Percentage problems

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

Given 30% of 500

$$\begin{gathered} = 30 100 x 500 \\ = 30 x 5 \\ = 150 \end{gathered}$$

Hence, 30% of Rs. 500 = Rs. 150.

           2000 x X 100 + 4000 x Y 100 = 320

and      2000 x X 100 + 10000 x Y 100 = 680

x = 4  and  y = 6

x-y = -2

Let the number be 'x'

Then, according to the given data,

  5 2 x - 2 5 x 5 2 x x 100

=  21 25 x 100

= 84%

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 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

$$\begin{gathered} 20 1 - 4 20 2 = 20 1 - 1 5 2 \\ = 20 4 5 2 \\ = 20 16 25 \\ = 64 5 \\ = 12 . 8 \end{gathered}$$