MATHS MODEL TEST PAPER FOR SUMMATIVE ASSESSMENT – 3

7/29/2019 MATHS MODEL TEST PAPER FOR SUMMATIVE ASSESSMENT 3

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Model Test Paper-6 (Term-I) 1

Model Test Paper - 3 (Solved)

[For Summative Assessment-1 (Term - I)]

Time : 3 hours - 123 hours M.M. : 80

General Instructions : Same as in CBSE Sample Question Paper.

SECTION A

(Question numbers 1 to 10 are of 1 mark each.)

1. The HCF of 52 and91 is :

(a) 13 (b) 17 (c) 19 (d) 23

Sol. (a) By Euclids division lemma, we have91 = 52 1 + 39, 52= 39 1 + 13, 39 = 13 3 + 0

HCF of 52 and 91 = 13

2. The graph ofy = 3x 1 intersects :

(a) y-axis at1

3(b) x-axis at

1

3(c) x-axis at 3 (d) y-axis at 3

Sol. (b) As, zero of 3x 1 is1

3, it means the graph of the line y = 3x 1 meets thex-axis

at x =1

3.

3. If a system of simultaneous linear equations has infinitely many solutions, then the

given equations are :(a) consistent (b) inconsistent (c) dependent (d) both (a) and (c)

Sol. (d) If the system of simultaneous linear equations has infinitely many solutions, it

means they are dependent and consistent.

4. IfABCDEF such that their areas are 64 cm2 and144 cm2respectively and AB

= 4 cm, then the length of DE is :

(a) 12 cm (b) 8 cm (c) 6 cm (d) none of these

Sol. (c) Since ABC ~ DEF

( )

( )

Area ABC

Area DEF=

2

2

AB

DE

64

144=

2

2

4

DE

DE2 =144 16

64= 36 DE = 6 cm

5. Which of the following is not true?

(a) sin2 + cos2 = 1 (b) sin2 = 1 cos2(c) cos2 = 1 sin2 (d) sin2 cos2 = 1

Sol. (d) sin2 cos2 = 1 is not an identity, therefore it is not true.


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2 Sample Papers in Mathematics-X (Term -I)

6. If tanA = 1, then 22

1

tan A

tan A is :

(a) 0 (b) 2 (c) 1 (d) not defined

Sol. (d) We have, tanA = 1

Then,2

2 tanA

1 tan A=

( )2

2 1 2=

01 1, which is not defined.

7. sin 45 + cos 45 is equal to :

(a)1

2(b) 2 (c)

1

2(d) none of these

Sol. (b) We have, sin 45 + cos 45 =1 1 2

+ =2 2 2

= 2

8. The correct relation between mean, median and mode is :

(a) mode = 3 median 2 mean (b) mean = 3 mode 2 median(c) median = 2 mode 3 mean (d) none of these

Sol. (a) The correct relation between mean, median and mode is :

mode = 3 median 2 mean

9. In ABC, AB = 6 3 cm, AC = 12 cm and BC = 6 cm. The angle B is :

(a) 90 (b) 60 (c) 45 (d) 120

Sol. (a) We have AC2 = 144, AB2 = 108, BC2 = 36

AB2 + BC2 = 108 + 36 = 144 = AC2

B = 90 [By converse of Pythagoras theorem]

10. In ABC, DE || BC and2

=3

AD

ABand AE= 12 cm,

then value of EC is :

(a) 8 cm (b) 6 cm

(c) 4 cm (d) none of these

Sol. (b) In ABC, DE || BC. Then, by basic proportionality

theorem, we have,AD

AB=

AE

AC

2

3=

12

AC AC =

12 3

2= 18 cm

Hence, EC = AC AE = (18 12) cm = 6 cm

SECTION B

(Question numbers 11 to 18 carry 2 marks each.)

11. Given that HCF(306, 657) = 9, find LCM (306, 657)

Sol. We know that for any two numbers HCF LCM = their product


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9 LCM = 306 657 LCM =306 657

9= 22338

12. Divide the polynomial f(x) = 14x3 5x2 + 9x 1 by the polynomialg(x) = 2x 1.Also

find the quotient and remainder.

Sol. We have,

Clearly, quotient q(x) = 7x2 + x + 5

and remainder r(x) = 4

OR

If the sum of the zeroes of the polynomial 2x2 kx +9

2is twice the product of zeroes,

find the value of k.

Sol. We have product of the zeroes =9 9

2 4

2

=

Sum of the zeroes =2

k

We have,9

2 92 4

kk= =

13. For what value of k will the system of equations x + 2y + 7 = 0

and 2x + ky + 14 = 0 represent coincident lines?

Sol. The given system of equations will represent coincident lines if they have infinitely

many solutions. The condition for which is

1

2

a

a=

1 1

2 2

=b c

b c

1

2=

2 7=

14k k = 4

14. In the figure, findF.

Sol. In triangles ABC and DEF, we have

AB BC CA 1= = =

DF FE ED 2

Therefore, by SSS-criterion of

similarity, we have


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ABC DFE

A = D, B = F and C = E

D = 80, F = 60

Hence, F = 60.

15. Prove that sin 35 sin 55 cos 35 cos 55 = 0

Sol. LHS = sin 35 sin 55 cos 35 cos 55

= sin (90 55) sin (90 35) cos 35 cos 55

= cos 55 cos 35 cos 35 cos 55 = 0 = RHS

16. In a ABC right angled at A, if AB = 5 cm,AC= 12 cm, and BC= 13 cm, find sinB

and cosC.

Sol. With reference to B, we have

Base, AB = 5 cm, perpendicular AC = 12 cm and hypotenuse, BC = 13 cm

sinB =AC 12

=BC 13

With reference to C, we have

Base AC = 12 cm, perpendicular AB = 5 cm and

hypotenuse BC = 13 cm

cosC =AC 12

=BC 13

.

17. Calculate the mean for the following distribution :

Variable 5 6 7 8 9Frequency 4 8 14 11 3

Sol. To calculate the mean we first prepare the following table

xi

fi

fix

i

5 4 20

6 8 48

7 14 98

8 11 88

9 3 27

Total fi

= 40 fix

i= 281

Mean =281

=40

i i

i

f x

f= 7.025


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18. For the following distribution, write the modal class :

Marks Below 10 Below 20 Below 30 Below 40 Below 50 Below 60

No. of students 3 12 27 57 75 80

Sol. We first convert the given frequency distribution into a simple frequency distribution.

Marks 010 1020 2030 3040 4050 5060

No. of students 3 9 15 30 18 5

Here the class 3040 has maximum frequency, so modal class is 3040.

SECTION C

(Question numbers 19 to 28 carry 3 marks each)

19. Show that 2 3 is irrational.

Sol. If possible, let 2 3 be rational.

Let its simplest form be 2 3 =a

b, where a and b are positive integers having no

common factor other than 1. Then,

2 3 = 3 =2

a a

b b (i)

Since, a and 2b are non-zero integers, so

2

a

b

is rational.

Thus, from (i), it follows that 3 is rational.

This contradicts the fact that 3 is irrational.

This contradiction arises by assuming that 2 3 is rational.

Hence, 2 3 is irrational. Proved.

OR

Use Euclids division algorithm to find the HCF of 441, 567, 693.

Sol. First we find the HCF of 567 and 693

693 = 1 567 + 126567 = 4 126 + 63

126 = 2 63 + 0

HCF of 567 and 693 = 63

Now, 441 = 7 63

HCF of 63 and 441 is 63

Hence, HCF of 441, 567 and 693 is 63.


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20. Five years hence, fathers age will be three times the age of his son. Five years ago,

father was seven times as old as his son. Find their present ages.

Sol.Let the present age of father be x years and the present age of son be y years.Five years hence, fathers age = (x + 5) years

Five years hence, sons age = (y + 5) years

According to condition, x + 5 = 3(y + 5) = x 3y 10 = 0 (i)

Five years ago, fathers age = (x 5) years

Five years ago, sons age = (y 5) years

According to condition, (x 5) = 7 (y 5) x 7y + 30 = 0 (ii)Subtracting equation (ii) from equation (i), we get, 4y 40 = 0 y = 10Putting y = 10 in equation (i), we get, x 30 10 = 0 x = 40Hence, present age of father is 40 years and present age of son is 10 years.

21. If , and are the zeroes of the quadratic polynomialf(x) = ax2 + bx + c, then

evaluate4 + 4.

Sol. Since and are the zeroes of the quadratic polynomial f(x) = ax2 + bx + c.

+ = b

aand =

c

a

Now, 4 + 4 = (2 + 2)2 222

4 + 4 = {( + )2 2}2 2()2

4

+ 4

=

22 2

2 2 + = , =

b c c b c

a a a a a

4 + 4 =

22 2

2 2

2 2

b ac c

a a=

( )2

2 2 2

4

2 2b ac a c

a

22. In the figure, ifACB = CDA,AC= 8 cm and AD = 3 cm, find BD.

Sol. In ACB and CDA, we haveACB = CDA [Given]BAC = CAD [Common]

ACB ~ ADC [AA similarity]

AC

AD

AB

AC=

8

3

AB

8= AB =

64

3

BD = AB AD =64

33

cm =55

cm3


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OR

Any point X inside DEF is joined to its vertices. From a point

P in DX, PQ is drawn parallel to DE meeting XE at Q and QRis drawn parallel to EF. Meeting XF in R. Prove that PR || DF.

Sol. Join PR

In XED, we have PQ || DE

XP XQ

=PD QE

(i) [BPT]

In XEF, we have QR || EF

XQ XR

QE RF= (ii) [BPT]

From (i) and (ii), we get,XP XR

PD RF=

Thus, in XFD, points P and R are dividing sides XD and XF in the same ratio.Therefore, by converse of BPT, we have PR || DF. Proved.

23. ABC is an isosceles triangle right-angled at B. Similar triangles ACD and ABE are

constructed on sides AC and AB. Find the ratio between the areas of ABE and

ACD.

Sol. Let AB = BC = x.

It is given that ABC is right-angled at B.

AC2 = AB2 + BC2

AC2 = x2 + x2 AC = 2x

It is given that, ABE ACD

( )

( )

2

2

Area ABE AB=

Area ACD AC

( )

2

2=

2

x

x

1=

2

24. ABC is a right triangle, right angled at C. IfA = 30 and AB = 40 units, find the

remaining two sides andB ofABC.

Sol. We have, A + B + C = 180

30 + B + 90 = 180 [ A = 30 and C = 90]B = 180 120 = 60

Now, cosA =AC

AB cos 30 =

AC

40

3

2=

AC

40 AC =

3 40

2 20 3 units


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and, sinA =BC

AB

sin 30 =BC

40

1

2=

BC

40

BC = 40 1

2= 20 units.

Hence, AC = 20 3 units, BC = 20 units and B = 60

25. Prove that : tan2A tan2B =

2 2 2 2

2 2 2 2

cos B cos A sin A sin B=

cos B cos A cos A cos B

Sol. LHS = tan2A tan2B

=2 2

2 2sin A sin Bcos A cos B

=2 2 2 2

2 2sin A cos B cos A sin B

cos A cos B

=( ) ( )2 2 2 2

2 2

1 cos A cos B cos A 1 cos B

cos A cos B

=

2 2 2 2 2 2

2 2

cos B cos A cos B cos A + cos A cos B

cos A cos B

=

2 2

2 2

cos B cos A

cos A cos B[It proves Ist part]

= ( ) ( )2 2

2 2

1 sin B 1 sin A

cos A cos B

=

2 2

2 2

sin A sin B

cos A cos B= RHS Proved.

26. If sec 5A = cosec (A 36), where 5A is an acute angle, find the value of A.

Sol. We have, sec 5A = cosec (A 36)

sec 5A = sec{90 (A 36)} sec 5A = sec (126 A) 5A = 126 A

6A = 126 A = 21

OR

Find the value of : cosec(65 + ) sec(25 ) tan(55 ) + cot(35 + )Sol. We have, cosec(65 + ) sec(25 ) tan(55 ) + cot(35 + )

= cosec{90 (25 )} sec(25 ) tan{(90 (35 + )} + cot(35 + )


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Model Test Paper-6 (Term -I) 9

= sec(25 )} sec(25 ) cot(35 + ) + cot(35 + )

[ cosec(90 ) = sec and tan (90 ) = cot]

= 0 0 = 0

27. The table below shows the daily expenditure on food of 25 household in a locality.

Daily Exp. (in Rs.) 100-150 150-200 200-250 250-300 300-350

No. of households 4 5 12 2 2

Find the mean daily expenditure on food by a suitable method.

Sol. Let assumed mean be 225.

Classes fi

xi

di= x

i 225 u

i=

50

idf

iu

i

100-150 4 125 100 2 8

150-200 5 175 50 1 5

200-250 12 225 0 0 0

250-300 2 275 50 1 2

300-350 2 325 100 2 4

fi

= 25 fiu

i= 7

Here, fi

= 25,fiu

i= 7, and h = 50

Now, x = A +7

= 225 + 5025

i i

i

f uh

f= 225 14 = 2111

Hence, the mean daily expenditure is 211.

28. Calculate the median from the following data :

Marks 0-10 10-20 20-30 30-40 40-50

No. of students 5 15 30 8 2

Sol. To calculate the median, we first prepare the cumulative frequency table :

Marks No. of students Cumulative frequency

(Frequency)

0-10 5 5

10-20 15 20

20-30 30 50

30-40 8 58

40-50 2 60

Here, N = 60 N/2 = 30

The cumulative frequency just greater than N/2 = 30 is 50 and the corresponding

class is 20-30. Hence, 20-30 is the median class.


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l = 20, f = 30, c = 20, h = 10

Now, Median =2

N

cl h

f

+

Median = 20 +30 20 10

10 = 20 + 23.330 3

=

SECTION D

(Question numbers 29 to 34 carry 4 marks each.)

29. Show thatn2 1 is divisible by 8, if n is an odd positive integer.

Sol. We know that any odd positive integer is of the form 4q + 1 or 4q + 3 for someinteger q.

So, we have the following cases :

Case I : When n = 4q + 1

In this case, we have : n2 1 = (4q + 1)2 1 = 16q2 + 8q + 1 1

= 16q2 + 8q = 8q (2q + 1) = 8 m

n2 1 is divisible by 8

Case II : When n = 4q + 3

In this case, we have : n2 1 = (4q + 3)2 1 = 16q2 + 24q + 9 1

= 16q2 + 24q + 8

= 8 (2q2 + 3q + 1) = 8 m

n2 1 is divisible by 8

Hence, n2 1 is divisible by 8. Proved.

30. Solve : x + y = a + b,ax by = a2 b2

Sol. The given system of equations may be written as

x + y (a + b) = 0

ax by (a2 b2) = 0

By cross-multiplication, we get

( ) ( ) ( ){ } ( ) ( ){ }

( )

2 2 2 2

11 1

yx

a b b a b a b a a b

b a

=

+ +

=

2 2 2 2 2 2 1

= = + + + +

yx

b aa b ab b a b a ab


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( ) ( ) ( ) 1

= = + + +

x y

a a b b a b a b

( )

( )( )

( )

+ += = and = =

+ +

a a b b a bx a y b

a b a b

Hence, the solution of the given system of equations is x = a, y = b

OR

A fraction becomes9

,11

if 2 is added to both the numerator and the denominator. If

3 is added to both the numerator and the denominator it becomes5

.6

Find the

fraction.

Sol. Let the numerator be x and denominator be y. Then, according to the question,

Case I. 2 92 11

x

y

+=

+ 11( 2) 9( 2)x y+ = +

11 22 9 18x y+ = + 11 9 4x y = (i)

Case II.3 5

3 6

x

y

+=

+ 6 ( 3) 5 ( 3)x y+ = +

6 18 5 15x y+ = + 6 5 3x y =

5 36yx = (ii)

Putting5 3

6

yx

= in (i), we get

5 311 9 4

6

yy

= (iii)

Multiplying (iii) by 6, we get 11(5 3) 54 24y y =

55 33 54 24y y =

24 33 9y = + =

Putting y = 9 in (i), we get 11 9 9 4x = 11 4 81 77x = + = x = 7

Hence, the required fraction is

7

.9

31. Find the zeroes of the polynomial p(x) = x3 5x2 2x + 24, if it is given that the

product of its two zeroes is 12.

Sol. Let , , be the zeroes ofp(x).

+ + = 5 (i)

+ + = 2 (ii)


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= 24 (iii)

Now, 12 + + = 2 [ = 12, given]

+ = 14 ( + ) = 14 (iv)

Also, 12 = 24 = 2

From (iv), + = 7 = 7 (v)

From (ii), (7 ) + ( + ) = 2 7 2 14 = 2

2 7 + 12 = 0 2 4 3 + 12 = 0 ( 4) 3( 4) = 0

( 4)( 3) = 0 = 4 or = 3

= 7 4 or = 7 3 = 3or = 4

Hence, zeroes the polynomial are 3, 4, 2.32. If two triangles are equiangular, prove that the ratio of the corresponding sides is

same as the ratio of the corresponding altitudes.

Sol. Given : Two triangles ABC and DEF in which

A = D, B = E, C = F and AL BC, DM EF

To ProveBC AL

=EF DM

Proof Since equiangular triangles are similar.

ABC ~ DEF

AB BC=DE EF ........... (i)

In triangle ALB and DME, we have

ALB = DME [Each equal to 90]B = E [Given]

ALB ~ DME [AA similarity]

AB AL

=DE DM

............. (ii)

From (i) and (ii), we getBC AL

=EF DM

33. If sec + tan = p, show that2

2

1

+ 1

p

p= sin

Sol. LHS =( )

( )

22

2 2

sec + tan 1 1=

+ 1 sec + tan + 1

p

p

=

2 2

2 2

sec + tan + 2sec tan 1

sec + tan + 2sec tan + 1


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=

( )

( )

2 2

2 2

sec 1 + tan + 2sec tan

sec + 2sec tan + 1 + tan

=

2 2

2 2

tan + tan + 2sec tan

sec + 2sec tan + sec

=

2

2

2tan + 2tan sec

2sec + 2sec tan

=

( )

( )

2tan tan sec

2sec sec + tan

+

=tan sin

=sec cos . sec

= sin = RHS Proved

OR

If tan + sin = m and tan sin = n, show that m2 n2 = 4 mn .

Sol. LHS = m2 n2

= (tan + sin )2 (tan sin )2

= (tan + sin + tan sin ) (tan + sin tan + sin )= 4 tan sin .

And, RHS = 4 mn = 4 (tan sin ) (tan sin ) + = 4 2tan sin2

= 4

22

2

sinsin

cos

= 4

2 2 2

2

sin sin cos

cos

= 4

2 2

2

sin (1 cos )

cos

4

2

sin4

cos

=

= 4

2sin sin

4sincos cos

=

= 4 sin .tan

Hence, LHS = RHS. Proved.

34. The following distribution gives the daily incomes of 50 workers of a factory.

Daily income (in Rs.) 100-120 120-140 140-160 160-180 180-200

Number of workers 12 14 8 6 10

Convert the above distribution to a less than type cumulative frequency distribution,

and draw its ogive.

Sol. We first prepare the cumulative frequency distribution table by less than method as

given below :

Daily income (in Rs.) No. of workers Income less than Cumulative frequency

100-120 12 120 12

120-140 14 140 26

140-160 8 160 34

160-180 6 180 40

180-200 10 200 50


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We plot the points (120, 12), (140, 26), (160, 34), (180, 40), (200, 50) and join these

points by a free hand curve to get the required ogive.

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MATHS MODEL TEST PAPER FOR SUMMATIVE ASSESSMENT – 3