Quantitative Aptitude - Arithmetic Ability Questions

u = k , v= 3k 

∴2uvu+v ⇒2×k×3kk+3k=12  => k = 8km/h

side of first square = 40/4 =10cm

side of second square = 32/4 = 8cm

area of third squre = 10² - 8² = 36 sq cm

side of third square = 36 = 6 cm

required perimeter = 6*4 = 24cm

The possible outcomes are as follows :

5H, 5T, (H, 4T), (T, 4H), (2H, 3T) (3H, 2T), i.e. 6 outcomes in all.

Therefore the probability that head appears an odd number of times = 3/6 =1/2 (In only three outcomes out of the six outcomes, head appears an odd number of times).

First,we count the number of odd days for the left over days in the given period.Here,given period is 09-07-2013 to 07-01-2014

 Therefore, No. of Odd Days =  1 + 3 + 2 + 3 + 2 + 3 + 0 = 14 = 0 odd days

So, given day Sunday + 0 = Saturday is the required result.

A : B : C = (16000 * 3 + 11000 * 9) : (12000 * 3 + 17000 * 9) : (21000 * 6)

= 147 : 189 : 126

= 7 : 9 ; 6.

Difference of B and C's shares = Rs.  ( 26400 * 9/22 - 26400 * 6/22 ) = Rs. 3600.

Let the beds be numbered 1 to 7.

Case 1 : Suppose Anju is allotted bed number 1. 

Then, Parvin cannot be allotted bed number 2. 

So Parvin can be allotted a bed in 5 ways. 

After alloting a bed to Parvin, the remaining 5 students can be allotted beds in 5! ways.

So, in this case the beds can be allotted in 5´5!ways = 600 ways.

Case 2 : Anju is allotted bed number 7. 

Then, Parvin cannot be allotted bed number 6 

As in Case 1, the beds can be allotted in 600 ways.

Case 3 : Anju is allotted one of the beds numbered 2,3,4,5 or 6. 

Parvin cannot be allotted the beds on the right hand side and left hand side of Anju’s bed. For example, if Anju is allotted bed number 2, beds numbered 1 or 3 cannot be allotted to Parvin.

Therefore, Parvin can be allotted a bed in 4 ways in all these cases.

After allotting a bed to Parvin, the other 5 can be allotted a bed in 5! ways.

Therefore, in each of these cases, the beds can be allotted in 4´ 5! = 480 ways. 

The beds can be allotted in (2x 600 + 5 x 480)ways = (1200 + 2400)ways = 3600 ways

Let 1kg of Rs. 100 then 840gm is of Rs. 84.
Now (label on can 1kg but contains 840kg ) so for customer it is of Rs. 100 and further gives 4% discount [he sells his article on 4% loss on cost price.]
So now S.P = Rs. 96
But actually it contains 840 gm so C.P for shopkeeper = Rs. 84
S.P = Rs. 96
C.P = Rs. 84
Profit% = {(S.P-C.P)/C.P}x100
{(96-84)/84} x 100 = 14.28571429% PROFIT.