MAT 2001 Question Paper

AD and CE are two trees of height 28 m and 20 m respectively. AC = 17 m

In right $$\triangle$$ ABC,

=> $$(BC)^2=(AC)^2-(AB)^2$$

=> $$(BC)^2=(17)^2-(8)^2$$

=> $$(BC)^2=289-64=225$$

=> $$BC=\sqrt{225}=15$$ m

=> Ans - (C)

Let B and C together have = Rs. $$9x$$

=> A's share = $$\frac{2}{9}\times9x=Rs.$$ $$2x$$ ------------(i)

Also, amount that A receive = $$(770-9x)$$ -----------(ii)

Comparing equations (i) and (ii), => $$2x=770-9x$$

=> $$11x=770$$

=> $$x=\frac{770}{11}=70$$

$$\therefore$$ A's share = $$2\times70=Rs.$$ $$140$$

=> Ans - (A)

Let $$x$$ must be subtracted.

=> $$\frac{14-x}{17-x}=\frac{34-x}{42-x}$$

=> $$(14-x)(42-x)=(34-x)(17-x)$$

=> $$588-14x-42x+x^2=578-34x-17x+x^2$$

=> $$56x-51x=588-578$$

=> $$5x=10$$

=> $$x=2$$

=> Ans - (C)

Expression : $$\frac{3\pi}{5}$$ radians

= $$\frac{3}{5}\times180$$

= $$3\times36=108^\circ$$

=> Ans - (A)

Let $$sin A=4$$ and $$cos A=7$$

Expression : $$\frac{(7 \sin A - 3 \cos A)}{(7 \sin A+ 2 \cos A)}$$

= $$\frac{7(4)-3(7)}{7(4)+2(7)}=\frac{28-21}{28+14}$$

= $$\frac{7}{42}=\frac{1}{6}$$

=> Ans - (D)

Given : BC = 10 m and $$\angle ACB=30^\circ$$

To find : AC + AB = ?

Solution : In right $$\triangle$$ ABC

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

=> $$\frac{1}{\sqrt3}=\frac{AB}{10}$$

=> $$AB=\frac{10}{\sqrt3}$$ ---------------(i)

Similarly, $$cos(30^\circ)=\frac{BC}{AC}$$

=> $$\frac{\sqrt3}{2}=\frac{10}{AC}$$

=> $$AC=\frac{20}{\sqrt3}$$ ------------------(ii)

$$\therefore$$ Height of tree = $$AB+AC$$

= $$\frac{10}{\sqrt3}+\frac{20}{\sqrt3}$$

= $$\frac{30}{\sqrt3}=10\sqrt3$$ m

=> Ans - (B)

Given : $$pqr=1$$

Expression : $$\left(\left(\frac{1}{(1 + p + q^{-1})}\right) + \left(\frac{1}{(1 + q + r^{-1})}\right) + \left(\frac{1}{(1 + r + p^{-1})}\right)\right)$$

= $$\frac{1}{(1+p+\frac{1}{q})}+\frac{1}{(1+q+\frac{1}{r})}+\frac{1}{(1+r+\frac{1}{p})}$$

= $$\frac{q}{(1+q+pq)}+\frac{1}{(1+q+pq)}+\frac{1}{(1+\frac{1}{pq}+\frac{1}{p})}$$

= $$\frac{q}{(1+q+pq)}+\frac{1}{(1+q+pq)}+\frac{pq}{(1+q+pq)}$$

= $$\frac{1+q+pq}{1+q+pq}=1$$

=> Ans - (A)

Edge of cube = 10 cm

Radius of the largest possible cone = 5 cm and height of cone = 10 cm

Volume of cone = $$\frac{1}{3}\pi r^2h$$

= $$\frac{1}{3}\times\frac{22}{7}\times(5)^2\times10$$

= $$\frac{25\times220}{21}=261.9$$ $$cm^3$$

=> Ans - (C)

Let radius of each small spherical ball = $$r$$ cm

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

=> $$8\times\frac{4}{3}\pi (r)^3=\frac{4}{3}\pi (10)^3$$

=> $$r^3=\frac{(10)^3}{8}$$

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

$$\therefore$$ Surface area of each ball = $$4\pi r^2$$

= $$4\times\pi \times(5)^2=100\pi$$ $$cm^2$$

=> Ans - (D)

Equation : $$x^2 - 6x + 6 = 0$$

Sum of roots = $$a+b=6$$ and Product of roots = $$ab=6$$ --------------(i)

Now, squaring both sides, => $$(a+b)^2=(6)^2$$

=> $$a^2+b^2+2ab=36$$

=> $$(a^2+b^2)+2(6)=36$$

=> $$(a^2+b^2)=36-12=24$$

=> Ans - (B)