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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.
C
$150000
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
The answer is 42
55967
B. 437950