CBSE Sample Papers for Class 10 Maths Standard Set 1 with Solutions - #NCSOLVE 📚

Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths Standard with Solutions Set 1 are designed as per the revised syllabus.

CBSE Sample Papers for Class 10 Maths Standard Set 1 with Solutions

Time: 3 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 ofl mark each.)

Question 1.
If a = 2² x 3^x,  b = 2² x 3 x 5, c = 2²    x  3 x 7 and LCM (a, b, c) = 3780 then x is equal  to
(a) 1
(b) 2
(c) 3
(d) 0
Answer:
(c) Given, a = 2² x 3^x, b = 2² x 3 x 5, c = 2² x 3 x 7
and LCM (a, b, c) =3780
We know that LCM is the product of the greatest power of each prime factor involved in the number.
∴ LCM (a, b, c) = 2² x 3¹ x 5 x 7
= 3780 = 2² x3^xx5x7
⇒ 3780 = 140 x 3^x
⇒  3^x = \(\frac{3780}{140}\) =27
⇒ 3^x = 3³ ⇒ x = 3

Question 2.
The shortest distance (in units) of the point (2,3) from Y-axis is
(a) 2
(b) 3
(c) 5
(d) 1
Answer:
(a) The shortest distance from (2, 3) to Y-axis is the x-coordinate i.e. 2.

Question 3.
If the lines given by 3x + 2ky = 2 and 2x + 5t + 1=0 are not parallel then k has to be
(a) \(\frac{15}{4}\)
(b) ≠ \(\frac{15}{4}\)
(c) any rational number.
(d) any rational number having 4 as denominator.
Answer:
(b) Given, lines are 3x + 2ky = 2 and 2x + 5y + 1 = 0.
The given equations can be rewritten as 3x + 2ky -2 = 0 and 2x + 5y + 1 = 0
On comparing the given equation with standard form i.e. a₁x + b₁y + c₁= 0 and a₂x + b₂y + c₂= 0.
We get,
a₁= 3, b₁ = 2k, c₁= – 2
and a₂ = 2, b₂ = 5, c₂ = 1
Also, given the lines are not parallel.
∴ \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2} \Rightarrow \frac{3}{2} \neq \frac{2 k}{5} \Rightarrow k \neq \frac{15}{4}\)

Question 4.
A quadrilateral ABCD is drawn to circumscribe a circle. If BC = 7 cm, CD = 4 cm and AD = 3 cm then the length of AB is
(a) 3 cm
(b) 4 cm
(c) 6 cm
(d) 7 cm
Answer:
(c) For a quadrilateral ABCD circumscribe a circle, the sum of ihe lengths of opposite sides are equal.
∴ AB + CD = BC+ AD ⇒ AB + 4 = 7 + 3 ⇒ AB = 6 cm

Question 5.
If sec0 + tan0 = x then sec0 – tan0 will be
(a) x
(b) x²
(c) \(\frac{2}{x}\)
(d) \(\frac{1}{x}\)
Answer:
(d) Given, sec θ + tan θ = x
We know that sec²θ-tan²θ = 1
⇒ (sec θ – tan θ) (sec θ + tan θ) = 1
⇒ (sec θ – tan θ) × x = 1
⇒ sec θ – tan θ = \(\frac{1}{x}\)

Question 6.
Which one of the following is not a quadratic equation?
(a) (x + 2)² = 2(x + 3)
(b) x² + 3x= (-1)(1 – 3x)²
(c) x³ – x² + 2x + 1 = (x + 1)³
(d) (x + 2) (x + 1) = x² + 2x + 3
Answer:
(d) From option (d), we have
(x + 2) (x + 1) = x² + 2x + 3
⇒ x² + 2x + x + 2 = x² + 2x + 3
⇒ x + 2 = 3 ⇒ x – 1 = 0
Which is not a quadratic equation.

Question 7.
Given below is the picture of the Olympic rings made by taking five congruent circles of radius lcm each, intersecting in such a way that the chord formed by joining the point of intersection of two circles is also of length 1 cm.Total area of all the dotted regions (assuming the thickness of the rings to be negligible) is

Answer:
(d) Given, radius of each circle = 1cm and length of the chord joining the point of intersection of two circle = 1 cm.
Since, the chord length is equal to the radius, the triangle formed by centre and the two intersection points is an equilateral triangle with side length 1 cm.
The angle subtended by the chord at the centre is 60°.

Question 8.
A pair of dice is tossed. The probability of not getting the sum eight is……….
(a) \(\frac{5}{36}\)
(b) \(\frac{31}{36}\)
(c) \(\frac{5}{18}\)
(d) \(\frac{5}{9}\)
Answer:
(b) When two dices are tossed, the total number of possible outcomes are 6× 6. i.e. 36.
Let E = Event of getting the sum eight then E would consist of 5 outcomes namely (2,6), (3,5), (4,4), (5,3) and (6, 2).
∴ Number of outcomes favourable of E = 5.
Now, probability of getting the sum eight,
P(E) = \(\frac{5}{36}\)
∴ Probability of not getting the sum eight,
\(P(\bar{E})=1-P(E)=1-\frac{5}{36}=\frac{31}{36}\)

Question 9.
If 2 sin 5x = √3,0° ≤ x ≤ 90° then x is equal to
(a) 10°
(b) 12°
(c) 20°
(d) 50°
Answer:

Question 10.
The sum of two numbers is 1215 and their HCF is 81 then the possible pairs of such numbers are
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(c) Given, HCF of two numbers = 81
∴ The numbers can be 81x and 81y.
Also, given 81x + 81y = 1215 ⇒ x + y= 15
Which given four pair as (1,14), (2,13), (4,11) and (7, 8).

Question 11.
If the area of the base of a right circular cone is 51cm² and it’s volume is 85cm² then the height of the cone is given as
(a) \(\frac{5}{6}\)
(b) \(\frac{5}{3}\)
(c) \(\frac{5}{2}\)
(d) 5 cm
Answer:
(d) Given, area of the base of a right circular cone = 51cm²
⇒ πr² = 51    …. (i)
Also, given volume of right circular cone = 85 cm³
⇒ \(\frac{1}{3}\) πr² h = 85
⇒ \(\frac{1}{3}\)  x 51 x h = 85 [From eq –(i)]
⇒ 17h = 85
⇒ h = 5 cm

Question 12.
If zeroes of the quadratic polynomial  a x²+b x+c(a, c≠0) are equal then
(a) c and b must have opposite signs
(b) c and a must have opposite signs
(c) c and b must have same signs
(d) c and a must have same signs
Answer:
(d) Given, quadratic polynomial is ax² + bx + c.
We know that if the zeroes are equal then the value of discriminant is zero.
∴ b² – 4ac = 0
⇒ b² = 4 ac
⇒ac = \(\frac{b^2}{4}\)
∴ ac ≥ 0 [∵ \(\frac{b^2}{4}\) is always positive
Hence, the product is positive if a and c must have the same sign (either both positive or both negative).

Question 13.
The area (in cm²) of a sector of a circle of radius 21cm cut off by an arc of length 22 cm is
(a) 441
(b) 321
(c) 231
(d) 221
Answer:
(c) Given, length of the arc, l = 22 cm
and radius of the circle, r = 21 cm
We know that
Area of the sector of a circle = \(\frac{1}{2}\) lr
= \(\frac{1}{2}\) x 22 x 21 = 231cm² 

Question 14.
If ΔABC – ΔDEF, AB = 6 cm, DE- 9 cm, EF = 6 cm and FD = 12 cm then the perimeter of ΔABC is
(a) 28 cm
(b) 28.5 cm
(c) 18cm
(d) 23 cm
Answer:
(_C) Given, ΔABC ~ ΔDEF
⇒ \(\frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}\) —–(i)
[ ∵ corresponding sides of two similar triangles are proportional]
Also given, AB = 6 cm, DE = 9 cm, EF = 6 cm and FD= 12 cm.
On taking first two terms of Eq. (i),

On taking first and third term of Eq. (i), we get

Now, perimeter of
ΔABC = AB + BC+ CA
= 6 cm + 4 cm + 8 cm
= 18 cm

Question 15.
If the probability of the letter chosen at random from the letters of the word ‘Mathematics’ to be a 2 vowel is \(\frac{2}{2 x+1}\) then x is equal to
(a) \(\frac{4}{11}\)
(b) \(\frac{9}{4}\)
(c) \(\frac{11}{4}\)
(d)\(\frac{4}{9}\)
Answer:
(b) Total number of letters in the word Mathematics = 11
∴Total number of possible outcomes =11
and number of vowel in the Mathematics = 4
∴ Probability of getting vowels in the word 4
Mathematics = \(\frac{4}{11}\)
Also, given probability of getting vowel in the word 2
Mathematics = – \(\frac{2}{2 x+1} \)

Question 16.
The points A(9,0), B(9, -6) ,C(-9,0) and D(-9,6) are the vertices of a
(a) square
(b) rectangle
(c) parallelogram
(d) trapezium
Answer:

Since, the opposite sides are equal and the diagonals are not equal.
Therefore, A, B, C and D are the vertices of a parallelogram.

Question 17.
The median of a set of 9 distinct observation is 20.5. If each of the observations of a set is increased by 2 then the median of a new set
(a) is increased by 2
(b) is decreased by 2
(c) is two times the original number
(d) remains same as that of original observations
Answer:
(a) We know that if each of the observations of a set is increased by a constant value say k, then the median of the new set is also increased by k. Hence, if each of the observation of a set is increased by 2 then the median is also increased by 2.

Question 18.
The length of a tangent drawn to a circle of radius 9 cm from a point at a distance of 41cm from the centre of the circle is
(a) 40 cm
(b) 9 cm
(c) 41 cm
(d) 50 cm
Answer:
(a) Given, radius of the circle, OA=9 cm
and distance from the centre of the circle to the external point, OP=41cm

DIRECTIONS: In the question number 19 and 20, a statement of Assertion (A) is followed by a statement of Reason (R).
Choose the correct option:
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A)
(b) Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A)
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true.

Question 19.
Assertion (A) The number 5ⁿ cannot end with the digit 0, where n is a natural number.
Reason (R) A number ends with 0, if its prime factorisation contains both 2 and 5.
Answer:
(a) Prime factorisation of
5ⁿ = 5 x 5 x…. x 5 (n times)
Here, prime factorisation of 5ⁿ contains only 5.
We know that a number ends with 0, If its prime factorisation contains both 2 and 5.
Hence, both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Question 20.
Assertion (A) If cos A + cos² A = 1 then sin² A+ sin⁴ A = 1.
Reason (R) sin² A + cos² A = 1.
Answer:
(a) Given, cos A+ cos² A = 1 ….(i)
⇒ cos A = 1 – cos² A
⇒ cos A = sin² A   ….(ii) [∵ sin² A + cos² A = 1]
On substituting the value of cos A from Eq. (ii) in Eq. (i), we get
sin² A + (sin² A)² = 1
⇒ sin² A + sin⁴ A – 1
Hence, both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

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

Question 21
(a) The AP 8,10,12,………. has 60 terms. Find the sum of last 10 terms.
Or
(b) Find the middle term of AP 6,13, 20,………….  230.
Answer:
(a) Given AP is 8,10,12,… upto 60 terms.
Here, first term, a = 8,
common difference, d = 10-8 = 2
and total number of terms, n = 60
a₆₀ = a + (60 – 1) d [∵ = a + (n- 1) d]
= 8+59 x2 = 8+118 = 126
and a₅₁ = a + (51 – 1 )d = 8+ 50 x 2
= 8 + 100 = 108
The last 10 terms of given AP are
a₅₁,a₅₂, ………….a₅₃
∴ Sum of last ten term = \(\frac{10}{2}\left[a_{51}+a_{60}\right]\)
= 5 [108+ 126]
= 5 x 234 = 1170
Or
(b) Given, AP is 6,13, 20,….,230
Here, first term, a = 6
common difference, d = 13 – 6 = 7
and last term, l = 230
∴ a_n = l = 230
⇒ a + (n – 1) d = 230
⇒ 6 + (n – 1) 7 = 230
⇒ (n- 1)7 = 224
⇒ n – 1 = 32
⇒ n = 33
We know that if number of terms in an AP is \(\frac{n+1}{2}\) th term
Here, n = 33
∴ Middle term =\(\frac{33+1}{2}=\frac{34}{2}\) =17th term
∴ a₁₇ = a + (17- 1) d = 6 + 16×7
= 6+ 112 = 118
Hence, the middle term is 118.

Question 22.
If sin(A+ B) = land cos(A- B) = \(\frac{\sqrt{3}}{2}\) ,0° < A, B< 90°, find the measure of angles A and B
Answer:
Given, sin( A + B)= 1
⇒ sin( A + B) = sin 90°
⇒ A + B = 90°…………..(i)
and ( A – B) = \(\frac{\sqrt{3}}{2}\)
cos( A – B) = cos 30°
A – B = 30°……….(ii)
On adding Eq. (i) and (ii), we get
2A = 120³ ⇒ A = 60°
On putting value of A in Eq. (i), we get 60P + B = 90°
⇒ B-30°
Hence, the measure of angle A and B are 60° and 30°, respectively.

Question 23.
If AP and DQ are medians of triangles ABC and DEF respectively, where ΔABC – ΔDEF then prove that  \(\frac{A B}{D E}=\frac{A P}{D Q}\)
Answer:
Given, AP and DQ are the medians of the triangles ABC and DEF,

Question 24
(a) A horse, a cow and a goat are tied, each by ropes of length 14m, at the comers A, B and C respectively, of a grassy triangular field ABC with sides of lengths 35 m, 40 m and 50 m. Find the area of grass field that can be grazed by them.
Or
(b) Find the area of the major segment (in terms of n) of a circle of radius 5cm, formed by a chord subtending an angle of 90° at the centre.
Answer:
(a) The length of the rope is the radius of the circular area each animal can graze, so r=14m
Now, area of grass filed that can be grazed by them

Or
(b) Given, radius of the circle, r = 5 cm
and angle subtended by arc at the centre,
θ=90°
Now, area of the minor sector

Question 25.
A ΔABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC are of lengths 10 cm, and 8 cm respectively. Find the lengths of the sides AB and AC, if it is given that ar(ΔABC) = 90 cm².

Answer:
Given, BD = 10 cm, DC = 8 cm
and ar(ΔABC) = 90 cm².
Let r be the radius of the inscribed circle we know that the length of tangents from an external point to the circle are equal.
.’. BD= BE= 10 cm CD = CF = 8 cm and AF = AE = rcm (let)
Therefore, AC = AF+ FC = (x + 8) cm
BC = BD+ CD= 10+8= 18cm
and AB= AE+ BE = (x + 10) cm
Now, ar(ΔABC) = ar(ΔAOC) + ar(ΔBOC) + ar Δ(AOB)
= \(\frac{1}{2}\) x r x AC +\(\frac{1}{2}\) x r x BC+\(\frac{1}{2}\)  x r x.AB
[∵OF=OD=OE =\(\vec{r}\)]
⇒ 90 = \(\frac{1}{2}\) x 4 (x + 8+ 18 + x + 10) [r = 4cm]
⇒ 45 = 2x + 36
⇒ 2x = 9 ⇒ x = 45cm
∴ AF = AE = 45 cm
Now, AB = AE + BE = 45+ 10= 14.5cm
and AC = AF + CF = 45 + 8 = 12.5 cm

SECTION-C
(Section B consists of 5 questions of 3 marks each.)

Question 26.
In the given figure, XY and X’ Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X’ Y’ at B. Prove that ∠AOB = 90°.

Answer:
Given XY and X’Y’ are two parallel tangents.
Another tangent AB touches the circle at C and intersect XY at A and X’Y’ at B.
To prove ∠AOB = 90°
Proof We know that tangents drawn from an external point to a circle are equal in length.
∴ AP = AC [∵ A is an external point],..(i)
Thus, in ΔAPO and SACO, AP = AC [from Eq. (i)]
AO=AO    [common sides]
OP = OC   [radii of a circle]
ΔAPO ≅ SACO   [by SSS congruence rule]
Then, ∠OAP = ∠OAC [by CPCT]… (ii)
⇒ ∠PAC = 2 ∠CAO  …………..(iii)
Similarly, we can prove that ∠CBO= ∠OBQ
⇒ ∠CBQ= 2∠CBO  ………. (iv)
Since,   XY || X’Y’ [given]
∴ ∠PAC + ∠QBC= 180°
[∵ sum of interior angles on the same side of transversal is 180°]
⇒ 2 ∠CAO+ 2 ∠CBO = 180° [from Eqs. (iii) and (iv)]
⇒ ∠CAO + ∠CBO=  90° …………..(v)
Now, in ΔAOB,
∠CAO+ ∠CBO+ ∠AOB = 180° [by angle sum property of triangle]
⇒ ∠CAO + ∠CBO = 180° – ∠AOB ………..(vi)
From Eqs. (v) and (vi), we get 180° – ∠ AOB = 90°
⇒ ∠AOB = 90°   Hence proved.

Question 27.
In a workshop, the number of teachers of English, Hindi and Science are 36, 60 and 84, respectively. Find the minimum number of rooms required, if in each room the same number of teachers are to be seated and all of them being of the same subject.
Answer:
Given, number of teachers of English = 36,
number of teachers of Hindi = 60
and number of teachers of Science = 84
Prime factorisation of 36, 60 and 84 are
36 = 2² x 3²
60 = 2² x 3 x 5
and 84 = 2² x 3 x 7
∴ HCF = (36, 60, 84) = 2² x 3 = 12
[∵ HCF is the product of the smallest power of each common prime factor]
So, 12 teachers can be seated in each room.
Required number of rooms = \(\frac{36}{12}+\frac{60}{12}+\frac{84}{12}\)
= 3+5+ 7=15

Question 28.
Find the zeroes of the quadratic polynomial 2x² – (1 + 2-√2 ) x + √2 and verify the relationship between the zeroes and coefficents of the polynomial.
Answer:

So, the relationship between the zeroes and its coefficients is verified.

Question 29.
(a) If sinθ + cosθ = √3 then prove that tan0 + cot9 = 1.
Or
Prove that \(\frac{\cos A-\sin A+1}{\cos A+\sin A-1}\) +cosec A+cot A.
Answer:

Question 30.
On a particular day, Vidhi and Unnati couldn’t decide on who would get to drive the car. They had one coin each and flipped their coin exactly three times. The following was agreed upon:
1. If Vidhi gets two heads in a row, she would drive the car.
2. If Unnati gets a head immediately followed by a tail, she would drive the car.
Who has greater probability to drive the car that day? Justify your answer.
Answer:
When a coin flipped exactly three times then the possible outcomes are
HHH, HHT, HTH, THH, HIT, THT, TTH, TTT.
∴ Total number of possible outcomes = 8
The outcomes where Vidhi gets two heads in a row
are HHH, HHT and THH.
So, the number of favourable outcomes for Vidhi = 3
∴ Required Probability, P(Vidhi drives the car) = \(\frac{3}{8}\)
The out comes where Unnati gets a head immediately followed by a tail
are THH, HTH, THT and TTH.
So, the number of favourable outcomes for Unnati = 4
∴ Required Probability, P(Unnati drives the car) = \(\frac{4}{8}\)
Since \(\frac{4}{8}\) >\(\frac{3}{8}\)
∴  Unnati has greater probability to drive the car.

Question 31
(a) The monthly income of Aryan and Babban are in the ratio 3:4 and their monthly expenditures are in ratio 5 : 7. If each saves ₹15000 per month, find their monthly incomes.
Or
(b) Solve the following system of equations graphically:
2x + y = 6,2x – y-2 = 0. Find the area of the triangle so formed by two lines and X-axis.
Answer:
(a) Let the income of Aryan and Babban be 3x and Ax, respectively.
and their expenditure be 5y and 7y.
Since, each save ₹ 15000.
3x-5y= 15000                            …(i)
4x-7y = 15000                           ….(ii)
On multiplying Eq. (i) by 4 and Eq. (ii) by 3, we get
2x – 20y = 60000                         …(iiii)
12x-21y = 45000                         …(iv)
On subtracting Eq. (iv) from Eq. (iii), we get y =15000
On puting y = 15000 in Eq. (i),
we get 3x – 5 x 15000 = 15000
⇒ 3x = 15000 + 75000 = 90000
⇒ x = 30000
Hence, income of Aryan
= 3x = 3 x 30000 = ₹ 90,000 and income of Babban
= 4x = 4 x 30000 = ₹ 1,20,000
Or
(b) Given, system of equations is 2x + y = 6 and 2x – y – 2 = 0
Now, let us find at least two solutions of each of the above equations, as shown in the following tables.
Table for line 2x + y= 6 or y = 6- 2x is

Now, plot the points A (0,6), B(2,2) and C(3,0) on a graph paper and join them to get a straight line AC. Similarly plot the points D(1,0), B(2,2) and E(4,6) on same graph paper and join them to get a straight line DE.

It is clear from the graph that the lines AC and DE intersects each other at B(2,2). So, the solution of the given system of equations is (2,2).
The coordinates of triangle BCD formed by two lines and X-axis are B(2,2), C(3,0) and D(1,0).
∴ Area of Δ BCD =\(\frac{1}{2}\) x BM x DC= \(\frac{1}{2}\) x 2 x 2 = 2 sq units.

SECTION-D
(Section D consists of 4 questions of 5 marks each.)

Question 32.
A train travels at a certain average speed for a distance of 63 km and then travels a distance of 72 km at an average speed of 6 km/hr more than its original speed. If it takes 3 hours to complete the total journey, what is the original average speed?
Answer:
Let the original average speed of the train be x km/hr.
The time taken to travel the first 63 km is t₁=\(\frac{63}{x}\) hour

⇒ 63(x + 6) + 72x = 3x(x + 6)
⇒ 63+ 378 + 72x = 3x² + 18x
⇒ 3x² + 18x – 135x – 378 = 0
⇒ 3x² – 117x – 378 = 0
⇒ x² – 39x- 126 = 0
⇒ x² – 42x + 3x – 126 = 0  [by splitting the middle term]
⇒ x(x – 42) + 3(x – 42) = 0
⇒(x – 42) (x +3) = 0
⇒ x = 42, – 3
Since, the speed cannot be negative, so the original average speed is 42 km/hr.

Question 33.
Prove that 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. Hence, in ΔPQR, prove that a line l intersects the sides PQ and PR of a ΔPQR at L and M respectively such that LM||QR. If PL = 5.7 cm, PQ = 15.2 cm and MR = 5.5 cm then find the length of PM (in cm).
Answer:
Let a ΔABC is which a line DE parallel to BC intersect AB to D and AC to E
Construction Join BE, CD and draw EF ⊥ AB and DG⊥AC.
To prove DE divides the two sides in the same ratio.
i..e \(\frac{A D}{D B}=\frac{A E}{EC}\)
Proof:



Hence, the length of PM is 3.3 cm.

Question 34.
(a) From a solid right circular cone, whose height is 6 cm and radius of base is 12 cm, a right circular cylindrical cavity of height 3 cm and radius 4 cm is hollowed out such that bases of cone and cylinder form concentric circles. Find the surface area of the remaining solid in terms of π.
Or
(b) An empty cone of radius 3 cm and height 12 cm is filled with ice-cream such that the lo wer part of the cone, which is \(\left(\frac{1}{6}\right) \mathrm{th}\) of the volume of the cone is unfilled (empty) but a hemisphere is formed on the top. Find the volume of the ice-cream.
Answer:
(a) Given, the height of a solid circular cone (h₁) = 6 cm
and the radius (r₁) = 12 cm.
So, the slant height

Now, curved surface area of cone
= πr₁= π x 12 x 6√5
= 72√5π cm²
and the area of the base of cone
= π₁² = π×(12)²
= 144π cm²
Also, given the height of right circular cylinder cavity (h₂)= 3 cm and radius (r₂) = 4 cm
Now, curved surface area of cylinder = 2πr₂h₂ = 2π × 4 × 3
= 24π cm²
and area of the circular base of cylinder = πr₂²
= π x 4² = 16π cm²
Now, remaining area of base = Area of base of cone – Area of base of cylinder
= 144π – 16π = 128π cm²
So, total surface area of remaining solid
= Curved surface area of cone + Remaining area of base + Curved surface of cylinder + Area of base of cylinder
= 72√5π + 128π + 24π + 16π
= 72√5π + 168π
= (72√5 + 168)π cm²
Or
(b) Given, the radius of cone (i;) = 3 cm
and height of cone (h) = 12 cm
Since, a hemisphere is formed on the top.
The radius of hemisphere (r₂) = 3
Now, the volume of cone = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3}\) × π × (3)² x 12
= 36π cm³
Since, the lower part, which is \(\left(\frac{1}{6}\right)\)th of the volume of the cone, is unfilled.
∴ The filled part of the cone is
1 – \(\frac{1}{6}=\frac{5}{6}\) of the total cone volume.
So, the volume of filled part of cone
= \(\frac{5}{6}\) x volume of cone
= \(\frac{5}{6}\)  x 36π = 30π cm³
Now, the volume of hemisphere = \(\frac{2}{3}\) πr³
= \(\frac{2}{3}\)  x 71 x (3)³
= 18π cm³
∴ The total volume of the ice-cream
= Volume of cone + Volume of hemisphere
= 30π + 18π = 48π
= 48 × 3.14= 150.72 cm³

Question 35.
(a) If the mode of the following distribution is 55 then find the value of x. Hence, find the mean.

Class interval	0-15	15-30	30-45	45-60	60-75	75-90
Frequency	10	7	X	15	10	12

Or
(b) A survey regarding heights (in cm) of 51 girls of class X of a school was conducted and the following data was obtained:

Heights (in cm)	Number of girls
less than 140	04
less than 145	11
less than 150	29
less than 155	40
less than 160	46
less than 165	51

Find the median height of girls. If mode of the above distribution is 148.05, find the mean using empirical formula.
Answer:
(a) Here, the given mode is 55, which lies between 45-60.
Therefore, the modal class is 45-60.
∴  I = 45, f₁= 15, f₀ = x, f₂ = 10 and h = 15.

(b) To calculate the median height, we need to convert the given data in the continuous grouped frequency distribution. Given distribution is of less than type and 140,145, 150, ….165 given the upper limits of the corresponding class intervals. So, the classes “should be below 140, 140-145, 145-150,…160-165.

Clearly, the frequency of class interval below 140 is 4, since there are 4 girls with height less than 140. For the frequency of class interval 140 – 145 subtract the number of girls having height less than 140 from the number of girls having height less than 145.
Thus, the frequency of class interval 140-145 is 11-4 = 7.
Similarly, we can calculate the frequencies of other class intervals and get the following table.

Since, the cumulative frequency just greater than 25.5 is 29 and the corresponding class interval is 145-150.
.’. Median class = 145-150

Hence, the required median height of girl is 149.03 cm.
Since, mode = 148.05 and Median = 149.03
∴ Mode = 3 Median -2 Mean
⇒ 148.05 = 3 x 149.03 – 2 Mean
⇒ 148.05 – 447.09 = – 2 Mean
⇒ 299.04 = 2 Mean
⇒ Mean =\(\frac{299.04}{2}\) = 149.52 cm

SECTION-E
(Section E consists of 3 case study-based questions of 4 marks each.)

Question 36.
In a class, the teacher asks every student to write an example of AP. Two boys Aryan and Roshan writes the progression as -5, -2,1,4,………….. and 187,184,181,…….. , respectively. Now the teacher asks his various students the following questions on progression.
Help the students to find answers for the following.
(i) Find the sum of the common difference of two progressions.
(ii) Find the 34th term of progression written by Roshan.
(iii) (a) Find the sum of first 10 terms of the progression written by Aryan.
Or
(b) Which term of the progressions will have the same value?
Answer:
(i) Given, AP by Aryan and Roshan are -5, -2, 1, 4,…
Here, first term, a = – 5 and common difference,
d = – 2 – (-5) = -2+5 = 3 and 187,184,181,..
Here, first term, A = 187 and common difference,
D = 184 – 187 = – 3
∴ Sum of the common difference of two progression
= d+ D = 3 + (-3) = 3-3 = 0

(ii) For Roshan’s progression, A = 187 and D = – 3.
∴ A₃₄= a + (34 – 1)D [∵ = A + (n-1)D]
= 187 + 33 x (-3)= 187 – 99 = 88
Hence, the 34th term of progression written by Roshan is 88.

(iii) (a) For Aryan’s progression,
a = – 5 and d = 3
∴ S₁₀=\(\frac{10}{2}\) [2x(-5) + (10-1)3]
[∵  S_n = \(\frac{n}{2}\) 2a+(n-1)d]
= 5(-10+ 27)
= 5x 17 = 85
Hence, sum of first 10 terms of the progression written by Aryan is 85.
Or
(b) Let nth term of both the progression are equal.
∴ a_n = A_n
⇒ a + (n – 1)d = A + (n – 1)D
⇒ – 5 + (n – 1) (3) = 187 + (n- 1)(-3)
⇒ -5 + 3n – 3 = 187 – 3n + 3
⇒ – 8+3 = 190 – 3n
⇒ 6n= 198
⇒ n = 33
Hence, 33th term of both the progressions will have the same value.

Question 37.
A group of class X students goes to picnic during winter holidays. The position of three friends Aman, Kirti and Chahat are shown by the points P, Q and R.

(i) Find the distance between P and R.
(ii) Is Q, the midpoint of PR? Justify by finding midpoint of PR.
(iii) (a) Find the point on X-axis, which is equidistant from P and Q.
Or
(b) Let S be a point, which divides the line joining PQ in ratio 2:3. Find the coordinates of S.
Answer:
The coordinates of Aman are P( 2, 5),
the coordinates of kirit are Q(4,4)
and the coordinates of Chahat are R(8,3).
(i) Distance between P and R is given by

(iii) (a) Let the point on X-axis be A(x, 0). Here, the point A is equidistant from P and Q.
∴ PA = QA
⇒ \(\sqrt{(x-2)^2+(0-5)^2}=\sqrt{(x-4)^2+(0-4)^2}\)
⇒ (x – 2)² + 25 = (x – 4)² + 16
⇒ x² + 4 -4x + 25 = x² + 16-8x + 16
⇒ 29 – 4x = 32 – 8x
⇒ 8x – 4x = 32 – 29
⇒ 4x = 3 ⇒ x = \(\frac{3}{4}\)
Hence, the required point is, (\(\frac{3}{4}\),o)
Or
(b) We know that if a point A divides the line segment joining by points B(x_v y₂) and C (x₂, y₂) in ratio m: n
Coordinates of \(A=\left(\frac{m x_2+n x_1}{m+n}, \frac{m y_2+n y_1}{m+n}\right) \)
∴ Coordinates of S, which divides the line segment joining P(2, 5) and Q(4, 4) in the ratio 2 : 3.

Question 38.
India gate (formerly known as all India war memorial) is located near Karthavya path, (formerly Rajpath) at New Delhi. It stands as a memorial to 74187 soldiers of Indian Army, who gave their life in the first world war. This 42m tall structure was designed by sir Edwin Lutyens in the style of Roman triumphal arches. A student Shreya of height 1 m visited India Gate as a part of her study tour.
(i) What is the angle of elevation from Shreya’s eye to the top of India Gate, if she is standing at a distance of 41m away from the India Gate?
(ii) If Shreya observes the angle of elevation from her eye to the top of India Gate to be 60° then how far is she standing from the base of the India Gate?
(iii) (a) If the angle of elevation from Shreya’s eye changes from 45° to 30°, when she moves some distance back from the original position. Find the distance she moves back.
(b) If Shreya moves to a point, which is at a distance of—m from the India Gate then find the \(\frac{41}{\sqrt{3}}\)m angle of elevation made by her eye to the top of India Gate.
Answer:
Given, height of India Gate, AE = 42 m
and distance between India Gate and Shreya,
BC = 41 m and height of Shreya DC = 1 m

(i) Let θ be the angle of elevation from Shreya eye in the top of India Gate.
Here, AB = AE – BE = AE – DC [∵BE = DQ]
= 42 – 1 = 41m
Now, in ΔABC tan θ = \(\frac{A B}{B C}=\frac{41}{41}\)
⇒ tanθ = 1 ⇒  tan θ = tan 45°
⇒ θ = 45°
Hence, the angle of elevation is 45°

(ii) Given, angle of elevation, θ=60°
Let the distance from the India Gate, BC=xm

The post CBSE Sample Papers for Class 10 Maths Standard Set 1 with Solutions appeared first on Learn CBSE.

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