[Exercise- Page 174]
E.8.1. Prepare a table and determine which of the followings are equations and which are not. Present with reasons.
(a) 15 = x + 5
(b) (y - 6) ＜ 3
(c) 63 = 2
(d) z - 4 = 0
(e) (4 ✕ 3) - 12 = 0
(f) 2x + 3 = x - 15
(g) y + 25 ＞ 30
(h) 8 - x = 11
(i) 20 - (10 - 5) = 3 ✕ 5
(j) 50 = 5
(k) 15y = 45
(l) 7 = (11 ✕ 2) + x
Solution:
E.8.2. Express the following problems in the table below as equations:
Solution:
E.8.3. Pick out the correct root from the values in the column beside each equation. Explain why the remaining values are not roots.
Solution:
E.8.4. Mina went to the market with a 100 taka note. She bought one dozen pens, each costing x taka from a shop. The shopkeeper returned her 40 taka. Mina bought y exercise books, each costing 12 taka from another shop and was left with 4 taka.
a) Find the cost of each pen.
b) How many exercise books did Mina buy?
Solution:
E.8.5. Mr Karim invested some of his tk 56000 at 12% profit per annum and the remaining at 10% per annum. After one year he received total profit of tk 6400. How much money did he invest at 10% profit?
Solution:
E.8.6. Shakib scored double the runs of Mushfiqur Rahim in a cricket match. Total runs of both were 2 short of double century. Who scored how many runs?
Solution:
E.8.7. Fill up the empty squares
a)
b)
Solution:
E.8.8. A water bottle weighs 150 gm. Mina put some water bottles in a bag which weighs 50 gm. The number of water bottles is denoted by x and the weights of the water bottles plus the weight of the bag is denoted by y.
a) Write down the relation between x and y using an equation.
b) Find the value of y when x = 15
c) Find the value of x when y = 1100
Solution:
E.8.9. The cost of x packets of biscuits and 1 bottle of drink together is y Taka. The cost of 1 packet of biscuit is tk 20 and cost of 1 bottle of drink is tk 15.
a) Write down the relation between x and y using an equation.
b) Find the value of y when x = 25.
c) Find the value of x when y = 255.
Solution:
E.8.10. The length of the playground of your school is 16 m more than the width.
(a) If the width of the field is x m, find the perimeter of the field in terms of x.
(b) If the perimeter of the field is 120 m, find the area of the field.
Solution:
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