Students can access the CBSE Sample Papers for Class 10 Maths Basic with Solutions and marking scheme Set 6 will help students in understanding the difficulty level of the exam.
CBSE Sample Papers for Class 10 Maths Basic Set 6 with Solutions
Maximum Marks: 80
Time Allowed: 3 hours
General Instructions:
1. This Question Paper has 5 Sections A, B, C, D and E.
2. Section A has 20 Multiple Choice Questions (MCQs) carrying 1 mark each.
3. Section B has 5 Short Answers-I (SA-I) type questions carrying 2 marks each.
4. Section C has 6 Short Answers-IT (SA-II) type questions carrying 3 marks each
5. Section D has 4 Long Answers (LA) type questions carrying 5 marks each.
6. Section E has 3 source based/case based/passage based integrated units of assessment (4 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 2 marks, 2 Qs of 3 marks and 2 Qs of 5 marks has been given provided. An internal choice has been provided in 2 marks questions of Section E.
8. Draw neat figures wherever required. Take π = 22/7, wherever required if not stated.
Section – A(20 Marks)
Question 1.
HCE of 92 and 152 is
(A) 4
(B) 19
(C) 23
(D) 57
Solution:
(A) 4
Explanation: Prime factorisation of 92
= 2 × 2 × 23
Prime factorisation of 152 = 2 × 2 × 2 × 19
To find HCF, we multiply all the prime factors common to both numbers:
Therefore, HCF = 2 × 2 = 4
Question 2.
The product of a non-zero rational and an irrational number is
(A) always irrational
(B) always rational
(C) rational or irrational
(D) one
Solution:
(A) always irrational
Explanation: The product of a non-zero rational and an irrational number is always an irrational number.
Question 3.
A quadratic polynomial having sum and product of its zeroes as 5 and 0 respectively, is
(A) x2 + 5x
(B) 2x(x – 5)
(C) 5x2 – 1
(D) x2 – 5x + 5
Solution:
(B) 2x(x – 5)
Explanation: Let the zeroes be α and β
According to question, α + β = 5 and αβ = 0
Now, p(x) = k (x2 – (α + β)x + αβ), where k is any real number
⇒ p(x) = k (x2 – 5x + 0),
When k = 2
⇒ p(x) = 2 (x2 – 5x) = 2x (x – 5)
Question 4.
The solution of the pair of linear equations x = -5 and y = 6 is
(A) (-5, 6)
(B) (-5, 0)
(C) (0, 6)
(D) (0, 0)
Solution:
(A) (-5, 6)
Explanation: Given that x = –5 and y = 6
The lines drawn for the given equations meet at (–5, 6) and Thus (–5, 6) is the solution of the given equations.
Question 5.
Which of the following equations has the sum of its roots as 3?
(A) 2x2 – 3x + 6 = 0
(B) -x2 + 3x – 3 = 0
(C) √2x2 – \(\frac{3}{\sqrt{2}}\)x + 1 = 0
(D) 3x2 – 3x + 3 = 0
Solution:
(B) -x2 + 3x – 3 = 0
Explanation: –x2 + 3x – 3 = 0
On comparing with ax2 + bx + c = 0
a = –1, b = 3, c = –3
∴ Sum of the roots = -b/a = -3/-1 = 3
Question 6.
X-axis divides the join of (2, -3) and (5, 6) in the ratio
(A) 1:2
(B) 2:1
(C) 2:5
(D) 5:2
Solution:
(A) 1:2
Explanation: Let P(x, 0) be a point on x-axis which divides the join of A(2, –3) and B(5, 6) in the ratio m:n, then using section formula.
⇒ 6m – 3n = 0
⇒ 2m – n = 0
⇒ 2m = n
⇒ \(\frac{m}{n}\) = \(\frac{1}{2}\)
i.e., m : n = 1 : 2
Question 7.
If ΔABC ~ ΔEDF and ΔABC is not similar to ΔDEF, then which of the following is not true?
(A) BC × EF = AC × FD
(B) AB × EF = AC × DE
(C) BC × DE = AB × EF
(D) BC × DE = AB × FD
Solution:
(C) BC × DE = AB × EF
Explanation: Since, ΔABC ~ ΔEDF, then we get
\(\frac{A B}{E D}=\frac{A C}{E F}=\frac{B C}{D F}\)
From first two, AB × EF = AC × DE. Option (B) is correct.
From last two, BC × EF = AC × FD. Option (A) is correct.
From first and last, BC × DE = AB × FD. Option (D) is correct
Question 8.
In figure, a quadrilateral ABCD is drawn to circumscribe a circle such that its sides AB, BC, CD and AD touch the circle at P, Q, R and S respectively. If AB = x cm, BC = 7 cm, CR = 3 cm and AS = 5 cm, the x =
(A) 10
(B) 9
(C) 8
(D) 7
Solution:
(B) 9
Explanation:
AS = AP = 5 cm, CQ = CR = 3 cm
BP = BQ, BQ + CQ = BC = 7 cm
BQ = 7 – 3 = 4 cm ⇒ BP = 4 cm
AB = x, AP + BP = x
⇒ 5 + 4 = x
⇒ x = 9.
Question 9.
In the figure of ABC, the points D and E are on the sides CA, CB respectively such that DE || AB, AD = 2x, DC = x + 3, BE = 2x – 1 and CE = x. Then, x is
(A) \(\frac{3}{5}\)
(B) \(\frac{2}{5}\)
(C) \(\frac{1}{5}\)
(D) \(\frac{4}{5}\)
Solution:
(A) \(\frac{3}{5}\)
Explanation:
Question 10.
If sin A + cos B = 1, A = 30° and B is an acute angle, then find the value of B.
(A) 45°
(B) 30°
(C) 90°
(D) 60°
Solution:
(D) 60°
Explanation: sin 30° + cos B = 1
\(\frac{1}{2}\) + cos B = 1
∴ cos B = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
i.e., cos B = cos 60° [∴ cos 60° = \(\frac{1}{2}\)]
Hence, ∠B = 60°.
Question 11.
If the height of a vertical pole is √3 times the length of its shadow on the ground, then the angle of elevation of the Sun at that time is
(A) 30°
(B) 60°
(C) 45°
(D) 75°
Solution:
(B) 60°
Explanation: Let the length of shadow is x,
Then height of pole = √3 x
Now,
Question 12.
If sin A + sin2 A = 1, then the value of the expression (cos2 A + cos4 A) is
(A) 1
(B) \(\frac{1}{2}\)
(C) 2
(D) 3
Solution:
(A) 1
Explanation: Given, sin A + sin2 A = 1
sin A = 1 – sin2 A = cos2 A [∵ cos2 θ + sin2 θ = 1]
On squaring both sides, we get
sin2 A = cos2 A
1 – cos2 A = cos4 A
cos2 A + cos4 A = 1
Question 13.
In a right circular cone, the cross-section made by a plane parallel to the base is a
(A) circle
(B) frustum of a cone
(C) sphere
(D) hemisphere
Solution:
(A) circle
Explanation: In a right circular cone, if any cut is made parallel to its base, we get a circle.
Question 14.
If the circumference of a circle and the perimeter of a square are equal, then
(A) Area of the circle = Area of the square
(B) Area of the circle > Area of the square
(C) Area of the circle < Area of the square
(D) Nothing definite can be said about the relation between the areas of the circle and square.
Solution:
(B) Area of the circle > Area of the square
Explanation: According to question,
Circumference of a circle = Perimeter of square
Let ‘r’ and ‘a’ be the radius of circle and side of square.
2πr = 4a
A1 > A2
Area of circle is greater than the area of square.
Question 15.
Romy is blind folded and asked to pick one ball from each of the jars
The chance of Romy picking a red ball is same in
(A) jars 2 and 3
(B) jars 1 and 3
(C) jars 1 and 2
(D) all the three jars
Solution:
(C) jars 1 and 2
Explanation:
For jar 1, probability picking a red ball = \(\frac{4}{8}\) = \(\frac{1}{2}\)
For jar 2, probability picking a red ball = \(\frac{3}{6}\) = \(\frac{1}{2}\)
For jar 3, probability picking a red ball = \(\frac{4}{6}\) = \(\frac{2}{3}\)
Question 16.
If xi‘s are the mid-points of the class intervals of grouped data, fi’s are the corresponding frequencies and x̄ is the mean, then (fixi – x) is equal to
(A) 0
(B) -1
(C) 1
(D) 2
Solution:
(A) 0
Question 17.
Slant height of a cone if the radius is 7 cm and the height is 24 cm _______.
(A) 26 cm
(B) 25 cm
(C) 52 cm
(D) 62 cm
Solution:
(B) 25 cm
Explanation: Slant height of cone,
l = \(\sqrt{r^2+h^2}\)
Hence, r = 7 cm and h = 24 cm
∴ l = \(\sqrt{(7)^2+(24)^2}\)
= √625 = 25 cm
Question 18.
Consider the following frequency distribution of the heights of 60 students of a class
Heights (in cm) | No. of students |
150 – 155 | 15 |
155 – 160 | 13 |
160 – 165 | 10 |
165 – 170 | 8 |
170 – 175 | 9 |
175 – 180 | 5 |
Find the upper limit of the median class in the given data.
(A) 165
(B) 170
(C) 175
(D) 155
Solution:
(A) 165
Explanation:
Heights (in cm) | No. of students | Cumulative frequency |
150 – 155 | 15 | 15 |
155 – 160 | 13 | 15 + 13 = 28 |
160 – 165 | 10 | 28 + 10 = 38 |
165 – 170 | 8 | 38 + 8 = 46 |
170 – 175 | 9 | 46 + 9 = 55 |
175 – 180 | 5 | 55 + 5 = 60 |
Since, total frequency is 60.
\(\frac{N}{2}\) = 30
And cumulative frequency greater than or equal to 30 lies in class 160-165.
So, median class is 160-165.
∴ Upper limit of median class is 165.
Directions: In the following questions, A Statement of Assertion (A) is followed by a statement of Reason (R).
Mark the correct choice as.
Question 19.
Assertion (A): Co prime numbers are those numbers that have only one common factor, which is 1.
Reason (R): 0 is a rational number.
(A) Both (A) and (R) are true and (R) is the correct explanation for (A).
(B) Both (A) and (R) are true but (R) is not the correct explanation for (A).
(C) (A) is true but (R) is false.
(D) (A) is false but (R) is true.
Solution:
(B) Both (A) and (R) are true but (R) is not the correct explanation for (A).
Explanation: Here, Both Assertion and Reason are true.
But Reason is not the correct explanation of assertion.
Question 20.
Assertion (A): The point A is on the y-axis at a distance of 4 units above from the origin. If the coordinates of the point B are (-3, 0). Then length of AB is 5 units.
Reason (R): Distance between points A(x1, y1) and B(x2, y2) is \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\).
(A) Both (A) and (R) are true and (R) is the correct explanation for (A).
(B) Both (A) and (R) are true but (R) is not the correct explanation for (A).
(C) (A) is true but (R) is false.
(D) (A) is false but (R) is true.
Solution:
(A) Both (A) and (R) are true and (R) is the correct explanation for (A).
Explanation: If the point A is above the origin then is coordinates are (0, 4)
To find the distance AB we need to apply formula
Therefore, Both A and R are true and R is the correct explanation for A.
Section – B(10 Marks)
Question 21.
For what value of k, 2x + 3y = 4 and (k + 2)x + 6y = 3k + 2 will have infinitely many solutions ?
Solution:
We have, for the equation 2x + 3y – 4 = 0
a1 = 2, b1 = 3 and c1 = -4
and for the equation (k + 2) x + 6y – (3k + 2) = 0
a2 = k + 2, b2 = 6 and c2 = –(3k + 2)
Question 22.
In the given figure, two tangents TP and TQ are drawn to circle with centre O from an external point T. Prove that ∠PTQ = 2∠OPQ.
Solution:
Let ∠OPQ be θ, then
∠TPQ = 90° – θ
Since, TP = TQ
∠TQP = 90° – θ [opposite angles of equal sides]
Now, ∠TPQ + ∠TQP + ∠PTQ = 180° [Angle sum property of a triangle]
⇒ 90° – θ + 90° – θ + ∠PTQ = 180°
⇒ ∠PTQ = 180° – 180° + 2θ
⇒ ∠PTQ = 2θ
Hence, ∠PTQ = 2 ∠OPQ
Hence Proved.
OR
In given figure, AB is the diameter of a circle with center O and AT is a tangent. If ∠AOQ = 58°, ∠ATQ.
Solution:
∠AOQ = 58° (Given)
∠ABQ = \(\frac{1}{2}\)∠AOQ
[Angle on the circumference of the circle by the same arc]
= \(\frac{1}{2}\) × 58°
= 29°
∠BAT = 90° [∵ OA ⊥ AT]
∠ATQ = 90° – 29°
= 61°
Question 23.
In ABC, if X and Y are points on AB and AC respectively such that \(\frac{A X}{X B}\) = \(\frac{3}{4}\), AY = 5 and YC = 9, then state whether XY and BC parallel or not.
Solution:
Hence, XY is not parallel to BC.
Question 24.
If tan A = \(\frac{3}{4}\), find the value of \(\frac{1}{\sin A}+\frac{1}{\cos A}\).
Solution:
Question 25.
What is the volume of a right circular cylinder of base radius 7 cm and height 10 cm? [Use π = 22/7]
Solution:
Here, r = 7 cm, h = 10 cm
Volume of cylinder = πr2h
= \(\frac{22}{7}\) × (7)2 × 10
= 1540 cm3
OR
The rain water from 22 m × 20 m roof drains into cylindrical vessel of diameter 2 m and height 3.5 m. If the rain water collected from the roof fills (\(\frac{4}{5}\)) of cylindrical vessel then find the rainfall in cm.
Solution:
Volume of water collected in cylindrical vessel
Section – C(18 Marks)
Question 26.
144 cartons of Coke cans and 90 cartons of Pepsi cans are to be stacked in a canteen. If each stack is of the same height and if it contain equal cartons of the same drink, what would be the greatest number of cartons each stock would have ?
Solution:
Since, the greatest number of cartons is the HCF of 144 and 9o.
144 = 24 × 32
90 = 2 × 32 × 5
Therefore, HCF = 2 × 32 = 18
Thus the greatest number of cartons = 18.
Question 27.
If α and β are the zeroes of the polynomial f(x) = 5x2 – 7x + 1 then find the value of \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\).
Solution:
Since, α and β are the zeroes of the polynomial f (x) = 5x2 – 7x + 1
Question 28.
In an A.E if sum of its first n terms is 3n2 + 5n and its kth term is 164, find the value of k.
Solution:
Here, Sn = 3n2 + 5n
⇒ S1 = 3.12 + 5.1 = 8 = a1
S2 = 3.22 + 5.2
= 22 = a1 + a2
a2 = 22 – 8 = 14
⇒ d = 6
ak = 164
⇒ 8 + (k – 1)6 = 164
⇒ k = 27
OR
The sum of ages (in years) of a son and his father is 35 years and product of their ages is 150 years, find their ages.
Solution:
Let the age of father be x years and age of son be y years.
∴ x + y = 35 and xy = 150
Or y = 35 – x
Putting the value of y in xy = 150
x(35 – x) = 150
⇒ x2 – 35x + 150 = 0
⇒ (x – 30)(x – 5) = 0
⇒ x = 30 or x = 5 (rejected)
⇒ y = 5
Hence, the age of father = 30 years and the age of son = 5 years.
Question 29.
In the figure, PQ is a tangent to a circle with centre O. If ∠OAB = 30°, find ∠ABP and ∠AOB.
Solution:
In ΔOAB, OB = OA (radius of a circle)
Then, ∠OAB = ∠OBA
∠OBA = 30°
∴ ∠AOB = 180° – (30° + 30°)
∴ ∠AOB = 120°
∠ABP = ∠OBP – ∠OBA
= 90° – 30° = 60°
Question 30.
Evaluate : sin2 30° cos2 45° + 4 tan2 30° + \(\frac{1}{2}\)sin2 90° – 2 cos2 90° + \(\frac{1}{24}\)
Solution:
OR
sin θ + cos θ = √3 , then prove that tan θ + cot θ = 1
Solution:
sin θ + cos θ = 3 ⇒ (sin θ + cos θ)2 = 3
⇒ 1 + 2 sin θ cos θ = 3 ⇒ sin θ cos θ = 1
∴ tan θ + cot θ = \(\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}\) = 1
Detailed Solution:
Given sin θ + cos θ = √3
Squaring on both sides,
(sin θ + cos θ)2 = (√3)2
sin2 θ + cos2 θ + 2 sin θ cos θ = 3
1 + 2 sin θ cos θ = 3
2 sin θ cos θ = 2
sin θ cos θ = 1 … (i)
Hence Proved.
Question 31.
A box consists of 100 shirts of which 88 are good, 8 have minor defects and 4 have major defects. Ramesh, a shopkeeper will buy only those shirts which are good but ‘Kewal’ another shopkeeper will not buy shirts with major defects. A shirt is taken out of the box at random. What is the probability that:
(i) Ramesh will buy the selected shirt?
(ii) ‘Kewal’ will buy the selected shirt?
Solution:
(i) Number of good shirts = 88
P (Ramesh buys a shirt) = \(\frac{88}{100}\) or \(\frac{22}{25}\)
(ii) Number of shirts without major defect = 96
P (Kewal buys a shirt) = \(\frac{96}{100}\) or \(\frac{24}{25}\)
Section – D(20 Marks)
Question 32.
Two water taps together can fill a tank in 1\(\frac{7}{8}\) hours. The tap with longer diameter takes 2 hours less than the tap with smaller one to fill the tank separately. Find the time in which each tap can fill the tank separately.
Solution:
Let the smaller tap fills the tank in x hours.
Therefore, the larger tap fills the tank in x – 2 hours.
Therefore, x = 5
Smaller and larger taps can fill the tank separately in 5 hours and 3 hours respectively.
OR
Speed of a boat in still water is 15 km/h. It goes 30 km upstream and returns back at the same point in 4 hours 30 minutes. Find the speed of the stream.
Solution:
Let the speed of stream be x km/hr.
∴ Speed of boat upstream = (15 – x) km/hr
Speed of boat downstream = (15 + x) km/hr
⇒ 200 = 225 – x2
x = 5 (Rejecting – 5)
⇒ Speed of stream = 5 km/hr
Question 33.
In Fig. O is the centre of a circle of radius 5 cm. T is a point such that OT = 13 cm and OT intersects circle at E. If AB is a tangent to the circle at E, find the length of AB, where TP and TQ are two tangents to the circle.
Solution:
PT = \(\sqrt{169-25}\) = 12 cm
and TE = OT – OE = 13 – 5
= 8 cm
Let PA = AE = x. (Tangents)
Then, TA2 = TE2 + EA2
Or, (12 – x)2 = 82 + x2
24x = 80
Or, x = 3.3 cm.
Thus AB = 2 × x = 2 × 3.3
= 6.6 cm
Question 34.
Isha is 10 years old girl. On the result day, Isha and her father Suresh were very happy as she got first position in the class. While coming back to their home, Isha asked for a treat from her father as a reward for her success.
They went to a juice shop and asked for two glasses of juice.
Aisha, a juice seller, was serving juice to her customers in two type of glasses. Both the glasses had inner radius 3 cm. The height of both the glasses was 10 cm.
First type: A Glass with hemispherical raised bottom.
Second type: A glass with conical raised bottom of height 1.5 cm.
Isha insisted to have the juice in first type of glass and her father decided to have the juice in second type of glass. Out of the two, Isha or her father Suresh, who got more quantity of juice to drink and by how much?
Solution:
Capacity of first glass = πr2H – \(\frac{2}{3}\)πr3
= π × 9(10 – 2)
= 72π cm3
Capacity of second glass = πr2H – \(\frac{1}{3}\)πr3
= π × 3 × 3(10 – 0.5)
= 85.5π cm3
∵ Suresh got 42.43 cm3 more quantity of juice.
Detailed Solution:
In case of first type glass, we have
radius (r) = 3 cm
Height (H) = 10 cm
Capacity of Juice in cylindrical part = πr2H
= π × (3)2 × 10
= 226.28 cm3 (Approx.)
In case of second glass, we have
radius of base (r) = 3 cm
height of bottom part (h) = 1.5 cm
= 268.71 cm3 (Approx.)
Since, Suresh used second glass for drinking juice,
so he got more quantity of juice.
i.e., Suresh got 268.71 – 226.28 = 42.43 cm3 more juice than Isha.
OR
From a solid cylinder of height 5.6 cm and diameter 2.8 cm, a conical cavity of same height and same diameter is hollowed out. Find the volume of the remaining solid.
Solution:
= 11.498 cm3
Volume of remaining solid = Volume of cylinder – Volume of cone
= 34.496 – 11.498
= 22.998 = 23 cm3 (approx.)
Question 35.
On annual day of a school, 400 students participated in the function. Frequency distribution showing their ages is as shown in the following table:
Find mean and median of the above data.
Solution:
Section – E(12 Marks)
Question 36.
An asana is a body posture, originally and still a general term for a sitting meditation pose, and later extended in hatha yoga and modern yoga as exercise, to any type of pose or position, adding reclining, standing, inverted, twisting, and balancing poses. In the figure, one can observe that poses can be related to representation of quadratic polynomial.
Based on above situation answer the following questions.
(i) Name the shape of the poses represented in the figures.
(ii) The graph of parabola represented by quadratic polynomial ax2 + bx + c will face ……. if a < 0?
(iii) In the graph, how many zeroes are there for the polynomial? Find the zeroes.
Solution:
(i) Since, the posture of the body during asana is in U shaped so it represents a parabola.
(ii) We know that,
In ax2 + bx + c for the graph of parabola opening downwards, the value of must be negative or a < 0.
(iii) Since, the graph in the given figure is of a parabola which cuts the x-axis at two points,
So, it has two zeroes.
Since, the graph in the given figure is of a parabola which cuts the x-axis at two points -2 and 4.
So, the zeroes are -2 and 4.
OR
The zeroes of the quadratic polynomial 4√3x2 + 5x – 2√3.
Solution:
On factorizing the given polynomial,
⇒ 4√3x2 + 5x – 2√3 = 0
⇒ 4√3x2 + 8x – 3x – 2√3 = 0
⇒ 4x(√3x + 2) – √3(√3x + 2) = 0
⇒ (4x – √3) (√3x + 2) = 0
On equating to zero we get,
x = \(\frac{-2}{(\sqrt{3})}, \frac{(\sqrt{3})}{4}\)
Question 37.
Shreya purchased a new table from market with her mother. The top of the table is hexagonal in shape as shown in the given figure.
(i) Find the coordinates of points H and G.
(ii) Write one pair of points which have same ordinate.
(iii) Find the distance between the points A and B.
OR
Find the coordinates of the line joining points M and Q.
Solution:
(i) Coordinates of point H = (1, 5)
Coordinates of point G = (5, 1)
(ii) Points T and O has same ordinates i.e., T = (3, 9)
and O = T = (11, 9)
(iii) Coordinates of points A = (1, 9)
Coordinates of point B = (5, 13)
Distance, d = \(\sqrt{(1-5)^2+(9-13)^2}\)
= \(\sqrt{16+16}\)
= √32
= 4√2 units.
OR
Coordinates of point M = (5, 11)
Coordinates of point Q = (9, 3)
Mid-point = \(\left(\frac{5+9}{2}, \frac{11+3}{2}\right)\)
= (7, 7)
Question 38.
Read the following text and answer the following questions on the basis of the same:
In geometry, an equilateral triangle is a triangle in which all three sides have the same length. In the familiar Euclidean geometry, an equilateral triangle are also equiangular; that is, all three internal angles area also congruent to each other and are each 60°. Here we have an equilateral triangle which is inscribed in a circle of radius 6 cm.
(i) What is the length of BD?
Solution:
ABC be an equilateral triangle.
∴ OA = OB = OC = 6 cm (given)
D is the mid-point of BC and BO & CO are bisectors of ∠B and ∠C.
Hence, the length of BD is 3√3 cm.
(ii) Find the side of this equilateral triangle.
Solution:
We have proved that
BD = 3√3 cm
Then, BC = 2 × BD
= 2√3 × 3
= 6√3 cm.
(iii) Find the length of OD.
OR
Find the length of AD.
Solution:
Here, OB = 6 cm and BD = 3√3 cm
⇒ 2AD = 18
⇒ AD = 9 cm.