Question

estion In a race of 500 meters, A beats B by 50 meters but in a race of 1 km, B beats C by 250 meters. Quantity : In a race of 800 meters, by how much distance will A beat C? Quantity 11 : 250 meters

Quantity: l is greater than Quantity: II
Quantity: lis greater than equal to Quantity : Il Quantity: l is less than Quantity : II
Quantity : Il is greater than equal to Quantity: 1 ​

Answer 1

Final Answer:

The correct answer is Quantity l is greater than Quantity II.

Given:

In a race of 500 meters, A beats B by 50 meters but in a race of 1 km, B beats C by 250 meters.

Quantity I : In a race of 800 meters, by how much distance will A beat C?

Quantity II : 250 meters

Options:

Quantity: l is greater than Quantity: II

Quantity I : is greater than equal to Quantity : Il

Quantity: l is less than Quantity : II

Quantity : Il is greater than equal to Quantity: 1 ​

To Find:

The distance by which A will beat C in a race of 800 meters is to be found out. Accordingly the correct options is to be asserted depending on the relationship that will be evaluated between Quantity I  and Quantity II.

Explanation:

The following concepts will be used for solving this problem.

• Speed is equal to the distance covered divided by the time elapsed.
• The Ratio of speeds of A, B, and C is equal to the Ratio of distances covered in the same time by  A, B, and C.

Step 1 of 5

Suppose the race of 500 meters is completed in time "t minutes". Note here the following observations.

• A beats B by 50 meters.

• The distance covered by A in "t minutes"$=500$ meters

• The distance covered by B in "t minutes" $=500-50=450$ meters

Step 2 of 5

Suppose the race of 1 km or 1000 meters is completed in time "t minutes". Here observe the following.

• B beats C by 250 meters.
• The distance covered by B in "t minutes"$=1000$ meters

• The distance covered by C in "t minutes" $=1000-250=750$ meters

Step 3 of 5

Assume the speeds of A, B, and C are $a$ meters/minute, $b$ meters/minute and $c$ meters/minute, respectively.

As per the data available from the race of 500 meters, write the following

$\frac{a}{b} =\frac{500}{450} =\frac{10}{9}$

As per the data available from the race of 1 km or 500 meters, write the following

$\frac{b}{c} =\frac{1000}{750} =\frac{4}{3}$

Step 4 of 5

Rearrange the above ratios in the following way.

$\frac{a}{b} \\=\frac{10}{9} \\=\frac{10*4}{9*4} \\=\frac{40}{36}$

$a:b=40:36$

$\frac{b}{c} \\=\frac{4}{3}\\=\frac{4*9}{3*9}\\=\frac{36}{27}$

$b:c=36:27$

Combine the above ratios to obtain the following.

$a:b:c=40:36:27$

Step 5 of 5

Now for the race of 800 meters, deduce the following.

$a:c=40:27\\\frac{a}{c} =\frac{40}{27}$

Rewrite the above ratio in the following way.

$\frac{a}{c}\\ =\frac{40}{27}\\=\frac{40*20}{27*20\\}=\frac{800}{540}$

So, it is clear by the time A finished 800 meters, C could finish 540 meters only.

So, The distance by which A will beat C in a race of 800 meters is $=800-540=260$ meters.

Obviously, $260 > 250$ is true.

Therefore, , Quantity: l is greater than Quantity: II.

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