l.k. gupta (mathematics classes) www ... adc ar adp ar adq ar adp( ) − = −( ) ( ) ( ) ar apc ar...

L.K. Gupta (Mathematics Classes) wwwwww..ppiioonneeeerrmmaatthheemmaattiiccss..ccoomm MOBILE: 9815527721, 4617721

PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH

SOLUTIONS ON www.pioneermathematics.com →→→→ latest updates

1111

PIONEER GUESS PAPER

9th CBSE (SA-II)

MATHEMATICS

“Solutions” TIME: 2:30 HOURS MAX. MARKS: 80

GENERAL INSTRUCTIONS & MARKING SCHEME •••• All questions are compulsory.

•••• The questions paper consists of 34 questions divided into four sections A, B, C and D.

•••• Section A contains 10 questions of 1 mark each, which are multiple choice type questions,

Section B contains 8 questions of 2 mark each, Section C contains 10 questions of 3 marks each, Section D contains

6 questions of 4 marks each.

•••• There is no overall choice in the paper. However, internal choice is provided in one question of 2 marks, 3

questions of 3 marks and two questions of 4 marks.

•••• Use of calculators is not permitted.

NAME OF THE CANDIDATE PHONE NUMBER

L.K. Gupta (Mathematics Classes)




2222

Section−−−−A

1. x = 5, y = 2 is a solution of the linear equation

(A) x + 2 y = 7 (B) 5x + 2y = 7 (C) x + y = 7 (D) 5 x + y = 7

Sol: (C)

2. The equation x = 7, in two variables, can be written as

(A)1 . x + 1. y = 7 (B) 1. x + 0. y = 7 (C) 0. x + 1. y = 7 (D) 0. x + 0. y = 7

Sol: (B)

3. Three angles of a quadrilateral are 750, 900 and 750. The fourth angle is

(A) 900 (B) 950 (C) 1050 (D) 1200

Sol: (D)

4. The figure obtained by joining the mid-points of the sides of a rhombus, taken in order,

is

(A) a rhombus (B) a rectangle (C) a square (D) any parallelogram

Sol: (B)

5. ABCD is a trapezium with parallel with sides AB = a cm & DC = b cm. E & F are the mid-

points of the non-parallel sides. The ratio of ar (ABFE) & ar (EFCD) is

(A) a : b (B) (a+3b) : (3a+b)

(C) (3a+b) : (a+3b) (D) (2a+b) : (3a+b)

Sol: (C)




3333

( )

( )

( )

( )

1 ha b 2lar ABFE a b 2l2 2

1 har EFCD 2b 2lb b 2l

2 2

× + + ×+ +

= =+

× + + ×

….. (1)

As ∆ AOD and ∆ EMD are similar

⇒ l h / 2

x 2lx h

= ⇒ =

Put in (1)

( )

( )

a ba b

ar ABFE a b x 3a b2

a bar EFCD 2b x a 3b2b

2

− + +

+ + + = = =

−+ + +

[ ∵ a = b + 2x ⇒ a b

x2

−= ]

6. Diagonals of a parallelogram ABCD intersect at O. If

0 0BOC 90 and BDC 50 , then OAB is∠ = ∠ = ∠

(A) 900 (B) 500 (C) 400 (D)100

Sol: (C)

7. AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, the distance of

AB from the centre of the circle is :

(A) 17cm (B) 15cm (C) 4cm (D) 8cm

Sol: (D)

8. With the help of a ruler and a compass, it is possible to construct an angle of :

(A)350 (B)400 (C)37.50 (D)47.50




4444

Sol: (C)

9. The radius of a sphere is 2r, then its volume will be

(A) 34 /3 rπ (B) 34 rπ (C) 38 r /3π (D) 332/3 rπ

Sol: (D)

10. The class-mark of the class 130−150 is :

(A)130 (B)135 (C)140 (D)145

Sol: (C)

SECTION B

Question numbers 11 to 18 carry 2 marks each.

11. Determine the point on the graph of the equation 2x + 5y = 20 whose x-coordinate is 5

2

times its ordinate.

Sol:

As the x-coordinate of the point is 5

2 times its ordinate, therefore,

5x y

2= .

Now, putting 5

x2

= y in 2x + 5y = 20, we get, y = 2. Therefore, x = 5. Thus, the required point is

(5, 2).

12. Three angles of a quadrilateral ABCD are equal. Is it a parallelogram?

Sol:

It need not be a parallelogram, because we may have 0A B C 80∠ = ∠ = ∠ = and

0D 120 . Here, B D∠ = ∠ ≠∠ .

13. ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ (Fig.). If AQ

intersects DC at P, show that ar (BPC) = ar (DPQ)




5555

Sol:

( ) ( )ar ACP ar BCP= (1)

[Triangles on the same base and between same parallels]

( ) ( )ar ADQ ar ADC= (2)

( ) ( ) ( ) ( )ar ADC ar ADP ar ADQ ar ADP− = −

( ) ( )ar APC ar DPQ= (3)

From (1) and (3), we get

ar ( ) ( )BCP ar DPQ=

14. In the Fig., OD is perpendicular to the chord AB of a circle whose centre is O.

If BC is a diameter, show that CA = 2OD.

Sol:

Since OD AB⊥ and the perpendicular drawn from the centre to a chord bisects the chord.

∴ D is the mid-point of AB

Also, O being the centre, is the mid-point of BC.

Thus, in ABC,D∆ and O are mid-points of AB and BC respectively. Therefore,




6666

OD AC�

1and OD CA

2=

Segment joining the mid point s of two

sides of a triangle is half of the third side

∴ −

CA 2OD⇒ =

15. Draw a line segment AB of length 5.8 cm. Draw the perpendicular bisector of this line

segment.

Sol:

Steps of Construction:

(i) Draw a line segment AB = 5.8 cm.

(ii) With A is centre and radius more than half of AB, draw arcs, one on each side of AB.

(iii) From B (with the same radius) cut both arcs at P and Q.

(iv) Join PQ to meet AB at C.

(v) AC= BC = 2.9cm.

C⇒ bisects AB.

16. A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the

curved surface of the pillar at the rate of 12.50 per m2.




7777

Sol:

Radius = 25 cm = 25

m100

Height 3.5m=

C.S.A. 2 rh= π22 25 35

27 100 10

= × × ×11

m2

=

Cost = 12.50 per 2m

11 11 1250 275m Rs.68.75

2 2 100 4= × = =

17. A soft drink is available in two packs –

(i) a tin can with a rectangular base of length 5 cm and width 4 cm, having of 15 cm and

(ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container

has greater capacity and by how much?

Sol:

Volume of rectangular tin can l b h= × ×

5 4 15= × ×2300 cm=

Volume of plastic cylinder 2r hπ

22 7 710

7 2 2= × × ×

3385cm=

The plastic cylinder has greater capacity by 85 3cm [=385–300]

18. Calculate the mean for the following distribution:

x: 5 6 7 8 9

f: 4 8 14 11 3

Sol:




8888

x: f: xf

5 4 20

6 8 48

7 14 98

8 11 88

9 3 27

Total 40 281

if 40=∑

i ix f 281=∑

i i

i

x fx

f=∑∑

281

40= = 7.025

SECTION−−−−C

Question numbers 19 to 28 carry 3 marks each.

19. Draw a graph of each of the following equations :

(i) x + 2 = 0

(ii) 2y + 3 = 0

(iii)4x − 6 = 0

Sol:

(i) x + 2 = 0 ⇒x = −2

Clearly, it does not contain y. So, its graph is a line parallel to y-axis passing through the point

(−2, 0). In fact it passes through every point whose x-coordinate is −2.

Plotting any two points, say (−2, 0) and (2, 1) given by above table on the graph paper and

joining them, we obtain the straight line as shown.




9999

(ii) 2y + 3 = 0⇒ 2y=−3

⇒y = 3

2− =−1.5

The equation y=−1.5 does not contain x. So, its graph is a line parallel to x-axis passing through

the point (0, −1.5). Clearly, y=−1.5 means that for all values of the abscissa x, the ordinate y is

−1.5

Thus, we have the following table exhibiting the abscissa and ordinate of the points lying on

the line represented by given equations.




10101010

Plotting the points (0, −1.5), (1, −1.5) and (2, −1.5) on the graph paper and joining them by a

line, we obtain the graph of the line represented by the given equation as shown in graph.

(iii) 4x – 6 = 0

⇒4x = 6

⇒x = 6

1.54

=

20. If the point (3, 4) lies on the graph of 3y = ax + 7, then find the value of a.

Sol:

If point (3, 4) lies on the graph of 3y = ax + 7, then it must satisfy it.

( ) ( )3 4 a 3 7⇒ = +

512 3a 7 3a 5 a

3⇒ = + ⇒ = ⇒ =

21. In a quadrilateral ABCD, CO and DO are the bisectors of C and D∠ ∠ respectively.

Prove that 1

COD ( A B)2

∠ = ∠ + ∠ .

Sol:




11111111

Given: Quadrilateral ABCD , CO and DO are the bisectors ∠C and ∠D.

Proof: In COD∆

∠COD = 180o

– ( ∠ OCD + ∠ ODC)

∠COD = 180o 1

( )2

− ∠ + ∠C D

∠COD = 180o

− o1

[360 [( )]2

− ∠ + ∠A B

∠COD = 180o

– 180o

+ 1

( )2

A B∠ + ∠

∠ COD = 1

( )2

A B∠ + ∠ Hence proved.

22. If ABCD is a parallelogram, the prove that

ar ( ) ( ) ( ) ( ) ( )gm1ABD ar BCD ar ABC ar ACD ar ABCD

2∆ = ∆ = ∆ = ∆ = �

Sol:

To prove: ar gm1ΔABD ar(ΔBCD) ar(ΔABC) ar(ΔACD) ar( ABCD)

2= = = = �

Constructions: Draw DP ⊥ AB, BQ ⊥ CD.




12121212

Proof:

In ∆ABD, 1

ar ΔABD .AB.DP2

=

…. (1)

In ∆BCD, 1

ar ΔBCD DC.BQ2

=

….. (2)

As AB = CD, PD = BQ = distance between parallel line.

So (1) and (2) are equal.

ar ΔABD ar ΔBCD∴ = …. (3)

Similarly ar ∆ABC = ar ∆CDA

ar || ABCD ar ΔABD arΔBCD= + (but they are equal)

ar || ABCD 2 ar ΔABD=

1ar || ABCD ar ΔABD

2= …. (4)

From (3) and (4)

ar gm1ΔABD ar(ΔBCD) ar(ΔABC) ar(ΔACD) ar( ABCD)

2= = = = �

Hence proved.

23. Find the length of a chord which is at a distance of 5 cm from the centre of a circle of

radius 10 cm.

Sol:

Let AB be the chord and O is the centre of the circle

from O draw OL AB⊥




13131313

Join OA. Since , perpendicular from the centre of the circle bisects the chord.

1AL LB AB

2∴ = =

In ΔOAL ,

2 2 2OA OL AL= +

2 2 2(10) (5) AL⇒ = +

⇒ 2AL 100 25 75= − =

AL 75 3 5 5 5 3 cm⇒ = = × × =

So AB = 2 AL 2 5 3 10 3 cm× = × =

24. How many liters of water flow out of a pipe having an area of cross-section of 5 cm2 in

one minute, if the speed of water in the pipe is 30 cm/sec?

Sol:

Speed of water in the pipe = 30 cm / sec

volume of flowing water in 1 second 2r h= π35 30 150cm= × =

volume of water flowing in 1 minute

3150 60 9000 cm= × = =9L

[ 31cm 1000L= ]

25. A well with 10 m inside diameter is dug 8.4 m deep. Earth taken out of it is spread all

around it to a width of 7.5 m to form an embankment. Find the height of the embankment.

Sol:

Volume of earth dug out 2r h= π = 22 84

57 10

× ×5× 660= m3

Area of embankment= 2 2(R r )π −

= ( ) ( )( )2 22212.5 5

7× −




14141414

=22

7×(156.25 − 25)

=22 13125

7 100×

5775412.5

14= = m2

Height volume

Area= =

660

412.5=

660 10

4125

×1.6 m=

26. The ages of ten students of a group are given below. The age have been recorded in

years and months:

8 – 6, 9 – 0, 8 – 4, 9 – 3, 7 – 8, 8 – 11, 8 – 7, 9 – 2, 7 – 10, 8 – 8

(i) What is the lowest age?

(ii) What is the highest age?

(iii) Determine the range?

Sol:

(i) 7 years 8 months

(ii) 9 years 3 months

(iii) Range =Max − Min= 9 years 3 months –7 years 8 months = 1 year 7 months

27. Following are the ages of 360 patients getting medical treatment in a hospital on a day:

Construct a Cumulative frequency distribution

Age (in years): 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70

No. of Patients : 90 50 60 80 50 30

Sol:




15151515

Marks No of students Cumulative Frequency

10–20 90 90

20– 30 50 140

30– 40 60 200

40–50 80 280

50–60 50 330

60–70 30 360

28. The percentage of marks obtained by a student in monthly unit tests are given below :

Unit test : I II III IV V

Percentage of marks obtained: 69 71 73 68 76

Find the probability that the student gets:

(i) more than 70% marks

(ii) less than 70% marks

(iii) a distinction.

Sol:

(i) P(more than 70% marks)

= Total no.of percentagesabove70%

P(A)Total numberof pecentages

=3

0.65

= =

(ii) P(less than 70% marks) = Total no.of percentagesbelow70%

P(A)Total numberof pecentages

=2

0.45

= =

(iii) A distinction 1

0.25

= =




16161616

Section−−−−D

Question numbers 29 to 34 carry 4 marks each

29. Draw the graphs of the lines represented by the equations x + y = 4 and 2x −y = 2 in the

same graph. Also, find the coordinates of the point where the two lines intersect.

So, the common point is (2, 2) which is the point of intersection of both lines.




17171717

30. ABCD is a square E, F, G and H are points on AB, BC, CD and DA respectively, such that

AE = BF = CG = DH. Prove that EFGH is a square.

Sol:

Given ABCD is a square ,E,F,G,H are the mid points of AB , BC ,CD and DA

AE = BF = C G = DH

To prove : - EFGH is a square .

Proof : AE = BF = CG = DH = x

∴ BE = CF = DG = AH = y

In ΔAEHandΔBFE

A E = BF (given) , ∠A = ∠B = [ each 90 o , AH = BF [ given]

By SAS ≅ , triangles are ≅

ΔAEH BFE≅ ∆

∠1 = ∠2 and ∠3 = ∠4

But, ∠1 + ∠3 = 90 o

and ∠2 + ∠ 4 = 90 o

∠1 + ∠3 + ∠2 + ∠4 = 90 o + 90 o

∠1 + ∠4 +∠1 + ∠4 = 180 o

2 ( ∠1 + ∠4) = 180 o




18181818

∠1 + ∠4 = 90 o

∠HEF = 90 o

Similarly , ∠F = ∠G = ∠H = 90 o

Hence, EFGH is a square

31. In a ∆ ABC, P and Q are respectively the mid-points of AB and BC and R is the mid-point

of AP. Prove that:

(i) ( ) ( )ar PBQ ar ARC∆ = ∆

(ii) ( ) ( )1

ar PRQ ar ARC2

∆ = ∆

(iii) ( ) ( )3

ar RQC ar ABC8

∆ = ∆ .

Sol:

Given : P and Q are the mid points of AB and BC R is the midpoint of AP

To prove: (i) ar (PBQ ) = ar ( ARC∆ ) (ii) ar ( PRQ∆ ) = 1

ar(ARC)2

(iii) ar ( RQC∆ ) = 3

ar( ABC)8

∆

Proof : We know that each Median of a triangle divides it into 2

triangles of equal area

(i) Since CR is the median of CAP∆

1ar( CRA) ar( CAP) ..(i)

2∴ ∆ = ∆

also , CP is a median of of CAB∆

ar( CAP) ar(CPB)..(ii)∴ ∆ =

from (i) and (ii) we get

1ar( ARC) ar( CPB)..(iii)

2∴ ∆ = ∆




19191919

PQ is the median of PBC∆

ar( CPB) 2ar( PBQ)..(iv)∴ ∆ = ∆

from (iii) and (iv) we get

ar( ARC) ar( PBQ)..(v)∴ ∆ = ∆

(ii) Since QP and QR medians of triangles QAB and QAP

ar( QAP) ar( QBP)..(vi)∴ ∆ = ∆

and ar (QAP) 2ar( QRP) ..(vii)∆ = ∆

from (vi) and (viii) we get

ar 1

( PRQ ar( PBQ) .. .(viii)2

∆ = ∆

ar (1

PRQ ar( ARC)2

∆ = ∆ …(viii)

(iii) since CR is a median of CAP∆

1ar( ARC) ar( CAP)

2∴ ∆ = ∆

= 1 1

ar(ABC) [ CPis themedianof ABC]2 2

∆

∵

= 1

ar( ABC) ...(ix)4

∆

Since RQ is median of RBC∆

1ar( RQC) ar( RBC)

2∴ ∆ = ∆

= 1

{ar(ABC) ar( ARC)}2

− ∆

= 1 1

{ar( ABC) ar[ ABC}[from(ix)]2 4

∆ − ∆

= 3

ar( ABC)8

∆

Hence proved.




20202020

32. Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F

respectively. Prove that the angles of ∆ DEF are 0 0 0A B C90 , 90 and 90

2 2 2− − − .

Sol:

We have,

BED BAD∠ =∠

ABED

2

∠⇒ ∠ =

Also, BEF BCF∠ = ∠

1BEF C

2⇒ ∠ = ∠

1 1BED BEF A C

2 2∴ ∠ + ∠ = ∠ + ∠

( )1

DEF A C2

⇒ ∠ = ∠ + ∠

( )01DEF 180 B

2⇒ ∠ = −∠ 0A B C 180 ∴∠ +∠ +∠ =

0 1DEF 90 B

2⇒ ∠ = − ∠

We observe that AF subtends angles ADF and ACF∠ ∠ at points D and C in the same segment.

ADF ACF∴ ∠ = ∠

1ADF C

2⇒ ∠ = ∠

Also,

ADE ABE∠ = ∠

1ADE B

2⇒ ∠ = ∠

1 1ADE ADF B C

2 2∴ ∠ +∠ = ∠ + ∠




21212121

( )1

EDF B C2

⇒ ∠ = ∠ + ∠

( ) 01 1EDF 180 A 90 A

2 2⇒ ∠ = − ∠ = − ∠

Similarly, we have

0 1DEF 90 C

2∠ = − ∠

33. A circus tent is cylindrical to a height of 3 metres and conical above it. If its diameter is

105 m and the slant height of the conical portion is 53 m, calculate the length of the canvas 5 m

wide to make the required tent.

Sol:

Slant height of cone = 53m

Height of cylinder = 3m

Diameter = 105 m

Radius = 105

52.52

=

C.S.A. of conical portion πrl=

C.S.A. of cylindrical portion 2πrh=

Total C.S.A of circus tent = πrl 2πrh+

2

πr(l 2h)

2252.5(53 2(3))

7

2252.5(59)

7

9735m

= +

= × +

= ×

=

Area of canvas = l b×

So, 9735 l 5

9735l

5

= ×

⇒ =




22222222

So, length of the canvas =1947m .

34. The time taken, in seconds, to solve a problem by each of 25 pupils is as follows:

16, 20, 26, 27, 28, 30, 33, 37, 38, 40, 42, 43, 46, 46, 46, 48, 49, 50, 53, 58, 59, 60, 64, 52, 20

(A) Construct a frequency distribution for these data using a class interval of 10 seconds.

(B) Draw a histogram to represent the frequency distribution.

Sol:

(a) Frequency distribution

(b)




23232323

We represent the class limit along X-axis on a suitable scale and the frequencies along y-axis

on a suitable scale.

Since the scale on X-axis starts at 15, a kink (break) is indicated near the origin to signify that

the graph is drawn to scale beginning at 15, and not at the origin.

l.k. gupta (mathematics classes) www ... adc ar adp ar adq ar adp( ) − = −( ) ( ) ( ) ar apc ar...

Documents