Aptitude and Reasoning Questions

In one day work done by K = 1/20

Work done by K in 2 days = 1/10

Work done by K,L,M on 3rd day =1/20 + 1/30 + 1/60 = 1/10

Therefore, workdone in 3 days is = 1/10 + 1/10 = 1/5

1/5th of the work will be completed at the end of 3 days.

The remaining work = 1 - 1/5 = 4/5

1/5 of work is completed in 3 days

Total work wiil be done in 3 x 5 = 15 days.

Let the number be x. Then, x - 3x/5 = 50  <=>  x = (50 * 5) / 2 = 125..

Ans.
Given that length of first train + length of second train = 660
Speed of first train : Speed of second train = 5:8 =5x : 8x
Time taken to cross the poll by two trains = 4:3 =4y:3y
Now ,
(5x*4y) + (8x*3y) = 660
44xy = 660
xy = 15
L1 - L2 = 24xy -20xy = 4xy = 60 m

Here to say which fraction is greater from the given options, the fraction with greater numerator with same denominator or smaller denominator with same numerator.

To find the greater fraction, among the given options, make the denominators same for the options by finding the LCM of all the denominators.

7/12, 9/16, 13/20, 32/80

Now, LCM of 12, 16, 20, 80 = 240.

Then,

140/240, 135/240, 156/240, 96/240

Here 

$$\begin{gathered} 156240 > 140240 > 135240 > 96240 \\ => 1320 > 712 > 916 > 3280 \end{gathered}$$

Hence, largest fraction = 13/20.

Here asked for the value of 28 tens

=> 28 x 10 = 280.

Let the speed of the trains be x and y respectively
length of train1 = 47x
length of train2 = 31y
Relative speed= x + y
Time taken to cross each other = 33 s
=> 47x+31yx+y= 33
=> (47x + 31 y) = 33(x + y)
=> 14x = 2y
=> x/y = 2/14 = 1/7

 = 1:7

The simplest triangles are ABG, BCG, CGE, CDE, AGE and AEF i.e. 6 in number.

The triangles composed of two components each are ABE, ABC, BCE and ACE i.e. 4 in number.

There are 6 + 4 = 10 triangles in the figure.

Time required will be in ratio = 5:4

B alone will do it in =16 days