MAT 2001 Question Paper

Let $$a=2$$ and $$b=5$$

To find = $$(3a + 4b) : (5a + 6b)$$

= $$\frac{3(2)+4(5)}{5(2)+6(5)}=\frac{(6+20)}{(10+30)}$$

= $$\frac{26}{40}=\frac{13}{20}$$

=> Ans - (A)

Sum of $$n$$ consecutive natural number cubes = $$[\frac{n(n+1)}{2}]^2$$

and sum of $$n$$ consecutive natural numbers = $$\frac{n(n+1)}{2}$$

Expression : $$(1^3 + 2^3 + 3^3 + ........ + 15^3) - (1 + 2 + 3 + ......... + 15)$$

= $$(\frac{15\times16}{2})^2-(\frac{15\times16}{2})$$

= $$(120)^2-120$$

= $$14400-120=14280$$

In the circle, OB=OC (radii), => $$\angle OBC=\angle OCB=37^\circ$$

In $$\triangle$$ OBC, using angle sum property

=> $$\angle BOC+\angle OBC+\angle OCB=180^\circ$$

=> $$\angle BOC+37^\circ+37^\circ=180^\circ$$

=> $$\angle BOC=180^\circ-74^\circ=106^\circ$$

Now, angle subtended by an arc at the centre is double the angle subtended by it at any point on the circle.

=> $$\angle BOC=2\times\angle BAC$$

=> $$\angle BAC=\frac{106^\circ}{2}=53^\circ$$

=> Ans - (C)

Area of the field that the horse can gaze = area of sector having radius $$r=14$$ m

= $$\frac{\theta}{360^\circ} \pi r^2$$

= $$\frac{90^\circ}{360^\circ} \times\frac{22}{7} \times14\times14$$

= $$\frac{1}{4}\times22\times28$$

= $$22\times7=154$$ $$m^2$$

=> Ans - (A)

Factors of 14 = $$1,2,7,14$$

Thus, only two digit numbers possible are 27 and 72

Now, when 45 is added, number will be reversed, => Number = 27

=> Ans - (B)

Let reduced price of the radio = Rs. $$x$$

=> Reduced Price + Sale tax = Rs. 2568

=> $$x+\frac{7x}{100}=2568$$

=> $$\frac{107x}{100}=2568$$

=> $$x=\frac{2568}{107}\times100$$

=> $$x=24\times100=2400$$

$$\therefore$$ Reduction needed = $$2568-2400=Rs.$$ $$168$$

=> Ans - (C)

Let speeds of train and car be $$x$$ km/hr and $$y$$ km/hr respectively.

According to ques,

=> $$\frac{160}{x}+\frac{600}{y}=8$$ ----------(i)

and $$\frac{240}{x}+\frac{520}{y}=8+\frac{12}{60}=8.2$$  -----------(ii)

Applying : 3 x (i)-2 x (ii)

=> $$\frac{1800}{y}-\frac{1040}{y}=24-16.4$$

=> $$\frac{760}{y}=7.6$$

=> $$y=\frac{760}{7.6}=100$$

Similarly, $$x=\frac{160}{2}=80$$

$$\therefore$$ Speeds of train and car are 80 km/hr and 100 km/hr respectively.

=> Ans - (A)

Let original duration of tour be $$x$$ days and expense per day = Rs. $$\frac{360}{x}$$

According to ques,

=> $$(x+4)(\frac{360}{x}-3)=360$$

=> $$360-3x+\frac{1440}{x}-12=360$$

=> $$-3x^2+1440-12x=0$$

=> $$x^2+4x-480=0$$

=> $$(x+24)(x-20)=0$$

=> $$x=20,-24$$

$$\because x$$ cannot be negative, thus number of days = 20

=> Ans - (C)

Let original number of persons be $$x$$ and amount received by each person = Rs. $$\frac{6500}{x}$$

According to ques,

=> $$(x+15)(\frac{6500}{x}-30)=6500$$

=> $$6500-30x+\frac{15\times6500}{x}-450=6500$$

=> $$-30x^2+(15\times6500)-450x=0$$

=> $$x^2+15x-3250=0$$

=> $$(x+65)(x-50)=0$$

=> $$x=50,-65$$

$$\because x$$ cannot be negative, thus number of persons are 50

=> Ans - (A)

Let time taken by A alone to fill the cistern = $$t$$ minutes

=> Time taken by B = $$(t+5)$$ minutes

According to ques, => $$\frac{1}{t}+\frac{1}{t+5}=\frac{1}{6}$$

=> $$\frac{t+t+5}{t(t+5)}=\frac{1}{6}$$

=> $$12t+30=t^2+5t$$

=> $$t^2-7t-30=0$$

=> $$(t-10)(t+3)=0$$

=> $$t=10,-3$$

$$\because$$ Time cannot be negative, => $$t=10$$

Thus, time taken by A and B separately to fill the cistern is 10 min and 15 min respectively.

=> Ans - (C)