Algebra: Bank Exam Practice Questions (SBI, IBPS, RRB, PO & Clerk)

Ans: C

Explanation: Let’s use variables to represent the lengths of the sections:
* Section 1: x
* Section 2: y
* Section 3: z
* Section 4: w

We are given the following information:
1. x + y + z + w = 60 (The total length is 60 meters)
2. x = y + 6 (The first section is longer than the second by 6 meters)
3. y = (5/12)z (The second section’s length is 5/12 of the third section’s length)
4. z = 2w + 4 (The third section is 4 meters longer than twice the length of the fourth section)

Now we can solve this system of equations. Let’s substitute equations 2, 3, and 4 into equation 1:
(y + 6) + y + z + w = 60
Substitute y = (5/12)z:
((5/12)z + 6) + (5/12)z + z + w = 60
Substitute z = 2w + 4:
((5/12)(2w + 4) + 6) + (5/12)(2w + 4) + (2w + 4) + w = 60
Simplify:
(10w/12 + 20/12 + 6) + (10w/12 + 20/12) + 2w + 4 + w = 60
(5w/6 + 5/3 + 6) + (5w/6 + 5/3) + 2w + 4 + w = 60
5w/6 + 5/3 + 6 + 5w/6 + 5/3 + 2w + 4 + w = 60
(5w/6 + 5w/6 + 2w + w) + (5/3 + 5/3 + 6 + 4) = 60
(5w/3 + 3w) + (10/3 + 10) = 60
(5w + 9w)/3 + 40/3 = 180/3
14w/3 = 140/3
14w = 140
w = 10

Now solve for z: z = 2w + 4 = 2(10) + 4 = 24
Now solve for y: y = (5/12)z = (5/12)(24) = 10
Now solve for x: x = y + 6 = 10 + 6 = 16

The length of the first section (x) is 16 meters.

Correct Option: C

Ans: B

Explanation: Let’s break down the problem:

1. **Total people:** 40
2. **Boys in each group are equal.** Let’s say the number of boys in group A = x, then number of boys in group B = x. So total boys = 2x.
3. **Group A:** 8 girls, and x boys. So group A has a total of 8 + x people.
4. **Group B:** Has 2 more people than group A. Therefore, group B has (8 + x + 2) = 10 + x people. Also, Group B has x boys.
5. **Total people equation:** (8 + x) + (10 + x) = 40
6. **Solve for x:** 18 + 2x = 40 => 2x = 22 => x = 11 (This is the number of boys in each group)
7. **Find the number of people in each group:**
Group A = 8 girls + 11 boys = 19 people.
Group B = 10 + 11 = 21 people.
8. **Girls in Group B:** Total people in group B is 21 and the number of boys is 11, so number of girls = 21 – 11 = 10 girls.
9. **Compare with 16:** Girls in Group B (10) vs 16.

Quantity A is the number of girls in group B, which is 10.
Quantity B is 16.
10 < 16.

Correct Option: B

Ans: A

Explanation: To find the share of each employee, divide the total sum of money (Rs. 12360) by the number of employees (60). 12360 / 60 = 206.

Ans: B

Explanation: First, solve for x in the equation x² + 13x + 42 = 0. We can factor this as (x + 6)(x + 7) = 0. Therefore, the solutions for x are x = -6 and x = -7.
Next, solve for y in the equation y² + 19y + 88 = 0. We can factor this as (y + 8)(y + 11) = 0. Therefore, the solutions for y are y = -8 and y = -11.
Now, compare the values of x and y:
If x = -6, then x > y (since -6 > -8 and -6 > -11).
If x = -7, then x > y (since -7 > -8 and -7 > -11).
Since x is always greater than y regardless of the pairing of the solutions, the relationship is x > y for all possible solutions.
Correct Option: B

Ans: B

Explanation:
Equation I: x² – 20x + 99 = 0. We can factor this as (x – 9)(x – 11) = 0. Therefore, x = 9 or x = 11.
Equation II: y² – 16y + 63 = 0. We can factor this as (y – 7)(y – 9) = 0. Therefore, y = 7 or y = 9.
Now we compare the values:
– If x = 9, then x = y or x > y (if y=7)
– If x = 11, then x > y (either y=7 or y=9)
Thus, x can be greater than or equal to y. If x = 11, then x > y.

If y = 9 and x = 9, x = y
If x = 11 and y = 9, x>y
If x=9 and y=7, x>y
Correct Option: B

Ans: B

Explanation:
Let’s solve the equations.

Equation I: 7x² – 4x – 51 = 0
We can factor this as (7x + 17)(x – 3) = 0
So, x = -17/7 or x = 3

Equation II: y² – 24y + 143 = 0
This factors as (y – 11)(y – 13) = 0
So, y = 11 or y = 13

Now let’s compare the values:
x = -17/7 ≈ -2.43
x = 3
y = 11
y = 13

Comparing the values:
-2.43 < 11, -2.43 < 13
3 < 11, 3 < 13

Since x can be less than or equal to y in all cases, we should choose an option that captures all possibilities.

Correct Option: B

Ans: C

Explanation:
**Equation I: 3x² – 14x + 15 = 0**
We can factor this quadratic equation:
(3x – 5)(x – 3) = 0
So, the roots are:
x = 5/3 ≈ 1.67
x = 3

**Equation II: 15y² – 34y + 15 = 0**
We can factor this quadratic equation:
(5y – 3)(3y – 5) = 0
So, the roots are:
y = 3/5 = 0.6
y = 5/3 ≈ 1.67

Comparing the values of x and y:
x = 3, y = 0.6 => x > y
x = 3, y = 5/3 => x > y
x = 5/3, y = 0.6 => x > y
x = 5/3, y = 5/3 => x = y
So, x ≥ y in most cases.

Correct Option: C

Ans: A

Explanation: First, solve the quadratic equation I: 2x² – 17x + 36 = 0. We can factor this as (2x – 9)(x – 4) = 0. So, the solutions for x are x = 9/2 = 4.5 and x = 4.

Next, solve the quadratic equation II: y² + 10y + 24 = 0. We can factor this as (y + 6)(y + 4) = 0. So, the solutions for y are y = -6 and y = -4.

Now, compare the values:
x can be 4.5 or 4.
y can be -6 or -4.
When x = 4.5, x > y.
When x = 4, x > y.
Therefore, x is always greater than y.

Correct Option: A

Ans: A

Explanation: Let’s use variables:
* Let ‘x’ be the unit five years ago.
* Let ‘y’ be the unit in ten years.
Five years ago:
* Father’s age: 7x
* Son’s age: 2x

In ten years (15 years from five years ago):
* Father’s age: 7x + 15
* Son’s age: 2x + 15
In ten years (from now):
* Father’s age: 2y
* Son’s age: y
Therefore:
* 7x + 15 = 2y — (1)
* 2x + 15 = y — (2)

Substitute equation (2) into (1):
7x + 15 = 2(2x + 15)
7x + 15 = 4x + 30
3x = 15
x = 5

Now, find the father’s age five years ago: 7x = 7 * 5 = 35.
The father’s current age is 35 + 5 = 40.

Correct Option: A

Ans: C

Explanation: First, solve the quadratic equations.

For equation (1):
$x^2 – 14x + 48 = 0$
Factoring: $(x – 6)(x – 8) = 0$
Solutions for x: $x = 6$ or $x = 8$

For equation (2):
$y^2 – 18y + 80 = 0$
Factoring: $(y – 8)(y – 10) = 0$
Solutions for y: $y = 8$ or $y = 10$

Now, compare the values of x and y:
– If $x = 6$ and $y = 8$, then $x < y$
– If $x = 6$ and $y = 10$, then $x < y$
– If $x = 8$ and $y = 8$, then $x = y$
– If $x = 8$ and $y = 10$, then $x < y$

The possible relationships between x and y are x < y or x = y.

Ans: E

Explanation: First, solve Equation I: x² – 208 = 233 => x² = 441 => x = ±21. Second, solve Equation II: y² + 47 – 371 = 0 => y² = 324 => y = ±18. Comparing the values, we have: x can be 21 or -21, and y can be 18 or -18. If x = 21, then x > y. If x = -21, then x < y. Therefore, we cannot establish a definitive relationship between x and y.

Correct Option: E

Ans: B

Explanation:
Solve equation I: x^2 – 5x + 6 = 0. Factoring, we get (x-2)(x-3) = 0. So, x = 2 or x = 3.

Solve equation II: y^2 – 7y + 12 = 0. Factoring, we get (y-3)(y-4) = 0. So, y = 3 or y = 4.

Now compare the values:
If x=2, and y=3 or 4, then x < y.
If x=3, and y=3 or 4, then x = y or x < y.

Since x can be less than, or equal to y, but is never greater than y, the relationship is x ≤ y. However, since the possible solutions for x don’t always create x ≤ y, we need to compare each option.
x=2, y=3: x < y
x=2, y=4: x < y
x=3, y=3: x = y
x=3, y=4: x < y
The only option that holds true for all situations is x ≤ y, however, option E has the least ambiguous meaning to solve the question.

Ans: E

Explanation:
Solve Equation I: x² – 13x + 40 = 0
This factors to (x – 8)(x – 5) = 0. So, x = 8 or x = 5.

Solve Equation II: y² – 13y + 42 = 0
This factors to (y – 7)(y – 6) = 0. So, y = 7 or y = 6.

Comparing the solutions:
If x = 8, then x > y (either 7 or 6).
If x = 5, then x < y (either 7 or 6).
Since we cannot definitively say x is always greater than y or less than y, or equal, we need to compare the solutions. The possible x values are 5 and 8 and the possible y values are 6 and 7. The smallest x (5) is less than both y values, and the largest x (8) is greater than both y values. Since there is not a consistent comparison across all possible values, the answer is E.

Correct Option: E

Ans: C

Explanation: First, solve Equation I: 2x² + 5x + 3 = 0. This factors to (2x + 3)(x + 1) = 0. Therefore, x = -3/2 or x = -1.
Next, solve Equation II: 2y² – 7y + 6 = 0. This factors to (2y – 3)(y – 2) = 0. Therefore, y = 3/2 or y = 2.
Now, compare the values:
– x = -3/2 = -1.5, and y can be 1.5 or 2. Thus x < y in these cases.
– x = -1, and y can be 1.5 or 2. Thus x < y in these cases.
Since x is always less than y in all the possible scenarios, we conclude that x < y.

Correct Option: C