Aptitude and Reasoning Questions

Length = 6 m 24 cm = 624 cm
Width = 4 m 32 cm = 432 cm
HCF of 624 and 432 = 48
Number of square tiles required = (624 x 432)/(48 x 48) = 13 x 9 = 117.

Let number of 20 ps coins = x and

number of 25 ps coins = y

Given total coins in the bag = 50

x + y = 50.......(1)

But the total money in the bag = Rs. 10.25

0.20x + 0.25y = 10.25

20x + 25y = 1025.........(2)

Now multiplying (1) by 25 we get

25x+25y=1250.............(3)

By solving (2) and (3)

20x + 25y = 1025;

=> x = 45;

Then, the no. of 20 ps coins are 45.

   Sum = [(B.D.xT.D.)/ (B.D.-T.D.)]

              = [(B.D.xT.D.)/B.G.]

T.D./B.G.  = Sum/ B.D.

   =1650/165

               =10

Thus, if B.G. is Re 1, T.D. = Rs. 10.

If B.D.is Rs. ll, T.D.=Rs. 10.

If B.D. is Rs. 165, T.D. = Rs. [(10/11)xl65]

            =Rs.150

 And, B.G. = Rs. (165 - 150) = Rs, 15.

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.

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.