(1) Few data are missing (indicated by --) in the table and you are expected to calculate them from the available data if required.
(2) There are in total 8 destinations (I, II, III, IV, V, VI, VII and VIII). If a car has to go from one destination to another destination it will have to travel through in between destinations. For ex. If car A travels from destination I to IV. it will have to travel through destinations II and III
(3) Time required column depicts time required by a car mentioned in the same row to cover the distance between destinations mentioned in the same row.
Name of the Car |
Speed of the Car (In Kmph) |
Distance between destinations (In Km) |
Time Required (In hours) |
A | 77 | Between I and II = 188 | -- |
B | -- | Between II and III = 254 | 5 12/23 |
C | -- | Between III and IV = 228 | 5 1/3 |
D | -- | Between IV and V =162 | 6 |
E | 36 | Between V and VI = -- | 8 2/3 |
F | 22 | Between VI and VII= -- | 6 7/11 |
G | 42 | Between VII and VIII= -- | 4 1/3 |
1. What is the respective ratio of the distance between destinations IV and VII and the distance between destinations V and VIII ? a. 19 : 22
b. 29 : 34
c. 33 : 38
d. 31 : 32
e. 29 : 30
b. 58
c. 32
d. 29
e. 27
b.1 pm
c.1.45 pm
d.10.30 am
e.12.30 am
b. 37
c. 26
d. 32
e. 28
5. If H ‘ s speed 10 more than than A.how many hours early than to reach a destination from I to VIII( approximately) ?a .4 hrs
b. 2 hrs
c. 5 hrs
d. 1 hrs
e. 6 hrs
ANSWERS:
1. c2. a
3. b
4. d
5. b
EXPLANATION:
According to the given informationUseful formulas is
d=s×t , t=d/s ; d=distance , t=time , s=speed
two cars or trains are moving to opposite towards eachother the time take they meet is =(total distance)/(sum of the speeds)
Distance Between V and VI= 36× 8 2/3
= 36×26/3=312 km
Between VI and VII=22×6 7/11
=73×22/11= 146 km
Between VII and VIII=42×13/3= 182 km
distance between destinations VII and VIII=182
ratio is 162 : 182 รจ81:91
2. a t =distance b\w IV and V/speed
=162/18=9 hrs
t= distance b\w V and VI/speed
=312/60=5 1/5 hrs
Avg speed= total dis/total time
=474/(9+5 1/5)
=38.85
Approximately 39 kmph
3. b t=( total distance I to V) /(sum of speeds of A&D)
total dis=(188+254+228+162)=832 km
D ‘s speed=162/6=27 kmph
total speeds=77+27=104 kmph
time taken by they meet=832/104= 8 hrs
that means they at 1 pm (5 am +8 hrs=1pm)
4.d Total distance b\w =188+254+228+162+312+146+182
=1472 km
B’s speed=254/ 5 12/23
=254×23/127
=46 kmph
Time taken by B to travel I to VIII=1472/46=32 hrs
5.b Total distance I to VIII =1472
Time taken by to reach I to VIII= 1472/77=19.11
=19
H’s speed =77+10=87
Time taken by to reach I to VIII=1472/87=16.88
=17
So H reach destination 2 hrs early than A (approx)
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