Given 40% minimum passing marks for boys = 483 + 117 = 600
=> 1% = 600/40
=> 100% = 600 x 100/40 = 1500
Minimum passing marks for girls = 35% of 1500 = 35 x 1500/100 = 525.
Let the number be 'x'
Then, according to the given data,
=
= 84%
and
x = 4 and y = 6
x-y = -2
Given 30% of 500
Hence, 30% of Rs. 500 = Rs. 150.
Given 45 kg for painting
And 6% 0f 45 for wasting => 6x45/100 = 2.7 kg
Now total paint required = 45+2.7 = 47.7 kgs
But tins are available in only 4kgs
Then, no. of tins req = 47.7/4 = 11.4 => 12 tins
Let the numbers be p & q
q-p/4 = 4q/6
=> q-4q/6 = p/4
=> 2q/6 = p/4
=> p/q = 4/3
Let 'M' be the maximum marks in the examination.
Therefore, Madhu got 32% of M = 32M/100 = 0.32M
And Kumar got 42% of M = 42M/100 = 0.42M.
In terms of the maximum marks Kumar got 0.42M - 0.32M = 0.1M more than Madhu. ---- (1)
The problem however, states that Kumar got 16 marks more than the cut-off mark and Madhu got 8 marks less than the cut-off mark. Therefore, Kumar has got 16 + 8 = 24 marks more than Madhu. ---- (2)
Now, Equating (1) and (2), we get
0.1M = 24 => M = 24/0.1 = 240
'M' is the maximum mark and is equal to 240 marks.
We know that Madhu got 32% of the maximum marks.
Therefore, Madhu got 32 x 240/100 = 76.8 marks.
We also know that Madhu got 8 marks less than the cut-off marks.
Therefore, the cut-off marks will be 8 marks more than what Madhu got
= 76.8 + 8 = 84.8.
80x900/100 + 256 x 4 x ? = 1331 - 99
720 + 1024 x ? = 1232
? = 1232 - 720/1024
? = 512/1024 = 0.5
? = 0.5
This can be solved as
Let number of boys = B
Number of girls = G
From the given data,
1B/3 + 1G/2 = 300
But given B/3 = 100
=> B = 300
G/2 = 300 - 100
G = 200 x 2
=> G = 400
Therefore, total number of students in the college = B + G = 700.
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