MAT 2001Question Paper

Let usual speed = $$4$$ m/min and usual time taken be $$t$$ min

=> New speed = $$\frac{3}{4}\times4=3$$ m/min and new time = $$(t+20)$$ min

Speed is inversely proportional to time

=> $$\frac{4}{3}=\frac{t+20}{t}$$

=> $$4t=3t+60$$

=> $$t=60$$

$$\therefore$$ Usual time = 60 min = 1 hour

=> Ans - (B)

Difference between SI and CI for 3 years = $$\frac{Pr^2}{100^2}(3+\frac{r}{100})$$ and for 2 years = $$\frac{Pr^2}{100^2}$$

Now, rate of interest = $$r=20\%$$ and time = $$t=3$$ years

Let principal amount be Rs. $$P$$

=> Difference = $$P\times\frac{(20)^2}{(100)^2}(3+\frac{20}{100})=48$$

=> $$P\times\frac{1}{25}\times\frac{16}{5}=48$$

=> $$P=\frac{48\times125}{16}=3\times125$$

=> $$P=Rs.$$ $$375$$

=> Ans - (C)

Let us assume all 90 are rabbits, thus there must be 360 legs (Rabbits have 4 legs and pigeons have 2)

But there are $$360-224=136$$ less legs, so rabbits must be less.

=> Number of pigeons = $$\frac{136}{2}=68$$

Alternate Method : Let number of pigeons be $$x$$ and number of rabbits be $$y$$

=> $$x+y=90$$ -----------(i)

and $$2x+4y=224$$ ------(ii)

Solving the two equations, we get : $$x=68$$ and $$y=22$$

=> Ans - (B)

Let sides of rectangular field be $$3x$$ and $$4x$$ m respectively.

=> Area = $$lb=7500$$

=> $$3x\times4x=7500$$

=> $$x^2=\frac{7500}{12}=625$$

=> $$x=\sqrt{625}=25$$

=> Perimeter of field = $$2(3x+4x)=14x$$

= $$14\times25=350$$ m

$$\therefore$$ Cost of fencing = $$350\times0.25=Rs.$$ $$87.50$$

=> Ans - (A)

Time required by B alone = $$\frac{1}{6}-\frac{1}{10}$$

= $$\frac{5-3}{30}=\frac{2}{30}=\frac{1}{15}$$

$$\therefore$$ B alone finishes the work in 15 days.

=> Ans - (B)

Let the perimeter of square and equilateral triangle be $$12x$$ cm

=> Each side of square = $$\frac{12x}{4}=3x$$ cm and each side of equilateral triangle = $$4x$$ cm

Diagonal of a square = $$d=\sqrt2\times$$ side

=> $$3x\times\sqrt2=12\sqrt2$$

=> $$x=\frac{12}{3}=4$$

=> Side of equilateral triangle = $$4\times4=16$$ cm

$$\therefore$$ Area of equilateral triangle = $$\frac{\sqrt3}{4} s^2$$

= $$\frac{\sqrt3}{4}\times(16)\times(16)$$

= $$64\sqrt3$$ $$cm^2$$

=> Ans - (C)

Let capacity of cistern = L.C.M. (5,6) = 30 units

Efficiency to fill it = $$\frac{30}{5}=6$$ units/hr

Let efficiency of leak = $$-x$$ units/hr

According to ques, => $$(6-x)\times6=30$$

=> $$6-x=5$$

=> $$x=1$$

$$\therefore$$ Time taken by leak to empty it = $$\frac{30}{1}=30$$ hours

=> Ans - (C)

Side of the equilateral triangle = $$s=12$$ cm

=> Area of triangle = $$\triangle=\frac{\sqrt3}{4}s^2$$

= $$\frac{\sqrt3}{4}\times12\times12=36\sqrt3$$ $$cm^2$$

$$\therefore$$ Circumradius = $$R=\frac{abc}{4\triangle}$$

= $$\frac{12\times12\times12}{4\times36\sqrt3}$$

= $$\frac{12}{\sqrt3}=4\sqrt3$$ cm

=> Ans - (B)

Sides of the triangle are $$a=6,b=11,c=15$$ cm

Area of a triangle = $$\triangle=rs$$, where $$r$$=in radius and $$s$$ is semi perimeter

=> Semi perimeter = $$s=\frac{a+b+c}{2}=\frac{6+11+15}{2}=16$$ cm

Area of triangle using Heron's Formula = $$\sqrt{s(s-a)(s-b)(s-c)}$$

= $$\sqrt{16\times10\times5\times1}$$

= $$\sqrt{2^5\times5^2}=20\sqrt2$$ $$cm^2$$

$$\therefore$$ $$r=\frac{\triangle}{s}$$

= $$\frac{20\sqrt2}{16}=\frac{5\sqrt2}{4}$$ cm

=> Ans - (A)

Let the number of sides of the polygon be $$n$$

Sum of interior angles of any polygon = $$(n-2)\times180=1620$$

=> $$n-2=\frac{1620}{180}=9$$

=> $$n=9+2=11$$

=> Ans - (B)