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.

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.

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.

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

If a card hand that consists of four Queens and an Ace is rearranged, nothing has changed.

The hand still contains four Queens and an Ace. Thus, use the combination formula for problems with cards.

We have 4 eights and 4 sevens.

We want 3 eights and 2 sevens.

C(have 4 eights, want 3 eights) x C(have 4 sevens, want 2 sevens) 

C(4,3) x C(4,2) = 24

Therefore there are 24 different ways in which to deal the desired hand.

10C5*28C1*10C6 = 7266

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 :

Area of the park = (60 x 40) = 2400 sq.m  

Area of the lawn = 2109  sq.m 

Area of the crossroads = (2400 - 2109) = 291 sq.m   

Let the width of the road be x metres. Then,   

60x + 40x - (x * x) = 291  

=>(x * x) - 100x + 291 = 0    

=>(x - 97)(x - 3) = 0  

=>x=3