Bank PO Questions

2 engines of former type for one hour consumes (2 x 24)/(6 x8) = 1 metric ton 

i.e. 3 engines of latter type consumes 1 ton for one hour

Hence 9 engines consumes 3 tons for one hour

For 15 hours it is 15 x 3 = 45 metric tonnes.

The given series follows the rule,
+(11x1), +(11x3), +(11x5), +(11x7), +(11x9),....

So starting from 7, 7 + 11 = 18, 18 + 33 = 51, 51 + 55 = 106, ...
So wrong number is 40 which is to be 51.

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.

Second is the tool used by the first.

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.

Let the Number be x. 

Given that one third of a Number is 96. 

i.e x = 3 x 96 = 288.

 67% of  x = x×67100=288×67100=192.96

Interest received by L from K = 8% of half of Rs.40,000  

⇒8100×20000=1600  

Amount received by L per annum for being a working partner = 1200 x 12 = Rs.14,400

Let 'A' be the part of remaing profit that 'L' receives as his share.

 Total income of 'K' = only his share from the reamaing profit

                              = 'A', as both share equally.

Given income of L = Twice the income of K

--> (1600 + 14400 + A ) = 2A

--> A= Rs.16000

Thus total profit = 2A + Rs.14,400= 2(16000) + 14400 

                                                  = 32000 +14400 = Rs.46,400.

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.