Students can access the CBSE Sample Papers for Class 10 Maths Standard with Solutions and marking scheme Set 1 will help students in understanding the difficulty level of the exam.
CBSE Sample Papers for Class 10 Maths Standard Set 1 with Solutions
Time Allowed: 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 02 marks each.
- Section C has 6 questions carrying 03 marks each.
- Section D has 4 questions carrying 05 marks each.
- Section E has 3 case based integrated units of assessment (04 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. Qs of 5 marks, 2.Qs 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 n =22/7 wherever required if not stated.
Section – A ( 20 Marks)
Section A Consists of Multiple Choice Type questions of 1 mark each
Question 1.
If 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
(A) xy
(B) xy2
(C) x3y3
(D) x2y2
Solution:
(B) xy2
Detailed Answer:
Given, a = x3y2 and b = xy3
LCM (a, b) = x3y2
Product (A) and (B) = x3y2 x xy3
Dividing product (a, b) by LCM (a, b)
\( \frac{\text { Product } a \text { and } b}{\text { LCM }} \) = \( \frac{x^3 y^2 \times x y^3}{x^3 y^3} \) = xy2
Question 2.
The given linear polynomial y = f(x) has
(A) 2 zeros
(B) 1 zero and the zero is ‘3’
(C) 1 zero and the zero is ‘4’
(D) No zero 1
Solution:
(B) 1 zero and the zero is ‘3’
Detailed Answer:
Polynomial y = f(x) is intersect x-axis at only one point, so the zero is one and intersect x-axis at point (3), so the zero of the polynomial y = f(x) is 3.
Question 3.
The lines representing the given pair of linear equations are non intersecting. Which of the following statements is true?
(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) \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
Solution:
(B) \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
Detailed Answer:
Two pair of linear equations is non intersecting so there line is parallel lines, there are no solution
\(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
Question 4.
The nature of roots of the quadratic equation 9x2 – 6x – 2 = 0 is:
(A) No real roots
(B) 2 equal real roots
(C) 2 distinct real roots
(D) More than 2 real roots 1
Solution:
(C) 2 distinct real roots
Detailed Answer:
Given quadratic equation is 9x2 – 6x – 2 = 0
9x2 – 6x – 2=0
a = 9
b = -6
c = -2
Now, b2 – 4ac
⇒ (-6) 2 – 4 × 9 × (-2)
⇒ 36 + 72
b2 – 4ac = 0
So the roots is distinct real. Here, the given equation is of order 2, so there are two roots.
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. The between their 4th terms is:
(A) 1
(B) -7
(C) 7
(D) 8
Solution:
(C) 7
Detailed Answer:
Given two APs and their first term are -1 and -8 and have same common difference.
Since, nth term is given as Tn = a + (n – 1)d
Now, 4th term of first AP
T4 = (-1) + (4 – 1)d
T4 = -1 + 3d …(i)
4th term of second AP
T4’ = -8 + 3d …(ii)
Subtract eq (ii) from eq (i)
T4 – T4’ = (-1 + 3d) – (-8 + 3d)
T4 – T4’ = 7
Hence the difference between their 4th term is 7.
Question 6.
What is the ratio in which the line segment joining (2,-3) and (5,6)is divided by x-axis?
(A) 1:2
(B) 2:1
(C) 2:5
(D) 2:1
Solution:
(A) 1:2
Detailed Answer:
Question 7.
A point (x, y) is at a distance of 5 units from the origin. pane such points lie in the third quadrant?
(A) 0
(B) 1
(C) 2
(D) infinitely many
Solution:
(D) infinitely many
Detailed Answer:
(x, y) is 5 units from the origin. There are infinity many point in from origin between point (x, y).
Question 8.
In ΔABC, DE || AB. If AB = a, DE = x, BE = b and EC = c. Then x expressed in terms of a, b and c is
(A) \(\frac{a c}{b}\)
(B) \(\frac{a c}{b+c}\)
(C) \(\frac{a b}{c}\)
(D) \(\frac{a b}{b+c}\)
Solution:
(B) \(\frac{a c}{b+c}\)
Detailed Answer:
Using basic proportional theorem.
Question 9.
If O is centre of a circle and chord PQ makes an angle 50° with the one a tangent PR at the point of contact P, then the angle subtended by the chord at the centre is:
(A) 130°
(B) 100°
(C) 50°
(D) 30°
Solution:
(B) 100°
Detailed Answer:
Given ∠QPR = 50°
PQ is a tangent, so ∠OPR is 90°.
Then, ∠OPR = ∠OPQ + ∠QPR
90° = ∠OPQ + 50°
40° = ∠OPQ = ∠OQP (radii of the circle)
In ΔOQP
∠POQ + ∠OQP + ∠OPQ = 180°
∠POQ + 40° + 40° = 180°
∠POQ = 100°
Question 10.
A quadrilateral PQRS is drawn to circumscribe a circle.
If PQ = 12 cm, QR = 15cm and RS = 14 cm, find the length of SP is:
(A) 15cm
(B) 14cm
(C) 12cm
(D) 11cm
Solution:
(D) 11cm
Detailed Answer:
PQ + RS = PS + QR
12 + 14 = x + 15
26 – 15 = x
x = 11cm
Question 11.
Given that sin θ = \(\frac{a}{b}\), find cos θ is:
(A) \(\frac{b}{\sqrt{b^2-a^2}}\)
(B) \(\frac{b}{a}\)
(C) \(\frac{\sqrt{b^2-a^2}}{b}\)
(D) \(\frac{a}{\sqrt{b^2-a^2}}\)
Solution:
(C) \(\frac{\sqrt{b^2-a^2}}{b}\)
Detailed Answer:
Question 12.
(sec A + tan A) (1 – sin A) equals:
(A) sec A
(B) sin A
(C) cosec A
(D) cos A
Solution:
(D) cos A
Detailed Answer:
Since, (sec A + tan A) (1 – sin A)
= cos A
Question 13.
If a pole 6 m high casts a shadow 23 m long on the ground, then the Sun’s elevation is:
(A) 6°
(B) 45°
(C) 30°
(D) 90°
Solution:
(A) 6°
Detailed Answer:
Let the pole AC is 6 m and shadow AB is 2√3 m.
tan θ = √3
tan θ = tan 60°
θ = 60°
Question 14.
If the perimeter and the area of a circle are numerically equal, then the radius of the circle is:
(A) 2 units
(B) π units
(C) 4 units
(D) 7 units
Solution:
(A) 2 units
Detailed Answer:
Let radius of the circle be r
Given, perimeter of circle = Area of circle
2πr = πr2
2 = r
r = 2 units
Question 15.
It is proposed to build a new 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:
(A) 10 m
(B) 15 m
(C) 20 m
(D) 24 m
Solution:
(A) 10 m
Detailed Answer:
Let radius of new park be R.
r1 = 8 cm
r2 = 6 m
Now, Area of new park = Sum of area of two circular parks πR2 = πr12 + πr22
πR2 = π(r12 + r22)
R2 = 64 + 36
R = 10 m
Question 16.
There is a complete shaded square board of side ‘2a’ units circumscribing a shaded circle. Jayadev is asked to keep a dot on the above said board. The probability that he keeps the dot on the complete shaded region.
(A) \(\frac{\pi}{4}\)
(B) \(\frac{4-\pi}{4}\)
(C) \(\frac{\pi-4}{4}\)
(D) \(\frac{4}{\pi}\)
Solution:
(B) \(\frac{4-\pi}{4}\)
Detailed Answer:
Since, Total area of board = Area of complete shaded + Area of shaded region
= Area of square board (2a)2 = 4a2 unit2
Radius of shaded circle = 2a/2 = 2
Now Area of complete shaded region = Area of square board – Area of shaded circle
= 4a2 – πa2
= a2(4 – π)
Required probability = \(\frac{\text { Area of complete shaded region }}{\text { Total area }}\)
= \(\frac{a^2(4-\pi)}{4 a^2}=\frac{4-\pi}{4}\)
Question 17.
2 cards of hearts and 4 cards of spades are missing from a pack of 52 cards. A card is drawn at random from the remaining pack. What is the probability of getting a black card?
(A) \(\frac{22}{52}\)
(B) \(\frac{22}{46}\)
(C) \(\frac{24}{52}\)
(D) \(\frac{24}{46}\)
Solution:
(B) \(\frac{22}{46}\)
Detailed Answer:
Given, 2 cards of hearts and 4 cards are missing then,
Remaining total cards = 52 – 6
n(S) = 46
Total black cards = 26
remaining black cards n(E) = 26 – 4
= 22
Probability (getting black cards)
= \(\frac{n(E)}{n(S)}\)
= \(\frac{22}{46}\)
Question 18.
The upper limit of the modal class of the given distribution is:
(A) 165
(B) 160
(C) 155
(D) 150
Solution:
(D) 150
Detailed Answer:
Class | Frequency |
135-140 | 4 |
140-145 | 7 |
145-150 | 18 |
150-155 | 11 |
155-160 | 6 |
160-165 | 5 |
Higher frequency is 18 then modal class is 145 – 150.
The upper limit at modal class is 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.
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 joining the plane surfaces of the hemisphere and cone together.
(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.
Solution:
(A) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)
Detailed Answer:
Total surface area of top = Curved surface of cone + Curved surface area of hemisphere
Assertion is true.
Reason: Top is obtained by fixing the plane surface of hemisphere and cone together The flat surface of the cone and hemisphere will be contact with each other when they are fixed together to form the top.
Reason is true.
Both assertion and reason are true and reason (R) is 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.
(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.
Solution:
(C) Assertion (A) is true but reason (R) is false.
Detailed Answer:
Hence, common difference is equal each term so this series is A.P. So, assertion is true.
Reason: The term of A.E have also negative and positive rational number, so reason is false.
Section – B (10 Marks)
Section B consists of 5 questions of 2 marks each.
Question 21.
Prove that √2 is an irrational number.
Solution:
Let us assume, to the contrary, that √2 is rational.
So, we can find integers a and b such that √2 = 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.
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.
Solution:
ABCD is a parallelogram.
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.
Solution:
PA ⇒ PB; CA = CE; DE = DB
[Tangents to a circle]
Perimeter of ΔPCD = PC + CD + PD
= PC + CE + ED + PD
= PC + CA + BD + PD
= PA + PB
Perimeter of APCD = PA + PA = 2PA
= 2(10) = 20 cm
Detailed Answer:
P is a external point and PA and PB are drawn
then PA = PB = 10 cm
C, D is also external point and drawn CA, CE and ED, BD
Then ED = BD
AC = CE
Now, Perimeter of ΔPCD = PC + CD + PD
= PC + CE + ED + PD
= (PC + AC) + (BD + PD)
= PA + PB
= PA + PA
= 2PA
= 2 × 10cm
= 20cm
Question 24.
(A) If tan (A + B) = √3 and tan (A-B) = = \(\frac{1}{\sqrt{3}}\); 0° < A + B < 90°; A>B, find A and B.
Solution:
∵ tan(A + B) = √3
∴ A + B = 60° … (i)
∵ tan(A – B) = \(\frac{1}{\sqrt{3}}\)
∴ A – B = 30 … (ii)
Adding (i) & (ii), we get 2A = 90°
⇒ A = 45°
Also (i) – (ii), we get 2B = 30°
⇒ B = 15°
OR
(B) Find the value of x if
2cosec2 30° + x sin2 60° – \(\frac{3}{4}\) tan2 30° = 10
Solution:
2cosec2 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 + x(3) – 1 = 40
⇒ 3x = 9
⇒ x = 3
Question 25.
(A) With vertices A, B and C of ΔABC as centres, arcs are drawn with radii 14cm and the three portions of the triangle so obtained are removed. Find the total area removed from the triangle.
Solution:
Total area removed
Detailed Answer:
OR
(B) Find the area of the unshaded region shown in the given figure.
Solution:
The side of a square = Diameter of the semi-circle = a
Area of the unshaded region = Area of a square of side ‘a’ + 4(Area of a semi-circle of diameter ‘a’)
The horizontal/vertical extent of the white region
= 14 – 3 – 3 = 8cm
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
= 8cm
2a = 8cm
⇒ a = 4cm
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 (18 Marks)
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 art form 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?
Solution:
Number of students in each group subject to the given condition = HCF (60, 84, 108)
HCE (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 quadratic polynomial 5x2 + 5x + 1, find the value of
(1) α2 + β2
(2) α-1 + β-1
Solution:
P(x) = 5x2 + 5x + 1
Question 28.
(A) 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?
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 digit.
So the obtain by reversing the digits = 10y + x
According to the given condition.
(10x + y) + (10y + x) = 66
ie, 11(x + y) = 66
ie, 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
(B) Solve: \(\frac{2}{\sqrt{x}}\) + \(\frac{3}{\sqrt{y}}\) = 2; \(\frac{4}{\sqrt{x}}\) – \(\frac{9}{\sqrt{y}}\) = -1 x,y>0
Solution:
Let \(\frac{1}{\sqrt{x}}\) be ‘m’ and \(\frac{1}{\sqrt{y}}\) be ‘n’
Then the given equations become
2m + 3n = 2
4m – 9n = -1
(2m + 3m = 2) × – 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 = \(\frac{1}{2}\)
√x = 2
x = 4 and n = \(\frac{1}{3}\)
√y = 3
y = 9
Question 29.
(A) 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 P radius OA.
Solution:
∠OAB = 30°
∠OAP = 90° [Angle between the tangent and the radius at the point of contact]
∠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°
[Angle Sum Property]
60° + 60° + ∠APB = 180°
∠APB = 60°
:. ΔABPis an equilateral triangle, where AP = BP = AB.
PA =6cm
In Right ΔOAP, ∠OPA = 30°
tan 30° = \(\frac{O A}{P A}\)
\(\frac{1}{\sqrt{3}}\) = \(\frac{O A}{6}\)
OA = \(\frac{6}{\sqrt{3}}\)
= 2√3 cm
OR
(B) Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that ∠PTQ = 2∠OPQ.
Solution:
Let ∠TPQ= 0
∠TPO = 90° [Angle between the tangent and
the radius at the point of contact]
∠OPQ = 90° – θ
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° – 20
∠PTQ = 2(90°- θ)
∠PTQ = 2∠OPQ [using (1)]
Question 30.
IF 1 + sin2 θ = 3 sin θ cos θ then prove that tan θ = 1 or 1/2.
Solution:
Given, 1 + sin2 θ = 3 sinθ cos θ
Dividing both sides by cos2 θ,
= \( \frac{1}{\cos ^2 \theta} \) + tan2 θ = 3 tan θ
sec2 θ + tan2 θ = 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 = 0
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 lor \(\frac{1}{2}\)
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.
Length [in mm] | Number of leaves |
118-126 | 3 |
127-135 | 5 |
136-144 | 9 |
145-153 | 12 |
154-162 | 5 |
163-171 | 4 |
172-180 | 2 |
Find the mean length of the leaves.
Solution:
= 149 – 2.025 = 146.975
Average length of the leaves = 146.975
Section – D (20 Marks)
Section D consists of 4 questions of 5 marks each.
Question 32.
(A) A motor boat whose speed is 18 km/h in still water takes 1 h. more to go 24 km upstream than to return downstream to the same spot. Find the speed of stream.
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 + 48x – 324 =0
x = 6o or -54
Since x is the speed of the stream, it cannot be negative.
Therefore, x = 6 gives the speed of the stream
= 6km/h
OR
(B) 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 time taken by the smaller pipe to fill the tank = x h
Time taken by the larger pipe = (x – 10) h
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, \(\frac{1}{x}\) + \(\frac{1}{x-10}\) = \(\frac{8}{75}\)
8x2 – 230x + 750 = 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.
Image
(a) State and prove Basic Proportionality theorem.
Solution:
Statement: If a line is drawn parallel to one side of a triangle intersecting the other two side of distinct point, then the other two sides are divided in the same ratio.
Let ΔABC in which a line DE parallel to BC intersects AB at D and AC at E
Since, ΔBDE and ΔDEC lie between the same parallel DE and BE and are on the same base DE.
We have area of ΔBDE = area of ΔDEC …(iii)
From eqs. (i), (ii) and (iii) we get
\(\frac{A D}{D B}\) = \(\frac{A E}{E C}\) Hence Proved
(b) In the given figure ∠CEF = ∠CFE. F is the midpoint of DC.
Prove that \(\frac{A B}{B D}\) = \(\frac{A E}{F D}\).
Solution:
Question 34.
(A) 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?
Solution:
Length of the pond, l = 50 m, width of the pond, b = 44m
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
= 462 m3
Diameter of the pipe = 14 cm
Radius of the pipe, r = 7 cm = \(\frac{7}{100}\) m
Area of cross-section of pipe = πr2
Speed of water if the rise in water level is to be attained in 1 hour = 30 km/h.
OR
(B) 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.
Solution:
Radius of the cylindrical tent (r) = 14 m
Total height of the tent = 13.5 m
Height of the cylinder = 3 m
Height of the Conical part = 10.5 m
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 = 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 of the data.
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
= 40 + 6.67
= 46.67
Section – E (12 Marks)
Case study based questions are compulsory
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 maximum distance for an Indian female athlete.
Keeping her as a role model, Sanjitha is determined to earn gold in Olympics one 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 many throws Sanjitha practiced on 11th day of the camp?
Solution:
Number of throws during camp, a = 40; d = 12
t11 = a + 10d
= 40 + 10 × 12
= 160 throws
(ii) What would be Sanjitha’s throw distance at the end of 6 weeks?
Solution:
a = 7.56 m; d = 9 cm = 0.09 m
n = 6 weeks
tn = a + (n – 1)d
t11 = 7.56 + 6(0.09)
= 7.56 + 0.54
Sanjitha’s throw distance at the end of 6 weeks = 8.1 m
OR
When will she be able to achieve a throw of 11.16 m?
Solution:
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) How many throws did she do during the entire camp of 15 days?
Solution:
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 time frame 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 mid-fielders and forward formed a parallelogram. Find the position of the central mid-fielder (D) if the position of other players who formed the parallelogram are: A(1, 2), B(4, 3) and C(6,6).
Solution:
Let D be (a, b), then
Mid point of AC = Midpoint of BD
Central mid-fielder is at (3, 5).
(ii) Check if the Goal keeper G(-3, 5), Sweeper H(3, 1) and Wing-back K(0, 3) fall on a same straight line.
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 J.
Solution:
C is NOT the mid-point of IJ.
(iii) If Defensive mid-fielder A(1, 4), Attacking mid-fielder 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.
Solution:
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
\(\left(\frac{1+a}{2}, \frac{4+b}{2}\right)\) = (2, -3)
1 + a = 4; a = 3;
4 + b = -6; b = -10;
E is (3, -10).
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 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?
Solution:
tan 45° = \(\frac{80}{CB}\)
CB = 80 m
(ii) How far did the bird fly in the mentioned time?
OR
After hitting the tree, how far did the ball travel in the sky when Kaushik saw the ball?
Solution:
(iii) What is the speed of the bird in m/min if it had flown 20(√3 +1) m?
Solution: