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.
This can be solved as
80x900/100 + 256 x 4 x ? = 1331 - 99
720 + 1024 x ? = 1232
? = 1232 - 720/1024
? = 512/1024 = 0.5
? = 0.5
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.
Let the numbers be p & q
q-p/4 = 4q/6
=> q-4q/6 = p/4
=> 2q/6 = p/4
=> p/q = 4/3
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 marks obtained by S = M
R's marks = 0.80M
Q's marks = 1.25 (0.80M)
P's marks = (0.90) (1.25) (0.80M)
= 0.9M
But 0.9M = 360
M = ?
M = 400
Hence % of marks obtained by S
= 400/500 x 100 = 80%.
Let number of pages in a book = p
According to questions
p - [3p/7 + (2/5 x 4p/7)] = 36
= p - 23p/35 = 36
=> 12p = 36 × 35
p = 105
Let C's salary be Rs. x
Then B's salary = 80x/100
=> A's salary = (50/100) * (80x/100) = 40x/100
But given total earnings in a month = 22000
=> x + 80x/100 + 40x/100 = 22000
=> x + 120x/100 = 22000
=> x = 10000
=> A's salary = 40 x 10000/100 = Rs. 4000
only (c) is correct since it is divisible by 4.
Let the original number of element be x then the new no.of elements will be
So K must be divisible by 4
Since, x = Kx5/4
The surface area of a cube = 6 = 6
New Surface area = 6 x 1.44
Therefore, %
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