CAT 2017 Question Paper Slot 1

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 $$(A+9.5) - A = 9.5$$