ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 8 Algebraic Expressions Check Your Progress

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 8 Algebraic Expressions Check Your Progress

Question 1.
Consider the expression \(\frac { 3 }{ 2 }\) x2y – \(\frac { 1 }{ 2 }\) xy2 + 6x2y2.
(i) How many terms are there? What do you call such an expression?
(ii) List out the terms.
(iii) In the term \(\frac { -1 }{ 2 }\) xy2, write down the numerical coefficient and the literal coefficient.
(iv) In the term \(\frac { -1 }{ 2 }\) xy2, what is the coefficient of x?
Solution:
\(\frac { 3 }{ 2 }\) x2y – \(\frac { 1 }{ 2 }\) xy2 + 6x2y2
(i) It has 3 terms : Trinomial
(ii) \(\frac { 3 }{ 2 }\) x2y, \(\frac { -1 }{ 2 }\) xy2, 6x2y2
(iii) In \(\frac { -1 }{ 2 }\) xy2,
numerical coefficient = \(\frac { -1 }{ 2 }\)
Literal coefficient = xy2
(iv) In the term \(\frac { -1 }{ 2 }\) xy2
coefficient of x = \(\frac { -1 }{ 2 }\) y2

Question 2.
Write the Degree of the following polynomials:
(i) \(\frac { 2 }{ 5 }\) x3 – 7x2 – \(\frac { 1 }{ 2 }\) x + 3
(ii) \(\frac { 2 }{ 3 }\) xy2 – 5xy + \(\frac { 3 }{ 5 }\) y2x2 + 2x
Solution:
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 8 Algebraic Expressions Check Your Progress Q2.1

Question 3.
Identify monomials, binomials and trinomials from the following algebraic expressions:
(i) 5x × y
(ii) 3 – 5x
(iii) \(\frac { 1 }{ 2 }\) (7x – 3y + 5z)
(iv) 3x2 – 1.2xy
(v) -3x3y4z5
(vi) 5x(2x – 3y) + 7x2
Solution:
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 8 Algebraic Expressions Check Your Progress Q3.1

Question 4.
Using horizontal method:
(i) Add x2 + y2 – 2xy, -2x2 – y2 – 2xy and 3x2 + y2 + xy
(ii) Subtract -x2 + y2 + 2xy from 2x2 – 3y2.
Solution:
(i) x2 + y2 – 2xy – 2x2 – y2 – 2xy + 3x2 + y2 + xy
= x2 – 2x2 + 3x2 + y2 – y2 + y2 – 2xy – 2xy + xy
= 2x2 + y2 – 3xy
(ii) (2x2 – 3y2) – (-x2 + y2 + 2xy)
= 2x2 – 3y2 + x2 – y2 – 2xy
= 3x2 – 4y2 – 2xy

Question 5.
Using column method, add ab + 2bc – ca and 2ab – bc – ca and subtract 4ab + 5bc – 3ca.
Solution:
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 8 Algebraic Expressions Check Your Progress Q5.1

Question 6.
The sides fo a triangle are 5a – 3b, 3a + 2b and 5b – 2a, find its perimeter.
Solution:
Sides of a triangle are 5a – 3b, 3a + 2b and 5b – 2a
Perimeter = 5a – 3b + 3a + 2b + 5b – 2a
= 8a – 2a + 4b
= 6a + 4b

Question 7.
If two adjacent sides of a rectangle are 4x +7y and 3y – x, find its perimeter.
Solution:
Two adjacent sides of a rectangle are 4x + 7y and 3y – x
Perimeter = 2(4x + 7y + 3y – x) = 2(3x + 10y) = 6x + 20y

Question 8.
Subtract the sum of 3x2 + 2xy – 2y2 and 5y2 – 7xy from 5x2 + 2y2 – 3xy.
Solution:
Sum of 3x2 + 2xy – 2y2 and 5y2 – 7xy
= 3x2 + 2xy – 2y2 + 5y2 – 7xy
= 3x2 – 5xy + 3y2
Now,
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 8 Algebraic Expressions Check Your Progress Q8.1

Question 9.
What must be added to 5x3 – 2x2 + 3x + 7 to get 7x3 + 7x – 5?
Solution:
Required expression
= 7x3 + 7x – 5 – (5x3 – 2x2 + 3x + 7)
= 7x3 + 7x – 5 – 5x3 + 2x2 – 3x – 7
= 2x3 + 2x2 + 4x – 12

Question 10.
How much is 3p – 4q + r less than 4p + 3q – 5r?
Solution:
Required expression
= (4p + 3q – 5r) – (3p – 4q + r)
= 4p + 3q – 5r – 3p + 4q – r
= p + 7q – 6r

Question 11.
How much is 3a2 – 5ab + 7b2 + 3 greater than 2a2 + 2ab + 5?
Solution:
Required expression
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 8 Algebraic Expressions Check Your Progress Q11.1

Question 12.
How much should 5x3 + 3x2 – 2x + 1 be increased to get 6x2 + 7?
Solution:
Required expression
= 6x2 + 7 – (5x3 + 3x2 – 2x + 1)
= 6x2 + 7 – 5x3 – 3x2 + 2x – 1
= -5x3 + 3x2 + 2x + 6

Question 13.
Subtract the sum of 12ab – 10b2 – 18a2 and 9ab + 12b2 + 14a2 from the sum of ab + 2b2 and 3b2 – a2.
Solution:
Sum of 12ab – 10b2 – 18a2
and 9ab + 12b2 + 14a2
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 8 Algebraic Expressions Check Your Progress Q13.1

Question 14.
when a = 3, b = 0, c = -2, find the values of:
(i) ab + 2bc + 3ca + 4abc
(ii) a3 + b3 + c3 – 3abc
Solution:
a = 3, b = 0, c = -2
(i) ab + 2bc + 3ca + 4abc
= 3 × 0 + 2 × 0 × (-2) + 3(-2)(3) + 4(3)(0)(-2)
= 0 + 0 – 18 + 0
= -18
(ii) a3 + b3 + c3 – 3abc
= (3)3 + (0)3 + (-2)3 – 3 × 3 × 0 × (-2)
= 27 + 0 – 8 – 0
= 19

Question 15.
Write the algebraic expression for the nth term of the number pattern 13, 23, 33, 43, ………..
Solution:
13, 23, 33, 43
13 = 10 × 1 + 3
23 = 10 × 2 + 3
33 = 10 × 3 + 3
43 = 10 × 4 + 3
10 × n + 3 = 10n + 3
Where n is a natural number.

ML Aggarwal Class 7 Solutions for ICSE Maths

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