**NOTE:**

(I) Please check that this question paper consists of 23 pages.

(II) 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.

(III) Please check that this question paper consists of 40 questions.

(IV)** Please write down the serial number of the question before attempting it.**

(V) 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.

**MATHEMATICS (STANDARD) – Theory**

—————————————————————————————————————————————–Time allowed: * 3 hours* Maximum Marks:

*—————————————————————————————————————————————–*

**80****General Instructions:**

**Read the following instructions very carefully and strictly follow them:**

(i) This question paper comprises four sections – A, B, C and D. This question paper carries 40 questions. All questions are compulsory.

(ii) **Section A** – Question numbers **1** to **20** comprises of **20** questions of **one** mark each.

(iii) **Section B** – Question numbers **21** to **26** comprises of **6** questions of **two** marks each.

(iv) **Section C** – Question numbers **27** to **34** comprises of **8** questions of **three** marks each.

(v) **Section D** – Question numbers **35** to **40** comprises of **6** questions of **four** marks each.

(vi) There is no overall choice in the question paper. However, an internal choice has been provided in 2 questions of one mark, 2 questions of two marks, 3 questions of three marks and 3 questions of four marks. You have to attempt only **one** of the choices in such questions.

(vii) In addition to this, separate instructions are given with each section and question, wherever necessary.

(viii) Use of calculators is not permitted.

**Section – A**

Question numbers **1** to **10** are multiple choice questions of **1** mark each. Select the correct option.

Question 1: On dividing a polynomial by , quotient and remainder are found to be and respectively. The polynomial is:

(A) (B) (C) (D)

Answer:

We know that

Where , and

Substituting all the above values we get

Question 2: In Figure – 1, is an isosceles triangle, right angled at . Therefore:

(A) (B) (C) (D)

Answer:

Given

Using pythagoras theorem

or

Question 3: The point on the axis which is equidistant from and is

(A) (B) (C) (D)

**(OR)**

The centre of a circle whose end points of a diameter are and is

(A) (B) (C) (D)

Answer:

Given points and

Let the equidistant point be as it is on x-axis.

Therefore

Therefore the required point is

**OR**

Given points and

Let the center is

and

Question 4: The value(s) of for which the quadratic equation has equal roots, is

(A) (B) (C) (D)

Answer:

Given

and

For equal roots,

Question 5: Which of the following is not A.P.?

(A) (B)

(C) (D)

Answer:

For

.

Therefore this is an A.P.

For

.

Therefore this is an A.P.

For

.

Therefore this is NOT an A.P.

For

.

Therefore this is an A.P.

Question 6: The pair of linear equations and is

(A) Consistent (B) Inconsistent

(C) Consistent with one solution (D) consistent with many solutions

Answer:

Given:

For

For consistent equations

In our case this is .

Hence the equations are inconsistent.

Question 7: In Figure – 2, is tangent to the circle with centre at , at the point . If , then is equal to:

(A) (B) (C) (D)

Answer:

is an isosceles triangle because (radius of the same circle)

Now

since

Question 8: The radius of a sphere (in cm) whose volume is , is

(A) (B) (C) (D)

Answer:

Volume of sphere

Let the radius cm

Question 9: The distance between the points and is

(A) (B) (C) (D)

Answer:

Given points and

Distance between points

Question 10: In Figure – 3, from an external point , two tangents and are drawn to a circle of radius cm with centre . If , then length of is:

(A) cm (B) cm (C) cm (D) cm

Answer:

Given

Also

Therefore is a square

Hence all sides are equal.

Therefore cm

In Question Nos. **11** to **15**, Fill in the blanks. Each question carries **1** mark.

Question 11: The probability of an event that is sure to happen, is

Answer:

The probability of an event is a number describing the chance that the event will happen. An event that is certain to happen has a probability of 1.

Question 12: Simplest form of is

Answer:

Question 13: is a rectangle whose three vertices are and . The length of its diagonals is

Answer:

units

Question 14: In the formula

Answer:

Given

where class size, assumed mean and class mark

Question 15: All concentric circles are to each other.

Answer:

All concentric circles are similar to each other.

Question Nos. **16** to **20** are short answer type questions of **1** mark each.

Question 16: Find the sum of the first natural numbers.

Answer:

First term Common difference Last term

Question 17: In Figure – 4, the angle of elevation of the top of a tower from a point on the ground, which is m away from the foot of the tower, is . Find the height of the tower.

Answer:

m

Question 18: The LCM of two numbers is and their HCF is . If one of the numbers is , find the other.

Answer:

We know

Product of two numbers

Therefore the other number

Question 19: Form a quadratic polynomial, the sum and product of whose zeroes are and respectively.

**OR**

Can be a remainder while dividing by ?

Answer:

Let and are two roots of equation

Therefore

and

A quadratic equation can be written in the form

**OR**

cannot be the remainder because the degree of the remainder should be less than the degree of the divisor

Long division

The remainder is

Question 20: Evaluate:

Answer:

**Section – B**

Question Nos. **21** to **26** carry **2** marks each.

Question 21: In the Figure – 5, and Prove that:

Answer:

Given and

To Prove:

Proof: In

Note: line drawn parallel to one side of the triangle, intersects the other two sides in distinct points, then it divides the other two sides in the same ratio.

… … … … … i)

Similarly, in

… … … … … ii)

From i) and ii)

Hence proved.

Question 22: Show that is an irrational number, where is given to be an irrational number

**OR**

Check whether can end with the digit for any natural number .

Answer:

Let be a rational number

, where and

is a rational number. But it is given that is an irrational number.

Hence our assumption that is a rational number is wrong.

Hence is an irrational number.

**OR**

For any number to end with a or , must be divisible by .

This factorization does not contain any term of .

Hence there is no value of for which ends with digit or

Question 23: If and are interior angles of a , then show that

Answer:

To prove:

Given

Therefore LHS RHS.

Hence proved.

Question 24: In Figure – 6, a quadrilateral is drawn to circumscribe a circle. Prove that .

**OR**

In Figure – 7, find the perimeter of , if cm.

Answer:

Given is a quadrilateral

The quadrilateral touches the circle at and

To prove:

Proof:

We know that the lengths of the tangents drawn from an external point to a circle are equal. Therefore

… … … … … i) … … … … … ii)

… … … … … iii) … … … … … iv)

Adding i), ii), iii) and iv) we get

Hence proved.

**OR**

… … … … … i)

… … … … … ii)

We know that the lengths of the tangents drawn from an external point to a circle are equal. Therefore

and . Also

Therefore we can write i) as

… … … … … iii)

and similarly ii) as … … … … … iv)

Adding iii) and iv) we get

Perimeter of the

Therefore the perimeter of cm

Question 25: Find the mode of the following distribution.

Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |

Number of Students | 4 | 6 | 7 | 12 | 5 | 6 |

Answer:

Marks | Number of Students | |

0-10 | 4 | |

10-20 | 6 | |

20-30 | 7 | |

30-40 | 12 | |

40-50 | 5 | |

50-60 | 6 |

Mode

Question 26: cubes, each of volume , are joined end to end. Find the surface area of the resulting cuboid.

Answer:

Volume of cube

Therefore the side of the cube

Dimension of the new cuboid

cm cm cm

Therefore surface area of cuboid

**Section – C**

Question Nos. **27** to **34** carry **3** marks each.

Question 27: A fraction becomes when is subtracted from the numerator and it becomes when is added to its denominator. Find the fraction.

**OR**

The present age of a father is three years more than three times the age of his son. Three years hence the father’s age will be years more than twice the age of the son. Determine their present ages.

Answer:

Let the fraction be

Therefore

… … … … … i)

Also

… … … … … ii)

Subtracting i) from ii) we get

Substituting in i)

Therefore the fraction is

**OR**

Let the present age of the son years

Therefore the present age of father

After years

years

Therefore present age of father years and the present age of son is years.

Question 28: Use Euclid Division Lemma to show that the square of any positive integer is either of the form for some integer .

Answer:

As per Euclid’s Division Lemma, if and are two positive integers, than

Let be the positive integer and

Therefore where and is any positive integer.

Therefore can be either or .

If , then

where

If , then

where

If , then

where

Therefore the square of any positive number is of the form or for some integer

Question 29: Find the ratio in which the axis divides the line segment joining the points and . Also find the point of intersection.

**OR**

Show that the points and are vertices of an isosceles right triangle.

Answer:

Given y-axis divides the line segment.

Therefore the point

Applying section formula

Comparing

Therefore the ratio is

Hence the point of intersection is

**OR**

Given three vertices of a triangle

Similarly,

And

Since is an isosceles triangle. … … … … … i)

Applying Pythagoras theorem

Therefore is a right angled triangle. … … … … … ii)

Hence from i) and ii), is an isosceles right angled triangle.

Question 30: Prove that

Answer:

LHS

RHS. Hence proved.

Question 31: For an A.P., it is given that the first term , common difference and the term . Find and sum of first terms of the A.P.

Answer:

Given: First term , Common difference

term

We know,

Therefore

We know

Hence

Question 32: Construct a with sides cm, cm and . Then construct a triangle whose sides are of the corresponding sides of .

**OR**

Draw a circle of radius cm. Take a point outside the circle at a distance of cm from the center of the circle and construct a pair of tangents to the circle from that point.

Answer:

Step 1: Construct

- Draw base of side cm
- Draw
- Taking as a center, cm as radius, draw an arc. Let the point where arc intersect the ray be point
- Join
- This completes the construction of

Now we need to make a triangle which is times its size. Therefore the scale factor

Step 2:

- Draw any ray making an acute angle with on the side opposite to vertex
- Mark 4 points – and so that
- Join and draw a line through parallel to to intersect at
- Draw a line through parallel to the line to intersect at
- Therefore is the required triangle

Now consider and

is common

Since

( AA similarity)

Therefore

By construction

Therefore our construction is justified.

**OR**

Draw a circle of radius cm

Mark any point at a distance of cm from the center

Join and locate the midpoint

Taking as a center draw a circle of radius cm

The two circles intersect at and

Join and and they would be the tangents to the circle from the point .

Question 33: Read the following passage and answer the questions given at the end :

Diwali Fair

A game in a booth at a Diwali Fair involves using a spinner first. Then, if the spinner stops on an even number, the player is allowed to pick a marble from a bag. The spinner and the marbles in the bag are represented in Figure – 8.

Prizes are given, when a black marbles is picked. Shweta plays the same once.

(i) What is the probability that she will be allowed to pick a marble from the bag?

(ii) Suppose she is allowed to pick a marble from the bag, what is the probability of getting a prize, when it is given that the bag contains balls out of which 6 are black?

Answer:

i) Even numbers in the spinner

Total numbers in the spinner

Therefore Probability to pick marble

ii) Number of black marbles

Number of white marbles

Therefore total number of marbles

Therefore the probability of picking a black marble

Therefore the probability of winning or

Question 34: In Figure-9, a square is inscribed in a quadrant of a circle. If the radius of circle is cm, find the area of the shaded region.

Answer:

Area of the quadrant of the circle

cm

Now

cm

Therefore area of square

Hence the shaded area

**Section – D**

Question Nos. **35** to **40** carry **3** marks each.

Question 35: Obtain other zeroes of the polynomial if two of its zeores are and .

**OR**

What minimum must be added to so that the resulting polynomial will be divisible by ?

Answer:

Given

Given that and are zeros of

This means that and are factors of .

Therefore is a factor of

Long division

Therefore the other two zeros are

**OR**

Long division

So add to the polynomial for it to be completely divisible by

Question 36: Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Answer:

Given

To prove:

Construction: Draw and

Area of … … … … … i)

Area of … … … … … ii)

Dividing i) by ii)

… … … … … iii)

Consider and

as

(By AA similarity)

From i)

… … … … … iv)

Since

Substituting in iv)

Similarly, we can prove

Hence proved.

Question 37: Sum of the areas of two squares is . If the difference of their perimeter is m, find the sides of the two squares.

**OR**

A motor boat whose speed is km/h in still water takes hour more to go km upstream than to return downstream to the same spot. Find the speed of the stream.

Answer:

Let the sides of the two square be and

… … … … … i)

Perimeter of the two square will be and

… … … … … ii)

Substituting ii) in i) we get

or (this is not possible)

m

Hence m

**OR**

Speed of the boat in still water km/hr

Let the speed of the stream

Therefore speed of the boat upstream km/hr

Similarly, the speed of the boat down stream km/hr

Time taken upstream Time taken down stream

(not possible) or km/hr

Therefore the speed of the stream km/hr

Question 38: A solid toy is in the form of a hemisphere surmounted by a right circular cone of same radius. The height of the cone is cm and the radius of the base is cm. Determine the volume of the toy. Also find the area of the coloured sheet required to cover the toy.

Answer:

Volume of toy

Area of colored sheet required to cover the toy

CSA of cone CSA of hemisphere

Question 39: A statue m tall, stands on the top of a pedestal.From a point on the ground, the angle of elevation of the top of the statue is and from the same point the angle of elevation of the top of the pedestal is . Find the height of the pedestal. Use

Answer:

Please refer to the adjoining diagram

In

… … … … … i)

Similarly, in

… … … … … ii)

Therefore from i) and ii) we get

m

Therefore the height of the pedestal m

Question 40: For the following data, draw a ‘less than’ ogive and hence find the median of the distribution.

Age in years | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |

Number of persons | 5 | 15 | 20 | 25 | 15 | 11 | 9 |

**OR**

The distribution given below shows the number of wickets taken by bowlers in one-day cricket matches. Find the mean and the median of the numbers of wickets taken.

Number of wickets | 20-60 | 60-100 | 100-140 | 140-180 | 180-220 | 220-260 |

Number of bowlers | 7 | 5 | 16 | 12 | 2 | 3 |

Answer:

Less than frequency distribution

Age | No of persons | Class | Cumulative Frequency |

0-10 | 5 | Less than 10 | 5 |

10-20 | 15 | Less than 20 | 20 |

20-30 | 20 | Less than 30 | 40 |

30-40 | 25 | Less than 40 | 65 |

40-50 | 15 | Less than 50 | 80 |

50-60 | 11 | Less than 60 | 91 |

60-70 | 9 | Less than 70 | 100 |

Now plotting less than ogive

Therefore Median (from graph)

**OR**

Number of wickets | Number
of bowlers |
|||

20-60 | 7 | 40 | -3 | -21 |

60-100 | 5 | 80 | -2 | -10 |

100-140 | 16 | 120 | -1 | 16 |

140-180 | 12 | 160 | 0 | 0 |

180-220 | 2 | 200 | 1 | 2 |

220-260 | 3 | 240 | 2 | 6 |

Assumed mean

Class size

Mean

To find median

Number of wickets | Number of bowlers | |

20-60 | 7 | 7 |

60-100 | 5 | 12 |

100-140 | 16 | 28 |

140-180 | 12 | 40 |

180-220 | 2 | 42 |

220-260 | 3 | 45 |

Median class

Median