Quantitative Aptitude - Arithmetic Ability Questions

Area of the park = (60 x 40) = 2400 sq.m  

Area of the lawn = 2109  sq.m 

Area of the crossroads = (2400 - 2109) = 291 sq.m   

Let the width of the road be x metres. Then,   

60x + 40x - (x * x) = 291  

=>(x * x) - 100x + 291 = 0    

=>(x - 97)(x - 3) = 0  

=>x=3

28 May, 2006 = (2005 years + Period from 1.1.2006 to 28.5.2006)

Odd days in 1600 years = 0

Odd days in 400 years = 0

5 years = (4 ordinary years + 1 leap year) = (4 x 1 + 1 x 2) 6 odd days

Jan. Feb. March April May

(31 + 28 + 31 + 30 + 28 ) = 148 days

148 days = (21 weeks + 1 day) 1 odd day.

Total number of odd days = (0 + 0 + 6 + 1) = 7 0 odd day.

Given day is Sunday.

Let the three numbers be x, y, z.

From the gien data,

x = 2y ....(1)

x = z/2 => z = 2(2y) = 4y .....(From 1)  ...........(2)

Given average of three numbers = 56

Then,

$$\begin{gathered} x + y + z3 = 56 \\ 2y + y + 4y3 = 56 (From 1 & 2) \\ 7y = 56 x 3 \\ y = 24 \end{gathered}$$

Now,

x = 2y => x = 2 x 24 = 48

z = 4y = 4 x 24 = 96

Now, the highest number is z = 96 & smallest number is y = 24

Hence, required sum of highest number and smallest number

= z + y

= 96 + 24

= 120.

Area =    = 96

Sum of Area =  

The possible outcomes that satisfy the condition of "at least one house gets the wrong package" are:
One house gets the wrong package or two houses get the wrong package or three houses get the wrong package.

We can calculate each of these cases and then add them together, or approach this problem from a different angle.
The only case which is left out of the condition is the case where no wrong packages are delivered.

If we determine the total number of ways the three packages can be delivered and then subtract the one case from it, the remainder will be the three cases above.

There is only one way for no wrong packages delivered to occur. This is the same as everyone gets the right package.

The first person must get the correct package and the second person must get the correct package and the third person must get the correct package.
 1×1×1=1

Determine the total number of ways the three packages can be delivered.
 3×2×1=6

The number of ways at least one house gets the wrong package is:
  6−1=5
Therefore there are 5 ways for at least one house to get the wrong package.

Let A be event of selecting a number divisible by 3. B be the event of selecting a number divisible by 7.

  

 A = { 3, 6, 9, 12, 15, 18, 21, 24, 27, 30 }, so n(A)=10
 B = { 7, 14, 21, 28 }, n(B)= 4 

 

Since A and B are not mutually exclusive So :