Tag: applications of simultameous equations

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Multiple choice maths unitary method time speed and distance problems involving speed word problems on speed time and distance time and distance word problems on simultaneous equations applications of simultameous equations unitary method idea of speed distance and time speed math time work and distance ratio and proportions linear equations in two variables maths

A cyclist, after riding a certain distance, stopped for half an hour to repair his bicycle, after which he completes the whole journey of 30km at half speed in 5 hours. If the breakdown had occurred 10 km farther off, he would have done the whole journey is 4 hours. Find where the breakdown occurred and his original speed.

  1. 5km/hr,10km

  2. 10km/hr,10km

  3. 10km/hr,5km

  4. 5km/hr,5km
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let original speed be v and distance to breakdown be x. The equations derived from the two scenarios (30km total) lead to v = 10 km/hr and x = 10 km.

Multiple choice time speed and distance problems involving speed word problems on speed time and distance time and distance word problems on simultaneous equations applications of simultameous equations unitary method idea of speed distance and time speed math time work and distance ratio and proportions linear equations in two variables maths

Ramu walks at 3kms per hour while his brother Somu walks at 4kms per hour. They both started to school, which is 2kms away from their house at 7:00 A.M. Then, which of these is are correct?

  1. Somu reaches school at 7:20 AM

  2. Somu reaches school 10 mins earlier than Ramu

  3. Ramu reaches school at 7:30 AM

  4. All the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Ramu =3 kmph Somu =4 kmph Both starts at 7 AM Distance =2km

Time taken by Ramu ⇒S=v×t

2=3×t

t=23hr=23×60

=40min

Time taken by Somu ⇒2=4×t

t=12hr=30 min.

∴ Somu reaches school 10 mins earlier than Ramu.

Multiple choice time speed and distance problems involving speed word problems on speed time and distance time and distance word problems on simultaneous equations applications of simultameous equations unitary method idea of speed distance and time speed math time work and distance ratio and proportions linear equations in two variables maths

A goods train of length 500 m transverse a 2-km-long tunnel at a constant speed of 36 km/hour. The time needed for the train to completely pass through the tunnel is

  1. 3 minutes 20 seconds

  2. 4 minutes 10 seconds

  3. 4 minutes 40 seconds

  4. 6 minutes 10 seconds
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Total distance to cover = 2000 m + 500 m = 2500 m = 2.5 km. Speed = 36 km/hr = 10 m/s. Time = distance / speed = 2500 / 10 = 250 seconds. 250 seconds = 4 minutes and 10 seconds.

Multiple choice ime speed and distance problems involving speed word problems on speed time and distance time and distance word problems on simultaneous equations applications of simultameous equations unitary method idea of speed distance and time speed math time work and distance ratio and proportions linear equations in two variables maths

The time taken by a person to cover 150 km was 2.5 hrs more than the time taken in the return journey. If he returned at a speed of 10km/hr more than the speed of going, what was the speed per hour in each direction?

  1. going speed =−30km/hr, returning=−20km/hr

  2. going speed =20km/hr, returning=30km/hr

  3. going speed =30km/hr, returning=40km/hr

  4. going speed =30km/hr, returning=20km/hr
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Letting the going speed be $v$, the returning speed is $v+10$. The time equation $150/v - 150/(v+10) = 2.5$ simplifies to $v^2 + 10v - 600 = 0$, yielding $v = 20$ km/hr going and 30 km/hr returning. Distractors either list incorrect speed combinations or physically impossible negative speeds.

Multiple choice time speed and distance problems involving speed word problems on speed time and distance time and distance word problems on simultaneous equations applications of simultameous equations unitary method idea of speed distance and time speed math time work and distance ratio and proportions linear equations in two variables maths

The present ages of Reena and Usha ar 24 and 36 yr respectively. What was the ratio between the ages of Usha and Reena respectively 8 yr ago?

  1. 7:4

  2. 4:7

  3. 11:8

  4. 8:11
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Age of Reena after 8yr=24−8=16yr
Age of Usha After 8yr=36−8=28
So, the ratio will be 1628=47

Multiple choice time speed and distance problems involving speed word problems on speed time and distance time and distance word problems on simultaneous equations applications of simultameous equations unitary method idea of speed distance and time speed math time work and distance ratio and proportions linear equations in two variables maths

A train is running at a uniform speed of 60km/h. How much time it will take to cover a distance of 300km?

  1. 5 h

  2. 4.5 h

  3. 6 h

  4. 7 h
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$$speed=\frac{distance}{time}$$ $$time=\frac{distance}{speed}$$ $$time=\frac{300}{60}h$$ $$time=5 hr$$

Multiple choice time speed and distance problems involving speed word problems on speed time and distance time and distance word problems on simultaneous equations applications of simultameous equations unitary method idea of speed distance and time speed math time work and distance ratio and proportions linear equations in two variables maths

If a man walks 17 km in 5 hours, how long(in hrs) would he take to walk 20.4 km

  1. 6

  2. 5

  3. 8

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Time taken to walk 17km=5hrs.
Time taken to walk $1km=\dfrac{5}{17}hrs.$
Time taken to walk $20.4km=\dfrac{5}{17} \times20.4$

$=6 hrs.$

Multiple choice time speed and distance problems involving speed word problems on speed time and distance time and distance word problems on simultaneous equations applications of simultameous equations unitary method idea of speed distance and time speed math time work and distance ratio and proportions linear equations in two variables maths

A man runs round a circular field of radius 50m at the speed of 12km/hr. What is the time taken by the man to take twenty rounds of the field?

  1. 30 min

  2. 32 min.

  3. 34 min.

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Circumference = 2 * pi * r = 2 * 3.14 * 50 = 314 m. Total distance for 20 rounds = 20 * 314 = 6280 m = 6.28 km. Speed = 12 km/hr. Time = 6.28 / 12 hours = 0.5233 hours = 31.4 minutes. 32 minutes is the closest approximation.

Multiple choice time speed and distance problems involving speed word problems on speed time and distance time and distance word problems on simultaneous equations applications of simultameous equations unitary method idea of speed distance and time speed math time work and distance ratio and proportions linear equations in two variables maths

Two boys P and Q walk at 6 km/hr and 6 km/hr respectively. If Q covers a certain distance in 10 minutes less than P. What is the time taken by Q to cover the same distance (in minutes)?

  1. 40

  2. 60

  3. 30

  4. 70
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice time speed and distance problems involving speed word problems on speed time and distance time and distance word problems on simultaneous equations applications of simultameous equations unitary method idea of speed distance and time speed math time work and distance ratio and proportions linear equations in two variables maths

A can run 1 km in 4 min 54 sec and B in 5 min. How many metres start can A can give B in a km race so that the race may end in a dead heat ?

  1. 16 m

  2. 18 m

  3. 20 m

  4. 22 m
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Distance covered by B in 6 sec $= \dfrac{(1000}{300}×6)=20 m$