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Hindi Course B

Hindi Course A

1. In the given figure if DE || BC, AE = 8 cm, EC = 2 cm and BC = 6 cm, then find DE. [1]Sol. In DADE and DABC,

ÐDAE = ÐBAC [Common]ÐADE = ÐABC [Corresponding angles]By AA axiom of similarityDADE ~ DABC

\AEAC

DEBC

= [CPCT]

Þ8

8 2+DE6

= Þ 10 × DE = 48

Þ DE = 4.8 cm Ans.

2. Evaluate :2

21 cot 45º10 · .1 sin 90º-+

[1]

Sol.2

21 cot 4510·1 sin 90- °+ °

2

21 (1)10·1 (1)-=+

010. 02

= = Ans.

3. If cosec q5= ,4 find the value of cot q. [1]

Sol. We know that,cot2 q = cosec2 q – 1

25 14æ ö= -ç ÷è ø

25 25 16 9116 16 16

-= - = =

Þ cot2 q9

16=

i.e. cot q34

= Ans.

Section – A

A

B

D E

C6 cm

8 cm

2 cm

MATHEMATICS 2014 TERM ISET I

Time allowed : 3 hours Maximum marks : 90

Arundeep’s Solved Papers Mathematics 2014 (Term I)1

https://www.arundeepselfhelp.info/index.php?route=product/product&path=180&product_id=395


4. Following table shows sale of shoes in a store during one month :

Size of shoe 3 4 5 6 7 8Number of pairs sold 4 18 25 12 5 1

Find the model size of the shoes sold. [1]Sol. Maximum number of pairs sold = 25 (size 5)\ Model size of shoes = 5 Ans.

Section – B

5. Find the prime factorisation of the denominator of rational number expressed as 6.12 insimplest form. [2]

Sol. Let x 6.1212...........= ...(i)Þ 100x 612.1212...........= ...(ii)

Subtracting eq. (i) from (ii), we get99x = 606

Þ x606 20299 33

= =

\ Denominator = 33and Prime factorisation of 33 be = 3 × 11 Ans.

6. Find a quadratic polynomial, the sum and product of whose zeroes are 3 and13

respectively. [2]

Sol. Given, sum of zeroes, (S) 3=

and Product of zeroes, (P)13

=

Thus, quadratic polynomial is given as f(x) = x2 – Sx + P

\ f(x) 2 133

x x= - +23 3 1

3x x- += 2

1 ( 3 3 1)3

x x= - + Ans.

7. Complete the following factor tree and find the composite number x. [2]

Sol. y = 5 × 13 = 65x = 3 × 195 = 585 Ans.

3 3311 11

1

3 5853 1955 65

13 131


Arundeep’s Solved Papers Mathematics 2014 (Term I)38. In a rectangle ABCD, E is middle point of AD. If AD = 40 m and AB = 48 m, then find EB.

[2]Sol. Given, E is the mid-point of AD

\ AE40 m 20 m

2= =

Also ÐA = 90º [Angle of a rectangle]\ In rt. DBAE, we have

EB2 = AB2 + AE2 [Pythagoras’ theorem]= (48)2 + (20)2 = 2304 + 400 = 2704

EB 2704= = 52 52´ = 52 m Ans.9. If x = p sec q + q tan q and y = p tan q + q sec q, then prove that x2 – y2 = p2 – q2. [2]

Sol. L.H.S. = x2 – y2

= (p sec q + q tan q)2 – (p tan q + q sec q)2

= (p2 sec2 q + q2 tan2 q + 2pq sec q tan q) – (p2 tan2 q + q2 sec2 q + 2pq sec q tan )= p2 sec2 q + q2 tan2 q + 2pq sec q tan q – p2 tan2 q – q2 sec2 q – 2pq sec q tan q= p2 (sec2 q – tan2 q) – q2 (sec2 q – tan2 q)

= p2 – q2 [Q sec2 q – tan2 q = 1]

= R.H.S. Hence Proved.10. Given below is the distribution of weekly pocket money received by students of a class.

Calculate the pocket money that is received by most of the students. [2]

Pocket Money (in ` ) 0–20 20–40 40–60 60–80 80–100 100–120 120–140No. of Students 2 2 3 12 18 5 2

Sol. Pocket Money Number of(in ` ) Students0–20 220–40 240–60 360–80 12 f080–100 18 f1 (Maximum)100–120 5 f2120–140 2

Here maximum frequency is 18 which lies in class 80 – 100.\ Modal class = 80 – 100

Thus, mode 1 01 0 22

f fl h

f f f-

= + ´- -

18 1280 2036 12 5-= + ´- -



680 2019

= + ´ 1208019

= + = 80 + 6.32

= 86.32 (approx.)\ Required pocket money = ` 86.32 (approx.)

Ans.

Section – C

11. Prove that 3 + 2 3 is an irrationalnumber. [3]

Sol. Let us assume to the contrary, that 3 2 3+is rational.So that we can find co-prime positiveintegers a and b (b ¹ 0), such that

3 2 3 ab

+ =

Rearranging the equation, we get

2 333a a b

b b-= - =

Þ 33 3

2 2 2a b a b

b b b-= = -

Þ 33

2 2ab

= -

Since a and b are integer, and b ¹ 0,

So3

2 2ab- is rational number and hence

3 is rational.

But this contradicts the fact that 3 is anirrational number.

So we conclude that 3 2 3+ is anirrational number.

Hence Proved.12. Solve by elimination :

3x = y + 55x – y = 11 [3]

Sol. Given equations are,3x = y + 5 ...(i)5x – y = 11 ...(ii)

On subtracting eq. (i) and (ii), we get

3x – y = 55x – y = 11– + –

– 2x = –6Þ x = 3

Putting the value of x in eq. (i) ; we have3 (3) – y = 5

Þ 9 – 5 = y Þ y = 4\ x = 3, y = 4 Ans.

13. A man earns ` 600 per month more thanhis wife. One-tenth of the man’s salaryand one-sixth of the wife’s salary amountto ` 1,500, which is saved every month.Find their incomes. [3]

Sol. Let wife’s monthly income = ` xThen man’s monthly income = ` (x + 600)According to the question, we have

1 1( 600) ( )10 6

x x+ + = 1,500

Þ3( 600) 5

30x x+ +

= 1,500

Þ 3x + 1,800 + 5x = 45,000Þ 8x = 45,000 –1,800

Þ x43, 200 5, 400

8= =

Thus, wife’s income = ` x = ` 5,400and Man’s income = ` (x + 600) = ` (5400 + 600)

= ` 6,000 Ans.14. Check whether polynomial x – 1 is a

factor of the polynomial x3 – 8x2 + 19x – 12.Verify by division algorithm. [3]

Sol. Let P (x) = x3 – 8x2 + 19x – 12Put x = 1, we haveP (1) = (1)3 – 8 (1)2 + 19 (1) – 12= 1 – 8 + 19 – 12= 20 – 20= 0

\ (x – 1) is a factor of P (x).


Arundeep’s Solved Papers Mathematics 2014 (Term I)5Verification :

Since remainder = 0.Thus, (x – 1) is a factor of P (x).

Hence Verified.15. If the perimeters of two similar triangles

ABC and DEF are 50 cm and 70 cmrespectively and one side of DABC = 20 cm,then find the corresponding side ofDDEF. [3]

Sol.

D

Given, DABC ~ DDEF,Perimeter of DABC = 50 cmPerimeter DDEF = 70 cmOne side of DABC = 20 cmLet AB = 20 cmDABC ~ DDEF [Given]

\Peri ( ABC)Peri ( DEF)

DD

ABDE

=

Þ5070

20DE

=

\ 50 × DE = 1400Þ DE = 28 m

The corresponding side of DDEF = 28 cm. Ans.

16. In the figure if DE || OB and EF || BC,then prove that DF || OC. [3]

Sol. Given, In DABC, DE || OB and EF || BCTo Prove : DF || OCProof : In DAOB, we haveDE || OB

\AEEB

AD=DO ...(i)

[Thales’ Theorem]Similarly, in DABC, we haveEF || BC

\AEEB

AF=FC ...(ii)

[Thales’ Theorem]From (i) and (ii), we have

ADDO

AF=FC

\ DF || OC[By Converse of Thales’ Theorem]

Hence Proved.17. Prove the identity :

(sec A – cos A) · (cot A + tan A) = tan A · sec A.[3]

Sol. L.H.S. = (sec A – cos A) (cot A + tan A)

1 cos A sin Acos Acos A sin A cos A

=æ ö æ ö- +ç ÷ ç ÷è ø è ø

2 2 21 cos A cos A sin Acos A sin A cos A

æ ö æ ö- += ç ÷ ç ÷è ø è ø

2sin A 1cosA sin A cos A

= ´

[Qcos2 A + sin2 A = 1]



sin A 1cos A cos A= ´

= tan A · sec A = R.H.S. Hence Proved.

18. Given 2 cos 3q = 3, find the value of q. [3]

Sol. Given, 2 cos 3q 3=

Þ cos 3q3

2= Þ cos 3q = cos 30º Þ 3q = 30º Þ q = 10º Ans.

19. For helping poor girls of their class, students saved pocket money as shown in the followingtable :

Money saved (in ) 5–7 7–9 9–11 11–13 13–15Number of students 6 3 9 5 7

Find mean and median for this data. [3]Sol. The table of values is given as under :

Money No. of x idi =

fidi c.f.

saved Studentsx – 10

2i

(in ` ) (fi)5–7 6 6 –2 –12 67–9 3 8 –1 –3 99–11 9 a=10 0 0 1811–13 5 12 1 5 2313–15 7 14 2 14 30

Sfi = 30 Sfidi = 4

where h = 2

(i) Then by step deviation method, we have,

Mean i ii

f da h

få

= + ´å

Þ x410 2

30= + ´ = 10 + 0.27 = ` 10.27Ans.

(ii) Now, N = Sfi = 30

\N2

30= 15,2

= Clearly the cummlative frequency just greater than 15 be 18 which lies in class

9 – 11.


Arundeep’s Solved Papers Mathematics 2014 (Term I)7\ Median class is 9–11.

Thus, median

N F2l h

f

-= +

15 99 29-= + ´

69 29

= + ´

= 9 + 1.33 = ` 10.33 Ans.20. Monthly pocket money of students of a class is given in the following frequency distribution:

[3]

Pocket money (in ) 100–125 125–150 150–175 175–200 200–225Number of students 14 8 12 5 11

Find mean pocket money using step deviation method.Sol. The table of values is given as under :

Pocket No. ofx i di = fidi

Money Students x – 162.525

i(in ` ) (fi)

100–125 14 112.5 –2 –28125–150 8 137.5 –1 –8150–175 12 0 0175–200 5 187.5 1 5200–225 11 212.5 2 22

Sfi = 50 Sfidi = –9

Then big step deviation method, we have

Mean i ii

f da h

få

= + ´å

9162.5 2550-= + ´ = 162.5 – 4.5 = ` 158 Ans.

Section – D

21. If two positive integers x and y are expressible in terms of primes as x = p2q3 and y = p3q,what can you say about their LCM and HCF. Is LCM a multiple of HCF ? Explain. [4]

Sol. Given, x = p2q3 = p × p × q × q × qAnd y = p3q = p × p × p × q

\ HCF = product of the smallest power of each common prime factor in the numbers x and y = p × p × q = p2q


Arundeep’s Solved Papers Mathematics 2014 (Term I)8and LCM product of the greatest power ofeach prime factor in the numbers x and y= p × p ×p × q × q × q = p3q3

Þ LCM = pq2 × p2q = pq2 × HCFYes, LCM is a multiple of HCF.Explanation :Let a = 12 = 22 × 3b = 18 = 2 × 32

\ HCF = 2 × 3 = 6 ...(i)LCM = 22 × 32 = 36LCM = 6 × 6LCM = 6 (HCF) [From (i)]Here LCM is 6 times the HCF. Ans.

22. Sita Devi wants to make a rectangularpond on the road side for the purpose ofproviding drinking water for streetanimals. The area of the pond will bedecreased by 3 square feet if its length isdecreased by 2 ft. and breadth isincreased by 1 ft. Its area will beincreased by 4 square feet if the lengthis increased by 1 ft. and breadth remainssame. Find the dimensions of the pond.What motivated Sita Devi to providewater point for street animals ? [4]

Sol. Let length of rectangular pond = x ft.and breadth of rectangular pond = y ft.

\ Area of rectangular pond = xyAccording to the question, we have(x – 2) (y + 1) = (xy – 3)

Þ xy + x – 2y – 2 = xy – 3Þ x – 2y = – 1 ...(i)

according to second given condition, wehave(x + 1) y = (xy + 4)

Þ xy + y = xy + 4i.e. y = 4 ...(ii)

Putting the value of y in eq. (i), we getx – 2 (4) = – 1

Þ x – 8 = – 1Þ x = – 1 + 8 = 7

Thus, length of rectangular pond = 7 ft.and Breadth of rectangular pond = 4 ft.

Values :1. Water is essential for the survival of all

living things including street animals.2. Water is the base of life and no one can

live without it.23. If a polynomial x4 + 5x3 + 4x2 – 10x – 12

has two zeroes as – 2 and – 3, then findthe other zeroes. [4]

Sol. Given, polynomial isf(x) = x4 + 5x3 + 4x2 – 10x – 12.Since two zeroes of f(x) are – 2 and – 3

\ (x + 2) (x + 3) = x2 + 3x + 2x + 6= x2 + 5x + 6

Dividing the polynomial f(x) withx2 + 5x + 6,

By division algorithm, we have

\ x4 + 5x3 + 4x2 – 10x – 12= (x2 + 5x + 6) (x2 – 2)

= (x + 2) (x + 3) ( 2) ( 2)x x- +

Other zeroes : 2 0x - = or 2 0x + =

2x = or 2x = -The zeroes of the polynomial are – 2, – 3,

2 and 2.- Ans.

24. Find all the zeroes of the polynomial8x4 + 8x3 – 18x2 – 20x – 5, if it is given

that two of its zeroes are52 and

5 .2

-

Sol. Given polynomial is given byf(x) = 8x4 + 8x3 – 18x2 – 20x – 5



Since two zeroes are52 and

52

-

\ 25 5 5( )2 2 2

x x x2æ ö æ ö æ ö

- ´ + = -ç ÷ ç ÷ ç ÷ç ÷ ç ÷ ç ÷è ø è ø è ø2 5

2x= -

Dividing the polynomial f(x) with 2 52

x -

By division algorithm, we have

\ f(x) = 8x4 + 8x3 – 18x2 – 20x – 5

2 25 (8 8 2)2

x x xæ ö= - + +ç ÷è ø

2 25 .2 (4 4 1)2

x x xæ ö= - + +ç ÷è ø

2 2(2 5) (4 2 2 1)x x x x= - + + +

2(2 5) [2 (2 1) 1(2 1)]x x x x= - + + +

f(x) 2(2 5) (2 1) (2 1)x x x= - + +

all zeroes of given polynomial x given byf(x) = 0

Þ 2x2 – 5 = 0 or 2x + 1 = 0 or 2x + 1 = 0

i.e. x =52

± or x =1

2-

or x =1

2-

All the zeroes are5 5 1 1, , and .2 2 2 2

- --

Ans.25. In the figure, there are two points D and

E on side AB of DABC such thatAD = BE. If DP || BC and EQ || AC, thenprove that PQ || AB. [4]

Sol.

Q

In DABC, we haveDP || BC (Given)

ÞADDB

AP=PC ...(i)

[Thales’ Theorem]Also, EQ || AC (Given)

ÞBEEA

BQ=QC [Thales’ Theorem]

ÞADDB

BQ=QC ...(ii)

[Q AD = BE ;\ EA = DB]From eq. (i) and (ii) ; we have

APPC

BQ=PC

\ PQ || AB (Inverse of Thales theorem)Hence proved.


Arundeep’s Solved Papers Mathematics 2014 (Term I)1026. In DABC, altitudes AD and CE intersect each other at the point P. Prove that

(i) DAPE ~ DCPD (ii) AP × PD = CP × PE(iii) DADB ~ DCEB (iv) AB × CE = BC × AD [4]

Sol. Given, In DABC, AD ^ BC and CE ^ AB(i) In DAPE and DCPD, we haveÐ1 = Ð4 [Each 90º]Ð2 = Ð3 [Vertically opposite angles]By AA axiom of similarity, we haveDAPE ~ DCPDHence proved.

(ii) DAPE ~ DCPD [Proved above]

\APCP

PE=PD [CPCT]

Þ AP × PD = CP × PEHence Proved

(iii) In DADB and DCEB, we haveÐ5 = Ð7 (Each 90º)Ð6 = Ð6 (Common)By AA axiom, of similarityDADB ~ DCEB Hence Proved.

(iv) DADB ~ DCEB [Proved Above]

ABCB

AD=CE [cpct]

Þ AB × CE = BC × AD Hence Proved27. Prove that :

(cot A + sec B)2 – (tan B – cosec A)2 = 2 (cot A . sec B + tan B . cosec A). [4]Sol. L.H.S.

= (cot A + sec B)2 – (tan B – cosec A)2

= (cot2 A + sec2 B + 2 cot A sec B) – (tan2 B + cosec2 A – 2 tan B cosec A)= cot2 A + sec2 B + 2 cot A sec B – tan2 B – cosec2 A + 2 tan B cosec A= (sec2 B – tan2 B) – (cosec2 A – cot2 A) + 2 (cot A sec B + tan B cosec A)= 1 – 1 + 2 (cot A sec B + tan B cosec A) [Q sec2 B – tan2 B = 1, cosec2 A – cot2 A = 1]= 2 (cot A sec B + tan B cosec A) = R.H.S. Hence Proved.

28. Prove that :(sin q + cos q + 1) · (sin q – 1 + cos q) · sec q · cosec q = 2. [4]

Sol. L.H.S. = (sin q + cos q + 1) · (sin q – 1 + cos q) · sec q cosec q= [(sin q + cos q) + 1] · [(sin q + cos q) – 1)] · sec q · cosec q

= [(sin q + cos q)2 – (1)2] sec q cosec q [Q (a + b) (a – b) = a2 – b2]= [sin2 q + cos2 q + 2 sin q cos q – 1] · sec q cosec q


Arundeep’s Solved Papers Mathematics 2014 (Term I)11= (1 + 2 sin q · cos q – 1) · sec q cosec q [Q sin2 q + cos2 q = 1]

= 2 (sin q cos q) ·1 1·

cos sinq q= 2 = R.H.S. Hence Proved.

29. If tan (20º – 3a) = cot (5a – 20º), then find the value of a and hence evaluate :sin a · sec a · tan a – cosec a · cos a · cot a. [4]

Sol. Given, tan (20º – 3a) = cot (5a – 20º)Þ tan (20º – 3a) = tan [90 – (5a – 20º)] [Q cot q = tan (90º – q)]Þ 20º – 3a = 90º – 5a + 20º Þ – 3a + 5a = 90º + 20º – 20ºÞ 2a = 90ºÞa = 45º

Now, sin a · sec a · tan a – cosec a · cos a · cot a= sin 45º · sec 45º · tan 45º – cosec 45º · cos 45º · cot 45º

1 12 1 2 12 2

= ´ ´ - ´ ´ = 1 – 1 = 0 Ans.

30. The frequency distribution of weekly pocket money received by a group of students isgiven below :

Pocket More More More More More More More More More Moremoney than than than than than than than than than thanin ( ) or or or or or or or or or or

equal equal equal equal equal equal equal equal equal equalto 20 to 40 to 60 to 80 to 100 to 120 to 140 to 160 to 180 to 200

Number

90 76 60 55 51 49 33 12 8 4of

students

Draw a ‘more than type’ ogive and from it, find median. Verify median by actualcalculations. [4]

Sol.

PQ

R


Arundeep’s Solved Papers Mathematics 2014 (Term I)12Taking pocket money (in Rs.) along x-axis and number of students along y-axis, plot the points(20, 90), (40, 76), (60, 60), (80, 55), (100, 51), (120, 49), (140, 33), (160, 12), (180, 8) and (200,4) on graph paper and join all these point by free hand gives the required more than type ogive.Now mark a point representing 45 on y-axis by P, from P draw a line || to x-axis meeting ogive atQ. From Q, draw QR perpendicular to x-axis.Then abscissa of point R gives the required median

\ Md = 125.The table of values is given as under :

Pocket money No. ofc.i. fi c.f(in ` ) Students

More than or equal to 20 90 20–40 14More than or equal to 40 76 40–60 30More than or equal to 60 60 60–80 60–55 = 5 35More than or equal to 80 55 80–100 55–51 = 4 39

More than or equal to 100 51 100–120 51–49 = 2 41More than or equal to 120 49 120–140 57More than or equal to 140 33 140–160 33–12=21 78

More than or equal to 160 12 160–180 82More than or equal to 180 8 180–200

12 – 8 = 486

More than or equal to 200 4 200–2208–4 = 4

90

N = 90 å i

49–33=16

4

Here, N = 90

\N2

90 452

= =

The cumulative frequency just greater than 45 be 51 and corresponding median class is 120–140

\ Median

N2

cfl h

f

-= + ´

45 41120 2016-= + ´ 4 20120

16´= + = 120 + 5 = ` 125

Hence Verified31. Cost of living Index for some period is given in the following frequency distribution :

Index 1500–1600 1600–1700 1700–1800 1800–1900 1900–2000 2000–2100 2100–2200No. of weeks 3 11 12 7 9 8 2

Find the mode and median for above data. [4]


Arundeep’s Solved Papers Mathematics 2014 (Term I)13The table of values is given as under :

Index Number of weeks (f i) cf1500–1600 3 31600–1700 11 f0 141700–1800 12 f1 261800–1900 7 f2 331900–2000 9 422000–2100 8 502100–2200 2 52

N = Sfi = 52

Here N = 52

\N2

52 262

= = , now commulative frequency which is just greater than or equal to 26 be

1700 – 1800.\ Median class is 1700–1800, Here l = 1700 ; c.f. = 14 ; f = 12 ; h = 100

\ Median

N2

cfl h

f

-= + ´

26 141700 10012-= + ´

121700 100 180012

= + ´ =

Also, Maximum frequency is 12 which lies in 1700 – 1800.\ Modal class is 1700–1800, Here l = 1700 ; f1 = 12 ; f0 = 11 ; f2 = 7

Thus, mode 1 01 0 22

f fl h

f f f-

= + ´- -

12 111700 10024 11 7-= + ´- -

11700 1006

= + ´ = 1700 + 16.67

= 1716.67 Ans.


General Instructions :(i) All questions are compulsory.

(ii) The question paper consists of 34 questions divided into four sections – A, B, C and D.(iii) Section A contains 8 questions of 1 mark each, which are multiple choice type questions, Section

B contains 6 questions of 2 marks each, Section C contains 10 questions of 3 marks each andSection D contains 10 questions of 4 marks each.

(iv) Use of calculators is not permitted.Section – A

Question numbers 1 to 8 carry 1 mark each. For each of these questions four alternative choices havebeen provided of which only one is correct. Select the correct choice.

1. The first three terms of an APrespectively are 3y – 1, 3y + 5 and 5y + 1.Then y equals :(a) – 3 (b) 4(c) 5 (d) 2

Sol. (c) Since 3y – 1, 3y + 5 and 5y + 1 are in A.P.\ 2 (3y + 5) = 3y – 1 + 5y + 1

[If a, b, c are in A.P., b – a = c – bÞ 2b = a + c]

Þ 6y + 10 = 8yÞ 10 = 2yÞ y = 52. In Fig. 1, QR is a common tangent to

the given circles, touching externally atthe point T.

The tangent at T meets QR at P. IfPT = 3.8 cm, then the length ofQR (in cm) is:(a) 3.8 (b) 7.6(c) 5.7 (d) 1.9

Sol. (b) Given, PT= 3.8 cmWe know that the lengths of the tangentsdrawn to a circle from a point outside thecircle are equal.

\ QP = PT = 3.8 cmand PR = PT = 3.8 cm

\ QR = QP + PR = (3.8 + 3.8) cm = 7.6 cm3. In Fig. given below PQ and PR are two

tangents to a circle with centre O. IfÐQPR = 46º, then ÐQOR equals :

MATHEMATICS 2014 TERM - II (OUTSIDE DELHI)SET I


Arundeep’s Solved Papers Mathematics 2014 (Outside Delhi)14


Arundeep’s Solved Papers Mathematics 2014 (Outside Delhi)15(a) 67º (b) 134º(c) 44º (d) 46º

Sol. (b) We know that tangent to a circle isperpendicular to the radius at the point ofcontact.

\ ÐOQP = Ð ORP = 90ºNow, in quadrilateral ORPQ, we haveÐQOR + 90º + 46º + 90º = 360º

Þ ÐQOR + 226º = 360ºÞ ÐQOR = 360º – 226º = 134º4. A ladder makes an angle of 60º with the

ground when placed against a wall. Ifthe foot of the ladder is 2 m away fromthe wall, then the length of the ladder(in metres) is :

(a)43 (b) 4 3

(c) 2 2 (d) 4

Sol. (d) Suppose AB is the ladder of length x m.\ OA = 2 m, ÐOAB = 60º

In right DAOB, sec 60º 2x=

Þ 2 2x= Þ x = 4 m

Hence the required length of ladder be4 metre.

5. If two different dice are rolled together,the probability of getting an evennumber on both dice, is :

(a)136 (b)

12

(c)16 (d)

14

Sol. (d) Number of possible even numbers onboth dice = 9, i.e. (2, 2), (2, 4), (2, 6),(4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6).Total possibilities when two dice are rolled = 36{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

\ Probability of getting an even number on

both dice9 1

36 4= =

6. A number is selected at random fromthe numbers 1 to 30. The probability thatit is a prime number is :

(a)23 (b)

16

(c)13 (d)

1130

Sol. (c) Total outcomes of selecting a numberfrom 30 numbers = 30Favourable numbers (prime numbers) = 10,i.e. (2, 3, 5, 7, 11, 13, 17, 19, 23, 29)

\ Required probability of selecting a prime

number10 1 .30 3

= =

7. If the points A (x, 2), B (– 3, – 4) andC (7, – 5) are collinear, then the valueof x is :(a) – 63 (b) 63(c) 60 (d) – 60

Sol. (a) If points A (x, 2), B (– 3, – 4) andC (7, – 5) are collinear, then area of triangle


Arundeep’s Solved Papers Mathematics 2014 (Outside Delhi)16formed by these points is 0.[if A(x1, y1) ; B(x2, y2) and C(x3, y3) arevertices of DABCThen area of DABC

=12 |x1(y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)|

Þ12 [x (– 4 + 5) – 3 (– 5 – 2) + 7 (2 + 4)] = 0

Þ12 [x + 21 + 42] = 0 Þ x + 63 = 0

Þ x = – 638. The number of solid spheres, each of

diameter 6 cm that can be made bymelting a solid metal cylinder of height45 cm and diameter 4 cm, is :(a) 3 (b) 5(c) 4 (d) 6

Sol. Given diameter of sphere = 6 cm

\ radius of sphere =62 cm = 3 cm = r

\ volume of each solid sphere = 343

rp

= 34 3 363p ´ = p

Also, diameter of solid cylinder = R =42 cm = 2cm

and height of solid cylinder = h = 45 cm\Volume of solid cylinder =pR2h =p × 22 × 45 =180p

\ Required no. of solid sphere made

=volume of cylinder 180

volume of each sphere 36p=p = 5

\ Ans (b)

Section – B

Question numbers 9 to 14 carry 2 marks each.9. Solve the quadratic equation

2x2 + ax – a2 = 0 for x.

Sol. Consider the equation2x2 + ax – a2 = 0

Þ 2x2 + 2ax – ax – a2 = 0Þ 2x (x + a) – a (x + a) = 0Þ (2x – a) (x + a) = 0Þ 2x – a = 0 or x + a = 0

Þ 2ax = or x = – a

10. The first and the last terms of an APare 5 and 45 respectively. If the sum ofall its terms is 400, find its commondifference.

Sol. Let total no. of terms of given A.P. be n.Given, a = 5, an = 45 and Sn = 400

We know, Sn ( )2 nn a a= +

Þ 400 (5 45)2n= + Þ 400 = 502

n ´

Þ 400 = 25n Þ n =40025

Þ n = 16\ a16 = 45Þ a + 15d = 45 Þ 5 + 15d = 45

Þ 15d = 40 Þ 40 815 3

d = =

Hence the required common difference of

given A.P. be83 .

11. Prove that the line segment joining thepoints of contact of two parallel tangentsof a circle, passes through its centre.

Sol. Given : PQ and RS are two parallel tangentsto a circle at B and A respectively. O is thecentre of the circle.To prove : AB passes through O.Construction : Join OA and OB.Proof : OB is perpendicular to PQ.[Tangent is perpendicular to radius at thepoint of contact.]



Now, PQ || RS Þ BO (Produced to RS) isperpendicular to RS. ...(i)[A line perpendicular to one of the twoparallel lines is perpendicular to other linealso]Also, OA is perpendicular to RS[Reason as above] ...(ii)From (i) and (ii), OA and OB must coincideas only one line can be drawn perpendicularfrom a point outside the line to the line.

\ A, O, B are collinear.Þ AB Passes through O, the centre of the

circle.12. If from an external point P of a circle

with centre O, two tangents PQ and PRare drawn such that ÐQPR = 120º, provethat 2PQ = PO.

Sol. Given : PQ and PR are tangents from pointP to circle with centre O.Also, ÐQPR = 120ºTo Prove : 2PQ = OPConstruction : Join OQ, OP and ORProof : In triangles OQP and ORP, we haveOQ = OR (radii)OP = OP (common)PQ = PR[tangents drawn from a point outside thecircle to the circle are equal in length]

\ DOQP @ DORP (SSS axiom of congruency)\ ÐOPQ = ÐOPR = 60º (c.p.c.t)

In right-angled triangle OQP, we have

OPPQ = sec 60º = 2 Þ OP = 2PQ

13. Rahim tosses two different coinssimultaneously. Find the probability ofgetting at least one tail.

Sol. When two different coins are tossedsimultaneously, then total possibilities = 4,i.e. (H, H), (H, T), (T, H), (T, T)Number of favourable outcomes for at leastone tail = 3, i.e. (H, T), (T, H), (T, T).

\ Required probability of getting at least one

tail3= .4

14. In given below, a square OABC isinscribed in a quadrant OPBQ of a circle.If OA = 20 cm, find the area of theshaded region. (Use p = 3.14)

Sol. OA = 20 cm.

\ OB 2= × side

(Q OB is a diagonal of the square)

Þ OB 20 2 cm=



\ Shaded area = Area of sector OPBQ– Area of square OABC

2 290 (20 2) (20)360º

°= ´ p -

3.14400 2 1 400 14 2pé ù é ù= ´ - = -ê ú ê úë û ë û

= 400 [1.57 – 1] = 400 × 0.57 = 228 cm2

Section – CQuestion numbers 15 to 24 carry 3 marks each.

15. Solve the equation4 53 ;

2 3x x- =

+

3 ,2

x ¹ 0, - for x.

Sol. Given equation be,4 53 ,

2 3x x- =

+

30,2

x ¹ -

Þ4 3x

x- 5

2 3x=

+Þ (4 – 3x) (2x + 3) = 5xÞ 8x + 12 – 6x2 – 9x = 5xÞ 6x2 + 6x – 12 = 0Þ 6 (x2 + x – 2) = 0Þ x2 + x – 2 = 0Þ x2 + 2x – x – 2 = 0Þ x (x + 2) – 1 (x + 2) = 0Þ (x + 2) (x – 1) = 0Þ Either x + 2 = 0 or x – 1 = 0Þ x = – 2, 1

16. If the seventh term of an AP is19 and

its ninth term is1 ,7 find its 63rd term.

Sol. Let a be the first term and d be the commondifference of the given AP.

Given, a719

= Þ a + 6d19

= ...(i)

[_ an = a + (n – 1) d]

Also, a917

= Þ a + 8d17

= ...(ii)

Subtracting (i) from (ii), we get

a + 8d – a – 6d1 17 9

= -

Þ 2d9 7 2

63 63-= = Þ

163

d =

Substituting the value of d in (i), we get

1663

a + ´ 19

=

Þ a 1 6 7 – 6 19 63 63 63

= - = =

\ a163

= and d163

=

Now, a63 = a + 62d1 16263 63

= + ´

1 62 63 163 63+= = =

17. Draw a right triangle ABC in whichAB = 6 cm, BC = 8 cm and ÐB = 90º.Draw BD perpendicular from B on ACand draw a circle passing through thepoints B, C and D. Construct tangentsfrom A to this circle.

Sol. Steps of construction :1. BC = 8 cm is drawn.2. ÐABC = 90º is drawn.3. AB = 6 cm is cut and AC is joined.4. Triangle ABC is the required triangle, right

angled at B.5. From point B, an arc is drawn cutting AC

at X and Y.6. Arc XY is bisected, by drawing two arcs

of same radius cutting at N.



7. BN is joined meeting AC at D.8. BD is perpendicular to AC.9. As ÐBDC = 90º, therefore, BC is

hypotenuse of DBDC.10. Take point O as mid-point of BC.11. With O as centre, a circle is drawn passing

through B, D and C.12. AO is joined.13. With AO as diameter, a circle is drawn

cutting the circle BDC at R and B

(Q ÐABO = 90°) join AR.

14. AB and AR are the tangents from A to circleBDC.

18. If the point A (0, 2) is equidistant fromthe points B (3, p) and C (p, 5), find p.Also find the length of AB.

Sol. Q A (0, 2) is equidistant from the pointsB (3, p) and C (p, 5).

\ AB = AC Þ AB2 = AC2

Þ (0 –3)2 + (2 – p)2 = (0 – p)2 + (2 – 5)2

Þ 9 + 4 – 4p + p2 = p2 + 9Þ 4p = 4 Þ p = 1\ Coordinates of point B be (3, 1)

\ AB 2 2(0 3) (2 1)= - + -

9 1 10= + = units

19. Two ships are there in the sea on eitherside of a light house in such a way thatthe ships and the light house are in thesame straight line. The angles ofdepression of two ships as observed fromthe top of the light house are 60º and45º. If the height of the light house is200 m, find the distance between the two

ships. [Use 3 1.73]=

Sol. Let AB be the light house of height 200 m.C and D are two ships on either sides oflight house with angles of depression 60ºand 45º respectively.

\ ÐACB = 60º and ÐADB = 45º[Alternate angles in both cases]

In right-angled triangle ABC,

BCAB = cot 60º

Þ BC1200×3

= 200 m3

= ...(i)

In right-angled triangle ABD,

BDAB = cot 45º

Þ BD = 200 × 1 = 200 m ...(ii)

\ Distance between ships = CD = CB + BD

200 200 3= 200 20033

+ = +



200 1.73 2003´= + 346 200

3= + = (115.33 + 200) m

= 315.33 m.20. If the points A (– 2, 1), B (a, b) and C (4, – 1) are collinear and a – b = 1, find the values

of a and b.

Sol. Q Points A (– 2, 1), B (a, b) and C (4, – 1) are collinear..

\ Area of triangle formed by these points is zero.

[if A(x1, y1) ; B(x2, y2) and C(x3, y3) are vertices of DABC

Then area of DABC =12

|x1(y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)|]

Þ12 [– 2 (b + 1) + a (– 1 – 1) + 4 (1 – b)] = 0

Þ – 2b – 2 – 2a + 4 – 4b = 0 Þ 2a + 6b = 2

Þ a + 3b = 1 ...(i)

Also, a – b = 1 Þ a = b + 1 ...(ii)

Putting the value of a from (ii) in (i), we get

b + 1 + 3b = 1 Þ 4b = 0 Þ b = 0

\ from eqn. (ii), a = 1

Thus, a = 1, b = 021. In Fig. given below a circle is inscribed in an equilateral

triangle ABC of side 12 cm. Find the radius of inscribedcircle and the area of the shaded region.

[Use p = 3.14 and 3 1.73]=

Sol. Given : An equilateral triangle of side 12 cm.Let radius of incircle be r. Join OA, OB and OC.Also, join OD, OE and OF.Here, AB, BC and AC are the tangents to the circle.

\ OD ^ BC, OE ^ AC and OF ^ AB.[Radius is perpendicular to the tangent at the point of contact]Now, ar (ABC) = ar (AOB) + ar (BOC) + ar (COA)

Þ 23 · (12)

41 112 122 2

r r= ´ ´ + ´ ´ 1 122

r+ ´ ´ [_ area of equilateral DABC =3

4 (side)2]

Þ 36 3 = 18r

Þ r 2 3 cm= = 2 × 1.73 = 3.46 cm


Arundeep’s Solved Papers Mathematics 2014 (Outside Delhi)21Thus, required radius of incircle = 3.46 cmArea of shaded portion = Area of equilateraltriangle DABC – area of circle

2 23 (12) (2 3)4

= - p

36 3 12= - p = 12 (3 × 1.73 – 3.14) cm2

= 12 (5.19 – 3.14) cm2 = 12 × 2.05 cm2

= 24.60 cm2

22. In Fig. given below PSR, RTQ and PAQare three semicircles of diameters 10cm, 3 cm and 7 cm respectively. Findthe perimeter of the shaded region.

[Use p = 3.14]

Sol. Diameters of Semicircle PSR, PAQ andQTR are 10 cm, 7 cm and 3 cm and hence

corresponding radii are102 cm,

72 cm and

32 cm i.e. 5 cm, 3. 5 cm and 1.5 cm

respectively.Perimeter of shaded region= length of arc PSR + length of arc PAQ +length of arc QTR= [p (5) + p (3.5) + p (1.5)] cm= p × 10 cm = 3.14 × 10 cm = 31.4 cm

23. A farmer connects a pipe of internaldiameter 20 cm from a canal into acylindrical tank which is 10 m indiameter and 2 m deep. If the waterflows through the pipe at the rate of 4km per hour, in how much time will thetank be filled completely ?

Sol. Internal diameter of pipe = 20 cm

\ Internal radius of pipe = 10 cm1 m

10=


Arundeep’s Solved Papers Mathematics 2014 (Outside Delhi)22In 1 h, 4 km length of water flows into thetank.

\ In 1 hour, the volume of water which flowsout

2110

æ ö= p ç ÷è ø × 4 × 1000 m3

[_ 1 km = 1000 m]Volume of cylindrical tank = p (5)2 × 2 m2

[_ diameter of tank = 10 m, height = 2 m]\ Required time taken to fill the tank

Volume of tankVolume of water flows in 1 hour

=

25 2 100 h4 1000

p ´ ´ ´=p ´ ´

5 h4

= = 1 h 15 min.

24. A solid metallic right circular cone 20cm high and whose vertical angle is 60º,is cut into two parts at the middle of itsheight by a plane parallel to its base. Ifthe frustum so obtained be drawn into

a wire of diameter1 cm,

12 find the

length of the wire.Sol. Cone is cut by plane PB and PQDB is a

frustrum of cone.OA = AC = 10 cm, AB = r1 m ; CD = r2

Vertical angle = 60º, so Semi vertical angle = 30º

In right-angled triangle OAB, we have

ABOA = tan 30º Þ

1 110 3r

=

Þ r110= cm.

3In right angled triangle OCD, we have

CDOC = tan 30º Þ

2 120 3r

=

Þ r220= cm

3

Frustum is drawn into wire of diameter1 cm

12 and length x cm (say)

\ Volume of wire = volume of frustum

Þ pr2x 2 21 2 1 21 [ · ]3

h r r r r= p + +

Þ21

24xæ öp ´ç ÷è ø

103

p ´=2 210 20 10 20·

3 3 3 3

é ùæ ö æ ö+ +ê úç ÷ ç ÷ê úè ø è øë û

Þ1

576x´ 10 100 400 200

3 3 3 3é ù= + +ê úë û

Þ 576x 10 700

3 3= ´

Þ x7000 576

9= ´ = 7000 × 64 cm

= 448000 cm

\ length of wire448000 km100000

= = 4.48 km

[_ 1 km = 1000 m = 1000 × 100 cm

Þ 1 cm =1 km

100000 ]



Section – AQuestion numbers 25 to 34 carry 4 marks each.

25. The difference of two natural numbersis 5 and the difference of their

reciprocals is1 .

10 Find the numbers.

Sol. Let numbers be x and y, such that x > y.According to question,x – y = 5 ...(i)

If x > y, then1 1y x

>

Now,1 1y x

- 110

= ...(ii)

From (i) and (ii), we have

1 15y y

-+

110

=

Þ5

( 5)y yy y

+ -+

110

=

Þ 50 = y2 + 5yÞ y2 + 5y – 50 = 0Þ y2 + 10y – 5y – 50 = 0Þ y (y + 10) – 5 (y + 10) = 0Þ (y + 10) (y – 5) =0Þ Either y + 10 = 0 or y – 5 = 0Þ y = – 10 (rejected)

Since y be any natural number.Thus, y = 5.

\ from (i), x = 10\ Required natural numbers are 10 and 5.

26. Prove that the length of the tangentsdrawn from an external point to a circleare equal.

Sol. Given : A circle C (O, r). P is a pointoutside the circle and PA and PB aretangents to a circle.To Prove : PA = PBConstruction : Draw OA, OB and OP.

Proof : Consider triangles OAP and OBP.ÐOAP = ÐOBP = 90º ...(i)[Radius is perpendicular to the tangent atthe point of contact]

OA = OB (radii) ...(ii)OP is common ...(iii)

\ DOAP @ DOBP (RHS)[from (i), (ii), (iii)]

Þ AP = BP (cpct)27. The angles of elevation and depression

of the top and the bottom of a tower fromthe top of a building, 60 m high, are 30ºand 60º respectively. Find the differencebetween the heights of the building andthe tower and the distance betweenthem.

Sol. Let AB be the height of the tower and CDbe the building of height 60 m.ÐACE = 30º, ÐECB = 60º



Let difference between heights of buildingand tower be y m and distance between thetower and building by x m.

\ CD = EB = 60 m, CE = BD = x m.In right-angled triangle CEB.

CEBE = cot 60º

Þ CE1= 60 m3

´ 60 3 m3

=

Þ BD = CE 20 3 m=\ Distance between tower and building is

20 3 m.

In right-angled triangle CEA,

AECE = tan 30º Þ y

120 3 20 m3

= ´ =

\ Difference between the height of tower andbuilding is 20 m.

28. A bag contains cards numbered from 1to 49. A card is drawn from the bag atrandom, after mixing the cardsthoroughly. Find the probability that thenumber on the drawn card is :

(i) an odd number(ii) a multiple of 5

(iii) a perfect square(iv) an even prime numberSol. Total cards in the bag = 49.

\ Total number of outcomes = 49, when onecard is drawn.

(i) No. of favourable outcomes for oddnumber = 25, i.e. 1, 3, 5, .... 49.

\ Probability of getting an odd number25 .49

=

(ii) No. of favourable outcomes for multipleof 5 = 9, i.e., 5, 10, 15, .... 45.

\ Probability of getting a multiple of 59 .49

=

(iii) Favourable outcomes for a perfect square= 7, i.e., 1, 4, 9, 16, 25, 36 and 49.

\ Probability of getting a card with perfect

square number7 1 .49 7

= =

(iv) Favourable outcomes for an even primenumber = 1, i.e., 2.

\ Probability of getting a card numbered even

prime1 .49

=

29. Find the ratio in which the point P (x, 2)divides the line segment joining thepoints A (12, 5) and B (4, – 3). Also findthe value of x.

Sol. Let P (x, 2) divides the join of A (12, 5) andB (4, – 3) in the ratio k : 1.Then by section formula, we have

\ Point of division is P

4 12 3 5,1 1

k kk k

+ - +æ öç ÷+ +è ø = (x, 2)

Þ4 12

1kk

++ = x and

3 51

kk

- ++ = 2 ...(i)

Consider,3 5

1k

k- +

+ = 2

Þ –3k + 5 = 2k + 2

Þ – 5k = – 3 Þ k35

=

\ Required ratio is k : 1, i.e., 3 :15

i.e. 3 : 5

Now, substituting35

k = in (i), we get

x34 12 12 60 725 9

3 3 5 815

´ + += = = =++


Arundeep’s Solved Papers Mathematics 2014 (Outside Delhi)2530. Find the values of k for which the

quadratic equation(k + 4) x2 + (k + 1) x + 1 = 0 has equalroots. Also find these roots.

Sol. Given equation is(k + 4) x2 + (k + 1) x + 1 = 0 ...(i)On comparing with ax2 + bx + c = 0, a ¹ 0Here, a = k + 4 ; b = k + 1 ; c = 1first we note k + 4 ¹ 0 Þ k ¹ – 4Here D = b2 – 4ac = (k + 1)2 – 4 · (k + 4)= k2 + 2k + 1 – 4k – 16= k2 – 2k – 15

Q roots are equal, therefore, D = 0Þ k2 – 2k – 15 = 0Þ k2 – 5k + 3k – 15 = 0Þ k (k – 5) + 3 (k – 5) = 0Þ (k + 3) (k – 5) = 0Þ k + 3 = 0 or k – 5 = 0 Þ k = – 3, 5

when k = – 3, then from (i), we get(– 3 + 4) x2 + (– 3 + 1) x + 1 = 0

Þ x2 – 2x + 1 = 0Þ (x – 1)2 = 0Þ (x – 1) · (x – 1) = 0Þ x = 1, 1

when k = 5, then from (i), we get(5 + 4) x2 + (5 + 1) x + 1 = 0

Þ 9x2 + 6x + 1 = 0Þ (3x + 1)2 = 0Þ (3x + 1) · (3x + 1) = 0

Þ x1 1,3 3

= - -

\ Roots are 1, 1 or1 1, .3 3

- -

31. In an AP of 50 terms, the sum of first10 terms is 210 and the sum of its last15 terms in 2565. Find the A.P.

Sol. Let a be the first term and d be the commondifference of the given AP.

\ According to question, we havea1 + a2 + .... + a10 = 210and a36 + a37 + .... + a50 = 2565

Consider, a1 + a2 + .... + a10 = 210

Þ 1 1010 [ ]2

a a+ = 210

[using S ( )2nn a l= + where l is the last term]

Þ 5 [a + a + 9d] = 210Þ 2a + 9d = 42 ...(i)

Again consider,a36 + a37 + .... + a50= 2575

Þ 36 5015 [ ]2

a a+ = 2565

Þ a + 35d + a + 49d = 171 × 2Þ 2a + 84d = 171 × 2Þ a + 42d = 171 ...(ii)

eqn. (i) – 2 × eqn. (ii), we have2a + 9d – 2a – 84d Þ 42 – 342

Þ – 75d = – 300 Þ d = 4\ From (ii), a + 42 × 4 = 171Þ a = 171 – 168 = 3\ a = 3 and d = 4.

Hence required AP is a + a + d, a + 2d,a + 3d, ......i.e. 3, 3 + 4, 3 + 8, 3 + 12, .....i.e. 3, 7, 11, 15, ....

32. Prove that a parallelogramcircumscribing a circle is a rhombus.

Sol. Given : A parallelogram ABCD, whichcircumscribes a circle.To prove : ABCD is a rhombus.Proof : We know that the lengths of thetangents drawn from a point outside thecircle to the circle are equal in length. Usingthis result, we getAP = AS, BP = BQ, CQ = CR and DR = DS

...(i)Consider, AB + CD = AP + BP + CR + DR= AS + BQ + CQ + DS [using (i)]= (AS + DS) + (BQ + CQ)= AD + BC



Þ AB + AB = BC + BC[Q AB = CD ; AD = BC, opp. sides of ||gmare equal]

Þ 2AB = 2BC Þ AB = BCAs adjacent sides of parallelogram ABCDare equal, hence, parallelogram ABCD is arhombus.

33. Sushant has a vessel, of the form of aninverted cone, open at the top, of height11 cm and radius of top as 2.5 cm and isfull of water. Metallic spherical ballseach of diameter 0.5 cm are put in the

vessel due to which2 th5 of the water in

the vessel flows out. Find how many ballswere put in the vessel. Sushant madethe arrangement so that the water thatflows out irrigates the flower beds. Whatvalue has been shown by Sushant ?

Sol. Radius of base of the cone = R = 2.5 cmand height of cone = h = 11 cm.

\ Volume of water in cone

= 2R3hp

2 31 (2.5) 11cm3

= p ´

2 th5 of the volume of water in cone

2 32 (2.5) 11cm5 3

p= ´ ´ ...(i)

Given diameter of spherical ball = 0.5 cm.

\ Radius of spherical ball = r =0.52 cm

\ Volume of spherical ball

= 343

rp3

24 0.5 cm3 2

æ ö= p ç ÷è øLet x be the number of spherical balls bedropped.

We know volume of water displaced isequal to volume of body immersed.

\3

24 0.5 2 (2.5) 113 2 5 3

x pæ ö´ p = ´ ´ç ÷è ø

Þ x 2 32 1 8(2.5) 115 4(0.5)

= ´ ´ ´

Þ x4 2.5 2.5 115 0.5 0.5 0.5

´ ´ ´=´ ´ ´

44 25 25 10 44025 25

´ ´ ´= =´

440 balls were put in the vessel. As the waterdisplaced was used for irrigation of flowerbed, it shows Sushant is concerned for theconservation and best use of naturalresources.

34. From a solid cylinder of height 2.8 cmand diameter 4.2 cm, a conical cavity ofthe same height and same diameter ishollowed out. Find the total surface area



of the remaining solid.22Take7

é ùp =ê úë ûSol. The shaded conical cavity is hollowed out.

For cylinder :

Radius of base (r)4.2 cm2

= = 2.1 cm

height (h) = 2.8 cm

For cone :

Radius of base (r)4.2 cm2

= = 2.1 cm,

height (h) = 2.8 cm

Slant height (l) = 2 2r h+ 2 2(2.1) (2.8)= +

4.41 7.84= + 12.25= = 3.5 cm

Total surface Area of remaining solid= curved surface area of cylinder + Areaof top circular base + curved surface areaof cone= 2prh + pr2 + prl = pr [2h + r + l]= p × 2.1 [2 × 2.8 + 2.1 + 3.5] cm2

22 2.1 [5.6 5.6]7

= ´ ´ +

227

= × 2.1 × 11.2 cm2 = 73.92 cm2

14. The first and the last terms of an APare 7 and 49 respectively. If sum of allits terms is 420, find its commondifference.

Sol. Let a be the first term, d be the commondifference and an be the last term of theAP, with n terms and Sn be the sum to nterms of given A.P.Given, a = 7, an = 49, Sn = 420

\ Sn ( )2 nn a a= + Þ 420 (7 49)2

n= +

Þ 840 = 56n Þ n = 15\ a15 = 49Þ a + 14d = 49 Þ 7 + 14d = 49Þ 14d = 42 Þ d = 3\ Required common difference of given A.P.

is 3.

22. Solve the equation3 1 2 ;

1 2 3 1x x- =

+ -11, ,3

x x¹ - ¹ for x.

Sol. Consider the equation3 1 2 ,

1 2 3 1x x- =

+ -

where x ¹ – 1,13

Þ6 ( 1)2( 1)

xx

- ++

23 1x

=-

Þ5

2 ( 1)x

x-+

23 1x

=-

Þ (3x – 1) (5 – x) = 4 (x + 1)Þ 15x – 3x2 – 5 + x = 4x + 4Þ 3x2 – 12x + 9 = 0

Set-II (Uncommon Questions to Set-I)NOTE : Except for the following questions, all the

remaining questions have been asked in previous set.



Þ 3 (x2 – 4x + 3) = 0Þ x2 – 4x + 3 = 0Þ x2 – 3x – x + 3 = 0Þ x (x – 3) – 1 (x – 3) = 0Þ (x – 1) (x – 3) = 0Þ Either x – 1 = 0 or x – 3 = 0Þ x = 1, 3

23. Points A (– 1, y) and B (5, 7) lie on acircle with centre O (2, – 3y). Find thevalues of y. Hence find the radius of thecircle.

Sol. As points A (– 1, y) and B (5, 7) lie on acircle with centre O (2, – 3y).

\ OA = OB (radii) Þ OA2 = OB2

Þ (2 + 1)2 + (– 3y – y)2 = (2 – 5)2

+ (– 3 y – 7)2

Þ 9 + 16y2 = 9 + 9y2 + 49 + 42yÞ 7y2 – 42y – 49 = 0Þ 7 (y2 – 6y – 7) = 0Þ y2 – 7y + y – 7 = 0Þ y (y – 7) + 1 (y – 7) = 0Þ (y + 1) (y – 7) = 0Þ Either y + 1 = 0 or y – 7 = 0Þ y =– 1 or 7

when y = – 1

then centre O is (2, 3) and point A is (– 1, – 1)

\ Radius of circle 2 2(2 1) (3 1)= + + +

9 16 25= + = = 5 units

When y = 7,then centre O is (2, – 21) and point A is(– 1, 7)

\ Radius of circle 2 2(2 1) ( 21 7)= + + - -

9 784 793 units= + =

24. If the points P (– 3, 9), Q (a, b) and R(4, – 5) are collinear and a + b = 1, findthe values of a and b.

Sol. Since the points P (– 3, 9), Q (a, b) and R(4, – 5) are collinear, therefore, area oftriangle formed by these points is zero.

[if A(x1, y1) ; B(x2, y2) and (x3, y3) arevertices of DABCThen area of DABC

=12 |x1(y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)|]

Þ12 [– 3 (b + 5) + a (– 5 – 9) + 4 (9 – b)] = 0

Þ –3b – 15 – 14a + 36 – 4b = 0Þ 14a + 7b – 21 = 0 ...(i)Þ 2a + b = 3 ...(ii)

Also, a + b = 1On subtracting, we get a = 2Substituting the value of a in (i), we get4 + b = 3 Þ b = – 1

\ a = 2 b = – 131. The difference of two natural numbers

is 3 and the difference of their

reciprocals is3 .

28 Find the numbers.

Sol. Let the numbers be x and y, where x > y.Given, x – y = 3 ....(i)

If x > y, then1 1 .y x

>

\1 1 3

28y x- =



Þ1 1

3y y-

+3

28= [using eqn. (i)]

Þ3

( 3)y yy y

+ -+

328

=

Þ 23

3y y+3

28=

Þ y2 + 3y = 28Þ y2 + 3y – 28 = 0Þ y2 + 7y – 4y – 28 = 0Þ y (y + 7) – 4 (y + 7) = 0Þ (y – 4) (y + 7) = 0Þ Either y – 4 = 0 or y + 7 = 0Þ y = 4 or – 7

Since – 7 is rejected as it is not a naturalnumber, therefore, y = 4Thus, from (i) ; x – 4 = 3 Þ x = 7

\ Required numbers are 7 and 4.32. Prove that the tangent at any point of a

circle is perpendicular to the radiusthrough the point of contact.

Sol. Given : A circle with centre O, line l istangent to the circle at A.To prove : Radius OA is perpendicular tothe tangent at A.Construct : Take a point P, other then A,on tangent l. Join OP, meeting the circle at R.Proof : We know that tangent to the circletouches, the circle at one point and all otherpoints on the tangent lie in the exterior of acircle.

\ OP > OR (radius of circle)Þ OP > OA

(Q OR = OA, radius of circle)

Þ OA < OPÞ OA is the smallest segment, from O to a

point on the tangent.We know that smallest line segment from a

point outside the circle to the line isperpendicular segment.

l

Hence, OA is ^ to tangent l.Thus, tangent at any point of a circle isperpendicular to the radius through the pointof contact.

33. All the black face cards are removedfrom a pack of 52 playing cards. Theremaining cards are well shuffled andthen a card is drawn at random. Findthe probability of getting a :(i) face card(ii) red card(iii) black card(iv) king

Sol. In total 52 cards, 6 cards are black facecards which have been removed. [Facecards are ; 2 black Jack, 2 black Queen, 2black King]

\ Remaining cards = (52 – 6) = 46Total possibilities of drawing a card = 46.

(i) Favourable outcomes for a face card (king,queen, jack) are 6 (2 red kings + 2 redqueens + 2 red jacks)

\ Required probability of getting a face card

6 346 23

= =

(ii) Favourable outcomes for a red card are 26.\ Required probability of getting a red card

26 13 .46 23

= =

(iii) Favourable outcomes for a black card are(26 – 6) = 20, as all black face cards havebeen removed.



\ Required probability of getting a black card

20 10 .46 23

= =

(iv) Favourable outcomes for a king are 2(2 redkings)

\ Probability of getting a king2 1 .

46 23= =

34. Find the values of k for which thequadratic equation(3k + 1) x2 + 2 (k + 1) x + 1 = 0 has equalroots. Also find the roots.

Sol. Consider the given equation(3k + 1) x2 + 2 (k + 1) x + 1 = 0 ...(i)

First we notice 3k + 1 ¹ 0 Þ13

k ¹ -

On comparing given eqn. with ax2 + bx + c = 0Here a = 3k + 1 ; b = 2(k + 1) ; c = 1Here discriminant = b2 – 4ac= [2 (k + 1)]2 – 4 (3k + 1)= 4 [k2 + 2k + 1 – 3k – 1] = 4 [k2 – k]

Q roots are equal, therefore, D = 0Þ 4 (k2 – k) = 0 Þ k2 – k = 0Þ k (k – 1) = 0Þ Either k = 0 or k – 1 = 0Þ k = 0 or k = 1

When k = 0, then from (i), we get(0 + 1) x2 + 2 (0 + 1) x + 1 = 0

Þ x2 + 2x + 1 = 0 Þ (x + 1)2 = 0Þ (x + 1) (x + 1) = 0 Þ x = – 1, – 1\ Required roots are – 1, – 1.

When k = 1, then from (i), we get(3 + 1) x2 + 2 (1 + 1) x + 1 = 0

Þ 4x2 + 4x + 1 = 0 Þ (2x + 1)2 = 0Þ (2x + 1) (2x + 1) = 0

Þ x1 1,2 2

= - -

\ Roots are1 1,2 2

- -

Hence, roots are – 1, – 1 or1 1, .2 2

- -

Set-III (Uncommon Questions to Set-I and Set-II)NOTE : Except for the following questions, all the

remaining questions have been asked in previous set.

14. The first and the last terms of an APare 8 and 65 respectively. If the sum ofall its terms is 730, find its commondifference.

Sol. Let a be the first term, d be the commondifference, an be the last term and Sn bethe sum to first n terms of the given AP.Given : a = 8, an = 65, Sn = 730

Now, Sn ( )2 nn a a= +

Þ 730 (8 65)2n= +

Þ 73073

2n= Þ n = 20

\ a20 = 65Þ a + 19d = 65 Þ 8 + 19d = 65Þ 19d = 57 Þ d = 3\ Required common difference of given A.P.

be 3.22. If the points A (– 1, – 4), B (b, c) and

C (5, – 1) are collinear and 2b + c = 4,find the values of b and c.

Sol. Since the points A (– 1, – 4), B (b, c) andC (5, – 1) are collinear, therefore, area oftriangle formed by these points is zero.



[if A(x1, y1) ; B(x2, y2) and (x3, y3) arevertices of DABCThen area of DABC

=12 |x1(y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)|]

Þ12 [– 1 (c + 1) + b (– 1 + 4) + 5 (– 4 – c)] = 0

Þ – c – 1 + 3b – 20 – 5c = 0Þ 3b – 6c = 21Þ b – 2c = 7 ...(i)

Also, 2b + c = 4 ...(ii) [given]Multiplying (ii) by 2 and then adding to (i),we getb – 2c + 4b + 2c = 7 + 8

Þ 5b = 15Þ b = 3

Substituting the value of b in (i), we get3 – 2c = 7 Þ 2c = – 4 Þ c = – 2

\ b = 3, c = – 223. If the point P (2, 2) is equidistant from

the points A (– 2, k) and B (– 2k, – 3),find k. Also find the length of AP.

Sol. Point P (2, 2) is equidistant from the pointsA (– 2, k) and B (– 2k, – 3).

\ AP = BPÞ AP2 = BP2

Þ (2 + 2)2 + (2 – k)2 = (2 + 2k)2 + (2 + 3)2

Þ 16 + 4 + k2 – 4k = 4 + 4k2 + 8k + 25Þ – 3k2 – 12k – 9 = 0Þ – 3 (k2 + 4k + 3) = 0Þ k2 + 4k + 3 = 0Þ k2 + 3k + k + 3 = 0Þ k (k + 3) + 1 (k + 3) =0Þ (k + 3) (k + 1) = 0Þ Either k + 3 = 0 or k + 1 = 0Þ k = – 3 or – 1

When k = – 1, then point A is (– 2, – 1)

\ AP 2 2(2 2) (2 1)= + + +

16 9 25= + = = 5 unitsWhen k = – 3, then point A is (– 2, – 3)

\ AP 2 2(2 2) (2 3)= + + +

16 25 41 units= + =

24. Solve the equation14 51 ;

3 1x x- =

+ +x ¹ – 3, – 1, for x.

Sol. Consider the given equation,14 1

3x-

+5 ,

1x=

+ where x ¹ – 3, – 1

Þ14 3

3x

x- -

+5

1x=

+

Þ11

3x

x-+

51x

=+

Þ (11 – x) (x + 1) = 5 (x + 3)Þ 11x + 11 – x2 – x = 5x + 15Þ – x2 + 5x – 4 = 0 Þ x2 – 5x + 4 = 0Þ x2 – 4x – x + 4 = 0Þ x (x – 4) – 1 (x – 4) = 0Þ (x – 1) (x – 4) = 0Þ Either x – 1 = 0 or x – 4 = 0 Þ x = 1, 4

31. Find the value of p for which thequadratic equation(2p + 1) x2 – (7p + 2) x + (7p – 3) = 0 hasequal roots. Also find these roots.

Sol. Given equation is(2p + 1) x2 – (7p + 2) x + (7p – 3) = 0 ...(i)

First we notice 2p + 1 ¹ 0 Þ12

p ¹ -

On comparing given eqn. with ax2 + bx + c = 0We have a = 2p + 1 ; b = – (7p + 2) and

c = 7p – 3\ Discriminant D = b2 – 4ac\ D = [– (7p + 2)]2 – 4 (2p + 1) (7p – 3)



= (7p + 2)2 – 4 (14p2 – 6p + 7p – 3) = 49p2 + 4 + 28p – 4 (14p2 + p – 3)= 49p2 + 4 + 28p – 56p2 – 4p + 12 = – 7p2 + 24p + 16For equal roots, D = 0 Þ – 7p2 + 24p + 16 = 0

Þ 7p2 – 24p – 16 = 0 Þ 7p2 – 28p + 4p – 16 = 0Þ 7p (p – 4) + 4 (p – 4) = 0 Þ (7p + 4) (p – 4) = 0

Þ Either 7p + 4 = 0 or p – 4 = 0 Þ p 47

= - or 4

When4 ,7

p = - then from eqn. (i), we get

28 28 281 2 37 7 7

x xæ ö æ ö æ ö- + - - + + - -ç ÷ ç ÷ ç ÷è ø è ø è ø = 0

Þ – x2 + 14x – 49 = 0 Þ x2 – 14x + 49 = 0 Þ (x – 7)2 = 0 Þ x = 7, 7When p = 4, then from (i), we gt(8 + 1) x2 – (28 + 2) x + (28 – 3) = 0

Þ 9x2 – 30x + 25 = 0 Þ (3x – 5)2 = 0 Þ x =5 5,3 3

\ Roots are 7, 7 or5 5, .3 3

32. Cards numbered from 11 to 60 are kept in a box. If a card is drawn at random from thebox, find the probability that the number on the drawn card is :

(i) an odd number(ii) a perfect square number

(iii) divisible by 5(iv) a prime number less than 20Sol. Cards numbered 11 to 60 are kept in a box. Total cards are 50.

\ Total no. of outcomes of drawing a card = 50.(i) Favourable outcomes for odd number = 25, (i.e., 11, 13, 15, ...., 59).

\ Probability of drawing an odd numbered card25 1 .50 2

= =

(ii) Favourable outcomes for a perfect square number = 4 (i.e. 16, 25, 36, 49).

\ Probability of drawing a perfect square numbered card4 2 .

50 25= =

(iii) Favourable outcomes for a card number divisible by 5 = 10, (i.e., 15, 20, 25, 30, .... 60).

\ Probability of drawing a card numbered divisible by 510 1 .50 5

= =

(iv) Favourable outcomes for a prime number less than 20 = 4 (i.e. 11, 13, 17, 19).



\ Probability of drawing a prime number less than 204 2 .

50 25= =

34. The difference of two natural numbers is 5 and the difference of their reciprocals is5 .

14Find the numbers.

Sol. Let two natural numbers be x and y, where x > y. Hence x – y = 5 ...(i)

If x > y, then1 1y x

>

Hence,1 1 5

14y x- =

Þ1 1

5y y-

+5

14= [from (i)]

Þ5

( 5)y yy y

+ -+

514

= Þ 25

5y y+5

14= Þ y2 + 5y = 14

Þ y2 + 5y – 14 = 0 Þ y2 + 7y – 2y – 14 = 0 Þ y (y + 7) – 2 (y + 7) = 0Þ (y – 2) (y + 7) = 0Þ y = 2, y = – 7 (rejected) as – 7 is not a natural number.\ y = 2

From (i), x – 2 = 5 Þ x = 7\ Required numbers are 7 and 2.


General Instructions :(i) All questions are compulsory.

(ii) The question paper consists of 34 questions divided into four sections – A, B, C and D.(iii) Section A contains 8 questions of 1 mark each, which are multiple choice type questions, Section

B contains 6 questions of 2 marks each, Section C contains 10 questions of 3 marks each andSection D contains 10 questions of 4 marks each.

(iv) Use of calculators is not permitted.Section – A

Question numbers 1 to 8 carry 1 mark each. For each of these questions four alternative choices havebeen provided of which only one is correct. Select the correct choice.

1. If k, 2k – 1 and 2k + 1 are threeconsecutive terms of an A.P., the valueof k is(a) 2 (b) 3(c) – 3 (d) 5

Sol. (b) If k, 2k – 1, 2k + 1 are three consecutiveterms of an A.P., then(2k – 1) – k = (2k + 1) – (2k – 1)

Þ 2k – 1 – k = 2 Þ k – 1 = 2 Þ k = 3[if a, b, c are in A.P. then b – a = c – b]

2. Two circles touch each other externallyat P. AB is a common tangent to thecircles touching them at A and B. Thevalue of ÐAPB is(a) 30º (b) 45º(c) 60º (d) 90º

Sol. (d) We have, AT = TP and TB = TP[Lengths of the tangents from ext. point Tto the circles]

Þ ÐTAP = ÐTPA = x (say)and ÐTBP = ÐTPB = y (say)[equal sides have equal angle opposite to it]Also, in triangle APB, x + x + y + y = 180º

Þ 2x + 2y = 180ºÞ x + y = 90ºÞ ÐAPB = 90º

3. In a right triangle ABC, right-angled atB, BC = 12 cm and AB = 5 cm. Theradius of the circle inscribed in thetriangle (in cm) is(a) 4 (b) 3(c) 2 (d) 1

Sol. (c) In DABC,

AC 2 2AB BC= + 2 25 12= +

25 144 169= + = = 13 cmNow, ar (ABC) = ar (AOB) + ar (BOC)

+ ar (COA)

MATHEMATICS 2014 TERM II ( DELHI)SET I


Arundeep’s Solved Papers Mathematics 2014 (Delhi)34



Þ1 5 122

´ ´ 1 15 122 2

r r= ´ ´ + ´ ´ 1 132

r+ ´ ´

[_ area of triangle =12 × base × altitude]

Þ 60 = r (5 + 12 + 13)Þ 30r = 60 Þ r = 2

Alternate method :BP = OR = r, BR = OP = r

[BPOR is rectangle as all angles are 90º]\ PC = 12 – r = CQ and AR = 5 – r = AQ

Also, AC 2 25 12= +

25 144 169= + = = 13

Þ AQ + QC = 13Þ 5 – r + 12 – r = 13Þ 2r = 4 Þ r = 24. In a family of 3 children, the probability

of having at least one boy is

(a)78 (b)

18

(c)58 (d)

34

Sol. (a) In a family of three children,Total possibilities are : BBB, BGB, BBG,GBB, GGB, GBG, BGG, GGG, therefore,total no. of possible outcomes = 8Favourable outcomes for at least one boyare :

BGG, GBG, GGB, BBG, BGB, GBB, BBB,i.e. 7.

\ Required probability of having at least one

boy7 .8

=

5. The angle of depression of a car parkedon the road from the top of a 150 m hightower is 30º. The distance of the car fromthe tower (in metres) is

(a) 50 3 (b) 150 3

(c) 150 2 (d) 75

Sol. (b) Let AB be the tower of height 150 m.C is car and angle of depression is 30º.Let the required distance of the car fromthe tower be x m.

Therefore, ÐACB = 30º (alternate angle)In right-angled triangle ABC, we have

BCAB = cot 30º

Þ 150x

= 3

Þ x 150 3 m.=

Thus the required distance of the car from

the tower is 150 3 m.

6. The probability that a number selectedat random from the numbers 1, 2, 3, ...,15 is a multiple of 4, is

(a)4

15 (b)2

15



(c)15 (d)

13

Sol. (c) Total no. of possible outcomes = 15

(Q 15 numbers are given)Favourable outcomes for a multiple of 4= 3 (i.e. 4, 8, 12)

\ Probability of selecting a number which is

a multiple of 43 1

15 5= =

7. ABCD is a rectangle whose threevertices are B (4, 0), C (4, 3) andD (0, 3). The length of one of itsdiagonals is(a) 5 (b) 4(c) 3 (d) 25

Sol. (a) Both the diagonals are equal\ Length of diagonal BD

2 2(4 0) (0 3)= - + -

16 9 25 5= + = =

8. A chord of a circle of radius 10 cmsubtends a right angle at its centre. Thelength of the chord (in cm) is

(a) 5 2 (b) 10 2

(c)52 (d) 10 3

Sol. (b) AB is a chord. ÐAOB = 90º,AO = OB = 10 cm

\ Triangle AOB is right angled at O.Using pythagoras theorem, we have

\ AB 2 2OA OB= + 2 2(10) (10)= +

= 100 100+ = 100 2´

10 2 cm.=

Section BQuestion numbers 9 to 14 carry 2 marks each.

9. Find the values of p for which thequadratic equation 4x2 + px + 3 = 0 hasequal roots.

Sol. Given quadratic equation be 4x2 + px + 3 = 0On comparing with ax2 + bx + c = 0,We have a = 4 ; b = p ; c = 3Since given quadratic eqn. has equal roots,therefore, D = 0 Þ b2 – 4ac = 0

Þ (p)2 – 4 × 4 × 3 = 0Þ p2 = 48

Þ p 48 4 3= ± = ±

10. Find the number of natural numbersbetween 101 and 999 which are divisibleby both 2 and 5.

Sol. Numbers between 101 and 999 which aredivisible by both 2 and 5 (i.e. by L.C.M. of2 and 5 i.e. 10) are 110, 120, 130, ... 990.This forms an A.P. with a = 110, d = 10and an = 990Now, an = a + (n – 1) d

Þ 990 = 110 + (n – 1) 10Þ 880 = (n – 1) 10Þ 88 = n – 1Þ n = 89\ There are 89 natural numbers which are

divisible both by 2 and 5.


Arundeep’s Solved Papers Mathematics 2014 (Delhi)3711. In given below common tangents AB and

CD to the two circles with centres O1and O2 intersect at E. Prove thatAB = CD.

Sol. In the given figure, AB and CD are commontangents to the two given circles withcentres O1 and O2.We know that the lengths of the tangentsdrawn from a point outside the circle tothe circle are equal in length.

\ AE = EC and EB = EDÞ AE + EB = CE + EDÞ AB = CD.

12. The incircle of an isosceles triangle ABC,in which AB = AC, touches the sides BC,CA and AB at D, E and F respectively.Prove that BD = DC.

Sol. We know that the lengths of the tangentsdrawn from a point outside the circle tothe circle are equal in length.

\ AF = AE, BF = BDand CD = CE ...(i)Given AB = AC

Þ AF + FB = AE + ECÞ FB = EC [using (i), AF= AE]

Þ BD = CD[using (i), BF = BD and CD = CE]

13. Two different dice are tossed together.Find the probability

(i) that the number on each die is even.(ii) that the sum of numbers appearing on

the two dice is 5.Sol. Two different dice are tossed. Therefore,

total outcomes are 36.Here simple space S = {(1, 1), (1, 2),(1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2),(2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2),(3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2),(4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2),(5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2),(6, 3), (6, 4), (6, 5), (6, 6)}

(i) Favourable outcomes for even number onboth dice = 9, {(2, 2) (2, 4), (2, 6), (4, 2),(4, 4), (4, 6), (6, 2), (6, 4), (6, 6)}

\ Probability of getting even number on both

dice9 1

36 4= =

(ii) Favourable outcomes that the sum of thenumbers appearing in two dice is 5 are(1, 4), (2, 3), (3, 2), (4, 1), i.e. 4.

\ Probability of getting sum of numbers

appearing on two dice is 54 1 .

36 9= =

14. If the total surface area of a solidhemisphere is 462 cm2, find its volume.

22Take7

é ùp =ê úë û


Arundeep’s Solved Papers Mathematics 2014 (Delhi)38Sol. Let r cm be the radius of hemisphere

Then total surface area of a solid hemisphere= 2pr2 + pr2 = 3pr2

Now, 3pr2 = 462 (given)

Þ r2462 7 493 22

´= =´ = 7

2 Þ r = 7

\ Required volume of hemisphere 323

r= p

32 22 (7)3 7

= ´ ´ 23

= × 22 × 49 cm3

= 718.67 cm3

Section – C

Question numbers 15 to 24 carry 3 marks each.15. Solve for x :

16 151 ;1x x

- =+ x ¹ 0, – 1

Sol. Given equation be,16 1x

- 15 ,1x

=+ where

x ¹ 0, – 1

Þ16 x

x- 15

1x=

+

Þ (16 – x) (x + 1) = 15xÞ 16x + 16 – x2 – x = 15xÞ x2 = 16 = (± 4)2

Þ x = ± 416. The sum of the 5th and the 9th terms

of an AP is 30. If its 25th term is threetimes its 8th term, find the AP.

Sol. Let a be the first term and d be the commondifference of a given A.P.Given, a5 + a9 = 30

Þ a + 4d + a + 8d = 30Þ 2a + 12d = 30Þ a + 6d = 15 ...(i)

Also, a25 = 3a8Þ a + 24d = 3 (a + 7d)Þ a + 24d = 3a + 21dÞ 3d = 2a ...(ii)

From (i) and (ii), we havea + 4a = 15 Þ 5a = 15 Þ a = 3

\ From (ii), we get, 3d = 2 × 3 Þ d = 2\ a = 3, d = 2

Hence required A.P. is a, a + d, a + 2d,a + 3d, .... i.e. 3, 3 + 2, 3 + 4, 3 + 6, .....i.e. 3, 5, 7, 9, ....

17. Construct a triangle with sides 5 cm,5.5 cm and 6.5 cm. Now construct

another triangle, whose sides are35

times the corresponding sides of thegiven triangle.

Sol.

Steps of construction :1. A triangle with sides BC = 6.5 cm, AB = 5

cm and AC = 5.5 cm is constructed.2. ÐCBX is drawn below BC.


Arundeep’s Solved Papers Mathematics 2014 (Delhi)393. On BX, A1, A2, ..., A5 are marked, such

that BA1 = A1A2 = ... = A4A5.4. A5 and C are joined.5. C¢A3 is drawn parallel to A5C which meets

BC at C¢.6. A¢C¢ is drawn parallel to AC meeting AB at

A¢.7. DA¢BC¢ is the required triangle.

18. The angle of elevation of an aeroplanefrom a point on the ground is 60º. Aftera flight of 30 seconds the angle ofelevation becomes 30º. If the aeroplaneis flying at a constant height of

3000 3 m, find the speed of theaeroplane.

Sol. From the point of observation (O),

plane is at A, AL 3000 3 m= andÐAOL = 60º.After 30 seconds, plane is at B, therefore,

BM 3000 3 m= and ÐBOM = 30º.

Distance AB is covered in 30 seconds.In right-angled triangle OLA, we have

OLAL = cot 60º Þ OL = AL cot 60°

Þ OL13000 3 3000 m3

= ´ = ...(i)

In right-angled triangle OMB, we have

OMBM = cot 30º Þ OM = BM cot 30°

Þ OM 3000 3 3= ´ = 9000 m ...(ii)\ AB = LM = OM – OL

= (9000 – 3000) m = 6000 m[from (i) and (ii)]

Now in 30 s, distance covered = 6000 m\ In 1 hour (3600 s), distance covered by

plane 6000 3600 km30 1000

= ´

= 720 km\ Required speed of the aeroplane = 720 km/h.

19. If the point P (k – 1, 2) is equidistantfrom the points A (3, k) and B (k, 5),find the values of k.

Sol. Point P (k – 1, 2) is equidistant from thepoints A (3, k) and B (k, 5).

\ AP = BPÞ AP2 = BP2

Þ (k – 1 – 3)2 + (2 – k)2 = (k – 1 – k)2

+ (2 – 5)2

Þ (k – 4)2 + (2 – k)2 = (– 1)2 + (– 3)2

Þ k2 – 8k + 16 + 4 – 4k + k2 = 1 + 9Þ 2k2 – 12k + 10 = 0Þ k2 – 6k + 5 = 0Þ k2 – 5k – k + 5 = 0Þ k (k – 5) – 1 (k – 5) = 0Þ (k – 1) (k – 5) = 0Þ Either k – 1 = 0 or k – 5 = 0Þ k = 1 or 5

20. Find the ratio in which the line segmentjoining the points A (3, – 3) and B (– 2, 7)is divided by x-axis. Also find thecoordinates of the point of division.

Sol. Let point P (x, 0) on x-axis divides the joinof A (3, – 3) and B (– 2, 7) in the ratio k : 1

Then by section formula, the coordinatesof P are

2 3 7 3,1 1

k kk k

- + -æ öç ÷+ +è ø ...(i)


Arundeep’s Solved Papers Mathematics 2014 (Delhi)40If point P lies on x-axis, then y = 0

Þ7 3

1kk

-+ = 0 Þ 7k – 3 = 0 Þ k

37

=

\ Required ratio is3 :1,7 i.e. 3 : 7.

Substituting37

k = in (i), we get the

coordinates of point of division are ;

6 37P , 0 ,3 1

7

æ ö- +ç ÷ç ÷ç ÷+ç ÷è ø

i.e. 3P , 0 .2

æ öç ÷è ø

21. In Figure given below two concentriccircles with centre O, have radii 21 cmand 42 cm. If ÐAOB = 60º, find the area

of the shaded region.22Use7

é ùp =ê úë û

Sol. Given, two concentric circles with centre Oand ÐAOB = ÐCOD = 60º = q (say)OC (r1) = 21 cm, OA (r2) = 42 cm

\ Area of shaded region

2 22 1

(360º ) ( )360º

r r- q= ´ p -

[_ area of sector of circle = pr2 × 360q

° ]

2 2(360º 60º ) 22 (42 21 )360º 7

-= ´ -

2300 22 (21) (4 1)360 7

°= ´ ´ -°

2 25 22 (21) 3cm6 7

= ´ ´ ´ = 3465 cm2.

22. The largest possible sphere is carved outof a wooden solid cube of side 7 cm. Findthe volume of the wood left.

22Use7

é ùp =ê úë ûSol. As largest possible sphere is carved out of

a solid wooden cube of side equal to sideof cube i.e. 7 cm, so, diameter of the sphereis eqnal to side of cube i.e. 7 cm.

\ Volume of the wood left = Volume of thecube – Volume of the sphere

33 34 7(7) cm

3 2

é ùæ ö= - pê úç ÷è øë û[_ volume of cube = a3

and volume of sphere = 343

rp ]

3 4 22 1(7) 13 7 8

é ù= - ´ ´ê úë û= 343 [1 – 0.52] = 343 × 0.48 cm3

= 164.64 cm3.23. Water in a canal, 6 m wide and 1.5 m

deep, is flowing at a speed of 4 km/h.How much area will it irrigate in 10minutes, if 8 cm of standing water isneeded for irrigation ?

Sol. Given ; width of canal = 6 m,and depth of canal = 1.5 m and in 1 hour,


Arundeep’s Solved Papers Mathematics 2014 (Delhi)41length of water flows out by canal = 4 km

\ Volume of water flows out in 1 hour= 6 × 1.5 × 4000 m3 = 36000 m3

\ Volume of water flows out in 10 minutes

36000 1060

= ´ = 6000 m3 ...(i)

Suppose this water irrigates x m2 of areaand we require 8 cm of standing water.

\ Volume of water required 38 m

100x= ´

From (i) and (ii) we get

8 6000100

x ´ =

Þ x6000 100

8´= = 75000 m2

\ 75000 m2 of area is irrigated.24. In Figure given below ABCD is a

trapezium of area 24.5 sq. cm. In it,AD || BC, ÐDAB = 90º, AD = 10 cm andBC = 4 cm. If ABE in a quadrant of acircle, find the area of the shaded

region.22Take7

é ùp =ê úë û

Sol. Area of trapezium = 24.5 sq. cmAD = 10 cm, BC = 4 cmLet AB = h cm

\ Area of trapezium =12

(AD + BC) × AB

Þ 24.51 (10 4)2

h= + ´

Þ h24.5 3.5cm

7= =

Now, area of quadrant ABE

290º (3.5) sq. cm360º

= ´ p

21 22 (3.5) sq. cm4 7

= ´ ´

= 9.625 sq. cm ...(ii)\ Area of shaded region

= area of trap – area of quadrantABE= (24.5 – 9.625) sq. cm= 14.875 sq. cm [from (i), (ii)]

Section – D

Question numbers 25 to 34 carry 4 marks each.25. Solve for x :

2 4 10 ;3 5 3

x xx x

- -+ =- - x ¹ 3, 5

Sol. Given equation be,

2 43 5

x xx x

- -+- -

10 ,3

= x ¹ 3, 5

Þ( 3) 1 ( 5) 1

3 5x x

x x- + - ++

- -103

=



Þ1 11 1

3 5x x+ + +

- -103

=

Þ1 1

3 5x x+

- -10 23

= -

Þ5 3

( 3) ( 5)x xx x

- + -- -

43

=

Þ 22 8

8 15x

x x-

- +43

=

Þ 4x2 – 32x + 60 = 6x – 24Þ 4x2 – 38x + 84 = 0Þ 2x2 – 19x + 42 = 0Þ 2x2 – 12x – 7x + 42 = 0Þ 2x (x – 6) – 7 (x – 6) = 0Þ (2x – 7) (x – 6) = 0Þ Either 2x – 7 = 0 or x – 6 = 0

Þ x7 , 6.2

=

26. In a school, students decided to planttrees in and around the school to reduceair pollution. It was decided that thenumber of trees, that each section ofeach class will plant, will be double of theclass in which they are studying. If thereare 1 to 12 classes in the school and eachclass has two sections, find how manytrees were planted by the students.Which value is shown in this question ?

Sol. According to question, each section of :Class 1 will plant 2 trees, class II will plant4 trees, class III will plant 6 trees and soon.. class 12 will plant 24 trees and eachclass has 2 sections.

\ Number of trees planted= 4 + 8 + 12 + ... + 48This forms an A.P. with a = 4, d = 8 – 4 = 4and n = 12

\ Required number of trees planted

S1212 (4 48)2

= + S ( )2nn a lé ù= +ê úë û

Q

= 6 × 52 = 312

Value promted :

Students are concerned about safe andpollution free environment.

27. The angle of elevation of the top of atower at a distance of 120 m from a pointA on the ground is 45º. If the angle ofelevation of the top of a flagstaff fixedat the top of the tower, at A is 60º, thenfind the height of the flagstaff.

[Use 3 = 1.73]

Sol. Let BC be the tower and BD is flagstaff ofheight h m.

Let BC = x m.

AC = 120 m, ÐBAC = 45º

and ÐDAC = 60º

In right-angled triangle ACB, we have

ACBC = cot 45º Þ

120 1x

=

Þ x = 120 ...(i)

In right angled triangle ACD, we have

CDAC = tan 60º Þ

3120h x+ =



Þ h + x 120 3= Þ h 120 3 120= - [using (i), x = 120]

Þ h 120[ 3 1]= - = 120 [1.73 – 1] m = 120 × 0.73 = 87.6 m

\ Required height of the flagstaff is 87.6 m.28. Red queens and black jacks are removed from a pack of 52 playing cards. A card is drawn

at random from the remaining cards, after reshuffling them. Find the probability thatthe drawn card is

(i) a king (ii) of red colour (iii) a face card (iv) a queenSol. Red queens and black jacks, i.e. 2 + 2 = 4 cards are removed from a pack of 52 playing cards.

Remaining cards are 52 – 4 = 48.\ Possible number of outcomes of drawing one card from 48 cards is 48.(i) Favourable outcomes for drawing a king are 4.

\ Probability of drawing a king4 1

48 12= =

(ii) Favourable outcomes for a card of red colour are 24 as 2 red queens have already been removedand in total there are 24 red cards.

\ Probability of drawing a red card24 1 .48 2

= =

(iii) Favourable outcomes for a face card (4 kings, 2 queens, 2 jacks) = 8

\ Probability of drawing a face card8 148 6

= =

(iv) Favourable outcomes for drawing a queen are 2, as 2 red queens have been removed.

\ Probability of drawing a queen2 1 .

48 24= =

29. If A (– 3, 5), B (– 2, – 7), C (1, – 8) and D (6, 3) are the vertices of a quadrilateral ABCD,find its area.

Sol. Area of quadrilateral ABCD= Area of triangle ABC + Area of triangle ACD ...(i)Since we know that if A(x1, y1), B(x2, y2) and C(x3, y3) are vertices of DABC

Then area of DABC =12

|x1(y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)|

Now, ar (ABC)12

= [– 3 (– 7 + 8) – 2 (– 8 – 5) + 1 (5 + 7)]

12

= [– 3 + 26 + 12]

35 sq. units2

= ...(ii)


Arundeep’s Solved Papers Mathematics 2014 (Delhi)44Also, ar (ACD)

12

= [– 3 (– 8 – 3) + 1 (3 – 5) + 6 (5 + 8)

12

= [33 – 2 + 78]

1092

= sq. units ...(iii)

from (i), (ii), (iii), we get

ar (ABCD)35 109 1442 2 2

= + = = 72 sq. units

30. A motorboat whose speed in still water is 18 km/h, takes 1 hour more to go 24 kmupstream than to return downstream to the same spot. Find the speed of the stream.

Sol. Speed of motor boat in still water = 18 km/hLet speed of the stream = x km/h

\ Speed upstream = (18 – x) km/hSpeed downstream = (18 + x) km/h

Thus, time taken by motor boat to covered a distance of 24 km stream =24 hr

18 x-

and Time taken by motor boat to covered a distance of 24 km downstream =24 km

18 x+According to question, we have

Þ24 24

18 18x x-

- + = 1 Þ24 (18 ) 24 (18 )

(18 ) (18 )x xx x

+ - -- + = 1

Þ 2432 24 432 24

324x x

x+ - +

- = 1 Þ 48x = 324 – x2

Þ x2 + 48x – 324 = 0 Þ x2 + 54x – 6x – 324 = 0Þ x (x + 54) – 6 (x + 54) = 0 Þ (x – 6) (x + 54) = 0Þ Either x – 6 = 0 or x + 54 = 0 Þ x = 6 or x = – 54 (rejected)\ Required speed of the stream is 6 km/h.

31. In Figure given below PQ is a chord of length 16 cm, ofa circle of radius 10 cm. The tangents at P and Q intersectat a point T. Find the length of TP.

Sol. Given : PQ is chord of length 16 cm, TP and TQ are thetangents to a circle with centre O, radius 10 cm.To find : TP.Solution : Join OP and OQ.In triangles OTP and OTQ, we have



OT is commonOP = OQ (radii)TP = TQ [length of the tangents drawnfrom a point outside the circle to the circleare equal]

\ DOPT @ DOQT(SSS axiom of congruency)

\ ÐPOT = ÐQOT (by cpct) ...(i)Consider, triangles OPR and OQR, we haveOP = OQ (radii)OR is commonÐPOR = ÐQOR [from (i)]

\ DOPR @ DOQR(SAS axiom of congruency)

\ PR = RQ1 162

= ´ = 8 cm ...(ii)

ÐORP = ÐORQ = 90º ...(iii)In right angled triangle TRP, we haveTR2 = TP2 – (8)2 = TP2 – 64Also OT2 = TP2 + (10)2

Þ (TR + 6)2 = TP2 + 100

[Q OR 100 64= - = 6]

Þ TR2 + 12TR + 36 = TP2 + 100Þ TP2 – 64 + 12TR + 36 = TP2 + 100Þ 12TR = 128

Þ TR32 cm3

=

From (iv) ; we have

2323

æ öç ÷è ø

= TP2 – 64

Þ TP21024 64

9= +

1024 576 16009 9+= = =

2403

æ öç ÷è ø

\ TP40 cm3

=

32. Prove that the tangent at any point of acircle is perpendicular to the radiusthrough the point of contact.

Sol. Given : A circle with centre O, line l istangent to the circle at A.To Prove : Radius OA is perpendicular tothe tangent at A.Construct : Take a point P, other then A, ontangent l. Join OP, meeting the circle at R.Proof : We know that tangent to the circletouches, the circle at one point and all otherpoints on the tangent lie in the exterior of acircle.

\ OP > OR (radius of circle)

Þ OP > OA (Q OR = OA, radius of circle)Þ OA < OPÞ OA is the smallest segment, from O to a

point on the tangent.We know that smallest line segment from apoint outside the circle to the line isperpendicular segment.



Hence, OA is ^ to tangent l.Þ tangent at any point of a circle is

perpendicular to the radius through the pointof contact.

33. 150 spherical marbles, each of diameter1.4 cm, are dropped in a cylindricalvessel of diameter 7 cm containing somewater, which are completely immersedin water. Find the rise in the level ofwater in the vessel.

Sol. Let rise in water in cylinder be h cm when150 spherical marbles, each of diameter1.4 cm are dropped and fully immersed.Then volume of 150 spherical marbles=

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