What is the answer to this and give me step by step explaining

Stevie and Michael share their profit in a ratio 3:5. Stevie gets £120 more than Michael. How much profit did they make altogeth

Airida [17]

Answer:

£480

Step-by-step explanation:

we can multiply both terms of ratio with same number without changing the ratio.

Example

a:b = ax: bx (here we have multiplied both terms with x)\

Given

Stevie and Michael share their profit in a ratio 3:5

Let the profit for Stevie and Michael be 3x and 5x

Given Stevie gets £120 more than Michael

Profit for Steve = 120 + profit for Michael

3x = 120 + 5x

3x-5x = 120

=> -2x = 120

=> x= -120/2 = -60

As we see profit is negative , hence it cannot be solution.

It can be solved if Stevie and Michael share their profit in a ratio 5:3

(note: in ratio , bigger number should have bigger term in ratio)

Let the profit for Stevie and Michael be 3x and 5x

Given Stevie gets £120 more than Michael

Profit for Steve = 120 + profit for Michael

5x = 120 + 3x

5x-xx = 120

=> 2x = 120

=> x= x120/2 = 60

Profit for stevei = 5x =5*60 = 300

Profit for Michael = 3x = 3*60 = 180

Total profit for them together = 300 + 180 = 480 Answer.

8 0

3 years ago

If you had 1052 toothpicks and were asked to group them in powers of 6, how many groups of each power of 6 would you have? Put t

sukhopar [10]

1052 toothpicks can be grouped into 4 groups of third power of 6 ( $6^{3}$ ), 5 groups of second power of 6 ( $6^{2}$ ), 1 group of first power of 6 ( $6^{1}$ ) and 2 groups of zeroth power of 6 ( $6^{0}$ ).

The number 1052, written as a base 6 number is 4512

Given: 1052 toothpicks

To do: The objective is to group the toothpicks in powers of 6 and to write the number 1052 as a base 6 number

First we note that, $6^{0}=1,6^{1}=6,6^{2}=36,6^{3}=216,6^{4}=1296$

This implies that $6^{4}$ exceeds 1052 and thus the highest power of 6 that the toothpicks can be grouped into is 3.

Now, $6^{3}=216$ and $216\times 5=1080, 216\times 4=864$ . This implies that $216\times 5$ exceeds 1052 and thus there can be at most 4 groups of $6^{3}$ .

Then,

$1052-4\times6^{3}$

$1052-4\times216$

$1052-864$

$188$

So, after grouping the toothpicks into 4 groups of third power of 6, there are 188 toothpicks remaining.

Now, $6^{2}=36$ and $36\times 5=180, 36\times 6=216$ . This implies that $36\times 6$ exceeds 188 and thus there can be at most 5 groups of $6^{2}$ .

Then,

$188-5\times6^{2}$

$188-5\times36$

$188-180$

$8$

So, after grouping the remaining toothpicks into 5 groups of second power of 6, there are 8 toothpicks remaining.

Now, $6^{1}=6$ and $6\times 1=6, 6\times 2=12$ . This implies that $6\times 2$ exceeds 8 and thus there can be at most 1 group of $6^{1}$ .

Then,

$8-1\times6^{1}$

$8-1\times6$

$8-6$

$2$

So, after grouping the remaining toothpicks into 1 group of first power of 6, there are 2 toothpicks remaining.

Now, $6^{0}=1$ and $1\times 2=2$ . This implies that the remaining toothpicks can be exactly grouped into 2 groups of zeroth power of 6.

This concludes the grouping.

Thus, it was obtained that 1052 toothpicks can be grouped into 4 groups of third power of 6 ( $6^{3}$ ), 5 groups of second power of 6 ( $6^{2}$ ), 1 group of first power of 6 ( $6^{1}$ ) and 2 groups of zeroth power of 6 ( $6^{0}$ ).

Then,

$1052=4\times6^{3}+5\times6^{2}+1\times6^{1}+2\times6^{0}$