first term (sa-i) mathematics class x i...u-like cce sample question paper 1 27 solution. we divide...

CCE SAMPLE QUESTION PAPER I FIRST TERM (SA-I)

MATHEMATICS (With Solutions) CLASS X

I lFme Allowed : 3 to 3% H o g s ] [Maximum Marks : 80

General Instructions: ( i ) All questions are compulsory. (ii) The question paper consists of 34 questions divided into four sections A, B, C and D.

Section A comprises of 10 questions of 1 mark each, ;Section B comprises of 8 questions of 2 marks each, Section C comprises of 10 questions of 3 marks each and Section D comprises of 6 questions o f 4 m a r k s each.

( i i i ) Question numbers I to 10 in Section A are multiple choice questions where you are to select one correct option out of the given four.

, ( iv) There is no overall choice. However, internal choice has been prooided in I question of two marks, 3 questions of three marks each and 2 questions offour marks each. You have to attempt only one of the alternatives i n all such questions.

( v ) Use of calculators is not permitted.

'Question numbers 1 to 10 are of one mark each.

23 win tkrminate after 1. The decimal expansion of the rational number - 2 3 ~ 2 ' - -

how many places of decimal ? ( a ) 1 (b) 3 1 (c) 1 (dl 5 Sylution. Choice (b) i s correct.

23 will terminate after Thus, shows that decimal expansion of the rational number - ~ 3 5 ~ '

three places of decimal. 2. I f HCF (105,120) = 15, then LCM (105,120) is

( a ) 210 . (b) 420 (c) 840 ( d ) 1680

Solution. Choice (c) i s correct. '

HCF x LCM = Product of two numbers 15 x LCM = 105 x 120

a LCM = 105 x 120

15 + LCM = 105 x 8 = 840.

trainer

Rectangle

3. In figure, the graph of a polynomial p(x) is shown. The number of zeroes of p(x) is

Y t

Y" (a) 1 (b) 2 . . (c) 3 (d) 4

Solution. Choice (b) is correct. The number of zeroes ofp(x) is 2 as the graph intersects the x-axis at two points A and B in

figure. 4. The value of k for which 62,- 3y + 10 = 0 and 2r + ky + 9 = 0 has no solution is

(a) - 1 (b) -2 (c) 1 (dl 3

Solution. Choice (a) is correct. Here al = 6, bl = - 3, cl = 10 and az = 2, bz = k, c2 = 9 The given system of equations will have no solution if

- k = - 1 5. Let AABC - APQR, area of AABC - 81 cm2, area of APQR - 144 cm2 and QR =

6 cm, then length of BC is (a) 4 cm (b) 4.5 cm (c) 9 cm (d) 12 cm

Solution. Choice (b) is correct. Since AABC - APQR, therefore

.: area (AABC) = 81 cm2 (given) and area (APQR) = 144 em2 (given) 1

U-Like CCE Sample Question Paper 1 25

+ 3 x 3 BC = - 2

* BC = 4.5 cm cos (90" - 0) COs 0 - is 6. The value of

tan 0 (a) - sin2 0 (c) - cos2 0

Solution. Choice (a) is correct. cos (90" - 0) cos 0 - 1 - s i n e c o s e

t an 0 ' t an 0

- - s i n e c o s e sin 01cos 0

2 = cos 0 - 1 = - (1 -?sZ 0) =-sin 8

[.; QR = 6 cm (given)]

(b) - cosec2 0 (d) - cot 0

1 2 1 7. If tan 8 + - = 2, then the value of tan 0 + 7 is . tan 0 tan 0

(a) 3 (b) 4 (c) 2 (d) - 4

Solution. Choice ( c ) is correct.

Given, t a n 0 + - I - 2 t a n 0

1 + 2 t a n 0 . = 4 * t d 0 + 7 [.; (a + b12 = a2 + b2 + 2abl t an 0 t a n 0

* t a ~ P e + ~ + 2 = 4 ' t a n 0

* 1 t a n 2 0 + T = 4 - 2 t a n 0

* 1 t a n 2 0 c 7 = 2 tan 0

5sina-3cosrr is 8. If 5-tan a = 4, then the value of

. 5sinac2cosa

1 2 (a) - (b) 3

3

Solution. Choice (c) is correct. 4 5 t a n a = 4 * t a n a = - 5

- - (5 sin a - 3 cos a)/cos a [Dividing numerator and denominator by cos a]

(5 sin a + 2 cos a)/cos a

- - 5 tan a - 3 5(4/5) - 3 - - 5 tan a + 2 5(4/5) + 2 . 4 - 3 1 =-=- ~. 4 + 2 ' 6

9. If sec 4A - cosec (A - 209, where 4.4 is an acute angle, then the value of A is (a) 21" (b) 22" (c) 23" (d) 24"

Solution. Choice (b) is correct. Given, see 4.4 = cosec (A - 20") * cosec (90" - 4A) = cosec (A - 20") t.; cosec ($0' - 0) = sec 81 * 90"-4A-A-20" =e. - 4A +A - 90" + 20"

5A = 110" * - A-22"- 10. Which measure of central tendency is given.by the x-coordinate of the point

of intersection of 'more than ogive' and 'less than ogive' ? (a) Mean (b) Mode (c) Median (d) Mean and Median both.

Solution. Choice (c ) is correct. '

The median of a grouped data of central tendency is given by the x-coordinate of the point of intersection of the 'more than ogive' and 'less than ogive'.

-1 . .

Question numbers I1 to I8 carry 2 marks each. 11. Write 21975 as a product of its prime factors. Solution. We have

21975 = 5 x 4395 = 5 x 5 x 8 7 9 = 5 x 5 x 3 x 2 9 3 -3~5~x293. .

12. If the polynomial x4 + b3 + 8x2 + 122 + 18 is divided by another polynomial 2 + 5, the remainder comes out to be px + q. Fiqd the values ofp and q. . .


Solution. We divide the given polynomial by x2 + 5.

x2+2X+3

x2 + 5 x4 + 2x3 + 8x2 + 12% + 18 First term of qiotient is = x x4 + 5x2 x - -

2.2 2X3 + 3x2 + + of quotient is = 2x

- 2 ~ 3 + IOX . X . - 3x2 + 2x + 18 Third term ofquotient is = 3

I 3x2 - + 15 - X

2x + 3 ~.

3x2 I So, x4 + 2x3 + 8x2 + 12x + 18 = (x2 + 5)(x2 + 2x + 3) + (2x + 3) --

Quohent Remamder

But the given remainder is px + q. Therefore, 2 x + 3 = p x + q

Comparing both sides, we get : p = 2 and q = 3. Hence p = 2 and p - 3. 13. Solve the following system of equations :

Solution. The given system of equations can be written as :

--- = - l 2 2 ... (1)' x Y [Multiplying both sides of both the given equations by 21 2 1 and - + - = 16 ... (2) x Y

Now, multiplying (2) by 2 and adding in ( I ) , we get

1 . Substituting x = - in (11, we get 6

1 1 Thus, the solution of the given system of equations is x = - and y = -.

6 4

5 1-tana 14. If sec a = -, evaluate 4 l+tana

Solution. Let us draw a right W C (see figure)

Hypotenuse We have : sec a = , Base

:. AC = 5k. BC = 4k. where k is a ~ositive number . Using the grUlagorHs theorem, we have

AB2 = A C ~ - BC' - . AB2 = (5k12 - (4k12 * AB2 = 25k2 - 16k2 = 9k2 - AB = 3k

* AB 3k 3 tans=-=--=- BC 4k 4

l - t a n a 1-(314) (4-3114 4 - 3 1 Now, - - - -- -

l + t a n a l+(3/4)- (4+3) /4 4 + 3 = 7 Alternative Method :

Given, 5 sec a = - 4

* z 25 sec a = - 16

* z 25 l + t a n a = - 16

* . 25 t a n 2 a = - - 1 16

* 2 25- 16 tan a = - 16

* 16

* 3 t a n a = - 4

U-Like' CCE Sample Question Paper 1 29

l - t a n a - 1-(314)-(4-3)/4 Now, - l + t a n a 1+(3/4)-(4+3)/4

Prove that : (sin 0 + cosec 0)' + (cos 0 + sec 0)' = 7 + t a d 0 + cot2 8.

Solution. We have L.H.S. = (sin 0 + cosec 0)' + (cos 0 + sec 0)'

= (sin2 0 + cosec2 0 + 2 sin 0. cosec 0) + (cos2 0 + sec2 0 + 2 cos 0. sec 0) 1

=(s in20+cos20)+co~ec20+sec20+2s in0~& +2cose.-- sin 0 cos 0

2 =1+(1+co t20)+(1+ tan20)+2+2 [:. sec20=l+tan28;cosec20=1+cot 01

= RH.S. 15. Prove that.in the figure if ABCD is a trapezium with AB 11 DC 11 EF, then

Solution. Given : Atrapezium ABCD in which AB 1 1 DC 11 EF: AE BF To prove : - = - ED FC

Proof: In U C , we have EP 1 1 DC

Also, in W C , we have PFIIAB

[.: EF I / DC (given)]

... (1) [using BPTl

From (1) and (21, we have

Hence, proved.

16. In AABC, LA = 90" and AD I BC. Prove that AB2 + CD2 = BD' +A@. Solution. Let ABC be a triangle such that L A = 90" and AD I BC. We have to prove that

AB2 + C D ~ = B D ~ + A C ~ c In right AABD, we have

AB2 = AD2 + BD2 ... (1) [using Pythagoras Theorem] In right AACD, we have

AC2 =All2 + CD2 . . .(2) [using Pythagoras Theorem] NOW, AB2 -Ac2 = (AD2 + BD2) - (AD2 + CD2) [using (1) and (2)]

= B D ~ - A h. - AB2 + C D ~ = B D ~ + A C ~ . 17. The following distribution gives the scores of 230 students of a school : *

Scores

450 - 500 500 - 550 550 - 600 600 - 650 650 - 700 700 - 750 750 - 800

lbtal L

No. of students

35 40 32 24 27 18 34

230

. -

Scores No. of students 1 Score more than I Cumulatiue frequency

Write the above distribution as more than type cumulative frequency distribution.

Solution. The cumulative frequency table by more than type method :

400 - 450 450 - 500 500 - 550 550 - 600 600 - 650 650 - 700 700 - 750 750 - 800

400 - 450 20

20 35 40 32

- 24 27 18 34

We cgculate more than cumulative frequencies by adding successive classes from bottom to top.

18. For the following grouped frequency distribution find the mode :

I Class 1 3 - 6 1 6 - 9 1 9-12 112-15 1 15-18 1 18-21 121-24

400 450 500 550 600 650 700 750

Frequency

230 210 175 135 103 79 52 34

Solution. Here the maximum frequency is23 and the class corresponding to this frequency is 12 - 15.

2 5 1 10 23 21 12 3


So, the modal class is 12 - 15. Lower limit (1) of the modal class = 12. Class size (h) = 15 - 12 = 3 Frequency (fi) of the modal class = 23 Frequency (fo) of the class preceeding the modal class = 10 Frequency (fi) of the class following the modal class = 21 Using the formula :

Mode =' 1 + fl - fo h 2fl - fo - f 2

Question numbers 19 to 28 carry 3 marks each. 19. Show that any positive odd integer is of the form 69 + 1 or 69 + 3 or 6q + 5,

where q is a positive integer. . Solution. Let a be any positive integer, when a is divided by 6. Then, by Euclid's division

lemna, we get a = 6q + r, 0 a r < 6, where q and r are whole numbers.

So,.r=O,1,2,3,4,5. When r = 0 or 2 or 4, then a becomes an even integer, i.e., a = 6q or 6q + 2 or 6q + 4. When r = 1 or 3 or 5. then a becomes a positive odd integer, i.e., a = 6q + 1 or 6q + 3 or

6q + 5. Hence any positive odd integer is of the form 69 + 1 or 6q + 3 or 69 + 5, where q is a positive

integer. .

20. Prove that & + & is irrational.

Solution. Let us assume, to the contrary, that & + & is a rational number. That is we can find coprimep and q (q 0) such that

=> [Squaring both sides1

2 2 Since, p and q are integers, - 3q is rational, and so & is rational.

2pq

But this contradicts the fact that fi is irrational.

So, we conclude that f i + & is irrational. Or

Prove that - 5fi is irrational. 7

Solution. Let us assume to the contrary, that is a rational number. 7

. . 5f i - , wherep and q are coprime, i.e., their HCF is 1. - 7 q

7 P . Since P is a rational number, therefore, - = - 1s also rational number. 4 5 q

* & is a rational number.

But this contradicts the fact that is irrational.

5& . . This contradiction has arises because of our incorrect assumption that - 1s rational. 7

So, we conclude that - 5fi is irrational. 7

21. A two digit number is four times the sum of its digits and twice the product of the digits. Find the number.

Solution. Let the unit place digit be x and ten's place digit bey. Then, Original number = IOy + x It is given that a two digit number is four times the sum of its digits. . . lOy+x=4Cy+x) =, lOy-4y=42-x


* 6y = 3x * x 7 2y ... (1) Further, it is also given that a two digit number is twice the product of the digits.

1oy + x = 2(xy) * 1oy + 2y = 2[(2y)yI =z$ 12y = 4y2 * 3 = y Substituting y = 3 in ( I ) , we get

x = 2(3) = 6 Hence, the original number is 36.

Or Solve for x a i ~ d y :

3 ' - 2 a+b

a x - b y - a 2 - b 2 Solution. The given.system of equations can be written as

Y - 2 j b x = 2 = z $ b x + a y = 2 a b - + - - a b ab

and az -by=a2-b2 Multiplying (1) by b and (2) by a, we have

b2x + aby = 2ab2 and a2x - aby = a3-ab2 . '

Adding (3) and (41, we get b2x + a2x = 2ab2 + a3 - ab2

+ (b2 + a2)x = ab2 + a3 * (b2 + a2)x = a(b2 + a2)

+ x = a Substituting x = a in (2), we get

- by = - b2 + by=a2-a2+b2 * by = b2 + y = b Hence, x - a, y - b is the solution of the given system of equations. 22. I f a and p are the zeroes of the quadratic polynomial f ix) = & - &v + 7, find the

quadratic polynomial whose.zeroes are 2a + 38 and 3 a + 28. Solution. Since a and p are zeroes of the polynomial 2x2 - 5x i 7 , therefore

Let S and P denote respectively the sum and product of zeroes of the required polynomial, then

A A

and P = (2a + 3p)(3a + 2p) * P = 6a2 + 4ap + gap + 6p2 - P = 6a2 + 6p2 + 13up * P = 6(a2 + p2 + Zap) + up * P = 6(a + p12 + ap

=). . 75 7 82 p = - + - = - -41 2 2 2

Hence, the required polynomial p(x) is given by : p(x) = k(x2 - Sx + P)

or

or p(x) = k(@ - 252 + 82), where k is any non-zero real number.

l + c o s A + s i n A . .

23. Prove that : - 2 cosec A sin A 1 + cos A

Solution. We have

L.H.S. = l+cosA sinA

t sinA l+cosA

- - (1 + cos A)' + sin2 A sinA(l+cosA) .

- - ~ + ~ C O S A + C O S ~ A + S ~ ~ ~ A sin A (1 + cos A)

- - l + 2 c o s A + 1 sin A (1 + cos A)

- - .2+2cosA - sin A (1 + coa A)

= 2(l+cosA) sin A (1 + cos A)

.2 =- sin A

' =2cosecA = R.H.S.


24. Prove that sin 0

= 2 + sin 0 cot 0 + cosec 0 cot 0 - cosec 0

Solution. We have

R.H.S. = 2 + sin 0 cot 0 - cosec 0

= 2 + sin 0 (cot 0 + cosec 0) (cot 0 - cosec 0)(cot 0 + cosec 8)

= 2 + sin 0 (cot 0 + cosec 0) cot2 0 - cosec2 0

sin 0 cot 0 + sin 0 cosec 0 = 2 + cot2 0 - (1 + cot2 0)

sin 0 cot 0 + sin 0 cosec 0 = 2 +

-1

= -2 '+s ineco tO+l

- 1

- - - 1 + s i n 0 c o t 0 -1

- - sin 0 cosec 0 - sin 0 cot 0 (cosec 0 + cot 0)(cosec 0 - cot 0)

- - sin 0 (cosec 0 -cot 0) (cosec 0 + cot O)(cosec 0 -cot 0)

- - sin e cosec 0 + cot 0

[.; cosec2 0 = 1 + cot2 01

[.; sin 0.cosec 0 = 11

= L.H.S. 25. If ABC is an equilateral triangle with AD BC, then prove that m2 = ~DC'. Solution. Let ABC be a n equilateral triangle and AD I BC. I n AADB and AADC, we ha<e

AB =AC [given] LB = LC [Each = 6Oo1

and LADB = LADC [Each = 907 . . AADB - AADC ==s BD = DC ... (1) . . BC = BD + DC = DC + DC = 2DC ... (2) [using (i)]

In right angled AADC, we have AC2 = AD2 + DC'

j B C ~ = AD^ + DC' [.; AC = BC, sides of an equilateral A1 [using (2)l

* AD2 = 4DC2 - DC' =) AD' = 3Dc2 26. In figure, DEFG is a square and LBAC - 90". Show that DEZ = BD x EC.

C

Solution. In AAGF and ADBG, we have LGAF = LBDG

LAGF = LDBG So, by AA-criterion of similarity of triangles, we have

AAGF - ADBG .In AAGF and AEFC, we have

LFAG = LCEF LAFG = LECF

So, by AA-criterion of similarity of triangles, we have AAGF - AEFC -

From (1) and (2), we obtain ADBG - AEFC

=) BD DG - = - FE ' E C

[.; DEFG is a square.

[Each = 9Oo1 [Corresponding Lsl

[Each = 90°1 [Corresponding Lsl

:. EF = DE, DG = DE]

27. Find the mean of the following distribution, using step-deviation method :

Classes 4 - 8 8 - 12 12 - 16 16 - 20 20 - 24 24 - 28 28 - 32 32 - 36

Number of students 2

12 15 25 18 12 13

3

+ X = 18 + 1.92 6 X = 19.92 Hence, mean is 19.92.

s or


Solution. Let the assumed mean be a = 18 and h = 4. Calculation of Mean

fiui

-6 - 24 - 15

0 18 24 39 12

Zfiui = 48

-. The mean of the following distribution is 196.8. Find the value ofp.

From the table, n = 2fi = 100 and Zfiui = 48 Using the formula :

h ~ e a n ( X ) = a+- Zf i

xi - 18 u.=---- 4

- 3 - 2. - 1

0 1 2 3 4

d i=xi -18

- 12 - 8 - 4

0 4 8 .

12 16

r

class-marh

xi

6 10 14 18 22 26 30 34

-

Classes

4 - 8 8 -12

12 - 16 16 - 20 20 - 24 24 - 28 28 - 32 32 - 36

Total

160-240 1 240-320 1 320-400

Number of students

V) 2 l2 15 25 18 12 13 3

n=Zfif i-100

Classes I 0-80 1 80 - 160 24 44 P 35 Frequency 22

Mean

Class-mark (x,) 40

120 200 280 360

Solution. Calculation of

fFx, 880

4200 8800 280p

8640 Zfx, = 22520 + 280p

Classes 0 -80

80 - 160 160 - 240 240 - 320 320 - 400

Total

Frequency (f,) 22 35 . 44 P

24 n = Z f = 1 2 5 + p

38 U-Like Mathematics-X

From the table, n = Zfi = 125 + y, Z f x i = 22520 +280p . . Using the formula :

* 196.8 (given) = 22520+280p 125 + p

a p .= 25 Hence, the value o f p is 25.

' 28. Find .themedian of the' following distribution :

Marks 0 - 100

100 - 200

Frequency 2 5

1 . 200 - 300 300 - 400 400 - 500 500 - 600 600 - 700 700 - 800 800 - 900 900 - 1000 -

9 12 17 20 15 9 7 4

. .

Solution. Calculation of Median

Marks

0 - 100 '. 100 -200

200 - 300 300 - 400 400 - 500 500 - 600 600 - 700 700 - 800 800 - 900 900 - 1000

lbtal

Frequency cf, .

2 5 9

12 17 20 cf, 15 9 ' ~

7 4 .

n=Xfi=lOO

Cumulative frequency 8

(cf, 2 .

. 7: "~ .

16 28' , ,

45 (cfl 65 80 89 96

100


Now, 500 - 600 is the class whose cumulative frequency is 65 is greater than = 50. 7

From the table, f = 20, cf = 45, h = 100 Using the formula :

' n - - cf Median = 1 + - x h

f

Question numbers 29 to 34 carry 4 marks each.

29. If two zeroes of the polynomial z4 + 3r3 - 20.2 - 6.x + 36 are & and - 4, find the other zeroes of the polynomial.

Solution. Since two zeroes are & and - & , therefore (x - & )(x + 4) = x2 - 2 is a factor of the given polynomial.

Now, we divide the given polynomial by x2 - 2.

x2+3x-18 x2-2 x 4 + 3 x 3 - 2 0 x 2 - 6 ~ ~ 3 6 I First term of quotient is 7 = x

x4 - 2x2 X - + 3x3 - 18x2 - 6x + 36 Second term of quotient is 3x3- - 6x

I - 18x2 Third term of quotient is - = - 18

x2 I 0

So, x4 + 3x3 - 20x2 - 6x + 36 = (x2 - 2)(x2 + 3x - 18)

= (X - &)(x + &)[x2 + 6x - 3x - 181

= (x - &)(x + &)[x(x + 6) - 3(x + 611 =(x-&)(x+ &)(x+6)(x-3)

So, the zeroes of x2 + 3x - 18 = (x + 6)(x - 3) are given by x = - 6 and x = 3. Hence, the other zeroes of the given polynomial are - 6 and 3. 30. Prove that in a right triangle, the square of the hypotenuse is equal to the

sum of the squares of the other two sides. Solution. Given :A righ't triangle ABC, right angled a t B. To prove : ( ~ ~ ~ o t e n u s e ) ~ = ( ~ a s e ) ~ + (~erpendiculad~

40 U-Like Mathematics-X

= A B ~ + B C ~ . L.e., Construction : Draw BD I AC proof: AADB - AABC. [If a perpendicular is drawn from the vertex of the .A

right angle of a triangle to the hypotenuse then triangles on both sides of thepeaendicular are similar to the whole A D . C triangle and to each other.]

so, A D A B - = -

AB AC [Sides are proportional]

a AD.AC = AB2 ... (1) Also, M D C 7 AABC [Same reasoning as above]

CD BC - = - .~~

so, BC AC

[Sides are proportional]

CD.AC=BC~ ... (2) Adding (1) and (2), we have ADAC.+ CD.AC = AB? + BC' - (AD + CD)AC = AB2 + B c 2

ACAC = AB2 + BC2 * Hence, AC? = -AB2 + B C ~

Or Prove that the ratio of areas of two similar triangles is equal to the square of

their corresponding sides., Solution. Given : AABC and APQR such that AABC - APQR.

To prove : ar(AABC) AB2 BC' C A ~ =-=-=- ar(APQR) PQ' Q R ~ p2

Construction : Draw AD I BC and PS I QR.

, . ' . 'A.A' B D c Q s R

J X B C X A D Proof: ar(AABC) - - 2

ar (APQR) 2 QR ps 2

a r (IIABC) - BC x AD - ar (APQR) QR x PS

Now, in M B and APSQ, we have LB = LQ

1 [Area of A = -(base) 2 x height]

... (1)

[As AABC - APQRI


LADB = LPSQ' [Each = 9Oo1 3rd B A D = 3rd LQPS

Thus, AADB and APSQ are equiangular and hence, they are similar. AD AB Consequently - = - ... (2) PS PQ

[Jf A's are similar, the ratio of their corresponding sides is same]

But AB BC' PQ - QR

Now, from (1) and (31, we get

ar(AABC) BC BC ar (APQR) = R &R - ar (AABC) -- B C ~ - ar(APQR) QR'

As AABC - APQR, therefore

Hence, ar f u c ) . A B ~ B C ~ C A ~ =-=-=- ar(APQR) P Q ~ QR2 Rp2

31. Ifx-tanA+sinAandy=tanA-sinA,showthat

2 - y 2 = 4 f i Solution. Given, tan A + sin A = x and tanA-sinA = y Adding (1) and (2), we get

2 t a n A = x + y Subtracting (2) from (I), we get

2 s i n A = x - y Multiplying (3) and (4), we get

(x + y)(x - y) = (2 tanA)(2 sin A) ==2. ~ ~ - ~ ~ = 4 t a n A s i n A

Now, 4& = 4,/(tan A + sin A)(tan A - sin A)

[..' AABC - APQR]

... (3) [using (211

[From (4) and (511

... (5)

[using (1) and (2)l

sin A.sin A = 4

cos A

= 4 tanA sinA From (5) and (61, we obtain

~ ~ - ~ ~ = 4 t a n A s i n A = 4 & '

* x2 -y2=4&

Or Evaluate :

2 2 5 -cosec2 58" --cot 58" tan 32" - -tan 13" tan 37" tan 45" tan 53" tan 77" , 3 3 3

Solution. We have 2 2 5 -cosec2 58' --cot 58" tan 32" - -tan 13" tan 37" tan 45" t an 53" tan 77" 3 3 3

2 = zcosecZ. 58" --cot 58" tan (90" - 58") 3 3

5 --tan 13" tan 37" tan 45" tan (90" - 37") tan (90" - 13") 3

2 2 5 = -cosecz 58" --cot 58" cot 58" - -tan 13" tan 37" tan 45" cot 37" cot 13"

3 3 3 [.; tan (90" - 0) = cot 01

2 2 5 = -cosec2 58" - -cot2 58" --(tan 13" cot 13") tan 45" (tan 37" cot 37")

3 3 3 2 2 5

= - ( I+ cot2 58") - - cot2 58" - -(l) tan 45: (1) [.; cosec2 0 = 1 + cot2 0 and tan @cot 0 = 11 3 3 3 2 2 . 2 5

= - + - cotz 58" - - cotZ 58" --(I) 3 3 3 3

[': tan 45" = 11


32. Prove that : sin 8 (1 + tan 8) + cos 8 (1 + cot 8) = sec 8 + cosec 8.

Solution. We have L.H.S. = sin 8 (1 +tan 0) + cos 0 (1 + cot 0)

= (sin 8 + sin 0 tan 0) +'(cos 0 + cos 8 cot 0)

sin 8 +sinecos 8

= (sin 0 + cos 0) + (sin2 8 cos2 01 (cos8+sine)

= (sin 8 + cos 8) + /sin3 e + cos3 e\ ( sinecose ) (sin 0 + cos 0)(sin2 8 + cos2 8 -sin 8 cos 0)

= (sin e + cos 0) + sin 0 cos 0

[.: a3 +.b3 = (a + b)(a2 + b2 - ab)]

sin2 e+cos2 8-siGecos8 = (sin 0 + cos 8)

sin 8 cos 0 I 1-sinecos0

= (sin 8 + cos 8) sin 8 cos 8 I

sinecose + 1-sinOcos0 = (sin 0 + cos 8)

sin 8 cos 8 I - - sin 6 + cos 8

sin 0 cos 8 (1)

- - sin 8 cos 0 + sin 0 cos 8 sin 8 cos 8

- 1 1 =-+- cos 8 sin 8 sec 0 + cosec 0

= R.H.S. 33. The coach of a cricket team buys 3 bats and 6 balls for f 3900. Later, he buys

another .bat and 3 more balls of the same kind for 4 1300. Represent this situation algebraically and geometrically.

Solution. Let the price of a bat be f x and that of a ball be 7 y . It is given that the coach buys 3 bats and 6 balls for 7 3900. . . 3x + 6y = 3900 Also, it is given that the coach buys another bat and 3 more balls for f 1300.

The algebraic representation of the given conditions is 3x + 6y = 3900 ... (1)

x + 3y = 1300 ... (2) Geometrical representation : Let us draw the graphs of the'ecpations (1) and (2). For

this, we find two solutions of each of the equations which are given in tables.

3x + 6y = 3900 x + 3y = 1300

Plot the points A(O,650), B(1300, O ) , C(100,400) and B(1300,O) on graph paper and join the points to form the lines AB and BC as shown in figure.

The two lines (1) and (2) intersect at the point B(1300,O). So, x = 1300, y = 0 is the required solution of the pair of linear equations.

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34. The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table :

Length (in mm) 118 - 126 127 - 135 136 - 144 145 - 153 154 - 162 163 - 171 172 - 180

Number of Zeaues 3 5 9

12 5 4 2

Length (in mm) -


Draw a more than type ogive for the given data. Hence, obtain the median length of the leaves from the graph and verify the result by using the formula.

Solution. The given frequency distribution is not continuous. So, we &st make it continuous and prepare the cumulative frequency table by more than type given below :

Length (W 1111111

117.5 - 126.5 126.5 - 135.5 135.5 - 144.5 144.5 - 153.5 153.5 - 162.5 162.5 - 171.5 171.5 - 180.5

Other than the given class-interval, we assume that the class-interval 180.5 - 189.5 with zero frequency. Now, we mark the lower class limits on x-axis and the cumulative frequencies along y-axis on suitable scales to plot the points (117.5,40), (126.5,37), (135.5,32), (144.5, 231, (153.5, 111, (162.5, 61, (171.5, 2). We join these points with a smooth curve to get the "more than" ogive as shown in the figure.

n 40 Locate - = - = 20 on the y-axis (see figure). 2 2

Number of leaves [Frequency 0 1

3 5 9

12 5 4 2

Length more tlian 117.5 126.5 135.5 144.5 153.5 162.5 171.5

Cumulative frequency (cf)

40 37 32 23 11 6 2 .

. From this point, draw a line parallel to the x-axis cutting the curve at a point. From this point, draw a perpendicular to the x-axis. The point of intersection of this perpendicular with the x-axis determine the median length of the data (see figure), i.e., median length is 146.75 mm.

To calculate the median length, we need the frequency distribution table with the given cumulative frequency.

Table

Class interval 117.5 - 126.5 126.5 - 135.5 135.5 - 144.5 144.5 - 153.5

-~

153.5 - 162.5 162.5 - 171.5 171.5 - 180.5

Now, 144.5 - 153.5 is the class whose cumulative frequency is 29 is greater than

-=- = 40 20 .2 2

:. 144.5 - 153.5 is the median class. From the table : f = 12, cf = 17, h = 9 Using the formula :

(; - cf) Median = 1 +

f x h

= 144.5 + (20 - 17) 12

27 - ='144.5 + -

12 = 144.50 + 2.25 = 146.75 mm

So, about half the leaves have length less than 146.75 mm and the other half have length more than 146.75 mm.

-

Frequency (f) 3 5 9

12 5 4 2

Cumulative frequency (cf) 3 8

17 29 34 38 40

first term (sa-i) mathematics class x i...u-like cce sample question paper 1 27 solution. we divide...

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