Quantitative Aptitude - Arithmetic Ability Questions

Given sum of present ages of A, B, C & D

A + B + C + D = 176

and also given that,

(A-4) : (B-4) : (C-4) : (D-4) = 11 : 9 : 6 : 14

Let the ratio = x

=> A-4 + B-4 + C-4 + D-4 = 11x + 9x + 6x + 14x

=> (A + B + C + D ) - 16 = 40x

=> 176 - 16 = 40x

=> x = 4

=> A = 44 + 4 = 48

=> B = 40

=> C = 28

=> D = 60

Therefore, from the above conclusions

D is eldest with 60 years and C is youngest with 28 years of ages.

6.72/100 = 672/10000 = 42/625

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 efficiency of every man and every woman be 'm' unit/day and 'w' unit/day respectively

15×12×m = 10×16×w

⇒ m/w = 8/9

Total work = 15 × 12 × 9 = 1620 units

In 2 days, total work done = 15 x 9 + 16 x 8 = 263 units

So, in 10 days work done will be = 263 × 5 = 1315 units

Remaining work will be done in = (1620-1315)/(15×8) = 5/2 days

Total days = 10 5/2 days.

Concentration of glucose are in the ratio = 12:35:45

Quantity of glucose taken from A = 1 liter out of 2

Quantity of glucose taken from B = 3/5 x 3 = 1.5 lit

Quantity of glucose taken from C = 0.8 lit

So, total quantity of glucose taken from A,B and C = 3.6 lit

So, quantity of alcohol = (2 + 3 + 1) - 3.6 = 2.4 lit

Ratio of glucose to alcohol = 3.6/2.4 = 3:2

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.