Quantitative Aptitude - Arithmetic Ability Questions

From March 6th to November 11th,
there are 250 days and
Thus 250 x 24 hours
So from Mar 6th 9AM to Nov 11th 8PM, there will be 250 x 24 + 11 hours = 6011 hrs.

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

Number of runs scored more to increse the ratio by 1 is 26 - 14 = 12
To raise the average by one (from 14 to 15), he scored 12 more than the existing average.
Therefore, to raise the average by five (from 14 to 19), he should score 12 x 5 = 60 more than the existing average. Thus he should score 14 + 60 = 74.

As now, the time 4 : 48 pm after 45 min it will be

4:48 + 45 min = (4 hrs : 48 + 12 min) pm + 33 min = 5 hrs : 33 min pm.

Hence, it is 5:33 pm.

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.

P=2(l+b)

l=4b

Let F and C denotes the temparature in Fahrenheit anid Celcsius respectively.

Then, (F - 32) /(212 - 32) = (C - 0)/(100 - 0),  if c = 35, then F = 95.