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3 years ago

Jaime has 5 over 11 spaceof a project completed while Tim has finished 7 over 13 spaceof the same project.

Mathematics

1 answer:

faust18 [17]3 years ago

4 0

Answer:

a. Tim has completed a greater amount of work.

b. Together they have completed $\frac{142}{143}$ part of two projects.

Step-by-step explanation:

Let us assume that the project on which Jaime and Tim are working has x amount of space.

Now, given that Jaime has 5 over 11 space of a project completed while Tim has finished 7 over 13 space of the same project.

Therefore, Jaime has completed $\frac{5x}{11}$ amount of space and Tim has finished $\frac{7x}{13}$ amount of space.

a. Now, (7x) \times 11 > (5x) \times 13

⇒ $\frac{7x}{13} > \frac{5x}{11}$

Therefore, Tim has completed the greater amount of work. (Answer)

b. Together they have completed $(\frac{7x}{13} + \frac{5x}{11}) = \frac{(7 \times 11 + 5 \times 13)}{13 \times 11} x = \frac{142}{143}x$ amount of space out of 2x amount of space. (Answer)

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Step-by-step explanation:

Now, Given that:-

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So, Radius (r) = 10/2 = 5cm

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$volume \: of \: the \: cylinder \: = \pi {r}^{2} h$

$= > \pi \times {5}^{2} \times 12 {cm}^{3} = 300\pi {cm}^{3}$

Radius of the cone = 5 cm.

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= curved surface area of cylinder + curved surface area of cone + area of (upper) circular base of cylinder

$= 120\pi {cm}^{2} + 65\pi {cm }^{2} + 25 \pi {cm}^{2} \\ = 210 \times 3.14 {cm}^{2} = 659.4 {cm}^{2}$

<h3>Hope it helps you!!</h3>

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