class 10 cbse maths sa1 solutions 1

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Class 10 - Sample Question Paper (Mathematics) I – SA-I

Blue Print:

Chapter MCQ VSAQ SAQ LAQ Total Number system 2(2) 1(2) 2(6) 5(10) Algebra 2(2) 2(4) 2(6) 2(8) 8(20) Geometry 1(1) 2(4) 2(6) 1(4) 6(15) Trigonometry 4(4) 1(2) 2(6) 2(8) 9(20) Statistics 1(1) 2(4) 2(6) 1(4) 6(15)

Total 10(10) 8(16) 10(30) 6(24) 34(80)

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. (v) Use of calculators is not permitted.

Section – A

Question numbers from 1 to 10 are of one mark each.

1. Let

be a rational number, such that the prime factorization of q is of the form

2n5m, where n, m are non-negative integers. Then the decimal expansion of x is

a) Terminating

b) Non-terminating recurring

c) Non-terminating non recurring

d) None

Solution:

Option (a) is correct.


2. Given positive integers a and b, there exist unique integers q and r satisfying a = bq

+ r, where r satisfies

a) 0 <r < b

b) 0 < r ≤ b

c) 0 ≤r < b

d) 0 ≤r ≤ b

Solution:

Option (c) is correct.

3. The sum of the zeroes of the quadratic polynomial is

a) –

b)

c) –

d) –

Solution:

Sum of zeroes

Hence option (d) is correct.

4. The lines are

a) Intersecting lines

b) Parallel lines

c) Coincident lines

d) Perpendicular lines

Solution:

Given lines are

Here

Consider

and

Hence


Therefore the lines are coincident.

Hence option (c) is correct.

5. The sides of two similar triangles are in the ratio then the ratio of their areas

will be

a) 2:5

b) 4:25

c) 25:4

d) 5:2

Solution:

Ratio of areas of two similar triangles is equal to the ratio of squares of their

corresponding sides.

Ratio of areas of two similar triangles

The ratio is 4:25.

Hence option (b) is correct.

6. A ladder 13 m long is placed against a vertical wall of height 12m. The distance

between the foot of the ladder and the wall is

a) 5 m

b) 10 m

c) 7 m

d) 8 m

Solution:

In right

(By Pythagoras Theorem)

Hence the distance between the foot of the ladder

and the wall is 5 m.

Hence option (a) is correct.


7. If where 2A is an acute angle then

a)

b)

c)

d)

Solution:

Consider,

Hence option (b) is correct.

8.

a) 1

b) 0

c)

d)

Solution:


9.

a)

b)

c)

d)


Solution:

(

)


10. If the mode and median of a data are 27.8 and 26.7 then its mean is

a) 26.51

b) 27.15

c) 26.15

d) 27.51

Solution:

Mode = 3 Median – 2 Mean

27.8 = 3(26.7) – 2 Mean

2 Mean =

Mean = 26.15

Hence option (c) is correct.

Section – B

Question numbers 11 to 18 carry 2 marks each.

11. Given that HCF (306, 657) = 9, find LCM (306, 657).

Solution:

Given HCF (306, 657) = 9

The numbers are 306 and 657.

Recall that HCF × LCM = Product of two numbers

Hence LCM of 306, 657 is 22338.


12. Find the zeroes of the quadratic polynomial, and verify the

relationship between the zeroes and the coefficients.

Solution:

Given quadratic polynomial is

and

The zeroes of the given quadratic polynomial are

and

Sum of zeroes

(

) (

)

Product of zeroes

(

) (

)

13. Solve the equations

using substitution method.

Solution:

Given pair of equations are

Put the value of in equation (1).

( )


Put in equation (2)

14. The diagonals of a quadrilateral intersect each other at O such that

. Show

that ABCD is a trapezium.

Solution:

Given ABCD is a quadrilateral and diagonals AC and BD intersect each other at O

such that

Draw EF||AB

In

(By Basic Proportionality theorem)

[From equation (1)]

[By converse of basic proportionality theorem]

Hence AB || CD

Thus ABCD is a trapezium.


15. A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same

time a tower casts a shadow 28 m long. Find the height of the tower.

Solution:

Given length of vertical pole,

Length of shadow of vertical pole,

Given length of shadow of the tower,

Let height of the tower be

[By AA similarity criterion]

(nearly)

Hence the height of the tower is 42 m (nearly).


16. Prove that

Solution:

Consider a right angled triangle ABC with

From the figure,

By Pythagoras theorem, we have

Divide both sides with

(

)

(

)

17. Write the formula to find the median for grouped data and explain each term.

Solution:

The formula to find median of grouped data is, (

)

Where is the lower limit of median class

is the number of observations

is the cumulative frequency of the class preceding the median class

is the frequency of the median class and

is the class size


18. Find the mode of the following data:

Class 0 – 9 10 – 19 20 – 29 30 – 39 40 – 49 50 – 59 Frequency 6 9 14 21 16 8

Solution:

Class Frequency 0 – 9 6

10 – 19 9 20 – 29 14 30 – 39 21 40 – 49 16 50 – 59 8

Here modal class is 30 – 39

(

)

From the table,

(

)

(

)

Mode = 35.33 (nearly)


Section – C


19. U Eu ’ v w qu y p v g

of the form 3m or 3m + 1 for some integer m.

Solution:

Let be any positive integer

W k w by Eu ’ g b w p v g there exist

unique integers q and r satisfying, where ≤ .

Take

≤

Case (i):

Squaring on both the sides, we get

, where

Hence the square of is of the form

Case (ii):


, where


Case (ii):


, where



20. Show that √ is irrational.

Solution:

Given √

Let us assume that √ is rational.

Hence we can find co-primes a and b such that √

where

√

√

Since a and b are integers, we get

is a rational number.

Hence √ is also a rational number.

This is a contradiction as √ is an irrational number.

Therefore our assumption that √ is a rational number is incorrect.

Thus (√ is irrational.

21. Divide by and verify the division algorithm.

Solution:

Divisor Dividend

Quotient = Remainder

Recall that, Dividend = Divisor × Quotient + Remainder

Hence verified


22. Solve

Solution:

v

–

–

u

Hence equations (1) and (2) becomes,

u v

u p y qu w ‘ ’ qu

Put

in (3), we get


23. State and prove basic proportionality theorem.

Solution:

Basic Proportionality Theorem: A line drawn parallel to one side of a triangle divides

the other two sides in the same ratio.

Given: In

To prove:

Construction: Join B, E and C, D. Draw and

Proof: are on the same base DE and between the same parallels

[Triangles on the same base and between the same

Parallel lines are equal in area]

Consider

Equation (1) becomes,

Hence proved


24. In rhombus ABCD, prove that

Solution:

Given ABCD is a rhombus. i.e.

We know that in a rhombus, diagonals bisect perpendicularly.

Hence

and

In right [By Pythagoras theorem]

(

)

(

)

[


25. Prove that

Solution:

–


26. Evaluate

Solution:


27. Consider the following distribution of daily wages of 50 workers of a factory.

Daily wages 100 – 150 150 – 200 200 – 250 250 – 300 300 – 350 Number of workers 10 15 22 17 6 Find the mean daily wages of the workers of the factory by using an appropriate

method.

Solution:

Daily wages No. of Workers Mid value

100 – 150 10 125 1250

150 – 200 15 175 2625

200 – 250 22 225 4950

250 – 300 17 275 4675

300 – 350 6 325 1950

∑ = 70 ∑ 15450

We know that Mean ∑

∑

y

28. Find the median weight of the following students shown in the distribution table.

Weight 35 – 40 40 – 45 45 – 50 50 – 55 55 – 60 60 – 65 No. of Students

2 8 14 21 4 1

Solution:

Weight No. of Students Cumulative frequency 35 – 40 2 2 40 – 45 8 10 45 – 50 14 24 50 – 55 21 45 55 – 60 4 49 60 – 65 1 50

n = 50


50 – 55 is the median class.

Median (

)

Here

(

)

Median = 50.24 (nearly).

Section – D


29. Obtain all the zeroes of the polynomial if two of its

zeroes are √

and √

.

Solution:

Given √

and √

are the two zeroes of the polynomial

Hence ( √

)( √

) (

) is a factor of the given polynomial

Divide by (

)


(

)

( √

)( √

)

( √

)( √

)

( √

)( √

)

( √

)( √

)

( √

)( √

)

( √

) ( √

)

√

√

Hence the other two zeroes are

30. A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km down-stream. Determine the speed of the stream and that of the boat in still water.

Solution:

Let the speed of the boat in still water be km/h and speed of the stream be km/h. Then the speed of the boat downstream = Speed of the boat upstream

We know that

Given that the boat goes 30 km upstream.

Hence


Also it is given that the boat goes 44 km downstream.

Hence

Similarly the boat can go 40 km upstream and 55 km downstream in 13 hours.

u

Equations (1) and (2) becomes, u p y w ‘ ’ w ‘ ’ ub

Put

in (3), we get


Add (5) and (6) Put in (5), we get Hence the speed of the boat in still water is km/h and speed of the stream is km/h.

31. BL and CM are medians of a right angled at . Prove that

Solution:

Given: BL and CM are medians of a right angled at . To Prove: In right In right

But

[ is midpoint of AC]

(

)


In right

But

[ is midpoint of ]

(

)

Add (2) and (3), we get [From (1)]

32. If then show that

.

Solution: Given

We know that

(2)

Add (1) and (2)

We know that √

√ (

)


√

√

√

√

√(

)

33. Prove that

Solution:

Consider LHS:

Consider RHS:


Hence LHS=RHS

34. The mean of the distribution of daily wages of 50 workers of a factory is 145.2, find the missing frequencies. Daily Wages (in Rs) 100 – 120 120 – 140 140 – 160 160 – 180 180 – 200 Number of Workers 12 8 10

Solution:

Given ∑

Hence 30+

We know that Mean ∑

∑

But mean = 145.2

Daily wages (Rs) No. of

Workers Mid value

100 – 120 12 110 1320

120 – 140 130 130

140 – 160 8 150 1200

160 – 180 170 170

180 – 200 10 190 1900

∑ 30+ 4420+130 +170


Multiply (1) with 13, we get

Solve (2) and (3)

Put in equation (1), we get


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class 10 cbse maths sa1 solutions 1

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