MAT 2009 Question Paper

Relative speeds of train (running in opposite directions) = $$90+60=150$$ km/hr = $$150\times\frac{5}{18}=\frac{125}{3}$$ m/s

Lengths of train = $$300+200=500$$ m

$$\therefore$$ Time taken = $$\frac{500}{\frac{125}{3}}=4\times3=12$$ sec

=> Ans - (C)

First we arrange the three person on left side and two on right side, it can be done only in 1 way.

Now for the remaining 3 persons, number of ways = $$C^3_2$$

Selecting 4 people for each side can be done in = $$4!\times4!$$ ways

=> Total number of ways = $$C^3_2\times4!\times4!$$

= $$3\times24\times24=1728$$

=> Ans - (D)

Let sides of sheet be $$x$$ and $$5x$$ cm and height = 0.5 cm

Volume of cuboid = Volume of cube

=> $$lbh=a^3$$

=> $$x\times5x\times0.5=(10)^3$$

=> $$5x^2=2000$$

=> $$x=\sqrt{400}=20$$

$$\therefore$$ Sides are 20, 100 cm

=> Ans - (A)

The height of tree (before it broke) = AB + AC = 30 m. Let AB = $$h$$ m

In right $$\triangle$$ ABC,

=> $$sin(30^\circ)=\frac{AB}{AC}$$

=> $$\frac{1}{2}=\frac{h}{30-h}$$

=> $$30-h=2h$$

=> $$h=\frac{30}{3}=10$$

$$\therefore$$ Height where the tree broke = 10 m

=> Ans - (A)

Quantity of water in 250 kg dry grapes = $$\frac{10}{100}\times250=25$$ kg

=> Pulp of grapes = 225 kg

Since, fresh grapes has only 20% pulp, => $$20\%\equiv225$$

=> Total weight = $$100\%\equiv\frac{225}{20}\times100=1125$$ kg

=> Ans - (B)

Let the dealer bought 12, 15 and 20 kg of each type respectively.

Total cost price = $$(12\times100)+(15\times80)+(20\times60)=1200+1200+1200=Rs.$$ $$3600$$

Also, total selling price (since there is no profit no gain) = Rs. 3600

Now, Selling price of first two types after 20% profit = $$\frac{120}{100}\times2400=Rs.$$ $$2880$$

=> Selling price of 3rd type = $$3600-2880=Rs.$$ $$720$$

$$\therefore$$ Loss % = $$\frac{(1200-720)}{1200}\times100=40\%$$

=> Ans - (A)

Speed of first train = $$50\times\frac{5}{18}=\frac{125}{9}$$ m/s and speed of second train = $$x$$ m/s

Relative speed = $$(x+\frac{125}{9})$$ m/s

Length of second train = $$187.5$$ m

Using, speed = distance/time

=> $$x+\frac{125}{9}=\frac{187.5}{9}$$

=> $$x=\frac{62.5}{9}$$

$$\therefore$$ Speed of goods train = $$\frac{62.5}{9}\times\frac{18}{5}=25$$ km/hr

=> Ans - (B)

Pass code : _ _ _ _ 

Now, all these four places can be filled with any of the 10 numbers, since repetition is allowed, and 0 can be the first digit.

Thus, maximum number of attempts = $$10\times10\times10\times10=(10)^4$$

=> Ans - (B)

Let height of ladder be $$AD=h$$ m, => AE = $$(h-3)$$ m and ED = 9 m

In right $$\triangle$$ ADE,

=> $$(AD)^2=(AE)^2+(ED)^2$$

=> $$(h)^2=(h-3)^2+(9)^2$$

=> $$h^2=h^2-6h+9+81$$

=> $$6h=90$$

=> $$h=\frac{90}{6}=15$$

$$\therefore$$ Length of ladder = Height of wall = 15 m

=> Ans - (A)