WBBSE Solutions For Class 8 Maths Algebra Chapter 1 Commutative Associative and Distributive Laws
Commutative Associate and Distributive Introduction
In class 7 you have studied commutative, associative, and distributive laws, the application of four basic operations on polynomials, the deduction of some formulae, simple factorization of quadratic expressions, and the formation of linear equations of one variable and their solution. Here we shall discuss those topics in brief for recapitulation.
Commutative, Associative, and Distributive laws
We know that if a and b are any two integers then
1. The commutative law of addition is, a+ b = b + a
2. The commutative law of multiplication is, a x b = b x a
3. Associative law of addition is, (a+ b) + c = a + (b + c)
4. The associative law of multiplication is, (a x b) x c = a x (b x c)
5. The distributive law of multiplication is
1. (a + b) x c = a x c + b x c
2. ax (b + c) = ax b + ax c
3. (a – b) x c = a x c – b x c
4. a x (b – c) – ax b~ ax c
6. The distributive law of division is
1. (a + b)÷ c = a + c + b ÷ c
2. (a-b) ÷ c = a + c-b ÷ c
But c ÷ (a + b) ≠ c + a + c ÷ b
and c ÷ (a-b) ≠ c + a- c ÷ b
Read And Learn More WBBSE Solutions For Class 8 Maths
Algebra Chapter 1 Commutative Associative and Distributive Laws Some problems with Commutative, Associative, and Distributive laws
Example 1
Rewrite 4 + 7 + 9 in two different ways such that a single bracket is used in each case.
Solution:
Given 4 + 7 + 9
Given Sum In Two Different Ways Such That A Single Bracket Is Used In Each Case:-
4 + 7 + 9 = (4+ 7)+ 9 and 4 + 7 + 9 = 4 + (7 + 9)
Example 2
Add the sum of 6 and 8 with 5 and find the result.
Solution :
Given Sum Of 6 And 8 With 5.
5 + (6 + 8)
= 5+ 14
= 19
The Result Is 19.
WBBSE Class 8 Commutative Law Notes
Example 3
Prove that: (a + b + c) x = ax + bx + cx.
Solution :
Given (a + b + c) x
(a + b + c)x
= (a + d)x [assuming b + c = d]
= ax + dx [by distributive law]
= ax + (b + c)x [putting d = b + c]
= ax + bx + cx (Proved).
(a + b + c) x = ax + bx + cx.
Example 4
Simplify a/ a-b + b / b-4.
Solution :
The given expression = a / a-b + b/b-a
= a / a-b + b / – (a-b)
= a/ a-b – b/ a-b
= a-b / a-b
= 1
a / a-b + b/b-a = 1
Example 5
Simplify : -2-[-2-{-2-(2-3-2)}].
Solution :
The given expression = – 2-[- 2-{- 2-(2-3-~2)}]
= – 2 -[-2 – {- 2 – (2 – 1)}]
= – 2 – [- 2 – {- 2 – 1}]
= – 2 – [- 2 – {- 3}]
= – 2 – [- 2 + 3]
= -2 – [1]
= – 2 – 1
= – 3
– 2-[- 2-{- 2-(2-3-~2)}] = – 3
Example 6
Simplify : x-[y + {x-(y-x-y)}]
Solution :
The given expression = x–[y + {x-{y-x-y)}]
= x-\y + {x -(y – x + y)}]
= x-\y+{x-{2y- *)}]
= x-[y + {x-2y + x}]
-x-{y + {2x- 2y}]
= x-\y + 2x-2y]
= x-[2 x-y]
= x-2x + y
= y-x
x–[y + {x-{y-x-y)}] = y-x
Polynomial
If there are many terms in an algebraic expression then, it is called a polynomial. For example, a + 5b – 8c + 9d + 7e + 8f – 11x is a polynomial.
If there is only one term in an expression, then it is called a monomial.
For example, 5x is a monomial.
If there are two or three terms, then they are called binomial and trinomial respectively.
For example, 5a + 6b is a binomial and 7a – 12b + 8c is a trinomial.
Example 1
Find the sum of 4x2, – 3x², 7x2.
Solution :
The required sum = 4x2, – 3x², 7x2
= (4x2 + 7x2) – 3x2
= 11x2 – 3x2
= 8x2
4x2, – 3x², 7x2 = 8x2
Example 2
Find the sum of : – 2xy, 5xy, 9.xy and -7xy.
Solution :
Here, the required sum
= – 2xy + 5xy + 9xy – 7xy
= (5xy + 9xy) – (2xy + 7xy)
= 14xy – 9xy
= 5 xy
– 2xy + 5xy + 9xy – 7xy = 5 xy
Understanding Associative Law in Algebra
Example 3
Find the sum of the following expressions:
5x – 8y + z,
– 2x + 7y – 5z,
3s + 5y + 3z.
Solution :
\(\begin{array}{r}5 x-8 y+z \\
-2 x+7 y-5 z \\
3 x+5 y+3 z \\
\hline 6 x+4 y-z \\
\hline
\end{array}\)
The required sum = 6x+4y-z
Example 4
Simplify : 5a2 + 2d2 + 2ab + 3a2 – 6b2 – 5ab + 3a.
Solution :
The given expression
5a2 + 2d2 + 2ab + 3a2 – 6b2 – 5ab + 3a
= 5a2 + 2b2 + 2ab + 3a2 – 6b2 – 5ab + 3a
= (5a2 + 3a2) + (2 b2 – 6 b2) + (2ab – 5ab) + 3a
= 8a2 – 4b2 – 3ab + 3a
5a2 + 2d2 + 2ab + 3a2 – 6b2 – 5ab + 3a. = 8a2 – 4b2 – 3ab + 3a
Subtraction
The quantity from which the subtraction is to be made is called minuend. The quantity which is to be subtracted is called subtrahend. The result obtained after subtraction is called the difference or remainder. By subtraction of b from a, we mean the addition of the negative of b with a. It means that a – b = a + (- b).
Example 1
Subtract 5xy from 12xy.
Solution :
Given 5xy And 12xy
12xy – 5xy = 7xy
Example 2
Subtract – 9xy from 25xy
Solution:
25xy – (- 9xy) = 25xy + 9xy = 34xy.
Example 3
Subtract 5a2 + 4b2 + 2c2 from 7a2 – 3b2 + 8c2.
Solution :
Given 5a2 + 4b2 + 2c2 And 7a2 – 3b2 + 8c2.
\(\begin{array}{r}7 a^2-3 b^2+8 c^2 \\
5 a^2+4 b^2+2 c^2 \\
-\quad- \\
\hline 2 a^2-7 b^2+6 c^2 \\
\hline
\end{array}\)
Example 4
What is to be added with a2 – ab + 2b2 to get a2 + b2?
Solution :
The required expression = (a2 + b2) – (a2 – ab + 2b2)
= a2 + b2 – a2 + ab – 2 b2
= (a2 – a2) + (b2 – 2b2) + ab
= – b2 + ab
(a2 + b2) – (a2 – ab + 2b2) = – b2 + ab
Multiplication
In the case of algebraic multiplication of two quantities, the sign of the product is V when the two quantities are of the same sign and the sign of the product is ‘—’ when the two quantities are of opposite signs.
This may be explained in brief as follows :
(+ x) x (+ y) = + xy
(+ x) x (- y) = – xy
(- x) x (+ y) = -xy
(- x) x (- y) = + xy
In case of finding the product of the same variable having different powers, we follow the rule xm x xn = xm+n
Example 1
Multiply : 7x3 by 2x4
Solution :
Given 7x3 by 2x4
7x3 x 2x4 = 7 x 2 x x3 x x4
= 14 x x3+4
= 14 x x7
= 14x7
7x3 x 2x4 = 14x7
Example 2
Find the product ;
(- 5p2q) x (3pq2) x (- 2p²q²).
Solution :
Given
(- 5p2q) x (3pq2) x (- 2p²q²)
The required product = (- 5p2q) x (3pq2) x (- 2p2q2)
= (- 5) x 3 x (-2) x p2+1+2 x q1+2+2
= 30p5q5
(- 5p2q) x (3pq2) x (- 2p2q2) = 30p5q5
Example 3
Multiply : – 3p2q5r by – 7p3g2r5.
Solution :
Given – 3p2q5r And – 7p3g2r5
The required product = (- 3p2g5r) x (- 7p3q2r5)
= (- 3) x (- 7) x p2+3 x g5+2 x r1+5
= 21p5q7r6
(- 3p2g5r) x (- 7p3q2r5) = 21p5q7r6
Distributive Law Explained with Examples
Example 4
Find the product: (1/2 xy²) x (- 2/3 xy4) x (3x³y).
Solution :
Given
(1/2 xy²) x (- 2/3 xy4) x (3x³y)
= (1/2 xy²) x (- 2/3 xy4) x (3x³y).
= 1/2 x (- 2/3) x 3 x x1+1+3 x y2+4+1
= (-1) x x² x y7
= -x5y7
(1/2 xy²) x (- 2/3 xy4) x (3x³y). = -x5y7
Example 5
Multiply :7a +3b by 2a – b.
Solution :
Given 7a +3b by 2a – b
\(\begin{aligned}& 7 a+3 b \\
& 2 a-b \\
& \hline 14 a^2+6 a b \\
& \quad-7 a b-3 b^2 \\
& \hline 14 a^2-a b-3 b^2 \\
& \hline
\end{aligned}\)
The required product = 14a²-ab-3b²
Division
In the case of algebraic division of the form
a/b = c, we call a dividend, b the divisor, and c the quotient.
If the dividend and the divisor are of the same sign, then the sign of the quotient will be + and if the dividend and the divisor are of opposite signs the sign of the quotient will be -.
This may be explained in brief as follows :
\(\frac{(+x)}{(+y)}=+\frac{x}{y}\) \(\frac{(+x)}{(-y)}=-\frac{x}{y}\) \(\frac{(-x)}{(+y)}=-\frac{x}{y}\) \(\frac{(-x)}{(-y)}=+\frac{x}{y}\)In the case of finding the quotient of the same variable having different powers, we follow the rule xm ÷ xn = xm+n.
We shall also take x° = 1.
Example 1
Divide 35a4b8 by 5a2b2.
Solution :
Given 35a4b8 And 5a2b2
=35a4b8 / 5a2b2
= 7a4-2b8-2
= 7a²b6
35a4b8 / 5a2b2 = 7a²b6
Example 2
Divide (- 81m5n6) by (- 27m2n2).
Solution :
Given (- 81m5n6) And (- 27m2n2)
The required quotient
=(- 81m5n6) / (- 27m2n2).
= 3m5-2n6-2
= 3m³n4
(- 81m5n6) / (- 27m2n2) = 3m³n4
Example 3
Divide 4x5 + 3x4 + 8x3 + 7x by x2.
Solution :
The required quotient 4x5 + 3x4 + 8x3 + 7x2 / x²
\(=\frac{4 x^5+3 x^4+8 x^3+7 x^2}{x^2}=\frac{4 x^5}{x^2}+\frac{3 x^4}{x^2}+\frac{8 x^3}{x^2}+\frac{7 x^2}{x^2}\) \(=4 x^{5-2}+3 x^{4-2}+8 x^{3-2}+7 x^{2-2}\) \(=4 x^3+3 x^2+8 x^1+7 x^0\) \(=4 x^3+3 x^2+8 x+7\)Example 4
Divide 13m²n4 + 16m³n³-20m4n² by 4m²n².
Solution:
Given 13m²n4 + 16m³n³-20m4n² And 4m²n²
= \(\frac{12 m^2 n^4+16 m^3 n^3-20 m^4 n^2}{4 m^2 n^2}\)
= \(\frac{12 m^2 n^4}{4 m^2 n^2}+\frac{16 m^3 n^3}{4 m^2 n^2}-\frac{20 m^4 n^2}{4 m^2 n^2}\)
= \(3 n^2+4 m n-5 m^2\)
Some important formulae
In the previous class, you learned the following three formulae :
1. (a + b)2= a2 + 2ab + b2.
2. (a – b)2– a2 – 2ab + b2.
3. a2– b2 – (a + b) (a – b).
Also using the first two formulae the following five formulae can be established:
1. a2+ b2 = (a + b)2 – 2ab
2. a2+ b2 = (a – b)2 + 2ab
3. (a + b)2= (a – b)2 + 4ab
4. (a – b)2= (a + b)2 – 4ab.
5. ab = (a+b / 2)² – (a-b / 2)²
Algebra Chapter 1 Commutative Associative and Distributive Laws Some Examples
Example 1
Find the square of (a + 2b).
Solution :
Given (a + 2b):-
Square of (a + 2b).
= (a + 2b)2
= (a)2 + 2.a.2b + (2b)2
= a2+ 4 ab+ 4 b2
Square of (a + 2b). = a2+ 4 ab+ 4 b2
Example 2
Find the square of 101.
Solution :
Given 101
Square of 101 = (101)2
= (100 + 1)2
= (100)2 + 2 x 100 x 1 + (1)2
= 10000 + 200 + 1
= 10201
Square of 101 = 10201
Example 3
Simplify: (7x + 4y)2 – 2(7x+4y)(7x-4y)+(7x-4y)2.
Solution:
Given (7x + 4y)2 – 2(7x+4y)(7x-4y)+(7x-4y)2.
Let, 7x+4y = a and 7x – 4y = b.
Hence, the given expression
\(=a^2-2 a b+b^2\) \(=(a-b)^2\) \(=\{(7 x+4 y)-(7 x-4 y)\}^2\) \(=(7 x+4 y-7 x+4 y)^2\) \(=(8 y)^2=64 y^2\)Example 4
Simplify : 0.82 x 0.82 + 2 x 0.82 x 0.18 + 0.18 x 0.18.
Solution :
Given 0.82 x 0.82 + 2 x 0.82 x 0.18 + 0.18 x 0.18.
Let, a = 0.82 and b = 0.18
Then the given expression
=a x a + 2 x a x b + b x b
= a2 + 2 ab + b2
= (a + b)2
= (0.82 + 0.18)2
= (1.00)2
= (1)2
= 1
0.82 x 0.82 + 2 x 0.82 x 0.18 + 0.18 x 0.18 = 1
Short Notes on Properties of Operations in Algebra
Example 5
Find the square of x + 2y – 3z.
Solution :
Given x + 2y – 3z.
Square of x + 2y – 3z
= (x + 2y – 3s)2
= (x + 2y)2 – 2x(;c + 2y)x3z + (32)2
= (x)2 + 2x x 2y + (2y)2 – 6z(x + 2y) + 9z²
= x2 + 4xy + 4y² – 6zx – 12yz + 9Z2
= x2 + 4y2 + 9Z2 + 4xy – 12yz – 6zx
x + 2y – 3z = x2 + 4y2 + 9Z2 + 4xy – 12yz – 6zx
Example 6
If x + y = 2 and x-y= 1, then find the value of 8xy (x² + y²).
Solution :
Given x + y = 2 and x-y= 1.
8xy(x² + y2) = 4xy x 2(x² + y²)
= {(x² + y)2 -(x- y)2} {(x + y)2 + (x- y)2}
= {(2)2 — (1)2}{(2)2 + (1)2}
= (4 – 1) (4 + 1)
= 3 x 5
= 15
8xy(x² + y2) = 15
Example 7
If 2x + 3y = 9 and xy = 3, find the value of 4x2 + 9y².
Solution :
Given 2x + 3y = 9 And xy = 3.
4x2 + 9y²
= (2x)2 + (3y)2
= (2x + 3y)2 – 2.2x3y
= (2x + 3y)2 – 12xy
= (9)2 – 12.3
= 81-36
=45
4x2 + 9y² =45
Example 8
Express 35 as the difference between two squares.
Solution :
Given Number 35:-
35 = 7×5
= (7+5 / 2)² – (7-5 / 2)2
= (12/2)2 – (2/2)2
= (6)2 – (1)2
35 = (6)2 – (1)2
Key Differences Between Commutative and Associative Laws
Expression For The Difference Between Two Squares 35 = (6)2 – (1)2.
Example 9
Find the continued product of (a + b), (a-b), (a2 + b2), (a4 + b4).
Solution :
The required product = (a + b) (a – b) (a2 + b2) (a4 + b4)
= (a2 – b²) (a2 + b2) (a4 + b4)
= {(a2)2 – (b2)2} (a4 + b4)
= (a4 – b4) (a4 + b4)
= (a4)2 – (b4)2
= a8 – b8
(a + b) (a – b) (a2 + b2) (a4 + b4) = a8 – b8
Example 10
Express as the product of two expressions: a2 – 4ab + 4b2 – 4.
Solution :
Given a2 – 4ab + 4b2 – 4.
a2 – 4ab + 4b2 – 4
= (a)2 – 2 x a x 2b + (2b)2 – 4
= (a – 2b)² – (2)2
= (a – 2b + 2) (a – 2b – 2)
a2 – 4ab + 4b2 – 4 = (a – 2b + 2) (a – 2b – 2)
Factor
If the product of two or more expressions is equal to another expression, then those expressions are called the factors of the product.
For example: If p x q x r = x, then the expressions p, q, and r are called the factors of x. Therefore, by factorization of the expression x, three factors p, q, and r are obtained.
Different methods of factorization
1. If a polynomial contains one or more common factors in each of its terms then the common factor (or factors) are taken outside the bracket according to the distributive law and the remaining portion is kept inside the bracket.
For Example: a2b + ab + ab2
= ab(a + 1 + b)
= a x b x (a + b + 1).
x3y2 + x2y3 = x2y2(x+y)
= x x x x y x y x (x+y).
a2b + ab + ab2 = x x x x y x y x (x+y).
2. Some quadratic expressions may be factorized by using the formulae ; (a + b)2 = a2+ 2ab + b2, (a – b)2 = a2 – 2ab + b2.
For Example: x2 + 4xyz + 4y2z2
Given (a + b)2 = a2+ 2ab + b2, (a – b)2
= (x)2 + 2x x 2yz + (2yz)2
= (x + 2yz)2
= (x + 2yz) (x+ 2yz).
Again, 4a2 – 12ab + 9b2
= (2a)2 – 2 x 2a x 3b + (3b)2
= (2a – 3b)2
= (2a – 3b) (2a – 3b).
(a + b)2 = a2+ 2ab + b2, (a – b)2 = a2 – 2ab + b2 = (2a – 3b) (2a – 3b).
3. Applying the formula a2 – b2 = (a + b) (a – b), some quadratic expressions may be factorized.
For Example 25X2 – 81y2
= (5x)2 – (9y)2
= (5x + 9y) (5x – 9y).
25X2 – 81y2 = (5x + 9y) (5x – 9y).
4. Some expressions may be factorized by simultaneous application of the formulae of (a +b)2 [or (a – b)2] and a2 – b2.
For Example a4+ 4
= (a2)2 + (2)2
= (a2)2 + 2 x a2 x 2 + (2)2 – 4a2
= (a2 + 2)2 – (2a)2
= (a2 + 2 + 2a) (a2 + 2 – 2a)
= (a2 + 2a +2) (a2 – 2a + 2).
a4+ 4 = (a2 + 2a +2) (a2 – 2a + 2).
Algebra Chapter 1 Commutative Associative and Distributive Laws Some examples of factorization
Example 1
Factorize : 4a4b- 6a3b2 + 12a2b³
Solution:
Given
4a4b- 6a3b2 + 12a2b³
4a4b – 6a3 b2+ 12a2 + b3
= 2a2b(2a2 – 3ab + 6b2)
4a4b- 6a3b2 + 12a2b³ = 2a2b(2a2 – 3ab + 6b2)
Example 2
Factorize : x2 – (a + b)x + ab.
Solution:
Given
x2 – (a + b)x + ab.
x2 -(a +b)x+ ab
= x2 – ax – bx + ab
= x(x – a) – b(x – a)
= (x – a) (x – b)
x2 -(a +b)x+ ab = (x – a) (x – b)
Examples of Distributive Law in Real Life
Example 3
Factorize : 9(x – y)2 – 25(y – z)2.
Solution :
Given
9(x – y)2 – 25(y – z)2.
9(x – y)2 – 25(y – z)2
= {3(x – y)}2 – {5(y – z)}2
= (3x – 3y)2 – (5y – 5z)2
= {(3x – 3y) + (5y – 5z)} {(3x – 3y) – (5y – 5z)}
= (3x – 3y + 5y – 5z) (3x – 3y – 5y + 5z)
= (3x + 2y – 5z) (3x – 8y + 5z)
9(x – y)2 – 25(y – z)2 = (3x + 2y – 5z) (3x – 8y + 5z)
Example 4
Factorize : x4 + x2y2 + y4.
Solution :
Given
x4 + x2y2 + y4.
x4 + x2y2 + y4
= (x2)2 + 2.x2.y2 + (y2)2 – x2y2
= (x2 + y2)2 – (xy)2
= (x2 + y2 + xy) (x2 + y2 – xy)
x4 + x2y2 + y4 = (x2 + y2 + xy) (x2 + y2 – xy)
Example 5
Resolve into factors : x2 – y2 – 6xa+ 2ya + 8a².
Solution :
Given
x2 – y2 – 6xa+ 2ya + 8a².
x2 – y2 – 6xa + 2ya + 8a2
= x2 — 6xa + 9a2 – y2 + 2ya – a2
= x2 – 2.x. 3a + (3a)2 – (y2 – 2ya + a2)
= (x – 3a)2 – (y – a)2
= {(x – 3a) + (y – a)} {(x – 3a) – (y – a)}
= (x – 3a + y – a) (x – 3a – y + a)
= (x + y – 4a) (x – y – 2a)
x2 – y2 – 6xa + 2ya + 8a2 = (x + y – 4a) (x – y – 2a)
Example 6
Three factors of an expression are a ,a+1/a, and a-1/a; find the expression.
Solution:
Given
Three factors of an expression are a ,a+1/a, and a-1/a
The required expression= a(a+1/a)(a-1/a)
= a(a²-1/a²)
= a³-1/a
Linear equations of a single variable
A linear equation is one in which the power of the variable is one. In the previous class, you learned the method of finding solutions of linear equations of one variable. The procedure followed in solving such an equation can be expressed in brief:
1. If x + a = b, then x = b
2. If x- a = b, then x = a + b.
3. If ax = b, then x = b/a
4. If x/a = b, then x = ab.
Algebra Chapter 1 Commutative Associative and Distributive Laws Some examples of equation
Example 1
Solve: 1/2(x+1) + 1/3(x+2) = 13(x-2) + 4(13-x).
Solution :
Given 1/2(x+1) + 1/3(x+2) = 13(x-2) + 4(13-x).
16 – 5(7x – 2) = 13(x – 2) + 4(13 – x)
or, 16 – 35x + 10 = 13x – 26 + 52 – 4x
or, 26 – 35x = 26 + 9x
or,-35x – 9x = 26-26
or, x = 0 / -44
∴ x = 0
1/2(x+1) + 1/3(x+2) = 13(x-2) + 4(13-x) = 0
Example 2
Solve : 3(x – 1) – (x + 2) = x + 2(x – 1).
Solution :
Given 3(x – 1) – (x + 2) = x + 2(x – 1).
3(x – 1) – (x + 2) = x + 2(x – 1)
or, 3x – 3 – x – 2 = x + 2x – 2
or, 2x – 5 = 3x – 2
or, 2x – 3x = 5 – 2
or, – x = 3
or, x = – 3
WBBSE Class 8 Algebra Chapter Summary
Example 3
Solve: 1/2(x + l) + 1/3(x + 2) + 1/4(x + 3) = 16.
Solution :
Given 1/2(x + l) + 1/3(x + 2) + 1/4(x + 3) = 16.
Multiplying both sides of the given equation by 12 we get,
6(x + 1) + 4(x + 2) + 3(x + 3) = 192
or, 6x + 6 + 4x + 8 + 3x + 9 = 192
or, 13x + 23 = 192 or, 13* = 192 – 23
or, 13x = 169
or, x = 169/13
∴ x = 13
Example 4
Solve: ax/b – bx/a = a²-b².
Solution :
Given ax/b – bx/a = a²-b².
\(\frac{a x}{b}-\frac{b x}{a}=a^2-b^2\) \(\text { or, } x\left(\frac{a}{b}-\frac{b}{a}\right)=a^2-b^2\) \(\text { or, } x\left(\frac{a^2-b^2}{a b}\right)=a^2-b^2\) \(\text { or, } x=a^2-b^2 \times \frac{a b}{a^2-b^2}\) \(\text { or, } x=a b\)Example 5
Solve: x+4 / 5 + x+3 / 4 = x + 11/6.
Solution :
Given x+4 / 5 + x+3 / 4 = x + 11/6.
Multiplying both sides of the given equation by 60 we get,
12(x+4) + 15(x+3) = 10(x+11)
or, 12x + 48 + 15x + 45 = 10x + 110
or, 27x + 93 = 10x + 110
or, 27x -10 x = 110 – 93
or, 17x = 17
or, x = 17/17
= 1
x = 1
Example 6
Of the total number of fruits with a fruit vendor, 1/5 was mango, 1/4 was an apple, 2/5 was lichi and the rest 60 were oranges. Find the total number of fruits with the vendor.
Solution :
Given
Total Number Of Fruits With A Fruit Vendor Are
1/5 Was Mango,
1/4 Was An Apple,
2/5 Was Lichi, And
Rest 60 Were Oranges.
Let the total number of fruits be x.
∴ x/5 + x/4 + 2x/5 + 60 = x
or, 4x + 5x + 8x / 20= x – 60
or, 17x = 20x – 1200
or, 17x – 20x = -1200
or, -3x = -1200
or, 3x = 1200
or, x = 1200/3
= 400
The total number of fruits is 400.
Example 7
Half of a number is greater than 1/5th of it by 6. Find the number.
Solution:
Given
Half of a number is greater than 1/5th of it by 6.
Let, the number be x.
.’. Half of the number =x/2 and 1/5th of the number = x/5.
According to the question,
x/2 = x/5 + 6
or, x/2 – x/5 = 6
or, 5x – 2x /10 = 6
or, 3x/10 = 6
or, x = 6 x 10/3
= 20
The number is 20.
Example 8
The present age of the father is 7 times that of the son. After 10 years, the age of the father will be 3 times that of the son. Find the present age of the son.
Solution :
Given
The present age of the father is 7 times that of the son.
After 10 years, the age of the father will be 3 times that of the son.
Let, the present age of the son be x years then the present age of the father is 7x years. After 10 years –
age of son will be (x + 10) years and age of father will be (7x + 10) years
According to the question,
7x + 10 = 3(x+10)
or, 7x + 10 = 3x + 30
or, 7x -3x = 30 – 10
or, 4x = 20
or, x = 20/4
= 5
The present age of the son is 5 years.
Conceptual Questions on Algebraic Laws
Example 9
If the sum of three consecutive numbers is 90, then find the numbers.
Solution:
Given The Sum Of Three Consecutive Numbers is 90
Let, the three consecutive numbers be x, x +1, x + 2.
According to the question,
x + x + 1 + x + 2 = 90
or, 3x + 3 = 90
or, 3x = 90 – 3
= 87
or, x = 87/3
= 29
∴ The numbers are:
29, 29+1,29+2, or 29, 30, 31.
The numbers are 29, 30, and 31.
Example 10
1/3rd of the bamboo is with mud, 1/4th of it is in water, and 5 meters above the water. What is the length of the bamboo?
Solution :
Given
1/3rd Of The Bamboo Is With Mud.
1/4th Of It Is In Water And 5 Meters Above The Water.
Let, the length of the bamboo be x meters.
∴ Length of the bamboo within mud = x/3
meters and length of the bamboo within water = x/4 meters.
According to the question,
x – (x/3 + x/4) = 5
or, x – (4x+3x / 12) = 5
or, x – 7x/12 = 5
or, 5x/12 = 5
or, x = 5 x 12/2
= 12
The length of the bamboo is 12 meters.
Example 11
The sum of the digits of a two-digit number is 10. If 18 is subtracted the digits are reversed. What is the number?
Solution:
Given
The sum of the digits of a two-digit number is 10.
If 18 is subtracted the digits are reversed.
Let, the digit in the units place be x then the digit in the tens place is 10 – x.
Therefore, the number is 10 x (10 -x) + x
If the digits are reversed the number = 10x + 10 -x = 9x + 10
According to the question.
100 – 9x – 18
or, x = 4
∴ The required number = 100 – 9 x 4
= 100 – 36
= 64
The number is 64.
Example 12
The denominator of a fraction is 2 more than its numerator and if 3 is added to the numerator and 3 is subtracted from the denominator, then the fraction becomes equal to 7/3. What is the fraction?
Solution :
Given
The denominator of a fraction is 2 more than its numerator and if 3 is added to the numerator and 3 is subtracted from the denominator, then the fraction becomes equal to 7/3.
Let, the numerator of the fraction = x,
then denominator = x + 2
Therefore, the fraction = x / x+2
According to the question,
x+3 / x+2-3 = 7/3
or, x+3 / x-1 = 7/3
or, 7x -7 = 3x+9
or, 7x -3x = 7+9
or, 4x = 16
or, x = 16/4
= 4
∴ The fraction = 4/4+2
= 4/6
The fraction is 4/6
Example 13
The present age of the father is 3 times that of the son. After 15 years the age of the father will be twice the age of the son. What are the present ages of the father and the son?
Solution :
The present age of the father is 3 times that of the son.
After 15 years the age of the father will be twice the age of the son.
Given
Let, the present age of the son be x years. Then the present age of the father is 3* years.
After 15 years age of the son will be (x + 15) years
After 15 years age of the father will be
(3x + 15) years
According to the question,
3x + 15 = 2(x+15)
or, 3x + 15 = 2xx+30
or, x = 15
∴ The present age of the son = is 15 years and the present age of the father = 15 x 3 years = 45 years.
The present age of the son is 15 years and the present age of the father is 45 years.
Example 14
Divide 830 into two parts such that 30% of one part is 4 more than 40% of the other.
Solution :
Given
Given Number 830
We Need To Divide 30% of one part is 4 more than 40% of the other:-
Let, two parts be x and (830 – x).
According to the question,
x x 30/100 = (830 – x ) x 40/100 + 4
or, 3x/10 = 2(830 – x ) / 5 + 4
or, 3x/10 – 1660 – 2x / 5 = 4
or, 3x-3320+4x / 10 = 4
or, 7x – 3320 = 40
or, 7x = 3360
or, x = 3360 / 7
= 480.
∴ One part is 480 and the other part = (830 – 480)
= 350
∴ 480 and 350.
Example 15
The numerator of a fraction is 2 less than its denominator. If 1 is added with the numerator and denominator then the fraction becomes 4/5. Find the fraction.
Solution :
Given
The numerator of a fraction is 2 less than its denominator. If 1 is added with the numerator and denominator then the fraction becomes 4/5.
Let, the denominator of the fraction = x
and numerator = x – 2
Therefore, the fraction = x – 2 / x
According to the question,
x-2+1 / x+1 = 4/5
or, x-1 / x+1 = 4/5
or, 5x – 5 = 4x + 4
or, x=9
∴ The fraction = 9-2 / 9
= 7/9.