivative of (x-1)*(x-5)/( (x+1)**2) is
( (x-1) + (x-5) ) ( x+1)**2 – (x-1)(x-5)( 2 (x+1) )
= (2 x – 6) (x+1)**2 – (2) (x-1)(x-5) (x+1)
= 0
Divide through by 2 (x+1)
(x-3)(x+1) – (x-1)(x-5) = 0
(x**2 – 2 x – 3 ) – (x**2 – 6 x + 5) = 0
x**2 – x**2 – 2 x + 6 x – 3 – 5 = 0
4 x – 8 = 0
x = 2
Plugging in x = 2 into the original
(x**2-6*x+5)/(x**2+2*x+1)
gives us (2**2 – 6 * 2 + 5)/(2**2 + 2*2 + 1)
= (4 – 12 + 5) / (4 + 4 + 1) = -3/9 = -1/3
Least value is -1/3

13. One group of each = 2+3+5 = 10 and 1 remainder for each = 1+1+1 = 3. 10+3 = 13

37 SEATS….
because if passengers will occupy 3 seats, per head amount
payable will be 26.89.80.67 can not be divisible other than
3 and 26.89. so the,
Number of occupied seats will be 3
Number of unoccupied seats will be 40-3=37

b

C. 20

name

Suppose 1 person consumes 1 unit food daily.
So, for 24 days, 1600 person will consume 1600×24 units.
So total units of food available = 1600×24

Now total number of person =1200

So, total food will be consumed by 1200 person in (1600×24)/1200 = 32 days.