Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths Basic with Solutions Set 10 are designed as per the revised syllabus.
CBSE Sample Papers for Class 10 Maths Basic Set 10 with Solutions
Time: 3 Hours
Maximum Marks: 80
General Instructions:
Read the following instructions very carefully and strictly follow them:
- All questions are compulsory.
- The question paper consists of 30 questions divided into four sections A, B, C and D.
- Section-A contains 6 question of 1 marks each, Section B contains 6 questions of 2 marks each, Section C contains 10 questions of 3 marks each and Section D contains 8 questions of 4 marks each.
- There is no overall choice. However, an internal choice has been provided in two questions of 1 mark each, two questions of 2 marks each, four questions of 3 marks each and three questions of 4 marks each. You have to attempt only one of the alternatives in all such questions.
- Use of calculators is not permitted.
Section – A (6 Marks)
Question 1.
Write the discriminant of the quadratic equation (x + 5)2 = 2(5x – 3). 1
Answer:
The quadratic equation can be rewritten as x2 + 31 = 0
Thus, discriminant = b2 – 4ac = 02 – 4 (1) (31),
i.e., -124
Question 2.
Express 429 as a product of its prime factors. 1
Answer:
429 = 3 × 11 × 13
Question 3.
Find the sum of first 10 multiples of 6. 1
Answer:
10 multiples of 6 are: 6,12,18 ………, 60
So, Sum = \(\frac{10}{2}\)[6 × 2 + 9 × 6]
= 5 (12 + 54) = 330
Question 3.
Find the value(s) of x, if the distance between the points A(0, 0) and B(x, -4) is 5 units. 1
Answer:
Here, AB = \(\sqrt{(x-0)^2+(-4-0)^2}\)
= \(\sqrt{x^2+16}\)
= \(\sqrt{x^2+16}\) = 5 i.e. x2 = 9
or x ± 3
Question 4.
Two concentric circles of radii a and b (a > b) are given. Find the length of the chord of the larger circle which touches the smaller circle. 1
Answer:
Let AB be the required chord.
Obviously, AM = \(\sqrt{a^2-b^2}\)
So, AB = 2\(\sqrt{a^2-b^2}\)
Question 5.
In figure, PS = 3 cm, QS = 4 cm, ∠PRQ = θ, ∠PSQ = 90°, PQ ⊥ RQ and RQ = 9 cm. Evaluate tan θ.
If tan α = \(\frac{5}{12}\), find the value of sec α. 1
Answer:
As PSQ is a right triangle, right angled at S
PQ = \(\sqrt{P S^2+Q S^2}\)
= \(\sqrt{3^2+4^2}\) = 5 cm
In ΔRQP, we have tan θ = \(\frac{P Q}{R Q}\) = \(\frac{5}{9}\)
OR
Given tan α = \(\frac{5}{12}\), sec α = \(\sqrt{1+\tan ^2 \alpha}\)
= \(\sqrt{1+\left(\frac{5}{12}\right)^2}\)
= \(\sqrt{1+\frac{25}{144}}\)
= \(\frac{169}{144}\) = \(\frac{13}{12}\)
Section – B (12 Marks)
Question 6.
Points A(3, 1), B(5, 1), C(a, b) and D(4, 3) are vertices of a parallelogram ABCD. Find the values of a and b.
OR
Points P and Q trisect the line segment joining the points A(-2, 0) and B(0, 8) such that, P is near to A. Find the coordinates of points P and Q. 2
Answer:
We know that the diagonals AC and BD bisect each other at O (say)
So, mid-point of AC = mid-point of BD
i.e., (\(\frac{3+a}{2}, \frac{1+b}{2}\)) = (\(\frac{5+4}{2}, \frac{1+3}{2}\))
⇒ 3 + a = 9 and 1 + b = 4
⇒ a = 6 and b = 3
OR
Here, P divides AB internally in the ratio 1 : 2
So, P(\(\frac{0-4}{3}\), \(\frac{8}{3}\)) i.e., P(\(\frac{-4}{3}\), \(\frac{8}{3}\))
Q divides AB internotly in the ratio 2 : 1
So, Q(\(\frac{0-2}{3}\), \(\frac{16}{3}\)) i.e., Q(\(\frac{-2}{3}\), \(\frac{16}{3}\))
Question 7.
Solve the following pair of linear equations: 2
3x – 5y = 4
2y + 7 = 9x
Answer:
We have, 3x – 5y – 4 = 0
⇒ 3x – 5y = 4 …… (i)
Again 9x = 2y + 7
⇒ 9x – 2y = 7 ……. (ii)
By elimination method:
Multiplying equation (i) by 3, we get
9x – 15y = 12 …….. (iii)
Subtracting (ii) from (iii), we get
⇒ y = –\(\frac{5}{13}\)
Putting the value of equation (ii), we get
9x – 2(-\(\frac{5}{13}\)) = 7
⇒ 9x + \(\frac{10}{13}\) = 7
9x = 7 – \(\frac{10}{13}\)
⇒ 9x = \(\frac{91-10}{13}\)
⇒ 9x = \(\frac{81}{13}\)
⇒ x = \(\frac{9}{13}\)
Hence, the required solution is x = \(\frac{9}{13}\), y = –\(\frac{5}{13}\)
Question 8.
If HCF of 65 and 117 is expressible in the form 65n – 117, then find the value of n.
OR
On a morning walk, three persons step out together and their steps measure 30 cm, 36 cm and 40 cm respectively. What is the minimum distance each should walk so that each can cover the same distance in complete steps? 2
Answer:
Here, 65 = 5 × 13
117 = 3 × 3 × 13
HCF (65, 117) = 13
So, 13 = 65n – 117
⇒ n = 2
OR
Required minimum distance = LCM (30, 36, 40)
Now, 30 = 2 × 3 × 5 = 21 × 31 × 51
36 = 2 × 2 × 3 × 3 = 22 × 32
40 = 2 × 2 × 2 × 5 = 23 × 51
So, LCM (30, 36, 40)
= 23 × 32 × 51
= 360
Question 9.
A die is thrown once. Find the probability of getting (A) a composite number, (B) a prime number. 2
Answer:
Total number of possible outcomes = 6
Total number of composite numbers = 2, say [4, 6]
Total number of prime number = 3, say {2, 3, 5}
(A) P(a composite number) = \(\frac{2}{6}\) i.e., \(\frac{1}{3}\)
(B) P (a prime number) = \(\frac{3}{6}\) i.e., \(\frac{1}{2}\)
Question 10.
Cards numbered 7 to 40 were put in a box. Poonam selects a card at random. What is the probability that Poonam selects a card which is a multiple of 7? 2
Answer:
Total number of possible outcomes = 34
Favourable number of outcomes = 5,
say {7, 14, 21, 28, 35}
∴ Required Probability = \(\frac{5}{34}\)
Section – C (30 Marks)
Question 11.
AD and PM are medians of triangles ABC and PQR respectively where ΔABC ~ ΔPQR.
Prove that: \(\frac{A B}{P Q}=\frac{A D}{P M}\) 3
Answer:
OR
As ΔABC ~ ΔPQR
\(\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A C}{P R}\)
⇒ \(\frac{A B}{P Q}=\frac{2 B D}{2 P M}\) or \(\frac{B D}{Q M}\)
Now in Δs ABD and PQM, we have
∠B = ∠Q and \(\frac{A B}{P Q}=\frac{B D}{Q M}\) [by (i)]
By SAS similarity criterion, we get
ΔABD ~ ΔPQM
⇒ \(\frac{A B}{P Q}=\frac{A D}{P M}\)
Question 12.
Draw the graph of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Using this graph, find the values of x and y which satisfy both the equations. 3
Answer:
Table of values for x – y + 1 = 0
x | 2 | 1 | 0 |
y | 3 | 2 | 1 |
Table of values for 3x + 2y – 12 = 0
x | 0 | 2 | 4 |
y | 6 | 3 | 0 |
From the graph, we find
x = 2, y = 3
Question 13.
Prove that √3 is an irrational number.
OR
Find the greatest number which on dividing 1251, 9377 and 15628 leaves remainders 1, 2 and 3 respectively. 3
Answer:
Let us assume that √3 be a rational number
Then, √3 = \(\frac{p}{q}\), where p and q are co-primes and q ≠ 0
⇒ P2 = 3q2 …….. (i)
3 divides p2 i.e., 3 divides p also ……. (ii)
Let p = 3m, for some integer m
From (i), 9m2 = 3q2 or 3m2 = q2
3 divides q2 i.e., 3 divides q also …….. (iii)
From (ii) and (iii), we get that
3 divides p and q both
Which is a contradiction to the fact that p and q are co-primes.
Hence, our assumption is wrong
∴ √3 is irrational.
OR
1251 – 1 = 1250; 9377 – 2 = 9375; 15628 – 3 = 15625
Required greatest number = HCF (1250, 9375, 15625)
Now, 1250 = 2 × 54
9375 = 3 × 55
5625 = 56
HCF (1250, 9375, 15625) = 54 i.e., 625
Question 14.
(A) A, B and C are interior angles of a triangle ABC. Show that sin (\(\frac{B+C}{2}\)) = cos (\(\frac{A}{2}\))
(B) If ∠A = 90°, then find the value of tan (\(\frac{B+C}{2}\)).
OR
If tan (A + B) = 1 and tan (A – B) = \(\frac{1}{\sqrt{3}}\), 0° < A + B < 90°, A > B, then find the values of A and B. 3
Answer:
As, A, B, C are the interior angles of ΔABC,
A + B + C = 180°
B + C = 180° – A
or \(\frac{B+C}{2}\) = 90° – \(\frac{A}{2}\)
(A) sin(\(\frac{B+C}{2}\)) = sin(90°- \(\frac{A}{2}\)) = cos \(\frac{A}{2}\)
(B) tan (\(\frac{B+C}{2}\)) = tan(90° – \(\frac{A}{2}\)) = cot\(\frac{A}{2}\)
= cot 45° = 1 (∵ A = 90°)
OR
Since tan (A + B) = 1
A + B = 45 …… (i)
As tan (A -B) = \(\frac{1}{\sqrt{3}}\)
A – B = 30° ……. (ii)
From (i) and (ii), we have
A = 37.5° and B = 7.5°
Question 15.
In Figure, PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T. Find the length TP.
OR
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle. 3
Answer:
Let TR be x cm long and TP be y cm long.
As To is perpendicular bisector of PQ, PR = 4 cm
In right-triangle ORP, we have
OP2 = PR2 + OR2
OR2 = OP2 – PR2
= 52 – 42 = 9
⇒ OR = 3 cm
In ΔPRT, we have
y2 = x2 + 42 ……. (i)
In ΔOPT, we have
(x + 3)2 = 52 + y2
i.e. (x + 3)2 = 52 + x2 + 42 [using (i)]
Solving this equation, we have
x = \(\frac{16}{3}\) cm
From eq. (i), we get
y2 = \(\frac{256}{3}\) + 16
= \(\frac{400}{9}\)
⇒ y = \(\frac{20}{3}\)
OR
Here, ΔROC ≅ ΔQOC
∴ ∠1 =∠2
Similarly, ∠4 = ∠3, ∠5 = ∠6, ∠8 = ∠7
Also,
∠ROQ + ∠QOP + ∠POS + ∠SOR = 360°
⇒ 2∠1 + 2∠4 + 2∠5 + 2∠8 = 360°
⇒ ∠1 + ∠4 + ∠5 + ∠8 = 180°
So, ∠DOC + ∠AOB = 180°
and ∠AOD + ∠BOC = 180°
Question 16.
Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes if 8 cm of standing water is needed? 3
Answer:
Volume of water flowing through canal in 30 minutes.
= (5000 × 6 × 1.5) cu. m, i.e., 45000 cu. m
Area that will be irrigated = 45000 ÷ \(\frac{8}{100}\)
= 562500 sq m.
Question 17.
A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent. 3
Answer:
No. of days | No. of Students (fi) | xi | fixi |
0 – 6 | 10 | 3 | 30 |
6 – 12 | 11 | 9 | 99 |
12 – 18 | 7 | 15 | 105 |
18 – 24 | 4 | 21 | 84 |
24 – 30 | 4 | 27 | 108 |
30 – 36 | 3 | 33 | 99 |
36 – 42 | 1 | 39 | 39 |
Total | 40 | 564 |
Mean = \(\bar{x}=\frac{\Sigma f_i x_i}{\Sigma f_i}\) = \(\frac{564}{100}\) = 14.1
Question 18.
A car has two wipers which do not overlap. Each wiper has a blade of length 21 cm sweeping through an angle 120°. Find the total area cleaned at each sweep of the blades. take (π = \(\frac{22}{7}\)) 3
Answer:
Total area cleaned = 2 × Area of Sector
= 2 × \(\frac{\pi r^2 \theta}{360^{\circ}}\)
= 2 × \(\frac{22}{7}\) × 21 × 21 × \(\frac{120^{\circ}}{360^{\circ}}\)
= 924 sq cm.
Section – D (32 Marks)
Question 19.
A pole has to be erected at a point on the boundary of a circular park of diameter 13 m in such a way that the difference of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 m. Is it possible to do so? If yes, at what distances from the two gates should the pole be erected? 4
Answer:
Here, PB – PA = 7
Let AP be x metre long.
Then, PB = (x + 7) m
Since ΔAPB is a right triangle
AB2 = AP2 + BP2
⇒ 132 = x2 + (x + 7)2
Solving this quadratic equation, we getx = 5 Hence, it is possibles to erect a pole.
and the distance of the pole from the two gates are 5 m and 12 m
Question 20.
If m times the mth term of an Arithmetic Progression is equal to n times its nth term and m ≠ n, show that the (m + n)th term of the AP is zero.
OR
The sum of the first three numbers in an Arithmetic Progression is 18. if the product of the first and the third term is 5 times the common difference, find the three numbers. 4
Answer:
Let a be first term and d be the common difference of the given AP. Then,
m[a + (m – 1) d] = n[a + (n – 1)d]
⇒ ma + m (m – 1) d = na + n (n – 1)d
⇒ (ma – na) + (m2 – m – n2 + n)d = 0
⇒ (m – n) a + [(m + n) (m – n) – (m – n)]d = 0
⇒ a + (m + n – 1 )d = 0 (∵ m ≠ n)
i.e. the (m + n)th term is zero
OR
Let first three terms of AP be a – d, a and a + d
Then, as per the question,
(a – d) + a + (a + d) = 18
⇒ 3a = 18, a = 6
Also,
(a – d) (a + d) = 5 d
⇒ a2 – d2 = 5 d
or 36 – d2 = 5 d
or d2 + 5d – 36 = 0
or (d + 9) (d – 4) = 0
⇒ d = -9, 4
So, the three number are 15, 6, -3 or 2, 6,10.
Question 21.
In figure, a decorative block is shown which is made of two solids, a cube and a hemisphere. The base of the block is a cube with edge 6 cm and the hemisphere fixed on the top has a diameter of 4.2 cm. Find:
(A) the total surface area of the block.
(B) the volume of the block formed take π = \(\frac{22}{7}\).
Answer:
(A) Total surface area of the block
= TSA of cube + CSA of hemisphere – Base area of hemisphere
= 6a2 + 2πr2 – πr2
= 6a2 + πr2
= (6 × 62 + \(\frac{22}{7}\) × 2.1 × 2.11 sq. cm
= (216 + 13.86) sq cm
= 229.86 sq cm
(B) Volume of the block
= 63 + \(\frac{2}{3}\) × \(\frac{22}{7}\) × (2.1)3 cu.cm
= (216 + 19.40) cu.cm
= 235.40 cu.cm
Question 22.
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, prove that the other two sides are divided in the same ratio. 4
Answer:
Here is a ΔABC in which a line parallel to side BC intersects other two sides AB and AC at D and E respectively.
We need to prove that \(\frac{A D}{D B}=\frac{A E}{E C}\)
Let us join BE and CD and then draw DM ⊥ AC and EN ⊥ AB
Now ar(ΔADE) = \(\frac{1}{2}\) × AD × EN
= \(\frac{1}{2}\) × AE × DM …… (I)
and ar(ΔBDE) = \(\frac{1}{2}\) × DB × EN ……. (ii)
and ar(ΔDEC) = \(\frac{1}{2}\) × EC × DM ……. (iii)
from (i) and (ii), we have
Since, ΔBDE and ΔDEC are on the same base DE and between the same parallels BC and DE, ar(ΔBDE) = ar(ΔDEC)
Therefore, from (iv), (v) and (vi), we have
\(\frac{A D}{D B}=\frac{A E}{E C}\)
Question 23.
If 1 + sin2 θ = 3 sin θ cos θ, then prove that tan θ = 1 or tan θ = \(\frac{1}{2}\). 4
Answer:
1 + sin2 θ = 3 sin θ cos θ
Given, sec2 θ + tan2 θ = 3 tan θ
[On dividing both sides by cos2 θ]
1 + 2 tan2 θ – 3 tan θ = θ
Solving this quadratic equation, we get
tan θ 1 or \(\frac{1}{2}\)
Question 24.
The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun’s altitude is 30° than when it was 60°. Find the height of the tower. (Given √3 = 1.732) 4
Answer:
Let AB be the tower whose height is h metres
In ΔABD, \(\frac{h}{x}\) = tan 60°
⇒ h = √3x …… (i)
In ΔABC, \(\frac{h}{x+40}\) = tan 30°
from (i) and (ii), we have
3x = x + 40
⇒ x = 20
So, height of the tower
h = 20√3 m
= 20 × 1.732 m
= 34.64 m