series hrk/2 h$mos> z§ 30/2/1 code no. amob z§....students will read the question paper only...

30/2/1 1 P.T.O.

narjmWu H$moS >H$mo CÎma-nwpñVH$m Ho$ _wI-n¥ð >na Adí` {bIo§ & Candidates must write the Code on the

title page of the answer-book.

Series HRK/2 H$moS> Z§. Code No.

amob Z§.

Roll No.

g§H${bV narjm – II

SUMMATIVE ASSESSMENT – II

J{UV

MATHEMATICS

{ZYm©[aV g_` : 3 KÊQ>o A{YH$V_ A§H$ : 90

Time allowed : 3 hours Maximum Marks : 90

H¥$n`m Om±M H$a b| {H$ Bg àíZ-nÌ _o§ _w{ÐV n¥ð> 11 h¢ &

àíZ-nÌ _| Xm{hZo hmW H$s Amoa {XE JE H$moS >Zå~a H$mo N>mÌ CÎma-nwpñVH$m Ho$ _wI-n¥ð> na {bI| &

H¥$n`m Om±M H$a b| {H$ Bg àíZ-nÌ _| >31 àíZ h¢ &

H¥$n`m àíZ H$m CÎma {bIZm ewê$ H$aZo go nhbo, àíZ H$m H«$_m§H$ Adí` {bI| &

Bg àíZ-nÌ H$mo n‹T>Zo Ho$ {bE 15 {_ZQ >H$m g_` {X`m J`m h¡ & àíZ-nÌ H$m {dVaU nydm©• _| 10.15 ~Oo {H$`m OmEJm & 10.15 ~Oo go 10.30 ~Oo VH$ N>mÌ Ho$db àíZ-nÌ H$mo n‹T>|Jo Am¡a Bg Ad{Y Ho$ Xm¡amZ do CÎma-nwpñVH$m na H$moB© CÎma Zht {bI|Jo &

Please check that this question paper contains 11 printed pages. Code number given on the right hand side of the question paper should be

written on the title page of the answer-book by the candidate.

Please check that this question paper contains 31 questions. Please write down the Serial Number of the question before

attempting it.

15 minute time has been allotted to read this question paper. The question paper will be distributed at 10.15 a.m. From 10.15 a.m. to 10.30 a.m., the students will read the question paper only and will not write any answer on the answer-book during this period.

SET-1

30/2/1

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30/2/1 2

gm_mÝ` {ZX}e :

(i) g^r àíZ A{Zdm`© h¢ &

(ii) Bg àíZ-nÌ _| 31 àíZ h¢ Omo Mma IÊS>m| A, ~, g Am¡a X _| {d^m{OV h¢ &

(iii) IÊS> A _| EH$-EH$ A§H$ dmbo 4 àíZ h¢ & IÊS> ~ _| 6 àíZ h¢ {OZ_| go àË`oH$ 2 A§H$m| H$m h¡ & IÊS> g _| 10 àíZ VrZ-VrZ A§H$m| Ho$ h¢ & IÊS> X _| 11 àíZ h¢ {OZ_| go àË`oH$ 4 A§H$m| H$m h¡ &

(iv) H¡$bHw$boQ>am| H$m à`moJ H$aZo H$s AZw_{V Zht h¡ &

General Instructions :

(i) All questions are compulsory.

(ii) The question paper consists of 31 questions divided into four sections A,

B, C and D.

(iii) Section A contains 4 questions of 1 mark each. Section B contains

6 questions of 2 marks each, Section C contains 10 questions of 3 marks

each and Section D contains 11 questions of 4 marks each.

(iv) Use of calculators is not permitted.

IÊS> A

SECTION A

àíZ g§»`m 1 go 4 VH$ àË oH$ àíZ 1 A§H$ H$m h¡ & Question numbers 1 to 4 carry 1 mark each.

1. `{X {ÛKmV g_rH$aU 6x2 – x – k = 0 H$m EH$ _yb 3

2 h¡, Vmo k H$m _mZ kmV H$s{OE &

If one root of the quadratic equation 6x2 – x – k = 0 is 3

2, then find the

value of k.

CoolGyan.Org | A Portal For Complete Education

30/2/1 3 P.T.O.

2. EH$ 15 _r. bå~r gr‹T>r Xrdma Ho$ gmW 60 H$m H$moU ~ZmVr h¡ & Cg {~ÝXþ H$s D±$MmB© kmV H$s{OE Ohm± gr‹T>r Xrdma H$mo ñne© H$aVr h¡ &

A ladder 15 m long makes an angle of 60 with the wall. Find the height

of the point where the ladder touches the wall.

3. 9 _r. 8 _r. 2 _r. {d_mAm| dmbo YmVw Ho$ EH$ R>mog KZm^ H$mo {nKbmH$a 2 _r. ^wOm Ho$ R>mog KZm| _| T>mbm J`m h¡ & Bg àH$ma ~Zo KZm| H$s g§»`m kmV H$s{OE &

A solid metallic cuboid of dimensions 9 m 8 m 2 m is melted and

recast into solid cubes of edge 2 m. Find the number of cubes so formed.

4. O Ho$ÝÐ VWm QOR ì`mg Ho$ EH$ d¥Îm na EH$ ~mø {~ÝXþ P go ñne©-aoIm PQ ItMr JB© h¡ & `{X POR = 120 h¡, Vmo OPQ H$s _mn Š`m h¡ ?

PQ is a tangent drawn from an external point P to a circle with centre

O, QOR is the diameter of the circle. If POR = 120, what is the

measure of OPQ ?

IÊS> ~

SECTION B

àíZ g§»`m 5 go 10 VH$ àË`oH$ àíZ 2 A§H$m| H$m h¡ & Question numbers 5 to 10 carry 2 marks each.

5. x Ho$ {bE hb H$s{OE :

2x3 + 10x – 8 3 = 0

Solve for x :

2x3 + 10x – 8 3 = 0

6. `{X EH$ g_m§Va lo‹T>r Ho$ 7d| nX H$m gmV JwZm CgHo$ 11d| nX Ho$ ½`mah JwZo Ho$ ~am~a h¡, Vmo CgH$m 18dm± nX Š`m hmoJm ?

If seven times the 7th term of an A.P. is equal to eleven times the 11th

term, then what will be its 18th term ?


30/2/1 4

7. Xmo {d{^Þ nmgm| H$mo EH$ gmW \|$H$m J`m & àmßV g§»`mAm| H$m JwUZ\$b 18 go H$_ hmoZo H$s àm{`H$Vm kmV H$s{OE &

Two different dice are thrown together. Find the probability that the

product of the numbers appeared is less than 18.

8. `{X EH$ g_m§Va MVw^w©O Ho$ Xmo AmgÞ erf© (3, 2) d (– 1, 0) h¢ VWm BgHo$ {dH$U© (2, – 5) na à{VÀN>oX H$aVo h¢, Vmo AÝ` Xmo erfm] Ho$ {ZX}em§H$ kmV H$s{OE &

If two adjacent vertices of a parallelogram are (3, 2) and (– 1, 0) and the

diagonals intersect at (2, – 5), then find the coordinates of the other two

vertices.

9. Xr JB© AmH¥${V _|, `{X AB = AC h¡, Vmo {gÕ H$s{OE {H$ BE = EC.

In the given figure, if AB = AC, prove that BE = EC.

10. EH$ A{Ydf© (brn df©) _| 53 _§Jbdma hmoZo H$s àm{`H$Vm kmV H$s{OE &

Find the probability that in a leap year there will be 53 Tuesdays.


30/2/1 5 P.T.O.

IÊS> g SECTION C

àíZ g§»`m 11 go 20 VH$ àË`oH$ àíZ 3 A§H$m| H$m h¡ &

Question numbers 11 to 20 carry 3 marks each.

11. `{X {ÛKmV g_rH$aU (a – b) x2 + (b – c) x + (c – a) = 0 Ho$ _yb g_mZ hm|, Vmo {gÕ

H$s{OE {H$ 2a = b + c.

If the roots of the quadratic equation (a – b) x2 + (b – c) x + (c – a) = 0 are

equal, prove that 2a = b + c.

12. {ZåZ{b{IV loUr H$m `moJ\$b kmV H$s{OE :

5 + (– 41) + 9 + (– 39) + 13 + (– 37) + 17 + ... + (– 5) + 81 + (– 3)

Find the sum of the following series :

5 + (– 41) + 9 + (– 39) + 13 + (– 37) + 17 + ... + (– 5) + 81 + (– 3)

13. EH$ _rZma H$s {H$gr g_` H$s N>m`m, Cg N>m`m H$s VrZ JwZr h¡ O~ gy`© H$m CÞ`Z H$moU

60 h¡ & bå~r N>m`m Ho$ g_` gy`© H$m CÞ`Z H$moU kmV H$s{OE &

The shadow of a tower at a time is three times as long as its shadow

when the angle of elevation of the sun is 60. Find the angle of elevation

of the sun at the time of the longer shadow.

14. {~ÝXþAm| (3, –2) VWm (– 3, – 4) H$mo {_bmZo dmbo aoImIÊS> H$mo g_{Ì^m{OV H$aZo dmbo

{~ÝXþAm| Ho$ {ZX}em§H$ kmV H$s{OE &

Find the coordinates of the points of trisection of the line segment joining

the points (3, – 2) and (– 3, – 4).


30/2/1 6

15. Xr JB© AmH¥${V _|, PA VWm PB EH$ ~mø {~ÝXþ P go d¥Îm H$s ñne©-aoImE± h¢, Ohm±

PA = 4 go_r VWm BAC = 135 h¡ & Ordm AB H$s bå~mB© kmV H$s{OE &

In the given figure, PA and PB are tangents to a circle from an external

point P such that PA = 4 cm and BAC = 135. Find the length of chord

AB.

16. {gÕ H$s{OE {H$ d¥Îm Ho$ n[aJV ~Zr MVw^w©O H$s Am_Zo-gm_Zo H$s ^wOmE± d¥Îm Ho$ Ho$ÝÐ na

g§nyaH$ H$moU A§V[aV H$aVr h¢ &

Prove that the opposite sides of a quadrilateral circumscribing a circle

subtend supplementary angles at the centre of the circle.


30/2/1 7 P.T.O.

17. Xr JB© AmH¥${V _|, ABCD EH$ g_b§~ MVw^w©O, {OgH$s ^wOmE± AB = 18 go_r,

DC = 32 go_r, AB DC VWm AB d AC Ho$ ~rM H$s Xÿar 14 go_r h¡ & `{X A, B, C d

D H$mo Ho$ÝÐ boH$a 7 go_r g_mZ {ÌÁ`m Ho$ Mmn ItMo JE h¢, Vmo N>m`m§{H$V ^mJ H$m joÌ\$b

kmV H$s{OE &

In the given figure, ABCD is a trapezium with AB DC, AB = 18 cm,

DC = 32 cm and the distance between AB and AC is 14 cm. If arcs of

equal radii 7 cm taking A, B, C and D as centres, have been drawn, then

find the area of the shaded region.

18. EH$ R>mog bå~-d¥Îmr` e§Hw$ H$s {ÌÁ`m Am¡a D±$MmB© H$m AZwnmV 5 : 12 h¡ & `{X BgH$m Am`VZ

314 KZ go_r hmo, Vmo BgH$m gånyU© n¥îR>r` joÌ\$b kmV H$s{OE & [ = 3.14 br{OE ]

The radius and height of a solid right circular cone are in the ratio of

5 : 12. If its volume is 314 cm3, find its total surface area. [ Take = 3·14 ]


30/2/1 8

19. Xr JB© AmH¥${V _|, ABC, 3 BH$mB© wOm H$m EH$ g_~mhþ {Ì^wO h¡ & BgHo$ AÝ` Xmo erfm]

Ho$ {ZX}em§H$ kmV H$s{OE &

In the given figure, ABC is an equilateral triangle of side 3 units. Find

the coordinates of the other two vertices.

20. {gÕ H$s{OE {H$ {Ì^wO ABC {OgHo$ erf© A (– 2, 0), B (0, 2) VWm C (2, 0) h¢,

DEF {OgHo$ erf© D (– 4, 0), F (4, 0) VWm E (0, 4) h¢, Ho$ g_ê$n h¡ &

Show that ABC with vertices A (– 2, 0), B (0, 2) and C (2, 0) is similar to

DEF with vertices D (– 4, 0), F (4, 0) and E (0, 4).


30/2/1 9 P.T.O.

IÊS> X

SECTION D

àíZ g§»`m 21 go 31 VH$ àË`oH$ àíZ 4 A§H$m| H$m h¡ & Question numbers 21 to 31 carry 4 marks each.

21. Xmo d¥Îm A§V:ñne© H$aVo h¢ & CZHo$ joÌ\$bm| H$m `moJ\$b 116 dJ© go_r h¡ VWm CZHo$ Ho$ÝÐm| Ho$ ~rM H$s Xÿar 6 go_r h¡ & d¥Îmm| H$s {ÌÁ`mE± kmV H$s{OE &

Two circles touch internally. The sum of their areas is 116 cm2 and the

distance between their centres is 6 cm. Find the radii of the circles.

22. `{X 1 + 4 + 7 + 10 + ... + x = 287 h¡, Vmo x H$m _mZ kmV H$s{OE &

If 1 + 4 + 7 + 10 + ... + x = 287, find the value of x.

23. 3 go_r d 5 go_r {ÌÁ`mAm| Ho$ Xmo g§H|$Ðr` d¥Îm It{ME & ~mø d¥Îm na EH$ {~ÝXþ boH$a A§V: d¥Îm na Xmo ñne©-aoImAm| H$s aMZm H$s{OE &

Draw two concentric circles of radii 3 cm and 5 cm. Taking a point on the

outer circle, construct the pair of tangents to the inner circle.

24. 7 _r. D±$Mo EH$ ^dZ H$s MmoQ>r go EH$ _rZma Ho$ {eIa H$m CÞ`Z H$moU 60 VWm CgHo$ nmX H$m AdZ_Z H$moU 45 h¡ & _rZma H$s D±$MmB© kmV H$s{OE & [ 3 = 1·732 à`moJ H$s{OE ]

From the top of a 7 m high building, the angle of elevation of the top of a

tower is 60 and the angle of depression of its foot is 45. Find the height

of the tower. [Use 3 = 1·732]

25. EH$ ~Šgo _| 90 {S>ñH$ (Discs) h¢, {OZ na 1 go 90 VH$ g§»`mE± A§{H$V h¢ (EH$ {S>ñH$ na EH$ g§»`m) & `{X Bg ~Šgo _| go EH$ {S>ñH$ `mÑÀN>`m {ZH$mbr OmVr h¡, Vmo BgH$s àm{`H$Vm kmV H$s{OE {H$ Bg {S>ñH$ na A§{H$V hmoJr (i) Xmo A§H$m| H$s EH$ g§»`m, (ii) 5 go {d^mÁ` EH$ g§»`m &

A box contains 90 discs which are numbered from 1 to 90. If one disc is

drawn at random from the box, find the probability that it bears (i) a

two-digit number, (ii) a number divisible by 5.


30/2/1 10

26. 3 _r. ì`mg H$m EH$ Hw$Am± 14 _r. JhamB© VH$ ImoXm J`m & ~mha {ZH$mbr JB© {_Å>r H$mo 5 _r. Mm¡‹S>r EH$ d¥ÎmmH$ma db` (Ring) ~ZmZo Ho$ {bE g_mZ ê$n go \¡$bm`m J`m VWm EH$ àH$ma H$m ~m±Y ~Zm`m J`m & Bg ~m±Y H$s D±$MmB© kmV H$s{OE & A well of diameter 3 m is dug 14 m deep. The soil taken out of it is spread

evenly all around it to a width of 5 m to form an embankment. Find the

height of the embankment.

27. {gÕ H$s{OE {H$ EH$ ~mø {~ÝXþ go d¥Îm na ItMr JB© Xmo ñne©-aoImAm| H$s bå~mB`m± g_mZ hmoVr h¢ & Prove that the lengths of two tangents drawn from an external point to a

circle are equal.

28. 50 _r. 40 _r. {d_mAm| dmbo EH$ Am`VmH$ma nmH©$ _| EH$ Am`VmH$ma Vmbm~ ~Zm h¡, {Oggo Vmbm~ Ho$ Mmam| Amoa g_mZ Mm¡‹S>mB© H$s ~Zr Kmg H$s nÅ>r H$m joÌ\$b 1184 dJ© _r. h¡ & Vmbm~ H$s bå~mB© VWm Mm¡‹S>mB© kmV H$s{OE &

In a rectangular park of dimensions 50 m 40 m, a rectangular pond is

constructed so that the area of grass strip of uniform width surrounding

the pond would be 1184 m2. Find the length and breadth of the pond.

29. EH$ nmH©$ H$s AmH¥${V 7 _r. ì`mg Ho$ d¥Îm H$s h¡ & `h 0·7 _r. Mm¡‹S>mB© Ho$ amñVo go {Kam hþAm h¡ & Bg amñVo na gr_|Q> H$aZo H$m IM© kmV H$s{OE, `{X BgH$s bmJV à{V dJ© _r. < 110 h¡ &

A park is of the shape of a circle of diameter 7 m. It is surrounded by a

path of width of 0·7 m. Find the expenditure of cementing the path, if its

cost is < 110 per sq. m.

30. 20 go_r D±$Mo EH$ YmVw Ho$ bå~-d¥Îmr` e§Hw$, {OgH$m erf© H$moU 60 h¡, H$mo BgHo$ AmYma Ho$ g_m§Va EH$ g_Vb Ûmam CgH$s D±$MmB© Ho$ _Ü` go Xmo ^mJm| _| H$mQ>m OmVm h¡ & `{X Bg

àH$ma àmßV hþE {N>ÞH$ H$mo 16

1 go_r EH$g_mZ ì`mg Ho$ Vma Ho$ ê$n _| ItMm OmE, Vmo Vma

H$s bå~mB© kmV H$s{OE &

A metallic right circular cone 20 cm high and whose vertical angle is 60

is cut into two parts at the middle of its height by a plane parallel to its

base. If the frustum so obtained be drawn into a wire of uniform diameter

16

1 cm, find the length of the wire.


30/2/1 11 P.T.O.

31. EH$ ~ƒr nhbo {XZ AnZr ~MV H$m EH$ nm±M-énE H$m {gŠH$m JwëbH$ _| S>mbVr h¡ & dh

à{V{XZ AnZr ~MV _| nm±M-énE Ho$ {gŠHo$ H$s am{e EH$-EH$ ~‹T>mVr h¡ & `{X JwëbH$ _|

nm±M-énE Ho$ Hw$b 190 {gŠHo$ Am gH$Vo hm|, Vmo kmV H$s{OE {H$ dh {H$VZo {XZ VH$

JwëbH$ _| nm±M-énE Ho$ {gŠHo$ S>mb gH$Vr h¡ VWm CgZo Hw$b {H$VZm YZ ~Mm`m &

~MV H$aZo H$s AmXV na AnZo {dMma {b{IE &

A child puts one five-rupee coin of her saving in the piggy bank on the

first day. She increases her saving by one five-rupee coin daily. If the

piggy bank can hold 190 coins of five rupees in all, find the number of

days she can continue to put the five-rupee coins into it and find the total

money she saved.

Write your views on the habit of saving.


QUESTION PAPER CODE 30/2/1EXPECTED ANSWER/VALUE POINTS

SECTION A1. 6x2 – x – k = 0

22 26 – – k3 3

= 012

k = 212

2. cos 60° = h

1512

h = 7.5 m12

3. No. of cubes = 9 8 22 2 2

12

= 1812

4. POR = OQP + OPQ

OPQ = 120° – 90°12

= 30°12

SECTION B

5. 23x 10x – 8 3 = 0

23x 12x – 2x – 8 3 = 012

3x (x 4 3) – 2(x 4 3) = 012

( 3 x – 2) (x 4 3) = 012

x = 2 , – 4 33 1

2

30/2/1 (1)

30/2/1

15 m h60°

120°O

R

Q P


(2) 30/2/1

30/2/1

6. 7a7 = 11a11

7(a + 6d) = 11(a + 10d)12

7a – 11a + 12d – 110d = 012

–4a – 68d = 0

a + 17d = 012

a18 = 012

7. Total number of outcomes = 3612

P(Product appears is less than 18) = 26 1336 18 1

12

8. Let other two coordinates are

(x, y) and (x, y)

2 = x 3

2

x = 112

and, –5 = 2 y

2

y = –1212

Again, –1 x

2

= 2

x = 512

and 0 y

2

= –5

y = –1012

Hence co-ordinates are (1, –12) and (5, –10)

A(3, 2)

B(–1, 0)

D(x , y ) D(x, y)

(2, –5)


30/2/1 (3)

30/2/1

9. AB = AC (Given)

AD = AF (tangents from external point)12

On subtracting,

BD = CF

BD = BE (tangents from external point) 1

and CF = EC

BE = EC

10. In leap year = 52 weeks + 2 days12

Two days may be, (M, Tu), (Tu, W), (W, Th), (Th, F), (F, Sat)

(Sat, Sun), (Sun, M) 1

Required probability = 27

12

SECTION C

11. For equal roots

D = 0

(b – c)2 – 4(a – b) (c – a) = 0 1

b2 + c2 – 2bc – 4ac + 4a2 + 4bc – 4ab = 0

(b + c – 2a)2 = 0 112

2a = b + c12

12. The series can be rewritten as,

(5 + 9 + 13 + ... + 81) + (–41 + (–39) + (–39) + ... + (–5) + (–3))12

for 5 + 9 + 13 + ... 81

a = 5

d = 4


(4) 30/2/1

30/2/1

an = 81

5 + (n – 1)4 = 81

n = 20

Sn = 20 (5 81) 8602

12

for (–41) + (–39) + (–37) ... + (–5) + (–3)

a = –41

d = 2

an = –3

–41 + (n – 1)2 = –3

n = 2012

Sn = 20 (–41 – 3) – 4402

12

Sum of series = 860 – 440

= 42012

13. Correct Figure12

tan 60° = hx 1

h = 3x

tan = h

3x

tan = 3x

3x

tan = 13 1

= 30°12

D A

B

h

C x3x

60°


30/2/1 (5)

14. Let the co-ordinates be (x, y) and (x, y)

x = 1(–3) 2(3) 1

1 2

y = 1(–4) 2(–2) –8

1 2 3

112

x = 2(–3) 1(3) –1

1 2

y = 2(–4) 1(–2) –10

1 2 3

112

15. PA = PB = 4 cm (tangents from external point)12

PAB = 180° – 135°

= 45°12

APB = 180° – 45° – 45°

= 90°12

ABP is a isosceles right angled triangle

AB2 = 2AP2

= 2(4)2 = 32 1

AB = 4 2 cm12

16. AOS AOP

1 = 2 1

Similarly 4 = 3

5 = 6

8 = 7

(1 + 8) + (4 + 5) = (2 + 3) + (6 + 7) = 180° 1

AOD + BOC = 180°

and AOB + COD = 180° 1

30/2/1

A(3, –2)

B(x, y)

C(x , y )

D(–3, –4)

D R C

A P B

QS

7 654

3218 o


(6) 30/2/1

17. 3 marks be given to every attempt 3

18. r : h = 5 : 12

Let r = 5x

h = 12x

Volume = 21 r h3

314 = 21 3.14 (5x) 12 x3

x = 1 1

r = 5 cm

h = 12 cm

l = 2 2(12) (5) 13 cm 1

TSA = r(l + r) = 3.14 × 5 (13 + 5)

= 282.6 cm2 1

19. Co-ordinates of B are (5, 0)

Let co-ordinates of C be (x, y)

AC2 = BC2 1

(x – 2)2 + (y – 0)2 = (x – 5)2 + (y – 0)2

x2 + 4 – 4x + y2 = x2 + 25 – 10x + y2

6x = 21

x = 72 1

(x – 2)2 + (y – 0)2 = 9

227 – 2 y

2

= 9

y2 = 99 –4

30/2/1


30/2/1 (7)

30/2/1

y2 = 274

y = 3 3

2 (+ve sign to be taken), Co-ordinate of 7 3 3C ,

2 2 1

20. AB = 2 2(–2 0) (0 – 2) 2 2 units

BC = 2 2(0 – 2) (2 – 2) 2 2 units

CA = 2 2(2 2) (0 – 0) 4 units 1

DE = 2 2(–4 0) (0 – 4) 4 2 units

EF = 2 2(0 – 4) (4 – 0) 4 2 units

DF = 2 2(–4 – 4) (0 – 2) 8 units 1

ABDE =

BC AC 1EF DF 2

ABC ~ DEF 1

SECTION D

21. Let radii of circles be x, y (x > y)

x – y = 6 ...(1) 1

and x2 + y2 = 116

x2 + y2 = 11612

x2 + (x – 6)2 = 116

x2 + x2 + 36 – 12x = 116

x2 – 6x – 40 = 0 1

(x – 10) (x + 4) = 0 1

x = 10 cm (rejecting –ve value)12

and y = 4 cm

yx


(8) 30/2/1

30/2/1

22. 1 + 4 + 7 + 10 + ... + x = 287

Sn = 287

n (2 (n – 1)3)2

= 287

3n2 – n – 574 = 0 112

3n2 – 42n + 41n – 574 = 0

(3n + 41)(n – 14) = 0

n = 14 112

x = a + 13d = 40 1

23. For constructing correct concentric circle 1

For constructing correct pair of tangents 3

24. Correct Figure 1

tan 45° = 7x

x = 7 1

tan 60° = h – 7

x

7 3 = h – 7

h = 7( 3 1) 1

= 7(1.732 + 1)

= 19.124 m 1

25. (i) P(bears two digit number) = 81 9or90 10 2

(ii) P(a number divisible by 5) = 18 1or90 5 2

26. Let height of embankment be h mts17(1.5)2 × 14 = [(6.5)2 – (1.5)2] × 2 22.25 × 14 = 5 × 8 × h 1 h = 0.7875 m 1

tower

45°60°

7 m

x

xD B

E

h h–7

A

C

Building


30/2/1 (9)

30/2/1

27. For correct given, To prove, construction, figure 4×12 =2

for correct proof 2

28. Let width of grass strip be x mts.

area of park – area of pond = 1184

(50 × 40) – (50 – 2x) (40 – 2x) = 1184 1

2000 – 2000 + 180x – 4x2 = 1184

x2 – 45x + 296 = 0 1

x2 – 37x – 8x + 296 = 0

x = 8, 37 (rejected) 1

Length of pond = 50 – 16 = 34 m

Breadth of pond = 40 – 16 = 24 m 1

29. r1 = 3.5m, r2 = 4.2 m12

area of path = (4.2)2 – (3.5)2 1

= [(7.7) × 0.7]

= 22 7.7 0.77

= 16.94 m2 112

Cost of cementing the path = 16.94 × 110

= ` 1863.40 1

50

40

x

3.5m

4.2m


(10) 30/2/1

30/2/1

30. tan 30° = 1r10

r1 = 10

3 1

tan 30° = 2r20

r2 = 20

3 1

13

2 210 20 2001033 3

=

21 h32

1

1 70010 32 323 3 = h

h = 796444.4 cm 1

or

7964.44 m

31. Total saving = 190 × 5 = 950 1

The series 5 + 10 + 20 + ....

Sn = 950

n (2(5) (n – 1)5)2

= 950 1

n (2 + (n – 1)) = 380

n2 + n – 380 = 0

n2 + 20n – 19n – 380 = 0

n = 19 1

Views on the habit of saving 1

r1

r2

10 cm

20 cmh

132

=


series hrk/2 h$mos> z§ 30/2/1 code no. amob z§....students will read the question paper only...

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