maths model test paper for summative assessment – 2

7/29/2019 MATHS MODEL TEST PAPER FOR SUMMATIVE ASSESSMENT 2

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Model Test Paper-4 (Term -I) 1

Model Test Paper - 2 (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. If the HCF of55 and 22 is expressed in the form 55m 22 2, then the value of m

is :

(a) 1 (b) 1 (c) 2 (d) 2

Sol. (a) By Euclids division lemma, we have,

55 = 22 2 + 11 and 22= 11 2 + 0

HCF of 55 and 22 is 11

11 = 55 1 22 2 m = 1

2. If one zero of the polynomialf(x) = 5x2 + 13x + k is the reciprocal of the other, then

the value of k is :

(a) 0 (b) 5 (c)1

6(d) 6

Sol. (b) Let the zeroes be and1

. Then

Product of roots 1 =5k .

5k = 1 k = 5

3. If a pair of equations is consistant, it means, the graphs of these equations are :

(a) parallel (b) coincident

(c) intersecting (d) either intersecting or coincident

Sol. (d) The equations are consistent means, they have solutions.

So, their graphs may be either intersecting or coincident

4. In the figure, D, E, F are the mid-points of sides AB, AC

and BC respectively, then( )

( )

Area ABC

Area DEFis :

(a) 3 : 1 (b) 4 : 1(c) 5 : 2 (d) 3 : 2

Sol. (b) By AA similarity we can prove that ABC ~ DEF

Now,( )

( )

area ABC

area DEF

=

2

2

BC

DE=

( )[ ]

2

2

2DE

DEBC = 2DE

2

2

4DE= = 4 : 1.

DE


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

5. cos2 + 21

1 + cot is equal to :

(a) 2 (b) 0 (c) 1 (d) none of these

Sol. (c) We have, cos2 + 21

1 + cot= cos2 + 2

1

cosec = cos2 + sin2 = 1

6. cos223 sin267 is equal to :

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

Sol. (b) cos223 sin267 = cos223 sin2 (90 23) = cos223 cos223 = 0

7. If A, B and C are interior angles of a triangle ABC, then tan2

B + Cis equal to :

(a) tanA

2(b) cot

A

2(c) sec

A

2(d) cos

A

2

Sol. (b) In ABC, we know that, A + B + C = 180

B + C =180 A 2

B + C= 90

A

2

tanB + C

2

= tan (90 A

2) = cot

A

2

8. If the median and the mode of a dala are 16 each, then its mean is :

(a) 16 (b) 32 (c) 12 (d) 24

Sol. (a) We know that, 3 median = 2 mean + mode

3 16 = 2 mean + 16 48 16 = 2 mean mean =

32

2 = 16

9. XYZ and PQY are two equilateral triangles such that P is the mid-point of YZ. Ratio

of the areas of triangles XYZ and PQY is :

(a) 1 : 2 (b) 2 : 1 (c) 1 : 4 (d) 4 : 1

Sol. (d) Since XYZ and PQY are equiangularSo, XYZ ~ PQY

2 2

2 2

area( XYZ) YZ YZ 4

area( PQY) 1PY YZ

2

= = =

= 4 : 1

10. In the quadrilateral ABCD, B = 90 andACD = 90, then AD2 is :(a) AC2 AB2 + BC2

(b) AC2 + CD2 + AB2

(c) AB2 + BC2 + CD2

(d) AB2 + BC2 + AC2


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Model Test Paper-4 (Term-I) 3

Sol. (c) In ACD, we have,AD2 = AC2 + CD2 [Pythagoras theorem]

= AB2 + BC2 + CD2 [ AC2 = AB2 + BC2]

SECTION B

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

11. n2 1 is divisible by 8 if n is an odd integer. Is it true?

Sol. Yes, the statement is true. We can easily verify it by putting n = 3, 5, 7,

12. Show that the system of equations 2x + 5y = 17, 5x + 3y = 14 has a unique solution.

Sol. The system of the equations will have a unique solution if1 1

2 2

a b

a b

Here a1 =2, b1 = 5, a2 = 5, b2 = 3

1

2

2=

5

a

aand

1

2

5=

3

b

b

1 1

2 2

a b

a b

Hence, the given system of equations has a unique solution. Proved.

OR

For what value of m, will the equations 3x 2y = 4 andmx + 4y + 8 = 0 have infinitenumber of solutions?

Sol. Here, a1

= 3, b = 2, c1

= 4, a2

= m, b2

= 4, c2

= 8.

For the system to have infinite number of solutions, we have 1 1 1

2 2 2

= =a b ca b c

3 2 4

4 8

= =

m m = 6

13. Find the zeroes of the quadratic polynomialf(x) = abx2 + (b2 ac) x bc.

Sol. We have, f(x) = abx2 + (b2 ac) x bc = abx2 + b2x acx bc

= bx (ax + b) c (ax + b) = (ax + b) (bx c)

The zeroes off(x) are given by f(x) = 0

(ax + b)(bx c) = 0 ax + b = 0 or, bx c = 0

x = b

a or, x =

c

b

Thus, the zeroes off(x) are b

aand

c

b

14. IfABC ~ DEF such that area ofABC is 9 cm2and the area ofDEF is 16 cm2

and BC= 2.1 cm. Find the length of EF.

Sol. We have,( )

( )

2

2

Area ABC BC=

Area DEF EF

( )

2

2

2.19=

16 EF


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3 2.1

=4 EF

EF =4 2.1

cm3

= 2.8 cm

15. Prove that1 1

1 + 1 +

sin sin= 2 sec2

Sol. LHS =1 1

+1 + sin 1 sin

= ( ) ( )

1 sin + 1 + sin

1 + sin 1 sin

= 2 22 2

=1 sin cos

= 2sec2 = RHS Proved.

16. In a rectangle ABCD, AB = 20 cm, BAC = 60. Calculate side BC.

Sol. In ABC, we have, AB = 20, BAC = 60

tan (BAC) =BC

AB tan 60 =

BC

20

3 =BC

20[ tan 60 = 3 ]

BC = 20 3 cm

17. The mean weight of a class of 35 students is 45 kg. If the weight of the teacher be

included, the mean weight increases by 500 grams. Find the weight of the teacher.

Sol. Let the mean weight of a class of 35 students be 1x and that of both students and

teacher be 2x . Then 1x = 45 kg and 2 500= 45 +1000

x kg = (45 + 0.5) kg = 45.5 kg

11

1

=x

xn

and

22

2

=x

xn

145 =35

xand

245.5 =36

x

x1

= 1575 kg and x2

= 1638 kg

Total weight = weight of students + weight of teacher Weight of teacher = Total weight weight of students

= x2

x1

= (1638 1575) kg = 63 kg

18. Find the mode of the following data :

25, 16, 19, 48, 19, 20, 34, 15, 19, 20, 21, 24, 19, 16, 22, 16, 18, 20, 16, 19

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

xi

15 16 18 19 20 21 22 24 25 34 48

fi

1 4 1 5 3 1 1 1 1 1 1


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We observe that the value 19 has maximum frequency i.e. it occurs maximum number

of times. Therefore, mode of the given data is 19.

SECTION C


19. Show that 3 5 is irrational.

Sol. If possible, let 3 5 be rational.

Let its simplest form be 3 5 =a

b, where, a and b are non-zero integers having no

common factor other than 1.

Now, 3 5 = 5 =

3

a a

b b

........... (i)

But, a and 3b are non-zero integers.

3

a

bis rational

Thus, from (i), it follows that 5 is rational.

This contradicts, the fact that 5 is irrational.

Hence, 3 5 is irrational. Proved.

OR

Show that12n cannot end with the digits 0 or5 for any natural number n.

Sol. We have 12n

= (2 2 3)n

= 22n

3n

We see that the only primes in the factorisation of 12 n are 2 and 3 and not 5.

As, we know the prime factorisation of a number is unique [By fundamental theoremof Arithmetic]

Hence, 12n cannot end with the digits 0 or 5. Proved.

20. The sum of the digits of a two digit number is 8 and difference between the number

and that formed by reversing the digits is 18. Find the number.

Sol. Let the digit at units place be x and the digit at tens place be y. Then,

Number = 10y + x

Number formed by reversing the digits = 10x + y

According to the given conditions, we havex + y = 8 ........... (i)

and, (10y + x) (10x + y) = 18

9(y x) = 18 y x = 2 .......... (ii)

On solving equations (i) and (ii), we get x = 3, y = 5

Hence, required number = 10y + x = 10 5 + 3 = 53


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21. Find a cubic polynomial whose zeroes are 3, 5 and 2

Sol. We know that a cubic polynomial whose zeroes are , and is given by

p(x) = x3 ( + + ) x2 + ( + + ) x .................. (i)Let = 3, = 5 and = 2. Then

+ + = 3 + 5 + (2) = 6 + + = 3 5 + 5 (2) + (2) 3 = 15 10 6 = 1

And, = 3 5 (2) = 30Substituting the above values in (i), we get

p(x) =x3 6x2 + (1) x (30) =x3 6x2 x + 30

Hence, required polynomial is x3 6x2 x + 30

22. In the figure, if PQ || BC and PR || CD, prove thatAR AQ

=AD AB

Sol. In ABC, we have, PQ || BC [Given]

AQ AP

=AB AC

(i) [BPT]

In ACD, we have, PR || CD

AP AR

=AC AD

(ii) [BPT]

From (i) and (ii), we get

AQ AR AR AQ= =

AB AD AD AB Proved.

23.Two triangles BAC and BDC, right angled at A and D respectively, are drawn on thesame base BC and on the same side of BC. If AC and DB intersect at P, prove that

AP PC = DP PB.

Sol. In APB and DPC, we have

A = D = 90

and, APB = DPC [Vertically opposite angles]APB ~ DPC [AA similarity]

AP PB

=DP PC

AP PC = DP PB Proved.

ORIn a triangle ABC, AD is a median and AE BC. Prove that

AC2 = AD2+ BC.DE +

2

2

BC .

Sol. In AED, we have, AD2 = AE2 + ED2 [Pythagoras theorem]

AE2 = AD2 ED2 (i)


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In AEC, AC2 = AE2 + EC2 [From (i)]= AD2 ED2 + (ED + DC)2

= AD2 ED2 + ED2 + DC2 + 2.ED.DC.

= AD2 + 2.DE

2BC BC

.2 2

+

= AD2 + DE.BC +

2BC

2

Proved.

24. Solve the equation 2sin2 = 3 (0 < < 90)

Sol. We have, 2sin2 = 3

sin2 =3

2

sin2 = sin 60 [ sin 60 =3

2]

2 = 60 = 30

25. Given that sin (A + B) = sinA cosB + cosA sinB, find the value of sin 75.

Sol. Putting A = 45 and B = 30 in sin (A + B) = sinA cosB + cosA sinB, we get

sin (45 + 30) = sin 45 cos30 + cos 45 sin 30

sin 75 =1 3 1 1

+ 2 22 2

=3 1

+2 2 2 2

=3 + 1

2 2

26. Prove that = 0

++ +

sinA sinB cos A cosB

cos A cosB sinA sinB

Sol. LHS =sinA sinB cosA cosB

+cosA + cosB sinA + sinB

=( ) ( ) ( ) ( )

( ) ( )

sinA sinB sinA + sinB + cosA + cosB cosA cosB

cosA + cosB sinA + sinB

=( ) ( )

2 2 2 2sin A sin B + cos A cos B

cosA + cosB sinA + sinB=

( ) ( )( ) ( )

2 2 2 2sin A + cos A sin B + cos B

cosA + cosB sinA + sinB

= ( ) ( )

1 1

cosA + cosB sinA + sinB= 0 = RHS Proved.

OR

Prove that : tan2A tan2B =

2 2

2 2

sin A sin Bcos A cos B


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Sol. LHS = tan2 A tan2 B =

2 2

2 2

sin A sin B

cos A cos B

2 2 2 2

2 2

sin A cos B sin B cos A

cos A cos B

=

2 2 2 2

2 2

sin A (1 sin B) sin B (1 sin A)

cos A cos B

=

2 2 2 2 2 2

2 2

sin A sin A sin B sin B sin B sin A

cos A cos B

+=

2 2

2 2

sin A sin B

cos A cos B

= = RHS Proved

27. Following table shows the weight of 12 students :

Weight (in kg) 67 70 72 73 75

Number of students 4 3 2 2 1

Find the mean weight of the students.

Sol. To calculate the mean we prepare the following table

Weight (in kg) Frequency fix

i

xi

fi

67 4 268

70 3 210

72 2 144

73 2 146

75 1 75

fi

= 12 fix

i= 843

Mean = 843= =12

i i

i

f xxf

= 70.25 kg.

28. Find the median for the following frequency distribution :

x 1 2 3 4 5 6 7 8 9

f 8 10 11 16 20 25 15 9 6

Sol. To find the median first we prepare cumulative frequency table.

x f cf

1 8 8

2 10 18

3 11 294 16 45

5 20 65

6 25 90

7 15 105

8 9 114

9 6 120

N = 120


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Here, N =120 N

2= 60 kg.

We find that the cumulative frequency just greater thanN

2i.e., 60 is 65 and the

value ofx corresponding to 65 is 5. Therefore, median = 5.

SECTION D


29. Prove that n2 n is divisible by 2 for every positive integer n.

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

So, following cases arise :

Case 1 :When n = 2q, we have

n2 n = (2q)2 2q = 4q2 2q = 2q (2q 1)

n2 n = 2r, where r= q(2q 1) n2 n is divisible by 2

Case II : When n = 2q + 1, we have

n2 n = 2q + 1)2 (2q + 1) = (2q + 1) (2q + 1 1) = 2q(2q + 1)

n2 n = 2r, where r= q(2q + 1) n2 n is divisible by 2Hence, n2 n is divisible by 2 for every positive integer n. Proved.

30.Solve the following system of equations graphically

x + 3y = 6, 2x 3y = 12

and hence find the value of a, if 4x + 3y = a.

Sol. We have, x + 3y = 6 : And, 2x 3y = 12

6

3

=

xy

2 12

3

=

xy

3 0 6

1 2 0

x

y 3 0 6

2 4 0 x

y

Points are (3, 1), (0, 2) and (6, 0) Points are (3, 2), (0, 4) and (6, 0).

The graph is shown below.

Clearly, two lines intersect at P(6, 0).

Hence, x = 6, y = 0 is the solution of the given system of equations.

Putting x = 6, y = 0 in a = 4x + 3y, we get a = (4 6) + (3 0) = 24


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OR

Taxi charges consist of fixed charges and the remaining depending upon the distance

travelled in kilometres. If a person travels 10 km, he pays Rs 68 and for travelling

15 km, he pays Rs 98. Find the fixed charges and the rate per km.

Sol. Let the fixed charges be Rs x and rate per km be Rs y.

Then, x + 10y = 68 (i)

And, x + 15y = 98 (ii)

Subtracting (ii) from (i) we get, 5y = 30 y = 6From (i), x = 68 60 = 8

Hence, fixed charges are Rs 8 and rate per km is Rs 6.

31. Find a cubic polynomial with the sum, sum of the products of its zeroes taken two at

a time, and product of its zeroes as 3, 1 and 3 respectively.

Sol. Let the required cubic polynomial be 3 2 , 0+ + + ax bx cx d a (i)

Let , and be its zeroes.Then,

2

3

(coefficient of )

coefficient of + + =

x

x 3 3

= =

bb a

a(ii)

3

coefficient of

coefficient of + + =

x

x 1 = =

cc a

a(iii)

3

(constant term)

coefficient of =

x 3 3

= =

dd a

a(iv)


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If we take a = 1, then from (ii), (iii) and (iv), we get, 3, 1, 3= = =b c d

Required polynomial is

3 21. ( 3) ( 1) 3

+ + +x x x [From (i)]

3 23 3+i.e., x x x .

32. Prove that in a right angled triangle, the square of the hypotenuse is equal to the sum

of the squares of the other two sides.

Sol. Given : A right-angled triangle ABC in which B = 90.To Prove : AC2 = AB2 + BC2

Construction : From B draw BD AC.

Proof : In triangles ADB and ABC, we have

ADB = ABC [Each equal to 90]

and, A = A [Common] ADB ~ ABC [AA-similarity criterion]

AD AB

=AB AC

[ ]In similar triangles corresponding sides are proportional

AB2 = AD AC (i)In triangles BDC and ABC, we have

CDB = ABC [Each equal to 90]

and, C = C [Common] BDC ~ ABC [AA-similarity criterion]

DC BC

=BC AC

[ ]In similar triangles corresponding sides are proportional

BC2 = AC DC (ii)Adding equations (i) and (ii), we get

AB2 + BC2 = AD AC + AC DC

= AC(AD + DC) = AC AC = AC2

Hence, AC2 = AB2 + BC2 Proved.

33. If5 tan = 4, show that5 3 1

=5 2 6

+sin cos

sin cos

Sol. We have, 5 tan = 4 tan =4

5

Now,5 sin 3 cos

5 sin + 2 cos

=

5 sin 3 cos

cos

5 sin + 2 cos

cos

[Dividing Nr and Dr by cos ]


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=

5 sin 3 cos

cos cos

5 sin 2 cos+cos cos

=5 tan 3

5 tan 2

+=

45 3

5

45 + 25

4tan =

5

=4 3 1

=4 + 2 6

.

OR

Prove that : sin8 cos8 = (sin2 cos2) (1 2 sin2 cos2)Sol. LHS = sin8 cos8

= (sin4)2 (cos4)2

= (sin4 + cos4)(sin4 cos4)= (sin2 + cos2)(sin2 cos2)(sin4 + cos4)= (sin2 cos2)[{(sin2) + (cos2)}2 2sin2 cos2]

= (sin2 cos 2)(1 2sin2 cos2 ) = RHS Proved.34. Find the missing frequencies in the following frequency distribution if it is known that

the mean of the distribution is 1.46.

Number of accidents (x) 0 1 2 3 4 5 Total

Frequency (f) 46 ? ? 25 10 5 200

Sol. Let the missing frequencies be f1

and f2.

xi

fi

fix

i

0 46 0

1 f1 f1

2 f2

2f2

3 25 75

4 10 40

5 5 25

N = 86 + f1

+ f2

fix

i= 140 + f

1+ 2f

2

We have, N = 200 200 = 86 + f1 + f2f1 + f2 = 114 (i)Also, Mean = 1.46

1.46 = Ni if x

1.46 = 1 2140 + + 2

200

f f

292 = 140 + f1

+ 2f2

= f1

+ 2f2

= 152 (ii)

Solving equaitons (i) and (ii), we get

f1

= 76 and f2

= 38

maths model test paper for summative assessment – 2

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