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127.179(app)
given distance of the train along the wind is 695
and againt the wind is 498
and time = distance/speed
as we know that time is equal in both the cases hence equate
695/s1=498/s2———–(1);
where s1=speed of the plane + speed of the wind
and s2=speed of the plane -speed of the wind
given that speed of the wind is 21k/h
s1=sp+21
s2=sp-21
substitu in eq 1
we get the answer as 27.17(app)
Total Pages = 42+28 = 70 pages
for 70 pages we give Rs.20/-
and for 42 pages we give x
thus x = (42×20)/70
x = 12
The Answer = 12/-
original cost is 100^2 / (100^2 – p^2) Rs
By their Ability to suck dick.
340
half and hour
15/46=0.32
GERMANY
Country Name..
1. 1g
2. 3g with 1g counter
3. 3g
4. 3g plus 1g
5. 3g plus 1 g plus 1g weighed medicine
6. 9g with 3g counter
7. 9g with 1g of counter and 1g weighed medicine
8. 9g with 1g counter
9. 9g
10. 9g plus 1g
11. 9g plus 3g with 1g on counter
12. 9g plus 3g
13. All 3 weights on one side.
Pfull = 4 Hrs; Qfull = 5 Hrs
Assume velocity Vp = x then Vq = 4/5 x –> avg velocity = (5/5 + 4/5) / 2 = 0.9
If fastest time = 4hrs then the time it would take to fill up tank by alternating is 4/0.9 = 4.44 Hrs
2* pi* r ( h + r )
r=3cm
h=7cm
2 * 22/7 *7 ( 10 ) = 440 sq cm
the answer is E
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.
Let’s say, length of train = x metres
Speed at pole = Speed at platform
x/15 = 100+x / 25
5x = 300 + 3x
2x = 300
x = 150m
So, train is 150 metres long.
its a tricky question and very funny too.
the answer is 45% only which is gained by the minute hand.
63:45:55