Ques:- Two trains of length 120 m and 280 m are running towards each other on parallel lines at 42 kmph and 30 kmph respectively. In what time will they be clear of each other from the moment they meet?
Recent Answer : Added by Admin On 2022-09-29 18:55:02:
I solve it as the 280m train being stationary… and the 120m train moving at (42+30) = 72kph…. or 20mps
The front of the train will have to go 280m to pass the stationary train plus another 120m for the backs to clear… that is 400m in all.
The time to travel 400m at 20mps is 20secs
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435
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Ques:- Five farmers have 7, 9, 11, 13 & 14 apple trees, respectively in their orchards. Last year, each of them discovered that every tree in their own orchard bore exactly the same number of apples. Further, if the third farmer gives one apple to the first, and the fifth gives three to each of the second and the fourth, they would all have exactly the same number of apples. What were the yields per tree in the orchards of the third and fourth farmers?
Recent Answer : Added by John Rony On 2021-06-23 17:56:01:
So we consider the 2nd statement first. We can form an equation out of it.
14x-6=13y+3=9z+3
Using this, we can understand that the multiple of 14 and the multiple of 13 and 9 must have a difference of 9. The easiest way to ensure that is multiplying it by 9
14*9=126
13*9=117
If the 5th farmer gives 3 apples to the 4th farmer, they would have 123 and 120 apples respectively. However, we also know that the 2nd farmer has 117 apples (13*9=117, and this is a multiple of 9) if the 5th farmer gives 3 apples too the 2nd farmer, the 3rd, 4th and 5th farmers would have 120 apples each.
Now that we got 120, we should check if the first part of the question makes sense along with it. The equation would be
7a+1=11b-1=120
We know that 11*11=121 and 7*17=119. When we add 1 to 119 and subtract 1 from 121, we get 120 for each. In this way, all the farmers have 120 apples each.
Therefore, the 3rd farmer had a yield of 11 per tree and the 4th farmer had a yield of 9 per tree.
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I solve it as the 280m train being stationary… and the 120m train moving at (42+30) = 72kph…. or 20mps
The front of the train will have to go 280m to pass the stationary train plus another 120m for the backs to clear… that is 400m in all.
The time to travel 400m at 20mps is 20secs