Please help solving this math question 😅?

Question

Car A is driving at an average speed of 80km/hr at 6am.Car B is driving at an average speed of 100km/hr at 6.15am. What time will Car B catch up with Car A?

Answer 1

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Car A is driving at an average speed of 80 km/hr at 6 am.

Car B is driving at an average speed of 100 km/hr at 6.15 am. 

Car A is at a distance of 20 km when Car A begins to move.

Their relative speed is 100 - 80 = 20 km/hr

Car B catches up with Car A after 1 hour which is at 7:15 am

Source(s): https://upscfever.com/upsc-fever/en/test/ssc-cgl/t...

Answer 2

Thanks for the help ;)

Answer 3

Assuming the 2 cars start from the same point

Let   t = 0   be at 6:15 am

let d1 be the distance travelled by car A and write it as follows:

d1 = d0 + 80 km/hr  * t      

where d0 is the distance traveled by car A during the first 15 minutes between 6 am and 6:15 am at the speed of 80km/hr 

d0 = (6:15 - 6:00) * 80 km/hr = 15 minutes * 80 km/hr 

= 15 minutes * 80 km/(60 minutes) = 20 km

hence 

d1 = 20 km + 80 km/hr  *t

Let d2 be the distance travelled by car B starting at t = 0 (6:15 am) and write it as follows:

d2 = 100  km/hr  * t

Car B catches up with Car A when d1 = d2

20km/hr  + 80km/hr  *t = 100km/hr  * t

Solve for t 

100km/hr   * t - 80km/hr   * t = 20 km

20 km/hr  * t = 20 km

t = 20 km / 20 km/hr  = 1 hr 

Car B catches up with car A 1 hour after t = 0 which corresponds to 6:15 am

Hence car B catches up with car A at  6:15 am + 1 hr  = 7:15 am

more at 

https://www.analyzemath.com/math_problems/rate_tim...

Answer 4

ure muum ,,,,,,,,,,,

Source(s): ur mum

Answer 5

In 15/60 hr, A has travelled 80(15/60)=20 km.

Thus when B catches up A, B needs

15/60+20/(100-80)=1.25 hrs

=>

B catches up A at 7:15 am.

Answer 6

Just to point out that if we only know the average speeds, we can't say where one will actually catch up with the other, we can only make the best guess.

Answer 7

At 6:15 A is ahead by 80 * (1/4) = 20 km

Their relative speed is 100 - 80 = 20 km/hr

Car B catches up with Car A after 1 hour which is at 7:15 am

Answer 8

Recall: s = d/t → where s is the speed, d is the distance, t is the time

The car A is driving at an average speed of 80 km/h at 6am.

The car B is driving at an average speed of 100 km/h at 6.15am.

So after 15 minutes, i.e. after (1/4) of an hour, the car A makes:

s₀ = d₀/t₀

d₀ = s₀.t₀

d₀ = 80 * (1/4)

d₀ = 20 km ← this is the distance between the car A and the car B (after 15 minutes)

After 6.15am, the car A makes a distance:

s₁ = d₁/t

d₁ = s₁.t

After 6.15am, the car B makes a distance:

s₂ = d₂/t

d₂ = s₂.t

The car B will catch up with the car A when:

d₂ = d₀ + d₁

s₂.t = 20 + s₁.t

s₂.t - s₁.t = 20

t.(s₂ - s₁) = 20

t = 20/(s₂ - s₁)

t = 20/(100 - 80)

t = 20/(20)

t = 1 hour ← this is the necessary time for the car B to cath up the car A

time = 6.15 + 1

time = 7.15am

Answer 9

80t + 20= 100t

20t= 20; t= 1 hour

7.15 am