2/9 of the pupils in Primary 5 scored an 'A' for a test. The ratio of the number of pupils who scored a 'B' to the number who scored an 'A' was 5 : 3. The rest of pupils scored a 'C' and a 'D' in the ratio 6 : 5. The pupils who scored a 'B' were 48 more than the pupils who scored a 'C'.
(a) How many pupils were there in Primary 5?
(b) How many pupils failed the test if the failure was a 'D' grade?
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
Hello, apsi! All that in red ink is wrong!
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Now A = 72
A/ B = 6/5 B = 120
B = 72 *5/6 Thi is incorrect!!! C = 72
B = 45 D = 60
So no of pupils who failed = 45 Total = 324
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