Slimesewer

Paul = 2/5 Peter

Paul/Peter = 2/5

2x/(5x - 210) = 11/10

20x = 55x - 2310

  35x = 2310

     x = 66

(a) Peter = 5x = 330

(b) Peter throw away = 210/330

                                  = 7/11

Total number of stickers

          = 2,950

  Number of stickers stuck

    = 1/5 of 2,950 = 590

  Number of stickers with no holes

     = 33

  Number of stickers with no glue

      = 1/10 of 2,950 = 295

Number of stickers that could be used

  = 2,950 - (590 + 33 + 295)
  = 2,950 - 918 = 2,032

Price of Workbook = $6

Price of Textbook = (3/5)(6) + 6 = $9.6

Let T = number of Textbooks

     W = number of Workbooks

      S = total number of textbooks and workbooks

               S = T + W --> equation (1)
    * 0.35S = T --> equation (2)

    * 6W = 108 + 9.6T --> equation (3)

 equation (2) in equation (1)

      S = T + W

        S = 0.35S + W

         0.65S = W --> equation (4)

Total sales:

        6W + 9.6T = (108 + 9.6T) + 9.6T

         6W = 108 + 9.6T

                    * T = 0.35 (equation (2))

          6W = 108 + 9.6 (0.35S)

          6W/6 = (108 + 3.36S)/6

            W = 18 + 0.56S --> equation (5)

   equation (4) = equation (5)

      0.65S = 18 + 0.56S

      0.65S - 0.56 S = 18

             0.09S/0.09 = 18/0.09

                  S = 200 books

                200 books was sold altogether

Adam = $x money

Firdaus = $y

(x - 9)/y = 2/7 -(1)

2y = 7(x - 9) = -(1)

    2y = 7x - 63 -(1)

 x/(y - 9) = 7/11

   11x = 7(y - 9)

   11x = 7y - 63 

2 (11x - 7y = -63)

7 (2y - 7x = -63)

  14y - 49x = -63 * 7

  22x - 14y = -63 * 2 

| - - - - - - - - - - - - - - - - - -
|       -27x = -567
|            x = 21
|            y = 42
|   x + y = $63
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Katherine divided her pie into eights.

Savannah divided her pie into fourths.

Katherune = 6/8 = 3/4 ate.

3 fourths of Savannah's pie have to be eaten.

No. of students liking peanut butter swirl brownies

                                       = 8/39 = 16/78 = 16 students

No. of students liking Rocky Road brownies

                                       = 3/13 = 18/78 = 18 students

No. of students liking chocolate chip brownies

                                       = 2/13 = 12/78 = 12 students

No. of students liking brownies with walnuts 

                                       = 7/39 = 14/78 = 14 students

No. of students liking butter scotch brownies

                                       = 78 - (16 + 18 + 12 + 4)

                                       = 18 students  

Let total money = 1x

Let cost of one mango = M

                         pear = P

4x/11 = 4M + 12P -- (1)

P = 1/3M => M = 3P -- (2)

Put in (1) => 4(3P) + 12P = 4/11x

=> 12P + 12P = 4/11x => P = 4/24x 6 * 11

P = x/66 -- (3)

Money left => x = 4/11x = 7/11x Multiply by 7/11 both sides

7/11x = 7/11x * 66 6P = 42

Thomas can buy 42 pears with cost of his money

Let x be the amount of money David has.

given data

Amount of money spent by David on a pair of jeans = 4/15x

Amount of money spent on watch = 4/15x + 80

4/15x + (4/15x + 80) + 130 = x

8/15x + 210 = x

=> 210 = x - 8/15x => 210 = 7x/15 

x = $410 (Total amount of money David has initially)

(a) So, ratio of amount of money spent on the watch to the amount of money spent on pair of jeans.

= (4/15) (450) + 80)/ 4/15(450)  = (120 + 80)/ 120

= 200/120 = 5/3 or 5:3

(b) David's required sum of money is x, which is $450

We need to find n such that choosing 3 people from that where order doesn't matter and the number of ways is greater than 365. We use the formula for combination:

ncr = (n!)/r!(n-r)! In this case we will --> (n!)/3!(n-3)!

                          choose 3 people 

The number of ways for this must be greater than 365:

(n!)/3!(n-3)! ≥ 364 --> (n(n-1)(n-2)(n-3)!)/6(n-3)! ≥ 365

                                    n(n-1)(n-2) ≥ 2190
Use trial and error. Use values starting from n = 11:

For n = 15, the number of ways of choosing 3 random people became larger than 2190.

n = 15

2 * 7 = 14

2 zelks = 14 zacks

14 / 2 * 6

= 7 * 6

= 42

So 14 zacks = 42 zinks