*Instructions:*

- Please check that this question paper consists of 11 pages.
- Code number given on the right hand side of the question paper should be written on the title page of the answer book by the candidate.
- Please check that this question paper consists of 30 questions.
**Please write down the serial number of the question before attempting it.**- 15 minutes times has been allotted to read this question paper. The question paper will be distributed at 10:15 am. From 10:15 am to 10:30 am, the students will read the question paper only and will not write any answer on the answer book during this period.

**SUMMATIVE ASSESSMENT – II**

**MATHEMATICS**

Time allowed: 3 hours Maximum Marks: 80

*General Instructions:*

*(i) All questions are compulsory*

*(ii) The question paper consists of 30 questions divided into four sections – A, B, C and D*

*(iii) Section A consists of 6 questions of 1 mark each. Section B consists of 6 questions of 2 marks each. Section C consists of 10 questions of 3 marks each. Section D consists of 8 questions of 4 marks each.*

*(iv) There is no overall choice. However, an internal choice has been provided in two questions of 1 mark, two questions of 2 marks, four questions of 3 marks each and three questions of 4 marks each. You have to attempt only one of the alternative in all such questions. *

*(v) Use of calculator is not permitted.*

**SECTION – A**

*Question number 1 to 6 carry 1 mark each.*

Question 1: Find the coordinates of a point , where is the diameter of a circle whose center is and is the point .

Answer:

Let the coordinates of be

Since is the center, it is also the midpoint of

Similarly,

Question 2: For what value of , the roots of the equation are real?

**Or**

Find the value of , for which the roots of the equation are reciprocal of each other.

Answer:

Given

For roots to be real,

**Or**

Given

Roots are :

Therefore Roots are and

Question 3: Find if

**Or**

Find the value of

Answer:

**Or**

Question 4: How many two digits number are divisible by ?

Answer:

Numbers divisible by are

Lowest digit number divisible by is

The highest digit number divisible by is

So the series starts with and ends with

Difference between the numbers is

So the AP will be

We need to find

Now

Therefore there are 30 two digit numbers divisible by 3

Question 5: In Fig. 1, cm and cm. What is the ratio of the to the ?

Answer:

Given

Consider and

(alternate angles)

is common

( alternate angles)

( AAA criterion)

Since , by Thales Theorem

Since triangles are similar,

Question 6: Find a rational number between and .

Answer:

We know and

Therfore is between and

Hence is between and

**Section – B**

*Question number 7 to 12 carry 2 mark each.*

Question 7: Find the HCF of and using Euclid’s algorithm.

**Or**

Show that every positive odd integer is of the form or , where is some integer.

Answer:

According to Eculid’s division theorem, any positive number can be expressed as where is the quotient, is the divisor and is the remainder and

So HCF of and is

**Or**

Let be any positive integer.

Eculid’s division theorem, any positive number can be expressed as where is the quotient, is the divisor and is the remainder and

Take

Since , the possible remainders are

That is can be or

Since is odd, cannot be or

Therefore any odd integer is of the form or .

Question 8: Which term of will be more than its term?

**Or**

If , the sum of the first terms of an is given by , find the term.

Answer:

Given is

Here

Therefore term is given by

Required term

Let it be the term, then

Hence term is the required term.

**Or**

Given

Put

Hence the common difference is

Therefore term

Question 9: Find the ratio in which the segment joining the points and is divided by . Also find the coordinates of this point on .

Answer:

Given Points and

Let the point be the point which divides the segment joining and in the ratio

The using the section formula, we get

Therefore divides and in the ratio

Therefore the point is

Question 10: A game consists of tossing a coin three times and noting the outcome each time. If getting the same result in all the tosses is success, find the probability of losing the game.

Answer:

Possible outcomes of tossing a coin three times is

i.e. Total number of events

The favorable events are and

Therefore Probability of success is

Hence the probability of losing is

Question 11: A die is thrown once. Find the probability of getting a number which i) is a prime number ii) lies between and .

Answer:

The possible outcomes when a dice is thrown one are

Therefore Total No. of events

No of prime events are prime numbers

No of events when the number lies between and

Therefore

i) Probability is a prime number

ii) Probability number lies between and

Note: 1 only has one positive divisor itself . So it is not a prime number. A prime number is a positive integer whose positive divisor are exactly and the number itself.

Question 12: Find if the system of equations has infinitely many solutions.

Answer:

Given: and

If the system of equations are and and they have infinitely many solutions then it satisfy the following:

From the given system

Therefore

Taking the first two

… … … … … (i)

Taking the last two

… … … … … (ii)

From (i) and (ii) we get

**Section – C**

*Question number 13 to 22 carry 3 mark each.*

Question 13: Prove that is an irrational number.

Answer:

Assume is a rational number

i.e. it can be expressed as a rational fraction of the form where are relatively prime numbers.

Since

we have or

must be even and since is even, should be even

Let

We have

Since is even, is even and since is even, must be even

However two even numbers cannot be relatively prime so cannot be expresses as a rational fraction.

Hence is irrational.

Question 14: Find the value of such that the polynomial has sum of its zeros equal to half of their products.

Answer:

Polynomial is

Let and be the zeros of the equation

On comparing with we get

Therefore

… … … … … i)

… … … … … ii)

Given

Question 15: A father’s age is three times the sum of the ages of his two children. After years his age will be two times the sum of their ages. Find the present age of the father.

**Or**

A fraction becomes when is subtracted from the numerator and it becomes when is subtracted from the denominator. Find the fraction.

Answer:

Let the ages of sons be and

Given that father’s age is times the ages of his two sons

Therefore present age of father … … … … … i)

Given years hence, father’s are will be twice the sum of the ages of his sons

Father’s age after years

Ages of his sons year hence and

… … … … … ii)

Using i) and ii) we get that the age of father is years.

**Or**

Let the fraction be

… … … … … i)

Also

… … … … … ii)

Subtracting ii) from i) we get

From ii)

Hence the fraction is

Question 16: Find the point on which is equidistant from the point and .

Or

The line segment joining the points and is trisected at points and such that is nearer to . If also lies on the line given by , find the value of .

Answer:

Let be equidistant from and

Since

Hence the point is

**Or**

(trisects)

Applying section formula

Coordinates of

Since lies on

Question 17: Prove that .

**Or**

Answer:

To Prove:

LHS

RHS. Hence proved.

Or

To Prove:

LHS

RHS. Hence proved.

Question 18: In Fig. 2, is a chord of length cm of a circle of radius cm and center . The tangent at and intersect at point . Find the length of .

Answer:

(Length of the tangents from an external point to a circle are equal)

In

is the bisector of

Since

( perpendicular from center to a chord bisects the chord)

cm

In Right triangle

cm

Let

In Right triangle

… … … … … i)

Since is a tangent,

In Right triangle

From i)

Hence is cm

Question 19: In Fig. 3, and , prove that

**Or**

If and are point of sides and respectively, of , right angled at , prove that

Answer:

Given:

In … … … … … i)

In … … … … … ii)

Adding i) and ii)

. Hence proved.

**Or**

… … … … … i)

… … … … … ii)

Adding i) and ii)

. Hence proved.

Question 20: Find the area of the shaded region in Fig. 4, if is a rectangle with sides cm and cm and is the center of the circle. (Take )

Answer:

(angle subtended by the diameter on the circumference is )

is a rectangle

cm

Therefore the radius of the circle cm

Area of the circle

Area of

Hence the shaded area

Question 21: Water in a canal, m wide and m deep, is flowing at the speed of km/hour. How much area will it irrigate in minutes; if $latex 8 cm standing water is needed?

Answer:

Speed of water

Rate of flow

Therefore volume of water in mins

Let Area of irrigation

Question 22: Find the mode of the following frequency distribution

Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |

Frequency | 8 | 10 | 10 | 16 | 12 | 6 | 7 |

Answer:

**Section – D**

*Question number 23 to 30 carry 4 mark each.*

Question 23: Two water taps can together can fill in a tank in hours. The tap with longer diameter takes hours less than the tap with the smaller one to fill the tank separately. Find the time in which each tap can fill the tank separately.

**Or**

A boat goes km upstream and km downstream in hours. In hours it can go km upstream and km downstream. Determine the speed of the stream and that of the boat in still water.

Answer:

Let the time taken by the smaller tap to fill up the tank completely hours

So the volume of tank filled by the smaller tap in hours

Also it is given that the time taken by the larger tap is hours less

Time taken by larger tap to fill the tank completely

So the volume of tank filled by the larger tap in hours

Given that both the taps fill the tank in hours hours

Tank filled by smaller tap in hours

Similarly, tank filled by larger tap in hours

Therefore

hours or hours

cannot be hours as the larger tap takes hours less.

Hence hours

Therefore small tap will fill the tank in hours and the larger tap will fill the tank in hours.

**Or**

Let the speed of the boat in still water km/hr

Let the speed of the stream km/hr

Speed of the boat downstream km/hr

Sped of the boat upstream km/hr

Boat goes km upstream and km downstream in hours

… … … … … … i)

Boat goes km upstream and km downstream in hours

… … … … … … ii)

Let and

Substituting in i) and ii)

… … … … … … iii)

… … … … … … iv)

From iii) we get

Substituting in iv) we get

Substituting in iii) we get

Now solving for and

… … … … … … v)

… … … … … … vi)

Adding v) and vi) we get km/hr

From vi) km/hr

Therefore speed o boat in still water is km/hr and speed of stream is km/hr

Question 24: If the sum of the first four terms of an is and that of the first terms is . Find the sum of the first terms.

Answer:

sum of terms of an A.P.

… … … … … i)

… … … … … ii)

solving i) and ii)

gives and

so

Question 25: Prove that

Answer:

LHS

RHS. Hence proved.

Question 26: A man in a boat is rowing away from the light house at m high takes minutes to change the angle of elevation of the top of the light house from to . Find the speed of the boat in meters per minute. [ Use ]

**Or**

Two poles of equal height are standing opposite each other on either side of the road, which is m wide. From a point between them on the road, the angle of elevation of the top of the poles are to respectively. Find the height of the poles and the distance of the point from the poles.

Answer:

Therefore Speed m/min

**Or**

Given

m

m

Hence m

Question 27: Construct a in which is cm, cm and . The construct a triangle whose sides are of the corresponding sides of .

Answer:

Question 28: A bucket open at the top is in the form of a frustum of a cone with a capacity . The radii of the top and bottom of the circular ends of the bucket and cm and cm respectively. Find the height of the bucket, and also the area of the metal sheet used in making it. ( Use )

Answer:

Volume

cm cm

Volume of the bucket

cm

Therefore height of bucket cm

cm

Therefore surface are of the metal sheet used

curved surface area base area

Hence the height of the bucket is cm and surface are of the metal sheet is

Question 29: Prove that in a right angle triangle, the square of the hypotenuse is equal to the sum of square of the other two sides.

Answer:

Given: A right angled , right angled at

To prove:

Draw:

Proof: In

(common angle)

( by AA criterion)

Therefore (corresponding sides are proportional)

… … … … … i)

Similarly,

and … … … … … ii)

Adding i) and ii)

. Hence proved.

Question 30: If the median of the following frequency distribution is , find the values of and .

Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | Total |

Frequency | 5 | 9 | 12 | 3 | 2 | 40 |

**Or**

The marks obtained by students in a class in an examination are given below.

Marks | No. of Students |

0-5 | 2 |

5-10 | 5 |

10-15 | 6 |

15-20 | 8 |

20-25 | 10 |

25-30 | 25 |

30-35 | 20 |

35-40 | 18 |

40-45 | 4 |

45-50 | 2 |

Answer:

Class Interval | Frequency | Cumulative Frequency |

0-10 | ||

10-20 | 5 | 5 |

20-30 | 9 | 14 |

30-40 | 12 | 26 |

40-50 | 26 | |

50-60 | 3 | 29 |

60-70 | 2 | 31 |

Given, Median

The median class

Median

Given sum of frequencies

Hence and

**Or**

i)

Marks | Number of Students |

Less than 5 | 2 |

Less than 10 | 7 |

Less than 15 | 13 |

Less than 20 | 21 |

Less than 25 | 31 |

Less than 30 | 56 |

Less than 35 | 76 |

Less than 40 | 94 |

Less than 45 | 98 |

Less than 50 | 100 |

ii) ,

Hence Median