Quantitative Aptitude - Arithmetic Ability Questions

Probability that only one of them is selected = (prob. that brother is selected) × (prob. that sister is not selected) +  (Prob. that brother is not selected) × (Prob. that sister is selected)

= 15*23+45*13= 25

Let the three parts be A, B, C. Then,

A : B = 2 : 3 and B : C = 5 : 8 =5*35:8*35=  3:245  

A : B : C = 2 : 3 : =>  = 10 : 15 : 24

 B =245 = 30

We know that perimeter of rectangle = 2(l+b) = 2(8+7)= 30cm
Given perimeter of square is twice the perimeter of rectangle = 2(30) = 60cm
Therefore, side of the square is = 1/4 x 60 = 15cm
Circumference of the required semicircle = πr + 2r = 227× 152 + 2×152 = 23.57 + 15 = 38.57cm.

Given 2000 ---- 54 days

The provisions for 2000 men for 39 days can be completed by 'm' men for only 20 days.

i.e, 2000 ----- 39 days == m ---- 20 days

=> m x 20 = 2000 x 39

m = 3900

So total men for 20 days is 3900

=> 2000 old and 1900 new reinforcement.

Hence, reinforcement = 1900.

The hour hand moves from pointing to 12 to pointing to half way between 2 and 3. The angle covered between each hour marking on the clock is 360/12 = 30. Since the hand has covered 2.5 of these divisions the angle moved through is 75.

Let total population = p
number of females = 5p/9
number of males =(p-5p/9) = 4p/9
married females = 30% of 5p/9 = 30x5p/100x9 = p/6
married males = p/6
unmarried males =(4p/9-p/6) = 5p/18
Percentage of unmarried males in the total males = {(5p/18)/4p/9}x100 = (5p/18)x(9/4p) = 125/2 % = 62.5%

let the inner and outer radii be r and R meters
then, 2πr = 352/7

=> r = (352/7) * (7/22) * (1/2) = 8m
2πR = 528/7

=> R= (528/7) * (7/22) * (1/2)= 12m
width of the ring = R-r = 12-8 = 4m

To find the greatest number which divides the numbers 964, 1238 and 1400 leaving the remainders 41, 31 and 51 is nothing but the HCF of (964 - 41), (1238 - 31), (1400 - 51).

Therefore, HCF of 923, 1207 and 1349 is 71.