Grade 9 Number System
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1 ID : ae-9-number-system [] Grade 9 Number System For more such worksheets visit Answer t he quest ions () Write a multiple of -5-7 rational number? (2) Express the f ollowing numbers in the f orm of p q and reduce it to the lowest terms. A) B) C) 2.75 D) -0.2 E) F) (3) Find the average of the middle two rational numbers if are arranged in ascending order. (4) Find a rational number between and 4. (5) Write the rational number that are equal to its negative. (6) Examine if the variable 'x' here represents a rational or an irrational number. A) x 2 = 0.09 B) x 2 = 0.4 C) x 2 =.6 D) x 2 = 3 E) x 2 =.6 F) x 2 = 4 (7) Find rational numbers between -5 and 3. (8) Is 0.6 the multiplicative inverse of 4 6? Why or why not? Choose correct answer(s) f rom given choice (9) The product of two irrational numbers is a. neither rational or irrational number b. a rational or an irrational number c. an irrational number d. a rational number (0) If =7.483 then the value of is close to a..069 b c d (C) 20 Edugain (
2 () Choose an irrational number among the f ollowing : a. -25 b. ID : ae-9-number-system [2] c d. (2) Irrational numbers can be represented on a number line. a. False b. True (3) Which of the f ollowing can be the value of : a. 2 3 b. 4 c. 2 d. (4) The sum of a rational and an irrational numbers is a. whole number b. a rational or an irrational number c. a rational number d. an irrational number (5) If a and b are rational numbers such that than the possible values of a and b are : a. a=0 b=0 OR a = b = b. a = b = c. a=0 b=0 d. a = b = 20 Edugain ( All Rights Reserved Many more such worksheets can be generated at (C) 20 Edugain (
3 Answers ID : ae-9-number-system [3] () 20 7 (Answers may vary) Step Multiples of rational number means the multiplication of the rational number with some given number or any number. For example suppose x is the rational number. x2 multiple of x x2 are 2x x2 3x x2 etc. Theref ore one example of multiple of -5-7 is (2) A) Step The simplest way to solve such problems is to multiple by a f raction n n such that the numerator becomes an integer. The next step is to reduce the f raction to the simplest f orm. Here we can see that if we multiply the numerator by 0000 it will become an integer. So we do the f ollowing = x This reduces to Which in it's simplest f orm is = (C) 20 Edugain (
4 B) 57 ID : ae-9-number-system [4] Step The simplest way to solve such problems is to multiple by a f raction n n such that the numerator becomes an integer. The next step is to reduce the f raction to the simplest f orm. Here we can see that if we multiply the numerator by it will become an integer. So we do the f ollowing = x This reduces to Which in it's simplest f orm is = C) 4 Step The simplest way to solve such problems is to multiple by a f raction n n such that the numerator becomes an integer. The next step is to reduce the f raction to the simplest f orm. Here we can see that if we multiply the numerator by 00 it will become an integer. So we do the f ollowing 2.75 = 2.75 x This reduces to Which in it's simplest f orm is = 4 (C) 20 Edugain (
5 D) -3 ID : ae-9-number-system [5] 25 Step The simplest way to solve such problems is to multiple by a f raction n n such that the numerator becomes an integer. The next step is to reduce the f raction to the simplest f orm. Here we can see that if we multiply the numerator by 00 it will become an integer. So we do the f ollowing -0.2 = -0.2 x This reduces to Which in it's simplest f orm is = E) Step The simplest way to solve such problems is to multiple by a f raction n n such that the numerator becomes an integer. The next step is to reduce the f raction to the simplest f orm. Here we can see that if we multiply the numerator by 000 it will become an integer. So we do the f ollowing = x This reduces to Which in it's simplest f orm is = (C) 20 Edugain (
6 F) -3 ID : ae-9-number-system [6] 000 Step The simplest way to solve such problems is to multiple by a f raction n n such that the numerator becomes an integer. The next step is to reduce the f raction to the simplest f orm. Here we can see that if we multiply the numerator by 0000 it will become an integer. So we do the f ollowing = x This reduces to Which in it's simplest f orm is = (C) 20 Edugain (
7 (3) -7 ID : ae-9-number-system [7] 0 Step To compare f ractions f irst we need to make sure that all denominators are same so we can just compare the numerators of f ractions. The LCM of the denominators and 2 = 0 Now divide the LCM by the denominators and multiply the result with the numerator and denominator as f ollowing: or Step 4 Let s arrange the given numbers in ascending order we get: Step 5 Now the average of the middle two rational numbers = = = -7 0 Step 6 Thus the average of the middle two rational numbers is (C) 20 Edugain (
8 (4) 2.5 ID : ae-9-number-system [8] Step A rational number is a number that can be described as p q where p and q are both integers and q 0. So take any integer (here q=) between these two values or take the average of these two numbers ((pq)/2) to get a rational number between and 4. (5) 0 Step Rational numbers: A rational number is any number that can be expressed as the quotient or f raction p/q of two integers p and q with the denominator q not equal to zero. Since q may be equal to every integer is a rational number. Rational number zero(0) is the only rational number that is equal to its negative. Theref ore zero(0) is the only rational number that is equal to its negative. (6) A) Rational Step A rational number is one that can be represented as p q where p q are integers and q 0. It is given that x 2 = 0.09 x = (0.09) = ±0.3 = ±0.3 Since the variable 'x' can be represented in the f orm of p q the variable 'x' represents a Rational number. (C) 20 Edugain (
9 B) Irrational ID : ae-9-number-system [9] Step A rational number is one that can be represented as p q where p q are integers and q 0. It is given that x 2 = 0.4 x = (0.4) = 2 0 For x to be a rational number there must exist two integers such that their ratio is (0.4). But since ratio of two integers also has to be an integer such two integers cannot exist. Theref ore we can say that x is not a Rational number. C) Irrational Step A rational number is one that can be represented as p q where p q are integers and q 0. It is given that x 2 =.6 x = (.6) = 4 0 For x to be a rational number there must exist two integers such that their ratio is (.6). But since ratio of two integers also has to be an integer such two integers cannot exist. Theref ore we can say that x is not a Rational number. (C) 20 Edugain (
10 D) Irrational ID : ae-9-number-system [0] Step A rational number is one that can be represented as p q where p q are integers and q 0. It is given that x 2 = 3 x = (3) For x to be a rational number there must exist two integers such that their ratio is (3). But since ratio of two integers also has to be an integer such two integers cannot exist. Theref ore we can say that x is not a Rational number. E) Irrational Step A rational number is one that can be represented as p q where p q are integers and q 0. It is given that x 2 =.6 x = (.6) = 4 0 For x to be a rational number there must exist two integers such that their ratio is (.6). But since ratio of two integers also has to be an integer such two integers cannot exist. Theref ore we can say that x is not a Rational number. (C) 20 Edugain (
11 F) Rational ID : ae-9-number-system [] Step A rational number is one that can be represented as p q where p q are integers and q 0. It is given that x 2 = 4 x = (4) = ±2 = ±2 Since the variable 'x' can be represented in the f orm of p q the variable 'x' represents a Rational number. (7) Step A rational number is a number of the f orm p q where p and q are integers and q 0. A simple way of getting rational numbers between -5 and 3 is to leave the denominator the same and use the integers between -5 and 3 f or the numerator. Theref ore the rational numbers between -5 and 3 are (C) 20 Edugain (
12 ID : ae-9-number-system [2] (8) Yes - The product or f raction and its multiplicative inverse must be equal to. Step When we multiply a number by its Multiplicative Inverse we get. For example x x = [It means x is a multiplicative inverse of x.] 4 According to the question we have to prove that 0.6 is the multiplicative inverse of 6 To prove this we have to multiply the given number to the given multiplicative inverse and check if result is or not. Product. = = = It satisf ies the condition of product being equal to. Theref ore 0.6 is the multiplicative inverse of 4 6. (9) b. a rational or an irrational number Step The right answer is that it could be either. This can be seen with the help of an example A) Consider the two irrational numbers and. Their product = x = is an irrational number B) Consider the two irrational numbers and. Their product = x = 2 2 = 9 = 0. 0 is a rational number (C) 20 Edugain (
13 (0) a..069 ID : ae-9-number-system [3] Step In such problems we try and reduce it to known quantities Here we are told = i.e. = Now we need to try and reduce to a f orm involving and other rational numbers This is so that the answer can be computed easily Step 4 We see that = Step 5 Simplif ying we get = Step 6 Multiplying the numerator and denominator of by we get x = 7 = =.069 () d. Step Irrational Number: A real number that can not be written as a simple f raction. Decimal numbers which are either terminating or repeating can be written as f raction and are not irrational e.g..23 = 23/ = 4/3 On the other hand numbers which are neither terminating nor repeating cannot be written as f raction and are irrational numbers..e.g. π = = Now if we look at the all options we notice that as a simple f raction and hence is an irrational number. can not be written (2) b. True (C) 20 Edugain (
14 (3) c. 2 ID : ae-9-number-system [4] Step There are two possible solutions but we'll only look at one here Let's consider the f irst term and multiply it with Obviously it doesn't change the value since = x = 2- = Similarly take the second term and multiply it with x = 3-2 = Step 4 We do the same f or all the other terms. The last term is and multiply it with x = 9-8 = Step 5 So the original equation now becomes Step 6 Simplif ying we get = 3 - = 2 (C) 20 Edugain (
15 (4) d. an irrational number ID : ae-9-number-system [5] Step Let's assume that the sum of rational and irrational number is a rational number. rational number irrational number = rational number m n x = a b x = a b - m n mb - na x = nb [nb are integer and product of two integers are must be an integer.] [mb and na are integer and dif f erence of two integer must be an integer.] Now x is a rational number but we assume that x is an irrational number. T heref ore our assumption that sum could be rational number is not correct Thus the sum of a rational number and an irrational number is an irrational number. (5) c. a=0 b=0 Step If a and b are non-zero than they have to satisf y f ollowing relationship a/b = We know that number on right hand side of above equation is an irrational number. Theref ore at least one of the number on lef t hand side should also be irrational number. It is given in question that a and b are rational numbers. Theref ore above relationship can never be satisf ied and a and b have to be 0. (C) 20 Edugain (
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