Students can access the CBSE Sample Papers for Class 10 Maths Standard with Solutions and marking scheme Set 3 will help students in understanding the difficulty level of the exam.
CBSE Sample Papers for Class 10 Maths Standard Set 3 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 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.
√n is a natural number such that n > 1
Which of these can DEFINITELY be expressed as a product of primes?
(i) √n
(ii) n
(iii) \(\frac{\pi}{2}\)
(A) Only (ii)
(B) Only (i) and (ii)
(C) All (i), (ii) and (iii)
(D) Cannot be determined with knowing n
Solution:
(B) Only (i) and (ii)
Explanation: Natural Prime number (n) which are perfect squares are product of prime numbers.
Question 2.
Amit designs a flower vase using a graph of polynomial equations. Equation of the curve is given in the graph.
The curve m is a mirror image of p(y) on the y-axis. Which polynomial represents curve m?
(A) p(y) = -0.25y3 – 0.1y2 + 0.3y – 1
(B) p(y) = -0.25y3 – 0.1y2 – 0.3y – 1
(C) p(y) = 0.25y3 + 0.1y2 – 0.3y + 1
(D) p(y) = 0.25y3 + 0.1y2 – 0.3y
Solution:
(C) p(y) = 0.25y3 + 0.1y2 – 0.3y + 1
Explanation: Since, the curve m is a mirror image of the curve l, so all the signs are reversed i.e, polynomial represents the curve m as
p(y) = 0.25y3 + 0.1y2 – 0.3y + 1
Question 3.
If a pair of linear equations given by a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 has a unique solution, then which of the following is true?
(A) a1a2 = b1b2
(B) a1b2 ≠ a2b1
(C) \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\)
(D) \(\frac{a_1}{b_2} \neq \frac{b_1}{a_2}\)
Solution:
(B) a1b2 ≠ a2b1
Explanation: For unique solution
\(\frac{a_1}{a_2}≠\frac{b_1}{b_2}\)
a1b2 ≠ a2b1
Question 4.
Digital images consist of pixels. A pixel can be considered as the smallest unit on a display screen in a mobile or a computer. The number of pixels, their size and colours depend on the display screen and its graphic card. Display screens are rectangular in shape and their size is defined as the length of the diagonal.
Anmol is designing a web page for a display on a screen whose size is 1000 pixels. The width of the screen is 800 pixels.
Which of the following equation can be used to calculate the height (h) of the screen?
(A) h2 + 200 × 1800 = 0
(B) h2 – 200 × 1800 = 0
(C) h2 – 200 = 0
(D) h2 – 1800 = 0
Solution:
(B) h2 – 200 × 1800 = 0
Explanation: In right angled ΔABC,
Size, AC is the diagonal.
∴ AC = 1000 pixels
and AB is the width, so
AB = 800 pixels.
Using Pythagoras theorem,
h2 = (AC)2 – (AB)2
⇒ h2 = (1000)2 – (800)2
⇒ h2 = (1000 + 800) (1000 – 800)
[Using a2 – b2 = (a + b) (a – b)]
⇒ h2 = 1800 × 200
⇒ h2 – 1800 × 200 =0
Question 5.
4 groups in a class were asked to come up with an arithmetic progression (AP). Shown below are their responses:
Group | Arithmetic progression |
M | 4, 2, 0,-2, … |
N | 41, 38.5, 36, 33.5, … |
O | -19,-21,-23,-25, … |
P | -3, -3, -3,-3, …. |
Which of these groups correctly came up with an AP?
(A) only groups M and O
(B) only groups N and O
(C) only groups M, N and O
(D) all groups – M, N, O and P
Solution:
(D) all groups – M, N, O and P
Explanation: Far Group M; a = 4
d = (2 – 4) = (0 – 2)
= (-2 – 0)
= -2 (constant)
For Group N; a =41
d = (385 – 41) = (36 – 38.5)
= (335 – 36)
= -25 (constant)
For Group O; a = -19
d = [-21 – (-19))
= [-23 – (-21)]
= [-25 – (-23)]
= -2 (constant)
For Group P; a = -3
d = [-3 – (-3)l
= [-3 – (-3)]
= [-3 – (-3)]
= 0 (constant)
Thus each term differs from preceding term by constant value
∴ M, N, O and P all group form A.P.
Question 6.
Δ ABC is a triangle such that AB : BC = 1 : 2. Point A lies on the Y-axis and the coordinates of B and C are known. Which of the following formula can DEFINITELY be used to find the coordinates of A?
(i) Section formula
(ii) Distance formula
(A) only (i)
(B) only (ii)
(C) both (i) and (ii)
(D) neither (i) or (ii)
Solution:
(B) only (ii)
Explanation: As A lies on the Y-axis therefore co-ordinates of A will be (0, y).
So, we can find value of y-co-ordinate by using distance formula.
Question 7.
If three points (0, 0), (3, √3) and (3, λ) form an equilateral triangle, then X equals:
(A) 2
(B) -3
(C) -4
(D) None of these
Solution:
(D) None of these
Explanation: Let the given points are A(0, 0), B(3, √3) and C(3, λ).
Since AABC is an equilateral triangle, therefore AB = AC.
⇒ \(\sqrt{(3-0)^2+(\sqrt{3}+0)^2}\) = \(\sqrt{(3-0)^2+(\lambda-0)^2}\)
⇒ 9 + 3 = 9 + λ2
⇒ λ2 = 3
⇒ λ = ± √3
Question 8.
Leela has a triangular cabinet that fits under his staircase. There are four parallel shelves as shown below.
(Note: The figure is not to scale)
The total height of the cabinet is144 cm. Whatis the maximum height of a book that can stand upright on the bottom-most shelf?
(A) 18 cm
(B) 36 cm
(C) 54 cm
(D) 86.4 cm
Solution:
(C) 54 cm
Explanation:
HB = 54 cm
Thus, the maximum height of a book is 54 cm.
Question 9.
In the figure below, Δ PXY is formed using three tangents to a circle centred at O.
(Note: The figure is not to scale.)
Based on the construction, the sum of the tangents PA and PB is ____________ the perimeter of ΔPXY.
(A) lesser than
(B) greater than
(C) equal to
(D) (cannot be answered without knowing the tangent lengths)
Solution:
(C) equal to
Explanation: XM = XA … (i)
and YM = YB … (ii)
(∵ Tangents drawn from an external point to a circle are equal)
Now, In APXY
Perimeter of ΔPXY = PX + XY + YP
= PX + XM + MY + YP
(∵ XV = XM + MY)
From (i) and (ii) we get
= PX + XA + YB + YP
Thus perimeter of ΔPXY = PA + PB
(∵ PX + XA = PA
YB + YP = PB)
Hence, perimeter of APXY is equal to the sum of tangents PA and PB.
Question 10.
Here is a circle with centre O.
Manu wants to draw a tangent RS to the circle.
What is the number of points at which the line RS will meet the circle?
(A) 0
(B) 1
(C) 2
(D) 3
Solution:
(B) 1
Explanation: As there is one and only one tangent to a circle passing through a point lying on the circle.
Question 11.
If cos y = 0, then what is the value of \(\frac{1}{2}\) cos \(\frac{y}{2}\)
(A) 0
(B) \(\frac{1}{2}\)
(C) \(\frac{1}{\sqrt{2}}\)
(D) \(\frac{1}{2 \sqrt{2}}\)
Solution:
(D) \(\frac{1}{2 \sqrt{2}}\)
Explanation: cos y = 0 (Given)
We know that, cos 90° = 0
∴ cos y = cos 90°
So, y = 90°
Putting value of y in \(\frac{1}{2}\) cos \(\frac{y}{2}\), we get
\(\frac{1}{2}\) cos \(\frac{90°}{2}\) = \(\frac{1}{2}\) cos 45°
= \(\frac{1}{2}\) × \(\frac{1}{\sqrt{2}}\) [∵ cos 45° = \(\frac{1}{\sqrt{2}}\)]
= \(\frac{1}{2 \sqrt{2}}\)
Question 12.
ABC is an isosceles right triangle, right-angled at B. What is the value of 2 sin A × cos A?
(A) \(\frac{1}{2}\)
(B) 1
(C) \(\frac{3}{2}\)
(D) 2
Solution:
(B) 1
Explanation: Since ABC is an isosceles triangle.
Let AB = BC = x.
Question 13.
A 9 m high street-light pole is broken during a storm. The top end of the pole touches the ground at 30°. At what height did the pole breaks?
(A) 3 m
(B) 3.75 m
(C) 4.5 m
(D) 9 m
Solution:
(A) 3 m
Explanation: Let the height of Pole = BD = 9 m
Pole is broken at point C
So, let AC = CD = x m
And BC = (9 – x) m
x = 18 – 2x
x + 2x = 18
3x = 18
x = 6m
Thus, BC = 9 – 6 = 3m
∴ Pole break at height of 3 m.
Question 14.
Shown below is a circle with centre O. The shaded sector has an angle of 280° and area A cm2.
(Note: The figure is not to scale.)
Which of these is the area of the unshaded sector?
(A) \(\frac{2}{7}\) A cm2
(B) \(\frac{1}{3}\) A cm2
(C) \(\frac{2}{3}\) A cm2
(D) \(\frac{7}{9}\) A cm2
Solution:
(A) \(\frac{2}{7}\) A cm2
Explanation: Angle of shaded sector = 280°
Area of shaded sector = A cm2
Question 15.
In the figure given below, O is the centre of the circle, PR and RQ are chords of the circle. The radius of the circle is 5 cm. PR = 8 cm, QR = 6 cm and ∠PRQ = 90°.
(Note: The figure is not to scale.)
What is the approximate are of the shaded region?
(A) \(\left(\frac{25}{4} \pi-24\right)\)
(B) \(\left(\frac{25}{2} \pi-24\right)\)
(C) \(\left(\frac{25}{4} \pi\right)\)
(D) \(\left(\frac{25}{2} \pi\right)\)
Solution:
(B) \(\left(\frac{25}{2} \pi-24\right)\)
Explanation: Area of semicircle with radius 5 cm
Question 16.
Romy is blind folded and asked to pick one ball from each of the jars.
The chance of Romy picking a red ball is same in
(A) jars 2 and 3
(B) jars 1 and 3
(C) jars 1 and 2
(D) all the three jars
Solution:
(C) jars 1 and 2
Thus chance of Romy picking a red ball is same in jars 1 and 2.
Question 17.
Two dice are rolled simultaneously. What is the probability that 6 will come up at least once?
(A) \(\frac{1}{6}\)
(B) \(\frac{7}{36}\)
(C) \(\frac{11}{36}\)
(D) \(\frac{13}{36}\)
Solution:
(C) \(\frac{11}{36}\)
Explanation: Total possible outcomes, when two dice are thrown together = 6 × 6
i.e., n(S) = 36
Favourable outcomes are (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
∴ n(E) = 11
i.e., P(6 will come up at least once) = \(\frac{n(E)}{n(S)}\)
= [\(\frac{11}{36}\)
Question 18.
In statistics, an outlier is a data point that differs significantly from other observations of a data set.
If an outlier is included in the following data set, which measure (s) of central tendency would change?
12, 15, 22, 44, 44, 48, 50, 51
(A) only mean
(B) only mean and median
(C) all-mean, median, mode
(D) cannot be said without knowing the outlier
Solution:
(A) only mean
Explanation:
Mean = \(\frac{\text { Sum of all observations }}{\text { Total number of values }}\)
∴ On including outlier sum as well total number of values both will change.
So, mean would also change.
DIRECTIONS: Two statements are given below – one labelled Assertion (A) and the other labelled Reason (R). Read the statements carefully and choose the option that correctly describes statements (A) and (R).
(A) Both (A) and (R) are true and (R) is the correct explanation of the (A).
(B) Both (A) and (R) are true but (R) is not the correct explanation of the (A).
(C) (A) is true but (R) is false.
(D) (A) is false but (R) is true.
Question 19.
Assertion (A): In a right circular cone, the cross-section made by a plane parallel to the base is a circle.
Reason (R): If the volume and the surface area of a solid hemisphere are numerically equal, then the diameter of hemisphere is 9 units.
Solution:
(B) Both (A) and (R) are true but (R) is not the correct explanation of the (A).
Explanation: In case of assertion:
In a right circular cone, if any cut is made parallel to its base, we get a circle.
∴ Assertion is true.
In case of reason:
Let radius of sphere be r.
Given; volume of hemisphere = Surfacearea ofhemisphere
or, \(\frac{2}{3}\)πr3 = 3πr2
or, r = \(\frac{9}{2}\) units
∴ Diameter = \(\frac{9}{2}\) × 2 = 9 units
Reason is also true.
Hence, Assertion and Reason both are true. But Reason is not an explanation of Assertion.
Question 20.
Assertion (A): If nth term of an A.P. is (2n + 1), then the sum of its first three terms is 15.
Reason (R): The sum of first 16 terms of the A.P. 10, 6, 2, … is 420.
Solution:
(C) (A) is true but (R) is false.
Explanation: In case of assertion:
∵ an = (2n + 1)
∴ a1 – 2 x 1 + 1 = 3
l = a3 =2 x 3 + 1 = 7
Since, Sn = \(\frac{n}{2}\)[a + l]
Hence, Sn = \(\frac{3}{2}\)[3 + 7]
Sn = 15
∴ Assertion is true.
In case of reason:
Here, a = 10, d = 6 – 10 = -4 and n = 16
Sn = \(\frac{n}{2}\)[2a + (n – 1)d]
Sn = \(\frac{16}{2}\)[2 × 10 + (16 – 1)(-4)]
= 8[20 + 15 × (-4)]
= 8[20 – 60]
= 8 × (-40)
= -320
∴ Reason is false.
Hence, assertion is true but reason is false.
Section – B (10 Marks)
Section B consists of 5 questions of 2 marks each.
Question 21.
M and N are positive integers such that M = p2q3r and N = p3q2, where p, q, r are prime numbers.
Find LCM (M, N) and HCF (M, N)
Solution:
M = p2q3r
N = p3q2
∴ LCM(M, N) = p3q3r
HCF (M, N) = p2q2
Question 22.
In the below figure, QR = 4 cm, RP = 8 cm and ST = 6 cm.
(Note: The figure is not to scale.)
If the perimeter of ΔSTU is 27 cm, find the length of PQ. Show your steps.
Solution:
In ΔPQR and ΔSTU
∠P = ∠U = a (Given)
∠Q = ∠T = b (Given)
∴ ΔPQR – ΔSTU (By AA Similarity)
\(\frac{\text { Perimeter of } \triangle \mathrm{PQR}}{\text { Perimeter of } \triangle \mathrm{STU}}=\frac{Q R}{S T}\)
(∵ Ratio of the perimeters of two similar triangles is the same as the ratio of their corresponding sides)
⇒ \(\frac{\text { Perimeter of } \triangle \mathrm{PQR}}{27}=\frac{4}{6}\)
= \(\frac{4 \times 27}{6}\)
= 18 cm
Now, In ΔPQR
Perimeter = PQ + QR + PR
18 = PQ + 4 + 8
PQ = 18 – 12
PQ = 6 cm
Question 23.
If AB is a tangent drawn from a point A to a circle with centre O and BOC is a diameter of circle such that ∠AOC = 105°, then find ∠OAB.
Solution:
Given, AB and BOC are the tangent and diameter of the circle with centre O, respectively.
We know that, tangent at any point on a circle is perpendicular to the radius through the point of contact.
∴ OB ⊥ AB
⇒ ∠OBA = 90°
∠AOC = 105°
In ΔABO,
∠AOC = ∠OAB + ∠OBA
[∵ External angle = Sum of opposite internal angles]
⇒ 105° = ∠OAB + 90°
⇒ ∠OAB = 105° – 90°
⇒ ∠OAB = 15°
Question 24.
(A) Show that sin θ = cos (90° – θ) is true using the definition of trigonometric ratios.
Solution:
In ΔPQR
∠P + ∠Q + ∠R = 180° (Angle sum property)
∠P = 180° – (90° + θ)
∠P = 90° – θ
From (i) and (ii)
sin θ = cos (90° – θ) Hence Proved
OR
(B) In the triangles shown below, ∠Q = ∠T.
Write an expression each for cos Q and sin T.
Solution:
In right ΔPQR
Question 25.
(A) The figure below is a part of a circle with centre O. Its area is \(\frac{1250 \pi}{9}\) cm2 and the 10 sectors are identical.
(Note: The figure is not to scale)
Find the value of p, in degrees. Show your steps
Solution:
= 12.5°
Thus, value of β = 12.5°
OR
(B) Avikant bought a pair of glasses with wiper blades. He was curious to know the area being cleaned by each of the wiper blades. With the help of a ruler and a protractor, he found the length of each blade as 3 cm and the angle swept as 60°.
Solution:
(Note: The figure is for visual representation only)
(i) Find the area that each wiper cleans in one swipe, in terms of n.
(ii) If the diameter of each circular glass is 5 cm, what percent of the area of the glass will be cleaned by the blade in one swipe?
Show your work.
Solution:
Section – C (18 Marks)
Section C Consists of 6 questions of 3 marks each
Question 26.
Find all pairs of positive integers whose sum is 91 and HCF is 13. Show your work.
Solution:
Let pair of numbers be and ‘b’
HCF (a, b) = 13 (Given)
So, a and b be will be of the form.
a = 13x
b = 13y
(where x and y are co-primes)
Now a + b = 91 (Given)
So, 13x + 13y = 91
x + y = 7
Now all possible values of x and y are:
1 and 6
2 and 5
3 and 4
So, all values of a and b are
13 and 78
26 and 65
39 and 52
Question 27.
If one root of the quadratic equation 3x2 + px + 4 = 0 is \(\frac{2}{3}\), then find the value of p and the other root of the equation.
Solution:
3x2 + px + 4 = 0
∵ \(\frac{2}{3}\) is a root so it must satisfy the given equation
3(\(\frac{2}{3}\)) + p(\(\frac{2}{3}\)) + 4 = 0
\(\frac{4}{3}\) + \(\frac{2p}{3}\) + 4 = 0
On solving, we get
p = -8
3x2 – 8x + 4 = 0
3x2 – 6x – 2x + 4 = 0
3x(x – 2) – 2(x – 2) = 0
x = \(\frac{2}{3}\) or x = 2
Hence, x = 2
So the other root is 2.
Question 28.
(A) The two circles represent the ordered pairs, (a, b), which are solutions of the respective equations. The circles are are divided into 3 regions P, Q and R as shown.
Write one ordered pair each belonging to P, Q and R. Show your work.
Solution:
From Figure given
P ⇒ (b = a + 4)
If a = 0, then b = 4
∴ Ordered pair of P (0, 4).
Q ⇒ (b = -3a – 4)
If a – 0, then b = -4
∴ Ordered pair of Q (0, -4).
R ⇒ Equate both P and Q equation as L.H.S. is equal
b = a + 4
b = -3a – 4
a + 4 = -3o – 4
a + 3a = -4 – 4
4a = -8
Now, If a = -2
b = -2 + 4 = 2
Thus Ordered pair of R = (-2, 2)
OR
(B) Shown below is a pair of linear equations.
x + 0.999 y = 2.999
0.999x + y = 2.998
(i) Without finding the values of x and y, prove that x – y = 1.
(ii) Find the values of x and y. Show your work.
Solution:
0.001 (x – y) = 0.001
Thus x – y = 1 … (i)
(ii) Add both the equation
x = 2
Equating value of x, we get
2 – y = 1
-y = 1 – 2
-y = -1
y = 1
Thus x = 2 and y = 1
Question 29.
(A) Given below is the diagram of pair of pulleys.
The length of AC is 12 cm.
In the given figure, ∠CAB = 20°.
What is the measure of ∠AOC?
Solution:
Angle between two tangents = 20°
As tangents are equally inclined to each other.
∴ ∠CAO = ∠BAO = 10°
Now, In ΔAOC
∠CAO + ∠AOC + ∠ACO = 180°
Here, ∠ACO = 90°
(Tangent at any point of a circle is perpendicular to the radius through the point of contact.)
So, 10° + 90° + ∠AOC = 180°
∴ ∠AOC = 180° – 100°
= 80°
Hence ∠AOC = 80°.
(B) In given figure, two circles touch each other at the point C. Prove that the common tangent to the circles at C, bisects the common tangent at P and Q.
Solution:
Since, PT = TC
⇒ QT = TC
[Tangents of circle from external point]
So, PT = QT
Now PQ = PT + TQ
⇒ PQ = PT + PT
⇒ PQ = 2PT
⇒ \(\frac{1}{2}\) PQ = PT
Hence, the common tangent to the circle at C, bisects the common tangents at P and Q.
Question 30.
Prove that:
\(\frac{{cosec}^2 x-\sin ^2 x \cot ^2 x-\cot ^2 x}{\sin ^2 x}\) = 1
Solution:
= cosec2 x – cot2 x
= 1
Thus, LHS = RHS Hence Proved
Question 31.
Arti owns a manufacturing company. She hires 5 supervisors and 20 operators of a 6-months project. The table given below shows their salary breakup.
Position | Salary for the two months | Salary for the remaining four months |
Supervisor | Between ₹18,000 to ₹20,000 | Between ₹22,000 to ₹25,000 |
Operator | Between ₹8,000 to ₹10,000 | Between ₹13,000 to ₹15,000 |
The mean salary of five supervisors for the first two months is ₹19,000.
The salary of three supervisors are ₹18,000, ₹18,500 and ₹20,000 respectively. Find the sum of other two supervisor’s salary for first two months.
Solution:
Mean salary of 5 supervisors for first two months = ₹19,000
⇒ 56,500 + 2 supervisor’s salary = 95,000
Hence, 2 supervisor’s salary = 95,000 – 56,500
= ₹38,500
Section – D (20 Marks)
Section D consists of 4 questions of 5 marks each
Question 32.
(A) In a flight of 600 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km/h from its usual speed and time of flight increased by 30 minutes. Find the scheduled duration of flight.
Solution:
Let usual speed of flight be x km/h, then according to question
= 600, -400
Since, speed cannot be negative, therefore original speed = 600 km/h
and original distance = 600 km
∴ Time = \(\frac{\text { Original distance }}{\text { Original speed }}\)
= \(\frac{600 \mathrm{~km}}{600 \mathrm{~km} / \mathrm{h}}\)
= 1 h
Hence, the scheduled duration of flight is 1 h.
OR
(B) A boat goes 30 km upstream and 44 km downstream in 10 h. In 13 h, it can go 40 km upstream and 55 km downstream. Determine the speed of the stream and that of the boat in still water.
Solution:
Let the speed of the boat in still water be x km/h and speed of the stream be y km/h.
Given, \(\frac{30}{x-y}+\frac{44}{x+y}\) = 10 …(i)
and \(\frac{40}{x-y}+\frac{55}{x+y}\) = 13 …(ii)
Solving Eqs. (i) and (ii), we get
x + y = 11 …(iii)
and x – y = 5 …(iv)
Solving Eqs. (iii) and (iv), we get
x = 8 and y = 3
Hence, speed of boat = 8 km/h and speed of stream = 3 km/h.
Question 33.
Shown below is circle with centre O. YX is the tangent to the circle at Y.
(Note: The figure is not to scale.)
(i) Prove that ΔZWY – ΔZYX.
(ii) Using part (i), find the length of ZY.
Show your steps and give valid reasons.
Solution:
(i) In ΔXYZ
∠ZYX = 90°
(∴ Tangent at any point of a circle is perpendicular to the radius through the point of contact.)
In ΔWYZ
∠ZWY = 90° (Angle is semicircle is right angle)
In ΔXYZ and ΔWYZ
∠ZYX = ∠ZWY = 90°
(Proved Above)
∠Z = ∠Z (common)
ΔXYZ ~ ΔWYZ
(By AA similarity criterion)
(ii) In ΔXYZ and ΔWYZ
\(\frac{Z X}{Z Y}\) = \(\frac{Z Y}{Z W}\)
(In similar triangles the ratio of their corresponding sides are equal.)
\(\frac{25+11}{Z Y}\) = \(\frac{Z Y}{25}\)
36 × 25 = ZY2
6 × 5 = ZY
∴ ZY = 30 cm
Question 34.
(A) A restaurant stores ice-cream in a box with a dispenser attached for filling ice-cream cones. The dimensions of the box and the ice-cream cones used by the restaurant are shown in Figure 1 below. To make each serving of dessert, the cone is first filled with ice-cream and then topped with a hemispherical scoop of ice-cream taken from the same box, as shown in Figure 2.
(Note: The figures are not to scale.)
Approximately how many desserts can be served out of a completely filled box of ice-cream?
Show your steps.
(Note: Take π as \(\frac{22}{7}\))
Solution:
Shape of dispenser is cuboid.
So, volume of dispenser = lxbxh = 40 × 30 × 115
= 138000 cm3
OR
(B) A right-circular cylindrical water taker supplies water to colonies on the outskirts of a city and to nearby village. Each colony has a cuboidal water tank. In villages, people come with matkas (spherical clay pots) to fill water for their household.
(Note: The figures are not to scale.)
(i) How many colonies in total would one full tanker be able to supply?
Solution:
Volume of right circular cylindrical water tanker = πr2h
As number of colonies cannot be in decimal,
∴ One full tanker will supply water to = 5 colonies.
(ii) If a tanker supplies water to 3 colonies and then goes to a village where 400 people fill their matkaes, roughly how much water is supplied by the tanker in all? Give your answer in m3.
Show your work.
(Note: Assume all the tanks/matkas are completely filled without any loss of water; Take π as \(\frac{22}{7}\); Use 1000000 cm3 = 1 m3.)
Solution:
Volume of one colony cuboidal tank = 42
Volume of 3 colonies cuboidal tank = 42 × 3
= 126 m3
Volume of one matka = \(\frac{4}{3}\)πr3
= \(\frac{4}{3}\) × \(\frac{22}{7}\) × 21 × 21 × 21
= 38808 cm3
1000000 cm3 = 1 m3
38808 cm3 = 0.038808 m3
= 0.04 m3
Volume of 400 matkas = 400 × 0.04 = 16 m3
Total water supplied by tanker = 126 + 16
= 142 m3
Question 35.
For the table given below, answer the following questions.
Salary (in ₹) | Number of operators |
₹13,000 | 5 |
₹13,500 | 3 |
₹14,000 | 2 |
₹14,500 | 4 |
₹15,000 | 6 |
(i) What is the median salary of the operators?
Solution:
= ₹14,250
Median salary of the operators = ₹14,250
(ii) What is the salary received by most of the operators?
Solution:
Salary received by most of the operators = ₹15000
Section – E (12 Marks)
Case study based questions are compulsory.
Question 36.
The school auditorium was to be constructed to accommodate at least 1500 people. The chairs are to be placed in concentric circular arrangement in such a way that the each succeeding circular row has 10 seats more than the previous one.
(i) If the first circular row has 30 seats, how many seats will be there in the 10th row?
Solution:
Since each row is increasing by 10 seats, so it is an A.P. with first term a = 30, and common difference d = 10.
So, number of seats in 10th row = = a10
= a + 9d
= 30 + 9 × 10 = 120
(ii) For 1500 seats in the auditorium, how many rows need to be there?
Solution:
Sn = \(\frac{n}{2}\)[2a + (n – 1)d]
1500 = \(\frac{n}{2}\)[2 × 30 + (n – 1)(10)]
3000 = 50n + 10n2
n2 + 5n – 300 = 0
n2 + 20n – 15n – 300 =0
(n + 20)(n – 15) = 0
Rejecting the negative value, n = 15
OR
If 1500 seats are to be arranged in the auditorium, how many seats are still left to be put after 10th row?
Solution:
No. of seats already put up to the 10th row = S10
S10 = \(\frac{10}{2}\){2×30 + (10 – 1)10)}
= 5(60 + 90) = 750
So, the number of seats still required to be put are 1500 – 750 = 750
(iii) If there were 17 rows in the auditorium, how many seats will be there in the middle row?
Solution:
Given, no. of rows = 17
Then the middle row is the 9th row.
a9 = a + 8d
= 30 + 80
= 110 seats
Question 37.
In order to conduct Sports Day activities in your School, lines have been drawn with chalk powder at a distance of 1 m each, in a rectangular shaped ground ABCD, 100 flowerpots have been placed at a distance of1 m from each other along AD, as shown in given figure below. Niharika runs 1/4th the distance AD on the 2nd line and posts a green (G) flag. Preet runs 1/5th distance AD on the eighth line and posts a red (R) flag.
(i) Find the position of green flag?
(ii) Find the position of red flag?
(iii) What is the distance between both the flags?
Solution:
(i) As Niharika runs \(\frac{1}{4}\)th distance of AD = y coordinate
= \(\frac{1}{4}\) × 100 = 25
And, x coordinate = 2
∴ Position of green flag = (2, 25)
(ii) As Preet runs l/5th distance of AD = y coordinate
= \(\frac{1}{5}\) × 100 = 20
And, x coordinate = 8
∴ position of red flag = (8, 20)
(iii) Distance between both the flags
OR
If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, Where should she post her flag?
Solution:
According to mid point formula
Question 38.
At a toll plaza, an electronic toll collection system has been installed. FASTag can be used to pay the fare. The tag can be pasted on the windscreen of a car.
At the toll plaza a tag scanner is placed at a height of 6 m from the ground. The scanner reads the information on the tag of the vehicle and debits the desired toll amount from a linked bank account.
For the tag scanner to function properly the speed of a car needs to be less than 30 km per hour. A car with a tag installed at a height of 1.5 m from the ground enters the scanner zone.
(i) The scanner gets activated when the car’s tag is at a distance of 5 m from it.
Give one trigonometric ratio for the angle between the horizontal and the line between the car tag and the scanner?
Solution:
One trigonometric ratio is
(ii) The scanner reads the complete information on the car’s tag while the angle between tag and scanner changes from 30° to 60° due to car movement. What is the distance moved by the car?
Solution:
Let the distance moved by the car
OR
A vehicle with a tag pasted at a height of 2 m from the ground stops in the scanner zone. The scanner reads the data at a angle of 45°. What is the distance between the tag and the scanner?
Solution:
Let the distance between the tag and the scanner be x m, then according to the figure
(iii) Which trigonometric ratio in a right triangle vary from 0 to 1?
Solution:
The values of sin and cos vary from 0 to 1.