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4 years ago

How can I find each quotient useing addition

Mathematics

1 answer:

LekaFEV [45]4 years ago

7 0

A quotient can only be figured out by division. Hope this helps!

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Find the surface area of the composite figure.

Vanyuwa [196]

Answer:

you need to provide an image

6 0

3 years ago

A pair of shoes which costs Rs 320 was sold at a loss of 5%what was the selling price if it is sold at profit of 20%find the gai

taurus [48]

Answer:

The Sell Price of the shoe at 5% loss is Rs 304.

The Sell Price of the shoe at 20% profit is Rs 384.

The Total Gain Percent is 12.5% when shoes are sold at Rs.360.

Step-by-step explanation:

Given,

Cost Price(C.P.) of shoe = Rs 320

Case 1.

Loss = 5%

C.P. = Rs 320

Here loss percent and cost price of the shoe is given, we have to find out the selling price(S.P.) of the shoe.

For this we use the formula of calculating S.P.

$S.P. = \frac{C.P.(100-loss\%)}{100}$

On substituting the given values, we get;

$S.P. = \frac{320(100-5)}{100}\\\\S.P. =\frac{320\times 95}{100} =\frac{30400}{100} =Rs\ 304$

The Sell Price of the shoe at 5% loss is Rs 304.

Case 2.

C.P. = Rs 320

Profit = 20%

Here profit percent and cost price of the shoe is given, we have to find out the selling price(S.P.) of the shoe.

For this we use the formula of calculating S.P.

$S.P. = \frac{C.P.(100+profit\%)}{100}$

On substituting the given values, we get;

$S.P. = \frac{320(100+20)}{100}\\\\S.P.=\frac{320\times 120}{100}=\frac{38400}{100}=Rs\ 384$

The Sell Price of the shoe at 20% profit is Rs 384.

Case 3.

C.P. = Rs 320

S.P. = Rs 360

Now we need to find the gain percent.

First we will find the gain.

Gain is the difference of S.P with C.P.

Gain = $360-320 = 40$

Now Gain Percent can be calculated by dividing gain with the cost price and then multiplying it with 100

Gain Percent = $\frac{40}{320}\times 100 = 12.5\%$

Hence The Total Gain Percent is 12.5% when shoes are sold at Rs.360.

4 0

3 years ago

Solve the system, using any method. y = x + 2 and x + 2y = 25

Leni [432]

$Elimination\ method\\\\\left\{\begin{array}{ccc}y=x+2&|subtract\ "x"\ from\ both\ sides\\x+2y=25\end{array}\right\\+\left\{\begin{array}{ccc}-x+y=2\\x+2y=25\end{array}\right|add\ the\ equations\ sides\ \\\----------\\.\ \ \ \ \ \ \ 3y=27\ \ \ \ \ |divide\ both\ sides\ by\ 3\\.\ \ \ \ \ \ \ \ \ \ y=9\\\\subtitute\ y=9\ to\ second\ equation\\x+2(9)=25\\x+18=25\ \ \ \ \ \ \ |subtract\ 18\ from\ both\ sides\\x=7\\\\Solution:x=7\ and\ y=9\to(7;\ 9)$