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SUMMATIVE ASSESSMEnT SAI)ModEl PrAcTIcE PAPEr dEMo
Key features of Model Practice Papers:
Summative Assessment Papers are based on the syllabus
prescribed by the CBSE Board for SA IExamination.
They are prepared exactly as per the sample papers suggested by
the CBSE Board.
The questions are selected in such a way that students get
acquainted with each and every concept of the prescribed syllabus
for SA I.
All the questions provided in the papers, are supported with detailed
and authentic solutions.
The Summative Assessment papers will give you chance to
practice more and more before the commencement of actual
examination, for obtaining high scores.
Class X MATHEMATICS
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Class X MATHEMATICS
SUMMATIVE ASSESSMENT (SA I)
MODEL PRACTICE PAPER - DEMO
2011 2012
Time allowed: 3 hours Maximum Marks: 80
General Instructions:
(i) All questions are compulsory.
(ii) The question paper consists of 34 questions divided into four sections A, B, C
and D.
(iii) Section A contains 10 questions of 1 mark each, which are multiple choice type
questions, Section B contains 8 questions of 2 marks each, Section C contains
10 questions of 3 marks each, and Section D contains 6 questions of 4 marks
each.
(iv) There is no overall choice in the paper. However, internal choice is provided in
one question of 2 marks. 3 questions of 3 marks and two questions of 4 marks.
(v) Use of Calculators is not permitted.
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SECTION A
Question numbers 1 to 10 carry 1 mark each. For each of the questions 1-10, four
alternative choices have been provided of which only one is correct. You have to
select the correct choice.
Q1. Which of the following relation is false?
(a) HCF ( ) 17 (b) LCM ( ) HCF ( )
(c) LCM ( ) 12121 (d) None of these.
Q2. Which one of the following is true?
(a) Sum of two irrational numbers is a rational number.
(b) Sum of rational and irrational numbers is a rational number.
(c) Product of two different irrational number is an irrational number.
(d) Product of rational and irrational numbers is a rational number.
Q3. Which of the following statement is false?
(a) A real number is zero of a polynomial if .
(b) A fourth degree polynomial is called a biquadratic polynomial.
(c) A Polynomial of degree zero is called a constant polynomial.
(d) None of these.
Q4. If the product of zeros of a polynomial is 4, then
value of ( is
(a) (b)
(c) (d) None of these.
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Q5. Sum of two numbers is 35 and their difference is 13, then the numbers are
(a) 34, 1 (b) 23, 12
(c) 24, 11 (d) None of these.
Q6. The given pair of linear equations are and , the value of
is
(a) (b)
(c) (d) None of these.
Q7. In a ABC, D and E are the points on the sides AB and AC respectively such that
AD , DB AE and EC , then relation between
DE and BC is
(a) Parallel (b) Perpendicular
(c) Intersecting (d) None of these.
Q8. In a ABC, D and E are the points on the sides AB and AC respectively such that
DE BC. If AD , DB AE and EC , then
value of is
(a) 1 (b) 2
(c) 3 (d) 4
Q9. If , and then is equal to
(a) (b)
(c) (d) None of these.
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Q10. Median of a given frequency distribution is found graphically with the help of
(a) Ogive (b) Histogram
(c) Frequency curve (d) None of these.
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SECTION B
Question numbers 11 to 18 carry 2 marks each.
Q11. Find the largest positive integer that will divide 396, 434 and 540 leavingremainders 5, 9 and 13 respectively.
Q12. If the zeros of the polynomial are prove
that .
Q13. Solve: and ; find if .
Q14. Evaluate:
OR
If are the interior angles of a triangle , prove that:
Q15. Given two triangles are , Find E
B
E
C4 cm
3.8 cm 3 cm
D
6 cm 7.6 cm
6
80
8 cmAF
B
E
C4 cm
3.8 cm 3 cm
D
6 cm 7.6 cm
6
80
8 cmAF
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SECTION C
Question numbers 19 to 28 carry 3 marks each.
Q19. Find the HCF and LCM of 288, 360 and 834 by prime factorization method.
Q20. Show that 7 is irrational.
OR
Prove that is irrational.
Q21. A man walks a certain distance with certain speed. If he walks an hour
faster, he takes 1 hour less. But, if he walks 1 an hour slower, he takes 3more hours. Find the distance covered by the man and his original rate of
walking.
OR
2 men and 7 boys can finish a piece of work in 4 days while 4 men and 4 boys
can finish it in 3 days. Find the time taken by one man along and that by one
boy alone to finish the work.
Q22. Find the value of and so that is divisible by
Q23. In a right angled at B, if AB 12 and BC 5 ,
Prove that
Q24. Given that , find the value of .
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Q25. is the bisector of If
, find
Q26. If and , prove that .
Q27. If the mean of the following distribution is 16, find the value of .
OR
4 6 10 14
1 3 4 2 2
A
15 cm
22 cm
B D C
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Find the median of the following data:
Q28. The mean of the following frequency distribution is 30. Find the value of .
Class interval Frequency
25.5 34.5 4
35.5 44.5 5
45.5 54.5 6
55.5 64.5 9
65.5 74.5 11
75.5 84.5 12
85.5 94.5 8
95.5 104.5 5
Class 0.5 19.5 20.5 39.5 40.5 59.5 60.5 79.5 80.5 99.5
Frequency 5 10 2 5
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SECTION D
Question numbers 29 to 34 carry 4 marks each.
Q29. Applying the division algorithm to find the quotient and remainder on dividing
by .
Q30. In an equilateral triangle the side BC is trisected at D.
Prove that
OR
In figure, D and E trisect BC. Prove that
Q31. Prove that
OR
Prove that
Q32. If and , prove that
A
DB E C
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Q33. Solve:
Q34. Compute the median from the following data:
Mid-value Frequency
115 6
125 25
135 48
145 72
155 116
165 60
175 38
185 22
195 3
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SECTION A
A1. Correct option:d
Explanation: Prime factorization are
HCF ( ) 17, LCM ( ) 12121
and LCM ( ) HCF ( ) are correct.
A2. Correct option: c
Explanation: Product of two different irrational number is irrational number.
A3. Correct option:d
A4. Correct option: c
Explanation: Given polynomial
product of zeros 4
4
A5. Correct option: c
Explanation: Let two numbers are and ,
Then + 35 (i)
and 13 (ii)
SOLUTION OF MODEL PRACTICE PAPER - DEMO
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Adding (i) and (ii) equations after that we get 2 48 or 24
Substituting the value of 24 in (i)
+ 35 or 11
A6. Correct option:a
Explanation: (i)
and (ii)
multiply (i) by , we get
(iii)
Adding equation (ii) and (iii), we get
or
A7. Correct option:a
Explanation: Given, and
and
Hence, (by basic proportional theorem)
A8. Correct option:d
A9. Correct option:b
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SECTION B
A11. According to question, largest positive integer is a factor of 396 5, 434 9 and
540 13.
Clearly, required number is the HCF of 391, 425 and 527.
Using the factor tree the prime factorizations of 391, 425 and 527 are as follows:
391 17 23, 425 17 and 527 17 31
HCF of 391, 425 and 527 is 17
Hence, required number is 17.
A12. Given are the zeros of the
Sum of the zeros
1 (i)
Product of the zeros (
(ii)
From (i) and (ii), we get
1 Proved.
A13. Given system of equation is
(i)
(ii)
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Substitute the value of from (ii) in (i), we get
Putting in , we get
Putting , in , we get
0
A14. We have,
OR
In triangle ,we have
Proved.
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A15. In triangles , we have
Therefore, by SSS-criterion of similarity, we have
A D, B F and C E, D 80,F 60
Hence, C E 60 (Since, A + B + C = 180)
A16. In , we have .
By Thales Theorem,
A
E
C
D B
B
E
C4 cm
3.8 cm 3 cm
D
6 cm 7.6 cm
6
80
8 cmAF
B
E
C
4 cm
3.8 cm 3 cm
D
6 cm 7.6 cm
6
80
8 cmAF
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A17. Less than type cumulative frequency distribution:
A18. Given frequency table
Marks obtained 0 15 15 30 30 45 45 60 60 75
Number of Students 10 13 18 5 14
Here, maximum frequency is 18 and the corresponding class is 30 45 .
So, 30 45 is the modal class. We have,
Lower limit of modal class
Size of the class interval
Frequency of the modal class
Frequency of the class preceding the modal class
Frequency of the class succeeding the modal class
Formula of obtain the mode
Daily income
(in rupees)
Number ofWorkers
or Frequency (
Less than
Daily income Number of Workers
or (
150 250 11 250 11
250 350 13 350 24
350 450 7 450 31
450 550 5 550 36
550 650 9 650 45
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SECTION C
A19. Using the factor tree for the prime factorization of 288, 360 and 834
288 ,
360 and
834
HCF of 288, 360 and 834 24 and
LCM of 288, 360 and 834 5760.
A20. Let us assume, on contrary, that 7 is rational.
Suppose 7 , where and are co-primes and .
Rearranging, we get
Since 7, and are integers, is rational, and so is rational. But this
contradicts the fact that is irrational. So we conclude that 7 is irrational.
OR
Assume that is rational.
Then there exist positive integers and such that = ,
Where and are co-prime and .
So, .(i)
Therefore, by theorem is divisible by 7 and so is divisible by 7.
So, we can write for some integer .
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Substituting in (i) we get;
.(ii)
So, by theorem is divisible by 7 and also is divisible by 7.
Therefore, and have at least 7 as a common factor. But this contradicts the
fact and are co-prime. So our assumption that is rational is incorrect
Hence. is irrational.
A21. Let the original speed be and total time taken be hrs. Then,
Distance
According to question, since distance covered in each case is same.
(i)
and, (ii)
adding (i) and (ii), we get
Substituting the value of in (ii), we get
OR
Suppose that one man alone can finish the work in days and one boy alone
can finish it in days. Then,
One mans one days work
One boys one days work
2 mens one days work
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7 boys one days work
Since 2 men and 7 boys can finish the work in 4 days.
(i)
Again, 4 men and 4 boys can finish the work in 3 days.
(ii)
Putting and in equations (i) and (ii), we get
(iii)
(iv)
Substituting the value of in (iii), we get
and
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Therefore, and
Thus, one man alone can finish the work in 20 days one boy alone can finish the
work in 30 days.
A22. If is exactly divisible by , then the remainder
should be zero.
On dividing, we get
Remainder 8
Now, Remainder 8
8)
and 8
and .
8
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A23. By Pythagoras theorem, we have
and .
LHS
RHS Proved.
A24. Let A and B
or
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A25. Let then
Since is the bisector of
A
15 cm
22 cm
B D C
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A26. Since and ; hence
(Alternate Angles)
Now in and
(Both 90)
Hence
From AA-criterion,
Proved.
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A27. Frequency distribution table:
We have 12, 92 + and mean 16
Mean
16
.
OR
65 75 is the median class. (where lise in of 35)
= 65, = 11, = 24, = 10
Median =
4 1 4
6 3 18
10 4 40
2
14 2 28
12 92 +
Classes
(Inclusive form)
Classes
(Exclusive form)
Frequency
( )
Cumulative
frequency ( )
25.5 34.5 25 35 4 4
35.5 44.5 35 45 5 9
45.5 54.5 45 55 6 15
55.5 64.5 55 65 9 24
65.5 74.5 65 75 11 35
75.5 84.5 75 85 12 47
85.5 94.5 85 95 8 55
95.5 104.5 95 105 5 60
= 60
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= 65 + 70.45 (approximate)
A28.
, , given
Class
(Inclusive form)
Class
(Exclusive form)
Frequency Class mark
0.5 19.5 020 5 10 50
20.5 39.5 2040 30 30
40.5 59.5 4060 10 50 500
60.5 79.5 6080 2 70 140
80.5 99.5 80100 5 90 450
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A30. Given: In , be a point on such that
To Prove:
Construction: Draw Join
Proof: In , and we have
and,
So, by RHS criterion of similarity, we have
Thus, we have
and (i)
Since . Therefore, is an acute triangle.
CB
A
D E
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OR
Given, D and E trisect BC
Therefore,
Let
Then, and
In right triangles and we have
, (i)
(ii)
DB E C
A
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and
(iii)
Now,
[From (i) (ii) and (iii)]
Proved.
A31. We have LHS
LHS
LHS
LHS
LHS
LHS
LHS
LHS RHS Proved.
OR
We have, LHS
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LHS
LHS
LHS
LHS (i)
and, RHS
RHS
RHS
RHS
RHS
RHS (ii)
From (i) and (ii)
LHS =RHS Proved.
A32. We have,
[where,
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Now,
Hence,
Proved.
A33. We have, (i)
(ii)
(iii)
From equation (i), we get
Substituting in equation (ii), we get
(iv)
Adding equation (iii) and (iv), we get
Putting in equation (iii), we get
Putting in equation (i), we get
Hence, and .
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