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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

Question numbers 1 to carry 1 mark each. For each of the questions 1-10, four

alternative choices have been provided of which only one is correct. You have to select

the correct choice.

1. The graph of the linear equation y = x passes through the point

(A) 3 3

,2 2

−

(B) 3

0,2

(C) (1, 1) (D) 1 1

,2 2

−

Sol: (C)

2. The graph of y = 6 is a line

(A) parallel to x-axis at a distance 6 units from the origin

(B) parallel to y-axis at distance 6 units from the origin

(C) making an intercept 6 on the x-axis.

(D) making an intercept 6 on both the axes.

Sol: (A)

3. In Fig., if AOB is a diameter of the circle and AC = BC then CAB∠ is equal to:

(A) 300 (B) 600 (C) 900 (D) 450

Sol: (D)

4. In Fig., AB and CD are two equal chords of a circle with centre O. OP and OQ are

perpendiculars on chords AB and CD, respectively. If 0POQ 150 , then APQ∠ = ∠ is equal to




3333

(A) 300 (B) 750 (C) 150 (D) 600

Sol: (B)

5. In Fig. the area of parallelogram ABCD is :

(A) AB × BM (B) BC × BN (C) DC × DL (D) AD × DL

Sol: (C)

6. A coin is tossed 1000 times with following frequencies Head : 455, Tail : 545, then

probability of a head is

(A) 0.455 (B) 0.547 (C) 0.545 (D) 0.575

Sol: (A)

7. In the given Fig. if 0P 40∠ = , then value of x is :

(A) 400 (B) 500 (C) 600 (D) 300

Sol: (B)




4444

8. If 0N 33∠ = , then the value of LOM∠ (Fig. ) is

(A) 570 (B) 330 (C) 900 (D) 660

Sol: (D)

9. The total surface area of a cube is 96 cm2. The volume of the cube is :

(A) 8 cm3 (B) 512 cm3 (C) 64 cm3 (D) 27 cm3

Sol: (C)

10. The marks obtained by 17 students in a mathematics test (out of 100) are given below:

91, 82, 100, 100, 96, 65, 82, 76, 79, 90, 46, 64, 72, 68, 66, 48, 49.

The range of data is:

(A) 46 (B) 54 (C) 90 (D) 100

Sol: (B)

Section B

Question numbers 11 to 18 carry 2 marks each.

11. The cost of a Video game is Rs 50 more than the twice the cost of a calculator write a

linear equation in two variables to represent this statement.

Sol:

Cost of Video game = 50 + 2 (cost of calculator)

Let cost of Video game = Rs x

Let cost of Calculator = Rs y




5555

∴ x = 50 = 2y

⇒ x − 2y − 50 = 0

12. Diagonal AC of a parallelogram ABCD bisects ( )A See fig.∠

Show that : -

(i) it bisects C∠

(ii) ABCD is a rhombus

Sol:

Given:- 1 2∠ = ∠ and ABCD is a � gm

To Prove : (i) AC bisects C∠

(ii) ABCD is a rhombus

Proof : -

(i) ABCD is a �gm

AB CD∴ � and AC is a transversal which intersects them

2 3 , 1 4∴ ∠ =∠ ∠ = ∠ (alternate interior s∠ )

But ( )1 2 given∠ =∠

3 4∴ ∠ =∠

Hence AC bisects C∠




6666

(ii) 1 2 and 1 4∠ =∠ ∠ =∠

2 4∴ ∠ =∠

AB BC⇒ = (Sides opposite to equal angles are equal )

Similarly AD = CD

Hence ABCD is a rhombus.

13. Let ABCD be a parallelogram of area 124 cm2. If E and F are the mid-points of sides AB

and CD respectively, then find the area of parallelogram AFED.

Sol:

Given AB = CD [ opposite sides of parallelogram are equal]

1CD[ E and F is midpoint ]

2∴ ∵ = AE = DF,

EB FC∴ = , Since , AEFD = EBCF

Since ar.(AEFD) ar.(EBCF)=

ar . (AEFD) = 1

ar.(ABCD)2

ar. (AEFD) = 21124 62cm

2× =

14. Find the length of a chord which is at a distance of 4 cm from the centre of the circle of

radius 6 cm.

Sol:

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

from O draw OL AB⊥ , join OA

since perpendicular from the centre of the circle bisects the chord.




7777

1AL LB AB

2∴ = =

In ΔOAL

2 2 2OA AL OL= +

2 2 2(6) AL (4)= +

2AL 36 16 20= − =

AL 20=

So AB = 2 AL 2 20× =

= 2 2 5 4 5 cm× =

15. In Fig., O is the centre of the circle. Find BAC∠ .

Sol:

OB = OA ( )OBA OAB x say∴ ∠ = ∠ =

OA = OC ( )OAC OCA y say∴ ∠ =∠ =

In OAB∆ , x + x + 80 = 1800

⇒2x = 100




8888

⇒x = 500

In 0 0OAC, y y 110 180∆ + + =

⇒2y = 700

⇒y = 350

0 0BAC x y 50 35∴ ∠ = + = + = 850

16. The areas of three adjacent faces of a cuboid are x, y and z. If the volume is V, prove that

V2 = xyz.

Sol:

Area of adjacent faces are x, y, z

l b x , b h y , h l z∴ × = × = × =

( ) ( ) ( )l b b h h l xyz× × × × × =

2 2 2l b h xyz= ……..(1)

Volume of cuboid = l b h = V

(1) ( )2

lbh xyz⇒ ∴ =

V2 = xyz

17. In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5

cm. Find the total radiating surface in the system.

Sol:

Length = 28 m =2800 cm

radius = 2.5 cm [r=diameter/2=5/2=2.5cm]

C.S.A. = 2 rhπ

22 252 2800

7 10= × × × = 44000 cm




9999

18. In a cricket match, a batsman hits a boundary 6 times out of 30 balls he plays, Find the

probability that on a ball played:

(i) he hits boundary (ii) he does not hit a boundary

Sol:

(i) P (batsman hits boundary) = 6

30

(ii) P (batsman does not hit boundary) 1 1 4

5 5

−= =

Section−−−−C

Question numbers 19 to 28 carry 3 marks each.

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

(i) −3x + 4y = 12 (ii) 3x − 2y = 0

Sol:

(i) – 3x + 4y =12

⇒4y=12+3x

⇒y=12 3x

4

+

When x=4, then y = 12 3(4) 24

64 4

+= =

When x=0, then y = 12 3(0) 12

34 4

+= =

When x=−4, then y = 12 3( 4)

04

+ −=




10101010

(ii) 3x – 2y = 0

⇒2y = 3x

⇒y = 3

x2

When x=6, then y = 3

(6) 92

=

When x=2, then y = 3

(2) 32

=

When x=−6, then y = 3

( 6) 92

− = −




11111111

20. The taxi fare in a city is as follows: -

For the first kilometer, the fare is Rs 8 and for the subsequent distance it is Rs 5 per km. Taking

the distance covered as x km and fare as Rs y, write a linear equation for this information and

draw its graph?

Sol:

The total distance covered = x km

For 1st km, the fare = Rs 8

1 km the fare = Rs 8

∴ For Remaining distance, the fare = Rs. 5

( )x 1− distance the fare = Rs. 5

Total fare = 8 × 1 + 5 ( )x 1−

y = 8 + 5x − 5

y = 3 + 5x

x –1 –0.5 0

y 2 0.5 3




12121212

21. ABCD is a parallelogram, AD is produced to E so that DE = DC and EC produced meets AB

produced in F. Prove that BF = BC.

Sol:

Given : - ABCD is a parallelogram , DE = DC,

To prove :- BF = BC

Proof: in ΔACE,DandOare the mid points of AE and AC




13131313

∴ DO � EC , OB � CF , AB = BF------- (i) , DC = BF

[ AB DC ABCDisaparl log ram=∵ ] ,

In ΔEDCandΔCBF we have

DC=BF

∠EDC = ∠CBF

and ∠ECD = ∠CFB, So By ASA ≅ rule Δ'sarecongruent,

we have

ΔEDCandΔCBF are congruent

⇒DE=BC

⇒DC=BC

⇒AB = BC

⇒BF = BC [ AB BFfrom(i)]=∵

Hence proved.

22. In Fig. PQRS is a square and T and U are, respectively, the mid-points of PS and QR.

Find the area of ∆ OTS if PQ = 8 cm.

Sol:

Given: PQRS is a square T and U are the mid- points PS and QR , PQ = 8 cm.

To find: area of OTS∆

Since T and U are the mid points of PS and QR

TU PQ∴ �

In PQS,Tis themidpoint∆ of PS and TO PQ�




14141414

1TO PQ 4 cm

2∴ = =

TS = 1

PS 4 cm2

=

1ar( OTS) (TO TS)

2∴ ∆ = ×

= 21(4 4)cm

2× = 21

16 8cm2

× =

23. An equilateral triangle of side 9 cm is inscribed in a circle. Find the radius of the circle.

Sol:

Let O be the centre of the circle and join OB and OC. draw an perpendicular

OL BC⊥

0OBL 30∠ = 0 0 0B 180 (60 50 ) ∠ = − + ∵

Now in , ΔOBL

0

9BL 2cos30OB OB

= =

3 9

2 2OB⇒ =




15151515

9OB 3 3 cm

3= =

24. A rectangular tank is 80 m long and 25 m broad. Water flows into it through a pipe

whose cross-section is 25 cm2, at the rate of 16 km per hour. How much the level of the water

rises in the tank in 45 minutes.

Sol:

Tank length = 80m, breadth = 25m, Let height = h

Volume = l b h× × = 80000 2500 h× ×

Area of cross section 225cm=

Length of cross section 45

16000 10060

= × ×

Volume of water in 45 min 45

25 16000 10060

= × × ×

458000 2500 25 16000 100

60h∴ × × = × × ×

4525 16000 100

60 1.58000 2500

h cm

× × ×

= =×

25. A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the

bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to

serve 250 patients?

Sol:

Radius 7

cm2

=

Height 4cm=

2 22 7 7r h 4

7 2 2π = × × ×




16161616

3154 cm=

3250 154 38500 cm× = =38.5L [∵1000cm3 =1L consult unit converter in the website]

26. The following table shows the daily production of T.V. sets in an industry for 7 days of a

week:

Day Mon Tue Wed Thu Fri Sat Sun

Number of T.V. Sets 300 400 150 250 100 350 200

Represent the above information by a pictograph.

Sol:

To represent above data by a bar graph, we graph, we first draw a horizontal & vertical line.

Since seven values of numerical data are given. So, we mark seven points on the horizontal line

at equal intervals and erect rectangles of same width at these points. The height of rectangles

are propotional to the numerical values of the data as shown in figure.

27. A die was rolled 10500 times the frequency of each outcome is shown in the table. What

was the empirical probability of each outcome?




17171717

Sol:

Total no. of trials = 10500

Empirical probability = No. of favourable events

Total no. of events

(i) P(getting a one on the throw of die) = 1675

10500=

67

420

(ii) P (getting two on the throw of die) = 1725

10500=

69

420

(iii) P(getting 3 on side) = 1642 821

10500 5250=

(iv) P(getting 4 on die) = 1768

10500=

442

2625

(v) P(getting 5 on die) = 1873

10500

(vi) P(getting 6 on die) = 1817

10500

28. The following table gives the life time of 400 neon lamps:

Life time

(in hours)

300−400 400−500 500−600 600−700 700−800 800−900 900−1000

Number

of lamps:

14 56 60 86 74 62 48

A bulb is selected at random. Find the probability that the life time of the selected bulb is:

(i) less than 400 (ii) between 300 to 800 hours (iii) at least 700 hours.

Sol:




18181818

Total no of trials = 400

(i) Let the event be A the trials in which event A happened = 14

Total no.of trials in whichevent A happenedP(A)

Total numberof trials=

14 7

400 200= =

(ii) Let the event be A The trials in which event A happened

14 56 60 86 74

290

= + + + +

=



290 29

400 40= =

(iii) Let the event be A The trials in which event A happened

74 62 48

184

= + +

=



184 92 23

400 200 50= = =

Section−−−−D

Question numbers 29 to 34 carry 4 marks each

29. The parking charges of a car on Mumbai Railway station for first two hours is Rs 50 and

Rs 10 for subsequent hours. Write down an equation and draw the graph of this. Read the

charges from the graph :

(i) for one hour (ii) for three hours (iii) for five hours

Sol:

NORMAL THINKING




19191919

Parking charges for first 2 hrs = 50 Rs.

Parking charges for above 2 hrs = 10 Rs./hrs

∴ Parking charges for x hrs. = 50 + 10 (x–2)

⇒ y = 30 + 10x where y = total charges, x = total hours.

x 1 3 5

y 50 60 80

Now read the charges from graph

(i) for one hour : (1, 40) , means for 1 hour, parking charges are Rs. 40 (Wrong)




20202020

(ii) for three hours : (3, 60), means for 3 hours, parking charges are Rs. 60.

(iii) for five hours: (5, 80), means for 5 hours, parking charges are Rs. 80.

PIONEER SOLUTION:

Parking charges for first 2 hrs = 50 Rs.

Parking charges for above 2 hrs = 10 Rs./hrs

∴ Parking charges for x hrs. = 50 + 10 (x–2) where x >2

⇒ y = 30 + 10x where y = total charges, x = total hours.

∴ y = 30 + 10x for x > 2

x 0 1 2 3 4 5

y 50 50 50 60 70 80

As y = 50 for 0 < x < 2, because parking charges for first 2 hrs are Rs 50.

Now read the charges from graph




21212121

(i) for one hour : (1, 50) , means for 1 hour, parking charges are Rs. 50

(ii) for three hours : (3, 60), means for 3 hours, parking charges are Rs. 60.

(iii) for five hours: (5, 80), means for 5 hours, parking charges are Rs. 80.

30. In Fig. , O is the centre of the circle, prove that x y z∠ = ∠ + ∠ .

Sol:

To prove ,

x y z∠ = ∠ + ∠

In figure 1 3 y∠ + ∠ = ∠ ………..(1)

[∵ external angle is equal to the sum of internal angles ]

Similarly 2 4 y .......(2)∠ + ∠ = ∠

and x 2 3 2 4∠ = ∠ = ∠ ……..(3)

therefore the angle subtended by on are of a circle at the centre is double the angle

subtended by it at any point on the remaining fact at the circle .

In quadrilateral D P R Q

0 0 0 0 0 0Z (180 13 ) y (180 4) 360∠ + − + ∠ + − ∠ =

0 02 y 3 4 2 3 .......(4)⇒∠ + ∠ = ∠ + ∠ = ∠

[ ]3 4∠ =∠∵

from (1 ) and (2) by adding




22222222

1 3 2 4 2 y∠ + ∠ + ∠ + ∠ = ∠

1 2 3 4 2 y⇒∠ + ∠ + ∠ + ∠ = ∠

2 1 2 3 2 y⇒ ∠ + ∠ = ∠ [ ]1 2 3 4∠ = ∠ ∠ = ∠∵

from (3)

02 1 x 2 y⇒ ∠ + = ∠

02(y 3) x 2 y⇒ − ∠ + = ∠

[ ]y 1 3∠ = ∠ + ∠∵

02 y 2 3 x 2 y∠ − ∠ + = ∠

0x 2 3= ∠ from (4 )

0 0 0x y z∠ = ∠ + ∠

31. ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD

at F.

(i) Prove that ar( ) ( )ADF ar ECF∆ = ∆

(ii) If the area of DFB∆ = 3 cm2, find the area of gm� ABCD.

Sol:

Given: ABCD is a parallelogram , P is point on BO

To prove : (i) ar ( ADO∆ ) = ar (CDO) (ii) ar ( ABP) ar( CBP).∆ = ∆

Proof : Since diagonals of gm� bisected each other




23232323

Ois∴ the midpoint of AC as well as BD

In ACD,DO is the median∆

ar( ADO) ar( CDO)∴ ∆ = ∆

(ii) Since o is the mid point of AC.

OP and OB are medians of TrianglesAPC and ABC∴

ar( AOP) ar( COP)andar( AOB) ar( COB)∴ ∆ = ∆ ∆ = ∆

⇒ ar ( AOB∆ ) - ar ( AOP ) = ar ( COB∆ ) - ar ( COP∆ )

⇒ ar ( ABP∆ ) = ar ( CBP∆ )

Hence proved.

32. Construct a triangle whose perimeter is 6.4 cm, and angles at the base are 600 and 450.

Sol:

To draw PQR∆ we follow the following steps

Steps of constructions:

Step- I: Draw a line segment XY = 4.6 cm

Step- II: Construct 0 0YXD P 45 and XYE 60∠ = ∠ = ∠ = ∠θ =

Step- III: Draw the bisectors of angles YXD and XYE∠ ∠ make their point of intersection as R

Step- IV: Draw right bisectors of RX and RY meeting XY at P and Q respectively .

Step- V: Join PR and QR to obtain the required PQR∆




24242424

33. A wooden toy is in the form of a cone surmounted on a hemisphere. The diameter of the

base of the cone is 16 cm and its height is 15 cm. Find the cost of painting the toy at Rs 7per

100 cm2.

Sol:

C.S.A of cone = πrl

l = 2 2h r+

2 2(15) (8)= +

225 64= + 289 17= =

C.S.A of cone = πrl

228 17

7= × ×

2992427.43

7= =

C.S.A of hemisphere = 22πr

= 22

2 8 87

× × ×2816

7= = 402.28

Total Area to be painted = 427 .43 + 402.28 = 829.71 2cm

Cost of painting 100 2cm = Rs. 7




25252525

Cost of painting 1 2cm = 7

100

Cost of painting 829. 71 2cm = ×

23(12) (6.75) 3

r4

=

= Rs 58 .08

34. The monthly profits (in Rs.) of 100 shops are distributed as follows:

Profits per shop : 0−50 50−100 100−50 150−200 200−250 250−300

No. of shops: 12 18 27 20 17 6

Draw a histogram for the data and show the frequency polygon for it.

Sol:

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

on a suitable scale.

Taking class-interval as bases and the corresponding frequencies as heights, we construct

rectangles to obtain histogram of the given frequency distribution as shown above.

l.k. gupta (mathematics classes) … sol(1).pdf · if e and f are the mid-points of sides ab and cd...

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