John runs to the market and comes back in 15 minutes. His speed on the way to the market is 5m/s and his speed on the way back is 4m/s. Find the distance to the market.

The distance to the market is 2000 m

Step-by-step explanation:

∵ John runs to the market and comes back in 15 minutes

→ Change the min. to the sec. because the unit of his speed is m/s

∵ 1 minute = 60 seconds

∴ 15 minutes = 15 × 60 = 900 seconds

→ Assume that t1 is his time to the market and t2 is his time from

  the market

∵  t1 + t2 = 15 minutes

∴ t1 + t2 = 900 ⇒ (1)

→ Assume that the distance to the market is d

∵ His speed on the way to the market is 5m/s

∵ Time = Distance ÷ Speed

∴ t1 = d ÷ 5 ⇒ (1 ÷ 5 = 0.2)

∴ t1 = 0.2d ⇒ (2)

∵ His speed on the way back is 4m/s

∴ t2 = d ÷ 4 ⇒ (1 ÷ 4 = 0.25)

∴ t2 = 0.25d ⇒ (3)

→ Substitute (2) and (3) in (1)

∵ 0.2d + 0.25d = 900

∴ 0.45d = 900

→ Divide both sides by 0.45

∴ d = 2000 m

∴ The distance to the market = 2000 m

true statements:

a) x + y = y + x   ⇒ commutative property of addition

b) (x * y) * z = x * (y * z) ⇒ associative property of multiplication

d) (x + y) + z = x + (y + z) ⇒ associative property of addition

false statements:

c) x - y = y - x   should be: x - y = - y + x

e) (x - y) - z = x - (y - z) should be: (x - y) - z = x - (y + z)

step-by-step explanation:

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