Sample Question Paper Mathematics First Term (SA - I) Class X Time: 3 to 3 ½ hours M.M.:90 General Instructions (i) All questions are compulsory. (ii) The question paper consists of 34 questions divided into four sections A, B, C and D. Section A comprises of 8 questions of 1 mark each, section B comprises of 6 questions of 2 marks each, section C comprises of 10 questions of 3 marks each and section D comprises of 10 questions of 4 marks each. (iii) Question numbers 1 to 8 in section A are multiple choice questions where you have to select one correct option out of the given four. (iv) There is no overall choice. However, internal choice has been provided in 1 question of two marks, 3 questions of three marks each and 2 questions of four marks each. You have to attempt only one of the alternatives in all such questions.
Transcript
Sample Question Paper Mathematics
First Term (SA - I) Class X
Time: 3 to 3 ½ hours M.M.:90 General Instructions (i) All questions are compulsory. (ii) The question paper consists of 34 questions divided into four sections A, B, C and D. Section A
comprises of 8 questions of 1 mark each, section B comprises of 6 questions of 2 marks each, section C comprises of 10 questions of 3 marks each and section D comprises of 10 questions of 4 marks each.
(iii) Question numbers 1 to 8 in section A are multiple choice questions where you have to select one
correct option out of the given four. (iv) There is no overall choice. However, internal choice has been provided in 1 question of two
marks, 3 questions of three marks each and 2 questions of four marks each. You have to attempt only one of the alternatives in all such questions.
Section A Question numbers 1 to 8 carry 1 mark each. For each question, four alternative choices have been provided of which only one is correct. You have to select the correct choice. Q. 1 Euclid’s Division Lemma states that for any two positive integers a and b, there exist unique integers q and r such that a = bq + r, where r must satisfy : (A) 1 < r < b (B) 0 < r < b (C) 0 ≤ r < b (D) 0 < r ≤ b Solution: Ans: (C) Q. 2 The graph of y = p(x) is given.
The number of zeroes of p(x) is : (A) 0 (B) 1 (C) 2 (D) 3 Solution: The graph intersects the x-axis at three points. The number of zeroes is 3. Ans: (D)
Q. 3 In the given figure if , then y + z is :
Solution:
Q. 4 Which of the following is not defined? Solution:
Ans: (C) Q. 5 If Solution:
Ans: (B) Q. 6
If sec A = cosec B = , then A + B
Solution:
Ans: (B)
Q. 7
The decimal expansion of the rational number will terminate after :
(A) 25 (B) 15 (C) 10 (D) 30 Solution: Class 10–15 has maximum frequency, i.e. 15
Lower Limit of modal class is 10. Ans: (C)
Section – B Question numbers 9 to 14 carry 2 marks each. Q. 9 Is a composite number? Justify your answer. Solution:
Q. 10 Find a quadratic polynomial, the sum of whose zeroes is 7 and their product is 12. Hence, find zeroes of the polynomial. Solution: Let the required quadratic polynomial be ax2 + bx + c ; a ≠ 0.
From (1) and (2), we get a = 1, b = – 7 and c = 12
Required quadratic polynomial = x2 – 7x +12 x2 – 7x + 12 = x2 – 4x – 3x +12 = x (x – 4) – 3 (x – 4) = (x – 3) (x – 4) Putting x – 3 = 0 and x – 4 = 0, we get x = 3 and x = 4
Q. 11 Solve for x and y : x + y = 8 2x – 3y = 1 Solution: x + y = 8 …(1) 2x – 3y = 1 …(2) Multiplying (1) by 3, we get
Q. 12 Prove that
OR Evaluate:
Solution:
OR
Q. 13
Solution:
Q. 14 Write a frequency distribution table for the following data :
Marks Above 0 Above 10 Above 20 Above 30 Above 40 Above 50 No. of students 40 38 31 25 20 0
Section C Question numbers 15 to 24 carry 3 marks each Q. 15 If x is rational and is irrational, then prove that is irrational.
OR Prove that is irrational. Solution: Let us assume, to the contrary, that is rational
OR
Q. 16 Find the zeroes of the quadratic polynomial 6x2 – 3 – 7x and verify the relationship between the zeroes and the coefficients.
OR
If are the zeroes of the polynomial p(x) = x2 + 5x + k such that , find the value of k. Solution: p(x) = 6x2 – 3 -7x = 6x2 – 7x – 3 = 6x2 – 9x + 2x – 3 = 3x (2x – 3) + 1 (2x – 3) = (3x + 1) (2x – 3) Putting 3x + 1 = 0 and 2x – 3 = 0, we get
OR
Q. 17 Find the value of sin geometrically. Solution: Consider an equilateral triangle PQR In equilateral ∆ PQR
PQ = QR = PR = a And
Draw the perpendicular PM from P to the side QR
In the right triangles PQM and PRM PQ = PR (Each = a) PM = PM (Common side)
Q. 18 The sum of the digits of a two digit number is 8. If 36 is added to the number, the digits interchange their places. Find the number. Solution: Let the digit at units place be x and tens place be y. Number = 10y + x and x + y = 8 … (1) Reverse of the number = 10y + x According to the given condition
Q. 19 Solution:
Q. 20 In the given figure, AB = x units, CD = y units and PQ = z units.
Prove that
Solution:
Q. 21 In the given figure, the perpendicular from A on side BC of a ∆ABC intersects BC at D such that DB = 3 CD. Prove that
Solution: In ∆ ABC CD + DB = BC
CD + 3CD = BC [ DB = 3 CD]
Q. 22 Find the unknown entries a, b, c, d, e and f in the following distribution of heights of students in a class.
Height (in cm) 150 - 155 155 - 160 160 – 165 165 - 170 170 - 175 175 – 180 Frequency 12 b 10 d e 2 Cumulative frequency a 25 c 43 48 f
Solution: a = 12 b = 25 – 12 = 13 c = 25 + 10 = 35 d = 43 – 35 = 8 e = 48 – 43 = 5 f = 48 + 2 = 50 Q. 23 Find the missing frequency f if the mode of the given data is 154
Class 120 - 130 130 - 140 140 - 150 150 - 160 160 - 170 170 – 180 Frequency 2 8 12 f 8 7
OR
If the mean of the following distribution is 6, find the value of P
2 4 6 10 P + 5 3 2 3 1 2
Solution: Mode = 154 …[Given]
Modal class = 150 – 160 Lower limit of modal class (l) = 150 Class size (h) = 160 – 150 = 10 Frequency of the modal class (f1) = f Frequency of class preceding the modal class (f0) = 12 Frequency of class succeeding the modal class (f2) = 8
OR
2 3 4 4 2 8 6 3 18
10 1 10 P + 5 2 2p + 10
= 11 = 52 + 2p
Q. 24
Solution:
Section D Question numbers 25 to 34 carry 4 marks each. Q. 25 Find all zeroes of 2x4 – 9x3 + 5x2 + 3x – 1, if two of its zeroes are . Solution: p(x) = 2x4 – 9x3 + 5x2 + 3x – 1 Two zeroes of p(x) are (2 + ) and (2 – ) Sum of the zeroes = 2 + + 2 – = 4 Product of the zeroes = (2 + ) (2 – ) = 22 – ( )3 = 4 – 3 = 1
A polynomial whose zeroes are (2 + ) and (2 – ) is = x2 – (sum of zeroes) x + product of zeroes = x2 – 4x + 1 On dividing 2x4 – 9x3 + 5x2 + 3x – 1 by x2 – 4x + 1, we get
Q. 26 If two scalene triangle are equiangular, prove that the ratio of the corresponding sides is same as the ratio of the corresponding angle bisector segments.
OR In the given figure, sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of ∆ PQR. Show that ∆ ABC ~ ∆ PQR
Solution:
OR
Given: AD is the median of ∆ ABC and PM is the median of ∆ PQR.
To Prove: ∆ ABC ~ ∆ PQR Construction: Produce AD to E such that AD = DE and produce PM to N such that PM = MN. Join BE, CE, QN, RN.
Proof: In quadrilateral ABEC BD = DC [ AD is the median] AD = DE [Construction] In parallelogram diagonals bisect each other.
ABEC is a parallelogram. Similarly, PQNR is also a parallelogram. BE = AC and QN = PR [In parallelogram opposite sides are eual]
From (1) and (2), we get
∆ ABC ~ ∆ PQR [ SAS similarity criterion]
Q. 27 Prove that
Solution: To prove :
Q. 28
OR
Determine the value of x such that
Solution:
OR
Q. 29 Determine graphically the coordinates of the vertices of a triangle, the equation of whose sides are y = x, 3y = x, x + y = 8 Solution: (i) y = x …(1)
x - 3 0 3 y - 3 0 3
(ii) 3y = x …(2)
x - 6 0 3 y - 2 0 1
(iii) x + y = 8 …(3) Or y = 8 – x
x -1 0 3 Y 9 8 5
The co-ordinates of the vertices of triangle ABC are (0, 0), (4, 4) and (6, 2) Q. 30 During the medical check-up of 35 students of a class, their weights were recorded as follows.
Cumulative frequency greater than 17.5 is 28 and corresponding class is 46 – 48 Median class is 46 – 48 l = 46, , cf = 14, f = 14, h = 2.
Q. 31 State and prove Thales Theorem Solution: Thales Theorem : If a line is drawn parallel to one side of a triangle, to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. Given: ∆ ABC and a line ‘l’ parallel to BC intersects AB at D and AC at E.
Construction: Join BE and CD. Draw EL AB and DM AC. Proof:
Q. 32 Use Euclid’s division lemma to show that the cube of any positive integer is of the form, 9m, 9m + 1 or 9m +8. Solution: Let a be any positive integer and b = 3. Then, by Euclid’s algorithm, a = 3q + r, for some integer q ≥ 0 and 0 ≤ r < 3. i.e., the possible remainders are 0, 1, 2. Thus, a can be of the form 3q or 3q + 1 or 3q + 2 If a = 3q, a3 = 27q3 = 9 × 3q3 = 9m where m = 3q3 If a = 3q + 1, a3 = (3q + 1)3 = 27q3 + 27q2 + 9q + 1 = 9 ( 3q3 + 3q2 + q) + 1 = 9m + 1 where m = 3q3 + 3q2 + q If a = 3q +2, a3 = (3q+2)3 = 27q3 + 54q2 + 36q + 8 = 9 (3q3 + 6q2 + 4q) + 8 = 9m + 8 where m = 3q3 + 6q2 + 4q a3 is either 9m or 9m + 1 or 9m + 8 Q. 33 The mean of the following frequency distribution is 62.8 and the sum of all the frequencies is 50. Find the missing frequencies f1 and f2
Class interval 0 – 20 20 – 40 40 – 60 60 – 80 80 – 100 100 – 120 Total Frequency 5 f1 10 f2 7 8 50
Q. 34 8 men and 12 boys can finish a piece of work in 10 days while 6 men and 8 boys can finish it in 14 days. Find the time taken by one man alone and that by one boy alone to finish the work. Solution: Let 1 man finish the work in x days.
In 1 day he finishes = work
8 men in 1 day finish work
Let 1 boy finish the work in y days.
In 1 day he does = work
12 boys in 1 day finish work
Since 8 men and 12 boys finish the work in 10 days.
Hence, one man alone can finish the work in 140 days and one boy alone can finish the work in 280 days.
