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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6. Correct answer: C
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
8. Correct answer: C
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.