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**SECTION - A**

**Questions 1 to 20 carry 1 mark each.****Q.1. The total number of terms in the expansion of (x + a) ^{501} + (x - a)^{501} is :**

Ans:

We know that the total number of terms in the expansion of (x + a)

∴ Total number of terms in the expansion of (x + a)

Total number of terms in the expansion of (x + a)

(a) -6

(b) 6

(d) -2

Explanation:

When x = -1, the given polynomial

5x-4

= 5(-l) - 4(-1)

= - 5 + 4 + 3

= - 9 + 3 = -6

OR

Correct option : (a)

Explanation: By Remainder Theorem, the remainder

P(1)=(1)

= 1 -3 + 2 + 1 = 1

(b) (0,-5)

(c) (5,0)

(d) (-5,0)

(a) 90°, 60°, 30°

(b) 80°, 70°, 30°

(d) None of these

(b) y = 0, z = 0

(d) None of these

Ans:

On y-axis, x = 0, z = 0

(d) diagonals bisect each other.

(b) 42 cm

(c) 36 cm

(d) 24 cm

Ans:

[ ∵ AC is the diagonal of parallelogram ABCD

∴ ...(i)

Now, L is the mid-point of DC

∴ ...(ii)

From (i) and (ii), we have

(b) 42°

(c) 28°

(d) 34°

∴ Other angle =

According to the question,

⇒

So the smaller angle is 36°.

b + c, c + a, a + b are in A.P

If b + c - (a + b + c)f c + a - (a - b + c), a + b - (a + b + c) are in A.P. [Subtracting (a + b + c)]

If - a, - b, - c are in A.P.

If a, b, c are in A.P., which is given to be true.

(b) 2.1 cm

(d) 1.6 cm

Explanation.

Volume of sphere = Volume of cone

⇒ r = 2.1 cm.

Ans:

Ans:

The first five prime numbers are 2, 3, 5, 7, 11.

Numbers are in ascending order

Explanation : v Area of two parallelograms between two parallel line and on same base and equal

∴ ar(parallelogram PQRS) = ar(parallelogram EFRS)

∵ Area of triangle is equal to half of parallelogram form on same base and lies between same two parallel lines

x =2y or 1.x - 2y + 0 = 0

Or

∵ (4,19) is a solution of y = ax + 3 ... (i)

On putting x = 4 and y = 19 in Eq. (i), we get

19 = a(4) + 3

⇒ 4a = 19-3 = 16

∴ a = 4

Parabola y

∴

Hence, the length of latus rectum

We know that the diagonals of a parallelogram bisect each other.

Therefore,

AC = 2 x OA = 2 x 3 cm = 6 cm

BD = 2 x OD = 2 x 2 cm = 4 cm

Therefore, AC = 6 cm and BD = 4 cm.

Height of the tent = 3.5 m

∴

Hence, 471.42 sq. m of canvas is required to obtain the required conical tent.

Ans:

Total surface area of a cone = π(L + R)

**SECTION - B**

**Question numbers 21 to 26 carry 2 marks each.****21. Find n[(A ∪ B ∪ C)'] , if n(U) = 6500, n(A) = 4000, n(B) = 2000, n(C) = 1000 and n( A ∩ B) = n(B ∩ C = n(A ∩ C) = 400; n( A∩ B ∩ C) = 200.****OR****Let A = {1, 2,3}, B = {3,4} and C = {4,5,6}. Find****(i) A x (B ∩ C)****(ii) ( A x B ) ∩ ( A x C)****Ans: **Given:

Given, n(U) = 6500, n(A) = 4000, n(B) = 2000, n(C) =1000, n(A **∩** B) = n(B **∩** C) = n(A **∩** C) = 400 and n(A **∩** B **∩** C) = 200

We know that

OR

We have**Q.22. Ram and Ravi have the same weight. If they each gain weight by 2kg, how will their new weights be compared?****Ans:** Let x kg be the weight each of Ram and Ravi, On gaining 2 kg weight of Ram and Ravi's will be (x + 2) each. According to Euclid's second axiom, when equals are added to equals, the wholes are equal. So, weight of Ram and Ravi are again equal.**Q.23. Ram and Ravi have the same weight. If they each gain weight by 2 kg, how will their new weights be compared ?Ans:** Let x kg be the weight each of Ram and Ravi.

On adding 2 kg,

Weight of Ram and Ravi will be (x + 2) kg each.

According to Euclid's second axiom, when equals are added to equals, the wholes are equal.

So, weight of Ram and Ravi are again equal.

Ans:

(i) the particular 3 students join excursion party.

(ii) the particular 3 students do not join excursion party.

In the first case, we have to choose 7 students out of the remaining 22 students and this can be done in

In the second case, we have to choose 10 students out of the remaining 22 students and this can be done in

∴ The required number of ways =

Ans:

**SECTION - C**

**Question 27 to 34 carry 3 marks each.****Q.27. Find the domain and range of the real function f(x) =****Ans:****Q.28.Find the points on the line 2x + 3y = 12, where it cut the X and Y axes.Ans: **When x = 0 , y = 4 and when y = 0 , x = 6

The line meets the X axis at (6,0)

The line meets the Y axis at (0,4)

Total distance covered= x km and total fare = Rs y, Since the fare for first kilometre is Rs 8 and fare for the remaining distance (x -1) km is Rs 5 per kilometre.

It can be written in the form of a linear equation in two variables as, 8 + 5(x- 1) = y.

Now, the equation becomes

y = 8 + 5x - 5

or y = 3 + 5x ...(i)

When x = 1

From equation (i) , we have

y = 3 + 5(1) =3 + 5 = 8

When x = 2

From equation (i), we have

y = 3 + 5(2) = 3 + 10 = 13

To draw the graph, we use the following table:

Ans:

Also, let AD meets BC at E.

Ans:

...(1)

...(2)

...(3)

[using (2) and (4)]

OR

Out of 52 cards, two cards can be drawn in

∴ Total number of elementary events =

Let A = event that both cards drawn are red

and B = event that both cards drawn are aces

There are 26 red cards, out of which 2 red cards can be drawn in

∴

There are 4 aces, out of which 2 aces can be drawn in

∴

There are 2 cards which are both aces and red, out of which 2 cards can be drawn in

∴

Required probability = Probability (both cards are red or both are aces) = P(A ∪ B)

1. Draw any line seomentAB = 13 cm.

2. Construct ∠PAB == 45° and ∠QBA - 15° and say AP and BQ meet at X.

3. Draw the perpendicular bisector of AX and let it intersect AB at Y

4. Draw the perpendicular bisector of BX and let it intersect AB at Z.

5. Join XY and XZ.

Thus, ΔXYZ is the required triangle.

So, height (length) cf the cylinder = 100 cm

Inner radius of the cylinder = r = cm = 27 cm.

Thickness of the iron sheet = 9 cm.

Outer radius of the cylinder= R = (27 + 9)cm=36cm

Thus, volume of the iron sheet used

**SECTION - D**

**Questions 35 to 40 carry 4 marks each****Q.35. Out of 100 students; 15 passed in English, 12 passed in Mathematics, 8 in Science, 6 in English and Mathematics, 7 in Mathematics and Science, 4 in English and Science, 4 in all the three. Find how many passed:(i) in English and Mathematics but not in Science ?(ii) in more than one subject only ?Ans: **Let E, M and S be the sets of students who passed English, Mathematics and Science respectively and U be set of all students.

According to given, we have

n(E) = 15, n(M) = 12, n(S) - 8, n(E ∩ M) = 6, n(M ∩ S) = 1, n(E ∩ S) = 4, n(E ∩ M ∩ S) = 4 and n(U) = 100.

∴ Number of students who passed in English and Mathematics but not in Science

∴ Number of students who passed in two subjects only

and the number of students who passed in all the three subjects = n(E ∩ M ∩ S) = 4

∴ Number of students who passed in more than two subjects

= Number of students who passed in two subjects only + Number of students who passed in all the three subjects

= 5 + 4 = 9

(ii) AD = BE

OR

AB and CD are respectively the smallest and longest sides of c quadrilateral ABCD (see fig.). Show that

∠A > ∠C and ∠B > ∠D.

**Find the value of n and the mean.OrFind the missing frequencies in the following frequency distribution, if it is known that the mean of the distribution is 1.46.**

...(i)

...(ii)

According to the given condition:

...(i)

and

...(ii)

From (2), we get

a

⇒ a = 1 (consider only real roots)

Substituting a = 1 in (1), we have

[∵ 10 x 10= 100 = 4 x 25]

Or

The given series is

1 + 3 + 7 + 15 + ...

The sequence o f differences between successive terms is 2, 4, 8 ,...

Clearly, it is an G.P.

Let T

S

S

Subtracting (2) from (1), we get

0 = 1 + {2 + 4 + 8 + ... + (Tn - Tn-1)} - T

⇒ T

⇒ T

Putting n = 1,2, 3,..., n, we get

T

T

T

...(7)

The histogram for given data is shown below

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