Quantitative Aptitude - Arithmetic Ability Questions

Let x, y, z be the number of balls received by the three persons, then

x≥5, y≥5, z≥5 and x+y+z=21

Let u≥0, v≥0, w≥0 then

∴ x + y + z =21

⇒ u + 5 + v + 5 + w + 5 = 21

⇒ u + v + w = 6 

∴Total number of solutions = C3-16+3-1=C28=28

Let us assume S as number of shirts and T as number of trousers
Given that each trouser cost = Rs.70 and that of shirt = Rs.30
Therefore, 70 T + 30 S = 810
=> 7T + 3S = 81......(1)
T = ( 81 - 3S )/7

We need to find the least value of S which will make (81 - 3S) divisible by 7 to get maximum value of T
Simplifying by taking 3 as common factor i.e, 3(27-S) / 7
In the above equation least value of S as 6 so that 27- 3S becomes divisible by 7

Hence T = (81-3xS)/7 = (81-3x6)/7 = 63/7 = 9

Hence for S, put T in eq(1), we get
S = 81-7(9)/3 = 81-63 / 3 = 18/3 = 6.
The ratio of T:S = 9:6 = 3:2.

We know that,

One square meter is equal to 10,00,000 sq mm.
Given, each mesh is 8mm length and 5mm wide
Therefore for each mesh : 8 x 5 = 40 sq mm.
So, number of meshes = 100000040 = 25000.

=> 60x5 : 36x6 : 75x3

=> 100 : 72 : 75

=> 72/247 x 7410 = Rs. 2160

Distance travelled by Ramu = 45 x 4 = 180 km

Somu travelled the same distance in 6 hours.

His speed = 180/6 = 30 km/hr

Hence in the conditional case, Ramu's speed = 45 - 9 = 36 km/hr and Somu's speed = 30 + 10 = 40km/hr.

Therefore travel time of Ramu and Somu would be 5 hours and 4.5 hours respectively.

Hence difference in the time taken = 0.5 hours = 30 minutes.

Let the distance traveled be x km.
Then, x/10 - x/15 = 2
3x - 2x = 60 => x = 60 km.
Time taken to travel 60 km at 10 km/hr = 60/10 = 6 hrs.
So, Robert started 6 hours before 2. p.m. i.e., at 8 a.m.
Required speed = 60/5 = 12 kmph.

The time taken by the leak to empty the tank = 18 - 110 = 5 - 440 = 140

Therefore, the leak empties the tank in 40 hours.

Let 138.009 + 341.981 - 146.305 = 123.6 + z

Then, z = (138.09+341.981) - (146.305+123.6)= 479.99 - 269.905 = 210.085