Aptitude and Reasoning Questions

After 10 days, the remaining food would be sufficient for the 1000 students for 20 more days

-->If 1000 more students are added, it shall be sufficient for only 10 days (as the no. of students is doubled, the days are halved).

Substitute sensible numbers and work out the problem. Then change the numbers back to letters. For example if the machine puts 6 caps on bottles in 2 minutes, it will put 6 /2 caps on per minute, or (6 /2) x 60 caps per hour. Putting letters back this is 60c/m.If you divide the required number of caps (b) by the caps per hour you will get time taken in hours. This gives bm/60c

Let the three consecutive odd numbers be x, x+2, x+4

Then,

x + x + 2 + x + 4 = 93

=> 3x + 6 = 93

=> 3x = 87

=> x = 29 => 29, 31, 33 are three consecutive odd numbers.

Therefore, the middle number is 31.

Upto first 5 minutes I, J and K will fill => 5[(1/20)+(1/30)+(1/40)] = 65/120

For next 6 minutes, J and K will fill => 6[(1/30)+(1/40)] = 42/120

So tank filled upto first 11 minutes = (65/120) + (42/120) = 107/120
So remaining tank = 13/120

Now at the moment filling with C and leakage @ 1/60 per minute= (1/40) - (1/70) = 3/280.
So time taken to fill remaining 13/120 tank =(13/120) /(3/280) = 91/6 minutes

Hence total time taken to completely fill the tank = 5 + 6 + 91/6 = 26.16 minutes.

35+125=17.88 ⇒35+55=17.88

⇒5=2.235

∴80+65=45+65=105=10×2.235=22.35

(20 x 18) men can complete the work in in one day.

one man's one day work = 1/360

(18 x 15) women can complete the work in 1 day

1 woman's one day work = 1/270

So, required ratio = 1270:1360= 4:3

let the side of the square be x meters
length of two sides = 2x meters
diagonal = 2x= 1.414x m

saving on 2x meters = .59x m

saving % = 0.59x2x*100%

= 30% (approx)

We have to find number of permutations of 4 objects out of 6 objects.

This number is 6P4= 360

Therefore, cards can be sent in 360 ways.