SOLUTION: there were 48 entries from Grade 1 and Grade 2 in a storytelling competition. The rest were from Grade 3. If 30 entries were not from Grade 1 and 28 entries were not from Grade 2

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Question 1121978: there were 48 entries from Grade 1 and Grade 2 in a storytelling competition. The rest were from Grade 3. If 30 entries were not from Grade 1 and 28 entries were not from Grade 2, how many entries were there altogether?
Found 3 solutions by Boreal, MathTherapy, ikleyn:
Answer by Boreal(15235)   (Show Source): You can put this solution on YOUR website!
x=from Grade 1
y-from Grade 2
z=from Grade 3
x+y=48
y+z=30, z=30-y
x+z=28
therefore, x+30-y=28
and x+y=48
2x+30=76
2x=46
x=23
y=48
z=5
There were 76 entries altogether.
Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!

there were 48 entries from Grade 1 and Grade 2 in a storytelling competition. The rest were from Grade 3. If 30 entries were not from Grade 1 and 28 entries were not from Grade 2, how many entries were there altogether?
Correct answer: 

Accept no other!! 

Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
 .


x = from Grade 1
y - from Grade 2
z = from Grade 3

x+y = 48    (1)     ("there were 48 entries from Grade 1 and Grade 2")

y+z = 30,   (2)     {"30 entries were not from Grade 1")

x+z = 28.   (3)     ("28 entries were not from Grade 2")


Add equations (1) and (2). You will get

x + 2y + z = 48 + 30 = 78.    (4)


In equation (4), replace (x+z) by 28, based on eq(3). You will get

28 + 2y = 78  ====>  2y = 78-28 = 50  ====>  y = 50/2 = 25.


Now from equation (1),  x = 48-25 = 23.

    From equation (2),  z = 30-25 =  5.


Answer.  x= 23;  y= 25;  z= 5.  x + y + z = 23 + 25 + 5 = 53.


         53 entries altogether.





 

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