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.

Since the numbers are co-prime, they contain only 1 as the common factor.
Also, the given two products have the middle number in common.
So, middle number = H.C.F of 551 and 1073 = 29;
First number = 551/29 = 19
Third number = 1073/29 = 37.
Required sum = 19 + 29 + 37 = 85.

From the given data,
(60 - 45)% = 12.5 + 4
15% = 16.5
=> 100% = ?
100% = 16.5 x 100/15 = 110

Hence, P = 45% of 110 = 45x110/100 = 49.5
Q = 50% of 110 = 55
R = 60% of 110 = 60 x 110/100 = 66
M = 12.5 + 49.5 = 62 or 66 - 4 = 62

Hence, no statement is required to answer.

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.

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.

10C5*28C1*10C6 = 7266

1st supporter recieve 440 calls
2nd supporter recieve 360 calls
3rd supporter recieve 300 calls

So total calls = 1100 calls ;
Calls this month= 1500
So remaining calls to be distributed is 400

So Now Ratio 1st:2nd:3rd ==> 440:360:300
=> 22:18:15

Now No. of More Calls 2nd supporter will get => [18/(22+18+15)] x 400
=> (18/55) x 400
=> 131 Calls
So 131 more Calls than last month.