700000562 cbse 10 maths termi samplesolution1

8/12/2019 700000562 CBSE 10 Maths TermI SampleSolution1

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CBSE X | MathematicsSample Paper 1 - Solution

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CBSE Board

Class X Mathematics

Term I

Sample Paper 1

Time: 3 hour Total Marks: 90

Solution

Section A

1. Correct answer: A

2. Correct answer: D

For an integer 'm' every odd integer is of form 2m + 1.

3. Correct answer: B

Since -3 is the root of quadratic polynomial, we have:

(k - 1)(-3)2+ 1 = 0

4. Correct answer: A

5. Correct answer: C

We know:

Mean =Sum of observations

Number of observations

Mean of 6 numbers = 16

Sum of the 6 observations = 16 6 = 96

Mean of 5 observations = 17

Sum of the 5 observations = 17 5 = 85

Number which is removed = 96 - 85 = 11


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x = 3 sec2- 1, y = tan2 - 2

x - 3y = 3sec2 - 1 - 3tan2 + 6

= 3(sec2 - tan2) + 5

= 3 + 5= 8

7. Correct answer: B

cos + cos2 = 1

1 - cos2 = cos

sin2 + sin4 = 2(1 cos ) + (1 - cos2 )2

= cos + cos2

= 1


Therefore, using angle sum property, we have:

P = 180o- (80o+ 60o) = 40o

Section B

9. Since, 870 = 225 3 + 195

225 = 195 1 + 30

195 = 30 6 + 15

30 = 15 2 + 0

HCF (870,225) = 15

10.

OR


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Subtracting equation (1) from 3x - = 5, we get

Now, from equation (1), we have:

x = 3

11. are roots of x2- (k + 6)x + 2(2k - 1)

1 1Now, k 6 2(2k 1)

2 2

k 6 2k 1

k 7

12.


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13.cot =7

8(given)

14.

Here, n = 40n

202

Median class is 145 - 150

Also, since highest frequency is 11, Modal class is 145 - 150.

C.I. f c.f.

135 - 140 4 4

140 - 145 7 11145 - 150 11 22

150 - 155 6 28

155 - 160 7 35

160 - 165 5 40

Section C

15.To prove 5 + 2 is irrational, let us assume 5 + 2 is rational.

We can find integers a and b where a, b are co-prime, b

Such that,

Now a, b are integers, is rational.

2 is rational.

Which is a contradiction. So 5 + 2 is irrational.

OR


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

is rational.

(n-1)+(n+1)+2 is rational

2n+2 is rational

But we know that is an irrational number

So 2n+2 is also an irrational number

So our basic assumption that the given number is rational is wrong.

Hence, is an irrational number.

16.f(x) = x2- 2x + 1

Zeroes of f(x) are and

Sum of zeroes = + = 2 andProduct of zeroes = . = 1

Required polynomial = k(x2- 4x + 4), where k is any integer.

17.

So the length and breadth of the rectangle are 23 metres and 11 metres respectively.

OR


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Let the fixed charge = Rs x

and the subsequent charge = Rs y

According to the question, we have:

x + 4y = 27 ... (i)

and x + 2y = 21 ... (ii)Subtracting (ii) from (i), we have:

2y = 6 or y = 3

So, from (i),

x = 27 - 12 = 15

Thus, the fixed charge is Rs 15 and the charge for each extra day is Rs 3.

18.The system has infinitely many solutions. Therefore,

Equating (1) and (2), we get:

2a + 2b = 3a - 3b

or, a = 5b ... (4)

Equating (2) and (3), we get:

9a + 3b - 6 = 7a + 7b

or, 2a - 4b = 6 ... (5)

On solving equations (4) and (5), we get,

10b - 4b = 6 or b = 1Thus from (4), we get, a = 5

19.Given, XY || QR

By Basic Proportionality theorem,


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20.

Given: ABCD is a rhombus

To prove: AB2+ BC2+ CD2+ AD2 = AC2+ BD2

Proof:We know, diagonals of a rhombus bisect at right angles.

Therefore, from triangle AOB,

AB2= AO2+ OB2

4AB2= AC2+ BD2

Thus,AB2+ BC2+ CD2+ DA2= 4AB2= AC2+ BD2

[As AB = BC = CD = DA]

21.

22.We have:


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23.

24.


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Section D

25.If the number 15n, where n N, was to end with a zero, then its prime factorisation

must have 2 and 5 as its factors.

But 15 = 5 x 315n= (5 x 3)n= 5nx 3n

So, prime factors of 15ninclude only 5 but not 2.

Also, from the Fundamental theorem of Arithmetic, the prime factorisation of a

number is unique.Hence, a number of the form 15n, where n N, will never end with a

zero.

26.To solve the equations, make the table corresponding to each equation.

x -1 -2 -3

y 4 2 0

x 4 -1

y 0 4

Now plot the points and draw the graph.


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Since, the lines intersect at the point (-1, 4), so x = -1 and y = 4 be the solution. Also, by

observation vertices of triangle formed by lines and x-axis are A (-1, 4), B (-3, 0)

and C (4, 0).

27.Let p(x) = x3+ 2x2+ kx + 3Using Remainder theorem, we have:

p(3) = 33+ 2 32+ 3k + 3 = 21

k = -9

Thus, p(x) = x3+ 2x2- 9x + 3

When p(x) is divided by (x - 3), the quotient is x2+ 5x + 6.

Value indicated:Promotion of cooperative learning among students and helpingeach other in the study, removal of gender bias.

28.AD is the median of triangle ABC, since D is mid-point of BC.


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OR

Given: A right triangle ABC right angled at B.

To prove: AC2= AB2+ BC2

Construction: Draw BD AC

Proof:

We know that: If a perpendicular is drawn from the vertex of the right angle of a

right triangle to the hypotenuse, then triangles on both sides of the perpendicular are

similar to the whole triangle and to each other.

ADB ABC

So, (Sides are proportional)

Or, AD.AC = AB2 ... (1)

Also, BDC ABC


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So,

Or, CD. AC = BC2 ... (2)

Adding (1) and (2),

AD. AC + CD. AC = AB2+ BC2AC (AD + CD) = AB2+ BC2

AC.AC = AB2+ BC2

AC2= AB2+ BC2

Hence Proved.

29.

In triangles ABC and ADE, we have:

A = A

ACB = AED

Therefore, ABC ADE (by AA similarity criterion)

Now, in right ABC ,AB2 = BC2+ AC2(by Pythagoras theorem)

AB2= 122+ 52= 144 + 25 = 169

AB = 13 cm

Substituting the lengths of the sides in (1), we get:

DE =36

13and AE =

15

13

30.To prove:

We will make use of the identity: cosec2A = 1 + cot2A

L.H.S

=


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= cosecA + cotA

= R.H.S

OR

=RHS


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31.

= sin = RHS

32.

Hence, LHS = RHS.

33.We first prepare the cumulative frequency distribution 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


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Now, we mark the upper class limits along x-axis bt taking a suitable scale and the

cumulative frequencies along y-axis by taking a suitable scale.

Thus, we plot the points (10,7),(20,17),(30,40),(40,91),(50,97)and(60,100).

Join the plotted points by a free hand to obtain the required ogive.

34.

It is given that total frequency N is 20So, 10+x+y = 20 i.e. x + y = 10 .(i)

Given 50% of the observations are greater than 14.4.

So median = 14.4, which lies in the class interval 12-18.

l = 12, cf = 4 + x, h = 6, f = 5, N = 20

x =4

Now using equation, 10+x+y = 20, we get y = 6.

Hence x =4 and y = 6.

700000562 cbse 10 maths termi samplesolution1

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