Algebra questions: Overview, Questions, Preparation

The branch of mathematics that deals with unknown quantities is known as algebra. In algebra, certain unknown quantities are represented in the form of letters; these are known as variables. An equation that involves variables and the relation between them is referred to as an algebraic equation. The expression that denotes variables is known as an algebraic expression.

Algebra:
Algebraic expressions consist of letters as variables. The term variable means that it has no fixed value. But a variable still denotes a quantity or a number. Therefore, just like any other number, various arithmetic operations are performed on variables. In an algebraic expression- 6x - 2y, 6x and 2y are the terms. Here, x and y are factors, and 6 and 2 are the coefficients.

Some algebraic identities are-
(m + n)² = m² + n² + 2mn
(m - n)² = m² + n² - 2mn
m² - n² = (m + n)(m – n)
m² + n² = (m + n)² – 2mn = (m - n)² + 2mn
m³ + n³ = (m + n)(m² – mn + n²)
m³ – n³ = (m – n)(m² + mn + n²)
(m + n)³ = m³ + 3mn(m+ n) + n³
(m – n)³ = m³ – 3mn(m – n) – n³

The topic- algebra problems is a part of the algebra section in the syllabus of class 10. This section carries a weightage of around 15 marks in the examination.

1: The number of red balls in a bag is three more than four times the number of green balls. If the number of red balls in the bag is 47, find the number of green balls.
Solution:
Let us denote the number of green balls by g.
According to the question, the number of red balls are three more than four times the number of green balls. So,
4g + 3= 47 => 4g= 47-3 => g= 44/4= 11.
Hence, the number of green balls in the bag is 11.

2: If the sum of two consecutive numbers is 41, find the numbers.
Solution:
Let us consider the two consecutive numbers as n and n+1.
According to the question, their sum is 41.
So, n + (n+1)= 41 => 2n= 41- 1 => 2n= 40 => n=20 and n+1= 21.
Hence the two numbers are 20 and 21.

3:The present age of A’s son is ⅓ rd of her present age. If her age was five times her son B’s age five years ago, find their present ages.
Solution:
Given, The present age of A’s son is ⅓ rd of her present age. So, B= ⅓ A.
Also, A - 6= 5( B - 6).
So, 3B- 6= 5x (B -6 ) => B= 12.
Now,  B= ⅓ A => A= 3B => A= 3 x 12= 36.
So, the mother's age is 36 years, and the son’s age is 12 years.
4: Solve (2r +s)².
Solution:
Using the identity, (m + n)2 = m2 + n2 + 2mn.
=> (2r + s)²= (2r)²+ (s)²+ 2(2r)(s)= 4r²+ s²+ 4rs.

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