Teach plus Interview Questions, Process, and Tips

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d

b=15,l=10 or b=10,l=15

find out what is the problem and taking the bad out always works.

Let cost for apple be a Cost for banana be b and Orange be c

So by first value expression becomes 17a + 13b + 9c =130 ———-1 therefore if you further solve a = (130 – 13b – 9c)/17 ———- 2 the second expression becomes: 13c + 7a + 10b = 100 ———- 3 If you put value of a in second expression it becomes: 13c + 7[(130 – 3b – 9c)/17] +10b = 100

Further if you solve you get value of b:

b = 10 – 2c ———-4

put value of b from 4 in 1

17a + 13 [10 – 2c] + 9c = 130

Further if you solve you find value of a

a = c ———-5

Put 5 in 3

13c + 7c + 10b = 100

further solve you get: c = 1 ———-6

from 5 and 6

a = c = 1 ———-7

Substitute value of c in expression 4

b = 10 – 2c b = 10 -2 * 1 b = 8 ———-8

therefore a + b + c = 10

Answer:50.24
Explanation:
Area of a circle =π*r*r
616=π*r*r
r*r= 616÷3.14
r*r =196
r =13
Circumference of a circle= 2*π*r
=2*π*13
=16*3.14=50.24

– 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

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