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2 years ago

Jose is now six years older than John. In two years, Jose will be twice as old as John. How old is John now ?

1 answer:

7 0

John is currently 6 years old.

You can find this by setting John's original age as x and Jose's original age as x + 6. Then we can set up an equation adding two to both.

John's new age = x + 2
Jose's new age = x + 8

Now we can multiply John's new age by two and set equal to Jose's age, since Jose's new age is twice as much as John's.

2(x + 2) = x + 8
2x + 4 = x + 8
x + 4 = 8
x = 4

This means John's original age is 4 and his new age is 6.

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3 years ago

Two cyclists, Alan and Brian, are racing around oval track of length 450m on the same direction simultaneously from the same poi

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Answer:

Alan: 200 m/min

Brian: 150 m/min

Step-by-step explanation:

Given : Two cyclists, Alan and Brian, are racing around oval track of length 450m on the same direction simultaneously from the same point. Alan races around the track in 45 seconds before Brian does and overtakes him every 9 minutes.

To find : What are their rates, in meters per minute?

Solution :

Let n represent the number of laps that Alan completes in 9 minutes.

Then n-1 is the number of laps Brian completes.

45 seconds = 3/4 minutes.

The difference in their lap times in minutes per lap is

$\frac{9}{(n -1)}-\frac{9}{n} =\frac{3}{4}$

Solving the equation we get,

$\frac{9n-9n+9}{n(n -1)} =\frac{3}{4}$

$9\times 4=3\times n(n -1)$

$36=3\times n(n -1)$

$n^2-n-12=0$

$n^2-4n+3n-12=0$

$n(n-4)+3(n-4)=0$

$(n-4)(n+3)=0$

$n=4,-3$

Neglecting n=-3

So, n=4

Then Alan's speed in m/min is

$S=\frac{D}{T}$

$S_a=\frac{4\times 450}{9}$

$S_a=200 m/min$

Brian completes 3 laps in that 9-minute time, so his rate is

$S_b=\frac{3\times 450}{9}$

$S_b=150 m/min$

Therefore, Alan: 200 m/min

Brian: 150 m/min

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