MAT 2005 Question Paper

Let time taken by waste pipe to empty the cistern be $$t$$ minutes

=> $$\frac{1}{12}+\frac{1}{15}-\frac{1}{t}=\frac{1}{20}$$

=> $$\frac{1}{t}=\frac{1}{12}+\frac{1}{15}-\frac{1}{20}$$

=> $$\frac{1}{t}=\frac{5+4-3}{60}$$

=> $$\frac{1}{t}=\frac{6}{60}=\frac{1}{10}$$

$$\therefore$$ The waste pipe takes 10 min to empty the cistern.

=> Ans - (B)

Let radius of third ball be $$r$$ cm

Volume of a sphere = $$\frac{4}{3}\pi R^3$$

=> $$[\frac{4}{3}\times\pi\times(\frac{3}{4})^3]+[\frac{4}{3}\times\pi\times(1)^3]+[\frac{4}{3}\times\pi\times r^3]=\frac{4}{3}\times\pi\times(\frac{3}{2})^3$$

=> $$\frac{27}{64}+1+r^3=\frac{27}{8}$$

=> $$r^3=\frac{27}{8}-\frac{27}{64}-1$$

=> $$r^3=\frac{216-27-64}{64}$$

=> $$r=\sqrt[3]{\frac{125}{64}}=\frac{5}{4}$$

$$\therefore$$ Diameter of third ball = $$2\times\frac{5}{4}=2.5$$ cm

=> Ans - (A)

Principal sum = Rs. 1950 and rate of interest = 10%

=> Amount after 1 year = $$1950+\frac{10}{100}\times1950$$

= $$1950+195=Rs.$$ $$2145$$

But, the man sold it for Rs. 2200, => Profit earned = $$2200-2145=Rs.$$ $$55$$

=> Ans - (A)

Given : Let O be the observer and XY is the pole = 50 m and YZ is the building = 250 m. $$\angle$$ XOY = $$\angle$$ YOZ = $$\theta$$

To find : XO = $$x$$ = ?

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

=> $$tan(\angle XOY)=\frac{YX}{XO}$$

=> $$tan\theta=\frac{50}{x}$$ -------------(i)

In right $$\triangle$$ XOZ,

=> $$tan(\angle XOZ)=\frac{ZX}{XO}$$

=> $$tan(2\theta)=\frac{300}{x}$$

=> $$\frac{2tan\theta}{1-tan^2\theta}=\frac{300}{x}$$

Substituting value from equation (i), we get :

=> $$\frac{2\times50/x}{1-(50/x)^2}=\frac{300}{x}$$

=> $$1-(\frac{50}{x})^2=\frac{1}{3}$$

=> $$(\frac{50}{x})^2=1-\frac{1}{3}=\frac{2}{3}$$

=> $$\frac{50}{x}=\sqrt{\frac{2}{3}}$$

=> $$x=\frac{50\sqrt3}{\sqrt2}$$

=> $$x=25\sqrt6$$ m

=> Ans - (D)

Let A = Event that husband is selected and B = Event that wife is selected.

=> $$P(A)=\frac{1}{7}$$ and $$P(\overline A)=1-\frac{1}{7}=\frac{6}{7}$$

Similarly, $$P(B)=\frac{1}{5}$$ and $$P(\overline B)=1-\frac{1}{5}=\frac{4}{5}$$

$$\therefore$$ Probability that only one is selected = $$P(A)\times P(\overline B)+$$ $$P(B)\times P(\overline A)$$

= $$(\frac{1}{7}\times\frac{4}{5})+(\frac{1}{5}\times\frac{6}{7})$$

= $$\frac{4}{35}+\frac{6}{35}$$

= $$\frac{10}{35}=\frac{2}{7}$$

If a container contains $$x$$ units of a liquid from which $$y$$ units are taken out and replaced by water. After $$n$$ operations, quantity of pure liquid

= $$x(1-\frac{y}{x})^n$$

$$\therefore$$ Milk in the container now = $$40(1-\frac{4}{40})^3$$

= $$40(\frac{9}{10})^3$$

= $$40\times\frac{729}{1000}$$

= $$\frac{729}{25}=29.16$$ liters

=> Ans - (D)

Using, $$M_1D_1H_1=M_2D_2H_2$$

3300 men having 850 gm per head have provisions for = $$32-7=25$$ days

Let total men after reinforcement be $$x$$

=> $$3300\times25\times850=x\times17\times825$$

=> $$x=\frac{3300}{825}\times\frac{850}{17}\times25$$

=> $$x=4\times50\times25=5000$$

$$\therefore$$ Strength of reinforcement = $$5000-3300=1700$$

=> Ans - (B)

Selling price of Rs. 5000 stock = $$5000\times\frac{156}{100}=Rs.$$ $$7800$$

Income from stock = $$\frac{12}{100}\times5000=Rs.$$ $$600$$

Now, let investment in 8% stock be Rs. $$x$$ and 9% stock be Rs. $$(7800-x)$$

According to ques,

=> $$\frac{8}{90}\times x+\frac{9}{108}\times(7800-x)=600+70$$

=> $$\frac{8x}{90}+650-\frac{x}{12}=670$$

=> $$\frac{16x-15x}{180}=20$$

=> $$x=3600$$

$$\therefore$$ Amount invested in 8% stock = Rs. 3600 and in 9% stock = Rs. 4200

=> Ans - (A)

1 anglet = 1% of 1$$^\circ$$

=> 100 anglets = 1$$^\circ$$

$$\therefore$$ 360$$^\circ$$ = $$360\times100=36000$$ anglets

=> Ans - (D)

In the word 'MATHEMATICS', we treat the vowels AEAI as one letter. Thus, we have MTHMTCS (AEAI)

Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different.

=> Number of ways of arranging these letters = $$\frac{8!}{(2!)(2!)}=10080$$

Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.

=> Number of ways of arranging these letters = $$\frac{4!}{2!}=12$$

$$\therefore$$ Required number of words = $$10080\times12=120960$$

=> Ans - (C)