ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 9 Algebra Check Your Progress

ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 9 Algebra Check Your Progress

Question 1.
Look at the following matchstick pattern of polygons. Complete the table. Also write the general rule that gives the number of matchsticks.

Solution:

Question 2.
Write an algebraic expression for each of the following:
(i) If 1 metre cloth costs ₹ x, then what is cost of 6 metre cloth?
(ii) If the cost of a notebook is ₹ x and the cost of a book is ₹ y, then what is the cost of 5 notebooks and 2 books?
(iii) The score of Ragni in Mathematics is 23 more than two-third of her score in English. If she scores x marks in English, what is her score in Mathematics?
(iv) If the length of a side of a regular pentagon is x cm, then what is the perimeter of the pentagon?
Solution:
(i) Cost of 1 metre cloth = ₹ x
Cost of 6 metre cloth = ₹ x × 6 = ₹6x
(ii) Cost of 1 notebook = ₹ x
Cost of 1 book = ₹y
Cost of 5 notebooks = 5 × (₹x) = ₹5x
Cost of 2 books = (₹y) × 2 = ₹2y
Total cost of 5 notebooks and 2 books = ₹(5x + 2y)
(iii) Score of Ragni in English = x marks
Score of Ragni in Mathematics = 23 more than two-third of her score in English
i.e. = \(23+\frac{2}{3}(x)\)
= \(23+\frac{2}{3} x\)
(iv) Side of regular pentagon = x cm
Perimeter of pentagon = 5 × Side = 5x cm

Question 3.
When x = 4 and y = 2, find the value of:
(i) x + y
(ii) x – y
(iii) x² + 2
(iv) x² – 2xy + y²
Solution:
(i) x + y
Put x = 4 and y = 2
We get,
⇒ 4 + 2 = 6

(ii) x – y
Put x = 4, y = 2
We get,
⇒ 4 – 2 = 2

(iii) x² + 2
Put x = 4 We get,
⇒ (4)² + 2
= 16 + 2= 18

(iv) x² – 2xy + y²
Put x = 4 and y = 2
We get,
⇒ (4)² – 2 × 4 × 2 + (2)²
= 16 – 16 + 4 = 4

Question 4.
When a = 3 and b = -1, find the value of 2a³ – b⁴ + 3a²b³.
Solution:
2a³ – b⁴ + 3a²b³
Put the value of a = 3, b = -1
= 2(3)³ – (-1)⁴ + 3(3)² (-1)³
= 2 × 3 × 3 × 3 – (-1) × (-1) × (-1) × (-1) + 3 × 3 × 3 × (-1) × (-1) × (-1)
= 54 – 1 – 27 = 26

Question 5.
When a = 3, b = 0, c = -2, find the values of:
(i) ab + 2bc + 3ca + 4abc
(ii) a³ + b³ + c³ – 3abc
Solution:
(i) ab + 2bc + 3ca + 4abc
Put the values of a = 3, b = 0, c = -2
3 × 0 + 2 × 0 × -2 + 3 × -2 × 3 + 4 × 3 × 0 × -2
= 0 + 0 – 18 + 0 = -18

(ii) a³ + b³ + c³ – 3abc
Put the values of a = 3, b = 0, c
= -2 (3)³ + (0)³ + (-2)³ – 3 × 3 × 0 × -2
= 27 + 0 – 8 – 0 = 19

Question 6.
Solve the following linear equations:
(i) 2x – \(1 \frac{1}{2}=4 \frac{1}{2}\)
(ii) 3(y – 1) = 2(y + 1)
(ii) n – 3 = 5n + 21
(iv) \(\frac{1}{3}(7 x-1)=\frac{1}{4}\)
Solution:

(ii) 3(y – 1) = 2(y + 1)
3y – 3 = 2y + 2
3y – 2y = 2 + 3 ⇒ y = 5
Verification
⇒ 3(5 – 1) = 2(5 + 1)
⇒ 3(4) = 2(6)
⇒ 12 = 12

(iii) n – 3 = 5n + 21
n – 5n = 21 + 3
⇒ -4n = 24
⇒ \(n=\frac{24}{-4}=-6\)
Verification
⇒ n – 3 = 5n + 21
⇒ -6 – 3 = 5(-6) + 21
⇒ -9 = -30 + 21
⇒ -9 = -9

(iv) \(\frac{1}{3}(7 x-1)=\frac{1}{4}\)
⇒ \(\frac{7 x-1}{3}=\frac{1}{4}\)
By cross multiplication,
4 (7x – 1) = 3 × 1 ⇒ 28x – 4 = 3
⇒ 28x = 3 + 4 ⇒ 28x = 7

ML Aggarwal Class 6 Solutions for ICSE Maths