Grade 9 Full Year 9th Grade Review

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1 ID : ae-9-full-year-9th-grade-review [1] Grade 9 Full Year 9th Grade Review For more such worksheets visit Answer t he quest ions (1) The average of Aden's marks in 5 subjects is 79. She got 85 marks in the 6 th subject. What is her average in all 6 subjects? (2) At what point does line represented by the equation 7x + 2y = 43 intersects a line which is parallel to the x-axis, and at a distance 3 units f rom the origin and in the negative direction of y- axis. (3) ABCD is a trapezium with sides AB and CD being parallel to each other. If P and Q are the midpoints of the diagonals of this trapezium, and AB = 13 cm and CD = 8 cm, f ind the length of segment PQ. (4) The perimeter of a triangular f ield is 90 meters and the ratio of the sides is 13:12:5. Find the area of the f ield. (5) If x + y - 4t = 0 then f ind the value of. (6) The area of a trapezium is 132 cm 2 and the height is 8 cm. Find the lengths of its two parallel sides if one side is 5 cm greater than the other. Choose correct answer(s) f rom given choice (7) Find the median of the f ollowing set 12, 20, 13, 11, 19, 12, 12, 13 a b c d (8) Find the surf ace area of the biggest sphere, which can f it in a cube of side 6a. a. 64 π a 2 b. 16 π a 2 c. 36 π a 2 d. 4 π a 2 (9) Express the number 0.1 in the f orm of p q and reduce it to the lowest terms. a. 1 9 b c d

2 ID : ae-9-full-year-9th-grade-review [2] (10) The number is a. Can't say b. Irrational c. Rational d. both (11) The numbers 1 to 17 are written on 17 pieces of paper and dropped into a box. Three of them are drawn at random. What is the probability that the three pieces of paper picked have numbers that are in arithmetic progression? a b c. 1 (12) If AC and EF are parallel, f ind ADB. d a. 120 b. 90 c. 100 d. 80 Fill in the blanks (13) ABCD is a square. Two arcs are drawn with A and B as centers, and radius equal to the side of square. If arcs intersects at point E, the angle BDE =. (14) The average of all possible f ive-digit numbers that can be f ormed by using each of the digits 3, 6, 5, 8, and 7 exactly once is.

3 ID : ae-9-full-year-9th-grade-review [3] Check True/False (15) The angle subtended by an arc at the centre is same as the angle subtended by it at any point on the remaining part of the circle, True False 2016 Edugain ( All Rights Reserved Many more such worksheets can be generated at

4 Answers ID : ae-9-full-year-9th-grade-review [4] (1) 80 We know that the average of marks in all subjects = Sum of marks in all subjects T otal number of subjects. The average of Aden's marks in 5 subjects = Sum of marks in 5 subjects = 5 79 = 395. Sum of marks in 5 subjects 5 = 79 Since she got 85 marks in the 6 th subject, the average of Aden's marks in 6 subjects = Sum of marks in 6 subjects 6 = Sum of marks in 5 subjects + Marks in the 6th subject 6 = = = 80 Theref ore, her average in all 6 subjects is 80. (2) (7, -3) Let's consider the second line f irst. The line which is parallel to the x-axis and is at a distance 3 units f rom the origin in the negative direction of the y-axis is def ined by the f ollowing equation y=-3 So, now we know that at the point of intersection, the value of y = -3 The equation of the f irst line is 7x + 2y = 43 Subtituting f or y with the value -3 in this equation, we get x = 7 So the answer is that the intersection is at the point (7, -3) (3) 2.5 cm

5 (4) 270 ID : ae-9-full-year-9th-grade-review [5] Since we know the perimeter, we can use Heron's f ormula to help us compute the area The f ormula states that the area of a triangle with sides a, b and c, and perimeter 2S = Let us assume the 3 sides are of length a=13x, b=12x and c=5x (we know this because the ratio of the sides is given as 13:12:5) We also know that a+b+c = 90. = 90 ( )x = 90 30x = 90 x = 90 = 3 30 From this we see that a = 39 m, b = 36 m and c=15 m. Also S=45 Step 5 Putting these values into Heron's f ormula, Area = Area = 270 m 2 (5) 1

6 (6) Lengths : 14 cm and 19 cm. ID : ae-9-full-year-9th-grade-review [6] T he f ollowing picture shows the trapezium ABCD, Let's assume, CD = x cm and CD AB. According to the question, one side of the trapezium is 5 cm greater than the other. Theref ore, AB = x + 5, Area(ABCD) = 132 cm 2, Height of the trapezium ABCD = 8 cm. The area of the trapezium ABCD = The height of the trapezium ABCD AB + CD 2 AB + CD = 2 Area(ABCD) T he height of the trapezium ABCD x x = x = x = 28 x = 14 cm. Now, CD = 14 cm, AB = = 19 cm. Thus, the lengths of its two parallel sides are 14 cm and 19 cm respectively.

7 (7) a ID : ae-9-full-year-9th-grade-review [7] If you look at the question caref ully, you will notice that the given data is 12, 20, 13, 11, 19, 12, 12, 13 To f ind the median f irst of all arrange the data in the ascending order, you get 11, 12, 12, 12, 13, 13, 19, 20 Total number of terms are 8 which is even. So, median is equal to the average of ( n 2 ) th and ( n 2 +1) th terms, (where n is the number of terms) ( n 2 ) th = ( 8 2 ) th = 4 th ( n 2 +1) th = ( ) th = 5 th 4 th and 5 th terms are 12 and 13 respectively median = 2 = 25 2 = 12.5 Theref ore the median of the data set is 12.5.

8 (8) c. 36 π a 2 ID : ae-9-full-year-9th-grade-review [8] The biggest sphere that can f it inside a cube of side 6a will have a diameter of 6a (anything larger will not f it in, as opposite sides are separated by a distance of 6a. This means that the radius of this sphere is 1 2 6a The surf ace area of a sphere of radius x is 4πx 2 Theref ore the surf ace area of this sphere is 4π( 1 2 6a) 2 Step 5 The answer is 36 π a 2

9 (9) ID : ae-9-full-year-9th-grade-review [9] a. 1 9 T he general method to solve such problems is to do some mathematical operations that help us remove the repeating part of the decimal Here we are given the decimal 0.1 Let's say p q = p q = p q - p q = p q = 1 Step 5 p q = 1 9

10 (10) b. Irrational ID : ae-9-full-year-9th-grade-review [10] We know that 3 is an irrational number as it cannot be represented in p/q f orm. 3 is used in a f raction in the given expression. Thus we cannot be sure if the given expression is rational or irrational until we simplif y it. Let's simplif y the given expression: = = = = Now we can see that the f irst part, 12, is rational as it is in p/q f orm. The second part, 6 on the other hand, has 3 which is in its simplif ied f orm. Step 5 Theref ore we are now certain that the given number is an Irrational number

11 (11) b This is a little complicated, so f ollow caref ully ID : ae-9-full-year-9th-grade-review [11] For making the explanation and the equations simpler, think of the number on the pieces of paper in the f orm of (2n+1) Here, we can see f rom the equation 2n+1 = 17, so n=8 The probability of getting 3 numbers in an A.P by selecting 3 numbers randomly between 1 and 17 is the ratio of - Number of ways we can get an A.P f rom 3 random numbers between 1 to 17, and - Number of ways to select 3 random numbers between 1 to 17 Let's look at the second part f irst. Three tickets can be drawn f rom (2n+1) numbers is in [(2 x n) + 1] C 3 ways i.e. Number of ways 3 tickets can be drawn = (2n+1)(2n)(2n-1) 3x2x1 Simplif ying this, we get the number of ways to draw 3 numbers between 1 and 17 = n(4n 2-1) 3 Here n = 8, so we can simplif y it as 680 Step 5 Now f or the ways we can get an A.P f rom 3 numbers bwetween Arithmetic Progressions of 3 numbers would be a sequence of 3 numbers that are separated by a common interval e.g. 1,2,3 or 3,5,7 etc. They are in the f orm (a, a+d, a + 2d), where a is an integer f rom 1 to (17-2), and d is another integer So it's helpf ul to think of the solution in terms of this interval. So we'll think of all the sequences that have an interval 1, then sequences with interval 2, and so on Step 6 So what are the possible sequences with interval 1. They are (1,2,3) (2,3,4)... (2n-1,2n,2n+1) T here are theref ore 2n-1 such possible sequences Step 7 Similarly, let's look at A.P with interval 2 between the terms. They are (1,3,5) (2,4,6)

12 ...<2n-3,2n-1,2n+1> T here are 2n-3 such possible sequences ID : ae-9-full-year-9th-grade-review [12] Step 8 We can generalize this to say that the number of such sequences with interval 'd' is (2n- (2d-1)) Obviously the largest possible integer is d=n, with just one sequence (1, n+1, 2n+1) Step 9 So the total number of such sequences is (2n-1) + (2n-3) + (2n-5) This is itself an AP with n terms and d=2 The sum of this sequence is n 2 [2 + (n-1)2] Simplif ying, we get n 2 Here n=8, so this is 64 0 So the probability is (12) c. 100 If you look at the f igure caref ully, you will notice that ADE = 50 and DBC = 150. According to question AC and EF are parallel. Theref ore we can say that EDB and DBC are alternate interior angles. Now EDB = DBC [Alternate interior angles] ADE + ADB = DBC [Since EDB = ADE + ADB] ADB = ADE [Since ADB = 150 ] ADB = [Since ADE = 50 ] ADB = 100 Theref ore ADB = 100

13 (13) 30 ID : ae-9-full-year-9th-grade-review [13] (14) On creating dif f erent possible f ive-digit numbers, each of the digits 3, 6, 5, 8, and 7 will occur the same number of times at each decimal place. Theref ore the sum of all one's digits of numbers will be the same as the sum of all ten's digits of numbers. This way, the sum of digits of all numbers will be the same at all decimal places. Theref ore the average of the f ace value at each decimal place will be: ( )/5 = 5.8 Now we know that the averages of f ace values of each decimal place are all equal to 5.8. The average of the numbers f ormed by such digits will be equal to: ( ) (average of f ace value) = = (15) False

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