Algebra Practice Questions Medium Level

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.

Answer key:

1. x = 10
2. x = 144, y = 72
3. ∣x − y∣ = 6
4. n = 93
5. x = −14/5
6. 0.9
7. x = ln(100)/2
8. a³ + b³ = 728​

Practice More -

• Algebra Practice Questions Easy Level
• Algebra Practice Questions Hard Level

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