Quantitative Aptitude - Arithmetic Ability Questions

Time taken by one tap to fill half of the tank = 3 hrs.

Part filled by the taps in 1 hour = 4 x 1/6 = 2/3

Remaining part = 1 - 1/2 = 1/2

2/3 : 1/2 :: 1 : p

p = 1/2 x 1 x 3/2 = 3/4 hrs. i.e., 45 min

So, total time taken = 3 hrs 45 min.

let original radius = r and  new radius = (50/100) r = r/2

original area = πr2 and new area = πr/22

decrease in area =  3πr2/4*1πr2 *100 = 75%

S = { 1, 2, 3, 4, .....18 } 

=> n(S) = 18

E1 = {2, 4, 6, 8, 10, 12, 14, 16, 18}

=> n(E1) = 9

E2 = {3, 6, 9, 12, 15, 18 }

=> n(E2) = 6

 E3 =E1∩E2={6, 12, 18} 

=> n(E3) = 3

∴E=E1 ∪ E2 = E1+E2-E3 

=> n(E) = 9 + 6 - 3 =12

 where E = { 2, 3, 4, 6, 8, 9, 10, 12, 12, 14, 15, 16, 18 }

∴P(E)=n(E)n(S)=1218=23

In the question it is asked for 5% of 1500

5 x 1500/100 = 15 x 5 = 75.

Hence, 5% of 1,500 is equal to 75.

Let the number of Rose plants be ‘a’.
Let number of marigold plants be ‘b’.
Let the number of Sunflower plants be ‘c’.
20a+5b+1c=1000; a+b+c=100

Solving the above two equations by eliminating c,
19a+4b=900

b = (900-19a)/4 

b = 225 - 19a/4----------(1)

b being the number of plants, is a positive integer, and is less than 99, as each of the other two types have at least one plant in the combination i.e .:0 < b < 99--------(2)

Substituting (1) in (2),

 0 < 225 - 19a/4 < 99

225 <  -19a/4 < (99 -225)

=> 4 x 225 > 19a > 126 x 4

=> 900/19 > a > 505

a is the integer between 47 and 27 ----------(3)
From (1), it is clear, a should be multiple of 4.

Hence possible values of a are (28,32,36,40,44)

For a=28 and 32, a+b>100
For all other values of a, we get the desired solution:
a=36,b=54,c=10
a=40,b=35,c=25
a=44,b=16,c=40

Three solutions are possible.

There are 13 spades ( including one king). Besides there are 3 more kings in remaining 3 suits

Thus   n(E) = 13 + 3 = 16

Hence nE¯=52-16=36 

Therefore, PE=3652=913

Given,
The ratio of the doctors to the nurses is 11 : 8
Number of nurses = 8/19 x 209 = 88.

25 Jan, 1981 = (1980 years + Period from 1.1.1981 to 25.1.1981)

Odd days in 1600 years = 0

Odd days in 300 years = 1

80 years = (60 ordinary years + 20 leap year) = (60 x 1 + 20 x 2)= 2 odd days

Jan = 25 days = (3 weeks + 4 days) = 4 odd days.

Total number of odd days = (0 + 1 + 2 + 4) = 7= 0 odd day.

Given day is Sunday.