**Directions (1 – 5): Read the following information and answer the questions accordingly.**

**1. Quantity I:**By selling 20 mangoes for Rs.520 a total profit of Rs. 40 is obtained. What is the cost price of one mango?

**Quantity II:**Total market price of 15 mangoes Rs.500 which was Rs.50 more than the total cost. What is the cost price of one mango?

a) Quantity I<Quantity II

b) Quantity I>Quantity II

c) Quantity I≤ Quantity II

d) Quantity I≥ Quantity II

e) Quantity I= Quantity II

**2. Quantity I :**A person marked the price of a refrigerator 60% above the cost price. By giving a discount of 25% the seller sold the refrigerator at the price of Rs.15000 . what is the cost price of the refrigerator?

**Quantity II:**The cost price of a refrigerator is Rs.7500, which is sold at a profit of 60%. What is the selling price of the refrigerator?

a) Quantity I< Quantity II

b) Quantity I> Quantity II

c) Quantity I≤ Quantity II

d) Quantity I≥ Quantity II

e) Quantity I= Quantity II

**3. Quantity 1:**Area of square having its perimeter is 2 times of circumference of circle having 21mtrs as its radius.

**Quantity 2:**Area of a rectangle whose length is 150% of its breadth and which perimeter is 270mtrs.

a) Quantity 1 > Quantity 2

b) Quantity 1 < Quantity 2

c) Quantity 1 ≥ Quantity 2

d) Quantity 1 ≤ Quantity 2

e) Quantity 1 = Quantity 2 or Relationship can’t be established.

**4. Quantity 1:**Perimeter of a Square having its side is equal to hypotenuse of a right-angled triangle having its sides 8mtrs and 6mtrs respectively.

**Quantity 2:**Perimeter of a Rhombus having its diagonals d1 and d2 are 12mtr and 16mtr respectively.

a) Quantity 1 > Quantity 2

b) Quantity 1 < Quantity 2

c) Quantity 1 ≥ Quantity 2

d) Quantity 1 ≤ Quantity 2

e) Quantity 1 = Quantity 2 or Relationship can’t be
established.

**5. Quantity 1:**The value of X from 1/(x-3) – 1/(x+5)=1/6

**Quantity 2:**The value of X from x

^{3}+ y

^{3}+ z

^{3}– 3xyz =0, if the value of y = 6 and z = -10.

a) Quantity 1 > Quantity 2

b) Quantity 1 < Quantity 2

c) Quantity 1 ≥ Quantity 2

d) Quantity 1 ≤ Quantity 2

e) Quantity 1 = Quantity 2 or Relationship can’t be
established.

**6. In a box carrying one dozen of oranges, one third have become bad. If 3 oranges are taken out from the box at random, what is the probability that at least one oranges out of the three oranges picked up is good?**

a) 1/55

b) 54/55

c) 45/55

d) 3/55

e) None of these

**7. A bag contains 4 red and 7 black balls. Two draws of three balls each are made, the ball being replaced after the first draw. What is the chance that the balls were red in the first draw and black in the second?**

a) 28/5445

b) 25/5448

c) 28/4554

d) 25/4554

e) None of these

**8. A basket contains 5 white and 9 black balls. There is another basket which contains 7 white and 7 black balls. One ball is to drawn from either of the two baskets. What is the probability of drawing a black ball?**

a) 3/7

b) 5/7

c) 4/7

d) 8/15

e) None of these

**9. A basket contains 5 white and 9 black balls. There is another basket which contains 7 white and 7 black balls. One ball is to drawn from either of the two baskets. What is the probability of drawing a white ball?**

a) 4/7

b) 6/7

c) 3/7

d) 2/7

e) None of these

**10. A bag contains 3 red, 5 yellow and 4 green balls. 3 balls are drawn randomly. What is the probability that the balls drawn contain balls of different colours?**

a) 4/15

b) 3/11

c) 1/12

d) 5/14

e) None of these

**Solutions:**

**1. A)**By using statement I,

Cost price = 520-40 = 480

Cost price per mango = 480÷20 = Rs.24

By using statement II,

Total cost price = 500-50=450

Cost price of per mango =
450÷15=Rs.30

So, Quantity I< Quantity II

**2. B)**By using statement I

Let the marked price be Rs.x

Then,(x×0.75)=15000

X =Rs.20000

Let cost price be y

1.60×y=20000

Y=Rs.12500

By using statement II

Profit%=profit/(cost price)×100

60%×7500= profit

Profit =Rs.4500

Hence, selling price is Rs.12000

So, Quantity I> Quantity II

**3. B)**Quantity 1:

Circumference of square = 2Ï€r =
2×22/7×21 = 132mtrs

Perimeter of square = 2 times of the
circumference of circle = 132 × 2 = 264mtrs

Side of Square a =264/4= 66mtrs

Area of square = a

^{2}= 66^{2}= 4356sq.mtrs
Quantity 2:

Perimeter = 2(L+ B)= 270mtrs

L+B = 135 mtrs

L =150% of B

Then L and B = 81, 54 respectively

Area of the rectangle = L×B = 81 × 54
= 4374sq.mtrs

So, Quantity 1 <Quantity 2.

**4. E)**Quantity 1:

Hypotenuse of a triangle = √(a

^{2}+b^{2}) = √(8^{2}+6^{2}) = √(64+36) = √100 = 10 mtrs
And Perimeter of the square = 4a = 4
× 10 = 40mtrs.

Quantity 2:

Diagonals of rhombus intersect at
90°angle.

The diagonals and the sides of the
rhombus form right triangle.

One side of the right triangle is
16/2= 8mtrs

Other side of the right triangle is
12/2= 6mtrs

Then the third side of the triangle =
√(a

^{2}+b^{2}) = √(8^{2}+6^{2}) = 10mtrs
And Perimeter of the rhombus = 4a = 4
× 10 = 40mtrs.

So Quantity 1 is equal to Quantity 2.

**5. E)**Quantity 1:

((x+5)–(x+3))/(x-3)(x+5) =1/6

8/(x

^{2}+2x-15) =1/6
48 = x

^{2}+ 2x – 15
48 + 15 = x

^{2}+2x
63 = x

^{2}+2x
(x+9)(x-7)= 0

Then x = 7 and -9

Quantity 2:

x

^{3}+ y^{3}+ z^{3}– 3xyz =0
x

^{3}+ y^{3}+ z^{3}– 3xyz = 0=(x+y+z)
x+6-10 = 0

x – 4 = 0

Then x = 4

So the Relationship can’t be
established.

**6. B)**No.of selections of 3 oranges from total 12 oranges n(S) =

^{12}C

_{3}= 220

No.of selections of 3 bad oranges
from 4 bad oranges =

^{4}C_{3}= 4
Therefore, n(E) no.of desired
selection = 220 – 4 = 21

p(E) = n(E)/n(S) = 216/220 = 54/55

**7. A)**Required Probability = [

^{4}C

_{3}/

^{11}C

_{3}] × [

^{7}C

_{3}/

^{11}C

_{3}] = 4/165 × 35/165 = 28/5445

**8. C)**Probability of choosing 1 basket = ½

Probability of black ball from 1

^{st}basket = ½ ×^{9}C_{1}/^{14}C_{1}= 9/28
Probability of black ball from 2

^{nd}basket = ½ ×^{7}C_{1}/^{14}C_{1}= 7/28
Required answer = 9/28 + 7/28 = 16/28
= 4/7

**9. C)**Probability of choosing 1 basket = ½

Probability of black ball from 1

^{st}basket = ½ ×^{5}C_{1}/^{14}C_{1}= 5/28
Probability of black ball from 2

^{nd}basket = ½ ×^{7}C_{1}/^{14}C_{1}= 7/28
Required answer = 5/28 + 7/28 = 12/28
= 3/7

**10. B)**

N(S) =

^{12}C_{3}= 220
Selection of one ball of each colour
=

^{3}C_{1}×^{5}C_{1}×^{4}C_{1}= 60
Therefore, P(E) = 60/220 = 3/11