Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths Standard with Solutions Set 7 are designed as per the revised syllabus.

## CBSE Sample Papers for Class 10 Maths Standard Set 7 with Solutions

Time: 3 hrs

Max. Marks: 80

Instructions

- This question paper has 5 Sections A-E.
- Section A has 20 MCQs carrying 1 mark each.
- Section B has 5 questions carrying 2 marks each.
- Section C has 6 questions carrying 3 marks each.
- Section D has 4 questions carrying 5 marks each.
- Section E has 3 Case Based integrated units of assessment (4 marks each).
- All questions are compulsory. However, an internal choice in 2 questions of 2 marks, 2 questions of 3 marks and 2 questions of 5 marks has been provided. An internal choice has been provided in the 2 marks questions of Section E.
- Draw neat figures wherever required. Take π = 22/7 wherever required if not stated.

Section A

(Section A consists of 20 questions of 1 mark each.)

Question 1.

If HCF(306, 657) = 9, then LCM (306, 657) is [1]

(a) 22338

(b) 23238

(c) 33228

(d) 32328

Answer:

(a) 22338

We know that

LCM (a, b) × HCF (a, b) = a × b

∴ LCM (306, 657) × HCF (306, 657) = 306 × 657

= \(\frac{306 \times 657}{9}\) [∵ HCF(306, 657)] = 9

= 22338

Question 2.

Check whether the given pair of triangles are similar. [1]

(a) Yes

(b) No

(c) Cannot say

(d) None of these

Answer:

(a) Yes

In ∆MNO and ∆DEF, \(\frac{M N}{E D}\) = \(\frac{2}{4}\) = \(\frac{1}{2}\)

\(\frac{N O}{E F}\) = \(\frac{5}{10}\) = \(\frac{1}{2}\)

and \(\frac{O M}{F D}\) = \(\frac{7}{14}\) = \(\frac{1}{2}\)

∴ \(\frac{M N}{E D}\) = \(\frac{N O}{E F}\) = \(\frac{O M}{F D}\)

Since, corresponding sides are in the same ratio.

∴ ∆MNO ~ ∆DEF [by SSS rule]

Question 3.

The value of

2 cos^{2} 30° + sec^{2} 30° + 2 cos0° + 3 sin 90° – tan^{2} 60° is [1]

(a) \(\frac{6}{29}\)

(b) \(\frac{4}{25}\)

(c) \(\frac{29}{6}\)

(d) \(\frac{3}{26}\)

Answer:

(c) \(\frac{29}{6}\)

We have,

2cos^{2}30° + sec^{2}30° + 2cos0°+ 3sin90° – tan^{2}60°

= 2 × \(\left(\frac{\sqrt{3}}{2}\right)^2\) + [atex]\left(\frac{\sqrt{3}}{2}\right)^2[/latex] + 2 × 1 + 3 × 1 – \((\sqrt{3})^2\)

= 2 × \(\frac{3}{4}\) + \(\frac{4}{3}\) + 2 + 3 – 3

= \(\frac{3}{2}\) + \(\frac{4}{3}\) + 2

= \(\frac{9+8+12}{6}\) = \(\frac{23}{9}\)

Question 4.

If one of the zeroes of the polynomial f(x) = (k^{2} + 8)x^{2} + 13x + 6x is reciprocal of the other, then the value of k is [1]

(a) 4, 2

(b) 6, 2

(c) 2, 3

(d) 4, 3

Answer:

(a) 4, 2

Let α and β be two zeroes of the given polynomial.

Then, α = \(\frac{1}{\beta}\) or β = \(\frac{1}{\alpha}\).

∴ α, \(\frac{1}{\alpha}\) are two roots/zeroes of the given polynomial.

By relationship between zeroes and coefficients of a polynomial, we have

α × \(\frac{1}{\alpha}\) = \(\frac{6 k}{k^2+8}\) ⇒ k^{2} + 8 = 6k

⇒ k^{2} – 6k + 8 = 0 ⇒ (k – 4)(k – 2) = 0

∴ k = 4, 2

Question 5.

The sum of probabilities of all the elementary events is [1]

(a) 0

(b) 1

(c) 1.5

(d) None of these

Answer:

(b) 1

The sum of probability of all the elementary events is always equal to one.

Question 6.

If two positive integer p and q can be expressed as p = ab^{2} and q = a^{3}b, where a, b being prime numbers, then LCM (p, q) is equal to [1]

(a) ab

(b) a^{2}b^{2}

(c) a^{3}b^{2}

(d) a^{3}b^{3}

Answer:

(c) a^{3}b^{2}

Given that p = ab^{2} = a × b × b

and q = a^{3}b = a × a × a × b

∴ LCM of p and q = LCM of ab^{2} and a^{3}b

= a × b × b × a × a = a^{3}b^{2}

[since, LCM is the product of the greatest power of each prime factor involved in the numbers]

Question 7.

If tan θ = \(\frac{12}{13}\), then the value of \(\frac{2 \sin \theta \cos \theta}{\cos ^2 \theta-\sin ^2 \theta}\) is [1]

(a) \(\frac{312}{25}\)

(b) \(\frac{25}{312}\)

(c) \(\frac{312}{15}\)

(d) \(\frac{15}{312}\)

Answer:

(a) \(\frac{312}{25}\)

Question 8.

The following pair of linear equations, x = y and x – 2 = y – 2 are [1]

(a) perpendicular

(b) coincident

(c) intersecting

(d) parallel

Answer:

(b) coincident

We have, x = y and x – 2 = y – 2 ⇒ x = y

On graphical representation, we get a pair of coincident lines.

Question 9.

The value of k, for which 2k + 7, 6k – 2 and 8k + 4 are 3 consecutive terms of an AP [1]

(a) \(\frac{2}{15}\)

(b) \(\frac{2}{17}\)

(c) \(\frac{15}{2}\)

(d) \(\frac{17}{2}\)

Answer:

(c) \(\frac{15}{2}\)

Here, 2k + 7, 6k – 2 and 8k + 4 will be consecutive terms of an AP, if

(6k – 2) – (2k + 7) = (8k + 4) – (6k – 2)

⇒ 4k – 9 = 2k + 6

⇒ 4k – 2k = 9 + 6 ⇒ 2k =15

∴ k = \(\frac{15}{2}\)

Question 10.

sin^{6}θ + cos^{6}θ is equal to [1]

(a) 3sin^{2}θcos^{2}θ

(b) (sin^{3}θ + cos^{3}θ)^{2}

(c)

(d) 1 – 3sin^{2}θcos^{2}θ

Answer:

(d) 1 – 3sin^{2}θcos^{2}θ

We have, sin^{6}θ + cos^{6}θ

= (sin^{2}θ + cos^{2}θ)^{3} – 3sin^{2}θcos^{2}θ(sin^{2}θ + cos^{2}θ) [∵ a^{3} + b^{3} = (a + b)^{3} – 3 ab(a + b)]

= (1)^{3} – 3sin^{2}θcos^{2}θ(1)

= 1 – 3sin^{2}θcos^{2}θ [∵ sin^{2}A + cos^{2}A = 1]

Question 11.

The longest diameter of the base of a cone, which can be fully fitted in a cube of edge 8 cm, is [1]

(a) 16 cm

(b) 4 cm

(c) 8 cm

(d) 6 cm

Answer:

(c) 8 cm

The longest diameter of the base of a cone, which can be fully fitted in a cube of edge 8 cm, is 8 cm.

Question 12.

The distance between the points (asinθ + bcosθ, 0) and (0, acosθ – bsinθ) is [1]

(a) a^{2} + b^{2}

(b) a^{2} – b^{2}

(c) \(\sqrt{a^2+b^2}\)

(d) \(\sqrt{a^2-b^2}\)

Answer:

(c) \(\sqrt{a^2+b^2}\)

Given points are

(asinθ + bcosθ, 0) and (0, acosθ – bsinθ)

By distance formula

d = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)

= \(\sqrt{(0-a \sin \theta-b \cos \theta)^2+(a \cos \theta-b \sin \theta-0)^2}\)

Question 13.

The algebraic sum of the deviations of a frequency distribution from its mean is [1]

(a) always positive

(b) always negative

(c) zero

(d) a non-zero number

Answer:

(c) zero

Let x_{1}, x_{2}, x_{3}, ……..x_{n} be the observations and \(\bar{x}\) its mean. Then,

\(\bar{Z}\)(\(\bar{x}\) – x_{i}) = \(\bar{Z}\)\(\bar{x}\) – \(\bar{Z}\)x_{i} = n\(\bar{x}\) – n\(\bar{x}\) = 0

Question 14.

In the given figure, PT is a tangent to a circle with centre O, at point R. If diameter SQ is produced, it meets with PT at point P with ∠SPR = x and ∠QSR = y, then the value of ∠x + 2∠y is [1]

(a) 60°

(b) 30°

(c) 0°

(d) 90°

Answer:

(d) 90°

In the given figure, OR= OS

⇒ ∠ORS = ∠OSR = y

[angles opposite to equal sides are equal]

Also, ∠ORT = 90°

[radius of a circle is perpendicular to the tangent at the point of contact]

Question 15.

If ∆ABC and ∆DEF are similar triangles, such that ∠A = 47°, ∠E = 83°, then the value of ∠C is [1]

(a) 0°

(b) 90°

(c) 50°

(d) 45°

Answer:

(c) 50°

Given, ∆ABC ~ ∆DEF

⇒ ∠A = ∠D, ∠B = ∠E, ∠C = ∠F

Given, ∠A = 47° and ∠B = 83°

∴ ∠C = 180° – (∠A + ∠B) = 180° – (47° + 83°) = 50°

Question 16.

If a number x is added to twice its square, then the resultant is 21, then, the quadratic representation of this statement is [1]

(a) 2x^{2} – x + 21 = 0

(b) 2x^{2} + x – 21 = 0

(c) 2x^{2} – x – 20 = 0

(d) None of these

Answer:

(b) 2x^{2} + x – 21 = 0

Let the number be x.

Then, according to the given condition,

2x^{2} + x = 21

⇒ 2x^{2} + x – 21 = 0

Question 17.

Tours for the national capital and the white house begin at 8:30 am from tour agency. Tours for the national capital leave every 15 min. Tours for the white house leave every 20 min. How many minutes after do the tours leave at the same time? [1]

(a) 60 min

(b) 30 min

(c) 80 min

(d) 100 min

Answer:

(a) 60 min

Required time LCM (15, 20)

By using prime factorisation method,

15 = 3 × 5

and 20 = 2 × 2 × 5 = 2^{2} × 5

∴ LCM (15, 20) = 2^{2} × 3 × 5 = 60 min

In every 60 min. tours leave at the same time.

Question 18.

If P is a point on F-axis, whose ordinate is 3 and Q is a point (-5, 2), then the distance PQ is [1]

(a) \(\sqrt{26}\) units

(b) \(\sqrt{24}\) units

(c) 5 units

(d) \(\sqrt{65}\) units

Answer:

(a) \(\sqrt{26}\) units

Since, P lies on Y-axis and has ordinate as 3

∴ The point P is (0, 3) and Q is (-5, 2).

Now, the distance of PQ is

PQ = \(\sqrt{(-5-0)^2+(2-3)^2}\)

= \(\sqrt{(-5)^2+(-1)^2}\) = \(\sqrt{25+1}\)

∴ PQ = \(\sqrt{26}\) units

Directions In the question number 19 and 20, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option.

Question 19.

Assertion (A) The class mark of the class 15-35 is 25.

Reason (R) The class mark is the difference of upper class limit and the lower class limit.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).

(c) Assertion (A) is true but Reason (R) is false.

(d) Assertion (A) is false but Reason (R) is true. [1]

Answer:

(c) Assertion (A) is true but Reason (R) is false.

We know that class mark of a class is the mid-point of a class i.e. Class mark

\(=\frac{\text { Upper class limit }+ \text { Lower class limit }}{2}\)

For class 15-35, class mark = \(\frac{35+15}{2}\) = \(\frac{50}{2}\) = 25

So, the given Assertion (A) is true.

We know that class size is the difference of upper class limit and lower class limit.

So, the given Reason (R) is false.

Hence, Assertion (A) is true but Reason (R) is false.

Question 20.

Assertion (A) The common difference of an AP, whose nth term is a_{n} = (3n + 7), is 3.

Reason (R) The nth term of an AP is a_{n} = a + (n – 1)d. [1]

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).

(c) Assertion (A) is true but Reason (R) is false.

(d) Assertion (A) is false but Reason (R) is true.

Answer:

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

The nth term of an AR

a_{n} = 3n + 7

∴ Common difference, d = a_{n} – a_{n – 1}

= 3n + 7 – [3(n – 1) + 7]

= 3n + 7 – [3n – 3+ 7]

= 3n + 7 – 3n – 4 = 3

So, the given Assertion (A) is correct.

We know that nth term of an AR

a_{n} = a + (n – 1 )d

⇒ a_{n} = a + nd – d

⇒ a_{n} = nd + (a – d)

So, the given Reason (R) is correct and it is the correct explanation of Assertion (A).

Hence, both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Section B

(Section B consists of 5 questions of 2 marks each.)

Question 21.

ABCD is a trapezium with AB || DC. If ∆AED is similar to ∆BEC, then prove that AD = BC. [2]

Answer:

Given Trapezium ABCD in which AB || DC and ∆AED ~ ∆BEC.

To prove AD = BC

Proof In ∆EDC and ∆EBA,

∠CED = ∠AEB [vertically opposite angles]

∠EDC = ∠EBA [alternate interior angles]

[∵ if triangles are similar, then their sides are proportional]

From Eqs. (i) and (ii), we get

\(\frac{E B}{E A}\) = \(\frac{E A}{E B}\)

⇒ (EB)^{2} = (EA)^{2}

⇒ EB = EA [taking positive square root]

From Eq. (ii), we get

\(\frac{E B}{E A}\) = \(\frac{A D}{B C}\) ⇒ \(\frac{A D}{B C}\) = 1 [∵ EB = EA]

∴ AD = BC Hence proved.

Or

Find the length of the median drawn through A on BC of a ∆ABC, whose vertices are A(7, – 3), B(5, 3) and C(3, -1).

Answer:

The median from the vertex of a triangle bisects the opposite side.

So, let AD be the median through A then D must be the mid-point of the side BC.

Question 22.

If two tangents inclined at an angle of 60° are drawn to a circle of radius 5 cm, then find the length of each tangent. [2]

Answer:

In the given figure, CA = 5 cm and ∠APB = 60°

In ∆APO, ∠PAO = 90°

[∵ the tangent at any point of a circle is perpendicular to the radius through the point of contact]

Since, ∠APB = 60°

⇒ ∠APO = \(\frac{60^{\circ}}{2}\) = 30° [∵ ∆APO ≅ ∆OBP]

∴ tan30° = \(\frac{A C}{A P}\) ⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{5}{A P}\) ⇒ AP = 5\(\sqrt{3}\)cm

Hence, the length of tangent is 5\(\sqrt{3}\) cm.

Or

In the given figure, if the angle between two radii of a circle is 130°, then find the angle between the tangents at the ends of the radii.

Answer:

From figure, ∠AOB = 130° ……. (i)

Clearly, ∠PAQ = ∠PBO = 90° ….. (ii)

[∵ radius is perpendicular to the tangent at the point of contact]

Now, applying angle sum property of quadrilateral in

∠PAO + ∠AOB + ∠OBP + ∠APB = 360°

⇒ 90° + 130° + 90° + ∠APR = 360°

[from Eqs. (1) and (ii)]

∴ ∠APB = 360° – 310° = 50°

Hence, the angle between the tangents at the ends of the radii is 50°.

Question 23.

Prove that

sec A (1 – sin A) (sec A + tan A) = 1 [2]

Answer:

LHS = sec A(1 – sin A)(sec A + tan A)

Question 24.

If the 3rd and 9th terms of an AP are 4 and – 8 respectively, which term of this AP is zero? [2]

Answer:

Let a be the first term and d be the common difference of the given AP

The nth term of the AP is a_{n} = a + (n – 1)d

∴ a_{3} = a + 2d = 4

[a_{3} = 4(given)] ……. (i)

and a_{9}= a + 8d = -8

[a_{9} = -8 (given)] ……. (ii)

On subtracting Eq. (i) from Eq. (ii), we get

6d = -12 ⇒ d = –\(\frac{-12}{6}\) = -2

∴ From Eq.(i), a + 2 × (-2) = 4

⇒ a – 4 = 4 ⇒ a = 4 + 4 = 8

Let the nth term of the AP is zero.

Then, a_{n} = 0 ⇒ a + (n – 1)d = 0

⇒ 8 + (n – 1)(-2) = 0 ⇒ (n – 1)(-2) = -8

⇒ n – 1 = \(\frac{-8}{-2}\) = 4 ⇒ n = 4 + 1 = 5

Hence, 5th term of the AP is zero.

Question 25.

5 yr hence, the age of Jacob will be three times that of his son. 5 yr ago, Jacob’s age was seven times that of his son. What are their present ages? [2]

Answer:

Let the present age of Jacob be x yr and the present age of his son be y yr.

5 yr hence, Jacob’s age =(x + 5) yr

and his son’s age = (y + 5) yr

By the first condition.

(x + 5) = 3(y + 5)

⇒ x + 5 = 3y + 15

⇒ x = 3y + 15 – 5

⇒ x = 3y + 10 …… (i)

5 yr ago, Jacob’s age = (x – 5) yr

and his son’s age = (y – 5)yr

By the second condition,

(x – 5) = 7 (y – 5)

⇒ x – 5 = 7y – 35

⇒ x = 7y – 35 + 5

⇒ x – 7y = -30 ……..(ii)

On substituting the value of x from Eq. (i) in Eq. (ii), we get

3y + 10 – 7y = -30

⇒ -4y = -30 – 10

⇒ -4y = -40

∴ y = \(\frac{-40}{-4}\) = 10

On putting the value of y in Eq. (i), we get

x = 3 × 10 + 10

⇒ x = 30 + 10 = 40

Hence, the present age of Jacob = 40 yr and the present age of his son = 10 yr

Section C

(Section C consists of 6 questions of 3 marks each.)

Question 26.

In given figure, PA and PB are tangents to the circle with centre O, such that ∠APB = 50°, then find the measure of ∠OAB. [3]

Answer:

Given, ∠APB = 50°

Since, P is an external point of a circle.

Therefore, PA =PB

[∵ tangents drawn from an external point to a circle are equal]

⇒ ∠PBA = ∠PAB

[∵ angles opposite to equal sides are equal] .. .(i)

In ∆APB,

∠APB + ∠PBA + ∠PAB = 180°

[∵ sum of all angles of a triangle is 180°]

50° + 2∠PAB = 180° [from Eq. (i)]

⇒ 2∠PAB = 130°

⇒ ∠PAB = 65° ………(ïi)

Also, radius OA is perpendicular to the tangent.

Therefore, ∠OAP = 90°

⇒ ∠OAB + ∠PAB = 90°

⇒ ∠OAB + 65° = 90° [from Eq. (ii)]

⇒ ∠OAB = 90° – 65° = 25°

Question 27.

Find the roots of the following equation

\(\frac{1}{x+4}\) – \(\frac{1}{x-7}\) = \(\frac{11}{30}\), x ≠ -4, 7. [3]

Answer:

Given, equation is

\(\frac{1}{x+4}\) – \(\frac{1}{x-7}\) = \(\frac{11}{30}\), x ≠ -4, 7

Hence, the roots of the given equation are 2 and 1.

Or

Solve for x, 2\(\left(\frac{2 x+3}{x-3}\right)\) – 25\(\left(\frac{x-3}{2 x+3}\right)\) = 5. [3]

Answer:

Let \(\frac{2 x+3}{x-3}\) = y

Then, \(\frac{x-3}{2 x+3}\) = \(\frac{1}{y}\)

Therefore, the given equation reduces to 2y-25— = 5

2y – 25\(\frac{1}{y}\) = 5

⇒ 2y^{2} – 25 = 5y

⇒ 2y^{2} – 5y – 25 = 0

⇒ 2y^{2} – 10y + 5y – 25 = 0

⇒ 2y(y – 5) + 5(y – 5) = 0

⇒ (y – 5)(2y + 5) = 0 ⇒ y = 5

or y = \(\frac{-5}{2}\)

Now, putting y = 5 in Eq. (i), we get

\(\frac{2 x+3}{x-3}\) = \(\frac{5}{1}\)

⇒ 5x – 15 = 2x + 3

⇒ 3x = 18 ⇒ x = 6

Again, putting y = –\(\frac{5}{2}\) in Eq. (i), we get

\(\frac{2 x+3}{x-3}\) = –\(\frac{5}{2}\)

⇒ -5x + 15 = 4x + 6

∴ 9x = 9 ⇒ x = 1

Hence, the values of x are 1 and 6.

Question 28.

Find the largest positive integer that will divide 398, 436 and 542 leaving remainders 7, 11 and 15, respectively. [3]

Answer:

It is given that when dividing 398 by the required number, then there is a remainder of 7.

This means that 398 – 7 = 391 is exactly divisible by the required number.

In other words, required number is a factor of 391.

Similarly, the required positive integer is a factor of 436 – 11 = 425 and 542 – 15 = 527.

Clearly, required number is the HCF of 391,425 and 527.

Using factor tree, we get the prime factorisations of 391,425 and 527 as follows:

391 = 17 × 23, 425 = 5^{2} × 17

527 = 17 × 31

∴ HCF of 391, 425 and 527 = 17

Hence, 17 is the required number.

Or

There are 156, 208 and 260 students in groups A, B and C, respectively. Buses are to be hired to take them for a field trip. Find the minimum number of buses to be hired, if the same number of students should be accommodated in each bus. [3]

Answer:

The prime factorisation of given numbers are

156 = 2^{2} × 3 × 13

208 = 2^{4} × 13

260 = 2^{2} × 13 × 5

and

The HCF of (156, 208, 260) = 2^{2} × 13 = 52

The minimum number of buses that are required

= \(\frac{156}{52}\) + \(\frac{208}{52}\) + \(\frac{260}{52}\) = 3 + 4 + 5 = 12

Question 29.

Prove that [3]

Answer:

Question 30.

A bag contains 18 balls out of which, x balls are red. [3]

(i) If one ball is drawn at random from the bag, then what is the probability’ that it is red ball?

Answer:

∵ Total number of balls in the bag = 18

Total number of red baUs in the bag = x

∴ P(drawing a red ball) = \(\frac{x}{18}\)

(ii) If 2 more red balls are put in the bag, then the probability of drawing a red ball will be 9/8 times that of probability of red ball coming in part (i). Find the value of x.

Answer:

∵ Number of red balls added to the bag = 2

∴ Total number of balls in the bag = 18 + 2 = 20

and total number of red balls in the bag = x + 2

Now, P (drawing a red ball) = \(\frac{x+2}{20}\)

According to the question,

\(\frac{x+2}{20}\) = \(\frac{9}{8}\left(\frac{x}{18}\right)\)

⇒ \(\frac{x+2}{20}\) = \(\frac{x}{16}\)

⇒ 16x + 32 = 20x

⇒ 4x = 32

∴ x = 8

Question 31.

8 chairs and 5 tables for a classroom cost ₹ 10500, while 5 chairs and 3 tables cost ₹ 6450. Find the cost of each chair and that of each table. [3]

Answer:

Let the cost of each chair be ₹ x and cost of each table be ₹ y.

Then, 8x + 5y = 10500 ……… (i)

and 5x + 3y = 6450 …….. (ii)

On multiplying Eq. (ii) by 5 and Eq. (i) by 3 and subtracting the results, we get

25x – 24x = 32250 – 31500

⇒ x = 750

On putting x = 750 in Eq. (i), we get

8 × 750 + 5y = 10500

⇒ 6000 + 5y = 10500

⇒ 5y = 4500

⇒ y = 900

∴ Cost of each chair = ₹ 750

and cost of each table = ₹ 900

Section D

(Section D consists of 4 questions of 5 marks each.)

Question 32.

The following table shows the age distribution of cases of a certain disease admitted during a year in a particular hospital.

Age(in years) | 5-14 | 15-24 | 25-34 | 35-44 | 45-54 | 55-64 | Total |

Number of cases | 6 | 11 | 21 | 23 | 14 | 5 | 80 |

Find the modal age. [5]

Answer:

As the distribution is discontinuous, so firstly we will convert it into a continuous distribution by using adjustment factor of \(\frac{15-14}{2}\) = 0.5

Hence, we get the following table

As, the class 34.5-44.5 has maximum frequency, so it is the modal class.

Here, l = 34.5, f_{1} = 23, f_{2} = 14, f_{0} = 21 and h = 10

where, l is the lower limit of class,

f_{1} is the frequency of the modal class,

f_{2} is the frequency of the class succeeding the modal class,

f_{0} is the frequency of the class preceding the modal class and h is the class size.

Hence, the modal age is 36.32 yr.

Or

If the median of the distribution given below is 30, then find the values of x and y. [5]

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

Number of students | 5 | X | 20 | 15 | Y | 5 | 60 |

Answer:

Cumulative frequency table

∵ Median = 30

So, it lies in the interval 30 – 40.

Thus, 30 – 40 is the median class.

∴ l = 30, h = 10, f = 15, cf = 25 + x and N = 60

where, l is the lower limit of median class,

N is the total number of observations,

cf is the cumulative frequency of class preceding the median class,

f is the frequency of the median class and h is the class size.

∵ Median = l + \(\frac{\frac{N}{2}-C}{f}\) × h

∴ 30 = 30 + \(\frac{\frac{60}{2}-(25+x)}{15}\) × 10

⇒ 15 = x + y ⇒ 15 = 5 + y ⇒ y = 10

Question 33.

From a solid cylinder whose height is 12 cm and diameter is 10 cm, a conical cavity of same height and same diameter is hollowed out. Find the volume and total surface area of the remaining solid. [5]

Answer:

Given, radius of the cylinder, r = \(\frac{10}{2}\) = 5 cm and height of the cylinder, h = 12 cm

Now, volume of remaining solid = Volume of the cylinder – Volume of the cone

= \(\frac{6600}{7}\) – \(\frac{2200}{7}\) = \(\frac{4400}{7}\) = 628.57 cm^{3}

Since, slant height of the cone,

l = \(\sqrt{r^2+h^2}\) = \(\sqrt{(5)^2+(12)^2}\)

= \(\sqrt{169}\) = 13 cm

∴ Curved surface area of the cone = πrl

= \(\frac{22}{7}\) × 5 × 13 = \(\frac{1430}{7}\)cm^{2}

Curved surface area of the cylinder

= 2πrh

= 2 × \(\frac{22}{7}\) × 5 × 12 = \(\frac{2640}{7}\)cm^{2}

and area of upper base of the cylinder = m2

= \(\frac{22}{7}\) × 5 × 5 = \(\frac{550}{7}\) cm^{2}

Now, total surface area of the remaining solid = Curved surface area of the cylinder + Curved surface area of the cone + Area of upper base of the cylinder

= \(\frac{2640}{7}\) + \(\frac{1430}{7}\) + \(\frac{550}{7}\)

= \(\frac{4620}{7}\) = 660 cm^{2}

Question 34.

If α, β are the zeroes of the polynomial f(x) = 2x^{2} + 5x + k, then satisfying the relation (α + β)^{2} – αβ = \(\frac{21}{4}\), then find the value of k. Also, find the zeroes of f(x). [5]

Answer:

Given that f(x) = 2x^{2} + 5x + k

Now, sum of zeroes, α + β = \(-\frac{\text { coefficient of } x}{\text { coefficient of } x^2}\)

= –\(\frac{5}{2}\)

and product of zeroes, αβ = \(\frac{\text { constant term }}{\text { coefficient of } x^2}\) = \(\frac{k}{2}\)

Also, given (α + β)^{2} – αβ = \(\frac{21}{4}\) ⇒ (-\(\frac{5}{2}\))^{2} – (\(\frac{k}{2}\)) = \(\frac{21}{4}\)

⇒ \(\frac{k}{2}\) = \(\frac{25}{4}\) – \(\frac{21}{4}\) ⇒ \(\frac{k}{2}\) = \(\frac{4}{4}\)

⇒ k = 2

∴ f(x) = 2x^{2} + 5x + 2

= 2x^{2} + 4x + x + 2 [splitting middle term]

= 2x(x + 2) + 1(x + 2)(2x + 1)(x + 2)

For finding the zeroes, put f(x) = 0

⇒ (2x + 1) = 0 and (x + 2) = 0

⇒ x = –\(\frac{1}{2}\) and x = -2

Hence, the zeroes of equation are –\(\frac{1}{2}\) and -2.

Or

Find the zeroes of the polynomial p(x) = 4\(\sqrt{3}\)x^{2} + 5x – 2\(\sqrt{3}\) and verify the relationship between the zeroes and its coefficients. [5]

Answer:

Hence, this verifies the relationship between the zeroes and coefficients.

Question 35.

In the given figure, CM and RN are respectively the medians of ∆ABC and ∆PQR. If ∆ABC ~ ∆PQR, then prove that

(a) ∆AMC ~ ∆PNR

(b) \(\frac{C M}{R N}\) = \(\frac{A B}{P Q}\)

(c) ∆CMB ~ ∆RNQ [5]

Answer:

Given ∆ABC ~ ∆PQR

Section E

(Case Study Based Questions)

(Section E consists of 3 questions. All are compulsory.)

Question 36.

Flight Features

The aviation technology has evolved many upgradations in the last few years. It has taken in account, speed, direction, and distance as well as other features of the flight. Even the wind plays a vital role, when a plane travels.

Angle of elevation The angle of elevation of an object viewed is the angle formed by the line of sight with the horizontal when it is above the horizontal level.

Based on the above information, answer the following questions

(i) If the point C moves towards the point B, then how does the angle of elevation vary? (1)

(ii) Find the distance of point C from the object. (2)

Answer:

(i) f the point C moves towards the point B, then angle of elevation increases.

(ii) In right angled ∆ACB,

Or

What is the value of distance BC? (2)

(iii) If angle of elevation changes from 60° to 45°, then what will be the new distance of BC? (1)

Answer:

In ∆ABC, tan 60° = \(\frac{A B}{B C}\) ⇒ \(\sqrt{3}\) = \(\frac{1500}{B C}\)

⇒ BC = \(\frac{1500}{\sqrt{3}}\) = 500\(\sqrt{3}\) m

If angle of elevation changes from 60° to 45, then in ∆ABC,

tan 45° = \(\frac{A B}{B C}\)

⇒ 1 = \(\frac{1500}{B C}\)

⇒ BC = 1500 m

Question 37.

A School Boy

Delhi Public School society, which has so many schools in different cities of India. One of the branch of Delhi Public School is in Ghaziabad. In that school thousand of students study in the classroom.

Out of them one of the boy is standing in the ground having coordinates (4, 1) facing towards East. He moves 4 units in straight line, then turns left and moves 3 units and stops. Now, he is at his home.

The representation of the above situation on the coordinate axes is shown below

Based on the above information, answer the following questions

(i) Find the area of ∆ABC. (1)

Answer:

∴ Area of ∆ABC = \(\frac{1}{2}\) × AB × BC

= \(\frac{1}{2}\) × 4 × 3

= 6 sq units

(ii) If we draw perpendicular lines from points A and B to the X-axis, then find the region covered by these perpendicular lines. (1)

Answer:

When, we draw perpendicular lines from the points, a rectangle is formed.

∴ Area of covered region = l × b

= 4 × 1 = 4 sq units

(iii) What is the shortest distance between his school and house? (2)

Answer:

From the given figure, shortest distance between school and house = AC

= \(\sqrt{(8-4)^2+(4-1)^2}\)

[from distance formula]

= \(\sqrt{4^2+3^2}\) = \(\sqrt{16+9}\)

= \(\sqrt{25}\) = 5 units

∴ Shortest distance = 5 units

Or

Suppose point D divides the line segment AB in the ratio 1:2, then find the coordinates of D. (2)

Answer:

By using internal section formula,

The coordinates of D

= \(\left(\frac{1 \times 8+2 \times 4}{1+2}, \frac{1 \times 1+2 \times 1}{1+2}\right)\)

= \(\left(\frac{8+8}{3}, \frac{1+2}{3}\right)\) = \(\left(\frac{16}{3}, \frac{3}{3}\right)\) = \(\left(\frac{16}{3}, 1\right)\)

Question 38.

Diwali Celebration

On a Diwali occassion, colourful rangoli is formed by using different colours, diya, candles and light etc. On this occassion, we fire out the crackers and eat sweets etc. Riya made a rangoli using two different colours red and green as shown in the figure.

Based on the above information, answer the following questions

(i) Find the area of rangoli covered by both red and green colours. (1)

Answer:

∴ Required region = Area of circle with radius 6 cm

= πr^{2} = \(\frac{22}{7}\) × 6 × 6

= \(\frac{792}{7}\) cm^{2}

(ii) Find the area of rangoli covered by green colour only. (2)

Answer:

Area of sector OCD, which makes an angle of 60° at the centre

Or

Find the length of are CD of green colour. (2)

(iii) If the cost of colour used in the rangoli is ₹ 1 per cm^{2}, then find the total cost of colour used in the rangoli. (1)

Answer:

Total cost of colour used = Area of circle covered by the colour × ₹ 1 per cm^{2}

= \(\frac{792}{7}\) × 1 = \(\frac{792}{7}\) = ₹ 113.4