apsiganocj

Let the total pens produced in 5 days be x.

Pens produced on the first day = x/5

second day = 28

third day = (x/5 + 28)/2

4th day = x/5 + 9

5th day = 64

∴, x/5 + 28 + (x/5 + 28)/2

          + x/5 + 9 + 64

          = x

=> x/5 + 1/2(x/5) + 1/2(28)

        + x/5 + 73 + 28 = x

=> x/5 + x/10 + x/5 + 101 + 14 = x

=> \( {2x + x + 2x \over 10} + 115 = x\)
=> \( {5x + 1150\over 10} = x\)

=> 5x + 1150 = 10x

=>  5x = 1150

=>    x = 230

∴, total pens = \(\boxed{230}\)

Let,

     a = Peter's sticker

     b = Jordan's sticker

     c = Albert's sticker

Given:

    1) \({a + b \over 2} = 924\) (average number of Peter's & Jordan's stickers)

         a + b = 2(924)

         a + b = 1848 (1)

     2)

         a = b + 232 => a - b = 232 (232)

     3) 

         c = a + 172 (3)

Solve for a using eq. (1) & (2), eliminate b

               a + b = 1848

            + a - b = 232

                    2a = 2080

                    2a/2 = 2080/2

                      a = 1040

Find b.                                    Find c.

        a + b = 1848                        C = a + 172

              b = 1848 - a                       = 1040 + 172

                 = 1848 - 1040             C = 1212

              b = 808

So,

      \( {Jordan’s stickers \over Albert’s stickers} = {b \over c} = {808 \over 1212}\)

                                         = 0.6666. . .

        Change 0.6666... to fraction.

                     x = 0.6666 . . .

                 10x = 6.6666 . . .

                -    x = 0.6666 . . .

                   9x = 6

                     x = 6/9 = 2/3

Therefore,

Jordan's stickers is 2/3 of Albert's stickers.

                    2/3 * 1212 = 808 ✓

Brand A = 2/5 of mobile 

Brand B = 4/5 of remaining

Brand C = Rest

After shipping

Brand A = 20% = 1/5

Brand B = 25% = 1/4

Brand C = 1/3

Left mobiles = 228

Let total phones are x

Brand A = 2/5x

Left = x - 2/5x = 3/5x

B = 4/5 * Left = 4/5 * 3/5x = 12/25x

C = x - (2/5x + 12/25x) = 3x/25

After shipping out

A = 1/5 * 2/5x = 2/25

B = 1/4 * 12/25x = 3/25x

C = 1/3 * 3x/25 = 1/25x

Total shipped = 2x/25 + 3/25x + 1/25x

Total shipped = 6/25x

Left = 228

x - 6/25x = 228

    19/25x = 228

      x = (228¹² * 25)/ 19

      x = 300

a) fraction shipped out = Shipped/Total

= 6/25x/300 = 6/25 = 6/25 

b) A left in warehouse = 2/5x - 2/25x

                                    = 120 - 24

                                    = 96 
    C left in warehouse = 3x/25 - 1/25x

                                    = 2/25 * 300

                                    = 24

A are more than C = 96 - 24

                              = 72

Let say Donna has x candies

Now, she kept half the number

of candies plus 3 candies

i.e. (x/2 + 3)

She gave remaining candies to Jane

i.e. x - (x/2 + 3) = x/2 - 3

Jane ate 1/3 of the candies 

plus 4 candies

i.e. 1/3 (x/2 - 3) + 4 = x/6 + 3

Jane gave remaining candies to

Kaye

i.e. (x/2 x 3) - (x/6 + 3)

= x/3 - 6

Kate ate 1/4 of the candies

ie 1/4 (x/3 - 6) = x/12 - 3/2

remaining candies are

(x/3 - 6) - (x/12 - 3/2)

= x/4 - 9/2 = 42

=> x/4 = 42 + 9/2

=> x = 4 (42 + 9/3)

     x = 186

Now Donna kept half the

number of candies plus 3 candies

=> x/2 + 3

=> 186/2 + 3

=> 96

Donna ate 96 candies

Total number of balls = 4

Number of balls selected = 3

Then total probability = 2/4 * 1/3 = 1/6

where 2/4 represents that out of 4 balls 2 of

Same color one selected and then as the selection

is without replacement so neat balls having 

1/3 probability. of selection on 1 ball is already selected

and not kept back in bag.

1 Can of Paint to Paint 1/18 of their house.

1/18 of house paint by 1 can

1 of house Paint by 18 can

4/9 of "     "     " = 18 * 4/9 = 8 can

5/6 "         "     " = 18 * 5/6 = 15 can

They buy total (8 + 15) can of paint

But it's required 18 can to paint the

whole house.

They should return = (23 - 18) = 5 can

5 can they should return to the store.

Let Dane has D, Enzo had E and

Fraser has F game cars at first.

(a) According to the question,

                  D = 2/5E     and E = 6F

             => D = \( {2\over 5}\) * 6F = \( {12 \over 5}F\)

Now the ratio of Dane's share to Enzo's share

       to Fraser's  share is

        D : E : F = 12/5 K: 6F : F

                      = 12/5 : 6 : 1

                      = 12 : 30 : 5

(b) Let Dane lost 'y' game cards in school.

    From the first         part (a)   Let Dane had 12x,

    Enzo had 30x and Fraser had 5x game card at

    first.

    According to the question,

           12x - y = 1/7 (30x + 5x)

         => 12x - y = 1/7 * 35⁵x

         => 12x - y = 5x

            => y = 7x    ------- (1)

     Also given three boys had 250 game cards altogether

            in the end.

        So,   (12x - y) + 300x + 5x = 250

           => (12x - 7x) + 35x = 250

           => 5x + 35x = 280

           => 40x = 280

            => x = 7

Dane lost y = 7x = 7 * 7 = 49 game cards in school.

Let, the number of toys Ahmad have at first = x

Number of toys Eddie have at first = y

System of equation are given by:

⇒ y = \( {5 \over 8}x………….(i)\)
and,

⇒ \( {0.7x \over y+17} = {2 \over 3}\)
⇒ 2.1x = 2y + 34

⇒ 2.1x = \(2 ({5 \over 8}x)+34\)

⇒ 2.1x = \( {5 \over 4}x + 34\)
⇒ 2.1x - \({5\over 4}x + 34\)

⇒ \( {8.4 - 5x\over 4} = 34\)
⇒ 3.4x = 136

⇒ x =\( {136 \over 3.4}\)
Put value of x in \((i):\)
⇒ y = \( {5 \over 8}(40)\)

⇒ y = 25

So, number of toys Ahmad have at first = 40

Number of toys Eddie have in the end = 25 + 17 = 42

its a tough one

60 ÷ (1+7) = 60 ÷ 8 = 7...... 4

{Rick take four first then Marty takes N, and Rick

takes (8-N), which can ensure that the last one is

taken by Rick.}

Dividing two fractions is the same as multiplying the first fraction by the reciprocal value of the second fraction.

Let x be the number of chocolate cakes bought by Jeff

The total number of cakes including chocolate cakes and vanilla cakes is 288.

∴ Number of vanilla cakes = 288 - x.

Out of total cakes 4/9 are chocolate & the rest were vanilla.

∴ Number of chocolate cakes is

            x = 4/9 x (total number of cakes)

               = 4/9 * 288 = 1152/9 = 128

∴ Number of vanilla cakes = 288 - 128 = 160

∴ Jeff bought 128 chocolate cakes & 160 vanilla cakes

Jeff ate some chocolate cakes, the number of chocolates left was 5/8 of the total number of chocolate at first

Let y be the number of chocolate left.

∴ y = 5/8 x (total number of chocolate at first)

      = 5/8 * 128 = 640/8 = 80

∴ (chocolate cake ate by Jeff) + y = Total number of chocolate cakes

    ∴ (Chocolate cake ate by Jeff) + 80 = 128

      => Choclate cake ate by Jeff = 128 - 80 = 48.

∴ Jeff eat 48 chocolate cakes.

This question is really hard. A math question with mixed numbers with this is really hard so it is easy to convert them to improper fractions.

Box A                Box B

W = x                      G

B = 8/5x                  BL

 
     White  marbles = x

    Black marbles B = 8/5x
 

\({X \over G + BL } = {8 \over 5} —- (1)\)

\({B\over G + BL + 138} = {4\over 3} —- (2)\)
Find B/G+BL = ?

B/G+BL = 8/5x/(G+BL) = 8/5(X/G+BL)

                                     = 8/5 * 8/5 (From (1)).

                                     = 64/25

So Ratio of Black marbles to the number of Blue and Green marbles = 64:25

X/G+BL = 8/5 ----(1)

So G+BL = 5x/8 --> Substitute in (2)

       B/G+BL+138 = 4/3

       B/5x/8 + 138 = 4/3

ie.   8/5x/8 + 138 = 4/3

         24/5x = 20x/8 + 552

       \( {192 - 100 \over 40}x = 552\)

             So x = 552 * 40/92 + 240

when x = 240

         B = 8/5x = 8/5 * 240

                       = 384

         G + BL = 5x/8 = 150

Ans:  Box A has 240 white marbles and 384 blue marbles So 384 - 240 = 144 more black marbles than white marbles.

Yes, this problem is really hard but it is solved.

Let,

      No of stickers Mark had =        'a'

      "    "    "           Nathan had =    'b'
      "    "    "           Owen had =      'c'

      Given,

              12.5% of Mark's stickers = 25% of Nathan's stickers

                   --->     a = 2 * b

                             a = 26    ---> b = a/2

No of stickers Owen had = 1/4th of Number of stickers Mark had

                              C = 1/4 * a

                             
                              a = 4c -->   C = a/4

a)   Ratio of all three is

                       a : b : c

                       a : a/2 : a/4

                       1 : 1/2 : 1/4

                    =    4 : 2 : 1

b)      Nathan gave 58 to Mark,

                then Owen had 1/5 of Mark's stickers

            -->           C = 1/5 * (a + 58)

                      -->    5c = a + 58
    we       (know)        a = 4c

                                  5c = 4c + 58

                                         C = 58

                                  --> a = 4 * 58 = 232

                                        b = a/2 = 232/2 = 116

        Total stickers = a + b + c

                              = 232 + 116 + 58

                              = 406

          
              Total = 406    

\([math]\left(x-\frac{1}{5}x-28\right)\left(1-\frac{7}{12}\right)=25\\ \left(\frac{4}{5}x- 28\right)\frac{5}{12}=25\\ \left(\frac{4}{5}x-28\right)=60\\ \frac{4}{5}x=88\\ x=110[/math]\)
Holt read 110 pages

420 green and red marbles Put into box A and

                 Box B respectively 

1 1/2 times as many red marble as green

                       marble

 -> Let green marble x

           3/2x + x = 420

              5x = 840

                x = 168 = green marble in Box A

                Red marble = 252 in Box B

    Some yellow marble in Box B

      Let's say y no. of yellow marble in B added

    for every 3 green marbles in A, 16 blue were

   added in Box A 

  -> green marble in A = 168

      for every 3, 16 blue added = 168/3 * 16

                                                 = 896 blue added 

Now Box A had four times as B

        896 + 168 = (252 + y) 4

         252 + y = 266

                 y = 14

         Yellow added were 14

    now, what is the ratio of number of yellow 

             to the number of blue marbles 

               =         14/896

                 
                   =     1/64

                     
                      =  1:64

∴  Joe's house is ( \(1{1 \over 4}\)-\( {5\over 6}\)) mile far

    from Ralph's house

  i.e (\(1{1 \over 4}\)-\( {5\over 6}\)) = ( \({4 + 1\over 4}\)-\( {5\over 6}\))

                    = ( \({5\over 4}\)-\( {5\over 6}\))

                    = ( \({5 * 3 - 5 * 2\over 12}\))

                    =  \({15 - 10\over 12}=\)-\( {5\over 12}\)
∴  Joe's house is \(({5 \over 12})\) mile far from

    Ralph's house

Chad has total 14 stickers and half of these

are silver.

So, number of silver stickers which Chad has

= 14 ÷ 2 = 7

Now, Sarah has 29 more stickers.

Total number of stickers Sarah has

=(14+29)=43

Number of silver stickers Sarah has

=(7−5)=2

To get same number of silver stickers as

Chad, Sarah needs 5 more silver stickers. 
Now, let x be the number of stickers Sarah

would bought. Then,

 40x/100 = 5

⇒ 2x = 25

⇒ x = 12.5

⇒ x ≈ 13

Then, total number of stickers

=43+13

=56

the required answer is 56.

https://www.jiskha.com/questions/1870671/a-baker-baked-240-tarts-the-ratio-of-the-number-of-chocolate-tarts-to-the-number-of