Marking Scheme

Secondary

October - November, 2000

Subject : Mathematics

Part -A

Maximum Marks 20

Q.No. Answers Q.No. Answer

Set A B C Set A B C

XXI/OS/O/2

1 32 36 A

2 31 35 B

3 30 34 D

4 29 33 B

5 28 32 C

6 27 31 D

7 26 30 B

8 25 29 A

9 24 28 D

10 23 27 C

11 16 18 D

12 15 17 A

13 14 16 D

14 13 15 D

15 12 14 B

16 11 13 B

17 10 12 C

18 9 11 B

19 8 10 C

20 7 9 C

21 6 8 C

22 5 7 B

23 22 6 C

24 21 5 B

25 20 4 D

26 19 3 B

27 18 2 B

28 17 1 A

29 40 40 C

30 39 39 B

31 38 38 C

32 37 37 D

33 36 26 C

34 35 25 D

35 34 24 C

36 33 23 A

37 4 22 C

38 3 21 C

39 2 20 B

40 1 19 B

(½ mark for each correct answer)

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PART -B

Maximum Marks : 80

Q. No Expected Value Points For dISTRIBUTION Total

a b c EAch Step OF MARKS

1 - 3 P + 2 = 12 or P = 10 , 1
2 - 4 Exp = X

1

3 - - x/3-x/5=8; x=60 + 1

4 3 5 800x65/100 = 520 marks 1 1

5 - 1 1 1

6 4 10 55º, 125º, 125º 1 1

7 5 2 90º 1 1

8 6 -

Verification 1

9 7 7 9-13, 13-17, 17-21, 21-25, 25-29 1(if any two are given) 1

10 8 - 9 1 1

11 - 13

1

Factors of () are

(4x-3) and (x-3) 1

= or 1 3

12 - 14 (A and B)'s one day work = 1/8

(B and C)'s one day work = 1/12

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a b c EAch Step OF MARKS

(C and A )'s one day work = 1/16

(A and B and C )'s one day work

= 1/2

C's one day work = 13/96-1/8= 1/96 1

A's one day work =13/96-1/12=5/96

B's one day work = 13/96- 1/16=7/96
Ratio of A's B's and C one day work is 5 : 7 : 1 1

A's share 5/13×1950 Rs. 750

B's share = 7/13×1950 = Rs. 1050 1 3

C's share = 1/13 × 1950 = Rs. 150

13 - 12 Draw perpendicular from O on 1

sides AB, BC and CD. O is

equidistant from AB and BC 1

OL= OM

Again O is equidistant from BC and CD,

OM = ON

Hence, OL = ON 1 3

14 12 11 tan 45º = 1; sec 60º = 2; cosec 30º = 2

cot 45º =1, sin 90º = 1, cos Oº =1 1

Exp = 1

= 1/2 3

15 13 - 1. Start with number (1.00)

2. Increase the number by 0.10

3. If the square of this number is less then 2, go to step 2

4. Decrease the number by 0.10 mark each for 6

5. Increase the number by 0.01 correct steps

6. If the square of the number is less then 2, go to the step 5 3

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a b c EAch Step OF MARKS

7. Decrease the number by 0.01 to arrive at the value of
up to two decimal places.

16 14 -

mark for each

rectangle (×5) =2
mark for polygon 3

17 18 16 1.Obtain a,b,c 2. Calculate D= mark for each

3. Is D O ? 4. Write. The equation has no real roots. box 1,2,4,5 3

5. Write : The equatuion has two real roots 1 mark for box 3

18 17 17 Area of base = 3/5 area of curved surface

1

2 3

19 - 21

1 mark for axes with
correct units

2 marks for each
line drawn

1 mark for solution 4

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a b c EAch Step OF MARKS

20 19 19 Suppose orignal annual milage

is A and orignal petrol price is B. If car goes M units in
1 unit of petrol.

Orignal bill = AB/M 1

'New milage =

New price of petrol =

New annual bill = (95× 110/100×100) AB/M 1

Percentage increase is

= (209/200×100)_100

i.e 209/2 _100 = 9/2 or 4% 1 4

21 20 - Present value of Scooter = Rs. 15,625

Rate of depreciation = 20%

time = 3 years

Let the price of Scooter after

3 years be = A

A = P (1-R/100)ⁿ 1

= 15625 [1-20/100]³ 2

Price after 3 years will be Rs. 8,000 1 4

22 24 20 Proof :

In ABD and CAD 1 mark for

ADB = CDA ----------(1) fig and given

BAD + CAD = 90º 1

and C + CAD = 90º

BAD + CAD = C + CAD

or BAD = C------------(2) 1

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From (1) and (2)

ABD CAD

BD/AD = AD/ DC

or = BD×DC 1 4

23 - 24 Side of 1st square = 72/4 = 18 cm

Side of 2nd Square = 96/4 = 24 cm

sum of areas of squares = 1

= 900 cm²

Side of the new square = 30 cm

Perimeter = 120 cm 1

Diagonal = cm 1 4

24 22 26 SNo. x_i f_i x_if_i

1. 3 6 18

2. 5 8 40 1 mark for

3. 7 15 105

4. 9 p 9p 1 mark for

5. 11 8 88

6. 13 4 52

———————————— —

Total 41+p 303+9p

—————————————

= (303 + 9p)/(41+p) 1

7.5 = (303 + 9p)/(41+p)

Solving p =3 1 4

25 23 - Fig.

In ABC , BC = AB=h 1

In ABD

BD = h cot 30º

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a b c EAch Step OF MARKS

= h

BD= BC +10 1

10= h-h

h=10/(-1)=10/0.73= 13.7 m (appox.) 1 4

26 - 23 Volume of cuboid = (100×80×64)cm³ 1

Side of the cube = 1

= 5×4×4 cm

Surface area = 6 (80)² 1 4

= 38400 cm² 1

27 28 27

Correct figures 5

Steps of construction : 2 7

Steps of the construction :

1. Draw BC = 3.5 cm

2. At B draw YBC = 45º

3. With B as centre and radius 8 cm draw arc which meets
BY at D

4. Join DC

5. Draw the right bisector of DC which meets BD in A

6. Join AC

Then ABC is the required triangle

28. 27 - Minimum balance between 11^th and last day of the month

January Rs. 100 1

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Q. No Expected Value Points For dISTRIBUTION Total

a b c EAch Step OF MARKS

February Rs. 6000 1

March Rs. 4000 1

Total : Rs. 10100

Interest @ 5% per annum

= Rs. 10100 × 5/100 × 1/12

= Rs. 505/12 = Rs. 42.08 2

She will get Rs. 6042.08 7

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Part B

Note : New Questions for Set B

XXI/OS/O/2 B

Q. No Expected Value Points For dISTRIBUTION Total

b EAch Step OF MARKS

1

x =1 1

2 Exp

1

9 Boys/ girls = 2/3 i.e 24/G =2/3

G = 24×3/2 = 36 1

10 Let the numbers be x and (24-x)

(24-x)= x/3 i.e x+x/3 =24

x=3×24/4 = 18 1

numbers are 18,6

11 2

3

15 DF =BC; DE= AC; EF=AB 1

Since BC=AC=AB 1

DF=DE=EF 1 3

16 Distance covered =(260+140)m

=400m

Relative speed =(40+50)km/hr 1

= 90km/hr

= 9000 m/hr. 1

Time required = 400/90000 hours

1 3

= 16 sec.

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Q. No Expected Value Points For dISTRIBUTION Total

b EAch Step OF MARKS

21 1 mark for arces
with correct units

1 mark for each
line drawn

1 mark for finding
the answer 4

25 The radius of circumscribed

circle =(diagonal of the square) 1

=×cm

Area of circumscribed circle = =22/7×

= 22/7 ×100/2

= 1100/7

=157.14 1

The radius of incribed

circle =(side the square)

=×10 cm 1

Area of incribed circle

=22/7×25

=550/7

= 78.5 1 4

26 Let the radius of culinder = r cm

Radius of the sphere =r/2

Volume of cylinder 1
Volume of sphere 1

Now

h = r/6 1

or h/r = 1/6; r : h = 6 : 1 1 4

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Part B

Note : New Questions for Set C

XXI/OS/O/2 C

Q. No Expected Value Points For dISTRIBUTION Total

C EAch Step OF MARKS

6.

a = 29.5 1

8

Exp. 1-2×1/2=0;2×1/2-1=0 1

9 2(5-2x) = 7 _ 3x

10 _ 4x = 7 _ 3x

x=3 1

15 1. Obtain the Values A,B and C

2. Is A>B?

3. If yes find whether A>C

4. If yes declare C is greatest and stop 1

5. If no, decle are C is greatest and stop

6. If B> A then find whether B>C 1

7. If yes, declare B is greatest and stop

8. If no, decleare C is greatest and stop 1 3

18. mark for each

correct rectangle
mark for polygon 3

22 Let the required numbers of years = n

Numbers of conversation periods = 2n

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Q. No Expected Value Points For dISTRIBUTION Total

C EAch Step OF MARKS

R = (20/2) =10% 1

A= P[1+R]²ⁿ 1

146410 = 100000[1+10/100]²ⁿ

146410/100000=[11/10]²ⁿ 1

4

n=2

25.

Figure 1

In

DE/AE = tan 30º

DE=AE tan 30º

In BCD; CD= BC tan 60º

BC= 60 cot 60º 1

= 20×1.73 1

= 34.60 m

Now AE

DE 1

Hence distance between tower and lamp post

= 34.60 m

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Q. No Expected Value Points For dISTRIBUTION Total

C EAch Step OF MARKS

Difference between heights 4

= 20m

28. Minimum balances for month

April Rs. 1800

May Rs. 1800 1 mark for

June Rs. 3000 each balance

July Rs. 1200

August Rs. 1200

September Rs. 3000

Total Rs. 12000

Interest = Rs. 12000×1/12 ×5/100 1 7

= Rs. 50

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