Quantitative Aptitude - Arithmetic Ability Questions

Let the length of the train be L metres

Speed of first train = 360018xL m/hour

Speed of secxond train = 360012xL m/hour

When running in opposite directions, relative speed = 200 L + 300 L m/hour

Distance to be covered = L + L = 2L metre

Time taken = 2L500Lx3600 sec

                 =14.4 sec

speed of the man = 4 x (5/18) m/sec = 10/9 m/sec

time taken = 3 x 60 sec = 180 sec

length of diagonal = speed x  time = (10/9) x 180 = 200m

Area of the field = (1/2) x (dioagonal)²

= (1/2) x 200 x 200 sq.m = 20000sq.m

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 age of A = A, B = B and C = C

From given

A = 3B

4 years ago,

C-4 = 2(A-4);

After 4 years

A + 4 = 31;

A = 27;
sub in above B, we get B = 9
C = 2(27 - 4) + 4 = 46 + 4 = 50

Hence B = 9yrs and C = 50yrs.

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.

NOTE: When a century year leaves a remiander 0, when divided by 400 then it is a leap year (366 days).

So, 1200 has 366 days.

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

From the given data,

12 children 16 days work,
One child’s one day work = 1/192.

8 adults 12 days work,
One adult’s one day’s work = 1/96.

Work done in 3 days = ((1/96) x 16 x 3) = 1/2

Remaining work = 1 – 1/2 = 1/2

(6 adults+ 4 children)’s 1 day’s work = 6/96 + 4/192 = 1/12

1/12 work is done by them in 1 day.

1/2 work is done by them in 12 x (1/2) = 6 days.