Quantitative Aptitude - Arithmetic Ability Questions

Let age of retirement = S
so according to the given condition 25000(x-20)
=10000 x 3 + 12000 + 14000 + 16000 + 18000 + 20000 + 22000 + 24000 + 26000 + 28000 + 30000 + 30000(S-33)
= 30000 + 210000 + 30000S-990000
= 30000S - 750000
or 25000S - 500000 = 30000x - 750000
or 5000x = 250000
or S = 50 yrs.

Total number of chairs = (3 + 5 + 4) = 12.

Let S be the sample space.

Then, n(s)= Number of ways of picking 2 chairs out of 12

= 12×11/2×1 = 66

Let n(E) = number of events of selecting 2 chairs for selecting no white chairs.

=> 8C2 = 8×7/2×1 = 28

Therefore required probability = 28/66 = 14/33.

To find the minimum distance, we have to get the LCM of 75, 80, 85

Now, LCM of 75, 80, 85 = 5 x 15 x 16 x 17 = 20400

Hence, the minimum distance each should walk so that thay can cover the distance in complete steps = 20400 cms = 20400/100 = 204 mts.

Number of words with 5 letters from given 9 alphabets formed = 95

Number of words with 5 letters from given 9 alphabets formed such that no letter is repeated is = 9P5

Number of words can be formed which have at least one letter repeated =  95 - 9P5

= 59049 - 15120

= 43929

1 boy's 1 day work = 1/70

1 girl's 1 day work = 1/140

(5 boys + 10 girls)'s 1 day work = (5/70 + 10/140) = (1/14 + 1/14) = 1/7

5 boys and 10 girls will complete the work in 7 days.

There are 7C2 ways of selecting two men, and 10C1 ways of selecting a woman.Since each position in the committee is different,arrange the three people in 3! ways.

So, 7C2*10C1*3! =1260

There are 8 students and the maximum capacity of the cars together is 9.

We may divide the 8 students as follows

Case I: 5 students in the first car and 3 in the second Or

Case II: 4 students in the first car and 4 in the second

Hence,     in Case I: 8 students are divided into groups of 5 and 3 in 8C3 ways.

Similarly, in Case II: 8 students are divided into two groups of 4 and 4 in 8C4 ways.

Therefore, the total number of ways in which 8 students can travel is

8C3+8C4 = 56 + 70 = 126.

The distance is constant in this case.

Let the time taken for travel with a speed of 10 kmph be 't'.

Now the speed of 15 kmph is 3/2 times the speed of 10 kmph.

Therefore, time taken with the speed of 15 kmph will be 2t/3 (speed is inversely proportional to time)

Extra time taken = t - 2t/3 = t/3

=> 1pm - 11am = 2hrs

=> t/3 = 2h

=> t = 6 hrs.

Now, Distance = speed x time = 10 x 6 = 60 kms

Time he takes to reach at noon = 6 - 1 = 5 hrs

Now, Speed = 60/5 = 12 kmph.