CAT Questions

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As the pipes are operating alternatively, thus their 2 minutes job is = 14 + 16 = 512 

In the next 2 minutes the pipes can fill another 512 part of cistern. 

Therefore, In 4 minutes the two pipes which are operating alternatively will fill 512 + 512 = 56Remaining part = 1 - 56 = 16 

Pipe A can fill 14 of the cistern in 1 minute 

Pipe A can fill 16 of the cistern in = 23 min 

Therefore, Total time taken to fill the Cistern  

4 + 23 minutes.

Work done by P alone in one day = 1/6th of the total work done by Q alone in one day = 1/3(of that done by P in one day) = 1/3(1/6 of the total) = 1/18 of the total. 

Work done by P and Q, working together in one day = 1/6 + 1/18  = 4/18 = 2/9 

They would take 9/2 days = 4 (1/2) days to complete the work working together.

Let ratio of the incomes of Pavan and Amar be 4x and 3x

and Ratio of their expenditures be 3y and 2y

4x - 3y = 1889 ......... I

and

3x - 2y = 1889 ...........II

I and II

y = 1889

and x = 1889

Pavan's income = 7556

An actor performs in a play. A musician performs at a concert.

Choices b, c, and d are incorrect because none is people who perform.

Hence, Play is to actor as concert is to Musician.

LCM of (80, 85, 90) can be found by prime factorizing them.

80 → 2 × 2 × 2 × 2 × 5

85 → 17 × 5

90 → 2 × 3 × 3 × 5

L.C.M of (80,85,90) = 2 × 2 x 2 × 2 × 3 × 3 × 5 × 17

= 16 x 9 x 85

= 144 x 85

= 12240

L.C.M of (80,85,90) = 12240.

1 man's 1 day work = 1/108
12 men's 6 day's work = 1/9 x 6 = 2/3

Remaining work = 1 - 2/3 = 1/3

16 men's 1 day work = 1/108 x 16 = 4/27
4/27 work is done by them in 1 day.

1/3 work is done by them in 27/4 x 1/3 = 9/4 days.

Let the salaries of Maneela and Shanthi one year before be M1, S1 & now be M2, S2 respectively.

Then, from the given data,

M1/S1 = 3/4  .....(1)

M1/M2 = 4/5 .....(2)

S1/S2 = 2/3  .....(3)

and M2 + S2 = 4160  .....(4)

Solving all these eqtns, we get M2 = Rs. 1600.