MAT 2008 Question Paper

Let the height of tower be h metres.

tan 30 = 1/root3 = (h-10)/base

base = (h-10) *root3

tan 60 = h/base

base = h/root3

So, (h-10) *root3 = h/root3

3h - 30 = h

h = 15 metres, therefore, height of tower is 15 metres.

tan angle = perpendicular/base

So, in right angled triangle, 

tan 45 = height of tower/breadth of river

Since tan 45 = 1

So, height of tower = breadth of river 

Hence B

Let the height of tower be h m.

Let x be the distance from the bottom of tower to the point where it makes 60 degrees angle [also breadth of the river]

Applying in right triangle:

tan 60 = h/x

x = h/root3

and

tan 30 = h/x+40

Putting value of x in above equation, we get

h = 20 root3

and x = 20 metres

So, breadth of the river = 20 metres.

Given:

Total students  = 100

Girls = 25 , Boys = 75

Rich = 20 Poor = 80

Fair complexion = 40, Dark = 60

Probability of selecting:

 a girl = 25/100 = 1/4

a rich person = 20/100 = 1/5

 a fair person = 40/100 = 2/5

So, Probability of selecting  a faircomplexioned rich girl is = 1/4*1*5*2/5 = 0.02

He can pick brown brown Or 

brown blue Or 

blue blue Or 

blue brown

Only the first and third choices are what we want

Brown is $\frac{\mathrm{5/}}{9}$ then $\frac{\mathrm{4/}}{8}$ as it is not replaced

Blue is $\frac{\mathrm{4/}}{9}$ then $\frac{\mathrm{3/}}{8}$ same reason

So P(2 brown)$=\frac{\mathrm{5/}}{9}\times \frac{\mathrm{4/}}{8}=\frac{\mathrm{20/}}{72}$

P(2 blue)$=\frac{\mathrm{4/}}{9}\times \frac{\mathrm{3/}}{8}=\frac{\mathrm{12/}}{72}$

P(2 same)$=\frac{\mathrm{20/}}{72}+\frac{\mathrm{12/}}{72}=\frac{\mathrm{32/}}{72}=\frac{\mathrm{4/}}{9}$

Total matches played = 4

Maximum points = 2 * 4 = 8

So to get atleast 7 points , India needs to win atleast 3 matches and one can be draw.

1. India wins all 4 matches: There is only one possible outcome

Hence, the probability of 4 wins = (0.5)*(0.5)*(0.5)*(0.5) = 0.0625

2. India wins 3 matches and draws 1: probability in that case = (0.5)*(0.5)*(0.5)*(0.05) = 0.00625

(since 4 cases, any one match an be draw)

P = 0.0625 + 0.025 = 0.0875

Let there be no women,only man so:

 Probability = 8C2/13C2 

= 14/39

So, Probability of at least 1 women = 1-14/39 

= 25/39

Hence C

Let the total candidates = t and  capable candidates be x and incapable be y.

Passing the tests = 0.8x + 0.25y

Fail the test = 0.2x + 0.75y

Given x = 40/100*t and y = 0.6t

Pass = 0.8*0.4t + 0.25*0.6t = 0.32t + 0.15t = 0.47t

 Proportion of capable college students who passed is = 32/47 = 68%

boys = 900

girls = 1100

Passed = 32% of 900 + 38% of 1100 = 288 + 418 = 706

Failed = 2000 - 706 = 1294

Total % of failed candidates = (2000 - 706) / 2000 = 1294/2000 =  64.7%

Let the salary be s.

Home rent = s/10

Education = 15% of 9s/10 = 135s/1000

Balance = s - (s/10 + 135s/1000) = s - 235s/1000 = 765s/1000

Clorhes = 765s/10000

Left salary = 765s/1000 * 9/10 = 6885/10000 s

Given : 6885s /10000 = 1377

s = 2000