MAT 2003 Question Paper

Let $$x=10$$

(A) : $$\frac{10}{10}=1$$

(B) : $$\frac{10+1}{10}=1.1$$     [MAX]

(C) : $$\frac{10}{11}<1$$

(D) : $$\frac{12}{13}<1$$

=> Ans - (B)

Expression : $$2x^2 + x - 3 < 0$$

=> $$2x^2-2x+3x-3<0$$

=> $$(2x+3)(x-1)<0$$

Now product of two numbers is negative only if one is positive and other is negative.

Case I : $$(2x+3)>0$$ => $$x>\frac{-3}{2}$$

and $$(x-1)<0$$ => $$x<1$$

$$\therefore$$ $$\frac{-3}{2}<x<1$$

Case II : $$(2x+3)<0$$ => $$x<\frac{-3}{2}$$

and $$(x-1)>0$$ => $$x>1$$, which is not possible.

=> Ans - (A)

Both the events are independent so if two events are independent say A and B then probability of both happening = P(A) $$\times$$ P(B)

P(A will live 15 more years) = $$\frac{7}{8}$$

P(B will live 15 more years) = $$\frac{9}{10}$$

P(A will live 15 more years and B will live 15 more years) = $$\frac{7}{8}\times\frac{9}{10}$$

= $$\frac{63}{80}$$

=> Ans - (B)

Selling price = Rs. 17,940

Marked price after discount of 8% = $$\frac{17940}{92}\times100=Rs.$$ $$19,500$$

Also, cost price = $$\frac{17940}{119.6}\times100=Rs.$$ $$15,000$$

If no discount is allowed, => selling price = Rs. 19,500

$$\therefore$$ Profit % = $$\frac{19500-15000}{15000}\times100$$

= $$\frac{4500}{150}=30\%$$

=> Ans - (D)

The number of tangents that can be drawn to two non-intersecting circles is 4.

=> Ans - (A)

Let number be 100

When it is increased by 10%, => new number = 110

When it is decreased by 10%, => new number = $$\frac{90}{100}\times110=99$$

Thus, original number decrease by 1%.

[Method 2]

% Change = $$10+(-10)+\frac{10\times(-10)}{100}$$

= $$-1\%$$

=> Ans - (B)

Average of 11 numbers = 10.9

=> Sum of 11 numbers = $$10.9\times11=119.9$$

Similarly, sum of first six numbers = $$10.5\times6=63$$

and sum of last six numbers = $$11.4\times6=68.4$$

Thus, middle number = $$(63+68.4)-119.9=11.5$$

=> Ans - (A)

Let cost price of the article = Rs. $$100x$$

Selling price after 5 % profit = Rs. $$105x$$

Now, new cost price = Rs. $$95x$$

and new selling price = Rs. $$(105x-1)$$

=> Profit % = $$\frac{105x-1-95x}{95x}\times100=10$$

=> $$\frac{10x-1}{95x}=\frac{1}{10}$$

=> $$100x-10=95x$$

=> $$x=\frac{10}{5}=2$$

$$\therefore$$ Cost price = Rs. 200

=> Ans - (A)

Let B's speed be $$x$$ km/hr and A's speed =  $$\frac{60}{18}=\frac{10}{3}$$ km/hr

Distance covered by B in 3 minutes = $$\frac{x}{20}$$ km

Now, time taken by A to travel 3.5 km = Time taken by B to travel 5.5 km - 3 minutes

=> $$3.5\times\frac{3}{10}=\frac{5.5}{x}-\frac{3}{60}$$

=> $$1.05+0.05=\frac{5.5}{x}$$

=> $$x=\frac{5.5}{1.1}=5$$ km/hr

=> Ans - (A)

Number of soap bars sold = $$\frac{20}{100}\times6435$$

= $$1287$$

=> Ans - (C)