apsiganocj

Total pupils = x 

2x/9 scored A grade. 

No. Of pupils who scored B = y 

Now y/ 2x/9 = 5/3

      9y/2x=5/3 

      y/x   =  10/27 - eq (1) 

Now assume 

   No of pupils who scored C = A

   No of pupils who scored D = B

Now 

   A/B  = 6/5 - eq(2) 

Now given

  y = A+48 -eq (3) 

Now we can write 

2x/9 + A+ B + Y = x

A+ B + Y = 7x/9

Put value of eq (3) 

B+ y-48+y= 7x/9

2y+B-48= 7x /9

Now put value of eq (2) 

5A/6 +2y -48 = 7x/9

(Put A = y-48) 

5(y-48) /6 +2y -48 = 7x/9

5y/6  - 40 +2y -48 = 7x/9

17y/6 -88 = 7x/9

Put value of eq(1)

17(10x/27) /6  -88 = 7x/9

170x/162  -7x/9  = 88  

  (170x - 126x)/162 = 88

  44x/ 162 = 88

    x = 162 * 2

   x = 324 

a)  total pupils (x) = 324

b) no. Of pupils of scores D grade = 

First of all  y/x = 10/27

                    y = 10*324/27  = 120

Now A = y- 48 = 120-48 = 72 

Now

 A/ B = 6/5

B = 72 *5/6

B = 45

So no of pupils who failed = 45

Total students enrolled in atleast any of subjects= students enrolled in algebra+students enrolled in mechanics+students enrolled in physics-students enrolled in algebra & physics-students enrolled in algebra & mechanics-students enrolled in mechanic & physics + students enrolled in all three 

= 48+39+35-20-19-22+10

=71

Therefore students not enrolled in any of three = 100-71

= 29

2/4x+45 = 120

1/2x+45=120

x+90=240

x=240-90

x=150

Was that what "word assignment" meant?
Since this task serves a statistical survey of the agriculture minister of our federal state, half cows and oxen are allowed and desired.

Let, Cayden had $x and Jayden had             $y.

According to the question,

                  x + y = 784          ...............(1)

and,         x - \({x \over 6}\frac{}{}\) = y - 124

                5x - 6y = -744        ................(2)

Solving equation (1) and (2)

eq(1) *6 + eq(2)

              6x + 6y = 4704

              5x - 6y = -744

         ...............................................

                   11x = 3960

                       x = 360

Therefore, Cayden had $360 at first.

10 ca we changed into 8 different ways:-

10

5,5

5,2,2,1

2,2,2,2,1,1

2,2,2,1,1,1,1

2,2,1,1,1,1,1,1

2,1,1,1,1,1,1,1,1

8 1s

Since 10,10 counts as 1

so we know that there will be a total of 8+7 different ways to change

so your answer is 15

Each ball has five options to go 

Therefore four balls  will be filled in =

5×5×5×5

= 5^4 ways = 625 ways

Each ball has five options to go 

Therefore four balls  will be filled in =

5×5×5×5

= 5^4 ways = 625 ways

That's me 

(1520 - 950) ÷ 950 = 570 ÷ 950

                               = 0.6

                               = 60%