MATHS MODEL TEST PAPER FOR SUMMATIVE ASSESSMENT– 1

7/29/2019 MATHS MODEL TEST PAPER FOR SUMMATIVE ASSESSMENT 1

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Model Test Paper-1 (Term-I) 1

Model Test Paper - 1 (Solved)

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

Time : 3 hours - 123 hours Maximum Marks : 80

General Instructions : Same as in in CBSE Sample Question Paper.

SECTION A

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

1. The decimal expansion of21

45is :

(a) terminating (b) non-terminating and repeating(c) non-terminating and non-repeating (d) none of these

Sol. (b)2

21 21 21

45 9 5 3 5= =

. Clearly, 45 is not of the form 2m 5n. So the decimal expansion

of21

45is non-terminating and repeating.

2. The zero of the polynomial px + q is :

(a) q (b)q

p(c)

p

q(d)

q

p

Sol. (b) Value ofpx + q is zero when x =q

p. Therefore, zero of the given polynomial is

qp

.

3. If the system of linear equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 are

inconsistent then :

(a)1 1

2 2

a b

a b (b)

1 1 1

2 2 2

a b c

a b c= =

(c)1 1 1

2 2 2

a b c

a b c= (d) none of these

Sol. (c) Condition for inconsistent equations is 1 1 1

2 2 2

a b c

a b c= .

4. The areas of two similar triangles are 169 cm2and121 cm2. If the longest side of thelarger triangle is 26 cm, then the longest side of the other triangle is :

(a) 12 cm (b) 14 cm (c) 19 cm (d) 22 cm

Sol. (d) Let x cm be the longest side of the other triangle.

Then, we have2

2

169 26 13 2622 cm.

121 11x

xx= = =


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

5. (1 cos2A) cosec2A is equal to :

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

Sol. (a) (1 cos2A) cosec2A = sin2A. cosec2A = 221sin A 1

sin A=

6. If cos =1

2, then the value of 2

2

1

sec

tan

+ is :

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

Sol. (a)2 2

2 sec 2sec 2

sec1 tan sec

= =

+

= 2 cos = 2 1

2= 1

7. The value of sin 39 cos 51 is :

(a) 0 (b) 1 (c) 1 (d) none of these

Sol. (a) sin 39 cos 51 = sin 39 cos (90 39) = sin 39 sin 39 = 08. The graphical representation of a cumulative frequency distribution is called :

(a) bar graph (b) histogram (c) ogive (d) frequency polygon

Sol. (c) The graphical representation of a cumulative frequency distribution is called ogive

or cumulative frequency curve.

9. In DEF, if DE2 = DF2 + EF2, then :

(a) E = 90 (b) F = 90 (c) D = 90 (d) none of these

Sol. (b) By converse of Pythagoras theorem, we get F i.e., angle opposite to side DE is

a right angle.

10. In the figure, line XY divides the ABC into two parts

of equal areas such that XY || AC, then

BX

AB is :

(a) 1: 2 (b) 2 : 1

(c) 2 :1 (d) 1 : 2

Sol. (d) Since, XY divides ABC into two parts of equal

areas.

ar ( BXY) =1

2ar ( ABC)

And, ABC ~ XBY

2 2

2 2

ar ( XBY) BX 1 BX BX 1=

ar ( ABC) 2 ABAB AB 2

= =

= 1: 2

SECTION B

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

11. Write whether the square of any positive integer can be of the form 3m + 2, where m

is a natural number. Justify your answer.

Sol. Any positive integer can be written as 3q, 3q + 1 or 3q + 2, for some integer q.


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Model Test Paper-1 (Term -I) 3

Therefore, square will be 9q2 = 3m, 9q2 + 1 + 6q = 3(3q2 + 2q) + 1 = 3m + 1,

9q2 + 12q + 4 = 3(3q2 + 4q + 1) + 1 = 3m + 1.

Hence, square of any positive integer cannot be of the form 3m + 2.

12. Find the quadratic polynomial, the sum and product of whose zeroes are 5 and 6.

Sol. Let and be the zeroes of the required polynomial p(x).

Then, + = 5 and = 6

p(x)= x2 ( + ) x + = x2 (5) x + 6 = x2 + 5x + 6

Hence, the required polynomial is f(x) = x2 + 5x + 6.

13. Find the values of and for which the following system of linear equations hasinfinite number of solutions : 2x + 3y = 7, 2x + ( + )y = 28

Sol. The given system of equations will have infinite number of solutions, if

2

2

3 7

28 =

+=

1 3 1

4 =

+=

= =1 1

4

3 1

4 and + = 4 and + = 12

= 4 and = 8 Ans.

OR

Solve for x and y :1

2

11

1 1

28

x y x y = + =,

Sol. Let1

x= A and

1

y= B. The given equations become

AB

21 = or A 2B = 2 (i)

And, A +B

2= 8 or 2A + B = 16 (ii)

Multiplying (ii) by 2 and adding to (i), we get

A 2B = 2

4A + 2B = 32

5A = 30 A = 6

Putting A = 6 in (ii), we get, 12 + B = 16 B = 4

Now,1

x= A x =

1

6and

1

y= B y =

1

4

Hence, x =1

6and y=

1

4.

14. In ABC, D and E are points on the sides AB and AC respectively. If AB = 12 cm,AD = 8 cm, AE = 12 cm and AC = 18 cm, then show that DE || BC.

Sol. We have, AB = 12 cm, AD = 8 cm, AE = 12 cm and AC = 18 cm


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AD 8 2 AE 12 2and

AB 12 3 AC 18 3 = = = =

AD AEAB AC

=

Hence, by converse of Thales theorem, DE || BC. Proved.

15. Prove that cos4A cos2A= sin4A sin2A

Sol. LHS = cos4A cos2A = cos2A (cos2A 1) = cos2A (1 cos2A)

= cos2A sin2A = sin2A (1 sin2A) = sin2A + sin4A

= sin4A sin2A = RHS Proved.

16. In a right triangle ABC, right angled at C if tan A = 1, then verify that 2sinA

cosA = 1

Sol. In ABC, we have, tan A = 1 BC

1AC

= BC = x and AC = x

By Pythagoras theorem, we have, AB2 = AC2 + BC2

AB2 = x2 + x2 AB = 2x

BC 1 AC 1

sin A = and cos A =AB AB2 2 2 2

x x

x x= = = =

2 sin A cos A = 2 1 1

12 2

= Verified.

17. Find the sum of the deviations of the variate values 3, 4, 6, 7, 8, 14 from their mean.

Sol. Variates are : 3, 4, 6, 7, 8, 14

Mean, 3 4 6 7 8 14 42 76 6

x+ + + + +

= = =

Sum of deviations from x = 7 is :

(3 7) + (4 7) + (6 7) + (7 7) + (8 7) + (14 7) = 4 3 1 + 0 + 1 + 7 = 0

18. Find the value of x, if the mode of the following data is 25.

15, 20, 25, 18, 14, 15, 25, 15, 18, 16, 20, 25, 20, x, 18

Sol. The frequency table of the given data is as given below :

Value (xi) : 14 15 16 18 20 25 x

Frequency (fi) : 1 3 1 3 3 3 1

It is given that the mode of the given data is 25. So, it must have the maximum

frequency. That is possible only when x = 25.Hence, x = 25

SECTION C


19. Prove that if x and y are odd positive integers, thenx2 + y2 is even but not divisible

by 4.

Sol. We know that, any odd positive integer is of the form 2q + 1 for some integer q.


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So, let x = 2m + 1 and y = 2n + 1 for some integers m and n.

x2 + y2 = (2m + 1)2 + (2n + 1)2

x2 + y2 = 4 (m2 + n2) + 4(m + n)+ 2

x2 + y2 = 4 {(m2 + n2) + (m + n)}+ 2

x2 + y2 = 4q + 2, where q = (m2 + n2) + (m + n)

x2 + y2 is even and leaves remainder 2 when divided by 4

x2 + y2 is even but not divisible by 4 Proved.

OR

Show that 2 3 is irrational.

Sol. If possible let 2 3 be rational. Let the simplest form of 2 3 bea

b, where a and b

are positive integers having no common factor other than 1. Then

2 3 32

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. The contradiction arises by assuming

that 2 3 is rational.

Hence, 2 3 is irrational. Proved

20. Verify that 3, 1 and 1

3are the zeroes of the cubic polynomial

p(x) = 3x3 5x2 11x 3 and then verify the relationship between the zeroes and itscoefficients.

Sol. We have, p(x) = 3x3 5x2 11x 3

p(3) = 3 33 5 32 11 3 3 = 81 45 33 3 = 0p(1) = 3 (1)3 5 (1)2 11 (1) 3 = 3 5 + 11 3 = 0

3 21 1 1 1 1 5 11

3 5 11 3 3 03 3 3 3 9 9 3

p = = + =

So, 3, 1 and 1

3are the zeroes of polynomial p(x).

Varification :

Let = 3, = 1 and = 1

3. Then,

+ + = 3 1 1

3=

2

3

5 5 Coeff. of

3 3 Coeff. of

x

x

= =

++ = 3 (1) + (1) 1 1

33 3

+

= 3 + 31 11 Coeff. of

13 3 Coeff. of

x

x= =


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= 3 (1)

= =

= 1

31

3

3

Constant term

Coeff. of3

x

21. Solve :2

x

+2

3

1

6y

= ,3

x+ =2

0

y and hence find a for which y = ax 4.

Sol. Taking1

xu v= =and

1

y. The given system of equations becomes

22

3

1

6

u v+ = 12u + 4v = 1 (i)

And, 3u + 2v = 0 (ii)

Multiplying equation (ii) by 2 and subtracting from equation (i), we get

6u = 1 u =1

6

Putting u =1

6

in (i), we get, 2 + 4v = 1 v = 1

4

Hence, x =1

61

4u

yv

= = =and

So, the solution of the given system of equations is x = 6, y = 4Putting x = 6, y = 4 in y = ax 4, we get, 4 = 6a 4 a = 0

22. In the given figure, if AB || CD, find the value of x.

Sol. Since, the diagonals of a trapezium divide each otherproportionally,

AO

OC=

BO

OD

3 1

5 3

2 1

6 5

x

x

x

x

=

+

(3x 1) (6x 5) = (2x+ 1) (5x 3) 18x2 15x 6x+ 5 = 10x2 6x+ 5x 3

18x2 21x+ 5 = 10x2 x 3 8x2 20x+ 8 = 0

2x2 5x+ 2 = 0 2x2 4xx+ 2 = 0

2x(x 2) 1(x 2) = 0 (x 2)(2x 1) = 0 x= 2 or x=1

2

x=1

2is rejected because forx=

1

2, OD becomes ve.

Hence,x= 2.OR

If A be the area of a right triangle and a one of the

sides containing the right angle, prove that the length

of the altitude on the hypotenuse is

2

44 2

Aa

a A+ .

Sol. In the figure, LMN is a right triangle right angled

at M and MP LN. Also MN = a.

Area ofLMN = A =1

2 a LM LM =

2A

a(i)


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OR

Prove thatcos

sin

sin

cossec

A

A

A

AA

1

12

++

+=

Sol. LHS = +cos

cos

A

1 + sinA

1 + sinA

A =+ +( )

cos ( sin )

cos

2 21A A

A 1+ sinA

=cos2 A + 1 + sin A + 2sinA

cosA (1 + sinA)

2

=+ +1 1 2 sin A

cos A (1 + sin A)

=+

= =2 1( sin )

cos

A

A(1 + sinA)

2

cosA2secA = RHS. Proved.

25. Prove thatsin cos cos

sin

cos sin sin

cos

(90 (90)

( )

)

( )90 90 +

= 11

Sol. LHS =sin cos ( ) cos

sin ( )

cos sin ( ) sin

cos ( )

90

90

90

90

+

=sin sin cos

cos

cos

cos sin

sin+ = sin2 + cos2 = 1 = RHS. Proved.

26. If tan =3

4, find the value of

1

1

+

cos

cos

Sol. We have, tan =3

4

sec = 1 2+ tan = 13

4

19

16

25

16

5

4

2

+

= + = =

Now, cos =1

seccos

=

45

+=

+

= =1

1

14

5

14

5

1

5

9

5

1

9

cos

cos

27. Find the mean of the following distribution :

x : 4 6 9 10 15

f : 5 10 10 7 8

Sol. To compute the mean we prepare the following table.

xi

fi

fix

i

4 5 206 10 60

9 10 90

10 7 70

15 8 120

Total fi

= 40 fix

i= 360


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Model Test Paper-1 (Term-I) 9

Mean x =360

940

i i

i

f x

f

= =

Thus, the mean of the given distribution is 9.28. Draw the less than ogive for the following frequency distribution :

Marks : 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60

Number ofstudents :

7 10 23 51 6 3

Sol. We first prepare the cumulative frequency table as given below :

Marks No. of students Marks less than Cumulative Frequency

0 - 10 7 10 7

10 - 20 10 20 17

20 - 30 23 30 40

30 - 40 51 40 91

40 - 50 6 50 97

50 - 60 3 60 100

Now, we plot the points (10, 7),

(20, 17), (30, 40), (40, 91),

(50, 97) and (60, 100) on a graph

paper.

Then, we join the plotted points

by a free hand curve to obtain the

required ogive as shown in figure.


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SECTION D


29. Given that x 5 is a factor of the cubic polynomial x3 3 5 x2 + 13x 3 5 ,find

all the zeroes of the polynomial.

Sol. p(x) = x 5 is a factor of the polynomial q(x) = x3 3 5x2 + 13x 3 5 , therefore,

we divide q(x) by p(x) to get other zeroes of q(x).

q(x) 3 2 23 5 13 3 5 ( 5)( 2 5 3)x x x x x x= + = +

( ) ( ) ( )25 5 2 5 2 3x x x x = + +

( ) ( ){ } ( ) ( ){ }5 5 2 5 2 5 2x x x x

= + +

( ) ( ){ } ( ){ }5 5 2 5 2x x x= +

( ) ( )( ) 0 5, 5 2 , 5 2q x x= = +

Hence, zeroes of the polynomial are ( ) ( )5, 5 2 and 5 2+ .

30. Let a, b, c, d, be positive rationals such that a b+ = c d+ , then show that either

(a = c) and (b = d) or b and d are squares of rationals.

Sol. Ifa = c, then, a b+ = c d+ b d b d = = .

So, let a c . Then, there exists a positive rational number x, such that a = c + x

Now, a b+ = c d+ c x b+ + = c d+ [ ]a c x= +

x b+ = d (i)

( )2

x b+ = ( )2

d

x2 + 2 bx + b = d d x2 b = 2x b b =2

2

d x b

x


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b is rational [d, x, b are rationals, d x b

x

2 2

2is rational]

b is the square of a rational number

From (i), we have, d = x+ b d is rational [ b is rational]

d is the square of a rational number.

Hence, either a = c and b = d or b and d are the squares of rationals. Proved.

31. In a two digit number, the tens digit is three times the units digit. When the number

is decreased by 54, the digits are reversed. Find the number.

Sol. Let the digit in the units place be x and the digit in the tens place be y.

Then, number = 10y + x

According to the given condition, we have

y = 3x (i)

Number obtained by reversing the digits = 10x + y

If the number is decreased by 54, the digits are reversed.

10y + x 54 = 10x + y 9x 9y = 54 x y = 6 (ii)

Putting y = 3x in equation (ii), we get, x 3x = 6 x = 3

Putting x = 3 in y = 3x, we get, y = 9

Hence, number = 10y + x = 10 9 + 3 = 93

OR

Solve :x

a

y

ba b

x

a

y

b+ = + + =,

2 22

Sol. The given system of equations may be written as

1 10

1 12 0

2 2

a

x

b

y a b

a

x

b

y. . . .+ +( ) = + =and

By cross-multiplication, we have

+ +

=

+

=

x

b

a

b b

y

a

b

a a ab a b

2 1 1 2

1

1 12 2 2

=

=

x

a

b b

y

a

b

a ab a b2 2 2 2 21 1

1

1 1

=

=

x

a b

b

y

a b

a

a b

a b2 2 2 2

1

=

= =

=xa b

b a b

a b

a ya b

a a b

a b

b2

2 2

2

2

2 2

21 1and

x = a2, y = b2


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MATHS MODEL TEST PAPER FOR SUMMATIVE ASSESSMENT– 1

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