Quantitative Aptitude - Arithmetic Ability Questions

C alone can finish the work in 40 days.
As given C does half as much work as A and B together
=> (A + B) can do it in 20 days
(A + B)s 1 days wok = 1/20.
A's 1 days work : B's 1 days Work = 1/2 : 1 = 1:2(given)
A's 1 day’s work = (1/20) x (1/3) = (1/60) [Divide 1/20 in the raio 1:2]
B's 1 days work = (1/20) x (2/3) = 1/30
(A+B+C)'s 1 day's work = (1/60) + (1/30) + (1/40) = 9/120 = 3/40

All the three together will finish it in 40/3 = 13 and 1/3 days.

ratio = a22a2

1/8 - 1/x = 1/10

 => x = 40 hrs

Let original Cost price is x
Its Selling price = (105/100) * x = 21x/20
New Cost price  = (95/100) * x  = 19x/20
New Selling price = (110/100 )* (19x/20 )= 209x/200
[(21x/20) - (209x/200)] = 1
=> x = 200

Speed of train relative to man = (60 + 6) km/hr = 66 km/hr.

[66*(5/18)] m/sec = (55/3) m/sec

Time taken to pass the man = [110*(3/55)]m/sec = 6 sec.

Since their HCFs are 7, numbers are divisible by 7 and are of the form 7x and 7y

Difference = 14
=> 7x - 7y = 14
=> x - y = 2

product of numbers = product of their hcf and lcm
=> 7x * 7y = 441 * 7
=> x * y = 63

Now, we have
x * y = 63 , x - y = 2
=> x = 9 , y = 7

The numbers are 7x and 7y
=> 63 and 49

Let the required number of revolutions made by larger wheel be x.

Then, More cogs, Less revolutions (Indirect Proportion)

  14 : 6 :: 21 : x => 14 x x = 6 x 21

x=6*21/14  =>  x=9

In one hour, the minute hand gains 330° over the hour hand. i.e., 60 minute, the minute hand gains 330° over the hour hand.

∴ In 16 minutes, the minute hand gain over the hour hand by