Grade 10 Full Year 10th Grade Review

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1 ID : ae-10-full-year-10th-grade-review [1] Grade 10 Full Year 10th Grade Review For more such worksheets visit Answer the questions (1) The sum of the n consecutive natural even numbers starting from 4 is 54. Find the value of n. (2) Find the zeros of the polynomial f(x) = x 3-10x x - 20, if it is given that product of its two zeros is 20. (3) If a prime number is less than 23, what is the probability that it is also less than 11. (4) Asma buys number of toys for Dhs60. If she had bought 3 more toys for the same amount, each toy would have cost Dhs1 less. How many toys did she buy? (5) Two dice are rolled. What is the probability that the two numbers add up to a prime number? (6) If the 20 th element of an arithmetic progression is 422 and the difference between the 42 nd element and the 32 nd element is 210, then what is the first element of the AP? (7) If 6 tanθ = 4, find the value of 6 sinθ + cosθ 6 sinθ - cosθ. (8) The distribution of IQ among a set of students is given below. What is their median IQ? Class interval Frequency (9) If α and β are the zeros of quadratic polynomial x 2-4x - 3, find the value of 1/α 3 + 1/β 3. (10) Simplify (11) Simplify. Choose correct answer(s) from the given choices (12) Find the next term of given sequence.. 14, 45, 144, 459, 1458,... a b c d (13) The value of tan x increases faster than sin x as x increases (x < 90 ). a. True b. False c. Depends on value of θ d. Depends on increment in θ

2 (14) Which of the following is the next term in the series: F, H, I, K, L, N, O, ID : ae-10-full-year-10th-grade-review [2] a. P b. Q c. O d. S Check True/False (15) Aisha standing on a platform 4 metres above the surface of a lake observes a cloud and its reflection in the lake, then the angle of elevation of the cloud is equal to the angle of depression of its reflection. True False 2017 Edugain ( All Rights Reserved Many more such worksheets can be generated at

3 Answers ID : ae-10-full-year-10th-grade-review [3] (1) n terms = 54 n( (n - 1))/2 = 54 [Using AP summation Sn = n/2(2a + d(n - 1) )] n 2 + 3n = 54 n 2 + 3n - 54 = 0 Step 5 (n - 6) (n + 9) = 0 Step 6 n = 6 or -9. Since n cannot be negative, n = 6

4 (2) 5, 4 and 1 ID : ae-10-full-year-10th-grade-review [4] Let α, β and γ be the zeros of polynomial αβ = 20 αβγ = 20 γ = 20/20 = 1 Step 5 Sum of zeros = α + β + γ = 10 Step 6 α + β + 1 = 10 Step 7 α + β = 10-1 = 9 Step 8 α + 20/α = 9 Step 9 α 2-9α + 20 = 0 0 (α - 5) (α - 4) = 0 1 α = 5 or 4 2 β = (9-5) = 4 or (9-4) = 5 3 Hence zeros are 5, 4 and 1

5 ID : ae-10-full-year-10th-grade-review [5] (3) 1 2 There are 8 prime numbers i.e. (2, 3, 5, 7, 11, 13, 17, 19) which are less than 23. Out of these 4 prime numbers are less than 11. Probability (The prime number is less than 11) = Number of prime numbers less than 11 Number of prime numbers less than 23 = 4 8 = 1 2

6 (4) 12 toys ID : ae-10-full-year-10th-grade-review [6] Let the number of toys bought = x Price of one toy = Dhs 60/x Price of one toy, if 3 more items were bought = Dhs 60/(x + 3) Since difference in price is Dhs1 Step 5 Step = x 2 + 3x Step 7 x 2 + 3x = 0 Step 8 x x - 12x = 0 Step 9 x (x + 15) - 12(x + 15) = 0 0 (x - 12) (x + 15) = 0 1 x = 12 or -15. Since number of toys cannot be negative, x = 12

7 ID : ae-10-full-year-10th-grade-review [7] (5) The two dice that are rolled can show any of these values Dice 1 : 1, 2, 3, 4, 5, 6 Dice 2 : 1, 2, 3, 4, 5, 6 So we can get a total of 36 combinations between them (6 x 6) If we take one value from the list of possible values from each Dice, we get numbers ranging from 2 (when both Dice show 1) to 12 (when both dice show 6). Let's enumerate the prime numbers between 2 and 12. They are 2, 3, 5, 7 and 11 We need to see in how many ways we can get each of these values Let's put the value rolled by the dice as (x,y), where x is the value rolled by Dice 1, and y the value rolled by Dice 2-2: The only way to get this is when we roll (1,1). 1 possibility - 3: We can get this by (1,2) or (2,1). 2 possibilities. - 5: We can get this by (2,3), (3,2), (1,4) or (4,1). 4 possibilities. - 7: We can get this by (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). 6 possibilities. - 11: We can get this by (5,6), or (6,5). 2 Possibilities This gives us a total of = 15 possible ways to get a prime number So the probability of getting the two numbers add up to a prime is 15 36

8 (6) 23 ID : ae-10-full-year-10th-grade-review [8] We know that, a n = a + (n - 1)d, Where, n = number of elements in an AP(arithmetic progression), a = first element of an AP, d = common difference According to the question, a 20 = 422, a 20 = a + (20-1)d Therefore, a + (20-1)d = 422 a + 19d = 422 a = d -----(1) a 42 = a + (42-1)d a 42 = a + 41d, a 32 = a + (32-1)d a 32 = a + 31d According to the question, the difference between the 42 nd element and the 32 nd element is 210. Therefore, a 42 - a 32 = 210 (a + 41d) - (a + 31d) = 210 a + 41d - a - 31d = d = 210 d = 21 By putting the value of 'd' in equation (1), we get: a = 23 Step 5 Thus, the first element of the AP is 23.

9 (7) 5/3 ID : ae-10-full-year-10th-grade-review [9] Lets 6 sinθ + cosθ S = 6 sinθ - cosθ Now divide numerator and denominator of this fraction by cosθ (6 sinθ + cosθ)/cosθ S = (6 sinθ - cosθ)/cosθ S = S = (6 sinθ/cosθ + 1) (6 sinθ/cosθ - 1) (6 tanθ + 1) (6 tabθ - 1) Now replace 6 tanθ with is value (6 tanθ = 4) (4 + 1) S = (4-1) S = 5/3

10 (8) 112 ID : ae-10-full-year-10th-grade-review [10] The distribution of IQ among the set of students can be re-arranged as shown in following table, Class interval Frequency(f i ) Cumulative frequency(cf) = = = = 108 From the given table we notice that, n = 108 and n/2 = 54. The Cumulative frequency(cf) just greater than or equal to the n/2 is 74, belonging to the interval Therefore, the median class = , Lower limit(l) of the median class = 112, Class size(h) = 8, Frequency(f) of the median class = 20, Cumulative frequency(cf) of the class preceding median class = 54 The median = l + ( n/2 - cf f ) h = ( = ) 8 Thus, the median IQ of the students is 112. (9) -100/27 Sum of zeros = α + β = -(-4/1) = 4 Product of zeros = αβ = -3/1 = -3 = -100/27

11 (10) 2 secθ ID : ae-10-full-year-10th-grade-review [11] = = = (11) sec θ + tan θ We know that 1 - sin 2 θ = cos 2 θ. Using this identity we can simplify denominator, if we multiply it by 1 + sinθ Now on multiplying numerator and denominator by 1 + sinθ, ( 1 + sinθ 1 - sinθ ) = ( 1 + sinθ 1 - sinθ 1 + sinθ 1 + sinθ ) = ( = ( (1 + sinθ)2 1 - sin 2 θ (1 + sinθ)2 cos 2 θ ) ) = 1 + sinθ cosθ = 1 cosθ + sinθ cosθ = secθ + tanθ (12) a Pattern - Multiply the number by 3 and add successive numbers with 3^n where n is the location of term in the series : 1458*3 + 3^5 = 4617

12 (13) a. True ID : ae-10-full-year-10th-grade-review [12] Let's observe this right angle triangle, sin x = BC AC tan x = BC AB Now let's increase the angle x such that height BC remains unchanged, By comparing above two pictures, we can notice that decrease in AB is larger than decrease in AC Now numerator of both the fractions are unchanged, but denominator of tan x is decreasing faster than denominator of sin x. Therefore tan x increases faster than sin x, as x increases Step 5 Hence given statement is True

13 (14) b. Q ID : ae-10-full-year-10th-grade-review [13] On finding out the numeral positions of the given alphabets in the series, we observe that the given members are following a pattern as given below: 6, 8, 9, 11, 12, 14, 15, The numeral position of the given series is increasing by 2 and then increasing by 1 in every consecutive term. By the same virtue, the next alphabet should be the one whose numeral position is = 17, or Q. Hence the correct option is b. (15) False Here S is the cloud and Q is it reflection in the water. Angle SAB is the angle of elevation of the cloud and angle QAB is the angle of depression of its reflection in the water. The angle of depression and angle of elevation will be equal only when SB = QB, which means the height of the pole is half of the height of the cloud from the surface of lake. This means the statement is false.