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

Ans : (B) Q. 2

: (A) a = 2, b = 5 (B) a = 5, b = 2 (C) (D) a = 2, b = Solution :

Q. 3 In the given figure, if (A) 3 cm (B) 6 cm (C) 4 cm (D) 6.7 cm

Solution : In the given figure,

Q. 4

Solution:

Q. 5

Solution:

Q. 6

(A) an even prime number (B) an even composite number (C) an odd prime number (D) an odd composite number Solution:

Q. 7 The number of solutions of the pair of linear equations x 2y 8 0 and 2x 4y 16 is : (A) 0 (B) 1 (C) infinitely many (D) 2 Solution:

Q. 8 If the mean of 5 observations x, x + 2, x + 4, x + 6 and x + 8 is 11, then the value of x is : (A) 4 (B) 7 (C) 11 (D) 6 Solution:

SECTION – B Question numbers 9 to 14 carry 2 marks each : Q. 9 Find the zeroes of the quadratic polynomial Solution:

Q. 10 Without using trigonometric table, evaluate Solution:

Q. 11 In the figure

Solution:

Q. 12 Given that the HCF(306, 657) = 9, find Solution: HCF (306, 657) = 9

Q. 13 In triangles PQR and MST

Solution:

Q. 14 Write the following distribution as less than type cumulative frequency distribution: C.I 0 - 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 Frequency 5 3 4 3 3 4 7 9 OR In a frequency distribution, the mode and mean are 26.6 and 28.1 respectively. Find out the median. Solution: Class – interval Cumulative frequency less than 10 5 = 5 less than 20 5 + 3 = 8 less than 30 8 + 4 = 12 less than 40 12 + 3 = 15 less than 50 15 + 3 = 18 less than 60 18 + 4 = 22 less than 70 22 + 7 = 29 less than 80 29 + 9 = 38 OR Mode = 26.6, Mean = 28.1

SECTION – C Question numbers 15 to 24 carry 3 marks each.

Q. 15 Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m OR Consider the number , where n is a natural number. Check whether there is any value of for which ends with the digit 0 Solution: Let n be a positive integer. Taking a = n and b = 3 in Euclid’s division lemma

OR If the number , for any n, were to end with the digit zero, then it would be divisible by 5. That is, the prime factorisation of would contain the prime 5 This is not possible because

Q. 16 Solve for x and y:

OR For what value of k, will the following system of equations have infinitely many solutions? Solution:

OR

The given system of linear equations is

Q. 17 On dividing by a polynomial g(x ), the quotient and remainder were

respectively. Find g(x) Solution:

Q. 18 Prove that

Solution:

Q. 19 . Find the length of PQ and PR Solution:

Q. 20 Use Euclid’s algorithm to find the HCF of 96 and 60. Solution:

Q. 21 In an isosceles triangle ABC with AB = AC, E is a point on side CB produced. If , prove that Solution:

Q. 22 is right angled at B. AD and CE are the two medians drawn from A and C, respectively.

If AC = 5 cm, find the length of CE

Solution: In right triangle ABD, right angled at B

Q. 23 The mean of the following frequency distribution is 52. Find the missing frequency. C.I 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 - 80 Frequency 5 3 4 F 2 6 13 Solution: C.I Frequency Class Mark

10 – 20 5 15 75 20 – 30 3 25 75 30 – 40 4 35 140 40 – 50 f 45 45f 50 - 60 2 55 110 60 – 70 6 65 390 70 – 80 13 75 975 = 33 + f

Q. 24 Find the mean of the following frequency distribution using assumed mean method: Classes 2 – 8 8 – 14 14 – 20 20 – 26 26 – 32 Frequency 6 3 12 11 8 OR 200 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in English alphabets in the surnames was obtained as follows No. of letters 1 – 5 5 – 10 10 – 15 15 – 20 20 – 25 No. of surnames 20 60 80 32 8 Find the median. Solution: Classes Frequency Class Mark

2 – 8 6 5 12 72 8 – 14 3 11 6 18 14 – 20 12 17 0 0 20 – 26 11 23 6 66 26 – 32 8 29 12 96

OR No. of letters No. of surnames Cumulative frequency 1 -5 20 20 5 – 10 60 80 10 – 15 80 160 15 – 20 32 192 20 – 25 8 200 n = 200

SECTION – D Question numbers 25 to 34 carry 4 marks each. Q. 25 Find all the zeroes of the polynomial if its two zeroes are Solution:

Q. 26 In the given figure, in and XY divides the into two regions such that

.

OR Two poles of height p and q metres are standing vertically on a levelled ground, ‘a’ metres apart. Prove that the height of the point of intersection of the lines joining the top of each pole to the foot of the opposite pole is given by .

Solution:

In the given figure

OR

Let the height of two poles AB and CD be p and q respectively. The height of the point of intersection of the lines joining the top of each pole to the foot of the opposite pole is EF i.e. x

Q. 27 OR Prove the identity

Solution:

OR

Q. 28 In an acute angled

Solution:

Q. 29 Draw the graphs of the following equations: x + y = 5, x – y = 5 (i) Find the solution of the equations from the graph. (ii) Shade the triangular region formed by the lines and the y-axis Solution:

3

8 5 2

3

(i) The two straight lines intersect each other at the point A(5, 0) (ii) The region bounded by the two lines and the y–axis is shaded in the graph

Q. 30 If the median of the following data is 525, find the values of x and y if the sum of the frequencies is 100 Class interval Frequency 0 – 100 2 100 – 200 5 200 – 300 x 300 – 400 12 400 – 500 17 500 – 600 20 600 – 700 y 700 – 800 9 800 – 900 7 900 – 1000 4 Solution: C.I Frequency Cumulative frequency 0 – 100 2 2 100 – 200 5 7 200 – 300 x 300 – 400 12 400 – 500 17 500 – 600 20 600 – 700 y 700 – 800 9 800 – 900 7 900 – 1000 4 n = 100

The sum of the frequencies = 100

Q. 31 Prove that is irrational Solutin: Let us assume to the contrary, that

Q. 32 State and prove Pythagoras theorem Solution: Pythagoras Theorem: In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Given:

Q. 33 The following table shows the marks obtained by 100 students of class X in a school during a particular academic session. Find the mode of this distribution. Marks Number of students less than 10 10 less than 20 15 less than 30 30 less than 40 50 less than 50 72 less than 60 85 less than 70 90 less than 80 95 less than 90 100 Solution: Class Interval

Cumulative Frequency

Frequency

0 – 10 10 10 10 – 20 15 5 20 – 30 30 15 30 – 40 50 20 40 – 50 72 22 50 – 60 85 13 60 – 70 9 5 70 – 80 95 5 80 – 90 100 5 Total 100

Q. 34 Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other they meet in 1 hour. What are the speeds of the two cars? Solution: Let X and Y be two cars starting from points A and B respectively. Let the speed of car X be x km/hr and that of y be y km/hr Case I : When two cars move in the same direction:

Let these cars meet at point C. Distance travelled by car X = AC and distance travelled by car Y = BC Two cars meet in 5 hours.

Case II : When two cars move in opposite direction:

Let these cars meet at point D. Distance travelled by car X = AD Distance travelled by car Y = BD Two cars meet in 1 hour.

Hence, speed of car X is 60 km/hr and speed of car Y is 40 km/hr