Quantitative Aptitude - Arithmetic Ability Questions

Let the average bill paid by twenty members = 'x'

But 19 men paid each = Rs. 70

20th man paid Rs. 90.25 more than the avg bill of 20 = x + 90.25

20x = 19(70) + x + 90.25
19x = 1330 + 90.25
19x = 1420.25
x = 1420.25/19 = Rs. 74.75

But the total bill = 20 x 74.75 = Rs. 1495.

Here in 100P2, P says that permutations and is defined as in how many ways 2 objects can be selected from 100 and can be arranged.

That can be done as,

100 P2 = 100!/(100 - 2)!

= 100 x 99 x 98!/98!

= 100 x 99 

= 9900.

As given Kapil's 8th birthday was celebrated 2 years ago,

Kapil’s present age = 8 + 2 = 10 years.

Father's age is twice of Kapil’s age after 10 years.
=> F + 10 = 2 (K + 10) = 2 (10+ 10)
F = 30years,

Avanthi's age =1/6(F) = 30/6 = 5 years.

Time from 12 p.m on monday to 2 p.m on the following monday = 7 days 2 hours = 170 hours

Therefore, The watch gains 2+445 min. or  34/5 min. in 170 hrs.

Now, 34/5 min. are gained in 170 hrs.

Therefore, 2 min are gained in 170×534×2 hrs = 50 hrs

Therefore, Watch is correct 2 days 2 hrs. after 12 p.m on monday i.e it will be correct at  2 p.m on wednesday.

Suppose A, B and C take x, x/2 and x/3 respectively to finish the work.
Then, (1/x + 2/x + 3/x) = 1/2
6/x = 1/2 => x = 12
So, B takes 6 hours to finish the work.

The average age of a couple = 26 years.

Total age of couples = 26 x 2 = 52 years

 Total age of couple after 11 years = (52 + 2 * 11 ) = 74 years

Suppose average age of 3 children after 11 years is 3x years.

The average age of children is 7 years.

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.

Let the Oranges be 'O', Bananas be 'B' and Apples be 'A'

From the given data,

2O + 3B + 4A = 15 ...... (I)

3O + 2B + A = 10 ..... (II)

On adding (I) and (II),

5O + 5B + 5A = 25

O + B + A = 5

4O + 4B + 4A = 20.