Sunday, 16 August 2015

Banking Pathway: Quant Quiz

Directions : In each of these questions, two equations numbered I and II with variables x and y are given. You have to solve both the equations to find the value of x and y. Give answer
1) if x > y 
2) if x >= y 
3) if x < y
4) if x <= y 
5) if x = y or relationship between x and y cannot be determined.

1. 
I. x^2 – 26x + 169 = 0 
II. y^2 – 28y + 196 = 0


2. 
I. 2x^2 + 6x – 36 = 0 
II. 2y^2 + 25y + 78 = 0

3. 
I. 7x^2 + 104x + 169 = 0 
II. y^2 + 10y – 39 = 0

4. 
I. x^2 – 6x – 16 = 0 
II. 3y^2 + 22y + 24 = 0

Directions: Each question below is followed by two statements A and B. You have to determine whether the data given in the statement is sufficient to answer the question. You should use the data and your knowledge of mathematics to choose between the possible answers. Give answer:

1) if the statement A alone is sufficient to answer the question, but the statement B alone is not sufficient.
2) if the statement B alone is sufficient to answer the question, but the statement A alone is not sufficient.
3) if the both statement A and B together are needed to answer the question.
4) if either the statement A alone or B alone is sufficient to answer the question.
5) if you cannot get the answer from the statements A and B together, but needed even more data.

5. What is the principal?
A. The simple interest earned for 8 year at the rate of 10% p.a is Rs. 960.
B. The simple interest is Rs. 480 for 4 year.

6. The sum of the ages of P, Q, R and S is 86 year, what is Q's present age?
A. The average age of P, Q and S is 24 year.
B. The average age of R and S is 28 year.

7. In how many days, can Q alone complete the work?
A. P does 3/5 of a work in 21 days.
B. Q does 3/4 of a work in 18 days.

8. Find the number of ways in which 10 players out of 14 players can be selected such that 3 particular player are always included and 2 particular players are always excluded?
1) 38 
2) 36 
3) 48 
4) 40 
5) 32

9. The ratio of the present age of a mother and daughter is 7 : 3. Four years ago the ratio of their ages was 5 : 2. What will be the mother's age five years from now? (in years)
1) 89 
2) 86 
3) 87 
4) 84 
5) 88

10. The sum of nine consecutive odd number of set A is 657. What is the sum of seven consecutive odd numbers whose lowest number is 18 more than the lowest number of set A?
1) 620 
2) 625 
3) 635 
4) 615 
5) 623



1. 3;
I. x^2 – 26x + 169 = 0
(x – 13)^2 = 0
x = 13
II. y^2 – 28y + 196 = 0
(y – 14)^2 = 0
y = 14
x < y

2. 2;
I. 2x^2 + 12x – 6x – 36 = 0
2x (x + 6) –6 (x + 6) = 0
(x + 6) (2x – 6) = 0
x = –6, 3

II. 2y^2 + 12y + 13y + 78 = 0
2y (y + 6) + 13 (y + 6) = 0
(y + 6 ) (2y + 13) = 0
y = –6, -13/2
x > y

3. 5;
I. 7x^2 + 13x + 91x + 169 = 0
(x + 13) (7x + 13) = 0
x = – 13, -13/7

II. y^2 + 13y – 3y – 39 = 0
(y – 3) (y + 13) = 0
y = 3, – 13
Relationship between x and y cannot be established.

4. 5;
I. x^2 – 8x + 2x – 16 = 0
(x + 2) (x – 8) = 0
x = – 2, 8
II. 3y^2 + 22y + 24 = 0
3y 2 + 4y + 18y + 24 = 0
(y + 6) (3y + 4) = 0
y = –6, -4/3
Relationship between x and y cannot be established.

5. 1;
From A, we get,
P = 100*SI/R*T = Rs. 1200

5. 5;
P + Q + R + S = 86
From A, P + Q + S = 24 × 3 = 72 R = 86 – 72 = 14

From B, R + S = 28 × 2 = 56
We cannot get Q’s present age from both statements.

6. 2;
From statement B, work done by Q = 18 × 4/3 = 24 days.

7. 2;
Out of 14 players, let us element 2 particular player
which are excluded. Now, there are 12 players for selection of these 12, three have to be included in team always. Thus remaining players are (12 – 3) = 9 and the required players for team (10 – 3) = 7. Now,
selection cannot be done in
9C7 ways.

9C7 = 9!/7!2! = 36 ways.

8. 1;
Let the present age of mother and daughter be 7x and 3x year.
7x – 4/3x – 4 = 5/2
14x – 8 = 15x – 20
x = 12

Mother’s age five years from now = 12 × 7 + 5
= 84 + 5 = 89 year

9. 5;
In Set A, 9x + 2(1 + 2 + ..... + 8) = 657
9x + 8*9*2/2 = = 657
9x + 36 × 2 = 657
x = 65

lowest number of another set = 65 + 18 = 83

Sum of seven consecutive number of another set
= 7x + (1 + 2 + 3 + 4 + 5 + 6) × 2
= 7x + 6*7*2/2
= 7x + 21 = 83 × 7 + 42 = 623

10. 4;
First arrange the 7 boys in a row in 7P7 = 7! ways.
Then, there are 8 gaps between them. Now 5 girls can be
arranged in 8 gaps in P5
8 ways.
Required number of ways of arrangement = 7! × 8P5 = 7 ! * 8 !/5!


= 8! × 7 × 6 = 42 × 8 !



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