\nQuestion 1.
\nFill in the blanks:
\n(i) In algebra, we use …………… to represent variables (generalized numbers).
\n(ii) A symbol or letter which can be given various numerical values is called a ……………
\n(iii) If Jaggu’s present age is x years, then his age 7 years from now is ……………
\n(iv) If one pen costs \u20b9X x, then the cost of 9 pens is ……………
\n(v) An equation is a statement that the two expressions are ……………
\n(vi) Trial an error is one of methods to obtain …………… of an equation.
\n(vii) 7 less than thrice a number y is ……………
\n(viii) If 3x + 4 = 19, then the value of x is ……………
\n(ix) The number of pencils bought for \u20b9 x at the rate of \u20b92 per pencil is ……………
\n(x) In the expression (-7)5<\/sup>, base = …………… and exponent = ……………
\n(xi) If base = 6 and exponent = 5, then the exponential form = …………… .
\nSolution:
\n(i) In algebra, we use letters to represent variables (generalized numbers).
\n(ii) A symbol or letter which can be given various numerical values is called a variable.
\n(iii) If Jaggu’s present age is x years, then his age 7 years from now is (x + 7) years.
\n(iv) If one pen costs \u20b9 x, then the cost of 9 pens is \u20b99x.
\n(v) An equation is a statement that the two expressions are equal.
\n(vi) Trial an error is one of methods to obtain the solution of an equation.
\n(vii) 7 less than thrice a num bery is 3y – 7.
\n(viii) If 3x + 4 = 19, then the value of x is 5.
\n(ix) The number of pencils bought for \u20b9 x at the rate of \u20b92 per pencil is \\(\\frac{x}{2}\\).
\n(x) In the expression (-7)5<\/sup>, base = -7 and exponent = 5.
\n(xi) If base = 6 and exponent = 5, then the exponential form = 65<\/sup>.<\/p>\n
Question 2.
\nState whether the following statements are ture (T) or false (F):
\n(i) If x is variable then 5x is also variable.
\n(ii) If y is variable then y – 5 is also variable.
\n(iii) The number of angles in a triangle is a variable.
\n(iv) The value of an algebraic expression changes with the change in the value of the variable.
\n(v) If the length of a rectangle is twice its breadth, then its area is a constant.
\n(vi) An equation is satisfied only for a definite value of the variable.
\n(vii) If x toffees are distributed equally among 5 children, then each child gets 5x toffees.
\n(viii) t minutes are equal to 60 t seconds.
\n(ix) If x is a negative integer, then -x is a positive integer.
\n(x) x = 5 is a solution of the equation 3x + 2 = 13.
\n(xi) 2y- 7 > 13 is an equation.
\n(xii) ‘One third of a number x added to itself gives 8’ can be expressed as \\(\\frac{x}{3}\\) + 8 = x.
\n(xiii)The difference between the ages of two sisters Lata and Asha is a variable.
\nSolution:
\n(i) If x is variable then 5x is also variable. True
\n(ii) If y is variable then y – 5 is also variable. True
\n(iii) The number of angles in a triangle is a variable. False
\n(iv) The value of an algebraic expression changes with the change in the value of the variable. True
\n(v) If the length of a rectangle is twice its breadth, then its area is a constant. False
\n(vi) An equation is satisfied only for a definite value of the variable. True
\n(vii) If x toffees are distributed equally among 5 children, then each child gets 5x toffees. False
\n(viii) t minutes are equal to 60 t seconds. True
\n(ix) If x is a negative integer, then -x is a positive integer. True
\n(x) x = 5 is a solution of the equation 3x + 2 = 13. False
\n(xi) 2y – 1 > 13 is an equation. False
\n(xii) ‘One third of a number x added to itself gives 8’ can be expressed as \\(\\frac{x}{3}\\) + 8 = x. False
\n(xiii)The difference between the ages of two sisters Lata and Asha is a variable. False<\/p>\n
Multiple Choice Questions<\/strong> Question 4. Question 5. Question 6. Question 7. Question 8. Question 9. Question 10. Question 11. Question 12. Question 13. Question 14. Question 15. Question 16. Question 17. Question 18. Question 19.
\nChoose the correct answer from the given four options (3 to 19):
\n<\/strong>Question 3.
\nI think of a number x, add 5 to it. The result is then multiplied by 2 and the final result is 24. The correct algebraic statement is
\n(a) x + 5 \u00d7 2 = 24
\n(b) (x + 5) \u00d7 2 = 24
\n(c) 2 \u00d7 x + 5 = 24
\n(d) x + 5 = 2 \u00d7 24
\nSolution:
\nLet number = x
\nAdd 5 to the number
\n\u21d2 i.e. x + 5
\nNow multiply result with 2
\ni.e. (x + 5) \u00d7 2
\nNow, final result is 24
\ni.e. (x + 5) \u00d7 2 = 24 (b)<\/p>\n
\nWhich of the following is an equation?
\n(a) x + 5
\n(b) 7x
\n(c) 2y + 3 = 11
\n(d) 2p < 1
\nSolution:
\n2y + 3 = 11 (c)<\/p>\n
\nIf each matchbox contains 48 matchsticks, then the number of matchsticks required to fill n such boxes is
\n(i) 48 + n
\n(b) 48 – n
\n(c) 48 \u00f7 n
\n(d) 48n
\nSolution:
\nMatchstick required to fill 1 matchbox
\n= 48 \u00d7 1 = 48
\nMatchstick required to fill 2 matchbox
\n= 48 \u00d7 2 = 96
\nMatchstick required to fill 3 matchbox
\n= 48 \u00d7 3 – 144
\n\u2234 Matchstick required to fill n matchbox
\n= 48 n (d)<\/p>\n
\nIf the perimeter of a regular hexagon is x metres, then the length of each of its sides is
\n(a) (x + 6) metres
\n(b) (x – 6) metres
\n(c) (x \u00f7 6) metres
\n(d) (6 \u00f7 x) metres
\nSolution:
\nPerimeter of hexagon = x metres
\n6(side) = x metres
\nSide = (x \u00f7 6) metres
\n\u2234 Side = (x \u00f7 6) metres (c)<\/p>\n
\nx = 3 is the solution of the equation
\n(a) x + 7 = 4
\n(b) x + 10 = 7
\n(c) x + 7 = 10
\n(d) x + 3 = 7
\nSolution:
\nWhen put the value of x = 3
\n3 + 7=10 (c)<\/p>\n
\nThe solution of the equation 3x – 2 = 10 is
\n(a) x = 1
\n(b) x = 2
\n(c) x = 3
\n(d) x = 4
\nSolution:
\n3x – 2 = 10
\n3x = 10 + 2
\n\\(x=\\frac{12}{3}=x=4\\) (d)<\/p>\n
\nThe operation not involved in forming the expression 5x + \\(\\frac{5}{x}\\) from the variable x and number 5 is
\n(a) addition
\n(b) subtraction
\n(c) multiplication
\n(d) division
\nSolution:
\nSubtraction (b)<\/p>\n
\nThe quotient of x by 3 added to 7 is written as
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-6-Solutions-for-ICSE-Maths-Chapter-9-Algebra-Objective-Type-Questions-1.png\")
\nSolution:
\n\\(\\frac{x}{3}+7\\) (a)<\/p>\n
\nIf there are x chairs in a row, then the number of persons that can be seated in 8 rows are
\n(a) 64
\n(b) x + 8
\n(c) 8x
\n(d) none of these
\nSolution:
\nLet the no. of chairs in a row = x
\n\u21d2 Number of persons that can be seated in a row = x
\nHence, number of persons that can be seoted in 8 row = 8x (c)<\/p>\n
\nIf Arshad earns \u20b9 x per day and spends \u20b9 y per day, then his saving for the month of March is
\n(a) \u20b9(31x – y)
\n(b) \u20b931(x – y)
\n(c) \u20b931 (x + y)
\n(d) \u20b931 (y – x)
\nSolution:
\nEarning of Arshad for 1 day = \u20b9 x
\nSpending of Arshad for 1 day = \u20b9 y
\nSaving for 1 day = \u20b9(x – y)
\nSaving for 31 days = \u20b931 (x – y) (b)<\/p>\n
\nIf the length of a rectangle is 3 times its breadth and the breadth is x units, then its perimeter is
\n(a) 4x units
\n(b) 6x units
\n(c) 8x units
\n(d) 10x units
\nSolution:
\nBreadth of rectangle = x units
\nLength of rectangle = 3(Breadth) = 3x
\nPerimeter of rectangle = 2(l + b)
\n= 2(3x + x)
\n= 2(4x) = 8x units (c)<\/p>\n
\nRashmi has a sum of \u20b9 x. She spend \u20b9800 on grocery, \u20b9600 on cloths and \u20b9500 on education and received as \u20b9200 as a gift. How much money (in \u20b9) is left with her?
\n(a) x – 1700
\n(b) x – 1900
\n(c) x + 200
\n(d) x – 2100
\nSolution:
\nTotal money = \u20b9 x
\nMoney spent = \u20b9800 on grocery
\nMoney spent = \u20b9600 on cloths .
\nMoney spent = \u20b9500 on education
\nMoney left with Rashmi
\n= x – \u20b9800 + \u20b9600 + \u20b9500
\n= x – 1900
\nShe received a gift of = \u20b9200
\n\u2234 Money left = x – 1900 + 200
\n= x – 1700 (a)<\/p>\n
\nFor any two integers a and b, which of the following suggests that the operation of addition is commutative?
\n(a) a \u00d7 b = b \u00d7 a
\n(b) a + b = b + a
\n(c) a – b = b – a
\n(d) a + b > a
\nSolution:
\na + b = b + a<\/p>\n
\nIn \\(\\left(\\frac{3}{4}\\right)^{5}\\), the base is
\n(a) 3
\n(b) 4
\n(c) 5
\n(d) \\(\\frac{3}{4}\\)
\nSolution:
\n\\(\\frac{3}{4}\\)<\/p>\n
\na \u00d7 a \u00d7 b \u00d7 b \u00d7 b can be written as
\n(a) a2<\/sup>b3<\/sup>
\n(b) a3<\/sup>b2<\/sup>
\n(c) a3<\/sup>b3<\/sup>
\n(d) a5<\/sup>b5<\/sup>
\nSolution:
\na \u00d7 a \u00d7 b \u00d7 b \u00d7 b
\n= a2<\/sup> \u00d7 b3<\/sup> = a2<\/sup>b3<\/sup> (a)<\/p>\n
\n(-5)2<\/sup> \u00d7 (-1)3<\/sup> is equal to
\n(a) 25
\n(b) -25
\n(c) 10
\n(d) -10
\nSolution:
\n(-5)2<\/sup> \u00d7 (-1)3
\n\u21d2 (-5) \u00d7 (-5) x (-1) \u00d7 (-1) \u00d7 (-1)
\n\u21d2 25 \u00d7 (-1) = -25 (b)<\/p>\n
\n(-2)3<\/sup> \u00d7 (-3)2<\/sup> is equal to
\n(a) 65<\/sup>
\n(b) (-6)5<\/sup>
\n(c) 72
\n(d) -72
\nSolution:
\n(-2)3<\/sup> \u00d7 (-3)2<\/sup>
\n\u21d2 (-2) \u00d7 (-2) \u00d7 (-2) \u00d7 (-3) \u00d7 (-3)
\n\u21d2 -8 \u00d7 9 = -72 (d)<\/p>\nML Aggarwal Class 6 Solutions for ICSE Maths<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"