MAT 2005 Question Paper

Given : $$\log2=0.30103$$

Let $$x=2^{56}$$

=> $$\log(x)=\log(2^{56})$$

=> $$\log(x)=56\log(2)$$

=> $$\log(x)=56\times0.30103=16.85$$

If characteristic is 16, then number of digits in $$2^{56}$$ is 17

=> Ans - (C)

Total number of ways of selecting 11 players out of 15 = $$C^{15}_{11}$$

= $$\frac{15\times14\times13\times12}{1\times2\times3\times4}=1365$$

There are 5 bowlers and 10 other players.

Now, favorable ways = (3 bowlers and 8 others) or (4 bowlers and 7 others) or (5 bowlers and 6 others)

=> Total favorable ways = $$(C^5_3\times C^{10}_8)+(C^5_4\times C^{10}_7)+(C^5_5\times C^{10}_6)$$

= $$(10\times45)+(5\times120)+(1\times210)$$

= $$450+600+210=1260$$

$$\therefore$$ Required probability = $$\frac{1260}{1365}=\frac{12}{13}$$

=> Ans - (D)

Given : AB = 60 m and $$\angle$$ ACB = $$15^\circ$$

To find : BC = ?

Solution : Using, $$tan(A-B)=\frac{tan(A)-tan(B)}{1+tan(A)tan(B)}$$

=> $$tan(15^\circ)=tan(60^\circ-45^\circ)$$

= $$\frac{tan(60^\circ)-tan(45^\circ)}{1+tan(60^\circ)tan(45^\circ)}$$

= $$\frac{\sqrt3-1}{1+\sqrt3}$$ ---------------(i)

In right $$\triangle$$ ABC,

=> $$tan(15^\circ)=\frac{AB}{BC}$$

=> $$\frac{\sqrt3-1}{\sqrt3+1}=\frac{60}{BC}$$

=> $$BC=60(\frac{\sqrt3+1}{\sqrt3-1})$$ m

Multiples of 3 are : 15,18,21,.....,102,105

The above series is an A.P. with first term = $$a=15$$ and common difference = $$d=3$$ and last term = $$l=105$$

Let number of terms be $$n$$

=> Number of terms = $$l=a+(n-1)d$$

=> $$15+(n-1)3=105$$

=> $$(n-1)3=105-15=90$$

=> $$n-1=\frac{90}{3}=30$$

=> $$n=31$$

=> Ans - (B)

L.C.M. (48,72,108) = 432

Thus, they will all change simultaneously after 432 seconds = 7 min 12 seconds

=> Time they will change  again = 8:20:00 + 00:07:12 = 8:27:12 hours

=> Ans - (C)

It is given that Temple AD = 54 m and

Solution : In right $$\triangle$$ ADE,

=> $$tan(60^\circ)=\frac{AD}{DE}$$

=> $$\sqrt3=\frac{54}{DE}$$

=> $$DE=\frac{54}{\sqrt3}$$

=> $$DE=18\sqrt3\approx31.18$$ m width

Similarly, in right $$\triangle$$ ABC,

=> $$tan(30^\circ)=\frac{AB}{BC}$$

=> $$\frac{1}{\sqrt3}=\frac{AB}{18\sqrt3}$$

=> $$AB=18$$

$$\therefore$$ Height of temple = CE = $$54-18=36$$ m height

Difference of heights=54-36=18

Internal and External radius of sphere are 2 cm and 4 cm respectively, and radius of cone = 4 cm.

Let height of cone = $$h$$ cm

Volume of hollow sphere = Volume of cone

=> $$\frac{4}{3}\pi\times[(r_1)^3-(r_2)^3]=\frac{1}{3}\pi R^2h$$

=> $$4\times(4^3-2^3)=(4)^2\times h$$

=> $$h=\frac{4\times56}{16}$$

=> $$h=14$$ cm

=> Ans - (B)

Sum of lengths of both trains = $$108+112=220$$ m

Speed of first train = 50 km/hr and let speed of second train = $$x$$ km/hr

=> Relative speeds of both trains (in m/s) = $$(50+x)\times\frac{5}{18}$$

Using, Speed = $$\frac{5}{18}(50+x)=\frac{220}{6}$$

=> $$50+x=\frac{220}{6}\times\frac{18}{5}$$

=> $$50+x=132$$

=> $$x=132-50=82$$

$$\therefore$$ Speed of second train = 82 km/hr

=> Ans - (D)

George takes 8 hours to copy a 50 page manuscript, => Efficiency of George = $$\frac{50}{8}=\frac{25}{4}$$ pages/hr

Similarly, Sonya's efficiency = $$\frac{50}{6}=\frac{25}{3}$$ pages/hr

=> Work done by them working together in 1 hr = $$\frac{25}{3}+\frac{25}{4}=\frac{175}{12}$$

$$\therefore$$ Time taken when they work together to finish a 100 page manuscript = $$\frac{100}{\frac{175}{12}}$$

= $$100\times\frac{12}{175}=\frac{48}{7}=6\frac{6}{7}$$ hours

=> Ans - (A)

Let Rajan's age when he was married be $$5x$$ years, => Rajan's present age = $$6x$$ years

According to ques, => $$5x+8=6x$$

=> $$x=8$$

=> Rajan's sister's age at the time of his marriage = $$5x-10$$

= $$5(8)-10=30$$ years

$$\therefore$$ Rajan's sister's present age (8 years after marriage) = $$30+8=38$$ years

=> Ans - (C)