Model 4 Time & Distance with Ratios Practice Questions Answers Test with Solutions & More Shortcuts
time & distance PRACTICE TEST [5 - EXERCISES]
Model 1 Basic Time & Distance using formula
Model 2 Vehicles in x/y of its usual speed
Model 3 Problems on average speed
Model 4 Time & Distance with Ratios
Model 5 Problems with Races
Question : 1 [SSC CPO S.I.2009]
Three cars travelled distance in the ratio 1 : 2 : 3. If the ratio of the time of travel is 3 : 2 : 1, then the ratio of their speed is
a) 1 : 3 : 9
b) 1 : 2 : 4
c) 4 : 3 : 2
d) 3 : 9 : 1
Answer »Answer: (a)
Required ratio = $1/3 : 2/2 : 3/1$
= $1/3$ : 1 : 3
$1/3$ × 3 : 1 × 3 : 3 × 3
[Since, Speed = $\text"Distance"/\text"Time"$]
= 1 : 3 : 9
Question : 2 [SSC CGL Prelim 1999]
In covering a certain distance, the speed of A and B are in the ratio of 3 : 4. A takes 30 minutes more than B to reach the destination. The time taken by A to reach the destination is :
a) 1$1/2$ hours
b) 2 hours
c) 2$1/2$ hours
d) 1 hour
Answer »Answer: (b)
Let the distance of destination be D km
Let the speed of A = 3x km/hr
then speed of B = 4x km/hr
According to question,
$D/{3x} - D/{4x}$ = 30 minutes = $1/2$ hr
= $D/{12x} = 1/2$
$D/{3x} = 4/2$ = 2 hours
Hence, time taken by A to reach destination = 2hrs.
Using Rule 9,Speed(s) ∝ $1/{time (t)}$ ⇒ s ∝ $1/t$$s_1t_1 = s_2t_2$(Provided distance is constant)
Here, $S_1 = 3x, S_2 = 4x$
$t_2 = y, t_1 = y + 30/60 = y + 1/2$
$S_1t_1 = S_2t_2$
$3x × (y + 1/2) = 4x × y$
$3y + 3/2 = 4y ⇒ y = 3/2$
Time taken by A
= $3/2 + 1/2$ = 2 hrs.
Question : 3 [SSC CGL Tier-II 2013]
A certain distance is covered by a cyclist at a certain speed. If a jogger covers half the distance in double the time, the ratio of the speed of the jogger to that of the cyclist is
a) 4 : 1
b) 1 : 2
c) 2 : 1
d) 1 : 4
Answer »Answer: (d)
Using Rule 1,
Let speed of cyclist = x kmph
& Time = t hours
Distance = ${xt}/2$ while time = 2t
∴ Required ratio = ${xt}/{2 × 2t}$ : x = 1 : 4
Question : 4 [SSC CGL Prelim 2007]
A train starts from A at 7 a.m. towards B with speed 50 km/h. Another train starts from B at 8 a.m. with speed 60 km/h towards A. Both of them meet at 10 a.m. at C. The ratio of the distance AC to BC is
a) 5 : 4
b) 6 : 5
c) 4 : 5
d) 5 : 6
Answer »Answer: (a)
AC = Distance covered by train starting from A in 3 hours
= 50 × 3 = 150 km
BC = Distance covered by train starting from B in 2 hours
= 60 ×2 = 120 km
∴ AC : BC = 150 : 120 = 5 : 4
Question : 5 [SSC CGL Tier-I 2016]
A truck covers a distance of 550 metre in one minute where as a bus covers a distance of 33 km in $3/4$ hour. Then the ratio of their speeds is :
a) 2 : 3
b) 3 : 4
c) 1 : 4
d) 1 : 3
Answer »Answer: (b)
Speed of truck
= ${550\text"metre"}/{60\text"second"} = (55/6)$ m./sec.
Speed of bus
= ${33 × 1000 \text"metre"}/{3/4 × 60 × 60 \text"second"} = 440/36$ m./sec.
∴ Required ratio = $55/6 : 440/36$
= 55 × 6 : 440 = 3 : 4
IMPORTANT quantitative aptitude EXERCISES
Model 4 Time & Distance with Ratios Shortcuts »
Click to Read...Model 4 Time & Distance with Ratios Online Quiz
Click to Start..time & distance Shortcuts and Techniques with Examples
-
Model 1 Basic Time & Distance using formula
Defination & Shortcuts … -
Model 2 Vehicles in x/y of its usual speed
Defination & Shortcuts … -
Model 3 Problems on average speed
Defination & Shortcuts … -
Model 4 Time & Distance with Ratios
Defination & Shortcuts … -
Model 5 Problems with Races
Defination & Shortcuts …
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