If one needs to score 34% to pass a test, the pass marks out of 40 would be:

 

1/4 of 12

= 1/4 x 12

= 3

Hence, 1/4 of 12 = 3.

From the given data,

$$\begin{gathered} 40100 x 75 + 80100 x 25 = K100 x 250 \\ 30 + 20 = 5K2 \\ 50 x 2 = 5K \\ K = 1005 = 20 \end{gathered}$$

From the given data,

80 x 15/100 of 80 + 5 x 80 x 25/100

= 960 + 100

= 1060.

Let the total amount he had with = Rs. p

From the given data,

p - 38460 - 24468 = 28p/100

p - 28p/100 = 24468 + 38460

72p/100 = 62928

72p = 62928 x 100

p = 62928 x 100/72

p = 87400

Hence, the total amount he had = p = Rs. 87400.

This can be solved as

$$\begin{gathered} 20 1 - 4202 = 20 1 - 152 \\ = 20452 \\ = 20 1625 \\ = 645 \\ = 12.8 \end{gathered}$$

Let p% of 50 is 2

=> p x 50/100 = 2

=> p = 4%

Hence, 2 is 4 percent of 50.

Given 21% of 500 + 31% of 1100 - 30% of 790 = ?

? = 21 x 5 + 31 x 11 - 3 x 79

? = 105 + 341 - 237

? = 446 - 237

? = 209.

Total engineers = 4800

Number of female engineers = 3/8 x 4800 = 1800

Then, number of male engineers = 4800 - 1800 = 3000

Now, required is female engineers is what percentage of the number of male engineers

=> 1800 is what % of 3000

=> p x 3000/100 = 1800

=> 30p = 1800

p = 60%

Hence, female engineers is 60 percentage of the number of male engineers.