Algebra Practice Questions Medium Level

Last Updated : 17 Feb, 2025

Algebra questions basically involve modeling word problems into equations and then solving them. Some of the very basic formulae that come in handy while solving algebra questions are :

(a + b) ² = a ² + b ² + 2 a b
(a - b) ² = a ² + b ² - 2 a b
(a + b) ² - (a - b) ² = 4 a b
(a + b) ² + (a - b) ² = 2 (a ² + b ² )
(a² - b² ) = (a + b) (a - b)
(a + b + c) ² = a ² + b ² + c ² + 2 (a b + b c + c a)
(a ³ + b ³ ) = (a + b) (a ² - a b + b ² )
(a ³ - b ³ ) = (a - b) (a ² + a b + b ² )
(a³ + b³ + c³ - 3 a b c) = (a + b + c) (a² + b² + c² - a b - b c - c a)
If a + b + c = 0, then a³ + b³ + c³ = 3 a b c
For a quadratic equation ax² + bx + c = 0, x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Check: Tips & Tricks for Algebra

Solved Questions on Algebra (Medium)

Question 1: Find a number such that when 5 is subtracted from 5 times that number, the result is 4, more than twice the number.

Solution:

Let us consider the number as ‘x’
Then, five times the number will be 5x
And, two times, the number will be 2x
So,
5x – 5 = 2x + 4
5x – 2x = 5 + 4
3x = 9
x = 9/3 = 3

Question 2: The sum of two numbers is 132. If one-third of the smaller exceeds one-sixth of the larger by 8, find the numbers.

Solution :

Let the two numbers be ‘x’ an ‘y’ such that x > y.
=> x + y = 132
and (y/3) = (x/6) + 8
=> x + y = 132
and 2 y – x = 48
=> x = 72 and y = 60

Question 3: The sum of two numbers is 24 and their product is 128. Find the absolute difference of numbers.

Solution:

Let the numbers be ‘x’ and ‘y’. => x + y = 24 and x y = 128
Here, we need to apply the formula (x + y)² – (x – y)²= 4xy
=> (24)²– (x – y)²= 4 x (128)
=> (x – y)²= (24)²– 4 x (128)
=> (x – y)² = 576 – 512
=> (x – y)²= 64
=> |x – y| = 8
Therefore, absolute difference of the two numbers = 8

Question 4: The sum of a two digit number ‘n’ and the number obtained by interchanging digits of n is 88. The difference of the digits of ‘n’ is 4, with the tens place being larger than the units place. Find the number ‘n’.

Solution :

Let the number be ‘xy’, where x and y are single digits.
=> The number is 10x + y
=> Reciprocal of the number = yx = 10y + x
=> Sum = 11 x + 11 y = 11 (x + y) = 88 (given)
=> x + y = 8
Also, we are given that the difference of the digits is 4 and x > y. => x – y = 4
Therefore, x = 6 and y = 2
Thus, the number is 62.

Question 5: (2x-1)/3 – (6x-2)/5 = 1/3

Solution:

We have,
(2x-1)/3 – (6x-2)/5 = 1/3
By taking LCM for 3 and 5, which is 15
((2x - 1) × 5)/15 – ((6x - 2) × 3)/15 = 1/3
(10x – 5)/15 – (18x – 6)/15 = 1/3
(10x – 5 – 18x + 6)/15 = 1/3
(-8x + 1)/15 = 1/3
By using cross-multiplication, we get,
(-8x + 1)3 = 15
-24x + 3 = 15
-24x = 15 – 3
-24x = 12
x = -12/24 = -1/2
Verification
LHS = (2x – 1)/3 – (6x – 2)/5
= [2(-1/2) – 1]/3 – [6(-1/2) – 2]/5
= (- 1 – 1)/3 – (-3 – 2)/5
= – 2/3 – (-5/5)
= -2/3 + 1
= (-2 + 3)/3 = 1/3
RHS

Question 6: Evaluate (2.3a⁵b²) × (1.2a²b²) when a = 1 and b = 0.5

Solution:

Let us simplify the given expression
=2.3a⁵b² × 1.2a²b²
=2.3 × 1.2 × a⁵ × a² × b² × b²
=2.76 × a⁵⁺² × b²⁺²
=2.76a⁷b⁴
Now let us substitute when, a = 1 and b = 0.5
For 2.76 a⁷ b⁴
= 2.76 (1)⁷ (0.5)⁴
= 2.76 × 1 × 0.0025
= 0.1725

Question 7: Solve 3e^x + 6 = 120

Solution:

Given,
3e^x + 6 = 120
3e^x = 120 – 6
3e^x = 114
e^x = 114/3
e^x = 38
x = ln 38

Question 8: If the sum of the square of two real numbers is 74 and their sum is 12. Then the sum of cubes of these two numbers is

Solution:

Let two numbers are a and b
Given, a² + b² = 74
a + b = 12
(a + b)² = a² + b² + 2ab
12² = 74 + 2ab
144 = 74 + 2ab
ab = 35
We get a = 7 and b = 5
Then, a³ + b³ = 7³ + 5³
= 343 + 125
= 468

Practice Problems on Algebra (Medium Level)

Question 1: Find a number such that when 7 is added to 4 times that number, the result is 3 less than five times the number.

Question 2: The difference of two numbers is 72. If half of the larger exceeds one-third of the smaller by 12, find the numbers.

Question 3: The product of two numbers is 144, and their sum is 30. Find the absolute difference of the two numbers.

Question 4: The sum of a two-digit number n and the number obtained by interchanging its digits is 132. The digits of n differ by 6, with the tens place being larger than the units place. Find the number n.

Question 5: Solve: 3x − 2)/4 − (7x + 1)/6 = 1/2

Question 6: Simplify and evaluate (1.5a⁴b³) × (2.4a²b⁵) when a = 2 and b = 0.5.

Question 7: Solve: 4e^2x− 8 =392

Question 8: If the sum of the square of two real numbers is 100 and their sum is 14, find the sum of the cubes of these two numbers.