Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths Standard with Solutions Set 1 are designed as per the revised syllabus.
CBSE Sample Papers for Class 10 Maths Standard Set 1 with Solutions
Time: 3 Hours
Maximum Marks: 80
General Instructions:
- This Question Paper has 5 Sections A, B, C, D and 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) with sub-parts of the values of 1, 1 and 2 marks each respectively.
- All questions are compulsory. However, an internal choice in 2 question of 5 marks, 2question of 3 marks and 2 questions of 2 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.
It two positive integers a and b are written as a = x3y2 and b = xy3, where x, y are prime numbers, then the result obtained by dividing the product of the positive integers by the LCM (a, b) is: [1]
(a) xy
(b) xy2
(c) x3y3
(d) x2y2
Answer:
(b) xy2
Explanation:
Product = a × b
= x3y2 × y3
= x4y5
LCM (a, b) = x3y3
= \(\frac{\text { Product }}{\text { LCM }}=\frac{x^4 y^5}{x^3 y^3}\)
= xy2
Question 2.
The given linear polynomial y = f(x) has: [1]
(a) 2 zeros
(b) 1 zero and the zero is ‘3
(c) 1 zero and the zero is ‘4‘
(d) No zero
Answer:
(b) 1 zero and the zero is ‘3‘
Explanation: The given polynomial y = f(x) has y = 0, at x = 3
Because, given polynomial f(x) and X-axis intersect at point (3, 0).
Question 3.
The given pair of linear equations is non-intersecting. Which of the following statements is true? [1]
(a) \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)
(b) \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
(c) \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}=\frac{c_1}{c_2}\)
(d) \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
Answer:
(b) \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
Explanation: If two lines in the same plane do not intersect.
Then they must be parallel. There exists no solution.
Algebraically for such a case.
\(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
Question 4.
Write the nature of roots of the quadratic equation 9x2 – 6x – 2 = 0.
(a) No real roots
(b) 2 equal real roots
(c) 2 distinct real roots
(d) More than 2 real roots
Answer:
(c) 2 distinct real roots
Explanation: The given quadratic equation:
9x2 – 6x – 2 = 0
Here, a = 9, b = -6, c = -2
Discriminant, D = b2 – 4ac
D = (-6)2 – 4(9) × (- 2)
D = 36 + 72 = 108
∵ D = 108 > 0. The roots are real and distinct.
Question 5.
Two APs have the same common difference. The first term of one of these is -1 and that of the other is -8. Then the difference between their 4th terms is: [1]
(a) 1
(b) -7
(c) 7
(d) 9
Answer:
(c) 7
Explanation: If two APs have the same common difference. Then,
Difference between their 4th terms = Difference between their 1st term
= – 1 – (- 8)
= – 1 + 8 = 7
Question 6.
Find the ratio in which the line segment joining (2, – 3) and (5, 6) is divided by X-axis. [1]
(a) 1 : 2
(b) 2 : 1
(c) 2 : 5
(d) 5 : 2
Answer:
(a) 1 : 2
Explanation: Let the line joining points A(2, – 3) and B(5, 6) be divided by X-axis at point P(x, 0) in the ratio (k : 1).
Here, x1 = 2, y1 = -3, x2 = 5, y2 = 6, m = k, n = 1
y = \(\frac{m y_2+n y_1}{m+n}\)
0 = \(\frac{k \times 6+1 \times-3}{k+1}\)
0 = 6k – 3
k = \(\frac{1}{2}\)
Thus, the required ratio is 1 : 2.
Question 7.
(x, y) is 5 unit from the origin. How many such points lie in the third quadrant? [1]
(a) 0
(b) 1
(c) 2
(d) infinitely many
Answer:
(d) infinitely many
Question 8.
In ΔABC, DE || AB. If AB = a, DE = x, BE = b and EC = c. Express x in terms of a, b and c. [1]
(a) \(\frac{a c}{b}\)
(b) \(\frac{a c}{b+c}\)
(c) \(\frac{a b}{b}\)
(d) \(\frac{a b}{b+c}\)
Answer:
(b) \(\frac{a c}{b+c}\)
Explanation: ΔCED and ΔCBA
∠CED = ∠CBA (DE || AB)
∠DCE = ∠ACB (Common)
∴ ΔCED ~ ΔCBA (By AA similarity)
⇒ \(\frac{\mathrm{EC}}{\mathrm{BC}}=\frac{\mathrm{DE}}{\mathrm{AB}}\) ….. (i)
According to question,
AB = a, DE = x, BE = b, EC = c
But BC = BE + EC
⇒ BC = b + c
Put the values in equation (i),
⇒ \(\frac{c}{b+c}=\frac{x}{a}\)
⇒ x = \(\frac{a c}{b+c}\)
Question 9.
If O is centre of a circle and chord PQ makes an angle 50° with the tangent PR at the point of contact P, find the angle made by the chord at the centre. [1]
(a) 130°
(b) 100°
(c) 50°
(d) 30°
Answer:
(b) 100°
Explanation: Since OP is perpendicular to PR.
∠OPR = 90°
∠OPR + ∠QPR =90°
∠OPQ + 50° = 90° (Given ∠QPR = 50°)
∠OPQ = 40°
∵ OP = OP = Radius of circle
∴ ∠OPQ = ∠OQP = 40° (Since, angles opposite to equal sides are equal)
Now in ΔOPQ,
∠POQ + ∠OPQ + ∠OQP = 180° (Angle sum property of triangle)
∠POQ + 40° + 40° = 180°
∠POQ = 100°
Question 10.
Quadrilateral PQRS is drawn to circumscribe a circle.
If PQ = 12 cm, QR = 15 cm and RS = 14 cm, find the length of SP: [1]
(a) 15 cm
(b) 14 cm
(c) 12 cm
(d) 11cm
Answer:
(d) 11 cm
Explanation: Quadrilateral PQRS is drawn to circumscribe a circle.
⇒ PQ, QR, RS and PS are tangent to circle.
∵ Length of tangents drawn to a circle from an external point are equal.
PA = PD …… (i)
AQ = QB …… (ii)
CR = BR ….. (iii)
SC = SD …… (iv)
Adding equations (i), (ii), (iii) and (iv), we get
(PA + AQ) + (CR + SC) = (PD + SD) + (BR + QB)
PQ + RS = PS + QR
Given, PQ = 12 cm, QR = 15 cm, RS = 14 cm, SP = ?
12 + 14 = SP + 15
26 – 15 = SP
SP = 11 cm
Question 11.
Given that sin θ = \(\frac{a}{b}\), find cos θ. [1]
(a) \(\frac{b}{\sqrt{b^2-a^2}}\)
(b) \(\frac{b}{a}\)
(c) \(\frac{\sqrt{b^2-a^2}}{b}\)
(d) \(\frac{a}{\sqrt{b^2-a^2}}\)
Explanation: sin θ = \(\frac{a}{b}\)
we know that, cos θ = \(\sqrt{1-\sin ^2 \theta}\)
cos θ = \(\sqrt{1-\frac{a^2}{b^2}}\)
cos θ = \(\sqrt{\frac{b^2-a^2}{b^2}}=\frac{\sqrt{b^2-a^2}}{b}\)
Question 12.
(sec A + tan A) (1 – sin A) = [1]
(a) sec A
(b) sin A
(c) cosec A
(d) cos A
Answer:
(d) cos A
Explanation:
(sec A + tan A) (1 – sin A) = (\(\frac{1}{\cos A}+\frac{\sin A}{\cos A}\)) (1 – sin A)
= \(\frac{(1+\sin A)}{\cos A} \times \frac{(1-\sin A)}{1}\)
= \(\frac{1-\sin ^2 A}{\cos A}=\frac{\cos ^2 A}{\cos A}\) = cos A (∵ 1 – sin2A = cos2A)
Question 13.
A pole 6 m high casts a shadow 2√3 m long on the ground, then the Sun’s elevation is : [1]
(a) 60°
(b) 45°
(c) 30°
(d) 90°
Answer:
(a) 60°
Explanation: Suppose that,
Pole = AB, Sun’s elevation = θ, Shadow = BC
Given: AB = 6 m, BC = 2√3
In ΔABC, tan θ = \(\frac{A B}{B C}\)
⇒ tan θ = \(\frac{6}{2 \sqrt{3}}\)
⇒ tan θ = \(\frac{3}{\sqrt{3}}\)
⇒ tan θ = \(\frac{\sqrt{3} \times \sqrt{3}}{\sqrt{3}}\)
⇒ tan θ = √3
⇒ tan θ = tan 60°
⇒ θ = tan 60°
Question 14.
If the perimeter and the area of a circle are numerically equal, then the radius of the circle is: [1]
(a) 2 units
(b) π units
(c) 4 units
(d) 7 units
Answer:
(a) 2 units
Explanation: According to question,
Area of circle = Perimeter of circle
πr2 = 2πr (∵ r = Radius of circle)
\(\frac{r^2}{r}=\frac{2 \pi}{\pi}\)
⇒ r = 2 units
Question 15.
It is proposed to build a single circular park equal in area to the sum of areas of two circular parks of diameters 16 m and 12 m in a locality. The radius of the new park is: [1]
(a) 10 m
(b) 15 m
(c) 20 m
(d) 24 m
Answer:
(a) 10 m
Explanation: Given:
Diameter of I circular park
2r1 = 16 m
r1 = 8m
Diameter of II circular park
2r2 = 12 m
r2 6 m
Let be radius of new circular park = r
According to question,
Area of new circular park = Area of I circular park + Area of II circular park
πr2 = πr12 + πr22
πr2 = π(r12 + r22
r2 = (82 + 62)
r2 = 64 + 36 = 100 m2
r = 10 m
Question 16.
There is a green square board of side ‘2a’ unit circumscribing a red circle. Jayadev is asked to keep a dot
(a) \(\frac{\pi}{4}\)
(b) \(\frac{4-\pi}{4}\)
(c) \(\frac{\pi-4}{4}\)
(d) \(\frac{4}{\pi}\)
Answer:
(b) \(\frac{4-\pi}{4}\)
Explanation:
Area of green square board = 2a × 2a = 4a2
Thus, n(s) = 4a2
A = Area of abovesaid board
n(A) = Area of green square – Area of red circle
= 4a2 – πa2
= (4 – π)a2
Required probability, P(A) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\)
P(A) = \(\frac{(4-\pi) a^2}{4 a^2}\)
P(A) = \(\frac{4-\pi}{4}\)
Question 17.
2 cards of hearts and 4 cards of spades are missing from a pack of 52 cards. What is the probability of
getting a black card from the remaining pack?
(a) \(\frac{22}{52}\)
(b) \(\frac{22}{46}\)
(c) \(\frac{24}{52}\)
(d) \(\frac{24}{46}\)
Answer:
(b) \(\frac{22}{46}\)
Explanation:
P = \(\frac{\text { Favourable outcome }}{\text { Total outcome }}=\frac{26-4}{25-6}\)
= \(\frac{22}{46}\)
Question 18.
Find the upper limit of the modal class from the given distribution. [1]
(a) 165
(b) 160
(c) 155
(d) 150
Answer:
(d) 150
Explanation: According to question, Frequencies distribution becomes.
Maximum frequency = 18
Hence, Modal Class = (145 – 150)
Upper limit of modal class = 150
DIRECTION: In the question number 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) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true and 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.
Question 19.
Statement A (Assertion): Total surface area of the top is the sum of the curved surface area of the hemisphere and the curved surface area of the cone.
Statement R (Reason): Top is obtained by fixing the plane surfaces of the hemisphere and cone together. [1]
Answer:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
Question 20.
Statement A (Assertion): -5, \(\frac{-5}{2}\), 0, \(\frac{5}{2}\) ….. is in Arithmetic Progression.
Statement R (Reason): The terms of an Arithmetic Progression cannot have both positive and negative rational numbers. [1]
Answer:
(c) Assertion (A) is true, but reason (R) is false.
Explanation: ∵ Common difference between the consecutive terms are equal in A.P.
T2 – T1 = T3 – T2 = T4 – T3
\(\frac{-5}{2}\) + 5 = 0 + \(\frac{5}{2}\) = \(\frac{5}{2}\) – 0
\(\frac{5}{2}\) = \(\frac{5}{2}\) = \(\frac{5}{2}\)
Hence, Statement A is true.
∵ The terms of an arithmetic progression can have both positive and negative rational number.
Hence reason R is false.
Section – B
Section B consists of 5 questions of 2 marks each.
Question 21.
Prove that √2 is an irrational number. [2]
Solution:
Let us assume, to the contrary, that 2√ is rational.
So, we can find integers a and b such that √2 = \(\frac{a}{b}\) where a and b are coprime.
So, b√2 = a
Squaring both sides, we get
2b2 = a2
Therefore, 2 divides a2 and so 2 divides a
So, we can write a = 2c for some integer c
Substituting for a, we get 2b2 = 4c2, that is, b2 = 2c2
This means that 2 divides b2, and so 2 divides b
Therefore, a and b have at least 2 as a common factor.
But this contradicts the fact that a and b have no common factors other than 1.
This contradiction has arisen because of our incorrect assumption that √2 is rational.
So, we conclude that √2 is irrational. Hence proved.
Question 22.
ABCD is a parallelogram. Point P divides AB in the ratio 2 : 3 and point Q divides DC in the ratio 4 : 1. Prove that OC is half of OA. [2]
Solution:
ABCD is a parallelogram.
AB = DC = a
Point P divides AB in the ratio 2 : 3
AP = \(\frac{2}{5}\)a, BP = \(\frac{3}{5}\)a
Point Q divides DC in the ratio 4 : 1
DQ = \(\frac{4}{5}\)a, CQ = \(\frac{1}{5}\)a
In ΔAPO and ΔCQO
∠APO = ∠CQO (AP || QC)
∠AOP = ∠COQ (Vertically opposite angles)
Then, ΔAPO ~ ΔCQO (AA similarity)
\(\frac{\mathrm{AP}}{\mathrm{CQ}}=\frac{\mathrm{PO}}{\mathrm{QO}}=\frac{\mathrm{AO}}{\mathrm{CO}}\)
\(\frac{\mathrm{AO}}{\mathrm{CO}}=\frac{\frac{2}{5} a}{\frac{1}{5} a}=\frac{2}{1}\) ⇒ OC = \(\frac{1}{2}\) OA Hence proved.
Question 23.
From an external point P, two tangents, PA and PB are drawn to a circle with centre O. At a point E on the circle, a tangent is drawn to intersect PA and PB at C and D, respectively. If PA =10 cm, find the perimeter of ΔPCD. [2]
Solution:
PA = PB; CA = CE; DE = DB [Tangents to a circle from an external point are equal]
Perimeter of ΔPCD = PC + CD + PD
= PC + CE + ED + PD
= PC + CA + BD + PD
= PA + PB
Perimeter of ΔPCD = PA + PA = 2PA
= 2(10) = 20 cm
Question 24.
If tan (A + B) = √3 and tan (A – B) = \(\frac{1}{\sqrt{3}}\); 0° < A + B < 90°; A > B, find A and B. [2]
OR
Find the value of x.
2 cosec230 + x sin2 60 – \(\frac{3}{4}\) tan2 30 = 10
Solution:
∵ tan (A + B) = √3
∴ A + B = 60° ……… (i)
∵ tan (A – B) = \(\frac{1}{\sqrt{3}}\)
∴ A – B = 30° ……… (ii)
Adding (i) and (ii), we get 2A = 90° ⇒ A = 45°
Also (i) – (ii), we get 2B = 30° ⇒ B = 15°
OR
2 cosec2 30 + x sin2 60 – \(\frac{3}{4}\) tan2 30 = 10
⇒ 2(2)2 + x(\(\frac{\sqrt{3}}{2}\))2 – \(\frac{3}{4}\) (\(\frac{1}{\sqrt{3}}\))2 = 10
⇒ 2(4) + x(\(\frac{3}{4}\)) – \(\frac{3}{4}\) (\(\frac{1}{3}\)) = 10
⇒ 8 + x (\(\frac{3}{4}\)) – \(\frac{1}{4}\) = 10
⇒ 32 + 3x – 1 = 40
⇒ 3x = 9 ⇒ x = 3
Question 25.
With vertices A, B and C of ΔABC as centres, arcs are drawn with radii 14 cm and the three portions of the triangle so obtained are removed. Find the total area removed from the triangle. [2]
OR
Find the area of the unshaded region shown in the given figure.
Solution:
OR
The side of a square ABCD = Diameter of the semi-circle = a
Area of the unshaded region = Area of a square ABCD + 4 (Area of a semi-circle of diameter ‘a’)
The horizontal/vertical extent of the white region = 14 – 3 – 3 = 8 cm
Radius of the semi-circle + Side of a square + Radius of the semi-circle = 8 cm.
2 (Radius of the semi-circle) + Side of a square = 8 cm
2a = 8 cm ⇒ a = 4 cm
Area of the unshaded region = Area of a square of side 4 cm + 4 (Area of a semi-circle of diameter 4 cm)
= (4)2 + 4 × \(\frac{1}{2}\) π(2)2 = (16 + 8π) cm2
Section – C
Section C consists of 6 questions of 3 marks each.
Question 26.
National Art convention got registrations from students from all parts of the country, of which 60 are interested in music, 84 are interested in dance and 108 students are interested in handicrafts. For optimum cultural exchange, organisers wish to keep them in minimum number of groups such that each group consists of students interested in the same artform and the number of students in each group is the same. Find the number of students in each group. Find the number of groups in each art form. How many rooms are required if each group will be allotted a room? [3]
Solution:
Number of students in each group subject to the given condition = HCF (60, 84,108)
HCF (60, 84, 108) = 12
Number of groups in Music = \(\frac{60}{12}\) = 5
Number of groups in Dance = \(\frac{84}{12}\) = 7
Number of groups in Handicrafts = \(\frac{108}{12}\) = 9
Total number of rooms required = 21
Question 27.
If α, β are zeroes of the quadratic polynomial 5x2 + 5x + 1, find the value of: [3]
1. α2 + β2
2. α-1 + β-1
Solution:
Question 28.
The sum of a two-digit number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2, find the number. How many such numbers are there? [3]
OR
Solve: \(\frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}\) = 2; \(\frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}\) = -1
Solution:
Let the ten’s and the unit’s digits in the first number be x and y, respectively.
So, the original number = 10x + y
When the digits are reversed, x becomes the unit’s digit and y becomes the ten’s digits.
So, the obtain by reversing the digits = 10y + x
According to the given condition,
(10x + y) + (10y + x) = 66
i.e., 11(x + y) = 66
i.e., x + y = 6 …….. (i)
We are also given that the digits differ by 2,
Therefore, either x – y = 2 …….. (ii)
or y – x = 2 ……… (iii)
If x – y = 2, then solving (i) and (ii) by elimination, we get x = 4 and y = 2
In this case, we get the number 42.
If y – x = 2, then solving (i) and (iii), by elimination, we get x = 2 and y = 4
In this case, we get the number 24.
Thus, there are two such numbers 42 and 24.
OR
Let \(\frac{1}{\sqrt{x}}\) be ‘m’ \(\frac{1}{\sqrt{y}}\) be ‘n’
Then the given equations become
2m + 3n = 2
4m – 9n = -1
– 2(2m + 3n = 2) ⇒ – 4m – 6n= – 4 ……… (i)
4m – 9n = – 1 ………. (ii)
Adding (i) and (ii), we get
– 15n = – 5 ⇒ n = \(\frac{1}{3}\)
Substituting n = \(\frac{1}{3}\) in 2m + 3n = 2, we get
2m + 1 = 2
2m = 1
m = \(\frac{1}{2}\)
m = \(\frac{1}{2}\) ⇒ √x = 2 ⇒ x = 4 and n = \(\frac{1}{3}\) ⇒ √y = 3 ⇒ y = 9
Question 29.
PA and PB are tangents drawn to a circle of centre O from an external point P. Chord AB makes an angle of 30° with the radius at the point of contact.
If length of the chord is 6 cm, find the length of the tangent PA and the length of the radius OA. [3]
Two tangents TP and TQ are drawn to a circle with centre O from an external point T.
Prove that ∠PTQ = 2 ∠OPQ.
Solution:
∠OAB = 30°
∠OAP = 90° [Angle between the tangent and the radius at the point of contract]
∠PAB = 90° – 30° = 60°
AP = BP [Tangents to a circle from an external point]
∠PAB = ∠PBA [Angles opposite to equal sides of a triangle]
In ΔABP, ∠PAB + ∠PBA + ∠APB = 180°
60° + 60° + ∠APB = 180°
∠APB = 60°
∴ ΔABP is an equilateral triangle, where,
AP = BP = AB.
PA = 6 cm
In right ΔOAP, ∠OPA = 30°
tan 30° = \(\frac{\mathrm{OA}}{\mathrm{PA}}\)
\(\frac{1}{\sqrt{3}}=\frac{\mathrm{OA}}{6}\)
OA = \(\frac{6}{\sqrt{3}}\) = 2√3 cm
Hence, length of PA is 6 cm and length of radius OA is 2√3 cm.
OR
Let ∠TPQ = θ
∠TPQ = 90° [Angle between the tangent and the radius at the point of contact]
∠OPQ = 90° – θ …….. (i)
TP = TQ [Tangents to a circle from an external point]
∠TPQ = ∠TQP = θ [Angles opposite to equal sides of a triangle]
In ΔPQT, ∠PQT + ∠QPT + ∠PTQ = 180° [Angle sum property]
θ + θ + ∠PTQ = 180°
∠PTQ = 180° – 2θ
∠PTQ = 2 (90° – θ)
∠PTQ = 2 ∠OPQ [using (i)] Hence proved
Question 30.
If 1 + sin2 θ = 3 sin θ cos θ, then prove that tan θ = 1 or \(\frac{1}{2}\). [3]
Solution:
Given, 1 + sin2 θ = 3 sin θ cos θ
Dividing both sides by cos2 θ
\(\frac{1}{\cos ^2 \theta}\) + tan2 θ = 3 tan θ
sec2 θ + tan2 = 3 tan θ
1 + tan2 θ + tan2 θ = 3 tan θ
1 + 2 tan2 θ = 3 tan θ
2 tan2 θ – 3 tan θ + 1 = θ
If tan θ = x, then the equation becomes 2x2 – 3x + 1 = 0
⇒ (x – 1) (2x – 1) = 0; x = 1 or \(\frac{1}{2}\)
tan θ = 1 or \(\frac{1}{2}\) Hence proved
Question 31.
The length of 40 leaves of a plant are measured correct to nearest millimetre, and the data obtained is represented in the following table: [3]
Length (in mm) | Number of leaves (f) |
118 – 126 | 3 |
127 – 135 | 5 |
136 – 144 | 9 |
145 – 153 | 12 |
154 – 162 | 5 |
163 – 171 | 4 |
172 – 180 | 2 |
Find the average length of the leaves.
Solution:
Mean = a + \(\frac{\Sigma f d}{\Sigma f}\) = 149 + \(\frac{-81}{40}\)
= 149 – 2.025 = 146.975
Average length of the leaves = 146.975
Section D
consists of 4 questions of 5 marks each.
Question 32.
A motor boat whose speed is 18 km/hr. in still water takes 1 hr. more to go 24 km upstream than to return downstream to the same spot. Find the speed of stream. [5]
OR
Two water taps together can fill a tank in 9\(\frac{3}{8}\) hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Solution:
Let the speed of the stream be x km/h.
The speed of the boat upstream = (18 – x) km/h
and The speed of the boat downstream = (18 + x) km/h.
The time taken to go upstream = \(\frac{\text { Distance }}{\text { Speed }}=\frac{24}{18-x}\) hours
The time taken to go downstream = \(\frac{\text { Distance }}{\text { Speed }}=\frac{24}{18+x}\) hours
According to the question,
\(\frac{24}{18-x}-\frac{24}{18+x}\) = 1
24 (18 + x) – 24 (18 – x) = (18 – x) (18 + x)
x2 + 48a – 324 = 0
x = 6 or – 54
Since x is the speed of the stream, it cannot be negative.
Therefore, x = 6 gives the speed of the stream = 6 km/h.
OR
Let the time taken by the smaller pipe to fill the tank = x hr.
Time taken by the larger pipe = (x – 10) hr.
Part of the tank filled by smaller pipe in 1 hour = \(\frac{1}{x}\)
Part of the tank filled by larger pipe in 1 hour = \(\frac{1}{x-10}\)
The tank can be filled in 9 \(\frac{3}{8}\) = \(\frac{75}{8}\) hours by both the pipes together.
Part of the tank filled by both the pipes in 1 hour = \(\frac{8}{75}\)
Therefore, \(\frac{1}{x}+\frac{1}{x-10}=\frac{8}{75}\)
75 (x – 10 + x) = 8x (x – 10)
75 (2x – 10) = 8x2 – 80x
150x – 750 = 8x2 – 80x
8x2 – 230x + 750 = 0 (x – 25) (8x – 30) = 0
x = 25, \(\frac{30}{8}\)
Time taken by the smaller pipe cannot be \(\frac{30}{8}\) = 3.75 hours, as the time taken by the larger pipe will become negative, which is logically not possible.
Therefore, the time taken individually by the smaller pipe and the larger pipe will be 25 and 25 – 10 = 15 hours, respectively.
Question 33.
(a) State and prove Basic Proportionality theorem.
(b) In the given figure ∠CEF = ∠CFE. F is the midpoint of DC. [5]
Prove that \(\frac{A B}{B D}=\frac{A E}{F D}\)
Solution:
(a) Statement: If a line is drawn parallel to one side of a triangle to intersect the other two side in distinct points, the other two sides are divided in the same ratio.
Given: ΔABC where DE || BC
To Prove: \(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\)
Construction: Join BE and CD.
Draw DM ⊥ AC and EN ⊥ AB
Proof: Now, Area of ΔADE = \(\frac{1}{2}\) × Base × Height
= \(\frac{1}{2}\) × AD × EN ……. (i)
Similarly, Area of ΔBDE = \(\frac{1}{2}\) × DB × EN ………. (ii)
Divide (i) by (ii),
\(\frac{{ar}(\mathrm{ADE})}{{ar}(\mathrm{BDE})}=\frac{\frac{1}{2} \times \mathrm{AD} \times \mathrm{EN}}{\frac{1}{2} \times \mathrm{DB} \times \mathrm{EN}}\)
= \(\frac{\mathrm{AD}}{\mathrm{DB}}\) ……. (A)
ar (ADE) \(\frac{1}{2}\) × AE × DM ……… (iii)
Similarly, ar (DEC) \(\frac{1}{2}\) × EC × EN ………. (iv)
Divide (iii) by (iv),
\(\frac{{ar}(\mathrm{ADE})}{{ar}(\mathrm{DEC})}=\frac{\frac{1}{2} \times \mathrm{AE} \times \mathrm{DM}}{\frac{1}{2} \times \mathrm{EC} \times \mathrm{DM}}\)
= \(\frac{\mathrm{AE}}{\mathrm{EC}}\) ……. (B)
Note that ΔBDE and ΔDEC are on the same base DE and between the same parallel lines BC and DE. ∴ ar (BDE) = ar (DEC)
Hence, \(\frac{{ar}(\mathrm{ADE})}{{ar}(\mathrm{BDE})}=\frac{{ar}(\mathrm{ADE})}{{ar}(\mathrm{DEC})}\)
\(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\)
[From (A) and (B)] Hence proved.
(b) Draw DG || BE
In ΔABE, \(\frac{\mathrm{AB}}{\mathrm{BD}}=\frac{\mathrm{AE}}{\mathrm{GE}}\) [BPT]
CF = FD [F is the mid-point of DC] …… (i)
In ΔCDG, \(\frac{D F}{C F}=\frac{G E}{C E}\) = 1 [Given DF = CF]
∠CEF = ∠CFE [Given]
GE = CE …….. (ii)
CF = CE [Sides opposite to equal angles] ……. (iii)
From (ii) and (iii), CF = GE …….. (iv)
From (i) and (iv), GE = FD
∴ \(\frac{\mathrm{AB}}{\mathrm{BD}}=\frac{\mathrm{AE}}{\mathrm{GE}}\) ⇒ \(\frac{\mathrm{AB}}{\mathrm{BD}}=\frac{\mathrm{AE}}{\mathrm{FD}}\) Hence proved.
Question 34.
Water is flowing at the rate of 15 km/h through a pipe of diameter 14 cm into a cuboidal pond which is 50 m long and 44 m wide. In what time will the level of water in pond rise by 21 cm?
What should be the speed of water if the rise in water level is to be attained in 1 hour?
OR
A tent is in the shape of a cylinder surmounted by a conical top. If the height and radius of the cylindrical part are 3 m and 14 m respectively, and the total height of the tent is 13.5 m, find the area
of the canvas required for making the tent, keeping a provision of 26 m2 of canvas for stitching and
wastage. Also, find the cost of the canvas to be purchased at the rate of 500 per m2. [5]
Solution:
Length of the pond, l = 50 m, width of the pond, b = 44 m
Water level is to rise by, h = 21 cm = \(\frac{21}{100}\) m
Volume of water in the pond= lbh =50 × 44 × \(\frac{21}{100}\)m3 = 462m3
Diameter of the pipe = 14 cm
Radius of the pipe, r = 7 cm = \(\frac{7}{100}\) m
Area of cross-section of pipe = πr2
= \(\frac{22}{7}\) × \(\frac{7}{100}\) × \(\frac{7}{100}\) × \(\frac{154}{10000}\) m2
Rate at which the water is flowing through the pipe, h = 15 km/h = 15000 m/h
Volume of water flowing in 1 hour = Area of cross-section of pipe x Height of water coming out
of pipe.
= (\(\frac{154}{10000}\) × 15000)m3
Time required to fill the pond = \(\frac{\text { Volume of the pond }}{\text { Volume of water flowing in } 1 \text { hour }}\)
= \(\frac{462 \times 10000}{154 \times 15000}\) = 2 hours
Speed of water if the rise in water level is to be attained in 1 hour = 30 km/h.
OR
Radius of the cylindrical tent (r) = 14 m
Total height of the tent = 13.5 m
Height of the cyllinder =3 m
Height of the conical part = 10.5 m
Slant height of the cone (l) = \(\sqrt{h^2+r^2}\)
= \(\sqrt{(10.5)^2+(14)^2}\)
= \(\sqrt{110.25+196}\)
\(\sqrt{306.25}\) = 17.5m
Curved surface area of cylindrical portion = 2πrh
= 2 × \(\frac{22}{7}\) × 14 × 3 = 264 m2
Curved surface area of conical portion = πrl
= \(\frac{22}{7}\) × 14 × 17.5 = 770 m2
Total curved surface area = 264 m2 + 770 m2 = 1034 m2
Provision for stitching and wastage = 26 m2
Area of canvas to be purchased = 1034 + 26 = 1060 m2
Cost of canvas = Rate × Surface area
= 500 × 1060 = ₹5,30,000
Question 35.
The median of the following data is 50. Find the values of ‘p’ and ‘q’, if the sum of all frequencies is 90. Also find the mode. [5]
Marks obtained | Number of students |
20 – 30 | p |
30 – 40 | 15 |
40 – 50 | 25 |
50 – 60 | 20 |
60 – 70 | q |
70 – 80 | 8 |
80 – 90 | 10 |
Solution:
Marks obtained | Number of students | Cumulative frequency |
20 – 30 | p | p |
30 – 40 | 15 | p + 15 |
40 – 50 | 25 | p + 40 |
50 – 60 | 20 | p + 60 |
60 – 70 | q | p + q + 60 |
70 – 80 | 8 | p + q + 68 |
80 – 90 | 10 | p + q + 78 |
90 |
p + q + 78 = 90
p + q = 12
Median = (l) + \(\frac{\frac{n}{2}-c f}{f}\) × h
50 = 50 + \(\frac{45-(p+40)}{20}\) × 10
\(\frac{45-(p+40)}{20}\) × 10 = 0
45 – (p + 40) = 0
p = 5
5 + q = 12
q = 7
Model = l + \(\frac{f_1-f_0}{2 f_1-f_0-f_2}\) × h
40 + \(\frac{25-15}{2(25)-15-20}\) × 10
= 40 + \(\frac{100}{15}\) = 40 + 6.67 = 46.67
Section – E
Question 36.
Manpreet Kaur is the national record holder for women in the shot-put discipline. Her throw of 18.86 m at the asian Grand Prix in 2017 is the biggest distance for an Indian female athlete.
Keeping her as a role model, Sanjitha is determined to earn gold in Olympics on day.
Initially her throw reached 7.56 m only. Being an athlete in school, she regularly practiced both in the mornings and in the evenings and was able to improve the distance by 9 cm every week.
During the special camp for 15 days, she started with 40 throws and every day kept increasing the
number of throws by 12 to achieve this remarkable progress.
(i) How may throws Sanjitha practiced on 11th day of the camp? [1]
(ii) What would be sanjitha’s throw distance at the end of 6 months? [2]
OR
When will she be able to achieve a throw of 11.16 m?
(iii) How many throws did she do during the entire camp of 15 days? [1]
Solution:
(i) Number of throws during camp, a = 40; d = 12
t11 = a + 10d
= 40 + 10 × 12 = 160 throws
(ii) a = 7.56 m; d = 9 cm = 0.09 m
n = 6 weeks
tn = a + (n – 1) d
= 7.56 + 6 (0.09) = 7.56 + 0.54
Sanjitha’s throw distance at the end of 6 weeks = 8.1 m
OR
a = 7.56 m; d = 9 cm = 0.09 m
tn = 11.16 m
tn = a + (n – 1) d
11.16 = 7.56 + (n – 1) (0.09)
3.6 = (n – 1) (0.09)
n – 1 = \(\frac{3.6}{0.09}\) = 40
n = 41
Sanjitha’s will be able to throw 11.16 m in 41 weeks.
(iii) a = 40; d = 12; n = 15
Sn = \(\frac{n}{2}\) [2a + (n – 1) d]
Sn = \(\frac{15}{2}\) [2(40)+ (15 – 1) (12)]
= \(\frac{15}{2}\) [80 + 168] = \(\frac{15}{2}\) [248] = 1860 throws
Question 37.
Tharunya was thrilled to know that the football tournament is fixed with a monthly timeframe from 20th July to 20th August, 2023 and for the first time in the FIFA Women’s World Cup’s history, two nations host in 10 venues. Her father felt that the game can be better understood if the position of players is represented as points on a coordinate plane.
(i) At an instance, the midfielders and forward formed a parallelogram. Find the postion of the i central midfielder (D) if the position of other players who formed the parallelogram are: A(1, 2), B(4, 3) and C(6, 6). [1]
(ii) Check if the Goalkeeper G(-3, 5), Sweeper H(3, 1) and Wing-back K(0, 3) fall on a same straight line. [2]
OR
Check if the full-back J(5, -3) and centre-back I (-4, 6) are equidistant from forward C(0, 1) and if C is the mid-point of IJ.
(iii) If Defensive midfielder A(1, 4), Attacking midfielder B(2, -3) and Striker E(a, b) lie on the same straight line and B is equidistant from A and E, find the position of E. [1]
Solution:
Full-back J (5, -3) and centre-back I (-4, 6) are equidistant from forward C (0, 1)
mid-point of IJ = (\(\frac{5-4}{2}\), \(\frac{-3+6}{2}\) ) (\(\frac{1}{2}\), \(\frac{3}{2}\))
C is NOT the mid-point of IJ.
(iii) A, B and E lie on the same straight line and B is equidistant from A and E.
⇒ B is the mid-point of AE.
(\(\frac{1+a}{2}, \frac{4+b}{2}\)) =(2, -3)
1 + a = 4 ⇒ a = 3 4 + b = -6 ⇒ b = -10
So position of E is (3, – 10) Ans.
Question 38.
One evening, Kaushik was in a park. Children were playing cricket. Birds were singing on a nearby tree of height 80 m. He observed a bird on the tree at an angle of elevation of 45°.
When a sixer was hit, a ball flew through the tree frightening the bird to fly away. In 2 seconds, he observed the bird flying at the same height at an angle of elevation of the 30° and the ball flying towards him at the same height at an angle of elevation of 60°.
(i) At what distance from the foot of the tree was he observing the bird sitting on the tree? [1]
(ii) How far did the bird fly in the mentioned time? [2]
OR
After hitting the tree, how far did the ball travel in the sky when Kaushik saw the ball?
(iii) What is the speed of the bird in m/min if it had flown 20(√3 + 1) m? [1]
Solution: