Transcript
Page 1: Solutions / CBSE 10th Mathematics Sample Paper

Solutions / CBSE 10th Mathematics Sample Paper

Section – A

1.We have ax by a b ... i

bx ay a b ... ii

Multiplying (i) by a and (ii) by b, we get 2 2a x aby a ab ... iii

2 2b x aby ab b ... iv (1 mark)Adding (iii) and (iv), we get

2 2 2 2a b x a b x 1 (½ mark)

Substituting x = 1 in (i), we geta 1 by a b

by b y 1 (½ mark)

OR

Let the father’s age be x years and son’s age be y years. Now according to problem

x 3y 3 x 3y 3 …(i) (½ mark) and …(ii) (½ mark) Subtracting (i) from (ii), we get

y 10 (½ mark)Substituting y = 10 in (i), we get

x 3 10 3 x 33 (½ mark) Hence, father’s age = 33 years, son’s age = 10 years

2. We have HCF = , LCM = and . We Know

(½ mark)

(1 mark)

(½ mark)2. We have

; a 0 , b 0, x 0

Page 2: Solutions / CBSE 10th Mathematics Sample Paper

(½ mark)

(½ mark)

(½ mark) (½ mark)

3.Let a be the first term and d be the common difference of an A.P., then 7a a 7 1 d 7a a 6d [Using an = a + (n – 1)d]

and 11a a 11 1 d 11a a 10d (½ mark) According to the problem

7 117a 11a

7 a 6d 11 a 10d (½ mark)

4 a 17d 0

a 18 1 d 0 (½ mark)

18a 0 (½ mark)

5. Cash price of the electric iron = Rs. 440 Cash down payment = Rs. 200 Balance to be paid = Rs. (440 – 200) = Rs. 240 (½ mark) Instalment paid = Rs. 244 Total price charged under the instalment plan = Rs. (200 + 244) = Rs. 444 Therefore, interest charged = Rs. 444 – Rs. 440 = Rs. 4 (½ mark)

Therefore, Rs. 4 is the interest charged on Rs. 240 for 1 month i.e. year

Int. 100rate %

Pr incipal Time

(½ mark)

4 1001

24012

4 10020

20

(½ mark)

Hence, rate 20% per annum

6.

Page 3: Solutions / CBSE 10th Mathematics Sample Paper

A

B CD

In and , we have(Given)

And [Common] [By AA criterion of similarity] (½ mark)

(½ mark)

(½mark)

(½ mark)

ORS R

Q

M

P O

40°65°

50°

[Angle in semi circle]

In triangle QRP, QRP QPR PQR 180

90 QPR 65 180

QPR 25

PQRS is a cyclic quadratic. QRS QPS 180

90 PRS 25 40 180 PRS 25 (1 mark)

PMQR is also a cyclic quadratic. MQR RPM 180

50 65 25 QPM 180

QPM 40 (1 mark)7. Total number of outcomes = 17

(i) Odd numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17 Number of favourable outcomes = 9 P(an odd number) = 9/17 (1 mark)

(ii) Numbers divisible by 3 or 5 are 3, 5, 6, 10, 12, 15 Number of favourable outcomes = 6 P(a number divisible by 3 or 5) =6/17 (1 mark)

Page 4: Solutions / CBSE 10th Mathematics Sample Paper

Section – B

8. We have y = x

x 0 1 4 y 0 1 4

y = 2x

x 0 1 2 y 0 2 4

and x + y = 6 y = 6 – x

x 0 1 3 y 6 5 3

(1 mark)

y

y

xxO

y = 2

x

(0,6

)(1

,5)

(1,1)

(0,0)

(4,4)B (2,4

) y = x

(1,2

)

x + y = 6

(3,3)

A

(1½ mark)

From figure, the three straight lines intersect at O(0, 0), A(3, 3) and B(2, 4). Vertices of the required triangle OAB are (0, 0), (3, 3) and (2, 4).

(½ mark)

9. We have

(1 mark)

Page 5: Solutions / CBSE 10th Mathematics Sample Paper

(1 mark)

(1 mark)

10. Let x be the larger number, then square of the smaller number will be 4x(1 mark)

2x 4x 45 0 x 9 x 5 0 (½ mark)

x = 9 or x = –5 (½ mark) The numbers are natural numbers

x = 9 (½ mark) Square of smaller number = 4 × 9 = 36 Smaller number = (½ mark) Hence, larger number = 9 and smaller number = 6

11. Let a be the first term and d be the common difference of the A.P. and (½ mark)

and (½ mark) and

and (½ mark)

(1 mark)

(½ mark)

22S 902

12. Let each instalment be Rs. x.r = Rate of interest = 5% per annum CI.

n = Time = 2 years For first instalment:

A = Rs.x, time = 1 year, r = 5% per annum and P = P11

15

x P 1100

nr

Using A P 1100

(½ mark)

120

P Rs. x21

For second instalment:A = Rs. x, time = 2 years, r = 5% per annum and P = P2

(½ mark)

2

220

P Rs. x21

Total sum borrowed = Rs. 4100 [Given]

Page 6: Solutions / CBSE 10th Mathematics Sample Paper

Therefore,

(1mark)

20 20x 1 Rs.4100

21 21

(½ mark)

(½ mark)Hence, annual instalment is Rs. 2,205.

13.

(1 mark)

(1 mark)

= 1 + 1 – 2 + 3 (1 mark)= 3

OR

LHS = (½ mark)

(½ mark)

(½ mark)

(½ mark)

(½ mark)

(½ mark)

= RHS

14. Radius of solid metallic sphere

Page 7: Solutions / CBSE 10th Mathematics Sample Paper

Volume of the sphere = (1 mark)

Radius of the base of the cone =

Volume of the cone = (1 mark)

Let n be the number of cones so formed.

n 3.0625 1543.5

(1 mark)

15.A

B C

PQ I

L5 cm

4 cm 3 cm

(2 marks)Steps of construction: (1 mark)1. Construct the ABC in which BC = 5 cm, CA = 3 cm and AB = 4 cm.2. Bisectand and . Let bisectors of these angles be BP and CQ respectively.

Let these intersect at I. 3. Draw the perpendicular IL on BC.4. Taking I as centre and IL as radius, draw the circle.

This is the required incircle of the ABC. Radius IL = 1 cm 16.

A(3, –4) P(p, –2) Q B(1, 2)

1 2

5, q

3

Let AB be the line-segment. Clearly P divides AB in the ratio of 1 : 2 internally. (½ mark)

[Using section formula]

(1 mark)

Now Q is mid-point of PB (½ mark)

[Using mid-point formula]

q = 0 (1 mark)

Page 8: Solutions / CBSE 10th Mathematics Sample Paper

OR

A(x , y )1 1

C(x , y )3 3B(x , y )2 2 D

1

2

Let D be the mid-point of BC,

(½ mark)

Let G (x, y) be the centroid of the triangle. Since, G divides AD in the ratio 2:1. (½ mark)

(1 mark)

and (1 mark)

Hence, the co-ordinates of the centroid of the triangle are

.

17.

A

T

C D B

AT is a tangent and TB is a chord of the circle. ATB = TCB ... (i) [Angles in the alternate segments] (½ mark) TD is the bisector of BTC.

... (ii) (½ mark)Adding (i) and (ii), we get

... (iii) (½ mark) As is the exterior angle of

... (iv) (½ mark) From (iii) and (iv), we get

Hence, is an isosceles triangle (1 marks)

18. Let the co-ordinates of point on the Y-axis is P(0, y). (½ mark) Also let A and B denote the points . Since AP = BP,

(½ mark)

Page 9: Solutions / CBSE 10th Mathematics Sample Paper

(½ mark)

(½ mark) (½ mark)(½ mark)

Hence, the point on Y-axis is (0, -2). .

19. The following table gives the share of each item as a component of 360°.

Items Expenditure (in percent) Share as a component of 360°

Food Entertainment Other expenditure Savings

40 25 20 15

040360 144

100

025

360 90100

020

360 72100

015

360 54100

Total

100 360°

(1½ mark)

Pie Chart:

Food144°

Entertainment90°

Other expenditure

72°

Savings54°

(1½ mark)

Section – C

20.

Page 10: Solutions / CBSE 10th Mathematics Sample Paper

hO

Q

P

Cloud

R

H

y

x

(1 mark)

According to figure, the cloud is at P and the image of the cloud P is at with respect to the lake.

PR P'R Let OP = x, OQ = y and PR = P’R = H In right triangle OPQ,

...(i) (1 mark)

In right triangle ,

… (ii) (1 mark)

Subtracting (ii) and (iii), we get

… (iii) (½ mark)

Again, in right triangle OPQ,

… (iv) (½ mark)

Dividing (v) and (iv), we get

(½ mark)

(½ mark)

OR

Page 11: Solutions / CBSE 10th Mathematics Sample Paper

45°

O Ax

60°

Q

P

4,0

00 m

(1 mark) Let P and Q be the positions of the two aeroplanes.In right-angled ,

tan60° =

... (i) (1½ mark)

In right-angled ,

tan45° =

[From (i)] (1½ mark)

Distance between two aeroplanes

= PQ = AP – AQ

(1 mark)21. Part – I

A

B D C

P

Q S RGiven:

To prove:

Construction: Draw and (½ mark)

Proof:

... (i) (½ mark)

In and

Page 12: Solutions / CBSE 10th Mathematics Sample Paper

... (ii) (1 mark)

But

[From (ii)] ... (iii) (½ mark)

[From (i) and (iii)] (½ mark)

Hence, (½ mark)

Part – II

(½ mark)

(½ mark)

(½ mark)

22. Part – I

A

Q

B

M

P

O

S T

Given: ST is a tangent to the circle with centre O. P and Q are points on

major and minor arc AB respectively.To prove: and Construction: Draw a diameter AM and join MB. (½ mark)Proof: [Angle in a semicircle]

MAT 90

In ,... (i)

Also ... (ii) (1 mark)From (i) and (ii),

Page 13: Solutions / CBSE 10th Mathematics Sample Paper

(½ mark)Also [Angle on the same segment AB]

(½ mark)AQBP is a cyclic quadrilateral.

(½ mark)

(½ mark) Hence, and

Part – II

A

B

C

S T

35°75°

Using part I,Since angles in the alternate segments are equal.

and (½ mark) and (1 mark)

ORPart – I

B

CA

D

1

2O

Given: ABCD is a cyclic quadrilateral inscribed in a circle with centre O.To prove: A + C = 180º and B + D = 180° Construction: Join AO and CO (½ mark)Proof: Arc ABC subtends angle 1 at the centre and subtends ADC at point D(Remaining part of the circle). ADC = ½ (1) … (i)Again arc ADC subtends 2 at the centre and subtends ABC at point B(Remaining part of the circle). ABC = ½ (2) ... (ii) (1½ mark)Adding (i) and (ii), we getADC + ABC = ½ (1 + 2) (½ mark) ADC + ABC = ½ (360°) = 180° (½ mark)B + D = 180°Similarly, A + C = 180° (½ mark)

Page 14: Solutions / CBSE 10th Mathematics Sample Paper

Part – II

x + 1 05 y + 5

4 y – 4

2 x + 4

D

A

B

C

Applying this property on given angles of circle, we get2x + 4 + 4y – 4 = 180°and x + 10 + 5y + 5 = 180° x + 2y = 90° … (iii) and x + 5y = 165° … (iv) (½ mark)Subtracting (iii) from (i), we get3y = 75° y = 25° (½ mark)Substituting y = 25° in (iii), we getx + 2 × 25° = 90° x = 40° (½ mark)Thus, values of x and y are 40° and 25° respectively.

23. Class Mid-value xi Frequency fi fixi

0-10 5 8 4010-20 15 10 15020-30 25 9 22530-40 35 12 42040-50 45 11 495

fi = 50 fixi = 1330 (3 mark)

(1 mark)

(1 mark)

24.

24 m

12 m

21 m

5 m

16 m

O

AB

Page 15: Solutions / CBSE 10th Mathematics Sample Paper

Radius of both conical and cylindrical part = r = 12 m, height of conical part = 16 m, height of cylindrical part (h) = 5 m.Lateral surface area of the tent … (i) (1 mark)

O

1 6 m

A 1 2 m B

2 2 212 16 2 144 256 400

(1 mark)Lateral surface area of tent

(1 mark)

(1 mark)

The cost of the cloth

(1 mark)

25. Annual income of Usha = Rs. 4,10,000Deduction on (P.M.’s Relief Fund + charitable society)

= Rs. 30,000 + Rs. 20,000

= Rs. 30,000 + Rs. 10, 000 = Rs. 40, 000 (1 mark)Total savings (PPF + LIC + NSCs)

= Rs. (60,000 + 4 × 4500 + 30,000)= Rs. 10,8,000

But 100% exemption for savings upto Rs. 1,00,000 (1 mark)Taxable income = Rs. (410000 – 40000 – 100000)

= Rs. 2,70,000 (1 mark)

Total income tax = Rs. [13000 + (270000 – 250000) ]

= Rs. [13000 + 6000]= Rs. 19,000 (1 mark)

Education Cess @ 2% = = Rs. 380 (½ mark)

Net tax payable = Rs. (19000 + 380) = Rs. 19,380 (½ mark)


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