Quantitative Aptitude - Arithmetic Ability Questions

Let the milk he bought is 1000 ml

Let C.P of 1000 ml is Rs. 100

Here let he is mixing K ml of water

He is getting 30% profit

=> Now, the selling price is also Rs. 100 for 1000 ml

=> 100 : K%

= 100 : 30

10 : 3 is ratio of milk to water

=> Percentage of milk = 10 x 100/13 = 1000/13 = 76.92%

Given that the month begins on a Friday and has 31 days

Sundays = 3rd, 10th, 17th, 24th, 31st
=> Total Sundays = 5

Every second & fourth Saturday is holiday.
2nd & 4th Saturday in every month = 2

Total days in the month = 31

Total working days = 31 - (5 + 2) = 24 days.

Let the sum be Rs. p, rate be R% p.a. and time be T years.

Then,

 P×T×(R+2)100-P×T×R100=108 ⇒2PT= 10800 ...... (1)

 And, 

 P×R×(T+2)100-P×R×T100=180 ⇒2PR = 18000 ......(2)

Clearly, from (1) and (2), we cannot find the value of p
So, the data is not sufficient.

Two men, three women and one child can be selected in ⁴C₂ x ⁶C₃ x ⁵C₁ ways = 600 ways.

Time ( when X was 30 km ahead of Y) = (120-30)/20 =4.5h

Time ( when Y was 30 km ahead of X) = (120+30)/20 = 7.5 h

Thus, required  difference in time = 3h

In 1hr, the bus covers 60 km without stoppages and 40 km with stoppages.

Stoppage time = time take to travel (60 - 40) km i.e 20 km at 60 km/hr.

stoppage time = 20/60 hrs = 20 min.

Let P's speed = x km/hr. Then, Q's speed = (7 - x) km/ hr.

So, 24/x + 24/(7 - x) = 14

x2 - 98x + 168 = 0

(x - 3)(x - 4) = 0 => x = 3 or 4.

Since, P is faster than Q,

so P's speed = 4 km/hr and Q's speed = 3 km/hr.

In this type of question, we will first calculate the distance covered in given time.

Distance covered will be : Number of revolutions x Circumference 

So we will be having distance and time, from which we can calculate the speed. 

Radius of wheel = 70 / 2 = 35 cm

Distance covered in 40 revolutions will be 40*circumferene=40*2πr=40*2*227*35=8800cm = 88 m

Distance covered in 1 sec  = 8810=8.8m/s=8.8*185km/hr=31.68km/hr