CBSE Sample Papers for Class 10 Maths Standard Set 6 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths Standard with Solutions Set 6 are designed as per the revised syllabus.

CBSE Sample Papers for Class 10 Maths Standard Set 6 with Solutions

Time: 3 hrs
Max. Marks: 80

Instructions:
1. This question paper has 5 Sections A-E.
2. Section A has 20 MCQs carrying 1 mark each.
3. Section B has 5 questions carrying 2 marks each.
4. Section C has 6 questions carrying 3 marks each.
5. Section D has 4 questions carrying 5 marks each.
6. Section E has 3 Case Based integrated units of assessment (4 marks each).
7. All questions are compulsory. However, an internal choice in 2 questions of 2 marks, 2 questions 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.
8. 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.
The HCF and LCM of two numbers are 33 and 264, respectively. When the first number is completely divided by 2, then the quotient is 33. The other number is [1]
(a) 66
(b) 130
(c) 132
(d) 196
Answer:
(c) 132

Given, HCF = 33 and LCM = 264
Let the other number be x.
First number =2 × 33=66
Product of numbers = HCF × LCM
∴ x × 66 = 33 × 264
⇒ x = \(\frac{33×264}{66}\) = 132

Question 2.
The HCF of 30, 72 and 432 is [1]
(a) 13
(b) 7
(c) 5
(d) 6
Answer:
(d) 6

By prime factorisation, we get
30 = 2¹ × 3¹ × 5¹
72 = 2³ × 3²
432 = 2⁴ × 3³
∴ HCF of 30, 72 and 432 =2 × 3 = 6.

Question 3.
If the probability that it will rain tomorrow is 0.75, then the probability that it will not rain tomorrow, is [1]
(a) 0
(b) 1
(c) 0.50
(d) 0.25
Answer:
(d) 0.25

We know that
P (tomorrow rain) + P (not tomorrow rain) = 1
⇒ 0.75 + P (not tomorrow rain) = 1
= P (not tomorrow rain) = 1 – 0.75 = 0.25
Hence, the probability that it will not tomorrow rain is 0.25.

Question 4.
Two tangents, drawn at the end points of diameter of a given circle are always
(a) parallel
(b) perpendicular
(c) intersect each other
(d) None of the above
Answer:
(a) parallel

Two tangents drawn at the end points of diameter are always parallel.

Question 5.
If the difference of mode and median of a data is 24, then the difference of median and mean is
(a) 12
(b) 24
(c) 8
(d) 36
Answer:
(a) 12

We have, Mode – Median = 24
We know, Mode = 3 Median – 2 Mean
∴ Mode-Median = 2 Median – 2 Mean
⇒ 24 = 2(Median – Mean)
⇒ Median – Mean = 12

Question 6.
If the distance between the points (x, -1) and (3, 2) is 5 units, then the value of x is
(a) – 7 or -1
(b) – 7 or 1
(c) 7 or 1
(d) 7 or – 1
Answer:
(d) 7 or – 1

Let P(x, -1) and Q (3,2) by the given point, then PO = 5,
∴ \(\sqrt{(x-3)^2 + (-1-2)^2}\) = 5
⇒ (x – 3)² + 9 = 25
⇒ x² – 6x + 9 + 9 = 25
⇒ x² – 6x – 7 = 0
⇒ (x – 7)(x + 1) = 0
⇒ x = 7 or x = -1

Question 7.
The perimeter of two similar triangles ∆ABC and ∆PQR are respectively, 48 cm and 36 cm. If PQ =12 cm, then the value of AB is
(a) 20 cm
(b) 16 cm
(c) 40 cm
(d) 25 cm
Answer:
(b) 16 cm

Given,

∆ABC ~ ∆PQR
∴ \(\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A C}{P R}\)
⇒ \(\frac{A B}{12}=\frac{48}{36}\)
∴ AB = 16 cm

Question 8.
The circumference of a circle is equal to the sum of the circumferences of two circles having diameters 34 cm and 28 cm. The radius of the new circle is
(a) 30 cm
(b) 31cm
(c) 32 cm
(d) 25 cm
Answer:
(b) 31cm

Let radius of new circle is r.
Then, 2πr = 2π(\(\frac{34}{2}\)) + 2π(\(\frac{28}{2}\))
⇒ r = 17 + 14 = 31cm

Question 9.
From a solid circular cylinder with height 10 cm and radius of the base 6 cm, a right circular cone of the same height and same base is removed, then the volume of remaining solid is
(a) 280 π cm³
(b) 330 π cm³
(c) 240 π cm³
(d) 440 π cm³
Answer:
(c) 240 π cm³

Volume of the remaining solid
= Volume of the cylinder – Volume of the cone
= π × 6² × 10 – π × 6² × 10
= (360π – 120π ) = 240π cm³

Question 10.
In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre, then the length of the arc is
(a) 22 cm
(b) 20 cm
(c) 21cm
(d) 19 cm
Answer:
(a) 22 cm

We have, r =21 cm and θ = 60° 0
∵ l = \(\frac{θ}{360°}\) × 2πr
∴ l = \(\frac{θ}{360°}\) × 2 × \(\frac{22}{7}\) × 21 = 22 cm

Question 11.
If x = \(\frac{1}{2}\) is a root of the equation x² + kx – \(\frac{5}{4}\) = 0, then the value of k is [1]
(a) 3
(b) 0
(c) 1
(d) 2
Answer:
(d) 2

Since, x = \(\frac{1}{2}\) is a root of the equation
x² + kx – \(\frac{5}{4}\) = 0.
On putting x = \(\frac{1}{2}\) in the given equation, we get
(\(\frac{1}{2}\))² + k(\(\frac{1}{2}\)) – \(\frac{5}{4}\) = 0
⇒ \(\frac{1}{4}\) + \(\frac{k}{2}\) – \(\frac{5}{4}\) = 0
⇒ \(\frac{k}{2}\) = \(\frac{5}{4}\) – \(\frac{1}{4}\)
⇒ k = 2(\(\frac{4}{4}\)) = 2.

Question 12.
In a ∆ABC, right angled at B, if base line is AB =12 and BC = 5, then the value of cos C is
(a) \(\frac{5}{13}\)
(b) \(\frac{13}{5}\)
(c) \(\frac{5}{17}\)
(d) \(\frac{17}{5}\)
Answer:
(a) \(\frac{5}{13}\)

In right angled ∆ABC, using Pythagoras theorem,

Question 13.
tan²θ . sin²θ is equal to
(a) tan²θ – sin²θ
(b) tan²θ + sin²θ
(c) \(\frac{tan²θ}{sin²θ}\)
(d) sin²θ . cot²θ
Answer:
(a) tan²θ – sin²θ

We have, tan²θ . sin²θ

Question 14.
The line segment joining the points P(3, -1) and Q(-6, 5) is trisected. The coordinates of point of trisection are
(a) (3, 3)
(b) (-3, 3)
(c) (3, -3)
(d) (- 3, – 3)
Answer:
(b) (-3, 3)

Since, the line segment AB is trisected.
∴ PB : BQ = 2 : 1

∴ Coordinates of B are
= (\(\frac{2(-6)+1(3)}{2+1}\), \(\frac{2(5)+1(-1)}{2+1}\))
= (\(\frac{-12+3}{3}\), \(\frac{10-1}{3}\))
= (\(\frac{-9}{3}\), \(\frac{9}{3}\))
= (-3, 3)

Question 15.
If PT is a tangent at T to a circle, where centre is O and OP =17 cm, OT = 8 cm, then the length of the tangent segment PT is
(a) 10 cm
(b) 20 cm
(c) 15 cm
(d) 25 cm
Answer:
(c) 15 cm

In right angled ∆OTP,
OP² = OT² + TP²

Hence, the length of tangent PT is 15 cm.

Question 16.
While computing the mean of grouped data, we assume that the frequencies are
(a) evenly distributed over all the classes
(b) centred at the class marks of the classes
(c) centred at the upper limits of the classes
(d) centred at the lower limits of the classes
Answer:
(b) centred at the class marks of the classes

While computing mean of grouped data, we assume that the frequencies are centred at the class marks of the classes.

Question 17.
If in two triangles ∆DEF and ∆PQR, ∠D = ∠Q and ∠R = ∠E, then which of the following is not true? [1]
(a) \(\frac{E F}{P R}=\frac{D F}{P Q}\)
(b) \(\frac{D E}{Q R}=\frac{E F}{Q P}\)
(c) \(\frac{D E}{Q R}=\frac{D F}{P Q}\)
(d) \(\frac{E F}{R P}=\frac{D E}{Q R}\)
Answer:
(b) \(\frac{D E}{Q R}=\frac{E F}{Q P}\)

In ∆DEF and ∆PQR,
∠D = ∠Q
∠E = ∠R [given]
[from AA criterion]
∆DEF ~ ∆PQR
⇒ ∠F = ∠P [by CPCT]
∴ \(\frac{E D}{R Q}=\frac{F E}{P R}=\frac{D F}{Q P}\)
Hence, option (b) is not true.

Question 18.
C is the mid-point of PQ is P is (4, x), C is (y,-1) and Q is (-2,4), then * and y respectively are
(a) – 6 and 1
(b) -6 and 2
(c) 6 and -1
(d) 6 and 2
Answer:
(a) – 6 and 1

Since, C(y, – 1)is mid-point of P(4, x) and Q(-2, 4).
So, y = \(\frac{4-2}{2}\) and -1 = \(\frac{4+x}{2}\)
∴ y = 1 and x = – 6

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.
Assertion (A) 5x² + 14x + 10 = 0 has no real roots.
Reason (R) ax² + bx + c = 0 has no real roots if b² < 4ac
(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 but 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
Answer:
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A)

Assertion (A) 5x² + 14x + 10 = 0
D = (14)² – 4.5.10 = 196 – 200 = – 4 < 0.
As, D < 0, no real roots.
Reason (R) True and it is of assertion correct explanation.

Question 20.
Assertion (A) Zeroes of f(x) = x² – 4x – 5 are 5, -1.
Reason (R) The polynomial whose zeroes are 2 + √3, 2 – √3 is x² – 4x + 7.
(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 but 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
Answer:
(c) Assertion (A) is true but Reason (R) is false

Assertion (A) Given, f(x) = x² -4x -5
On splitting the middle term
f(x) = x² – 5x + x – 5
= x(x – 5) + 1 (x – 5)
= (x + 1) (x – 5) f(x) = 0
x = -1 and x = 5 (true)

Reason (R) We know that the polynomial is
x² – (α + ß)x + αß
= x² – (2 + √3 + 2 – √3)x + (2 + √3) . (2 – √3)
= x² – 4x + (4 – 3)
= x² – 4x + 1
Hence, Assertion (A) is true but Reason (R) is false.

Section B
(Section B consists of 5 questions of 2 marks each.)

Question 21.
In the given figure, if ∠RPS = 25°, then value of ∠ROS.

Solution:
Since, OR ⊥ PR and OS ⊥ SP
∠ORP = ∠OSP = 90°
In quadrilateral ORPS,
∠ROS + ∠ORP + ∠RPS + ∠OSP = 360°
⇒ ∠ROS + 90° + 25° + 90° = 360°
⇒ ∠ROS = 360° – 205 = 155°

Question 22.
In given figure, DE || BC. If AD = x, DB = x – 2, AE = x + 2 and EC = x – 1, Find the value of x.

Solution:
In ∆ABC, we have
\(\frac{A D}{D B}=\frac{A E}{E C}\)
[by using basic proportionality theorem]
⇒ \(\frac{x}{x-2}\) = \(\frac{x+2}{x-1}\)
⇒ x(x – 1) = (x – 2) (x + 2)
⇒ x – x = x – 4
⇒ x = 4

Question 23.
If cos θ – sin θ = √2 sin θ, then prove that find the cos θ + sin θ = √ cos θ.
Solution:
Given, cos θ – sin θ = √2 sin θ

Question 24.
Solve the following quadratic equation 6x² + 7x – 10 = 0.
Or
Find the value of K for which the given equation has real and equal roots.
2x² – 10x + k = 0
Solution:
Given, quadratic equation is 6x² + 7x – 10 = 0
On comparing with ax² + bx + c = 0
a = 6, b = 7 and c = -10
D = b² – 4ac = (7)² – 4 × 6 × (-10) = 289 > 0
So, the given equation has real roots, given by

Or
Given, quadratic equation is
2x² – 10x + k = 0
On comparing with ax² + bx + c = 0
a = 2, b = -10 and c = k
D = b² – 4ac = (-10)² – 4 × 2 × k
= 100 – 8 k
The given equation has real and equal roots.
So, D = 0 ⇒ 100 – 8k =0
k = \(\frac{100}{8}\) = \(\frac{25}{2}\)

Question 25.
The short and long hands of a clock are 6 cm and 8 cm long, respectively. Then, find the sum of the distance travelled by their tips in 1 day. [take π = 22 /7]
Or
The radius of the wheel of a bus is 25 cm. If the speed of the bus is 33 km/h, then how many revolutions will the wheel make in 1 min?
Solution:
In 1 day i.e. 24 h, short (hour) hand of the clock make 2 revolutions and long (minute) hand make 24 revolutions.
In 1 revolution, distance travelled by tip of hour hand
= Circumference of circle of radius 6 cm
= 2 × \(\frac{22}{7}\) × 6
In 1 revolution distance travelled by tip of minute hand
= Circumference of circle of radius 8 cm
= 2 × \(\frac{22}{7}\) × 8
∴ Sum of distances travelled by tips of both hand in 1 day
= 2 × 2 × \(\frac{22}{7}\) × 6 + 24 × 2 × \(\frac{22}{7}\) × 8
= 2 × \(\frac{22}{7}\)(12 + 192)
= 2 × \(\frac{22}{7}\) × 204
= 1282.29 cm (approx).

Or

In 1 h, distance covered by wheel = 33 km
In 1 min, distance covered by wheel = \(\frac{33 \times 1000}{60}\) = 550 m
Now, number of revolutions made in 1 min
= \(\frac{\text { Distance covered by wheel }}{\text { Circumference of the wheel }}\)
= \(\frac{550}{2 \times \frac{22}{7} \times \frac{25}{100}}/latex] [∵ 25 cm = [latex]\frac{25}{100}\)m]
= \(\frac{550 \times 7 \times 100}{2 \times 22 \times 25}\) = 350

Section C
(Section C consists of 6 questions of 3 marks each.)

Question 26.
The cost of 2 kg of apples and 1 kg of grapes on a day was found to be ₹160. After a month, the cost of 4 kg of apples and 2 kg of grapes is ₹300. Represent the situation algebraically and geometrically.
Solution:
Let the cost of 1 kg of apples = ₹x
and the cost of 1 kg of grapes = ₹y
Algebraic representation
2x + y = 160 …….(i)
⇒ 4x + 2y = 300
2x + y = 150 ……..(ii)
Geometrical representation
We have, for Eq. (i)

The staight line l₁ and l₂ are the graphical representation of the Eq. (i) and Eq. (ii), respectively. The lines are parallel.

Question 27.
The vertices of ∆ABC are A (4, 6), 15(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E, respectively such that = \(\frac{A D}{A B}=\frac{A E}{A C}=\frac{1}{4}\). Determine the coordinates of D and E.
Solution:
Given,
\(\frac{A D}{A B}=\frac{1}{4}\) ⇒ \(\frac{A D}{A B}=\frac{A D}{AB-AD}\) = \(\frac{1}{4-1}\) = \(\frac{1}{3}\)
∴ D divides AB internally in the ratio 1 : 3

Question 28.
Prove the trigonometric identity
\(\sqrt{\frac{cosec A-1}{cosec A+1}}\) + \(\sqrt{\frac{cosec A+1}{cosec A-1}}\) = 2 sec A.
Or
If tanθ + sinθ = m and tanθ – sinθ = n, then show that (m² – n²)² = 16 mn or (m² -n²) = 4\(\sqrt{mn}\).
Solution:

Or

Given, tanθ + sinθ = m ………(i)
and tanθ – sinθ = n ………..(ii)
On adding Eqs. (i) and (ii), we get
2 tanθ = m + n
⇒ tanθ = \(\frac{m+n}{2}\)
∴ cotθ = \(\frac{1}{tanθ}\) = \(\frac{2}{m+n}\) ……….(iii)
On subtracting Eq. (ii) from Eq. (i), we get
2 sinθ = m – n
sinθ = \(\frac{m-n}{2}\)
∴ cosec θ = \(\frac{1}{sinθ}\) = \(\frac{2}{m-n}\) ……..(iv)
We know that cosec² θ – cot² θ = 1

Question 29.
In a trapezium, show that any line drawn parallel to the parallel sides of the trapezium, divides the non-parallel sides proportionally.
Or
∆ABC is an isosceles triangle in which AB = AC = 10 cm and BC = 12 cm. PQRS is a rectangle inside the isosceles triangle. If PQ = SR = y cm, PS = QR = 2x, then prove that x = 6 – \(\frac{3y}{4}\).
Solution:
Given In trapezium ABCD,
AB || DC and EF || AB
To prove \(\frac{A E}{E D}=\frac{B F}{F C}\)
Construction Join AC to intersect EF at G.

Proof Since, AB || DC and EF || AB
∴ EF || DC
[since, lines parallel to the same line are also parallel to each other]
In ∆ADC,
EG || DC [∵EF || DC]
By using basic proportionality theorem,
\(\frac{A E}{E D}=\frac{A G}{G C}\)
[by Thales theorem] …… (i)

In ∆ABC, GF || AB [∵ EF || AB]
By using basic proportionality theorem,
\(\frac{C G}{A G}=\frac{C F}{B F}\) or \(\frac{A G}{G C}=\frac{B F}{C F}\)
[by Thales theorem] … (ii)
and taking reciprocal of the terms]
From Eqs. (i) and (ii), we get
\(\frac{A E}{E D}=\frac{B F}{F C}\)
Hence proved.

OR
Given, AB = AC = 10 cm, BC = 12 cm and PORS is a rectangle in which PC = SR = y cm and PS = QR = 2x.

Question 30.
A bag contains 8 red balls and some blue balls. If the probability of drawing a blue ball is three times of a red ball, then find the number of blue balls in the bag.
Solution:
Let there be x blue balls in the bag.
∴ Total number of balls in the bag = (8 + x)
Now, P₁ = Probability of drawing a blue ball
= \(\frac{x}{8+x}\)
and P₂ = Probability of drawing a red ball
= \(\frac{8}{8+x}\)
It is given that
P₁ = 3P₂
⇒ \(\frac{x}{8+x}\) = 3 × \(\frac{8}{8+x}\)
⇒ \(\frac{x}{8+x}\) = \(\frac{24}{8+x}\)
⇒ x = 24
Hence, there are 24 blue balls in the bag.

Question 31.
If α and β are the zeroes of the quadratic polynomial f(x) = 3x² -4x + 1, then find a quadratic polynomial whose zeroes are
\(\frac{\alpha^2}{\beta}\) and \(\frac{\beta^2}{\alpha}\).
Solution:
Given that α and β are the zeroes of the polynomial
f(x) = 3x² -4x + 1
α + β = \(-\frac{\text { Coefficient of x }}{\text { Coefficient of x^2 }}\)
= \(-\frac{(-4)}{3}\) = \(\frac{4}{3}\)
and α . β = \(\frac{\text { Constant term }}{\text { Coefficient of x^2 }}\)
= \(\frac{1}{3}\)
Lets and P be sum and product of the zeroes of the
polynomial whose zeroes are \(\frac{\alpha^2}{\beta}\) and \(\frac{\beta^2}{\alpha}\).
Then,
S = \(\frac{\alpha^2}{\beta}\) + \(\frac{\beta^2}{\alpha}\) = \(\frac{\alpha^3+\beta^3}{\alpha\beta}\)

Section D
(Section D consists of 4 questions of 5 marks each)

Question 32.
The shadow of a vertical tower on a ground level increases by 10 m, when the angle of elevation of the Sun changes from 45° to 30°. Find the height of the tower correct to two decimal places.
Solution:
Let the height of vertical tower be AB = h m
and the length of shadow initially be BC = x m.

New length of shadow, BD = (10 + x) m
In ∆ABC, we have
tan45° = \(\frac{A B}{B C}\)
⇒ 1 = \(\frac{h}{x}\)
⇒ x = h …….(i)

In ∆ABD, we have

Question 33.
Find the missing frequencies in the following frequency distribution table, if n = 100 and median is 32.
Table 1
Solution:
Given, median = 32 and N = Σf_i = 100
Let f₁ and f₂ be the frequencies of the class interval 10-20 and 40-50, respectively.
Since, sum of frequencies = 100
∴ 10 + f₁ + 25 + 30 + f₂ + 10 = 100
⇒ f₁ + f₂ = 100 – 75
⇒ f₁ + f₂ = 25
⇒ f₂ = 25 – f₁ …(i)
Now, the cumulative frequency table for given distribution is

Here, N = 100 ⇒ \(\frac{N}{2}\) = 50
Given, median = 32, which belongs to the class 30-40,
So, the median class is 30 – 40.
Then, l = 30, h = 10, f = 30 and cf = 35 + f₁
∵ \(\text { Median }=l+\left[\frac{\frac{N}{2}-cf}{f}\right] \times h\)
32 = 30 + \(\left\{\frac{50-35-f_1}{30}\right\}\) × 10
32 – 30 = \(\frac{15-f_1}{30}\) × 10
⇒ 2 × 3 = 15 – f₁ ⇒ f₁ = 15 – 6 = 9
On putting the value of A, in Eq. (i), we get
f₂ = 25 – 9 = 16
‘Hence, the missing frequencies are f₁ = 9 and f₂ = 16.

Question 34.
A solid is composed of a cylinder with I hemispherical ends. If the whole length of the solid is 104 cm and the radius of each hemispherical end is 7 cm, then find the cost of polishing its surface at the rate of ₹2 per dm². [take π = 22/7]
Or
A gulabjamun contains sugar syrup upto about 30% of its volume. Find approximately how much syrup would be found in 45 gulabjamuns, each shaped like a cylinder with two hemispherical – ends with length 5 cm and diameter 2.8 cm (see figure).

Solution:
Given, whole length of the solid = 104 cm and
the radius of each hemisphere = 7 cm
Therefore, the length of the cylindrical part of the solid = (104 – 2 × 7) = 90 cm
For hemispherical portion
Radius, r = 7 cm
For cylindrical portion
Radius, r = 7 cm
Height, h = 90 cm
Total surface area of the solid
= 2 × Curved surface area of a hemisphere + Curved surface area of cylindrical part

= 2 [2πr²] + 2πrh
= 2 × [2π (7)²] + 2π (7) (90)
= 2 × [2 × \(\frac{22}{7}\) × (7)²] + 2 × \(\frac{22}{7}\) × (7) (90)
= 4 × 22 × 7 + 2 × 22 × 90
= 22 [28 + 180]
= 4576 cm²
Then, the cost of polishing at the rate of ₹2 per dm²
= ₹ 4576 × 2 = ₹ 91.52 [∵ dm² = 100 cm²]

Or

Given, one gulabjamun is a combination of a cylinder and two hemispheres.
Here, total length of one gulabjamun = 5 cm
and diameter = 2.8 cm

Radius of cylindrical part = Radius of hemispherical part
= r = \(\frac{2.8}{2}\) = 1.4 cm
Height of cylindrical part,
h =PQ – (PR + SQ) = 5 – (1.4 + 1.4)
= 5 – 2.8 = 2.2 cm
Volume of one gulabjamun
= 2 × Volume of hemispherical part + Volume of cylindrical part

Since, one gulabjamun contains sugar syrup upto about 30% of its volume.
Hence, the quantity of syrup found in 45 gulabjamuns
= 1127.28 × \(\frac{30}{100}\) = 1127.28 × \(\frac{3}{10}\)
= 338.184 = 338 cm²

Question 35.
In an AP, the sum of m terms is equal to n and the sum of n terms is equal to m, then prove that the sum of (m + n) terms is -(m + n).
Or
The sum of the third and seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of an AP.
Solution:
Let a and d be the first term and common difference of an AP.
Then, S_m = n
⇒ \(\frac{m}{2}\)[2a + (m – 1)d] = n
⇒ 2am + m(m – 1)d =2n …(i)
and S_n = m
⇒ \(\frac{n}{2}\)[2a + (n -1)d] = m
⇒ 2an + n(n – 1)d = 2m …..(ii)
On subtracting Eq. (ii) from Eq. (i), we get
2a(m-n) + {m(m -1)-n(n -1)}d = 2n – 2m
⇒ 2a(m – n)+{m² – m – n² + n}d = 2(n – m)
⇒ 2a(m – n) + {m² – n² -(m – n)}d = -2(m – n)
⇒ (m -n) [2a + (m + n – 1)d] = -2(m – n)
[∵ a² – b² = (a + b)(a – b)]
⇒ [2a+(m + n – 1)d] = -2
[dividing both sides by (m – n)]
⇒ 2a + (m + n – 1)d = -2 …………(iii)
Now, S_m+n = \(\frac{m+n}{2}\)[2a + (m + n – 1)d]
= \(\frac{m+n}{2}\)[-2] [from Eq. (iii)]
= -(m+n)
Hence proved.

Or

Let a and d be the f:rst term and common difference of an AP
Then, a₃ + a₇ = 6 and a₃ a₇ = 8
⇒ a + (3 – 1)d+ a + (7 – 1)d = 6
and [a + (3-1)c/][a-ri7-iiry] = 8
[∵ a_n = a + (n – 1)d]
⇒ a + 2d + a + 6d = 6
and [a + 2d][a + 6d] = 8
⇒ 2a + 8d = 6 and (a + 2d)(a+ 6d) = 8
⇒ a+ 4d = 3 and (a + 2d)(a + 6d) = 8
⇒ (3 – 4d + 2d )(3 – 4d + 6d) = 8
[∵ put a = 3 – 4d]
⇒ (3 – 2d)(3 + 2d) = 8
⇒ 9 – 4d² = 8
⇒ 4d² = 1
⇒ d² = \(\frac{1}{4}\)
d =±\(\frac{1}{2}\)
When d = \(\frac{1}{2}\)
a + 4d = 3 ⇒ a + 2 = 3 ⇒ a = 1

Section E
(Case study based questions are compulsory)

Question 36.
Book Seller
A book seller has 420 Science stream books and 130 Art stream books. He wants to stack them in such a way that each stack has the same number and they take up the least area of the surface.

On the basis of above information, answer the following questions.
(i) If number has no factors other than 1 and number it [1]
(ii) What is the maximum number of books that can be placed in each stack for this purpose? [2]
Or
If the book seller double the quantity, then what is the maximum number of books that can be placed in each stack? [2]
(iii) Find the LCM of the given book streams. [1]
Solution:
(i) If a number has no factors other than 1 and number itself is a prime number.

(ii) Given, number of Science books = 420
and number of Art books = 130

Maximum number of books that can be placed in each stack for the given purpose = HCF (420,130)
= 2¹ × 5¹ = 10
Or
If the book seller double the quantity, then the maximum number of books that can be placed in each stack is also doubled.

(iii) LCM of (420, 130) =2² × 3 × 5 × 7 × 13
= 4 × 15 × 91
= 5460

Question 37.
A ferris wheel (or a big wheel in the United Kingdom) is an amusement ride consisting of a rotating upright wheel with multiple passenger-carrying components (commonly referred to as passenger cars, cabins, tubes, capsules, gondoals, or pods) attached to the rim in such a way that as the wheel turn, they are kept upright, usually by gravity.
After taking a ride in Ferris wheel, Aarti came out from the crowd and was observing her friends who from the crowd and was observing her friends who were enjoying the ride. She was curious about the different angles and measures that the wheel will form. She forms the figure as given below.

(i) In the given figure, find ∠ROQ. [1]
(ii) Find ∠RSQ. [2]
Or Find ∠ORP [2]
(iii) Find ∠RQP [1]
Solution:
(i) In quadrilateral PQOR, we have

∠QPR + ∠PRO + ∠PQO + ∠ROQ = 360°
⇒ 30° + 90° + 90° + ∠ROQ = 360°
[∵ radius is always perpendicular to the tangent at point of contact]
⇒ ∠ROQ = 360° – 210° = 150°

(ii) We know that angle subtended by an arc at centre is double the angle subtended by it at any other part of the circle.
2 ∠RSQ = ∠ROQ
∠RSQ = \(\frac{1}{2}\) × 150° = 75°
Or
∠ORP = 90° as radius is always perpendicular to the tangent at the point of contact.
(iii) In AQOR, OQ = OR [radii]
Now, ∠ROQ + ∠ORQ + ∠OQR = 180°
⇒ 2∠OQR = 180° – 150°
⇒ 2∠OQR = 30°
⇒ ∠OQR = 15°
Again, ∠OQP = 90° [∵ OQ ⊥ OP]
⇒ ∠OQR + ∠RQP = 90°
⇒ ∠RQP = 90° – 15° = 75°

Question 38.
Your elder brother wants to buy a car and plans to take loan from a bank for his car. Fie repays his total loan of ₹118000 by paying every month starting with the first he increases the installment by ₹100 every month.

On the basis of above information, answer the following questions
(i) What amount does he still have to pay after 30th installment? [1]
(ii) Find the amount paid by him in 30th installment. [2]
Or Find the amount paid by him in the 30 installments. [2]
(iii) If total installments are 40, then what amount paid in the last installment? [1]
Solution:
(i) After 30th installment, he still have to pay
= 118000 – 73500 = ₹44500
(ii) Since, he pays first installment of ₹1000 and next consecutive months he pay the installment are 1100, 1200, 1300, …..
Thus, we get the AP sequence,
1000, 1100, 1200, …
Here, a = 1000 and d =1100 -1000 = 100
Now, T₃₀ = a + (30 – 1)d [∵ T_n = a + (n-1)d]
= 1000 + 29 × 100
= 1000 + 2900 = 3900
Hence, the amount paid by him in 30th installment is ₹3900.
Or
S₃₀ = \(\frac{30}{2}\){2a + {30-1)d}
[∵ S_n = \(\frac{n}{2}\) {2a + (n – 1) d}]
= 15(2 × 1000 + 29 × 100)
= 15(2000 + 2900)
= 15 × 4900
= ₹73500

(iii) The amount in last 40th installment is
T₄₀ = a + (40 – 1 )d
= 1000+ 39 × 100
= 1000 + 3900
= ₹4900