Quantitative Aptitude - Arithmetic Ability Questions

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.

Total money decided to contribute = 750 x 12 = 9000

Let 'b' boys dropped

The rest paid 150/- more

=> (12 - b) x 900 = 9000

=> b = 2

Hence, the number of boys who dropped out is 2.

From the given data,

let the initial quantity of the mixture = 5x

Then,

$$\begin{gathered} 2x - 163x - 24 + 40 = 14 \\ 8x - 64 = 3x + 16 \\ 5x = 80 \\ x = 16 lit \end{gathered}$$

Then the initial quantity of the mixture = 5x = 5 x 16 = 80 lit.

Volume of water flowing through 1 pipe of diameter x = Volume discharged by 6 pipes of diameter 12 cms.

As speed is same, area of cross sections should be same. 

Area of bigger pipe of diameter x = Total area of 6 smaller pipes of diameter 12 

i.e πR2 = 6πR12   

Here R = R and R1 = 12/2 = 6   

⇒R2 = 6×6×6   

R = 14.696   

=> D = X = 14.696 * 2 = 29.3938 cm.

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 :

Simple Interest on Rs. (260-20) for a gven time= Rs.20 

Simple Interest on Rs. (260-20) for a half time  = Rs.10 

True Discount on Rs.250 = Rs.10  

True Discount on Rs.260 = Rs.[(10/250) x 260] = Rs. 10.40

Given Word is PROMISE.

Number of letters in the word PROMISE = 7

Number of ways 7 letters can be arranged = 7! ways

Number of Vowels in word PROMISE = 3 (O, I, E)

Number of ways the vowels can be arranged that 3 Vowels come together = 5! x 3! ways

Now, the number of ways of arrangements so that three vowels should not come together

= 7! - (5! x 3!) ways = 5040 - 720 = 4320.