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mina [271]
3 years ago
8

Stephen & Richard share a lottery win of £2950 in the ratio 2 : 3. Stephen then shares his part between himself, his wife &a

mp; their son in the ratio 3 : 5 : 2. How much more does his wife get over their son?
Mathematics
1 answer:
horsena [70]3 years ago
3 0

Answer:

Stephen's wife got £354 more than his son.

Step-by-step explanation:

Given:

Amount of Lottery = £2950

Now Given:

Stephen & Richard share a lottery amount in the ratio 2 : 3

Let the common factor between them be 'x'.

So we can say that;

2x+3x=2950\\\\5x = 2950

Dividing both side by 5 we get;

\frac{5x}{5}=\frac{2950}{5}\\\\x = 590

So we can say that;

Stephen share would be = 2x =2\times 590 = \£1180

Now Given:

Stephen then shares his part between himself, his wife & their son in the ratio 3 : 5 : 2.

Let the common factor between them be 'y'.

So we can say that;

3y+5y+2y=1180\\\\10y=1180

Dividing both side by 10 we get;

\frac{10y}{10}=\frac{1180}{10}\\\\y=118

So Stephen's wife share = 5y = 5\times 118= \£590

And Stephen's son share = 2y=2\times118 =\£236

Now we need to find how much more her wife got then her son.

To find how much more her wife got than her son we will subtract Stephen's son share from Stephen's wife share.

framing in equation form we get;

Amount more her wife got than her son = 590-236 = \£354

Hence Stephen's wife got £354 more than his son.

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Water is added to a cylindrical tank of radius 5 m and height of 10 m at a rate of 100 L/min. Find the rate of change of the wat
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Answer:

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For this case we know that r=5m represent the radius, h = 10m the height and the rate given is:

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Step-by-step explanation:

For a tank similar to a cylinder the volume is given by:

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For this case we know that r=5m represent the radius, h = 10m the height and the rate given is:

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For this case we want to find the rate of change of the water level when h =6m so then we can derivate the formula for the volume and we got:

\frac{dV}{dt}= \pi r^2 \frac{dh}{dt}

And solving for \frac{dh}{dt} we got:

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We need to convert the rate given into m^3/min and we got:

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\frac{dh}{dt}=\frac{0.1 m^3/min}{\pi (5m)^2}= 0.0012732 \frac{m}{min}

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