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This page contains the most recently asked technical questions and answers in the MPHATEK Systems.
All of the questions listed below were collected by students recently placed at MPHATEK Systems.
c
each pencil 100/25 = 4
each book 100/15 = 6.66
100 – 15 = 85
85-5*4 = 65
65/6.66 = 9
he can buy 9 books that is not in given options currently so
answer is NONE
The ans is -20.
Solution:
A-1
B-2
C-3 D-4 …like wise till Z-26.
ELECTRICITY-5+12+5+3+20+18+9+3+9+20+25==129
GAS-7+1+9==27
ELECTRICITY-GAS=129-27-(Minus 2)=100
so
JACK-JILL=(10+1+3+11-(10+9+12+12)-(minus(2))==(-20)
Let ‘x’ be the number of perons in the group and let ‘y’ be
amount per head which they have to pay.
Then xy = 2400.
or y = 2400/x
Since two friends have forget the purse, x-2 persons should
share the total amount(2400).
If they share, they have to make an extra contribution of
Rs 100 to pay up the bill
That is x-2 persons should pay y+100 each
or (x-2)(y+100) = 2400
or y+100 = 2400/(x-2)
or y = 2400/(x-2) – 100
Therefore we have got two equations namely,
y= 2400/x and y = 2400/(x-2) – 100
Comparing these two, we get
2400/x = 2400/(x-2) – 100
Solving this we get
x^2 – 2x -48 = 0
or (x-8) (x+6) = 0
or x = 8 or x = -6
Since ‘x’ denote the number of persons, it should be
positive.
So, x = 8.
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1.5 hr
8*1024+3*64+3
8192+192+3
8387
1000/0011/1000/0111
So ie 7. 1s here
1/2:1/3:1/4 = 6:4:3
Ram = 6/13 * 3250 = 1500,
Shyam = 4/13 * 3250 = 1000,
Mohan = 3/13 * 3250 = 750
– While the train is moving, the jogger will also be running in the same direction.
– for the head(engine) of the train to get to the current position of the jogger 240m away, it will take:
45km/hr => 12.5m/s => 240/12.5 = 19.2 seconds.
– But in the same period of time, the jogger will still be running and will have moved to a new location by: 9km/hr => 2.5m/s => 2.5 * 19.2 = 48m
To get to the new location at the speed of 12.5m/s will take the train:
48/12.5 = 3.84sec
In this additional time, the jogger will move forward by:
3.84 * 2.5 = 9.6m
at a speed of 12.5m/s, it will take the train less than a second to cover the additional 9.6m
If we add the distance the jogger will cover in 1 second to 9.6, it is still less than what the train can cover per second. let us see (9.6 + 2.5 = 12.1)
Therefore, the head of the train will pass the runner at approximately: 19.2 + 3.84 + 1 => 24.04 seconds.
For the train to completely pass the runner, it will need its whole length of 120m to be in front of the runner.
This will take an additional (9.6 + 2) seconds.
Therefore for the length of the train to be ahead of the runner it will take approx. 35.65 (24.04 + 9.6 + 2) seconds
240-24
650-?
?=24*650/240
65SEC