Asked by Amponsah
A bag contains some balls of which 1/4 are red. Forty more balls of which 5 are red are added. If 1/5 of all the balls are red, how many balls was there originary?
Answers
Answered by
Reiny
original number of balls --- x
original number of red balls = x/4
number of red balls added = 5
total balls added = 45
new total = x+45
new red ball total = (x/4 + 5)
(1/5)(x + 45) = x/4 + 5
x/5 + 9 = x/4 + 5
times 20
4x + 180 = 5x + 100
x = 80
you had 80 balls
check:
you had 80 of which 20 were red
you added 45 of which 5 were red
new total 125
new number of red = 25
is 25 equal to 1/5 of 125 ? YES!!
original number of red balls = x/4
number of red balls added = 5
total balls added = 45
new total = x+45
new red ball total = (x/4 + 5)
(1/5)(x + 45) = x/4 + 5
x/5 + 9 = x/4 + 5
times 20
4x + 180 = 5x + 100
x = 80
you had 80 balls
check:
you had 80 of which 20 were red
you added 45 of which 5 were red
new total 125
new number of red = 25
is 25 equal to 1/5 of 125 ? YES!!
Answered by
Mathias
Let total number of balls be X
Number of red balls = 1/4x
Overall total num of balls=X +40
No of red balls = 1/4X +5
1/5(x +40) = 1/4 + 5
Times 20
4x +160 = 5x + 100
5x - 4x = 160 - 100
X = 60
60 balls were there originally
Prove
Red balls= 1/4 x 60 = 15
When 5 was added= 15+5= 20
When 40 added = 60+40=100
1/5 of all the balls are red=1/5 x100= 20.
Number of red balls = 1/4x
Overall total num of balls=X +40
No of red balls = 1/4X +5
1/5(x +40) = 1/4 + 5
Times 20
4x +160 = 5x + 100
5x - 4x = 160 - 100
X = 60
60 balls were there originally
Prove
Red balls= 1/4 x 60 = 15
When 5 was added= 15+5= 20
When 40 added = 60+40=100
1/5 of all the balls are red=1/5 x100= 20.
Answered by
Iddrisu Abdulai
Let y be the original number of balls.
Number of red balls originally there = (1/4)y or y/4
Since forty more balls are added, the new total number of balls becomes y+40.
Out of the forty balls added, 5 were red.
Hence total number of red balls = y/4 +5.
But 1/5 of all the balls are red.
Therefore, 1/5(y+40) = y/4 +5
Finding the LCM (i.e. 20) and solving the equation, y=60.
Hence 60 balls were originally there.
Number of red balls originally there = (1/4)y or y/4
Since forty more balls are added, the new total number of balls becomes y+40.
Out of the forty balls added, 5 were red.
Hence total number of red balls = y/4 +5.
But 1/5 of all the balls are red.
Therefore, 1/5(y+40) = y/4 +5
Finding the LCM (i.e. 20) and solving the equation, y=60.
Hence 60 balls were originally there.
Answered by
Agnes
Let x be number of balls
Red balls =¼
x of ¼= x/4
40 balls were added
5 balls out of the 40 are red
x/4+5
total number of new balls=40+ x
If ⅕ of (40+x) are red
⅕(40+x)=x/4+5
4(x+40)= 5(x+5)
4x + 160 = 5x +25
160-25= 5x-4x
x= 135
The originals ball =135 balls
Red balls =¼
x of ¼= x/4
40 balls were added
5 balls out of the 40 are red
x/4+5
total number of new balls=40+ x
If ⅕ of (40+x) are red
⅕(40+x)=x/4+5
4(x+40)= 5(x+5)
4x + 160 = 5x +25
160-25= 5x-4x
x= 135
The originals ball =135 balls
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