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

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

Time Allowed: 3 Hours
Maximum Marks: 80

General 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 02 marks each.
  4. Section C has 6 questions carrying 03 marks each.
  5. Section D has 4 questions carrying 05 marks each.
  6. 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.
  7. 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
  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.
If the sum of the zeros of the polynomial 2x2 + 3kx + 3 is 6, then the value of k is: (1)
(a) 5
(b) 2
(c) – 4
(d) 6
Answer:
(c) -4

Explanation:
Here, p(x) = 2x2 + 3kx + 3
On comparing it with ax2 + fax + c = 0, we get
a = 2, b = 3k and c = 3 .
sum of zeros = \(\frac{-b}{a}\)
= \(\frac{-3}{2}\)
And, sum of zeros = 6
∴ \(\frac{-3k}{2}\) = 6
⇒ k = – 4

Question 2.
The roots of the quadratic equation (3x – 5) (x + 3) = 0 is: (1)
(a) -3 , \(\frac{5}{3}\)
(b) 3 , \(\frac{-5}{3}\)
(c) -2 , \(\frac{5}{3}\)
(d) 1 ,\(\frac{3}{5}\)
Answer:
(a) -3 , \(\frac{5}{3}\)

Explanation:
The two roots of the given equation (3x – 5) (x + 3) = 0 are – 3 and -3 , \(\frac{5}{3}\).

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

Question 3.
The sum of natural numbers from 51 to 100 is: (1)
(a) 3005
(b) 2895
(c) 375
(d) 3775
Answer:
(d) 3775

Explanation:
Sum of numbers from 51 to 100
= \(\frac{50}{2}\) [2 × 51 + 49(1)]
= 25 [102 + 49]
= 25 × 151 = 3775

Question 4.
If A(6,2), B(4,2) and C(6,4) are the vertices of ∆ABC, then the length of the median through C is: (1)
(a) \(\sqrt{2}\) units
(b) \(\sqrt{3}\) units
(c) 5 units
(d) \(\sqrt{5}\) units
Answer:
(d) \(\sqrt{5}\) units

Explanation:
The coordinates of Z (the mid-point of AB) are (\(\frac{6+4}{2}\) , \(\frac{2+2}{2}\)) (5, 2)
So, length of CZ = \(\sqrt{(5-6)^{2}+(2-4)^{2}}\)
= \(\sqrt{1+4}\) = \(\sqrt{5}\) units

Question 5.
The base PQ of two equilateral triangles PQR and PQR’ with side ‘2a’ lies along g – axis such that the mid – point of PQ is at the origin. The coordinates of the vertices R and R’ of the triangles is: (1)
(a) (\(\sqrt{3}\)a, 0)(\(\sqrt{3}\)a, 0)
(b) (1, 0)(- 1, 0)
(c) (2, 0) (- 2, 0)
(d) (\(\sqrt{3}\)a, 0)(-\(\sqrt{3}\)a, 0)
Answer:
(d) (\(\sqrt{3}\)a, 0)(-\(\sqrt{3}\)a, 0)

Explanation:
Since, the mid-point of PQ is the origin and PQ = 2a.
∴ OP = OQ = a
Hence, the coordinates of P and Q are (0, a) and (0, – a) respecively.
Since, ∆PQRand PQR’ are equilateral triangle
∴ Their third vertices R and R’ lie on the ⊥r bisector of base PQ.
X’ OX is the perpendicular bisector of base PQ.
Thus, R and R’ lies on X-axis.
∴ Their Y – coordinates are 0.
In ∆ROP, OR2 + OP2 = PR2
CBSE Sample Papers for Class 10 Maths Standard Set 6 with Solutions - 1
⇒ OR2 + a2 = (2a)2
⇒ OR2 = 3a2
⇒ OR = \(\sqrt{3}\)a
Similarly, OR’ = \(\sqrt{3}\)a
Thus, the coordinates of vertices R and R’ are (\(\sqrt{3}\)a , 0) an (- \(\sqrt{3}\)a , 0) respectively.

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

Question 6.
In a ∆ABC, if DE is parallel to BC, \(\frac{AD}{DB}\) = \(\frac{3}{4}\) and AC = 15 cm, then the length AE is: (1)
(a) 45
(b) \(\frac{23}{7}\)
(c) 1
(d) \(\frac{45}{7}\)
Answer:
(d) \(\frac{45}{7}\)

Explanation:
Since, DE || BC,
CBSE Sample Papers for Class 10 Maths Standard Set 6 with Solutions - 2
By thales theorem
\(\frac{AD}{DB}\) = \(\frac{AE}{EC}\)
\(\frac{AE}{EC}\) = \(\frac{3}{4}\)
⇒ \(\frac{AE}{AC – AE}\) = \(\frac{3}{4}\)
⇒ \(\frac{AE}{15 – AE}\) = \(\frac{3}{4}\)
⇒ 7AE = 45
AE = \(\frac{45}{7}\) cm

Question 7.
If a right circular cylinder with a height of 7 cm has a volume of 448π cm3, then the radius is: (1)
(a) 8 cm
(b) 10 cm
(c) 11 cm
(d) 15 cm
Answer:
(a) 8 cm

Explanation:
Let the radius be r cm.
We know that,
Volume of cylinder = πr2h
448π = πr2h
\(\frac{448π}{π}\) = r2 × 7
448 = r2 × 7
\(\frac{448}{7}\) = r2
64 = r2
\(\sqrt{64}\)= r2
8 = r

Question 8.
If the bisector of an angle of a triangle bisects the opposite side, the triangles is: (1)
(a) isosceles
(b) scalene
(c) equilateral
(d) right angled triangle
Answer:
(a) isosceles

Explanation:
Given, ∆ABC, AD bisects ∠A and BC.
CBSE Sample Papers for Class 10 Maths Standard Set 6 with Solutions - 3
In ∆ABD and ∆ACD.
∠DAB = ∠DAC (AD bisects ∠A)
AD = AD (common)
BD = CD (AD bisects BC)
Thus, ∆ABD ~ ∆ACD (by SAS rule)
Thus, AB = AC (by cpct)
Hence, ∆ABC is an isosceles triangle.

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

Question 9.
Two concentric circles are of radii 5 cm and 3 cm. The length of the chord of the larger circle which touches the smaller circle is: (1)
(a) 5 cm
(b) 3 cm
(c) 8 cm
(d) 4 cm
Answer:
(c) 8 cm

Explanation:
Here, radius of circles are 3 cm and 5 cm i.e., OA = 3 cm and OB = 5 cm
CBSE Sample Papers for Class 10 Maths Standard Set 6 with Solutions - 4
Now, OA is ⊥r on BC and bisects BC
As, tangent is ⊥r to the radius and ⊥r from the centre bisects the chord.
∴ In ∆OAB, by pythagoras theorem
OB2 = AB2 + OA2
⇒ 52 = AB2 + 32
⇒ AB2 = 25 – 9 = 16
⇒ AB = 4 cm
and BC = 2AB
= 2 × 4
= 8 cm

Question 10.
The area of a sector of angle θ (in degrees) of a circle with radius ‘r’ is: (1)
(a) \(\frac{\theta}{360^{\circ}} \times \pi r^2\)
(b) πr2
(c) 2πr2
(d) \(\frac{\theta}{360^{\circ}} \times 2 \pi r\)
Answer:
(a) \(\frac{\theta}{360^{\circ}} \times \pi r^2\)

Explanation:
Area of sector of angle (θ) with radius (r) is = \(\frac{\theta}{360^{\circ}} \times \pi r^2\)

Question 11.
The perimeter of a semi – circular protractor is 36 cm. Find its diameter.
(a) \(\frac{72}{2+π}\)
(b) \(\frac{72}{π}\)
(c) \(\frac{36}{2+π}\)
(d) 0
Answer:
(a) \(\frac{72}{2+π}\)

Explanation:
Let the radius of protactor be V Then, its perimeter is = 2r + nr
CBSE Sample Papers for Class 10 Maths Standard Set 6 with Solutions - 5
and 36 = 2r + πr
⇒ r = \(\frac{36}{2+π}\)
∴ Diameter = 2r = \(\frac{2×36}{2+π}\)
= \(\frac{72}{2+π}\)

Question 12.
The modal class for the frequency distribution given below is: (1)

Class interval 0-20 20-40 40-60 60-80 80-100
Number of workers 15 18 21 29 17

(a) 20 – 40
(b) 0 – 20
(c) 60 – 80
(d) 80 – 100
Answer:
(c) 60-80

Explanation:
Since, the frequency of class 60-80 is maximum as 29.
Then, 60 – 80 is the modal class.

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

Question 13.
If P(E) = 0.005, then the probability of “not E” is: (1)
(a) 0.002
(b) 0.95
(c) 0.995
(d) 0.095
Answer:
(c) 0.995

Explanation:
P (not E) = 1 – P (E)
P (not E) = 1 – 0.005 = 0.995

Question 14.
The upper limit of the median class of the following frequency distribution is: (1)

Class 0-5 6-11 12-17 18-23 24-29
Frequency 13 10 15 8 11

(a) 17.5
(b) 18.5
(c) 19
(d) 19.5
Answer:
(a) 17.5

Explanation:
The classes in exclusive form are:
(-0.5)-5.5; 5.5-11.5 11.5-17.5; 17.5-23.5; 23.5 – 24.5 with cumulative frequencies of 13, 23, 38, 46 and 57.
Here, N = 57. So, \(\frac{N}{2}\) = 28.5
Cumulative frequency just greater than 28.5 is 38, which belongs to class 11.5 – 17.5.
Thus, median class is 11.5 – 17.5 whose upper limit is 17.5.

Question 15.
An unbiased die is rolled once. The probability of getting an even prime number is: (1)
(a) \(\frac{6}{5}\)
(b) \(\frac{1}{5}\)
(c) \(\frac{1}{6}\)
(d) 1
Answer:
(c) \(\frac{1}{6}\)

Explanation:
Possible outcomes on rolling a die are 1, 2, 3, 4, 5 and 6.
Out of these six numbers, only 2 is the even prime number.
So, the required probability is \(\frac{1}{6}\).

Question 16.
If sin B = 0.5, the value of 3 cos B – 4 cos3 B is: (1)
(a) 2
(b) 4
(c) -3
(d) 0
Answer:
(d) 0

Explanation: Here, sin B = 0.5 = \(\frac{1}{2}\)
CBSE Sample Papers for Class 10 Maths Standard Set 6 with Solutions - 6

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

Question 17.
When two hemispheres with the same radius are connected at their bases, then we get a: (1)
(a) cone
(b) cylinder
(c) sphere
(d) cuboid
Answer:
(c) Sphere

Explanation:
If we join two hemispheres of same radius along their bases, then we get a sphere.

Question 18.
The value 60° is: of cosec2 30° sin2 45° – sec2 60° is: (1)
(a) -2
(b) 1
(c) 2
(d) 3
Answer:
(a) -2

Explanation:
Here, cosec² 30° sin² 45° – sec² 60°
= (2)2 × (\(\frac{1}{\sqrt{2}}\))2 – 22
= 4 × \(\frac{1}{2}\) – 4 = 2 – 4 = – 2

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 as:
(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.
Statement A (Assertion): In the figure, if BC = 20 m, then height AB is 11.56 m.
CBSE Sample Papers for Class 10 Maths Standard Set 6 with Solutions - 7
Statement R (Reason): tan θ = \(\frac{AB}{BC}\) = \(\frac{Perpendicular}{base}\) where θ is angie ∠ACB. (1)
Answer:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)

Explanation:
tan30° = \(\frac{AB}{BC}\) = \(\frac{AB}{20}\)
AB = \(\frac{1}{\sqrt{3}}\) × 20
= \(\frac{20}{1.732}\) = 11.56 m

Question 20.
Statement A (Assertion): Total surface area of the cylinder having radius of the base 14 cm and height 30 cm is 3872 cm2.
Statement R (Reason): If r be the radius and h be the height of the cylinder, then total surface area = (2πrh + 2πr2). (1)
Answer:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)

Explanation:
Total surface area = 2πrh + 2πr2
= 2πr(h + r)
= 2 × \(\frac{22}{7}\) × 14(30 + 1 4)
= 88(44)
= 3872 cm2

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

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

Question 21.
Using prime factorisation, find the HCF and LCM of 150 and 240.
OR
Show that 3 + \(\sqrt{5}\) is an irrational number, assuming that \(\sqrt{5}\) is an irrational number. (2)
Answer:
Prime factorisation of 150 and 240.
CBSE Sample Papers for Class 10 Maths Standard Set 6 with Solutions - 8
Then, 150 = 2 × 3 × 5 × 5
240 = 2 × 2 × 2 × 2 × 3 × 5
HCF = 2 × 3 × 5 = 30
LCM = 2 × 3 × 5 × 5 × 2 × 2 × 2
= 1200

OR

If possible, let us assume that 3 + \(\sqrt{5}\) be a rational number. So, there exist positive integers a and b such that, 3 + \(\sqrt{5}\) = y where a and b are integers having no common factor other than 1.
⇒ \(\sqrt{5}\) = \(\frac{a}{b}\) – 3
⇒ \(\sqrt{5}\) = \(\frac{a – 3b}{b}\)
Since, \(\frac{a – 3b}{b}\) is a rational number, so \(\sqrt{5}\) must a rational number (LH.S. = R.H.S.) which is a contradiction to the fact that \(\sqrt{5}\) is irrational”.
Hence, 3 + \(\sqrt{5}\) is an irrational number.

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

Question 22.
Find the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).
OR
Using distance formula, show that the points A(1, -1), B(5, 2) and C( 9, 5) are collinear. (2)
Answer:
Let P(-1, 6) divide the join of A(-3, 10) and B(6, -8) in the ratio K : 1.
CBSE Sample Papers for Class 10 Maths Standard Set 6 with Solutions - 9
Then
P(-1, 6) = \(\left(\frac{6 K-3}{K+1}, \frac{-8 K+10}{K+1}\right)\)
⇒ \(\frac{6 K-3}{K+1}\) = -1 ; \(\frac{-8 K+10}{K+1}\) = 6
⇒ 6K – 3 = -K – 1 ; -8K + 10 = 6K + 6
⇒ 7K = 2 ; 14K = 4
⇒ K = \(\frac{2}{7}\)
Thus, the required ratio is 2 : 7.

OR

Given, points are A(l, – 1), B(5, 2) and C(9, 5)
Distance between AB = \(\sqrt{(5 – 1)^2+(2 + 1)^2}\)
= \(\sqrt{4^2+3^2}\)
= \(\sqrt{25}\)
= 5 units

Distance between BC = \(\sqrt{(9 – 5)^2+(5 – 1)^2}\)
= \(\sqrt{4^2+3^2}\)
= \(\sqrt{25}\)
= 5 units

Distance between AC = \(\sqrt{(9 – 1)^2 + (5 + 1)^2}\)
= \(\sqrt{8^2+6^2}\)
= \(\sqrt{64+36}\)
= \(\sqrt{100}\)
= 10 units
Then, AC = AB + BC = 10 units
Hence, the point A, B and C are collinear.

Question 23.
If \(\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}=\frac{1-\sqrt{3}}{1+\sqrt{3}}\) , then find the value of θ. (2)
Answer:
We have,
\(\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}=\frac{1-\sqrt{3}}{1+\sqrt{3}}\)
Dividing numerator and denominator of LH.S. by cos θ,
⇒ \(\frac{1-\tan \theta}{1+\tan \theta}=\frac{1-\sqrt{3}}{1+\sqrt{3}}\)
On comparing, we get tan θ = \(\sqrt{3}\)
⇒ θ = 60°

Question 24.
Raju got a playing lattu as his birthday present but it has surprisingly no colour on it. He wants to colour it with his crayons in a smart way . It is shaped like a cone surmounted by a hemi-sphere as shown. The entire lotto is 5 cm in height and the diameter of the hemi-sphere base is 3.5 cm.
CBSE Sample Papers for Class 10 Maths Standard Set 6 with Solutions - 10
Find:
(A) the surface area of the hemi-sphere.
(B) the area that he has to colour. (2)
Answer:
(A) Hemisphere is of radius 1.75 cm (i.e., 3.5/2)
CBSE Sample Papers for Class 10 Maths Standard Set 6 with Solutions - 11
So, its curved surface area = 2πr2
= (2 × \(\frac{22}{7}\) × 1.75 × 1.75) cm2
= 19.25 cm2.

(B) For Cone r = 1.75 cm
h = 5 – 1.75 = 3.25 cm
and l = \(\sqrt{r^2+h^2}\)
Area of the whole lattu = curved surface area of cone + curved surface area of hemisphere
= [π(1.75)\(\sqrt{(1.75)^2 + (3.25)^2}\) + 19.25] cm2
= [\(\frac{22}{7}\) × 1.75 × 3.691 + 19.25] cm2
= (20.30 + 19.25) cm2
= 39.55 cm2

Question 25.
For the following frequency distribution, determine the mean: (2)

Class 100-120 120-140 140-160 160-180 180-200
Frequency 12 14 8 6 10

Answer:

Class Frequency

(fi)

Mid Point (xi) xifi
100-120 12 110 1320
120-140 14 130 1820
140-160 8 150 1200
160-180 6 170 1020
180-200 10 190 1900
Total 50 7260

Then, Mean, \(\bar{x}=\frac{\Sigma f_i x_i}{\Sigma f_i}=\frac{7260}{50}=145.2\)

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

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

Question 26.
In an A.P., the last term is 28 and the sum of all the 9 terms of the A.P. is 144. Find the first term. (3)
Answer:
Let ‘a’ be the first term of AP and ‘d be the common difference. Here, total number of terms of AP is 9, i.e. n = 9
nth term = last term = an = a + 8d = 28 …(i)
Also,
Sn = S9= \(\frac{9}{2}\) [2a+(9-l)d] = 144
⇒ 9(a + 4d) = 144
⇒ 9a + 36d = 144 or a + 4d = 16 …..(ii)
Subtracting (ii) from (i), we get
(a + 8d) – (a + 4d) = 28 – 16
⇒ 4d =12
⇒ d = 3
Putting value of d in (i).
a + 8 × 3 = 28
⇒ a = 28 – 24 = 4
So, a = 4 and d = 3
Thus, the required first term is 4.

Question 27.
In what ratio does the x-axis is divide the line segment showing the points (-4, -6) and (-1, 7)? Find the coordinates of the point of division.
OR
Prove that the parallelogram circumscribing a circle is a rhombus. (3)
Answer:
Let the join of (-4, -6) and (-1, 7) be divided by a point P on x-axis in the ratio K : 1.
CBSE Sample Papers for Class 10 Maths Standard Set 6 with Solutions - 12
Then,
p(x,0) = p\(\left(\frac{-K-3}{K+1}, \frac{7 K-6}{K+1}\right)\)

⇒ \(\frac{7 K-6}{K+1}\) = 0
⇒ K = \(\frac{6}{7}\)
Hence, the required ratio is 6 : 7.

With K = \(\frac{6}{7}\), the point P is P(\(\frac{-\frac{6}{7}-4}{\frac{6}{7}+1}\),0) or P(\(\frac{-34}{13}\),0)
OR
Let ABCD be a parallelogram circumscribing a circle.
CBSE Sample Papers for Class 10 Maths Standard Set 6 with Solutions - 13
Let P, Q, R, S be the points where the circle touches the sides AB, BC, CD and DA respectively.
Now,
AB = DC and AB || DC
Also,
AD = BC and AD || BC (∴ ABCD is a parallelogram) …(i)
From the figure, we have:
AP = AS; BP = BQ; CR = CQ and DR = DS.
∵ AP + PB + CR + RD = AS + DS + CQ + BQ
⇒ AB + DC = AD + CB
⇒ 2 AB = 2 AD
⇒ AB = AD
Using (i), we have:
AB = BC = CD = DA
Thus, ABCD is a rhombus.

Question 28.
In ∆ABC right angles at B, if tan A = \(\frac{1}{\sqrt{3}}\), find value of: sin A cos C + cos A sin C (3)
Answer:
Consider a triangle ABC in which ∠B = 90°.
CBSE Sample Papers for Class 10 Maths Standard Set 6 with Solutions - 14

Question 29.
In the centre of a circle with a radius of 10 cm, a chord subtends a right angle. Find The area of the corresponding minor and major segment.
OR
A hollow cone is cut by a plane parallel to the base and upper portion is removed. If the curved surface area of the remainder is 0 \(\frac{8}{9}\) of the curved surface area of the whole cone, find the ratio of the line segments into which the cone’s altitude is divided by the plane. (3)
CBSE Sample Papers for Class 10 Maths Standard Set 6 with Solutions - 15
Answer:
In the mentioned circle,
O is the centre and AO = BO = Radius = 10 cm
AB is a chord which subtents 90° at centre 0, i.e., ∠AOB = 90°
Area of minor segment APB (Shaded region) = Area of sector ∆AOB – Area of ∆AOB
\(\left(\frac{\pi \times 10 \times 10}{4}\right)-(0.5 \times 10 \times 10)\)
= 78.5 – 50
= 28.5 cm2

Area of major sector = Area of circle – Area of sector AOB
= (π × 10 × 10) + \(\left(\frac{\pi \times 10 \times 10}{4}\right)\)
= 314 – 78.5
= 235.5 cm2

OR

In the figure, the smaller cone APQ has been cut off through the plane PQ || BC. Let r and R be the radii of the smaller and bigger cones and l and L be their slant heights respectively.
CBSE Sample Papers for Class 10 Maths Standard Set 6 with Solutions - 16
Here,
OQ = r, MC = R, AQ = l and AC = L
Now, ∆AOQ ~ ∆AMC
⇒ \(\frac{OQ}{MC}\) = \(\frac{AQ}{AC}\)
⇒ \(\frac{r}{R}\) = \(\frac{l}{L}\)
⇒ r = \(\frac{Rl}{L}\) ……..(i)
Since, curved surface area of the remainder = \(\frac{8}{9}\) of the curved surface area of the whole cone therefore, we get
CSA of smaller cone = \(\frac{1}{9}\) of the CSA of the whole cone
πrl = \(\frac{1}{9}\) πRL
⇒ π(\(\frac{Rl}{L}\))l = \(\frac{1}{9}\) (πRL)
⇒ l2 = \(\frac{L^2}{9}\)
⇒ \(\frac{l}{L}\) = \(\frac{1}{3}\)

Now, again in similar triangles, AOQ and AMC, we have
\(\frac{AO}{AM}\) = \(\frac{AQ}{AC}\)
\(\frac{AO}{AM}\) = \(\frac{1}{L}\) = \(\frac{1}{3}\)
AO = \(\frac{AM}{3}\)
OM = AM – OA = AM – \(\frac{AM}{3}\) = \(\frac{2}{3}\) AM
\(\frac{AO}{OM}\) = \(\frac{\frac{AM}{3}}{\frac{2AM}{3}}\) = \(\frac{1}{2}\)
Hence, the required ratio of the heights =1 : 2

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

Question 30.
A number ‘x’ is selected from the numbers 1, 2, 3 and then second number ‘y’ is selected from the numbers 1, 4, 9. Rnd the probability that the product ‘xy’ of the two numbers is less than 9. (3)
Answer:
Number Y can be selected in 3 ways and corresponding to each, such way there are 3 ways of selecting ‘y.
Therefore, 2 numbers can be selected in 9 ways as listed below:
(1, 1) (1, 4) (1, 9) (2, 1) (2, 4) (2, 9) (3, 1) (3, 4) (3, 9)
Total numbers of outcomes = 9
The product xy will be less than 9, if x and y are chosen in one of the following ways.
(1, 1) (1, 4) (2, 1) (2, 4) (3, 1)
Number of favourable outcomes = 5
P(product less than 9) = \(\frac{5}{9}\)

Question 31.
If the median of the following data is 32.5, find the missing frequencies. (3)i

Class interval 0-10 10-20 20-30 30-40 40-50 50-60 60-70 Total
Frequency fi 5 9 12 fi 3 2 40

Answer:
Median of data = 32-5
sum of frequency = 40
i.e., f1 + 31 + f2 = 40
∴ f1 + f2 = 9
Then, median class is 30-40

Class Interval fi c.f.
0-10 f1
10-20 5 5 + f1
20-30 9 14 + f1
30-40 12 26 + f1
40-50 f2 26 + f1 + f2
50-60 3 29 + f1 + f2
60-70 2 31 + f1+ f2
Total 40

Then,
\(\mathrm{M}_e=\frac{1+\left(\frac{\mathrm{N}}{2}-c f\right) \times h}{f}\)
⇒ \(32.5=30+\frac{\left(20-14-f_1\right)}{12} \times 10\)
⇒ 2.5 × 6 = (6 – f1) × 5
⇒ 15.0 = 30 – 5f1
⇒ 5f1 = 15
⇒ f1 = 3
and f2 = 6
Hence, the values of f1 and f2 are 3 and 6 respectively.

Section – D (20 Marks)
(Section D consists of 4 questions of 5 marks each.)

Question 32.
Two pipes together can start to fill a tank in \(9 \frac{3}{8}\) hours. The larger diameter pipe fills the tank separately in 10 hours less time than the smaller one. Determine the time at which each pipe can fill the tank on its own.
OR
A 2- digit number is four times the sum and three times the product of its digits. Find the number. (5)
Answer:
Let time taken by pipe of smaller diameter to fill the tan k = x hours
Let time taken by pipe of larger diameter to fill the tank = (x – 10) hours
In 1 hour, the pipe with a smaller diameter can fill \(\frac{1}{x}\) part of the tank.
In 1 hour, the pipe with larger diameter can fill \(\frac{1}{(x-10)}\) part of the tank.
The tank is filled up in \(\frac{75}{8}\) hours.

Thus, in 1 hour the pipe fill \(\frac{8}{75}\) part of the tank.
\(\frac{1}{x}\) + \(\frac{1}{(x-10)}\) = \(\frac{8}{75}\)
\(\frac{(x-10)+x}{x(x+10)}\) = \(\frac{8}{75}\)
\(\frac{2x-10}{x^2-10x}\) = \(\frac{8}{75}\)
75(2x – 10) = 8(x2 – 10x) by cross multiplication
150x – 750 = 8x2 – 80x
8x2 – 230x + 750 = 0
4x2 – 115x+ 375 = 0
4×2 – 100x -15x + 375 = 0
4x(x – 25) – 15(x – 25) = 0
(4x – 15)(x – 25) = 0
4x – 15 = 0 or x – 25 = 0
x = \(\frac{15}{4}\) or x = 25

Case 1:
When
x = \(\frac{15}{4}\)
Then x – 10 = \(\frac{15}{4}\) – 10
x = \(\frac{15-40}{4}\)
x = –\(\frac{25}{4}\)
Time can never be negative so x = 15/4 is not possible.

Case 2:
When x = 25 then
x – 10 = 25 – 10 = 15
∴ The pipe of smaller diameter can separately fill the tank in 25 hours, and the time taken by the larger pipe to fill the tank = (25 – 10) = 15 hours.

OR

Let, the number be 10x + y i.e., digit at units’s place is ‘y’ and digit at ten’s place is ’x’.
According to the question,
10x + y = 4(x + y)
⇒ 10x + y = 4x + 4y
⇒ 6x – 3y = 0
⇒ 2 x = y …(i)
and 10x + y = 3xy
From (i)
10x + 2x = 3 xy
⇒ 12x = 3 xy
⇒ y = 4
and x = 2
then, the numbers is 24.

Question 33.
In the given figure, ∆FEC is congruent to ∆GDB and ∠1 = ∠2. Prove that ∆ADE ~ ∆ABC. (5)
CBSE Sample Papers for Class 10 Maths Standard Set 6 with Solutions - 17
Answer:
Given : ∆FEC ≅ ∆GDB
CBSE Sample Papers for Class 10 Maths Standard Set 6 with Solutions - 17.1
and ∠1 = ∠2
To prove: ∆ADE – ∆ABC
Proof Since, ∆FEC = ∆GDB
Then, EC = BD (by cpct) …(i)
and ∠1 = ∠2
AE = AD
Then, AE _ AD EC ” BD (from equation (i) & (ii)]
∴ DE || BC (by converse of thales theorem)
∴ ∠1 = ∠ABC and ∠2 = ∠ACB
(corresponding pair of angles)

In ∆ADE and ∆ABC,
∠A = ∠A (common)
∠1 = ∠ABC (proved above)
∠2 = ∠ACB (proved above)
∴ ∆ADE ~ ∆ABC [by AA similarity]

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

Question 34.
If an isosceles triangle ABC in which AB = AC = 6 cm is inscribed in a circle of radius 9 cm, find the area of the triangle. (5)
Answer:
Let O be the centre of the circle and P be the mid-point of BC. Then, OP ⊥ BC.
CBSE Sample Papers for Class 10 Maths Standard Set 6 with Solutions - 18
Since, AABC is isosceles and P is the mid-point of BC.
Therefore, AP ⊥ BC as median from the vertex in an isosceles triangle is perpendicular to the base.
Let, AP = x and PB = CP = y
Applying pythagoras theorem in As APB and OPB, we have
AB2 = BP2 + AP2 and OB2 = OP2 + BP2
⇒ 36 = y2 + x2 …(i)
and 81 = (9 – x)2 + y2 …(ii)
⇒ 81 – 36 = (9 – x)2 + y2 – y2 + x2
(subtracting (i) from (ii))
⇒ 45 = 81 – 18x
⇒ x = 2 cm
Put x = 2 in equation (i), we get
36 = y2 + 4
⇒ y2 = 32
⇒ y = 4\(\sqrt{2}\) cm
BC = 2BP = 2y = 8\(\sqrt{3}\) cm
Hence, area of AABC = \(\frac{1}{2}\) BC × AP
= \(\frac{1}{2}\) × 8\(\sqrt{3}\) × 2 cm2
= 8\(\sqrt{3}\) cm2

Question 35.
Prove that \(\sqrt{a}\) is not a rational number, if ‘a’ is not perfect square.
OR
From an aeroplane vertically above a horizontal plane, the angles of depression of two consecutive kilometre stones on the opposite sides of the aeroplane are found to be α and β. Show that the height of the aeroplane is \(\frac{\tan \alpha \cdot \tan \beta}{\tan \alpha+\tan \beta}\).(5)
Answer:
Let \(\sqrt{a}\) be a rational number
\(\sqrt{a}\) = \(\frac{p}{q}\) where P and q are co-prime integers.
On squaring both side, we get q ≠ 0.
a = \(\frac{p^2}{q^2}\)
⇒ P2 = aq2 …(i)
⇒ a divides p2
⇒ a divides p
Let p be a prime number. If p divides n2, then p divides n, where n is a positive integer.
Let p = am, where m is any integer.
p2 = a2m2
⇒ aq2 = a2m2 [Using (i)]
⇒ q2 = am2
⇒ a divides g2
⇒ a divides q …(iii)
From (ii) and (iii), a is a common factor of both p and q which contradicts the assumption that p and q are co-prime integers.
So, our supposition is wrong.
Hence, \(\sqrt{a}\) is an irrational number.

OR

Let P be the position of plane, A and B be the positions of two stones one kilometre apart. Angles of depression of stones A and B are a and p respectively.
CBSE Sample Papers for Class 10 Maths Standard Set 6 with Solutions - 19
Let PC = h
In right-angled ∆ACP, we have
CBSE Sample Papers for Class 10 Maths Standard Set 6 with Solutions - 20

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

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

Question 36.
Mr. Naik is a paramilitary Intelligence Corps officer who is tasked with planning a coup on the enemy at a certain date. Currently he is inspecting the area standing on top of the cliff. Agent Vinod is on a hot air balloon in the sky. When Mr. Naik looks down below the cliff towards the sea, he has Ajay and Maran in boats positioned to get a good vantage point.
CBSE Sample Papers for Class 10 Maths Standard Set 6 with Solutions - 21
The main goal is to scope out the range and angles at which they should train their soldiers.
On the basis of the above information, answer the following questions:
(A) Write one pair of ‘angle of elevation’ and one pair of’angle of depression’. (1)
(B) If the vertical height of the balloon
from the top of the cliff is 12 m and ∠b = 30°, then find the distance between the Naik and vinod. (1)
(C) Ajay’s boat is 25 m away from the base of the cliff. If ∠d = 30° . What is the height of the cliff? (use \(\sqrt{3}\) = 1.73)
OR
If the height of the cliff is 30 m , ∠c = 45° and ∠d = 30°, then find the horizontal distance between the two boats (use \(\sqrt{3}\) = 1.73) (2)
Answer:
(A) One pair of angle of elevation is ∠b° and ∠e° and one pair of angle of depression is and ∠c° and ∠d°

(B) Then, sin 30° = \(\frac{Vertical height}{Distance between Naik and Vinod}\)
⇒ \(\frac{1}{2}\) = \(\frac{12}{D_NandV}\)
⇒ Distance = 24 m

(C) Here, ∠d° = ∠f° = 30°
Then, \(\frac{Height of cliff.}{Distance of Ajay’s boat from the base of cliff}\) = tan 30°
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{h}{25}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{25}{\sqrt{3}}\) × \(\frac{\sqrt{3}}{\sqrt{3}}\)
= \(\frac{25}{\sqrt{3}}\) × \({\sqrt{3}}\)
= 14.45 m

OR

Here, height of cliff = 30 m
Then, ∠c = ∠e = 45°
∴ tan 450 = \(\frac{height of cliff}{Distance of Maran’s boat}\)
⇒ 1 = \(\frac{30}{Distance of Maran’s boat}\)
⇒ Distance of maran’s boat = 30 m
And ∠d = ∠f= 30°
∴ tan 3o° = \(\frac{height of cliff}{Distance of Ajay’s boat}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{30}{D_A}\)
⇒ DA = 30\(\sqrt{3}\)

∴ Distance between boats = 30\(\sqrt{3}\) – 30
= 30 (1.73 – 1)
= 21.9 m

Question 37.
Prime Minister’s National Relief Fund (PMNRF) was established to help the families of earthquake affected village. The allotment officer is trying to come up with a method to calculate fair division of funds across various affected families so that the fund amount and amount received per family can be easily adjusted based on daily revised numbers.
CBSE Sample Papers for Class 10 Maths Standard Set 6 with Solutions - 22
The total fund allotted is formulated by the officer as: x3 -5x2 -2x- 6
The officer has also divided the fund equally among families of the village and each family receives an amount of x2 + 2x + 1. After distribution, an amount of 11x + 1 should be left to have some buffer for future disbursements.
On the basis of the above information, answer the following questions:
(A) If an amount of ₹ 540 is left after distribution, what is value of x? ?(1)
(B) How much amount (In rupees) does each family receive? (1)
(C) What is the amount of fund (In rupees) allocated ?
OR
If the sum of squares of zeroes of the polynomial x2 – 8x + k is 40, then find the value of k. ? (2)
Answer:
(A) Amount lift = 11x + 1
∴ 11x + 1 = 540
x = \(\frac{539}{11}\) = 49

(B) Since, x = 49
Amount received by each family is
x2 + 2x + 1 = (49)2 + 2(49) + 1
= 2401 + 98 + 1
= 2500

(C) Since, x = 49
∴ Fund alloted is-
x3 – 5x2 – 2x – 6
= (49)3 – 5(49)2 – 2(49) – 6
= 117649 – 12005 – 98 – 6
= 1,05,540

OR

If α and β are the zeroes of a quadratic polynomial p(x) = ax2 + bx + c,
a ≠ 0, then α + β = \(\frac{-b}{a}\) and αβ = \(\frac{c}{a}\)

Here,α + β = 8; αβ = k
It is given that
α2 + β2 = 40
α2 + β2 = (α + β)2 – 2αβ
40 = (8)2 – 2k
40 = 64 – 2k
2k = 24
k= 12

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

Question 38.
To celebrate Diwali festival among senior citizens of an old age home, four friends of a society, Rohan, Amar, Saran and Madhukar decided to pool some money to gift packs to every old man/woman staying in the neighbouring old-age home. They pooled money in the ratio 2:3:4:5. With the pooled money of ? 3500, they start preparing gift packs. In the preparation of one gift pack, Rohan, Amar, Saran and Madhukar spend 7, 6, 8 and 9 minutes respectively.
CBSE Sample Papers for Class 10 Maths Standard Set 6 with Solutions - 23
On the basis of the above information, answer the following questions:
(A) How much time (in minutes) was spent on one gift? (1)
(B) If each gift costs to them ₹ 70, how many senior citizens were given the cards? (1)
(C) How much amount was pooled by Rohan and Madhukar together for giving the gifts and how much time (in hours) was spent in preparing all the gift packs?
OR
How much amount was pooled by Saran? (2)
(A) Total time spent = 7 + 6 + 8 + 9
= 30 minutes

(B) Total money pooled = ₹ 3500
Cost of 1 gift = ₹ 70
N. of gifts = \(\frac{3500}{70}\) = 50

(C) Total money pooled = ₹ 3500
Money given by Rohan and Madhukar = \(\frac{2x + 5x}{14x}\) ×3500
= \(\frac{7}{14}\) × 3500
= ₹ 1750
No. of gifts are 50
Time taken for one gift = 30 minutes
.’. Total time taken = 50 × 30
= 1500 minutes
= \(\frac{1500}{60}\) = 25 h

OR

Total money polled = ₹ 3500
Let, the money contributed be 2x, 3x, 4x, 5x
2x + 3x + 4x + 5x = 3500
14x = 3500
x = \(\frac{500}{2}\) = 250
Then, contribution of saran = 4 × 250 = 1000