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Jack has twice as many dimes as quaters.

Total value = $6.30

Total value = $6.30 = 630 cents

Now Let the total number of dimes be x, and total number of quaters be y.

Hence it is given,

                         
                            \(x=2y\)
 

Further we know that the total amount is 630 cents.

Therefore we can write it as,

                       \(10x + 25y=630\)
Now we shall solve the two equations using substitution, put ^{\(x = 2y\)} in equation \(10x + 25y = 630\)

we get, 

10 (2y) + 25y = 630

   20y  +  25y = 630

               45y = 630

                   y = 14

Now we know ^{\(x = 2y,\)} Therfore \(x = 2(14)=28\)

The total number of dimes he has are 28.

Let x = number of brownies sold.

      y = number of cookies sold.

 : x/y = 3/5

 =>   5x - 3y = 0 -------- 1

Let $p be the price of each piece of cookies.

 Given that, each piece of bro is sold for $1

 more than each piece of cookie.

 : The price of each piece of brownie is $(8+1)

Eric collected $154 for all brownies and cookies sold.

  : (8+1)x + Py = 154 -------- 2

Also, Eric collected $44 more from the sale of brownies than the sale of cookies.

  :     (p+1)x = 44 + py

Substituting (p+1) x = 44+py in equation 2 we, get

              44 + py + py = 154

              =>   2py = 110

              =>   py = 55 ------ 3

 From equation 1, y = 5/3x.

  : Substituting value y = 5/3x in equation 3, we get

              P(5/3x) = 55 => px = \({55 * 3 \over 5} = 33{}{}\)

             => px * 33 -----4

 Substituting values of px = 33 of py = 55 in

 equation 2, we have

     (P+1)x + Py = 154

    => Px + x + Py = 154 

    => 33 + x + 55 = 154   =>  x = 154 - 33 - 55

                                         => x = 66

  Substituting  x = 66 in equation 1, we have

     5x - 3y = 0   =>  5(66)- 3y = 0 =>  3y = 330

                                                     => y = 110

    Substituting  x = 66 in equation 4, we have

         px = 33  =>  p (66) = 33 =>  P=0.5

: Each piece of cookie is sold at $0.5 and each piece of brownie is sold at $1.5

The number of cookie sold = y = 110.

The number of brownies sold is x = 66.

:  66 pieces of brownies were sold.

Let total number of strawberries = x.

Since 3 friends divided strawberries equally therefore

each friends get x/3 strawberries.

   Because           x/3 + x/3 + x/3 = x.

Since, after they each ate 4 strawberries, the total number of strawberries left was equal to the amount each friend had at the beginning.

That is (x/3-4) + (x/3-4) (x/3-4) = x/3 (because at the beginning each friend has x/3 strawberries).

=>  3x/3 - 12 = x/3

=>  n - 12 = x/3     =>  3(n - 12) = x

=>  3n - 36 = n   =>  2n = 36   =>  n = 18

The total number of strawberries = 18

Let's say the laptop costs 6x. the VR headset costs $y. The laptop cost $2.    So { x + y + z = 3100 1

                      { x - (y+z) = 1900 2

                      {    y = 32             3
3->2     x - 42 = 1900  4

3->1     x + 42 = 3100 5

 4+5      2x = 5000           x = 2500

x - 42 = 1900             z=150

y = 32                      y = 450

450 is the answer

Let the cost of 1 chair = x

∴ Cost of 3 chairs = 3x

Since Cost of 4 stools = Cost of 3 chairs.

∴ Cost of 4 stools = 3x

∴ Cost of 1 stool = \( {3x\over 4}{}{}\).........(1)

Cost of 2 chairs = 2x

Since the cost of 5 stools is $28 more than 2 chairs.

∴ Cost of 5 stools = 2x + 28

∴ Cost of 1 stool = \({2x+28\over 5}{}{}\)............(2)

From equations  (1) and (2) we have,

Cost of 1 stool = \( {3x \over 4}{}{}\)

Cost of 1 stool = \({2x+28 \over 5}\frac{}{}\)

∴ \( {3x\over 4}{}{} = {2x+28\over5}\)

⇒ 15x = 8x + 112

⇒ 7x = 112
⇒ x = 16
∴ Cost of 1 chair = $16

And Cost of 1 Stool = \( {3x \over 4}{}{} = {3\over4} * 16 = 12\)

Now we have,

Cost of 1 chair = $16

∴ Cost of 2 chairs = 2 × $16 = $32

Cost of 1 stool = $12

∴ Cost of 2 stools = 2 × $12 = $24

Hence the cost of 2 chairs and 2 stools = $32 + $24 = $56

∴ The cost of 2 chairs and 2 stools = $56

Let Patrick had 'x' number of Ducks.

Since he had 3 times as many chickens as ducks.

∴ Number of chickens Patrick had = 3x

Since he had a total of 2000 Ducks and chickens,

∴ 3x + x = 2000

⇒ 4x = 2000

⇒ x = 500

∴ Number of Ducks Patrick had = 500

And number of Chickens Patrick had = 3x = 3 × 500 = 1500

Now he sells some chicken and after selling he now has twice as many ducks as chickens.

Let he sells 'y' number of chickens,

∴ Number of chickens left = 1500 - y

And the number of Ducks = 500

he now has twice as many ducks as chickens.

∴ 2 × (1500 - y) = 500

⇒ 3000 - 2y = 500

⇒ 2y = 2500

⇒ y = 1250

this means that he sold 1250 chickens.

ANSWER :

Number of chickens Patrick sold = 1250

Let there were x boxes of apple at first.

 Monday sold :  \( {3\over7 }x + 7{}{}\)
 Remaining  :   \( x -({3 \over 7}x + 7) = {}{}\)\({4 \over 7}x - 7{}{}\)
 Tuesday sold:     \({3\over 5}({}{}\)\( {4 \over 7}x-7)+5 ={}{}\)\( {12 \over 35}x+{}{}\)\( {4 \over 5}{}{}\)
 Remaining:      13
 Set and equation :

        x - \( ({3 \over 7}x+7)-{}{}\)\( ({12 \over 35}x+{}{}\)\( {4 \over 5}){}{}\) =13

        x - \( {3 \over 7}x-7-{}{}\)\( {12 \over 35}x-{}{}\)\( {4\over 5}=13{}{}\)

                                      \({8 \over 35}x ={}{}\)\( {104\over 5}{}{}\)
                                           \(x\) = 91 (boxes)

There were 91 boxes of apples at first 

{  Total boxes -- Monday sold -- Tuesday sold = 13 }

The question easy by

includes apple:

 1 + 2 + 1 + 2 + 2 + 2 + 2 + 1 + 2 = 15

only banana:    1

 only  banana + cherry:   4

 =>  15+1+4=20

To calculate

the number of sports cards Miguel had.

Let y be the initial Corey's sports cards and x be the initial Miguel's sports cards then according to the question we have,

⇒ Initial Corey's sports cards= (\(1 {3 \over 4}{}{}\)) (Initial Miguel's sports cards)

⇒ y = (\( {7 \over 4}{}{}\))(x)

⇒ y =\({7x \over 4}{}{}\)

After giving half of Miguel's sports cards to his friend, Miguel's final sports cards are

\(x\)-\( {x \over 2}{}{}\) and Corey's sports card are \(y\)+\( {x \over 2}{}{}\).

According to the question,

⇒ Final Miguel's sports cards=(Final Corey's sports cards)−18

⇒ (x - \({x\over 2}{}{}\)) = (y + \( {x \over 2}{}{}\)) - 18

⇒ \(y\) = 18

Plug it above,

⇒ 18 = \( {7x \over 4}{}{}\)

⇒ \(x\) = \({18 * 4\over 7}{}{}\)

⇒ \(x\) = 10.29

⇒ \(x\) = 10

Thus, Miguel had 10 sports cards at first.

Sorry if this was a late response.

Set they had equal number which is x, and

 Dorothy left y paper hearts.

  So,   \( {1 \over 4}{}{}\)y + y + \( {1 \over 4}{}{}\)y - 5 = 85   
                                  y = 60

  and  Ellen (left)  =  \({1 \over 4}{}{}\)y  - 5  =  \( {1 \over 4}{}{}\) * 60  -5  =10

   So  the number of them had are x = 10 + 70  = 80

   So   Dorothy (sell)   = 80 - 60  =20.

20  paper hearts.