Exams

Given total applicants = 135
Given graduates are G=60
Otherthan graduates=135-60 = 75
Given experienced candidates = 80
1) For maximum number of graduates have experience
    Total graduates to have experience = 60

2) For minimum number of graduates have experience
    Remaining after taking other than graduates in experience= 80-75 = 5

Let l, b the sides of the base, so that the area of the base = lb
remains constant.
Let ‘h’ be the height of the solid.
Volume of the solid, V = l× b× h.
Since the volume is to be increased by 50%

It won't requires that much solving.

Here given the increase in weekly collection by Rs. 1,05,000

That is this increase denotes for all 7 days.

=> Gross Collection increase per day = Rs. 1,05,0007 = Rs. 15,000

Hence, the gross collection increase per day = Rs. 15,000.

The molar mass of H2O can be calculated as the following.

M(H2O)= [2 x M(H)] + [M(O)].

And we know that M(H) = 1g/mol, M(O) = 16g/mol.

So, M(H2O)= 2x1 + 16 = 18g/mol,

That means 1 mole of H2O contains 18g.

Hence, one mole of H2O corresponds to 18 grams.

Here total parts of milk and water in the solution is 6+2 = 8 parts.

1part = 640/8 = 80

old mixture contains 6parts of milk and 2 parts of water(6:2).

To get new mixture containing half milk and half water, add 4parts of water to the old mixture then 6:(2+4) to make the ratio same.

i.e, 4 x 80 = 320 ml.

We know that,

$$\begin{gathered} i2 = -1 \\ i3 = i2 x i = -1 x i = -i \\ i4 = i2 x i2 = -1 x -1 = 1 \\ Hence, i233 =i4 x 58 + 1 = i232 x i1 = 1 x i = i \end{gathered}$$

Speed of lorry = 36012 = 30 kmph

Speed of van =250100 x 30 = 75 kmph

Speed of bike = 35 x 75 = 45 kmph

Therefore, now required average speed of bike and van  = 75 + 452= 60 kmph.