Q5
Car A and Car B are travelling towards Town Y from Town X. Both
cars left Town X at different times. At 1030, Car A has travelled
3/8 of the distance between Town X and Town Y, while Car B has
travelled 2/9 of the distance between Town X and Town Y. At 1035,
the distance travelled by Car B is 3/5 of the distance travelled by
Car A. If the speed of both cars are the same, find the time Car B
reaches Town.
Q6
Sally arrives at the station at the same time every day. Her
husband drives from home to the station to pick her up. They always
arrive back at their house at the same time every day, driving at
the same speed. One day she arrives at the station one hour early
and starts to walk home. Her husband meets her along the road and
they drive home together. They arrived home 10 minutes earlier than
usual. Assuming all speeds of walking and driving are constant and
all transfers are immediate, how long had the woman been walking
when her husband met her? (answer in minutes)
Q5.
Both cars have the same speed.
At 1030, car A = 3/8 of distance = 54/144 of distance
and car B = 2/9 of distance = 32/144 of distance
At 1035, x/144 of the distance has been travelled, hence 5 mins
= x/144 of distance.
(32+x)/(54+x) = 3/5
= > x = 1
= > 5 mins = 1/144 of total distance.
Distance left for car B at 1030 = 1 - 32/144 = 112/144 of total
distance
Amount of time needed for car B
= [( 112/144 ) / ( 1/144 )] X 5 mins
= 560 mins
Time car B reaches Town Y
= 1030 + 560 mins
= 1030 + 9 hours 20 mins
= 1950 =
7.50
pm
Q6.
Interesting question, if you know what you are doing, it is
easy, if you don't, it is hard.
Time saved by husband = 10 mins
= > Distance between the point when husband meet wife and the
station
= 10/2 = 5 mins
Time wife arrived earlier = 1 hour = 60 mins
Time wife walked = 60 mins - 5 mins = 55 mins