Quantitative Aptitude - Arithmetic Ability Questions

Ratio of investments for 1 year

=> (P : Q : R) = (2x2 + 2.4x10) : (3x2 + 3.3x10) : (5x12)

=> (P : Q : R) = 28 : 39 : 60

Now R share = 190500 x 60/127 = Rs. 90000.

Let the cost price of the item = Rs. 50

Sold at 10% loss => for Rs. 50 loss  S.P = Rs. 45

From the  given data,

25/2 % gain if it is sold at Rs. 9 more

56.25 - 45 = 9

=> 11.25 = 9

=> 22.5 = 18

=> 45 = 36

=> 50 = ?

=> C.P = 50 x 3645 = 40

Hence, the Cost price of the item = Rs. 40.

As we know that a week has 7 days,

k weeks k days = (7k + k) days = 8k days.

To solve a pair of simultaneous equations such as those given we can add or subtract them.

Adding we get 4x + 4y = 20

Therefore 2x + 2y = 10

Let distance = x km and usual rate = y kmph.
Then, x/y - x/(y+3) = 40/60 --> 2y (y+3) = 9x ----- (i)
Also, x/(y-2) - x/y = 40/60 --> y(y-2) = 3x -------- (ii)
On dividing (i) by (ii), we get:

x = 40 km.

Let there are L men

job requires LK man hours.

job completed in first 4 hrs = L x 4 = 4L

job completed in next 4 hrs = 4 x L/2 = 2L

job completed in next 4 hrs = 4 x L/4 = L

job completed in last 4 hrs = 4 x L/8 = L/2

4L + 2L + L + L/2 = KL

K = 7+1/2 = 7.5 hours.

Given ratio of initial mixture of milk and water in Q = 5 : 3

Let the initial quantity of mixture in vessel Q = 8x

Let quantity of Milk = 5x and

Let quantity of water = 3x

According to the question,

5x + 10 x 353x + 10 x 25 = 85

=> 25x + 30 = 24x + 32

=> x = 2

Required Initial quantity of milk = 5x = 5 x 2 = 10 lit.

Let the distance covered be D km. 

 D10+4+ D10-4 = 5

 D14+D6=5

 10D = 42 x 5 = 210

=> D = 21 km