MAT 2013 Question Paper

If M1 men can do W1 work in D1 days working H1 hours per day and M2 men can do W2 work in D2 days working H2 hours per day (where all men work at the same rate), then

M1 D1 H1 / W1 = M2 D2 H2 / W2 .............(1)

So,given 15M=24W=36B...................(2)
where M,W,B are work done by one man, one woman and one boy respectively in one day.

Also given W2=9/4W1 ....(3)

for W2 let x Men(M) are required.
So total Human-power(for W2) = xM + 12W + 6B
now using relations from (2) Human-power(for W2) = xM + 15/2M + 15/6M = (10+x)M

So replacing all the values in eqn (1) and (3), we get 

=> 15M*12days*8Hrs/W1 = (10+x)M*30days*6Hrs/W2

=> 15M*12days*8Hrs/W1 = (10+x)M*30days*6Hrs/(9/4W1)

=> x = 8 Men

In triangle, 

tan 30 = 150-h/60

1/√3 = 150-h/60

20√3 = 150 - h

h = 150- 20√3

h = 150 - 34.64

h = 115.36m (nearly equal to 116m)

Speed of the boat upstream =9-3 =6 km/hr

Speed of the boat downstream=9+3=12 km/hr 

Let the distance between A and B be x kms.

So,time taken to travel x km upstream by boat= x/6 hrs.  

And time taken to travel x kms downstream = x/12 hrs.

According to the question ,

x/6 + x/12 =3

x[1/6 + 1/12]=3

x = 4×3 =12

So, distance between A and B is 12 kms.

let fixed unit be x and cost of each unit be y

540y + x =1800 .............(1)
620y + x = 2040.............(2)

Solving both the equation we will have y=3

Now placing y value in any of the equation give 

x=180 (fixed value)

So now we will have 

Bill = 180 +(500*3)

=1680

Let, the filling capacity of the pump be x m^3/min

Then, emptying capacity of the pump=(x+10) m^3/min.

So,2400/x - 2400/(x+10) = 8

On solving, we get

x=$$50 m^3/min$$

A does 1 round per 30 minutes

B does 1 round per 20 minutes

How many minutes would it take making a round together:

x/30 + x/20 = 1

2x + 3x = 60

5x = 60

x = 12

They meet every 12 minutes.

How many 12 minute periods are there from 8:00 am to 9:30 am 

90 minutes, so

90/12

Answer is 7 times they cross each other.

Area of 4 walls = 2(l+b)h =2(10+7) x 5 

= 170 sq m 

Area of 2 doors and 3 windows = 2(1x3)+(2x1.5)+2(1x1.5) 

= 12 sq m 

Area to be planted = 170 -12 = 158 sq m

Cost of painting = Rs. 158 x 3 = Rs. 474

Bag1 = 3 back, 4 white balls

Bag2 = 4 black, 3 white balls

Probability of selecting bag 1 = 2/6 = 1/3

Probability of selecting bag 2 = 1 - 1/3 = 2/3

Probability of choosing a black ball from bag 1 = 3/7

Probability of choosing a black ball from bag 2 = 4/7

Required probability = 1/3 * 3/7 + 2/3 * 4/7

= $$\frac{11}{21}$$

Ratio of the speeds of A and B = 5/3 : 1 = 5 :3

Thus, in race of 5 m, A gains 2 m over B.

2 m are gained by A in a race of 5 m.

80 m will be gained by A in race of (5/2*80) : 

=200 meters

So,

Winning post is 200 m away from the starting point.

Let the depth of the cistern be h meters 

Then 4.5×3×h = 50 

So, h = 50/13.5

h = 100/27

Area of sheet required = lb + 2 (bh + lh) 

= lb + 2h(l + b)

 = [4.5×3+2×100/27(4.5+3)] sq m 

 = (13.5+200/27×7.5)

 = (27/2+500/9)

 = 1243/18

Therefore weight of lead = (27×1243/18)

=  3729/2 kg 

= 1864.5