Quantitative Aptitude - Arithmetic Ability Questions

(P+Q)'s 1 day work = 1/24

P's 1 day work = 1/32

=> Q's 1 day work = 1/24 - 1/32 = 1/96

Work done by (P+Q) in 8 days = 8/24 = 1/3

Remainining work = 1 - 1/3 = 2/3

Time taken by Q to complete the remaining work = 2/3 x 96 = 64 days.

Relative speed = 60 + 90 = 150 km/hr.

= 150 x 5/18 = 125/3 m/sec.

Distance covered = 1.10 + 0.9 = 2 km = 2000 m.

Required time = 2000 x 3/125 = 48 sec.

Let a = 13, b = 14 and c = 15. Then, s=12a+b+c=21

 (s- a) = 8, (s - b) = 7 and (s - c) = 6.

Area =ss-as-bs-c =21×8×7×6 = 84 sq.cm   

A + B= 70%   

B + C =50%   

 ∵ (A+B)+(B+C)-(A+B+C)= B

=> B= 20%     A= 50%        and   C=30%

Hence A is most efficient

Find the number of cases in which none of the digits show a '6'.

i.e. all three dice show a number other than '6', 5×5×5=125 cases.

Total possible outcomes when three dice are thrown = 216.

The number of outcomes in which at least one die shows a '6' = Total possible outcomes when three dice are thrown - Number of outcomes in which none of them show '6'.

=216−125=91

The required probability = 91/256

Initial investment of Ram = Rs.2500.
After 2 months he withdraw Rs.1250 from his capital.
Therefore, we have, Ram invested Rs.2500 for 2 months and Rs.(2500-1250=) 1250 for 4 months.

Raj invested Rs. 2250 for 3 months and Rs.(2250-750=) 1500 for 3 months.

And, Rakesh invested Rs.3500 for 3 months;

Their investing ratio:
Ram:Raj:Rakesh = (2500x2 + 1250x4):(2250x3 + 1500x3):(3500x3)
= (10,000):(11,250):(10,500) = 1000:1125:1050 = 40:45:42

Total profit for 6 months = Rs.2540
Therefore, Ram's share = Rs.(2540 x 40/(40+45+42)) = Rs.(2540 x 40/127) = Rs.800
Raj's share = Rs.(2540 x 45/127) = Rs.900
Rakesh's share = Rs.(2540 x 42/127) = Rs.840

Let the total population be 'p'

Given ratio of male and female in a city is 7 : 8

In that percentage of children among male and female is 25% and 20%

=> Adults male and female % = 75% & 80%

But given adult females is = 156800

=> 80%(8p/15) = 156800

=> 80 x 8p/15 x 100 = 156800

=> p = 156800 x 15 x 100/80 x 8

=> p = 367500

Therefore, the total population of the city = p = 367500

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
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 in8C3 ways.

Similarly, in Case II: 8 students are divided into two groups of 4 and 4 in 8C4ways.

Therefore, the total number of ways in which 8 students can travel is:
8C3+8C4=56 + 70= 126