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.

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 :

Work done by Shyam and Rahim in 8 days = 8/32 = 1/4

Remaining work to be done by Shyam and Ram = 1 - 1/4 = 3/4

Given efficieny of Ram is half of Rahim i.e, as Rahim can do the work in 48 days, Ram can do the work in 24 days.

One day work of Ram and Shyam = (1/32 - 1/48) + 1/24 = 5/96

Hence, the total work can be done by Shyam and Ram together in 96/5 days.

Therefore, remaining work 3/4 can be done by them in 3/4 x 96/5 = 72/5 = 14.4 days.

   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.

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.

Given number is 987 = 3 x 7 x 47.

So, required number must be divisible by each one of 3, 7, 47.

None of the numbers in 553681 and 555181 are divisible by 3. While 556581 is not divisible by 7.
Correct answer is 555681.

10C5*28C1*10C6 = 7266

Hence the least value of is 4