Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths Basic with Solutions Set 6 are designed as per the revised syllabus.
CBSE Sample Papers for Class 10 Maths Basic Set 6 with Solutions
Time: 3 Hours
Maximum Marks: 80
General Instructions:
Read the following instructions very carefully and strictly follow them:
- This question paper contains 38 questions. All questions are compulsory.
- Question paper is divided into FIVE sections – Section A, B, C, D and E.
- In section A, question number 1 to 18 are multiple choice questions (MCQs) and question number 19 and 20 are Assertion – Reason based questions ofl mark each.
- In section B, question number 21 to 25 are very short answer (VSA) type questions of 2 marks each.
- In section C, question number 26 to 31 are short answer (SA) type questions carrying 3 marks each.
- In section D, question number 32 to 35 are long answer (LA) type questions carrying 5 marks each.
- In section E, question number 36 to 38 are case based integrated units of assessment questions carrying 4 marks each. Internal choice is provided in 2 marks question in each case study.
- There is no overall choice. However, an internal choice has been provided in 2 questions in Section B, 2 questions in Section C, 2 questions in Section D and 3 questions in Section E
- Draw neat figures wherever required. Take π = 22/7 wherever required if not stated.
- Use of calculators is not allowed.
Section – A (20 Marks)
Question 1.
Let E be an event such that P(not E) = \(\frac{1}{5}\), then P(E) is equal to: 1
(a) \(\frac{1}{5}\)
(b) \(\frac{2}{5}\)
(c) 0
(d) \(\frac{4}{5}\)
Answer:
(d) \(\frac{4}{5}\)
Explanation:
Given, P(not E) = \(\frac{1}{5}\)
then P(E) = 1 – \(\frac{1}{5}\)
= \(\frac{4}{5}\) [∴ P(E) + p(not E) = 1]
Question 2.
If p(x) = x2 + 5x + 6, then p(-2) is: 1
(a) 20
(b) 0
(c) -8
(d) 8
Answer:
(b) 0
Explanation: Given,
p(x) = x2 + 5x + 6
Then, p(-2) = (-2)2 + 5 × (-2) + 6
= 4 – 10 + 6 = 0
Question 3.
The mode of the numbers 2, 3, 3, 4, 5, 4, 4, 5, 3, 4, 2, 6, 7 is: 1
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(c) 4
Explanation: Given data 2, 3, 3, 4, 5, 4, 4, 5, 3, 4, 2, 6, 7.
In the given data; 2 is 2 times, 3 is 3 times, 4 is 4 times, 5 is 2 times, 6 is 1 time and 7 is 1 time.
So, mode of the data is 4.
Question 4.
How many tangents can be drawn to a circle from a point on it? 1
(a) One
(b) Two
(c) Infinite
(d) Zero
Answer:
(a) One
Explanation: Only 1 tangent can be drawn to a circle from a point on it.
Question 5.
A quadratic equation whose one root is 2 and the sum of whose roots is zero, is: 1
(a) x2 + 4 = 0
(b) x2 – 2 = 0
(c) 4x2 – 1 = 0
(d) x2 – 4 = 0
Answer:
(d) x2 – 4 = 0
Explanation: Let roots are α and β and α = 2
Give that sum of roots is 0.
⇒ α + β
⇒ 2 + β = 0
⇒ β = -2
Now, quadratic equation is
x2 – (α + β)x + αβ = 0
On putting values,
x2 -(2 – 2)x + 2x(-2) = 0
x2 – 4 = 0
Question 6.
Which of the following is not a quadratic equation? 1
(a) 2(x – 1) = 4x2 – 2x + 1
(b) 2x – x2 = x2 + 5
(c) (√2x + √3 )2 + x2 = 3x2 – 5x
(d) (x2 + 2x)2 = x4 + 3 + 4x3
Answer:
(c) (√2x + √3)2 + x2 = 3x2 – 5x
Explanation: (√2x + √3)2 + x2 = 3x2 – 5x
⇒ 2x2 + 3 + 2√6x + x2 =3x2 – 5x
⇒ 3x2 + 3 + 2√6x = 3x2 – 5x
⇒ (2√6 + 5)x + 3 = 0,
∴ This is not a quadratic equation
Question 7.
A quadratic polynomial whose sum and product of zeroes are 2 and -1 respectively is: 1
(a) x2 + 2x + 1
(b) x2 – 2x – 1
(c) x2 + 2x – 1
(d) x2 – 2x + 1
Answer:
(b) x2 – 2x -1
Explanation: A quadratic polynomial is written as x2 – (sum of roots)x + product of roots So, required equation is x2 – 2x – 1.
Question 8.
(HCF × LCM) for the numbers 30 and 70 is: 1
(a) 2100
(b) 21
(c) 210
(d) 70
Answer:
(a) 2100
Explanation: Product of two numbers
= (LCM × HCF) of the numbers
So, (HCF × LCM) of 30 and 70 = 30 × 70
= 2100
Question 9.
The length of the arc of a circle of radius 14 cm which subtands an angle of 60° at the centre of the circle is: 1
(a) \(\frac{44}{3}\) cm
(b) \(\frac{88}{3}\) cm
(c) \(\frac{308{3}\) cm
(d) \(\frac{616}{3}\) cm
Answer:
(a) \(\frac{44}{3}\) cm
Explanation: Since, length of the arc of centre angle θ = \(\frac{\theta}{360}\) × 2πr
Length of the arc = \(\frac{60}{360}\) × \(\frac{2 \times 22}{7}\) × 14
\(\frac{44}{3}\) cm
Question 10.
If the radius of a semi-circular protractor is 7 cm, then its perimeter is: 1
(a) 11 cm
(b) 14 cm
(c) 22 cm
(d) 36 cm
Answer:
(d) 36 cm
Explanation:
Perimeter of semi-circular protractor
Length of arc AB + Length of AB
Given, radius = 7 cm,
So, AB = 7 cm + 7 cm
= 14 cm
Now, circumference of circle = 2πr
= 2 × \(\frac{22}{7}\) × 7
= 44 cm
Length of arc
AB = \(\frac{44}{2}\) = 22 cm.
Hence, perimeter of semi-circular protractor
= 22 cm + 14 cm
= 36 cm
Question 11.
The angle of elevation of the top of a 15 m high tower at a point 15 √3 m away from the base of the tower is: 1
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Answer:
(a) 30°
Explanation:
Let AB be the tower of height 15 m and C be the point 15 √3 m away from the base of the tower, making elevation of the tower angle 0 of the top.
AB
Now, in triangle ABC, tanθ = \(\frac{A B}{B C}\)
⇒ tanθ = \(\frac{15}{15 \sqrt{3}}=\frac{1}{\sqrt{3}}\)
⇒ tanθ = tan 30°
⇒ θ = 30°
Question 12.
(\(\frac{2}{3}\)sin0°- \(\frac{4}{5}\)cos0°) is equal to: 1
(a) \(\frac{2}{3}\)
(b) \(\frac{-4}{5}\)
(c) 0
(d) \(\frac{-2}{15}\)
Answer:
(b) \(\frac{-4}{5}\)
Explanation:
(\(\frac{2}{3}\)sin0° – \(\frac{4}{5}\)cos0°) = \(\frac{2}{3}\) × 0 – \(\frac{4}{5}\) × 1) = \(\frac{-4}{5}\) [∵ sin0° = 0, cos0° = 1]
Question 13.
From a well-shuffled deck of 52 cards, a card is drawn at random . What is the probability of getting king of hearts? 1
(a) \(\frac{1}{52}\)
(b) \(\frac{1}{26}\)
(c) \(\frac{1}{13}\)
(d) \(\frac{12}{13}\)
Answer:
(a) \(\frac{1}{52}\)
Explanation: Total number of cards = 52
Number of king of hearts in the deck = 1
So, probability of getting a king of hearts = \(\frac{1}{52}\)
Question 14.
The number (5 – 3 √5 + √5 ) is: 1
(a) an integer
(b) a rational number
(c) an irrational number
(d) a whole number
Answer:
(c) an irrational number
Explanation: (5 – 3√5 + √5) = (5 – 2√5)
∴ √5 is an irrational number so, 2√5 is an irrational number.
Also, (5 – 2√5) or (5 – 3√5 + √5) is an irrational number.
Question 15.
If the pair of linear equations x – y = 1, x + ky = 5 has a unique solution x = 2, y = 1, then the value of k is: 1
(a) -2
(b) -3
(c) 3
(d) 4
Answer:
(c) 3
Explanation: Given equations are x – y = 1 and x + ky = 5.
Given solutions are x = 2, y = 1, these solutions must satisfy given equations.
So, putting x = 2, y = 1 in x + ky = 5,
we get;
2 + k = 5
⇒ k = 3
Also, equations has unique soLution, then:
\(\frac{1}{1} \neq \frac{-1}{k}\)
⇒ k ≠ -1, so k = 3
Question 16.
If ΔABC ~ ΔDEF and ΔA = 47°, ΔE = 83°, then ΔC is equal: 1
(a) 47°
(b) 50°
(c) 83°
(d) 130°
Answer:
(b) 50°
Explanation: Given,
ΔABC ~ ΔDEF
So, corresponding angles are equal, that is,
∠A = ∠D, ∠B = ∠E, ∠C = ∠F
Given, ∠A = ∠D = 47°
and ∠B = ∠E = 83°
In ΔABC,
∠A + ∠B + ∠C, = 180 [Property and triangle]
⇒ 47° + 83° + ∠C = 180°
⇒ 130° + ∠C = 180°
⇒ ∠C = 180° – 130° = 50°
Question 17.
The length of the tangent from an external point A to a circle of radius 3 cm, is 4 cm. The distance of A form the centre of the circle is: 1
(a) 7 cm
(b) 5 cm
(c) √7 cm
(d) 25 cm
Answer:
(b) 5cm
Explanation:
Let OB be the radius of circle and AB be the tangent to the circle from point A.
Tangent to a circle makes right angle with radius at the point of contant.
So, ∠ABO = 90°
In ΔABO, AB2 + OB2 = OA2
⇒ 42 + 32 = OA2
⇒ 16 + 9 = OA2
⇒ 25 = OA2
OA = 5 cm
Question 18.
The pair of linear equations x + 2y + 5 = 0 and – 3x – 6y + 1 = 0 has: 1
(a) a unique solution
(b) exactly two solutions
(c) infinitely many solutions
(d) no solution
Answer:
(d) no solution
Explanation: Given equations are
x + 2y + 5 = 0
and – 3x – 6y + 1 = 0
Here, a1 = 1, b1 = 2, c1 = 5
and a2 = -3, b2 = -6, c2 = 1
Now, \(\frac{a_1}{a_2}=\frac{1}{-3}=-\frac{1}{3}\);
\(\frac{b_1}{b_2}=\frac{2}{-6}=-\frac{1}{3}\);
\(\frac{c_1}{c_2}=\frac{5}{1}\) = 5
Since, \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\), so equations have no solution.
Assertion-Reason type questions
In question numbers 19 and 20, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option:
(a) Both Assertion (A) and Reason (R) are true and Reason (R) gives the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) does not give 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.
Question 19.
Assertion (A): If one root of the quadratic equation 4x2 – 10x + (k – 4) = 0 is reciprocal of the other, then value of k is 8.
Reason (R): Roots of the quadratic equation x2 – x + 1 = 0 are real. 1
Answer:
(b) Both Assertion (A) and Reason (R) are true but Reason (R) does not give the correct explanation of Assertion (A).
Explanation: Equation is 4x2 – 10 + (k – 4) = 0 Given that one root is reciprocal of the other.
Let one root be a then other root be \(\frac{1}{\alpha}\).
Since product of roots = \(\frac{\text { constant term }}{\text { coefficient of } x^2}\)
∴ α × \(\frac{1}{\alpha}=\frac{k-4}{4}\)
⇒ 1 = \(\frac{k-4}{4}\)
⇒ 4 = k – 4
⇒ k = 8
For equation
x2 – x + 1 = 0,
b2 – 4ac = (-1)2 – 4 × 1 × 1
= 1 – 4 = -3 < 0
so roots are imaginary.
Question 20.
Assertion (A): A tangent to a circle is perpendicular to the radius through the point of contact.
Reason (R): The lengths of tangents drawn from an external point to a circle are equal. 1
Answer:
(b) Both Assertion (A) and Reason (R) are true but Reason (R) does not give the correct explanation of Assertion (A).
Explanation: We know that a tangent to a circle is perpendicular to the radius through the point of contact and the lengths of tangents drawn from an external point to a circle are equal. Hence both statements are true.
Section – B (10 Marks)
Question 21.
Find the discriminant of the quadratic equation 3x2 – 2x + \(\frac{1}{3}\) = 0 and hence find the nature of its roots.
OR
Find the root of the quadratic equation x2 – x – 2 = 0. 2
Answer:
Given equation:
3x2 – 2x + \(\frac{1}{3}\) = 0
Discriminant D = b2 – 4ac
= (-2)2 – 4 × 3 × \(\frac{1}{3}\)
= 4 – 4
= 0
So, roots are real and equal.
OR
Given equation:
x2 – x – 2 = 0
⇒ x2 – 2x + x- 2= 0
⇒ x(x – 2)x + 1(x – 2) = 0
⇒ (x – 2) (x + 1) = 0
⇒ x = 2, -1
So, roots of the equation x2 – x – 2 = 0 are 2 and -1.
Question 22.
In the adjoining figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR. 2
Answer:
Question 23.
if sin α = \(\frac{1}{2}\), then find the value of (3 cos α – 4 cos3 α). 2
Answer:
Given, sinα = \(\frac{1}{2}\)
⇒ sin α = sin 30°
⇒ α = 30°
Now, (3cosα – 4cos3α) = (3cos 30° – 4cos3 30°)
Question 24.
Find the coordinates of the point which divides the join of A (-1, 7) and B (4, -3) in the ratio 2 : 3.
OR
If the points A (2, 3), B (-5, 6), C (6, 7) and D (p, 4) are the vertices of a parellelogram ABCD, find the value of p. 2
Answer:
Let C (x, y) divides the Line segment joining the points A (-1, 7) and B (4, – 3) in the ratio 2 : 3. Thus, the point C is,
[latex]\frac{2 \times 4+3 \times(-1)}{2+3}, \frac{2 \times(-3)+3 \times 7}{2+3}[/latex]
C[latex]\frac{8-3}{5}, \frac{-6+21}{5}[/latex] = C(1, 3)
So, coordinates of the point are (1, 3).
OR
Given that ABCD is a parallelogram.
AB and CD are parallel, and AB = CD.
Since, diagonals of a parallelogram, bisect each other.
∴ Mid-point of diagonal BD = Mid-point of diagonal.
∴ (\(\frac{-5+p}{2}, \frac{6+4}{2}\)) = (\(\frac{2+6}{2}, \frac{3+7}{2}\))
\(\frac{-5+p}{2}\) = \(\frac{2+6}{2}\)
⇒ p = 3
Question 25.
PA and PB are tangents drawn to the circle with centre O as shown in the figure .
Prove that √APB = 2 √OAB. 2
Answer:
Since, ∠OAP = 90° and ∠OBP = 90°
[∵ Angle between tangent and radius at the point of contact is 90°]
In quadrilateral OAPB,
∠OAP + ∠APS + ∠OBP + ∠AOB = 360°
[Angles sum property of quadrilateral]
⇒ 90° + ∠APB + 90° + ∠AOB = 360°
⇒ ∠APB + ∠AOB = 360° – 180°= 180° …….. (i)
In ΔOAB, we have OA = OB [Radii of circle]
⇒ ∠OAB = ∠OBA
[∵ Angles opposite to equal sides of a triangle are equal]
Also, ∠OAB + ∠OBA + ∠AOB = 180°
[Angle sum property of triangle]
⇒ 2∠OAB + ∠AOB = 180° ……. (ii)
Equating equations (i) and (ii),
∠APB + ∠AOB = 2∠OAB + ∠AOB
∠APB = 2∠OAB
Section – C 18 (Marks)
Question 26.
Find the area of the sector of a circle of radius 7 cm and of central angle 90°. Also, find the area of corresponding major sector. 3
Answer:
Radius of circle, OB = OA = 7 cm, and ∠AOB = 90°
AOB is a quadrant of circle.
So, area of sector AOB
= \(\frac{90^{\circ}}{360^{\circ}}\) × πr2
= \(\frac{1}{4}\)πr2
= \(\frac{1}{4}\) × \(\frac{22}{7}\) × 72
= \(\frac{77}{2}\) = 38.5 cm2
Area of major sector AOB = Area of circle – Area of quadrant
Area of circle = πr2 = \(\frac{22}{7}\) × 72
= 154 cm2
Now, area of major sector
= 154 cm2 – 38.5 cm2
= 115.5 cm2
Question 27.
If α, β are zeroes of the quadratic polynomial x2 – 5x + 6, form another quadratic polynomial whose zeroes are \(\frac{1}{\alpha}, \frac{1}{\beta}\). 3
Answer:
Given that α and β are zeroes of x2 – 5x + 6 = 0
x2 – 5x + 6 = 0
⇒ x2 – 3x – 2x + 6 = 0
⇒ x(x – 3) – 2(x – 3) = 0
⇒ (x – 3)(x – 2) = 0
⇒ x = 3, 2
Now, α = 3 and β = 2 then \(\frac{1}{\alpha}=\frac{1}{3}\) and \(\frac{1}{\beta}=\frac{1}{2}\)
Required equation: x2 – (\(\frac{1}{3}\) + \(\frac{1}{2}\))x + \(\frac{1}{2}\) × \(\frac{1}{2}\) = 0
⇒ x2 – \(\frac{5}{6}\)x + \(\frac{1}{6}\) = 0
⇒ 6x2 – 5x + 1 = 0
Question 28.
A die is rolled once. Find the probability of getting:
(A) an even prime number.
(B) a number greater than 4.
(C) an odd number. 3
Answer:
Numbers on a die: 1, 2, 3, 4, 5, 6
(A) Probability of getting an even prime number, that is, 2 = \(\frac{1}{6}\).
(B) Probability of getting a number greater than 4, that is, 5 and 6 = \(\frac{2}{6}\) = \(\frac{1}{3}\).
(C) Probability of getting a odd number, that is, 1, 3, 5= \(\frac{3}{6}\) = \(\frac{1}{2}\).
Question 29.
Prove that \(\frac{1+\tan ^2 \mathrm{~A}}{1+\cot ^2 \mathrm{~A}}\) = sec2 A – 1. 3
Answer:
Question 30.
Prove that the lengths of tangents drawn from an external point to a circle are equal.
OR
Two concentric circles with centre O are of radii 3 cm and 5 cm. Find the length of chord AB of the larger circle which touches the smaller circle at P. 3
Answer:
Let PQ and PR be two tangents, drawn from an external point P to a circle C (0, r).
To prove: PQ = PR.
Construction: Join OP, OQ and OR.
Proof: Since, the tangent at any point of a circle is perpendicular to the radius through the point of contact.
∴ ∠OQP = ∠ORP = 90°.
Now in right triangles OQP and ORP, we have
OQ = OR [Radii of the same circle]
∠OQP = ∠ORP [Each 90°]
OP = OP [Common]
ΔOQP ≅ ΔORP
[By RHS theorem of congruence]
⇒ PQ = PR
OR
In the figure, O is the common centre, of the given concentric circles.
AB is a chord of the bigger circle such that it is a tangent to the smaller circle at P. Since, OP is the radius of the smaller circle through P, the point of contact
∴ OP ⊥ AB
⇒ ∠APO = 90°
Also, a radius perpendicular to a chord bisects the chord.
∴ OP bisects AB
⇒ AP = \(\frac{1}{2}\) AB
Now, in right ΔAPO, OA2 = AP2 + OP2
⇒ 52 = AP2 + 32
⇒ AP2 = 52 – 32
⇒ AP2 = (5 – 3) (5 + 3) = 2 × 8
⇒ AP2 = 16 = (4)2
⇒ AP = 4 cm
⇒ \(\frac{1}{2}\)AB = 4
⇒ AB = 2 × 4 = 8 cm
Thus, the required length of the chord AB is 8 cm.
Question 31.
If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes \(\frac{1}{2}\) if we only add 1 to the denominator. What is the fraction?
OR
For which value of ‘k’ will the following pair of linear equations have no solution?
3x + y = 1
(2k – 1)x + (k – 1)y = 2k + 1. 3
Answer:
Let x be the numerator and y be the denominator of the fraction. Then the fraction x is \(\frac{x}{y}\).
Now, according to the given condition,
\(\frac{x+1}{y-1}\) = 1 ……….. (i)
Also, \(\frac{x}{y+1}\) = \(\frac{1}{2}\) ……… (ii)
From equation (i), x + 1 = y – 1
⇒ x – y = -2 ……… (iii)
From equation (ii), 2x = y + 1
⇒ 2x – y = 1 ……… (iv)
Subtracting equation (iii) from (iv), we getx
= 3
Putting value of x in (iii), we get
3 – y = – 2
⇒ y = s
∴ Required fraction is \(\frac{3}{5}\).
Given equations are 3x + y = 1
or 3x + y – 1 = 0
and (2k – 1) x + (k – 1) y + 1 = 2k + 1
or (2k – 1)x + (k – 1)y + 1 – (2k + 1) = 0
Here, a1 = 3, b1 = 1, c1 = – 1
and a2 = (2k – 1),
b2 = (k – 1), c2 = – (2k + 1)
For no solution,
Section – D (20 Marks)
Question 32.
Find the sum of first 51 terms of an A.P. whose second and third terms are 14 and 18, respectively.
OR
The first term of an A.P. is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference. 5
Answer:
Given, second and third terms of the A.P. are 14 and 18.
Let a be the first term and d be the common difference.
So, a + d = 14 …….. (i)
and a + 2d = 18 …….. (ii)
respectively subtracting equation (i) from (ii),
d = 4
Putting value of d in (i), we get
a + 4 = 14
⇒ a = 14 – 4
= 10
Sum of first n terms of an AP having first term a and common difference d, is:
Sn = \(\frac{n}{2}\)[2o + (n – 1)d]
Here, n = 51, a = 10 and d = 4
Putting values, Sn = \(\frac{51}{2}\)[2 × 10 + (51 – 1) × 4]
= \(\frac{51}{2}\)[20 + 200]
= \(\frac{51}{2}\) × 220
= 51 × 110
= 5610
OR
Given, a = 5 l = 45, and Sn = 400
We have. Sn = \(\frac{n}{2}\)(a + l)
Putting vaLues, 400 = \(\frac{n}{2}\)(5 + 45)
⇒ 400 = \(\frac{n}{2}\) × 50 = 25n
⇒ n = \(\frac{400}{25}\) = 16
Now, l = a + (n – 1)d
Putting values. 45 = 5 + (16 – 1)d
⇒ d = \(\frac{40}{15}\) = \(\frac{8}{3}\)
So, common difference = \(\frac{8}{3}\) and number of terms = 16.
Question 33.
The distribution below gives the weights of 30 students of a class. Find the median weight of the students: 5
Weight in kg | Number of Students |
40 – 45 | 2 |
45 – 50 | 3 |
50 – 55 | 8 |
55 – 60 | 6 |
60 – 65 | 6 |
65 – 70 | 3 |
70 – 75 | 2 |
Answer:
Weight (in kg) (Cl.) | Number of students (fi) | Cumulative frequency (cf) |
40 – 45 | 2 | 2 |
45 – 50 | 3 | 5 (= 2 + 3) |
50 – 55 | 8 | 13 (= 5 + 8) |
55 – 60 | 6 | 19 (= 13 + 6) |
60 – 65 | 6 | 25 (= 19 + 6) |
65 – 70 | 3 | 28 (= 25 + 3) |
70 – 75 | 2 | 30 (= 28 + 2) |
Here, Median number: \(\frac{N}{2}=\frac{30}{2}\) = 15
cf. just > median number (15) = 19
⇒ Median class = The class interval corresponding to c.f. of 19
⇒ Median class = 55 – 60
Lower limit of the median class, l = 55
Frequency of the median class, fm = 6
Size of the median class, h = 5 (= 60 – 55)
Cumulative frequency of the pre-median class, C = 13
We know that,
Median = l + \(\frac{h}{f_m}\)(\(\frac{N}{2}\) – C)
Median = 55 + \(\frac{5}{6}\) × (15 – 13)
= 55 + \(\frac{5}{6}\) × 2
= 55 + \(\frac{5}{3}\)
= 50 + 1.67 = 51.67
Thus, the required median weight of students is 51.67 kg.
Question 34.
The boilers are used in thermal power plants to store water and then used to produce steam. One such boiler consists of a cylindrical part in middle and two hemispherical parts at its both ends.
Length of the cylindrical part is 7 m and radius of cylindrical part is \(\frac{1}{2}\) m.
Find the total surface area and the volume of the boiler. Also, find the ratio of the volume of cylindrical part to the volume of one hemispherical part. 5
Answer:
Given: Radius of hemispherical ends, r = \(\frac{7}{2}\) m
We know that, surface area of each hemispherical end
= 2πr2
= 2 × \(\frac{22}{7}\) × \(\frac{7}{2}\) × \(\frac{7}{2}\) = 77 m2
∴ Surface area of both hemispherical ends
= 2 × 77 m2
= 154 m2
Length of cylindrical part = 7 m
Curved surface area of the cylindrical portion
= 2πrh
= 2 × \(\frac{22}{7}\) × \(\frac{7}{2}\) × 7
= 154 m2
Total surface area of the boiler
= 154 m2 + 154 m2
= 308 m2
Thus, the required surface area is 308 m2.
Volume of boiler = Volume of cylinder + Volume of 2 hemisphere
Question 35.
The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun’s altitude is 30° than when it was 60°. Find the height of the tower.
OR
From the top of a 7m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower. 5
Answer:
Putting value of h of equation (i) in equation (ii),
x + 40 = √3 × √3x
⇒ x + 40 = 3x
⇒ 2x = 40
⇒ x = 20 m
From (i), h = √3 × 20
= 20√3m
Thus, the height of the tower is 2o√3 m.
OR
Let the height of the tower = AB = (h + 7) m
CD = 7 m (height of building),
∠ACE = 60° and ∠ECB = 45°
In ΔCDB, \(\frac{\mathrm{CD}}{\mathrm{DB}}\) = tan 45°
⇒ \(\frac{7}{\mathrm{DB}}\) = 1
⇒ DB = 7 M
In ΔAEC, \(\frac{\mathrm{AE}}{\mathrm{CE}}\) = tan 60°
⇒ \(\frac{h}{7}\) = √3
⇒ h = 7√3 m
[Since, DB = CE = 7 m]
∴ Height of the tower = 7(√3 + 1) m.
Section – E (12 Marks)
Question 36.
Observe the figures given below carefully and answer the questions:
(A) Name the figure(s) wherein two figures are similar. 1
(B) Name the figure(s) where in the figures are congruent. 1
(C) Prove that congruent triangles are
also similar but not the converse.
OR
What more is least needed for two similar triangles to be congruent? 2
Answer:
(A) In figure A both quadriLateral are similar; and in figure C both triangles are similar.
(B) In figure C, both triangles are congruent.
(C) Congruent triangles are similar because for each side in one triangle, there is corresponding equal side in another. So, sides are in proportion, therefore congruent triangles are also similar.
But converse is not true. In similar triangles sides are in same proportion but not equal in length.
OR
Two pairs of corresponding angles are equal.
Three pairs of corresponding sides are proportional.
Two pairs of corresponding sides are equal and the corresponding angles between them are equal.
Question 37.
Ues of mobile screen for long hours makes your eye sight weak and give you headaches. Children who are addicted to play “PUBG” can get easily stressed out. To raise social awareness about ill effects of playing PUBG, a school decided to start BAN PUBG’ campaign, in which students are asked to prepare campaign board in the shape of a rectangle. One such campaign board made by class X student of the school is shown in the figure.
Based on the above information, answer the following questions:
(A) Find the coordinates of the point of intersection of diagonals AC and BD. 1
(B) Find the length of the diagonal AC. 1
(C) Find the area of the campaign Board ABCD.
OR
Find the ratio of the length of side AB to the length of the diagonal AC. 2
Answer:
Given coordinates are A (1, 1), B (7, 1), C (7, 5) and D (1, 5).
(A) Since ABCD are point of a rectangle, therefore point of intersection of diagonals AC and BD is mid-point of AC and BD.
Mid-point of AC = (\(\frac{1+7}{2}, \frac{1+5}{2}\)) = (4, 3)
Mid-point of BD = (\(\frac{7+1}{2}, \frac{1+5}{2}\)) = (4, 3)
Therefore, coordinates of the point of intersection of diagonals AC and BD are (4, 3)
(B) Using distance formula
Length of diagonal
AC = \(\sqrt{(7-1)^2+(5-1)^2}\)
= \(\sqrt{6^2+4^2}=\sqrt{36+16}\)
= \(\sqrt{52}\) units
= 2\(\sqrt{13}\) units
(C) Area of campaign board = 6 × 4
= 24 sq. units
OR
Length of side AB = \(\sqrt{(7-1)^2+(1-1)^2}\)
= \(\sqrt{6^2}\) = 6 units
Ratio of length of side AB to the length of diagonal
AC = 3 : \(\sqrt{13}\) = 3\(\sqrt{13}\) : 13
Question 38.
Khushi wants to organise her birthday party. Being health conscious, she decided to serve only fruits in her birthday party. She bought 36 apples and 60 bananas and decided to distribute fruits equally among all.
Based on the above information, answer the following questions:
(A) How many guests Khushi can invite at the most? 1
(B) How many apples and bananas will each guest get? 1
(C) If Khushi decides to add 42 mangoes, how many guests Khushi can invite at the most?
OR
If the cost of 1 dozen of bananas is ₹ 60, the cost of 1 apple is ₹ 15 and cost of 1 mango is ₹ 20, find the total amount spent on 60 bananas, 36 apples and 42 mangoes. 2
Answer:
Khushi bought 36 apples and 60 bananas and she decided to distribute fruits equally among all guest.
(A) We need to find HCF of 36 and 60.
36 = 2 × 2 × 3 × 3;
60 = 2 × 2 × 3 × 5
HCF of 36 and 60 = 2 × 2 × 3 = 12
Therefore, Khushi can invite at most 12 guests.
(B) Each guest will get \(\frac{36}{12}\) = 3 apples and \(\frac{60}{12}\) = 5 bananas.
(C) If khusi add 42 mangoes, then maximum number guests she can invite = HCF of (36, 60 and 42)
36 = 2 × 2 × 3 × 3;
60 = 2 × 2 × 3 × 5;
42 = 2 × 3 × 7
HCF of (36, 60 and 42) = 2 × 3 = 6
Therefore, Khushi can invite at most 6 guests.
OR
Cost of 1 dozen banana is ₹ 60, then cost of 5 dozen bananas = 60 × 5 = ₹ 300
Cost of 1 apple is ₹ 15, then cost of 36 apples
= 15 × 36 = ₹ 540
Cost of 1 mango is ₹ 20, then cost of 42 mangoes = 20 × 42 = ₹ 840
Hence, total amount spent
= 300 + 540 + 840
= ₹ 1680