CBSE SAMPLE PAPER 2009

CBSE SAMPLE PAPER 2009

CBSE SAMPLE QUESTION PAPERS

MATHEMATICS-CLASS X

Time : 3 Hours Max. Mark : 100

The weightage or the distribution of marks over different dimensions of the question paper

shall be as follows :

1. Weightage to Learning Outcomes

S. No. Learning Outcomes Marks

1. Knowledge 31

2. Understanding 45

3. Application 12

4. Skill 12

2. Weightage to content/subject Unit

S. No. Learning Outcomes Marks

1. Algebra 26

2. Commerical Mathematics 12

3. Mensuration 10

4. Trigonometry 10

5. Geometry 22

6. Statistics 12

7. Coordinate Geometry 8

Total : 100

3. Weightage to form of questions

S. No. Form of Marks for Number of Total

Question each question questions Marks

1. SA I 3 10 30

2. SA II 4 10 40

3. LA 6 05 30

4. The expected length of answer under different forms of questions and expected

time would be as follows :

S. No. Form of Questions No. of credit points Approx. Time

1. Short answer type (SA I) Upto 4 Credit Points 3-5 minutes

2. Short answer type (SA II) Upto 6 Credit Points 5-7 minutes

3. Long answer type (LA) Upto 8 Credit Points 8-10 minutes

137

These ranges of steps and time requirements for the answers are, however, suggestive. In

practice, actual number of steps and time needed may vary. As the total time is calculated on

the basis of the number of questions required to be answered and the length of their anticipated

answers, it would, therefore, be advisable for the candidates to budget their time properly by

cutting out the superfluous lengths and be within the expected limits.

5. Scheme of Options

All questions are compulsory i.e. there is no overall choice in the question paper. However,

internal choices have been provided in two questions of 3 marks each, two questions of 4

marks each and two questions of 6 marks each. These choices have been given from within the

same topic and in questions which test higher mental abilities of students.

6. Weightage to difficulty level of questions

S. No. Estimated Difficulty Level of Questions % of Marks

1. Easy 15%

2. Average 70%

3. Difficult 15%

A question may vary in difficulty level from individual to individual. As such, the assessment

in respect of each question will be made by the paper setter on the basis of general anticipation

from the group as whole taking the examination. This provision is only to make the paper

balanced in its weight, rather to determine the pattern of marking at any stage.

Based on the above design, there are two separate sample papers along with their Blue Prints

as well as questionwise analysis. For the examination of the Board, while the design of the

question papers will remain same, blue prints based on this design may change.

Note : Though weightages to content/subject units, objectives and forms of questions etc. have

been clearly assigned, yet depending on the exigencies of the paper, these can vary to some

extent in Board's examination.

138

üü

BLUE PRINT-I

Subject : Mathematics Class : X

Time : Three Hours Maximum Marks : 100

Objective Knowledge Understaning Application Skill Total

Form of LA SA SAI LA SA SAI LA SA SAI LA SA SAI

questions I II I II I II I II

Content Unit

Algebra

Linear Eqns - - 3(1) - - - - - - - 4(1) - 7(2)

Polynomials - - - - - 3(1) - - - - - - 3(1)

Rational Exp. - - 3(1) - - - - - - - - - 3(1)

Quadratic Eqns - - 3(1)
- 4(1) - - - - - - - 7(2)

Arith. Prog. - - 3(1)
- - 3(1) - - - - - - 6(2)

Sub Total - - 12(4) - 4(1) 6(2) - - - - 4(1) - 26(8)

Comm. Maths

Instalments - - 3(1) - - 3(1) - - - - - - 6(2)

Income Tax - - - 6(1) - - - - - - - - 6(1)

Sub-Total - - 3(1) 6(1) - 3(1) - - - - - - 12(3)

Geometry

Similar Î s - 4*(1) 2*(1) 9(2)

Circles 4 **(1) 3(1) 9(2)

Constructions 4(1) 4(1)

2(1)**

3(1)

4(2)*

Sub-Total - 8(2) - - - 6(2) - - - - 4(1) - 22(5)

Mensuration - - - - 4(1) - 6(1)
- - - - - 10(2)

Trigonometry - - - - 4(1)
6(1)
- - - - - 10(2)

Statistics - 4(1) - - 4(1) - - - - - 4(1) - 12(3)

Coordinate - 4(1)
- - 4(1) - - - - - - - 8(2)

Geometry

Sub-Total - 8(2) - - 16(4) - 12(2) - - - 4(1) - 40(9)

Total - 16(4) 15(5) 6(1) 20(5) 19(5)* 12(2) - - - 12(3) -

G. Total - 31(9) - - 45(11) 12(2) 12(3) 100(25)

}

}

139

SAMPLE QUESTION PAPER-I

Class X

Subject : Mathematics Time : 3 Hours

Max Marks : 100

General Instructions :

1. All questions are compulsory.

2. The question paper consists of 25 questions divided into three sections A, B and C.

Section A contains 10 questions of 3 marks each, Section B is of 10 questions of 4

marks each and Section C is of 5 questions of 6 marks each.

3. There is no overall choice. However, internal choice has been provided in two questions

of three marks each, two questions of four marks each and two questions of six marks

each.

4. In question on construction, the drawing should be neat and exactly as per the given

measurements.

5. Use of calculators is not permitted. However, you may ask for Mathematical tables.

SECTION A

Q1. Solve the following system of equations :

15x + 4y = 61

4x + 15y = 72

Q2. Reduce the following rational expression to its lowest terms :

÷

Q3. PQ and RS are two parallel chords of

a circle and the lines RP and SQ meet

at O on producing (as shown in the

given figure)

Prove that OP=OQ

x2 + 3x + 9

x2 — 25

x3 — 27

(x2 + 3x — 10)

O

P Q

R S

140

Q4. A suit is available for Rs. 1500 cash or for Rs. 500 cash down payment followed by 3

monthly instalments of Rs. 345 each. Find the rate of interest charged under the

instalment scheme.

Q5. A loan has to be returned in two equal annual instalments. If the rate of interest is 16%

per annum compounded annually and each instalment is of Rs. 1682, find the sum

borrowed and the total interest paid.

Q6. If (x — 2) is a factor of x2+ ax + b and a + b = 1, find the values of a and b.

Q7. Using quadratic formula, solve the following equation for x :

abx2 + (b2—ac) x — bc = 0

OR

The sum of the squares of two positive integers is 208. If the square of the larger number

is 18 times the smaller, find the numbers.

Q8. Which term of the A.P. 3, 15, 27, 39.... is 132 more than its 54th term ?

OR

Derive the formula for the sum of first n terms of an A.P. whose first term is 'a' and the

common difference is 'd'

Q9. Find the sum of the following arithmetic progression

1+3+5+7+................+199

Q10. Show that a line drawn parallel to the parallel sides of a trapezium divides the non nonparallel

sides proportionally.

SECTION B

Q11. Solve for x, + = , (x = -1, -2, -4)

Q12. Find graphically, the vertices of the triangle formed by the x-axes and the lines

2x — y + 8 = 0

8x + 3y — 24 = 0

Q13. Construct a triangle ABC in which BC = 13cm, CA = 5cm and AB = 12cm. Draw its

incircle and measure its radius.

Q14. The total surface area of a closed right circular cylinder is 6512 cm², and the

circumference of its base is 88 cm. Find the volume of the cylinder (use π = )

Q15. Prove the identity :

(1 + Cotθ - Cosecθ) (1 + tanθ + secθ) = 2.

1

x+1

4

x+4

22

7

2

x+2

141

OR

Without using trigonometric tables, evaluate :

+ -3tan² 60°

Q16. Show that the points (7, 10), (-2, 5) and (3, -4) are the vertices of an isosceles right

triangle.

OR

Using distance formula, show that the points (-1, -1), (2, 3) and (8, 11) are collinear.

Q17. Find the ratio in which the point (-3, p) divides the line segment joining the points

(-5, -4) and (-2, 3). Hence find the value of p.

Q18. Compute the missing frequencies 'f1' and 'f2' in the following data if the mean is

166 and the sum of observations is 52.

Classes 140-150 150-160 160-170 170-180 180-190 190-200 sum

Frequency 5 f120 f26 2 =52

Q19. An unbiased dice is tossed

i) Write the sample space of the experiment

ii) Find the probability of getting a number greater than 4

iii) Find the probability of getting a prime number.

Q20. The pie chart (as shown in the

figure) represents the amount

spent on different sports by a

sports club in a year. If the

total money spent by the

club on sports is Rs. 1,08,000/-,

find the amount spent on each sport.

cos 35°

sin 55°

tan 27° tan 63°

sin 30°

9

26

142

SECTION C

Q21. Prove that the angle subtended by an arc of a circle at its center is double the angle

subtended by it at any point on the remaining part of the circle.

Using the above result prove that the angle in a major segment is acute.

Q22. Prove that the ratio of the areas of two similar triangles is equal to the ratio of the

squares of their corresponding sides.

Using the above, prove that the area of an equilateral triangle described on the side of

a square is half the area of the equilateral triangle described on its diagonal.

Q23. From the top of a tower 60m. high, the angles of depression of the top and bottom of a

building whose base is in the same straight line with the base of the tower are observed

to be 30° and 60° respectively. Find the height of the building.

OR

An aeroplane flying horizontally at a height of 1.5km above the ground is observed at

a certain point on earth to subtend an angle of 60°. After 15 seconds, its angle of elevation

at the same point is observed to be 30°. Calculate the speed of the aeroplane in km/h.

Q24. A solid toy is in the form of a hemisphere surmounted by a right circular cone. If the

height of the cone is 4 cm and diameter of the base is 6 cm calculate :

i) the volume of the toy

ii) surface area of the toy (use π= 3.14)

OR

A bucket of height 8cm. and made up of copper sheet is in the form of frustrum of a

right circular cone with radii of its lower and upper ends as 3 cm and 9 cm respectively.

Calculate :

i) the height of the cone of which the bucket is a part

ii) the volume of water which can be filled in the bucket.

iii) the area of copper sheet required to make the bucket (Leave the answer in

terms of π)

143

Q25. Anil's total annual salary excluding HRA is Rs. 1,96,000. He contributes Rs., 5000 per

month in his G.P.F. How much he should invest in N.S.C. to get maximum rebate?

After getting maximum rebate he wants to pay income tax in equal monthly instalments.

Find the amount which should be deducted per month towards tax from his salary.

Assume the following for calculating income tax :

a) Standard deduction : (i) 40% of the total income subject to a

maximum of Rs. 30,000/- in case the total

annual income is up to Rs. 100,000.

(ii) Rs. 30,000/- in case the total annual

income is from Rs. 100,001 to Rs.

500,000.

b) Rate of income Tax :

Slab Income Tax

i) Up to Rs. 50,000 No tax

ii) From Rs. 50,001 to Rs. 60,000 10% of the amount exceeding Rs. 50,000

iii)From Rs. 60,001 to Rs. 1,50,000 Rs. 1000 + 20% of the amount exceeding Rs.

60,000

iv) Above Rs. 1,50,000 Rs. 19,000 + 30% of the amount exceding Rs.

1,50,000

c) Rebate in income tax : i) 20% of the amount of saving subject to

maximum Rs. 14,000/-, if gross income

is upto Rs. 1,50,000

ii) 15% of the amount of saving subject to a

maximum of Rs. 10,500/-if gross income

is above Rs. 1,50,000 but not exceeding

Rs. 500,000

144

MARKING SCHEME

SECTION A

Q1. 15x + 4y = 61

4x + 15y = 72

Adding the equations we get

x + y = 7 ................................ (i) 1

Subtracting we get

x - y = -1 ...............................(ii) 1

Solving (i) & (ii) 1

x = 3, y = 4

Q2. Writing as x 1

= x 1

= 1

Q3. ∠POQ = ∠RSQ - Ext. angle of eyclic quad PRSQ 1

∠OQP = ∠RSQ - ............. (PQ RS) 1

∴ ∠OPQ = ∠OQP ½

∴ OP = OQ ½

Q4. Cash Price = Rs. 1500

Price under Instalment Plan = Rs. 500 + Rs. 1035 = Rs. 1535

Interest Charged = Rs. 35 1

Principal for each month = Rs. 1000 + Rs. 655 + Rs. 310

∴ Total Principal = Rs. 1965 1

Rate = = = 21.31% approx 1

x2 + 3x +9

(x+5) (x-5)

(x+5) (x-2)

x3 - 3³

x2 + 3x +9

(x+5) (x-5)

(x+5) (x-2)

(x-3) (x²+3x+9)

x - 2

(x-5) (x-3)

35 x 100 x 12

1965 x 1

2800

131

Q. NO. VALUE POINTS Marks

O

P Q

R S

145

Q5. Principal of 1st instalment = 1682 ÷ (1 + ) = Rs. 1450 1

Principal of 2nd instalment = 1682 ÷ = Rs. 1250 1

Total Sum borrowed = Rs. 1450

+ Rs. 1250

= Rs. 2700 ½

Interest Charged = Rs. 3364 — Rs. 2700 ½

= Rs. 664

Q6. (x - 2) is a factor of x² +ax + b

∴ 4 + 2a + b = 0 Ü 2a + b = -4 1+1=2

also a+b=1

Solving to get a = -5

b = 6 1

Q7. x = ½

= ½

= 1

= or ½

= or ½

OR

Let two postive numbers be x & y and x > y

∴ x² + y² = 208 .....................................(i)

x² = 18y ................................................(ii) 1

Putting the value of (ii) in (i)

y² + 18y - 208 = 0 1

Ü (y+26) ( y-8)=0

Ü y= -26 or y =8 ½

Putting y = 8 in (ii) x = 12, x= -12 (false)

∴ x = 12, y = 8 ½

16

100

(29)

25

2

Q. NO. VALUE POINTS Marks

- (b² - ac) ± oe(b² - ac)² -4 (ab) (-bc)

2ab

- (b² - ac) ± oe(b² + ac)²

2ab

- (b² - ac) ± (b² + ac)

2ab

2ac

2ab

- 2b²

2ab

c

b

—b

a

146

Q8. Here a = 3 , d = 12

∴ t54= 3 + (54 —1).12 = 639 1

Let n be number of terms

∴ tn= 639 + 132 = 771 ½

Ü 3 + ( n—1).12 = 771 1

∴ n = 65 ½

OR

Writing Sn = a + (a+d) + (a+2d) + -------
. Where
=a +(n—1) d 1

∴ Sn =
+ (
—d) + (
—2d) + --------+a

∴ 2 Sn = (a+
) + (a+
) + (a+
) + -------+ (a+
) = n )a+
) 1

Sn = (a+
) = [2a +(n -1) d] 1

Q9. Here a=1, d=2

Let tn = 199

∴ 1 + (n—1).2 = 199 1

∴ n = 100 ½

∴ S100= . [2.1 + (100—1).2] 1

=50 [200]

= 10,000 ½

Q10. Correct figure ½

In Δ ABD, = -------- (i) [EOAB] 1

Similarly in Δ BCD, = ------ (ii) ½

(i) and (ii) Ü = 1

SECTION B

Q11. = 1

Ü 4 (x+1) (x+2) = ( x+4) (3x +4) ½

or 4x² + 12x + 8 = 3x² + 16x + 16 ½

or x² — 4x — 8 = 0 1

Solving to get x =2 + 2oe3, 2—2oe3, 1

n

2

100

2

DE

EA

DO

OB

DO

OB

CF

FB

DE

EA

CF

FB

3x + 4

(x+1) (x+2)

4

x + 4

n

2

Q. NO. VALUE POINTS Marks

147

Q12. 2x — y +8 = 0 ½

8x + 3y —24 = 0 ½

Correct graph of two lines with vertices

as (0, 8), (—4, 0) and (3, 0) 1

Q13. Correct Construction : 3 marks

Correct Measurement of radius : 1 mark

Q14. Let radius of base of cylinder = r cm.

∴ 2x r = 88

Ü r = 14 cm 1

Again 2πrh + 2πr² = 6512 cm²

∴ h = -14 = 60 cm 1½

Volume = x 14 x 14 x 60

= 36960 cm³ 1½

Q15. L.H.S.

1

= 1

= = 2 1

L.H.S. = R.H.S. 1

Q. NO. VALUE POINTS Marks

x -3 -4 0

y 2 0 8

x 0 3 6

y 8 0 -8

6512

88

22

7

22

7

sinθ + cosθ —1

sinθ

sinθ + cosθ + 1

cosθ ( )( )

(sinθ + cosθ)²—1

sinθ . cosθ

2sinθ cosθ

sinθ . cosθ

1+1=2

148

OR

+ - 3 tan²60° 2

= + - 3 tan²60° 1

= 1 + 2 -9

= -6 1

Q16. Let A = (7, 10) ; B = (—2, 5) ; C = (3, —4) ½

∴ AB = oe(—2 —7)² + (5—10)² ½

= oe106

BC = oe(3+2)² + (—4 — 5)²

= oe106 ½

CA = oe(7—3)² + (10+4)²

= oe16 + 196

= oe212 ½

Ü AB=BC ½

and CA² = AB² + BC² 1

∴ A, B & C are vertices of an isosceles rt. triangle ½

OR

Let A = (-1, -1); B = (2, 3) ; C=( 8, 11) ½

AB = oe(2 + 1)² + (3+1)²

= oe25 = 5 ½

BC = oe(8—2)² + (11—3)²

= oe36 +64 ½

= 10

CA = oe(—1—8)² + (—1—11)²

= oe225 ½

= 15

∴ CA = AB + BC 1

∴ (-1, -1) ; (2, 3) and (8, 11) are collinear 1

Q. NO. VALUE POINTS Marks

cos 35°

sin (90° - 35)°

tan 27° tan (90° - 27)°

sin 30°

cos 35°

cos 35°

tan 27° . cot 27°

sin 30°

149

Q17. Let the ratio be K : 1 in which x, y divides the join of (—5, —4) and (—2, 3)

∴ x = 1

y = 1

∴ = -3 (i) and = p (ii) 1

Ü K=2 ∴ Ratio is 2:1 ½

Putting value of K in (ii) we get p = ½

Q18. x : 145 155 165 175 185 195 sum ½

f : 5 f1 20 f26 2 52

f . x 725 155f13300 175f21110 390 5525+155f1+175f21

Mean = 166 = ∴ Ûfx = .52 = 8650 ½

Also f1 + f2= 52 — 33 = 19 Ü f2= 19 —f1

∴ 8650 = 5525 + 155f1+175 (19 —f1) 1

Ü f1= 10 ½

∴ f2= 19 — 10 = 9 ½

Q19. (i) Sample space = {1, 2, 3, 4, 5, 6 } 1

(ii) Numbers greater than 4 = 5, 6 ½

∴ Probability = = 1

(iii)Prime numbers = 2, 3, 5 ½

∴ Probability = = 1

Q20. For total expenditure on sports Rs. 108,000, Central angle = 360º 1

∴ Expenditure on Hockey = 108,000x = Rs. 30,000 1

Expenditure on - cricket = 108,000 x = Rs. 45,000 1

Expenditure on football = 108,000 x = Rs. 18,000 ½

Expenditure on Tennis = 108,000x = Rs.15000 ½

Q. NO. VALUE POINTS Marks

-2K -5

K+1

—2K —5

K+1

3K —4

K+1

9

26

4325

26

4325

26

2

6

1

3

3

6

1

2

100

360

150

360

60

360

50

360

3K -4

K+1

2

3

150

SECTION C

Q21. No Figure no marks

Correct, Fig. given, To prove and Construction ½ x 4=2

Correct Proof 2

Proof : 2 ∠APB = ∠AOB ½

( ∠AOB < 180°) Fig. ½

Ü ∠APB < 90° ½

½

Q22. No figure no marks

correct fig, given, to prove, construction 2 marks (½each)

correct proof 2

(ii) Proof Let side of square = a cm ∴ diagonal = oe2a cm ½

Δ APD Δ A QC (Equilateral) fig. ½

∴ = ½

= ½

Q23. Let Tower AB = 60 m and Building be DC Correct figure 1

In Δ ADB ------

= tan 60° 1

∴ BD = = 20oe3 m 1

∴ CP = 20oe3m ½

Again in Δ ACP---------

= tan 30° 1

Ü AP = 20m ½

Height of Building = CD = PB = AB — AP

= 60 — 20

= 40 m 1

Q. NO. VALUE POINTS Marks

area Δ APD

area Δ AQC

AD²

AC²

1

2

AB

BD

60

oe3

AP

CP

151

OR

Let A and B are two positions of the aeroplane. Let AB = d Correct fig 1

∴ = cot° 60° Ü OL = 1.5 ( ) = (0.5) oe3 km 1½

= cot 30° Ü OM = (1.5) (oe3) km 1½

∴ d = OM — OL = (1.5) oe3 — (0.5) oe3 = oe3 km 1

∴ speed = = = 240 oe3 km/hr 1

or 415.68 km/hr

Q24. Volume of toy = π(3)².4 + π(3)³ cm³ 1

= [12π + 18π] cm³ ½

= 30 x 3.14 = 94.20 cm³ 1

slant height of cone = oe3² + 4² = 5 cm 1

Total surface Area

= [π(3) (5) + 2π (3²)] cm² 1

= (15π + 18π) cm² ½

= 33 (3.14) = 103.62 cm² 1

OR

Let ABCD be the bucket which is the frustrum

of a cone with vertex O (as in fig.) ½

Let ON = x

Δ ONB ~ Δ OMC ∴ = Ü x = 4 1

∴ height of cone = 8 + 4 = 12 cm ½

Volume of bucket = [π(9)².12 — π(3)².4] cm³ 1

= 312 π cm³

Slant height of cone of radius 9cm = 9² + 12² cm ½

∴ L = 15 cm ½

Slant height of cone of radius 3cm = 3² + 4² cm

= 5 cm ½

Area of the copper sheet used to form bucket

= [π(9) (15) - π(3) (5) +π(3)² cm² 1

129π cm² ½

Q. NO. VALUE POINTS Marks

OL

AL

OM

BM

1

oe3

Distance

time

oe3

15

3600

1

3

2

3

4 cm

3 cm

x

x+8

3

9

[ ]

9 cm

8 cm

N 3cm

A B

O

D C

M

152

Q25. Taxable Income = Rs. [1,96,000 — 30,000] = Rs. 1,66,000 ½

Income Tax = Rs. [19,000 + 30% of 16,000] = Rs. 23,800 1

Savings in GPF = Rs. [12 x 5,000] Rs. 60,000 ½

∴ Amount to be invested in NSC for maximum rebate 1

= Rs. [70,000 — 60,000] = Rs. 10,000 1

∴ Maximum rebate availed = Rs. [70,000 x ] = Rs. 10,500 1

Net tax = Rs. [23800 — 10500] = Rs. 13300 1

Total tax to be paid per month = Rs. = Rs. 1108

Q. NO. VALUE POINTS Marks

15

100

13300

12

153

üü

BLUE PRINT-II

Subject : Mathematics Class : X

Time : Three Hours Maximum Marks : 100

Objective Knowledge Understaning Application Skill Total Grand

Form of

Question LA SA1 SA2 LA SA1 SA2 LA SA1 SA2 LA SA1 SA2 LA SA1 SA2 Total

Unit

Algebra

Linear Eqns - - 4(1) - - - - - - - - 3(1) - 4(1) 3(1) 7(2)

Polynomials - - - - 4(1) - - - - - - - - 4(1) - 4(1)

Rational Exp. - 4(1) - - - - - - - - - - - 4(1) - 4(1)

Quadratic Eqns - 4(1) - - 4(1) - - - - - - - - 8(2) - 8(2)

Arith. Prog. - - - - - 3(1) - - - - - - - - 3(1) 3(1)

Sub Total - 12(3) - - 8(2) 3(1) - - - - - 3(1) - 20(5) 6(2) 26(7)

Comm. Maths

Instalments - - 3(1) - - 3(1) - - - - - - - - 6(2) 6(2)

Income Tax - - - 6(1) - - - - - - - - 6(1) - - 6(1)

Sub-Total - - 3(1) 6(1) - 3(1) - - - - - - 6(1) - 6(2) 12(3)

Geometry 2* *

Similar Î s 4*(1) - - - - 3(1) - - - - - - 6(1)* - 3(1) 9(2)

** ** **

Circles 4(1) - - - 4(1) 2 - - - - - 6(1) 4(1) - 10(2)

Constructins - - - - - - - - - - - 3(1) - - 3(1) 3(1)

Sub-Total 8(2) - - - 4(1) 7(1) - - - - - 3(1) 12(2) 4(1) 6(2) 22(5)

Mensuration - - 4(1) - - - - - 6(2)- - - - 4(1) 6(2) 10(3)

Trigonometry - - - - 4(1)- 6(1)- - - - - 6(1) 4(1) - 10(2)

Statistics - - - - - 6(2)- - - 6(1) - - 6(1) - 6(2) 12(3)

Coordinate

Geometry - 4(1) - - 4(1)- - - - - - - - 8(2) - 8(2)

Sub-Total - 8(2) - - 8(2) 6(2) 6(1) - 6(2) 6(1) - - 12(2) 16(4) 12(4) 40(10)

Total - 8(2) - - 8(2) 6(2) 6(1) - 6(2) 6(1) - 6(2) 40(10) 30(10) 30(10)100(25)

G. Total 31(8) 45(11) 12(3) 12(3) 100(25)

154

Sample Question Paper-II

Class X

Subject : Mathematics Time : 3 Hours

Max Marks : 100

General Instructions :

1. All questions are compulsory.

2. The question paper consists of 25 questions divided into three sections A, B and C.

Section A contains 10 questions of 3 marks each, Section B is of 10 questions of 4

marks each and Sections C is of 5 questions of 6 marks each.

3. There is no overall choice. However, internal choice has been provided in two questions

of three marks each, two questions of four marks each and two questions of six marks

each.

4. In question on construction, the drawing should be neat and exactly as per the given

measurements.

5. Use of calculators is not permitted. However, you may ask for Mathematical tables.

SECTION A

Q1. Sove the following system of equations graphically

5x -y = 7

x - y = -1

Q2. Find the Arithmetic Progression whose third term is 16 and the seventh term exceeds

its fifth term by 12.

Q3. ABD is a triangle in which ∠DAB = 90°. AC is drawn perpendicular from A to DB.

Prove that :

AD² = BD x CD

Q4. A loan of Rs. 48,800/- is to be paid back in three equal annual instalments. If the rate of

interest is 25% per annum compounded annually, find the instalment.

Q5. A watch is available for Rs. 970 cash or Rs. 210 as cash down followed by three equal

monthly instalments. If the rate of interest is 16% per annum, find the monthly instalment.

Q6. Construct the pair of tangents drawn from a point, 5cm away from the centre of a circle

of radius 2cm. Measure the lengths of the tangents.

Q7. A solid metallic cylinder of radius 14cm and height 21 cm is melted and recast into 72

equal small spheres. Find the radius of one such sphere.

155

Q8. The rain water from a roof 22m x 20m drains into a conical vessel having diameter of

base as 2m and height 3.5m. If the vessel is just full, find the rainfall (in cm.)

OR

The largest sptere is carved out of a cube of side 7cm ; find the volume of the sphere.

Q9. The following table shows the marks secured by 100 students in an examination

Marks 0-10 10-20 20-30 30-40 40-50

Number 15 20 35 20 10

Find the mean marks obtained by a student.

Q10. A dice is thrown once. Find the probability of getting.

(i) a number greater than 3

(ii) a number less than 5

OR

A bag contains 5 red balls, 8 white balls, 4 green balls and 7 black balls. A ball is drawn

at random from the bag. Find the probability that it is.

(i) black

(ii) not green

SECTION B

Q11. Solve for x and y

(a—b)x + (a+b)y = a² —2ab —b²

(a+b) (x+y) = a² + b²

Q12. If (x+3) (x —2) is the G.C.D. of

f(x) = (x+3) (2x²—3x+a)

and g(x) = (x—2) (3x² + 10x—b)

find the value of a and b

Q13. If A = , B = , find

+

Q14. Solve for x :

+ = (x= 2,x=4)

Q15. A passenger train takes 2 hours less for a journey of 300 km if its speed is increased by

5 km/h from its usual speed. Find the usual speed of the train.

2x+1

2x—1

2x—1

2x+1

A+B

A—B

A—B

A+B

x—1

x-2

x—3

x—4

10

3

156

Q16. AB is a diameter of a circle with

centre O and chord CD is equal to

radius of the circle. AC and BD are

produced to meet at P. Prove that ∠CPD = 60°.

Q17. A circus tent is in the shape of a cylinder surmounted by a cone. The diameter of the

cylindrical part is 24m and its height is 11 m. If the vertex of the tent is 16m above the

ground, find the area of canvas required to make the tent.

Q18. Prove that :

+ = 1 + secθ cosecθ

OR

Evaluate :

+ 2tan 11° tan 31° tan 45° tan59°. tan79° —3 (sin²21° + sin²69°)

Q19. Find a point on the x-axis which is equidistant from the points (7, 6) and (—3, 4)

Q20. Three consecutive vertices of a parallelogram ABCD are A(1, 2), B(1, 0) and

C (4, 0). Find the fourth vertex D.

OR

If A (4, -8), B (-9, 7) and C (18, 13) are the vertices of a triangle ABC, find the length

of the median through A and coordinates of centroid of the triangle.

SECTION C

Q21. The number of hours spent by a school boy on various activities on a working day are

given below :

Activity Number of Hours

Sleep 7

School 8

Homework 4

Play 3

Others 2

Present the above information by a pie-chart.

tanθ

1—cotθ

cotθ

1—tanθ

sin39°

cos 51°

157

Q22. A vertical tower is surmounted by a flagstaff of height h metres. At a point on the

ground, the angles of elevation of the bottom and top of the flagstaff are α and β

respectively. Prove that the height of lower is :

OR

If the angle of elevation of a cloud from

a point h meters above a lake is αand

the angle of depression of its reflection

in the lake is β, prove that the distance

of the cloud from the point of observation is

Q23. If a line is drawn parallel to one side of a triangle, prove that the other two sides are

divided in the same ratio. Using the above result, prove the following :

The diagonals of a trapezium divide each other in the same ratio.

Q24. Prove that the sum of either pair of the opposite angles of a cyclic quadrilateral is 180°.

Using the above result, determine as under :

ABCD is a cyclic trapezium with AD BC. If ∠B=70°, determine the other three

angles of the trapezium.

OR

If two circles touch each other internally or externally, prove that the point of contact

lies on the line joining their centers.

Using the above result prove the following :

Two circles with centers O and O' and radii r1and r2touch each other externally at P.

AB is a line through P intersecting the two circles at A & B respectively. Prove that

OA OB'.

2h sec α

tan β — tan α

h tan α

tan β — tan α

158

Q25. Ramlal has a total annual income of Rs. 1,45,000/-. He contributes Rs. 2000 per month

in his GPF and pays and annual LIC premium of Rs. 15,000. If he pays Rs. 250 per

month for first 11months as advance income tax, find the income tax liability for the

last month. Use the following for calculating income tax :

a) Standard Deduction (i) 40% of the total income subject to a

maximum of Rs. 30,000/- in case the total

annual income is upto Rs. 100,000/-

(ii) Rs. 30,000/- in case the total annual income

is from Rs. 100,001 to Rs.500,000/-

b) Rates of Income tax

i) Upto Rs. 50,000 No tax

ii) Rs. 50,001 to Rs. 60,000 10% of the amount exceeding Rs. 50,000

iii)Rs. 60,0001 to Rs. 1,50,000 Rs. 1000 + 20% of the amount exceeding

Rs. 60,000.

c) Rebate on Savings 20% of the total savings if the gross income

is upto, 150,000 subject to a maximum of

Rs. 14,000.

159

MATHEMATICS

Marking Scheme II

SECTION A

Q1. Forming the table of values :

5x —y = 7 Ü ½

x — y +1 = 0 Ü ½

Graph of lines 1½

Getting the solution x = 2 , y = 3 ½

Q2. Let a be the first term and d, the common difference 1

∴ Third term = t3= a + 2d = 16 .................................................(i)

Also, t7 — t5= 12 or (a+6d) — (a+4d) = 12 Ü d = 6 ...................(ii) ½+½

From (i) and (ii), getting a = 4 ½

∴ The arithmetic progression is 4, 10, 16, 22, 28.................. ½

Q3. Correct Figure ½

Showing Δ DCA ~ Δ DAB

∴ = 1

Ü AD² = BD . CD ½

Q4. Let the instalment be Rs x

Present values of 1st, 2nd and 3rd instalments are

are x, x, x1½

∴ x [1 + + ] = 48800 1

OR x = 25000

∴ each instalment = Rs. 25000 ½

x 1 0 2

y -2 -7 3

x -1 0 2

y 0 1 3

AD

CD

4

5 (4) ( )

5

4

5

2 3

4

5

4

5

16

25

BD

AD

Q. No. Value Points Marks

160

Q5. Cash price of watch = Rs. 970

Cash down payment= Rs. 210

∴ Payment to be made in instalments = Rs. (970-210) = Rs 760 ½

Let Rs. x be each instalment

∴ + x = Rs. 760 1½

or, 3x + 3 = 760

or, x = 760 Üx = 250 1

Q6.. Correct construction 3

Q7. Volume of metallic cylinder = [π(14)2.21] cm3 1

This has been melted to form 72 spheres

Let r be the radius of the sphere

∴ x π r3= π 196.21 1

r3 = (196) (21)

24 x 4

= 3 ½

Ü r = 3.5 cm ½

Q8. Let h cm be the rainfall on the roof

∴ volume of water collected on roof = m3 = . h m³ 1

Voume of water in conical vessel = π(1)² x m³

Q. No. Value Points Marks

[ ][ ]

16 x

1200

76

25

72

4

3

24

( ) 7

2

x + x x 16 x 2

1200

+ x + x x 16 x 1

1200

x

(22 x 20 x h )

100

22

5

1

3

7

2

161

= x x m³ = m³ 1

Ü h =

Ü h = x = ∴ rainfall = cm 1

OR

The diameter of sphere = side of cube 1

∴ Radius of sphere = cm ½

Volume = πr3 ½

= x x x x = 179 cm³ 1

Q9. C.I xi fi fixi

0 -10 5 15 075

10-20 15 20 300

20-30 25 35 875

30-40 35 20 700

40-50 45 10 450

Correctly finding

Ífi ü 100 2400 ö Ífixi Í fixi 1

Í fi ½

x = 1

= = 24 ½

Q10. Total possible cases = 6 ½

Numbers greater than 3 on the die = 3 (4,5,6) ½

∴ (i) Probability of getting a number > 3 =3/6=1/2 1

(ii) Numbers less than 5 = 4 [1,2,3,4] ½

∴ Required probability = or ½

Q. No. Value Points Marks

1

3

22

7

7

2

11

3

22

5

11

3

5

22

11

3

2

1 5

6

7

2

4

3

4

3

22

7

7

2

7

2

7

2

2

3

11

Í fixi

Í fi

2400

100

4

6

2

3

5

6

162

OR

Total no. of balls in the bag = 24 ½

(i) Numbers of black balls = 7

∴ Required probability = 1

(ii) Number of balls which are not green = Total - green = 24 - 4 =20 1

∴ Required probability = = ½

SECTION – B

Q11. (a-b)x + (a+b)y = a2-2ab-b2 (i)

(a+b)x +(a+b)y = a2+b2 (ii) 1

(i) — (ii) Ü — 2bx = — 2b(a+b)

Ü x = ( a+b) 1½

substituting in (i) or (ii) to get y= — 1 ½

Q12. (x+3)(x-2) divides f(x)

∴ 2x2—3x+a has a factor (x—2) 1

∴ 2(2)2—3(2) + a = 0

8 — 6 + a = 0 Ü a = —2 1 ½

Similarly, (x + 3) divides 3x2+10x-b ½

∴ 3(—3)2—30 — b = 0

Ü b = —3 1

Q13. A+B = (2x+1)2+(2x—1)² 2(4x2+1) 1

4x2—1 4x2-1

(2x+1)2— (2x — 1)2 8x ½

4x2 — 1 4x2-1

A+B 4x2+1 4x2—1 4x2+1 1

A—B 4x2—1 8x 4x

Similarly, A—B 4x ½

A+B 4x2+1

1

Q. No. Value Points Marks

7

24

20

24

5

6

2ab

a+b

=

A — B = =

∴ =2 x =

=

∴ A+B

A—B

A—B

A+B

+ =4x²+1

4x

4x

+4x²+1=

(4x² + 1)² + 16x²

4x (4x² + 1)

16x4 + 24x²+1

= 16x³ + 4x

163

Q14. 1 + +1 + = 1

Ü

Ü = ½

Ü 4x2—30x+50=0 1

Ü 2x2—10x—5x+25=0,Ü 2x(x—5)—5(x—5)=0 Ü (x—5)(2x—5)=0 1

Ü x = 5, ½

Q15. Let the usual speed of train be x km/hour ½

According to the problem

— = 2 1

OR = 2 Ü x2+ 5x —750=0 1

or (x+30)(x—25)=0

Ü x=25 [Rejecting x = –30 as speed cannot be negative] 1

∴ The usual speed of train = 25 km/hour ½

Q16. OC = CD = OD Ü OCD is an equilateral traiangle 1

∴ ∠1 =∠2 =∠3=600

Again OA = OC and OB=OD

∴ ∠OAC = ∠OCA =β and ∠OBD= ∠ODB= α 1

∠5 = 1800—2 β

∠4 = 180 —2 α

1800— ∠1 = ∠5 +∠4 = 1200 1

1200= 3600—2(α + β)Ü α + β =1200 ½

∴∠6 = 60° i.e, ∠CPD=600 ½

Q. No. Value Points Marks

1

x—2

1

x—4

10

3

1

x—2

1

x—4

10

+ = 3 —2

2x —6

x²—6x+8

4

3

5

2

300

x

300

x+5

1500

x (x+5)

4

3 =

164

Q17. Area of canvas required to build the tent

= curved surface area of cylindrical

part + curved surface of conical part 1

OA2=52+122=169 Ü OA = 13 m 1

∴ Required area = 2 πrh + πrl = πr(2h+l) ½

= x 12(22+13) m2 = 1320 m2 1 ½

Q18. tan θ cot θ

1-cot θ 1-tan θ

L.H.S ½+1

= = ½+1

cotθ + 1+tanθ = 1 + + =1 + secθ cosecθ = R.H.S. ½+½

OR

cos 510 = cos(90-39)0 = sin390

tan 790 = tan(90—11)0 =

tan 590 = tan (90-31)0 =

tan 450 = 1

sin 690 = sin(90-21)0 = cos 2102½

∴Given expression becomes

+ 2 . tan 11° tan31° .1 . —3 (sin² 21° + cos² 21°) 1

= 1+2 —3 (1) = 0 ½

Q. No. Value Points Marks

22

7

1

tan 11°

1

tan 31°

sin

39°

sin 39°

1

tan

31°

1

tan 11°

+ = —

+ = 1 +sec θ cosec θ

tan θ

1— 1

tan

θ

1

tan θ (1 -tan θ)

1

tan θ (1 -tan θ)

tan² θ

1- tan θ

1 -tan³ θ

tan θ (1 - tan θ)

(1 — tan θ) (1 + tan θ + tan²θ

tan θ (1 — tan θ)

cos θ

sin θ

sin θ

cos θ

165

Q19. Any point P on x axis is given by (x,0) ½

(Distance) between (x, 0) and (7, 6) is given by oe (x — 7)² + 6² ……(i) 1

(Distance) between (x, 0) and (—3, 4) is given by oe (x + 3)² + 4² ……(ii) 1

(i) = (ii) Ü x2—14x + 49 + 36 = x2 + 6x + 9 + 16 1

OR, 20x = 60

x=3

∴ The point is (3,0) ½

Q20. Let the point D be (x, y)

∴ mid point of BD = (x+1), y 1 ½

Mid point of AC = (5/2, 1) 1

This is the same point

∴ = Ü x=4

and =1 Ü y = 2 ½ +½

∴ The co-ordinates of D are (4, 2) ½

OR

Co-ordinates of D are ( ,10) 1

∴ The length of AD

= oe (4- )2 + (—8 —10)2

= oe + 324 = oe 1

= oe 1297 ½

Co-ordinates of centroid

= 4 – 9+18 , —8+7+13 1

3 3

= ( , 4) ½

x+1

2

5

2

y

2

2 2 ( )

9

2

9

2

1

4

1297

4

( )

1

2

13

3

Q. No. Value Points Marks

166

SECTION C

Q21. Making the table:

Correct Central angles 2

Activity Duration in hours Central angle

Sleep 7 1050

School 8 1200

Home work 4 600

Play 3 450

Others 2 300

Drawing correct Pie chart with markings 4

Q22. figure 1

Writing the trignometric equation

= tan α Ü x = b cot α 1 ½

Again = tan β Ü = tan β 1

Ü (b+h) = 1

Ü b tan α + h tan α = b tan β

Ü h tan α = b(tan β - tan α) 1

Ü b = ½

b

x

b + h

x

b + h

b cot α

b tan β

tan α

h tan α

tan β —tan α

Q. No. Value Points Marks

167

OR

We have to find AD,

Let AC = A'C = x

∴ AB=x-h , A'B = x+h 1

Let BD = y

∴ = = tan α Ü x = h + y tan α 1 ½

= tan β

= tan β Ü x = y tan β — h 1

∴ h + y tan a =y tan β — h Ü = y 1

= cos α Ü AD = y sec α 1

AD = ½

Q23. Given, to prove, construction and correct figure ½ x 4 = 2

Correct proof 2

Draw OE AB ½

In Δ DAB, OE AB Ü= (i) ½

Similarly, In Δ ADC, EO AB DC

∴ = (ii) ½

From (i) and (ii), we get = ½

Q24. Given, to prove, construction and correct figure ½x4=2

Correct proof 2

ABCD is cyclic, therefore ∠D = 180° — 70° =110° ½

Also ∠C + ∠D=180°Ü ∠C = 180°– 110° = 70° 1

∴ ∠A = 180 — 70 = 110° ½

Q. No. Value Points Marks

AB

BD

x—h

y

A'B

BD

x + h

y

2h

tan β — tanα

BD

AD

2h sec

tan β — tan α

AE

ED

BO

OD

AE

ED

AO

OC

BO

DO

AO

OC







168

OR

Given ,to prove construction & correct figure 1/2 x 4 = 2

Correct proof 2

Figure ½

OPO' is a straight line

Since OA = OP = r

1

∴ ∠A = ∠∠1, Similarly ∠B = ∠2 ½

But ∠1 = ∠2 (vert. Opp. ∠s)∴ A = ∠B ½

But these are alternate angles ∴OAO' B ½

Q25. Taxable income = Rs. 145000 - 30,000 = Rs. 1,15,000 ½

Income tax = Rs. [1000+ ] = Rs. 12,000 1

Annual savings = Rs [2000 x 12 + 15000] = Rs. 39,000 1

Rebate = 20% of Rs. 39000 = Rs. 7800 1

∴ Tax = Rs. (12000 — 7800) = Rs. 4200 1

Income tax paid for first 11 months = Rs. (250 x 11) = Rs. 2750 1

∴ Income tax to be paid in the last month = Rs. (4200— 2750) = Rs. 1450 ½

Q. No. Value Points Marks

55000 x 20

100