f.4 maths

17. Let s be the original speed of the man.

Time required for the man to travel from A to B at original speed, t1 = 40/s

Time required for the man to travel from A to B at speed (s-2), t2 = 40/(s-2)

Since, t2 - t1 = 1
=> 40/(s-2) - 40/s = 1
=> [40s - 40(s-2)]/[s(s-2)] = 1
=> 80 = s(s-2)
=> s^2 - 2s - 80 = 0
=> (s - 10)(s + 8) = 0
=> s = 10 or s = -8 (rejected since speed must be > 0)
Therefore, the original speed of the man is 10km/h

18. Let m be the rowing speed of the man in still water.

Then, the rowing speed of the man against the current = (m - 2), and the rowing speed of the man with the current = (m + 2).

The time required for the man to row against the current = 9/(m-2), and the time required for the man to row with the current = 9/(m+2).

Since the total time required is 6 hours, we have
9/(m-2) + 9/(m+2) = 6
=> 3/(m-2) + 3/(m+2) = 2
=> [3(m+2) + 3(m-2)]/[(m-2)(m+2)] = 2
=> 6m = 2(m^2 - 4)
=> 3m = m^2 - 4
=> m^2 - 3m - 4 = 0
=> (m-4)(m+1) = 0
=> m = 4 or m = -1 (rejected since speed must be > 0)
Therefore the man's rowing speed in still water = 4 km/h

17.Let the original speed of the car is a km/h
the original time required for the journey: 40/a hours
the new time required for the journey: 40/(a﹣2) hours
40/a = 40/(a﹣2) - 1
40(a﹣2)/a = 40﹣(a﹣2)
40a﹣80 = 42a﹣a2
a2﹣2a - 80 = 0
(a + 8)(a﹣10) = 0
a = -8(rejected) or a = 10
∴his original speed is 10km/h.

18.Let the speed of the man in still water be x km/h
time required by the man to swim against the flow is 9/(x﹣2) h
time required by the man to swim along the flow is 9/(2+x) h
9/(x﹣2)+9/(2+x) = 6
[9(x+2)+9(x﹣2)]/(x﹣2) (x+2) = 6
9(x+2)+9(x﹣2) = 6(x﹣2) (x+2)
18x = 6x2﹣24
x2﹣3x﹣4 = 0
(x﹣4)(x+1)=0
x = 4 or x = -1(rejected)
∴the speed is 4km/h

cannot see the question ???