CBSE Sample Papers for Class 10 Maths Basic Set 8 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths with Solutions Set 8 are designed as per the revised syllabus.

CBSE Sample Papers for Class 10 Maths Set 8 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 Multiple Choice Questions (MCQs) carrying 1 mark each.
• Section B has 5 Short Answer-I (SA-I) type questions carrying 2 marks each.
• Section Chas 6 Short Answer-II (SA-II) type questions carrying 3 marks each.
• Section D has 4 Long Answer (LA) type questions carrying S 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.
• AU Questions are compulsory. However an internal choice in 2 Qs of 2 marks, 2 Qs of 3 marks and 2
  Questions of 5 marks has been provided. An internal choice has been provided in the 2 marks questions
  of Section E
• Draw neat figures wherever required. Take π = 22/7 wherever required if not stated.

SECTION – A (20 marks)
(Section – A consists of 20 questions of 1 mark each)

Question 1.
Two positive integers p and q are expressible as p = a³b and q = ab². The HCF (p, q) and LCM (p, q) is: [1]
(a) ab²
(b) a³b²
(c) a²b³
(d) a³b
Answer:
(b) a³b²

Explanation:
HCF (p, q) = ab;
LCM (p, q) = a³b²

Question 2.
If given A.P. is 11, 8, 5, 2, …………………, then the sum of 10th term is: [1]
(a) -23
(b) 28
(c) 24
(d) -25
Answer:
(d) -25

Explanation:
Given AP is 11, 8, 5, 2, ……..
a = 11
d = 8 – 11 = -3
n = 10
S₁₀ = \(\frac{10}{2}\)(2 × 11 + (10 – 1) × – 3)
= 5(22 – 27)
= 5 × -5
= -25

Question 3.
The ideal times of year to have chilLed shakes and ice creams are during the summer. During Lockdown, Saumyo was eager to try the watermeLon sharbat she had recently Learned to make from her online culinary classes. She cut a watermelon slice with a semi-circuLar cross section. If the perimeter of a semi- circuLar portion Is 36cm, then its radius is: [1]
(a) 12 cm
(b) 15 cm
(c) 7 cm
(d) 14 cm
Answer:
(c) 7 cm

Explanation:
Perimeter of semicircular = 36 cm
πr + 2r = 36
⇒ (π + 2)r = 36
⇒ (\(\frac{1}{2}\) + 2)r = 36
⇒ \(\frac{36}{7}\)r = 36
⇒ r = \(\frac{36 \times 7}{36}\)
⇒ r = 7

Question 4.
The solution of the pair of equations: [1]
2x + 3y = 9; 3x + 4g = 5 is
(a) 21, -17
(b) 20, 14
(c) -21, 17
(d) 20, 3
Answer:
(c) -21, 17

Explanation:
2x + 3y = 9 …(i)
3x + 4y = 5 …(ii)
Multiplying eq. (i) and (ii) by 3 and 2, respectively and then subtracting them, we get

From eq. (i)
2x + 51 =9
⇒ 2x = -42
⇒ x = -21
Thus, x = -21, y = 17 is the required solution.

Question 5.
Two vertices of a triangle are (4, -5) and (-5, -2). If the centroid of the triangle is the origin, the third vertex of the triangle is: [1]
(a) (1, 5)
(b) (2, 4)
(c) (1,4)
(d) (1,7)
Answer:
(d) (1,7)

Explanation:
Let the third vertex be (x, y). Then,
\(\left(\frac{4-5+x}{3}, \frac{-5-2+y}{3}\right)\) = (0, 0)
⇒ \(\frac{x-1}{3}\) = 0, \(\frac{y-7}{3}\) = 0
⇒ x = 1, y = 7
Thus, the third vertex is (1, 7).

Question 6.
In the adjoining figure, if PA = 10 cm, then the perimeter of ΔPCD is: [1]

(a) 16 cm
(b) 21 cm
(c) 18 cm
(d) 20 cm
Answer:
(d) 20 cm

Explanation:
We know lengths of tangents drawn from an external point to a circle are equal.
∴ From the figure, we have
PA = PB, CA = CE and DE = DB.
Now,
Perimeter of ΔPCD = PC + CE + ED + DP
= (PC + CE) + (ED + DP)
= (PC + CA) + (BD + DP)
= PA + PB
= 2 PA .
= 2 × 10 cm = 20 cm

Question 7.
What is mid-point of the line segment AB where A (- 5, 0) and B (0, 5) ? [1]
(a) \(\left(-\frac{5}{2}, \frac{5}{2}\right)\)
(b) (3, 5)
(c) \(\left(\frac{5}{2}, \frac{3}{2}\right)\)
(d) (2, 4)
Answer:
(a) \(\left(-\frac{5}{2}, \frac{5}{2}\right)\)

Explanation:
The mid-point of AB is
\(\left(\frac{-5+0}{2}, \frac{0+5}{2}\right)\) i.e, \(\left(-\frac{5}{2}, \frac{5}{2}\right)\)

Question 8.
If x sec 45° = 2, then what is the value of x ? [1]
(a) \(\frac{\sqrt{3}}{2}\)
(b) 2√2
(c) √2
(d) \(\frac{1}{\sqrt{2}}\)
Answer:
(c) √2

Explanation:
Given, x sec 45° = 2
⇒ x(√2) = 2
⇒ x = √2

Question 9.
If tan θ + cot θ = 4, then the value of tan⁴θ + cot⁴θ is: [1]
(a) 196
(b) 200
(c) 194
(d) 198
Answer:
(c) 194

Explanation:
Given, tan θ + cot θ = 4
Squaring on both sides,
We get,
tan² θ + cot² θ + 2 tan θ cot θ = 16
tan² θ + cot² θ + 2 = 16
tan² θ + cot² θ = 14

Again, squaring on both sides, we get
tan⁴ θ + cot² θ + 2 tan² θ cot² θ = 196
⇒ tan⁴θ + cot⁴θ + 2 = 196
⇒ tan⁴θ + cot⁴θ = 194

Question 10.
In an A.P., if a = 3.5, d = 0, n = 101, then the value of a_n is: [1]
(a) 2.5
(b) 3
(c) 4
(d) 3.5
Answer:
(d) 3.5

Explanation:
In the given A.P., d = 0
So, all its terms are same as a = 3.5,
∴ a₁₀₁ = 3.5

Question 11.
If A = 900, If Σf_id_i = – 400 and Σf_i = 100, then what is the value of x? [1]
(a) 890
(b) 986
(c) 465
(d) 896
Answer:
(d) 896

Explanation:
We know that
x̄ = A + \(\frac{\Sigma f_i d_i}{\Sigma f_i}\)
= 900 + \(\frac{(-400)}{100}\)
= 900 – 4 = 896

Question 12.
A 6 faced cube has letters A, B, C, D, A and C on its six faces. This cube is rolled once. What is the probability of getting B or C? [1]
(a) \(\frac{1}{2}\)
(b) \(\frac{1}{3}\)
(c) \(\frac{2}{3}\)
(d) \(\frac{1}{4}\)
Answer:
(a) \(\frac{1}{2}\)

Explanation:
Total number of outcomes = 6
∴ Number of favourable outcomes = 3
P (B or C) = \(\frac{3}{6}\) i.e. \(\frac{1}{2}\)

Question 13.
Which criterion of similarity will be used in proving that ΔABD – ACE ? [1]

(a) SAS
(b) AA
(c) RHS
(d) SSS
Answer:
(b) AA

Explanation:
In As ABD and ACE, we have AD = AE (given)
So, ∠D = ∠E (Angles opposite to equal sides are equal)
∠A = ∠A (Common)
So, by AA similarly criterion, ΔABD ~ ΔACE.
Hence, AA similarity criteria will be used.

Question 14.
A letter is chosen from the letters of the word MAINTENANCE The probability that it is N is: [1]
(a) \(\frac{1}{11}\)
(b) \(\frac{2}{11}\)
(c) \(\frac{3}{11}\)
(d) \(\frac{4}{11}\)
Answer:
(c) \(\frac{3}{11}\)

Explanation:
In the given word, there are in all eleven letters, of which three are N.
So, the required probability is \(\frac{3}{11}\)

Question 15.
If the equation x² + 4x + k = 0 has real and distinct roots, then the value of ‘k’ is: [1]
(a) k > 4
(b) k > 4
(c) k < 4
(d) k = 4
Answer:
(c) k < 4

Explanation:
As the given equation has real and distinct roots,
∴ Discriminant = (4)² – 4(1)(k) > 0
i.e. 16 – 4k > 0
or k < 4

Question 16.
Which term of the A.P.: -2, -7, -12,… will be -77? [1]
(a) 16th
(b) 10th
(c) 15th
(d) 12th
Answer:
(a) 16th

Explanation:
Let, the nth term of the A.P. be -77.
Then, a + (n – l)d = -77
For the given A.P., a = -2 and d = -5.
So, a + (n – 1)d = (-2) + (n – 1)(-5) = -77
-5(n – 1) = -75
or n- 1 = 15 or n = 16
So, the 16th term of the A.P. is (-77).

Question 17.
What type of lines are represented by the pair of equations 10x + 6y = 9 and 5x + 3y + 4 = 0 ? [1]
(a) Straight lines
(b) Intersecting lines
(c) Parallel lines
(d) None of these
Answer:
(c) Parallel lines

Explanation:
The pair of equations satisfy the relation
\(\frac{a_1}{a_2}=\frac{b_1}{b_2} \uparrow \frac{c_1}{c_2}\)
as \(\frac{10}{5}=\frac{6}{3} \neq \frac{9}{-4}\)
Hence, these equations represent parallel lines.

Question 18.
If an event is sure to occur, then what is its probability of occurrence ? [1]
(a) 0
(b) 1
(c) 2
(d) \(\frac{1}{2}\)
Answer:
(b) 1

Explanation:
The probability of occurrence of a sure event is 1.

Direction for questions 19 and 20: In 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.
Assertion (A): The length of the tangent to a circle from a point P, which is 25 cm away from the centre is 24 cm. The radius of the circle is 17 cm
Reason (R): Tangent is perpendicular to radius at the point of contact. [1]
Answer:
(d) Assertion (A) is false but reason (R) is true.

Explanation:
OQ is perpendicular to PQ.

∴ In ΔPOQ,
PQ² + OQ² = OP²
25² = OQ² + 24²
OQ² = 625 – 576 = 49
OQ = 7 cm

Question 20.
Assertion (A): If circumference of two circles are equal, then their areas are also equal.
Reason (R): Two circles are congruent if their radii are equal 1. [1]
Answer:
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A)

Explanation:
Let’s consider 2 circles of radii r₁ and r₂.
Then, 2πr₁ = 2πr₂
r₁ = r₂ = r
Then, A₁ = πr₂² = πr²
A₁ = A₂

SECTION – B (10 marks)
(Section – B consists of 5 questions of 2 marks: each)

Question 21.
Assuming that √2 is irrational, show that 5√2 is an irrational number. [2]
Answer:
Let 5√2 be rationaL Then,
5√2 = \(\frac{p}{q}\), where p and q are co-prime and q ≠ 0
⇒ √2 = \(\frac{p}{5q}\)
Here, \(\frac{p}{5q}\) is rational which implies √2 is rational which is a contradiction, as it is given than √2 is irrational
⇒ 5√2 is irrational.

Question 22.
Find the greatest number that divides 45 and 210 completely. [2]
Answer:
The greatest number that divides 45 and 240 completely is the HCF of 45 and 210.
Now, 45 = 3 × 3 × 5, or 3² × 5¹
210 = 2 × 3 × 5 × 7,
So, HCF (45, 210) = 3¹ × 5¹, i.e. 15
Hence, the required number is 15.

Question 23.
If x = a cos³θ and y = b sin³θ, then prove that: [2]
\(\left(\frac{x}{a}\right)^{\frac{2}{3}}+\left(\frac{y}{b}\right)^{\frac{2}{3}}\) = 1
OR
Prove that \(\sqrt{\sec ^2 \theta+{cosec}^2 \theta}\) = tan θ + cot θ.
Answer:

Question 24.
The largest possible sphere is carved out of wooden solid cube of side 7 cm. What is the radius of this sphere?
OR
A line intersect the y-axis and x-axis at the points P and Q respectively. If (2, – 5) is the mid point of PQ, then find the coordinates of the points P and Q. [2]
Answer:
The largest possible sphere that can be carved out of a sphere, is equal to diameter of the sphere.
∴ Diameter of sphere = 7 cm.
So, the radius of the largest possible sphere is 3.5 cm.
OR
Let the coordinates of P and Q be (0, y) and (x, 0) respectively.
Here, the mid-point of PQ is (2, – 5)
So, \(\left(\frac{x+0}{2}, \frac{0+y}{2}\right)\) = (2, -5)
⇒ x = 4, y = -10

Thus, the coordinates of P and Q are (0, – 10) and (4, 0) respectively.

Question 25.
In figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. [2]

Answer:
OB = \(\sqrt{O A^2+A B^2}=\sqrt{O A^2+O A^2}\)
= √2 OA = √2 × 20 = 20√2 cm
Area of shaded region = Area of quadrant OPBQ – Area of square OABC
= \(\frac{90^{\circ}}{360^{\circ}}\) × 3.14(20√2)² – 20 × 20
= \(\frac{1}{4}\) × 3.14 × 800 – 400
= 200 × 3.14 – 400
= 228 cm²

SECTION – C (18 marks)
(Section – C consists of 6 questions of 3 mark each)

Question 26.
Solve for x and y:
x + \(\frac{y}{4}\) = 11; \(\frac{5 x}{6}-\frac{y}{3}\) = 7
OR
A 2-digit number is such that the product of the digits is 20. If 9 is subtracted from the number, the digits interchange their places. Find the number. [3]
Answer:
The given equations are rewritten as:
4x + y – 44 = 0 ;
5x – 2y – 42 = 0
From 4x + y – 44 = 0, we have
y = 44 – 4x …..(i)
Substituting this value of y in 5x – 2y – 42 = 0, we have:
5x – 2 (44 – 4x) – 42 = 0
=> 13x – 88 – 42 = 0
=> x = 10
From (i) y = 44 – 40 = 4 Thus, x = 10, y = 4 is the required solution.
OR
Let ten’s digit and one’s digit of the two digit number be a and b respectively. Then, the number is 10a + b.
Here, ab = 20 ……..(i)
and (10a + b) – 9 = 10b + a
i.e 9a – 9b = 9
or a – b = 1
Solving (i) and (ii) simultaneously, we get
a = 5, b = 4
Thus, the number is 54.

Question 27.
Show that ΔABC with vertices A (- 2, 0), B (2, 0) and C(0, 2) is similar to ΔDEF with vertices D (- 4, 0), E (4, 0) and F (0, 4). [3]
Answer:
Let’s say that ΔABC ~ ΔDEF.
If \(\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{\mathrm{CA}}{\mathrm{FD}}\)

Now, using distance formula,
AB = 4, BC = √8 = 2√2
CA = √8, 2√2
DE = 8, EF = √32 = 4√2
FD = √32 , i.e 4√2
Here \(\frac{A B}{D E}=\frac{B C}{D E}=\frac{C A}{F D}=\frac{1}{2}\)
So, ΔABC ~ ΔDEF.

Question 28.
Prove that the lengths of tangents drawn from an external point to a circle are equal.
OR
In the figure, PQ and RS are the common tangents to two circles intersecting at O. [3]

Prove that PQ = RS
Answer:
Let AP and AQ be the two tangents drawn
to the circle from an external point A.
We need to show that AP = AQ.
Join OA, OP and OQ.
We know that tangent is perpendicular to radius at the point of contact.
∴ OP ⊥ AP and OQ ⊥ QA.

Consider ΔOPA and ΔOQA.
Here, OQ = OP (radii of the circle)
OA = OA (common)
∠OPA = ∠OQA
So, ΔOPA ≅ ΔOQA
⇒ PA = QA or AP = AQ
OR
We know that lengths of tangent drawn from an external point to a circle are equaL

So, from the figure, OP = OR and OS = OQ
Now, PQ = PO + OQ
= OR + OS
= RS

Question 29.
A number x is selected from the numbers 1, 2, 3 and then a second number y is selected randomly from the numbers 1, 4, 9. What is the probability that the product xy of the two numbers will be less than 9? [3]
Answer:
Total number of possible outcomes of
product xy = {1, 4, 9, 2, 8,18, 3,12, 27}.
Of these, 5 are less than 9.
So, P(xy < 9) = \(\frac{5}{9}\)

Question 30.
Find the value of:
\(\frac{5 \sin ^2 30^{\circ}+\cos ^2 45^{\circ}-4 \tan ^2 30^{\circ}}{2 \sin 30^{\circ} \cdot \cos 30^{\circ}+\tan 45^{\circ}}\) + cos 0° [3]
Answer:

Question 31.
The first and the last terms of an A.P. are 17 and 350 respectively. If the common difference is 9, how many terms are there in the A.P. ?
Answer:
Let the A.P. contains ‘n’ terms. Then,
a_n = l = 350
⇒ a + (n – 1)d = 350
⇒ 17 + (n – 1)(9) = 350
⇒ 9(n – 1) = 333
⇒ n – 1 = 37
⇒ n = 38
Thus, A.R contains 38 terms.

SECTION – D (20 marks)
(Section – D consists of 4 questions of 5 mark each)

Question 32.
From the top of a building 60 m high, the angle of depression of the top and bottom of a vertical lamp-post are observed to be 30° and 60° respectively. Find the height of the lamp-post, and the distance between the top of building and the top of lamp-post. [5]
Answer:
Let AB be the building and XY be the lamp post.
∴ AB = 60 m, ∠BYM = 30° and ∠BXA = 60°
Let ‘h‘ metres be the height of the lamp post and ‘d’ metres be the horizontal distance between the lamp post and the building.
Then,
From right ΔBMY, we have:
\(\frac{B M}{Y M}\) = tan 30°
⇒ \(\frac{60-h}{d}=\frac{1}{\sqrt{3}}\)

From (i) and (ii)
√3(60 – h) = 20√3
⇒ 60 – h = 20
⇒ h = 40
Hence the height of lamp post is 40 m
Now, BY = \(\sqrt{Y M^2+B M^2}\)
= \(\sqrt{d^2+20^2}=\sqrt{1200+400}\)
= \(\sqrt{1600}\) = 40 m
Thus, the distance between the top of the building and the top of lamp-post is 40 m.

Question 33.
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. Prove it. [5]
OR
BL and CM are medians of ΔABC, right¬angled at A.
Prove that: 4(BL² + CM²) = 5 BC²

Answer:
ABC is a triangle in which DE is a line parallel to side BC which cuts AB at D and AC at E.
We need to prove that:

Join BE and CD and draw EF ⊥ AB and DN ⊥ AC.
Now,

But ΔBDE and ΔCDE are on the same base DE and between the same parallels DE and
So, ar(ΔBDE) = ar(ΔCDE) … (iii)
⇒ \(\frac{{ar}(\triangle \mathrm{ADE})}{{ar}(\triangle \mathrm{BDE})}=\frac{{ar}(\mathrm{CDE})}{{ar}(\mathrm{CDE})}\)
Hence, by (i), (ii) and (iii), we have
\(\frac{A D}{B D}=\frac{A E}{E C}\)
or
\(\frac{A D}{D B}=\frac{A E}{E C}\)
OR
Since BL and CM are medians, So L is the mid-point of CAç and M is the mid-point of AB.
i.e. AL = \(\frac{1}{2}\)CA and AM = \(\frac{1}{2}\)AB
Using Pythagoras theorem in right triangle LAB, we have:
RL² = LA² + AB²
BL² = \(\frac{\mathrm{CA}^2}{4}\) + AB² ………..(i)
SimilarLy, in right triangLe CAM, we have:
CM² = CA² + AM²
⇒ CM² = CA² + \(\frac{\mathrm{AB}^2}{4}\)

Adding (i) and (ii), we get

(Using Pythagoras theorem in ΔABC)
⇒ 4(BL² + CM²) = 5 BC²

Question 34.
Find the median marks for the following frequency distribution : [5]

Answer:

Marks	Frequency	Cumulative Frequency
0-20	7	7
20-40	12	19
40-60	23	42
60-80	18	60
80-100	10	70

Here, N = 70, ∴ \(\frac{N}{2}\) = 35
Cumulative frequency just greater than 35 is 42 which belongs to class 40 – 60.
So, the median class is 40-60.
For this class,
l = 40,h = 20, c.f. = 19 and f = 2
So, Median = l + \(\frac{\frac{N}{2}-c . f}{f}\) × h
= 40 + \(\frac{35-19}{23}\) × 20
= 40 + 13.9 (approx)
= 53.9 (approx)

Question 35.
Between Mysore and Bangalore, 132 km apart, an express train travels in 1 hour less time than a passenger train (without taking into consideration the time they stop at intermediate stations). Find the average speed of the two trains if the express train’s average speed is 11 km/h higher than the passenger train’s.
OR
Form a pair of linear equations for the following problems and find their solution by substitution method.
(A) The cost of a taxi in a city consists of a fixed charge and a charge for the distance travelled. The cost for a 10 km travel is ₹ 105, while for a 15 km journey, the cost is ₹ 155. What are the fixed charges and the km charged? How much will it cost someoneto drive 25 Km?
(B) For ₹ 3800, the cricket team’s coach purchases 6 balls and 7 bats, he then spends 1750 for 3 bats and 5 balls. Find out how much each ball and bat costs. [5]
Answer:
Let average speed of passenger train = x km/h
Let average speed of express train = (x + 11) km/h
Time taken by passenger train to cover 132 km = \(\frac{132}{x}\)hours
Time taken by express train to cover 132 km = \(\left(\frac{132}{x+11}\right)\)hours

According to the given conditions,
\(\frac{132}{x}=\frac{132}{x+11}\) + 1
⇒ \(\left(\frac{1}{x}-\frac{1}{x+11}\right)\) = 1
⇒ 132\(\left(\frac{x+11-x}{x(x+11)}\right)\) = 1
⇒ 132 (11) = x(x + 11)
⇒ 1452 = x² + 11x
⇒ x² + 11x – 1452 = 0
Comparing equation x² + 11x – 1452 = 0 with general quadratic equation ax² + bx + c = 0, we get a = 1, b = 11 and c = -1452

Applying Quadratic Formula

As speed cannot be in negative. Therefore, speed of passenger train = 33 km/h And, speed of express train = x + ll = 33 + ll = 44 km/h
OR
(A) Let fixed charge = ₹ x and let charge for every km = ₹ y
According to given conditions, we have
x + 0y = 105 …(i)
x + 15 = 155 …(ii)
Using equation (i), we can say that
x = 105 – 10y
Puttinmg this in equation (ii), we get
105 – 10y + 15y = 155
⇒ 5x = 50
⇒ y = 10
Putting value of y in equation (i), we get x + 10(10) = 105
⇒ x = 105 – 100
⇒ x = 5
Therefore, fixed charge = ₹ 5 and charge per km = ₹ 10
To travel distance of 25 km, person will have to pay
= ₹ (x + 25y)
= ₹ (5+ 25 × 10)
= ₹ (5 + 250)
= ₹ 255

(B) Let cost of each bat = ₹ x
and let cost of each ball = ₹ y
According to given conditions, we have 7x + 6y = 3800 _(i)
And, 3x + 5y = 1750 …(ii)
Using equation 0), we can say that
7x = 3800 – 6y
x = \(\frac{3800-6 y}{7}\)

Putting this in eq. (ii), we get

⇒ 17y = 850
⇒ y = 50
Putting value of y in (ii). we get 3x + 250 = 1750
⇒ 3x = 1500
⇒ x = 500
Therefore, cost of each bat = ₹ 500 and cost of each ball = ₹ 50.

SECTION – E (12 marks)
(Case Study-Based Questions)
(Section – E consists of 3 questions. All are compulsory)

Question 36.
‘Origami’ is the art of paper folding, which is often associated with Japanese culture. Gurmeet is trying to learn Origami using paper cutting and folding technique. A square base is sometimes referred to as a “preliminary” base or preliminary fold.

Here ¡s a 20 cm × 20 cm square. Gurmeet wants to first cut the squares of integral Length from the corners and by foLding the flaps along the sides.

On the basis of the above information, answer the following questions
(A) How many different sizes of boxes Gurmeet can make? [1]
(B) How many different sizes of boxes Gurmeet can make if sides of the squares are not integral length? [1]
(C) Find the equation relating the size of the square cut out and volume of the box.
OR
Find the dimension of the box with maximum volume and minimum volume. [2]
Answer:
(A) Different size of squares are = 18 × 18 × 1,16 × 16 × 2,14 × 14 × 3 ,12 × 12 × 4, 10 × 10 × 5, 8 × 8 × 6
(B) As the side length of any value could be cut out from the square and it could be infinite in number.
(C) Let the width of square of each side be
Then, sides of box are 20 – 2x, 20 – 2x and x
Volume = lbh
= (20 – 2x) (20 – 2x)x
= (400 – 40x – 40x + 4x²)x
= 4x³ – 80x² + 400x
OR
On calculating the volume of the boxes given in the options. The box with dimension 14 × 14 × 3 has maximum volume as 588.
On calculating the volume of the boxes given in the option. The box with dimensions 18 × 18 × 1 has minimum volume.

Question 37.
The students of a school decided to beautify the school on the Annual day by fixing colourful flags on the straight passage of the school. They have 27 flags to be fixed at intervals of every 2 m. The flags are stored at the position of the middle most flag.

Ruchi was given the responsibility of placing the flags Ruchi kept her books where the flag were stored She could carry only one flag at a time.
On the basis of the above information, answer the following questions
(A) What is the position of the middle flog? [1]
(B) Find totaL distance travelled for placing all the flogs. [1]
(C) Find total distance travelled for placing 13 flags on left
OR
Find the maximum distance she travelled carrying a flag. [2]
Answer:
(A) There are 27 flags. So the middle most flag is 14th flag.

(B) Total distance travelled = 13 flags on left side +13 flags on right side
= 364 + 364
= 728 m

(C) For placing first placing she go 2 m and come back 2 m. Then for second flag, she goes 4 m and come back 4 m and so one….
Distance travailed = 4,8,12
Then it forms an A.P. with a = 4, d = 4 and n = 13
Then S₁₃ = \(\frac{13}{2}\) [8 + 12 × 4]
= \(\frac{13}{2}\) (56) = 364 m

Then, a₁₃ = a + (n – 1)d
= 4 + (13 – 1) × 4
= 4 + 48 = 52
∴ From carrying the flag to its position 52 she covers distance = \(\frac{52}{2}\)
= 26 m

Question 38.
Soumya throws a ball upwards, from a rooftop, 80 m above. It will reach a maximum height and then fall back to the ground. The height of the ball from the ground at time ‘t’ is ‘h’ which is given by h = -16t² + 64t + 80.

On the basis of the above information, answer the following questions:
(A) What is the height reached by the ball after 1 second? [1]
(B) How long will the ball take to hit the ground? [1]
(C) What are the two possible times to reach the ball at the same height of 128 m?
OR
What is the maximum height reached by the ball? [2]
Answer:
(A) In the basis of given equation, h = – 16t² + 64t + 80
when, t = 1 second
h = – 16(1)² + 64(1) + 80
= -16 + 144 = 128 m

(B) When ball hits the ground, h = 0
-16t² + 64t + 80 = 0
∴ t² – 4t – 5 = 0
(t – 5) (t + 1) = 0
t = 5 or t = – 1
Since, time cannot be negative, so the time = 5 seconds.

(C) Since, h = -16t² + 1612 + 80
⇒ 128 = -16t² + 64t² + 80
⇒ 16t² + 64t + 80 – 128 = 0
⇒ 16t² + 64t – 48 = 0
⇒ t² – 4t + 3 = 0
⇒ t² + 3t – t + 3 = 0
⇒ (t – 3) (t – 1) = 0
⇒ t = 3,1
OR
Rearrange the given equation, by completing the square, we get
h = -16 (t² – 4t – 5)
= -16[(t – 2)² – 9]
= – 16(t- 2)² + 144
Height is maximum, when t = 2
∴ Maximum height = 144 m