CAT 2017 Question Paper Quant Slot 1

Let Arun's current age be A. Hence, Barun's current age is 2.5A
Let Arun's age be half of Barun's age after X years.
Therefore, 2*(X+A) = 2.5A + X
Or, X = 0.5A
Hence, Barun's age increased by 0.5A/2.5A = 20%

Let the rate of work of a person be x units/day. Hence, the total work = 120x.

It is given that one first day, one person works, on the second day two people work and so on.

Hence, the work done on day 1, day 2,... will be x, 2x, 3x, ... respectively.

The sum should be equal to 120x.

$$120x = x* \frac{n(n+1)}{2}$$

$$n^2 + n - 240 = 0$$

n = 15 is the only positive solution.

Hence, it takes 15 days to complete the work.

It is given that the maximum weight limit is 630. The lightest person's weight is 53 Kg and the heaviest person's weight is 57 Kg.

In order to have maximum people in the lift, all the remaining people should be of the lightest weight possible, which is 53 Kg.

Let there be n people.

53 + n(53) + 57 = 630

n is approximately equal to 9.8. Hence, 9 people are possible.

Therefore, a total of 9 + 2 = 11 people can use the elevator.

We see that the man saves 20 minutes by changing his speed from 12 Km/hr to 15 Km/hr.

Let d be the distance

Hence,

$$\frac{d}{12} - \frac{d}{15} = \frac{1}{3}$$

$$\frac{d}{60} = \frac{1}{3}$$

d = 20 Km.

Let his total savings be 100x.

He invests 50x in fixed deposits. 30% of 50x, which is 15x is invested in stocks and 35x goes to savings bank.

It is given 85x = 59500

x = 700

Hence, 100x = 70000

Let the retail price be M and cost price be C.

Given,

0.85 M = 1.02 C

M = 1.2 C

If he wants 20% profit he has to sell at 1.2C, which is nothing but the retail price.

Let the speed of the river be $$x$$ and the speed of the boat be $$u$$. Let $$d$$ be the one way distance and $$t$$ be the initial time taken.

Given,

$$t = \frac{d}{u - x} + \frac{d}{u + x}$$ ... i

Also,

$$\frac{t}{4} = \frac{d}{2u - x} + \frac{d}{2u + x}$$

$$t = \frac{4d}{2u - x} + \frac{4d}{2u + x}$$ ... ii

Equating both i and ii,

$$\dfrac{d}{u - x}$$+ $$\dfrac{d}{u + x}$$ =  $$\dfrac{4d}{2u - x} + \dfrac{4d}{2u + x}$$

$$\dfrac{2u}{u^2 - x^2} = \dfrac{16u}{4u^2 - x^2}$$

$$4u^2 - x^2 = 8u^2 - 8x^2$$

$$\frac{u^2}{x^2} = \frac{7}{4}$$

$$\frac{u}{x} = \frac{\sqrt{7}}{2}$$

Given,

C1 : C2 : C3 = 9 : 10 : 8 ... i

C2 : C4 : C5 = 18 : 19 : 20 ... ii

Let's multiply i by 9 and ii by 5

C1 : C2 : C3 = 81 : 90 : 72

C2 : C4 : C5 = 90 : 95 : 100

Therefore, C1 : C2 : C3 : C4 : C5 = 81 : 90 : 72 : 95 : 100

Given,

100x - 81x = 19

x = 1

Hence, total profit = 100 + 95 + 72 + 90 + 81 = 438

Let the number of girls be 2x and number of boys be x.

Girls getting admission = 0.6x

Boys getting admission = 0.45x

Number of students not getting admission = 3x - 0.6x -0.45x = 1.95x

Percentage = (1.95x/3x) * 100 = 65%

The ratio of L, S, J for popcorn = 7 : 17 : 16

Let them be 7$$x$$, 17$$x$$ and 16$$x$$

The ratio of L, S, J for chips = 6 : 15 : 14

Let them 6$$y$$, 15$$y$$ and 14$$y$$

Given, 40$$x$$ = 35$$y$$, $$x = \frac{7y}{8}$$

Jumbo popcor = 16$$x$$ = 16 * $$\frac{7y}{8}$$= 14$$y$$

Hence, the ratio of jumbo popcorn and jumbo chips = 1 : 1

Let the cost of good mangoes be 2x per kg. The cost of medium mangoes be x per kg. 

CP of good mangoes = 160x

CP of medium mangoes = 40x

His selling price = 0.9*2x = 1.80x

Therefore, total revenue generated by selling all the mangoes = 120*1.8x = 216x

Hence, the profit % = $$\frac{16x}{200x} * 100 $$ = 8%

Let the cost price be C and the marked price be M.

Given,

0.6 M = 1.2 C

M = 2C

CP of 60 toys = 60C

Now only 50 are remaining.

Hence, 

M (1 - d) * 50 = 72C

1- d = 0.72

d = .28

Hence 28%

Since the square root can be positive or negative we will get two cases for each of the equation.

For the first one,

a + 3 = 3b .. i

a + 3 = -3b ... ii

For the second one,

a - 1 = 2(b -1) ... iii

a - 1 = 2 (1 - b) ... iv

we have to solve i and iii, i and iv, ii and iii, ii and iv.

Solving i and iii,

a + 3 = 3b and a = 2b  - 1, solving, we get a = 3 and b = 2, which is not what we want.

Solving i and iv

a + 3 = 3b and a = 3 - 2b, solving, we get b = 1.2, which is not possible.

Solving ii and iii

a + 3 = -3b and a = 2b - 1, solving, we get b = 0.4, which is not possible.

Solving ii and iv,

a + 3 = -3b and a = 3 - 2b, solving, we get a = 15 and b = -6 which is what we want.

Thus, $$\frac{a^2}{b^2} = \frac{25}{4}$$

Let, the average score of boys in the mid semester exam is A.
Therefore, the average score of girls in the mid semester exam be A+5.
Hence, the total marks scored by the class is $$20\times (A) + 30\times (A+5) = 50\times A + 150$$

The average score of the entire class is $$\dfrac{(50\times A + 150)}{50} = A + 3$$

wkt, class average increased by 2, class average in final term $$= (A+3) + 2 = A + 5$$

Given, that score of girls dropped by 3, i.e $$(A+5)-3 = A+2$$
Total score of girls in final term $$= 30\times(A+2) = 30A + 60$$

Total class score in final term $$= (A + 5)\times50 = 50A + 250$$

the total marks scored by the boys is $$(50A + 250) - (30A - 60) = 20A + 190$$
Hence, the average of the boys in the final exam is $$\dfrac{(20G + 190)}{20} = A + 9.5$$

Hence, the increase in the average marks of the boys is $$(A+9.5) - A = 9.5$$

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,

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

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

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

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

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

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

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

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

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

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

x - y - z = 25 and $$x\leq40,y\leq12$$, $$z\leq12$$
If x = 40 then y + z = 15. Now since both y and z are natural numbers less than 12, so y can range from 3 to 12 giving us a total of 10 solutions.Similarly, if x = 39, then y + z = 14. Now y can range from 2 to 12 giving us a total of 11 solutions.
If x = 38, then y + z = 13. Now y can range from 1 to 12 giving us a total of 12 solutions.
If x = 37 then y + z = 12 which will give 11 solutions.
Similarly on proceeding in the same manner the number of solutions will be 10, 9, 8, 7 and so on till 1.
Hence, required number of solutions will be (1 + 2 + 3 + 4 . . . . + 12) + 10 + 11
= 12*13/2 + 21
78 + 21 = 99

$$(n - 5) (n - 10) - 3(n - 2)\leq0$$
=> $$ n^2 - 15n + 50 - 3n + 6 \leq 0$$
=> $$n^2 - 18n + 56 \leq 0$$
=> $$(n - 4)(n - 14) \leq 0$$
=> Thus, n can take values from 4 to 14. Hence, the required number of values are 14 - 4 + 1 = 11.

$$f_{1}(x)=x^{2}+11x+n$$ and $$f_{2}(x) = x$$
$$f_{1}(x)=f_{2}(x)$$
=> $$x^{2}+11x+n = x$$
=> $$ x^2 + 10x + n = 0 $$
=> For this equation to have distinct real roots, b$$^2$$-4ac>0
$$ 10^2 > 4n$$
=> n < 100/4 
=> n < 25
Thus, largest integral value that n can take is 24.

For the value of given expression to be minimum, the values of $$a, b, c$$ and $$d$$ should be as close as possible. 30/4 = 7.5. Since each one of these are integers so values must be 8, 8, 7, 7. On putting these values in the given expression, we get
 $$(8 - 8)^{2} + (8 - 7)^{2} + (8 - 7)^{2}$$
=> 1 + 1 = 2

The total number of given points are 11. (10 on circumference and 1 is the center)
So total possible triangles = 11C3 = 165.
However, AOB, COD, EOF, GOH, JOK lie on a straight line. Hence, these 5 triangles are not possible. Thus, the required number of triangles = 165 - 5 = 160

The graph of the given function is as shown below.

We can see that the shortest distance of the point (1/2, 1) will be 1 unit.

The seventh term of an AP = a + 6d. Third term will be a + 2d and second term will be a + 16d. We are given that
$$ (a + 6d)^2 = (a + 2d)(a + 16d)$$
=> $$ a^2 $$ + $$36d^2$$ + 12ad = $$ a^2 + 18ad + 32d^2 $$
=> $$4d^2 = 6ad$$
=> $$ d:a = 3:2$$
We have been asked about a:d. Hence, it would be 2:3

We have been given that a + b + c + d = 7
Total ways of distributing 7 things among 4 people so that each one gets at least one = $$^{n-1}C_{r-1}$$ = 6C3 = 20
Now we need to subtract the cases where any one person got more than 3 erasers. Any person cannot get more than 4 erasers since each child has to get at least 1. Any of the 4 childs can get 4 erasers. Thus, there are 4 cases. On subtracting these cases from the total cases we get the required answer. Hence, the required value is 20 - 4 = 16

$$f(3) = \frac{15 + 2}{9 - 5} = \frac{17}{4}$$
$$f(f(3)) = \frac{5*17/4 + 2}{3*17/4 - 5} = \frac{93/4}{31/4} = \frac{93}{31} = 3$$
$$g(f(f(3))) = 3^2 - 3*2 - 1 = 2$$

$$a_{1}$$ = 3 and $$a_{2}$$ = 7. Hence, the common difference of the AP is 4. If we assume, k=3n 
We have been given that the sum up to 3n terms of this AP is 1830. Hence, $$1830 = \frac{k}{2}[2*3 + (k - 1)*4$$
=> 1830*2 = k(6 + 4k - 4)
=> 3660 = 2k + 4k$$^2$$
=> $$2k^2 + k - 1830 = 0$$
=> (k - 30)(2k + 61) = 0
=> k = 30 or k = -61/2
Since k is the number of terms so k cannot be negative. Hence, must be 30
So, 3n = 30
n = 10
Sum of the first '10' terms of the given AP = 5*(6 + 9*4) = 42*5 = 210
m($$a_1$$+ $$a_{2}$$ +...+ $$a_n$$) > 1830
=> 210m > 1830
=> m > 8.71
Hence, smallest integral value of 'm' is 9.