Quant Practice Questions for SBI PO 2017

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 x3+ y3+ z3– 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 = 2πr = 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 = a2= 662  = 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 = √(a2+b2) =  √(82+62 ) =  √(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 = √(a2+b2 ) =  √(82+62 ) = 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/(x2+2x-15)  =1/6
48 = x2+ 2x – 15
48 + 15 = x2+2x
63 = x2+2x
(x+9)(x-7)= 0
Then x = 7 and -9
Quantity 2:
x3+ y3+ z3– 3xyz =0
x3+ y3+ z3– 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) = 12C3 = 220
No.of selections of 3 bad oranges from 4 bad oranges = 4C3 = 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 = [4C3/11C3] × [7C3/11C3] = 4/165 × 35/165 = 28/5445
8. C) Probability of choosing 1 basket = ½
Probability of black ball from 1st basket = ½ × 9C1/14C1 = 9/28
Probability of black ball from 2nd basket = ½ × 7C1/14C1 = 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 1st basket = ½ × 5C1/14C1 = 5/28
Probability of black ball from 2nd basket = ½ × 7C1/14C1 = 7/28
Required answer = 5/28 + 7/28 = 12/28 = 3/7
10. B)
N(S) = 12C3 = 220
Selection of one ball of each colour = 3C1 × 5C1 × 4C1 = 60
Therefore, P(E) = 60/220 = 3/11