CAT 2017 Question Paper Slot 1

The following equation will form a square of side $$ 2\sqrt{2} $$.

The area of the square = $$(2\sqrt{2})^2$$ = 8 units.

The lengths are given as 40, 25 and 35.

The perimeter = 100

Semi-perimeter, s = 50

Area = $$ \sqrt{50 * 10 * 25 * 15}$$ = $$250\sqrt{3}$$

The triangle formed by the centroid and two vertices is removed.

Since the cenroid divides the median in the ratio 2 : 1

The remaining area will be two-thirds the area of the original triangle.

Remaining area = $$\frac{2}{3} * 250\sqrt{3}$$ = $$\frac{500}{\sqrt{3}}$$

The image of the figure is as shown. 

AB = AC = 6cm. Thus, BC = $$\sqrt{6^2 + 6^2}$$ = 6√2 cm

The required area = Area of semi-circle BQC - Area of quadrant BPC + Area of triangle ABC

Area of semicircle BQC

Diameter BC = 6√2cm

Radius = 6√2/2 = 3√2 cm

Area = $$\pi r^2 $$/2 = $$ \pi$$ * $$(3 \sqrt{2})^2 $$/2 = 9$$\pi$$

Area of quadrant BPC

Area = $$\pi r^2$$/4 = $$\pi*(6)^2$$/4 = 9$$\pi$$

Area of triangle ABC

Area = 1/2 * 6 * 6 = 18

The required area = Area of semi-circle BQC - Area of quadrant BPC + Area of triangle ABC

= 9$$\pi$$ - 9$$\pi$$ + 18 = 18

Let the volumes of the five cubes be x, x, 8x, 27x and 27x.

Let the sides be a, a, 2a, 3a and 3a. ($$x = a^3$$)

Let the side of the original cube be A.

$$A^3 = x + x + 8x + 27x + 27x$$

$$A = 4a$$

Original surface area = $$96a^2$$

New surface area = $$6(a^2 + a^2 + 4a^2 + 9a^2 + 9a^2)$$ = $$144a^2$$

Percentage increase = $$\frac{144 - 96}{96}*100 =$$ 50%

The volume of a cylinder is $$\pi * r^2 * 3 = 9 \pi$$

r = $$\sqrt{3}$$ cm.

Radius of the ball is 2 cm. Hence, the ball will lie on top of the cylinder.

Lets draw the figure.

Based on the Pythagoras theorem, the other leg will be 1 cm.

Thus, the height will be 3 + 1 + 2 = 6 cm

The length of the altitude from A to the hypotenuse will be the shortest distance.

This is a right triangle with sides 3 : 4 : 5. 

Hence, the hypotenuse = $$\sqrt{20^2+15^2}$$ = 25 Km.

Length of the altitude = $$\frac{15 * 20}{25}$$ = 12 Km

(This is derived from equating area of triangle, $$\frac{15\cdot20}{2}\ =\ \frac{25\cdot\text{altitude}}{2}$$)

The time taken = $$\frac{12}{30} * 60$$ = 24 minutes

We know that $$\log_3 x = a$$ and $$\log_{12} y=a$$
Hence, $$x = 3^a$$ and $$y=12^a$$
Therefore, the geometric mean of $$x$$ and $$y$$ equals $$\sqrt{x \times y}$$
This equals $$\sqrt{3^a \times 12^a} = 6^a$$

Hence, $$G=6^a$$ Or, $$\log_6 G = a$$

We know that $$x^2 - x - 1=0$$
Therefore $$x^4 = (x+1)^2 = x^2+2x+1 = x+1 + 2x+1 = 3x+2$$
Therefore, $$2x^4 = 6x+4$$

We know that $$x>0$$ therefore, we can calculate the value of $$x$$ to be $$\frac{1+\sqrt{5}}{2}$$
Hence, $$2x^4 = 6x+4 = 3+3\sqrt{5}+4 = 3\sqrt{5}+7$$

$$\log_{0.008}\sqrt{5}+\log_{\sqrt{3}}81-7$$

$$81 = 3^4$$ and $$0.008 = \frac{8}{1000} = \frac{2^{3}}{10^{3}} = \frac{1}{5^{3}} = 5^{-3} $$

Hence,

$$\log_{0.008}\sqrt{5}+ 8 -7 $$

$$ \log_{5^{-3}}5^{\frac{1}{2}}+ 8 -7 $$

$$\frac{log 5^{0.5}}{log 5^{-3}} + 1$$

$$ - \frac{1}{6} + 1$$

= $$\frac{5}{6}$$

$$\frac{81^x}{9} - \frac{81^x}{81} = 1944$$

$$81^x * [\frac{1}{9}- \frac{1}{81}] = 1944$$

$$81^x * [\frac{1}{81}] = 243$$

$$3^{4x} = 3^9$$

$$x = \frac{9}{4}$$