CBSE Class 10 Math Basic Sample Paper 2022 – 23 Solutions
CBSE Class 10 Math Basic Sample Paper 2022: Central Board of Secondary Education(CBSE) has released the CBSE Class 10 Math Basic Sample Paper 2022 – 23 on its official website on 16th September 2022.
(1) If two positive integers p and q can be expressed as p = ab2 and q = a 3b; a, b being prime numbers, then LCM (p, q) is
(a) ab
(b) a 2b 2
(c) a 3b 2
(d) a 3b 3
Ans: (c) a 3b 2
(2) What is the greatest possible speed at which a man can walk 52 km and 91 km in an exact number of hours?
(a) ab
(b) a 2b 2
(c) a 3b 2
(d) a 3b 3
Ans: (c) 13 km/hours
(3) If one zero of the quadratic polynomial x 2 + 3x + k is 2, then the value of k is
(a) 10
(b) -10
(c) 5
(d) –5
Ans: (b) -10
(4) Graphically, the pair of equations given by 6x – 3y + 10 = 0 2x – y + 9 = 0 represents two lines which are
(a) intersecting at exactly one point.
(b) parallel.
(c) coincident.
(d) intersecting at exactly two points.
Ans: (b) Paralle.
(5) If the quadratic equation x2 + 4x + k = 0 has real and equal roots, then
(a) k < 4
(b) k > 4
(c) k = 4
(d) k ≥ 4
Ans: (c) k = 4
(6) The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is
(a) 5 units
(b) 12 units
(c) 11 units
(d) (7 + √5) units
Ans: (b) 12
(7) If in triangles ABC and DEF, AB /DE = BC /FD , then they will be similar, when
(a) ∠B = ∠E
(b) ∠A = ∠D
(c) ∠B = ∠D
(d) ∠A = ∠F
Ans: (c) ∠B = ∠D
(8) In which ratio the y-axis divides the line segment joining the points (5, – 6) and (–1, – 4)?
(a) 1 : 5
(b) 5 : 1
(c) 1 : 1
(d) 1 : 2
Ans: (b) 5 : 1
(9) In the figure, if PA and PB are tangents to the circle with centre O such that ∠APB = 50°, then ∠OAB is equal to
(a) 25°
(b) 30°
(c) 40°
(d) 50°
Ans: (a) 25°
(10) If sin A = 1 2 , then the value of sec A is :
(a) √3 2
(b) 1 √3
(c) √3
(d) 1
Ans: (a) √3/ 2
(11) √3 cos2A + √3 sin2A is equal to
(a) 1
(b) 1 √3
(c) √3
(d) 0
Ans: (c) √3
(12) The value of cos1° cos2° cos3° cos4°…………..…..cos90° is
(a) 1
(b) 0
(c) – 1
(d) 2
Ans: (b) 0
(13) If the perimeter of a circle is equal to that of a square, then the ratio of their areas is
(a) 22 : 7
(b) 14 : 11
(c) 7 : 22
(d) 11: 14
Ans: (b) 14 : 11
(14) If the radii of two circles are in the ratio of 4 : 3, then their areas are in the ratio of :
(a) 4 : 3
(b) 8 : 3
(c) 16 : 9
(d) 9 : 16
Ans: (c) 16 : 9
(15) The total surface area of a solid hemisphere of radius 7 cm is :
(a) 447π cm2
(b) 239π cm2
(c) 174π cm2
(d) 147π cm2
Ans: (d) 147π cm2
(16) For the following distribution:
Class | 0 – 5 | 5 – 10 | 10 – 15 | 15 – 20 | 20 – 25 |
Frequency | 10 | 15 | 12 | 20 | 9 |
the upper limit of the modal class is
(a) 10
(b) 15
(c) 20
(d) 25
Ans: (c) 20
(17) If the mean of the following distribution is 2.6, then the value of y is
Variable (x) | 1 | 2 | 3 | 4 | 5 |
Frequency | 4 | 5 | y | 1 | 2 |
(a) 3
(b) 8
(c) 13
(d) 24
Ans: (b) 8
(18) A card is selected at random from a well shuffled deck of 52 cards. The probability of its being a red face card is
(a) 3 /26
(b) 3/ 13
(c) 2 /13
(d) 1 /2
Ans: (a) 3/ 26
(19) Assertion: If HCF of 510 and 92 is 2, then the LCM of 510 & 92 is 32460
Reason: as HCF(a,b) x LCM(a,b) = a x b
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true.
Ans: (d) Assertion (A) is false but Reason (R) is true.
(20) Assertion (A): The ratio in which the line segment joining (2, -3) and (5, 6) internally divided by x axis is 1:2.
Reason (R): as formula for the internal division is ( 2 + 1 + , 2 + 1 + )
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true.
Ans: (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Section B
(21) For what values of k will the following pair of linear equations have infinitely many solutions?
kx + 3y – (k – 3) = 0
12x + ky – k = 0
Ans: For a pair of linear equations to have infinitely many solutions :
a1/ a2 = b1/ b2 = c1 /c2 ⇒ k /12 = 3/ k = k−3/ k
/12 = 3/ ⇒ k 2 = 36 ⇒ k = ± 6
Also, 3 / = −3/ ⇒ k 2 – 6k = 0 ⇒ k = 0, 6.
Therefore, the value of k, that satisfies both the conditions, is k = 6.
(22) In the figure, altitudes AD and CE of Δ ABC intersect each other at the point P. Show that:
(i) ΔABD ~ ΔCBE
(ii) ΔPDC ~ ΔBEC
Ans: (i) In ΔABD and ΔCBE
∠ADB = ∠CEB = 90º
∠ABD = ∠CBE (Common angle)
⇒ ΔABD ~ ΔCBE (AA criterion)
(ii) In ΔPDC and ΔBEC
∠PDC = ∠BEC = 90º
∠PCD = ∠BCE (Common angle)
⇒ ΔPDC ~ ΔBEC (AA criterion)
[OR]
In the figure, DE || AC and DF || AE. Prove that BF/ FE = BE/ EC
Ans:
In ΔABC, DE || AC
BD/AD = BE/EC ………(i) (Using BPT)
In ΔABE, DF || AE
BD/AD = BF/FE ……..(ii) (Using BPT)
From (i) and (ii)
BD/AD = BE/EC = BF/FE
Thus, BF /FE = BE /EC
(23) Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Ans: Let O be the centre of the concentric circle of radii 5 cm and 3 cm respectively. Let AB be a chord of the larger circle touching the smaller circle at P
Then AP = PB and OP⊥AB
Applying Pythagoras theorem in △OPA, we hav
OA2=OP2+AP2 ⇒ 25 = 9 + AP2
⇒ AP2= 16 ⇒ AP = 4 cm
∴ AB = 2AP = 8 cm
(24) If cot θ = 7 8 , evaluate (1 + sin θ) (1− sin θ) (1 + cos θ) (1− cos θ)
Ans: Now, (1 + sinθ)(1 − sinθ) (1 + cosθ)(1 − cosθ) = (1 – sin2θ) (1 – cos2θ)
= cos2θ sin2θ = ( cosθ sinθ ) 2
= cot2θ
= ( 7 /8 ) 2 = 49/ 64
(25) Find the perimeter of a quadrant of a circle of radius 14 cm.
Ans: Perimeter of quadrant = 2r + 1 4 × 2 π r
⇒ Perimeter = 2 × 14 + 1/ 2 × 22/ 7 × 14
⇒ Perimeter = 28 + 22 =28+22 = 50 cm
[OR]
Find the diameter of a circle whose area is equal to the sum of the areas of the two circles of radii 24 cm and 7 cm.
Area of the circle = Area of first circle + Area of second circle
⇒ πR2 = π (r1) 2 + π (r1) 2
⇒ πR2 = π (24)2 + π (7)2 ⇒ πR2 = 576π +49π
⇒ πR2 = 625π ⇒ R2 = 625 ⇒ R = 25 Thus, diameter of the circle = 2R = 50 cm.
Section C
(26) Prove that √5 is an irrational number.
Ans: Let us assume to the contrary, that √5 is rational. Then we can find a and b ( ≠ 0) such that √5 = (assuming that a and b are co-primes).
So, a = √5 b ⇒ a 2 = 5b 2
Here 5 is a prime number that divides a 2 then 5 divides a also (Using the theorem, if a is a prime number and if a divides p 2 , then a divides p, where a is a positive integer)
Thus 5 is a factor of a
Since 5 is a factor of a, we can write a = 5c (where c is a constant). Substituting a = 5c We get (5c)2 = 5b 2 ⇒ 5c 2 = b
This means 5 divides b2 so 5 divides b also (Using the theorem, if a is a prime number and if a divides p 2 , then a divides p, where a is a positive integer).
Hence a and b have at least 5 as a common factor.
But this contradicts the fact that a and b are coprime. This is the contradiction to our assumption that p and q are co-primes.
So, √5 is not a rational number. Therefore, the √5 is irrational.
(27) Find the zeroes of the quadratic polynomial 6x 2 – 3 – 7x and verify the relationship between the zeroes and the coefficients.
Ans: 6×2 – 7x – 3 = 0 ⇒ 6×2 – 9x + 2x – 3 = 0
⇒ 3x(2x – 3) + 1(2x – 3) = 0 ⇒ (2x – 3)(3x + 1) = 0
⇒ 2x – 3 = 0 & 3x + 1 = 0
x = 3/2 & x = -1/3 Hence, the zeros of the quadratic polynomials are 3/2 and -1/3.
For verification
Sum of zeros = – coefficient of x /coefficient of x 2 ⇒ 3/2 + (-1/3) = – (-7) / 6 ⇒ 7/6 = 7/6
Product of roots = constant/ coefficient of x 2 ⇒ 3/2 x (-1/3) = (-3) / 6 ⇒ -1/2 = -1/2
Therefore, the relationship between zeros and their coefficients is verified.
(28) A shopkeeper gives books on rent for reading. She takes a fixed charge for the first two days, and an additional charge for each day thereafter. Latika paid Rs 22 for a book kept for six days, while Anand paid Rs 16 for the book kept for four days. Find the fixed charges and the charge for each extra day.
Ans: Let the fixed charge by Rs x and additional charge by Rs y per day
Number of days for Latika = 6 = 2 + 4
Hence, Charge x + 4y = 22
x = 22 – 4y ………(1)
Number of days for Anand = 4 = 2 + 2
Hence, Charge x + 2y = 16
x = 16 – 2y ……. (2)
On comparing equation (1) and (2), we get,
22 – 4y = 16 – 2y ⇒ 2y = 6 ⇒ y = 3
Substituting y = 3 in equation (1), we get,
x = 22 – 4 (3) ⇒ x = 22 – 12 ⇒ x = 10
Therefore, fixed charge = Rs 10 and additional charge = Rs 3 per day
[OR]
Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 3 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?
Ans: AB = 100 km. We know that, Distance = Speed × Time
AP – BP = 100 ⇒ 5x − 5y = 100 ⇒ x−y=20…..(i)
AQ + BQ = 100 ⇒ x + y = 100….(ii)
Adding equations (i) and (ii), we get,
x − y + x + y = 20 +100 ⇒ 2x = 120 ⇒ x = 60
Substituting x = 60 in equation (ii), we get, 60 + y = 100 ⇒ y = 40
Therefore, the speed of the first car is 60 km/hr and the speed of the second car is 40 km/hr.
(29) In the 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.
Ans: Since OT is perpendicular bisector of PQ. Therefore, PR=RQ=4 cm
Now, OR = √ − = √ − =3cm
Now, ∠TPR + ∠RPO = 90° (∵TPO=90°)
& ∠TPR + ∠PTR = 90° (∵TRP=90∘)
So, ∠RPO = ∠PTR
So, ⍙TRP ~ ⍙PRO [By A-A Rule of similar triangles]
So, TP/ PO = RP /RG
⇒ TP/ 5 = 4 /3 ⇒ TP = 20/ 3 cm
(30) Prove that
tan θ 1 − cot θ + cot θ 1 − tan θ = 1 + sec θ cosec θ
Ans: LHS = tan θ 1−cot θ + cot θ 1−tan θ = tan θ 1− 1 tanθ + 1 tanθ 1−tan θ
= tan2θ tan θ−1 + 1 tan θ (1−tan θ
= tan3θ−1 tan θ (tan θ−1)
= (tanθ −1) (tan3θ + tanθ+1 ) tan θ (tan θ−1)
= (tan3θ + tanθ+1 ) tan θ
= tanθ + 1 + sec = 1 + tanθ + secθ
= 1 + sin θ cos θ + cos θ sin θ
= 1 + sin2θ+ cos2θ sin θ cos θ
= 1 + 1 sin θ cos θ = 1 + sec θ cosec θ
[OR]
If sin θ + cos θ = √3, then prove that tan θ + cot θ = 1
Ans: If sin θ + cos θ = √3, then prove that tan θ + cot θ = 1
⇒ sin2θ + cos2θ + 2sin θ cos θ = 3
⇒ 1 + 2sin θ cos θ = 3 ⇒ 1 sin θ cos θ = 1
Now tanθ + cotθ = sin θ /cos θ + cos θ/ isn θ
= sin2θ+ cos2θ/ sin θ cos θ
= 1/ sin θ cos θ = 1/ 1 = 1
(31) Two dice are thrown at the same time. What is the probability that the sum of the two numbers appearing on the top of the dice is
(i) 8?
(ii) 13?
(iii) less than or equal to 12?
Ans: (i) P(8 ) = 5 36
(ii) P(13 ) = 0 36 = 0
(iii) P(less than or equal to 12) = 1
Section D
(32) An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11km/h more than that of the passenger train, find the average speed of the two trains.
Ans: Let the average speed of passenger train = x km/h.
and the average speed of express train = (x + 11) km/h
As per given data, time taken by the express train to cover 132 km is 1 hour less than the passenger train to cover the same distance. Therefore,
132/ − 132 /+11 = 1
⇒ 132 (+11−) (+11) = 1 ⇒ 132 11 (+11) = 1
⇒ 132 × 11 = x(x + 11) ⇒ x 2 + 11x – 1452 = 0
⇒ x 2 + 44x -33x -1452 = 0
⇒ x (x + 44) -33(x + 44) = 0 ⇒ (x + 44)(x – 33) = 0
⇒ x = – 44, 33
As the speed cannot be negative, the speed of the passenger train will be 33 km/h and the speed of the express train will be 33 + 11 = 44 km/h.
[OR]
A motor boat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.
Ans: Let the speed of the stream be x km/hr
So, the speed of the boat in upstream = (18 – x) km/hr
& the speed of the boat in downstream = (18 + x) km/hr
ATQ, distance/ upstream speed – distance/ downstream speed = 1
⇒ 24 /18 − – 24/ 18 + = 1
⇒ 24 [ 1 /18 − – 1/ 18 + ] = 1 ⇒ 24 [ 18 + −(18−)/ (18 − ).(18 + ) ] = 1
⇒ 24 [ 1/ 18 − − 1 /18 + ] = 1 ⇒ 24 [ 18 + −(18−) /(18 − ).(18 + ) ] = 1
⇒ 24 [ 1 /18 − − 1 /18 + ] = 1 ⇒ 24 [ 18 + −(18−)/ (18 − ).(18 + ) ] = 1
⇒ 48x = 324 – x 2 ⇒ x 2 + 48x – 324 = 0
⇒ (x + 54)(x – 6) = 0 ⇒ x = -54 or 6
As speed to stream can never be negative, the speed of the stream is 6 km/hr.
(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. In the figure, find EC if AD /DB = AE/ EC using the above theorem.
Ans: Figure
Given, To prove, constructions Proof
Application —-
(34) A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand.
Ans: Volume of one conical depression = 1/ 3 x π r 2 h
= 1 /3 x 22 /7 x 0.5 2 x 1.4 cm3 = 0.366 cm3
Volume of 4 conical depression = 4 x 0.366 cm3
= 1.464 cm3
Volume of cuboidal box = L x B x H
= 15 x 10 x 3.5 cm3 = 525 cm3
Remaining volume of box = Volume of cuboidal box –
Volume of 4 conical depressions
= 525 cm3 − 1.464 cm3 = 523.5 cm3
[OR]
Ramesh made a bird-bath for his garden in the shape of a cylinder with a hemispherical depression at one end. The height of the cylinder is 1.45 m and its radius is 30 cm. Find the total surface area of the bird-bath.
Ans: Let h be height of the cylinder, and r the common radius of the cylinder and hemisphere.
Then, the total surface area = CSA of cylinder + CSA of hemisphere
= 2rh + 2r 2 = 2 r (h + r)
= 2 x 22/ 7 x 30 (145 + 30) cm2
= 33000 cm2 = 3.3 m2
(35) A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 years.
Age (in years) | Number of policy holders |
Below 20 | 2 |
20-25 | 4 |
25-30 | 18 |
30-35 | 21 |
35-40 | 33 |
40-45 | 11 |
45-50 | 3 |
50-55 | 6 |
55-60 | 2 |
Ans: n = 100 ⇒ n/2 = 50, Therefore, median class = 35 – 40,
Class size, h = 5, Lower limit of median class, l = 35,
frequency f = 33, cumulative frequency cf = 45
⇒Median = l + [ n /2 − cf f ] × h
⇒Median = 35 + [ 50 – 45/ 33 ] × 5
= 35 + 25 /33 = 35 + 0.76
= 35.76 Therefore, median age is 35.76 years
Section E
(36) Case Study – 1
In the month of April to June 2022, the exports of passenger cars from India increased by 26% in the corresponding quarter of 2021–22, as per a report. A car manufacturing company planned to produce 1800 cars in 4th year and 2600 cars in 8th year. Assuming that the production increases uniformly by a fixed number every year.
Based on the above information answer the following questions.
(1) Find the production in the 1st year.
Ans: Since the production increases uniformly by a fixed number every year, the number of Cars manufactured in 1st, 2nd, 3rd, . . .,years will form an AP.
So, a + 3d = 1800 & a + 7d = 2600
So d = 200 & a = 1200
(2) Find the production in the 12th year.
Ans: t12 = a + 11d ⇒ t30 = 1200 + 11 x 200
⇒ t12 = 3400
(3) Find the total production in first 10 years.
Ans: Sn = 2 [2 + ( − 1)] ⇒ S10 = 10 2 [2 1200 + (10 − 1) 200]
⇒ S10 = 13 2 [2 1200 + 9 x 200]
⇒ S10 = 5 x [2400 + 1800 ]
⇒ S10 = 5 x 4200= 21000
[OR]
In which year the total production will reach to 15000 cars?
Ans: Let in n years the production will reach to 31200
Sn = /2 [2 + ( − 1)] = 31200 ⇒ 2 [2 1200 + ( − 1)200] = 31200
⇒ / 2 [2 x 1200 + ( − 1)200] = 31200 ⇒ [ 12 + ( − 1) ] = 312
⇒ n 2 + 11n -312 = 0
⇒ n 2 + 24n – 13n -312 = 0
⇒ (n +24)(n -13) = 0
⇒ n = 13 or – 24. As n can’t be negative. So n = 13
(37) Case Study – 2
In a GPS, The lines that run east-west are known as lines of latitude, and the lines running north-south are known as lines of longitude. The latitude and the longitude of a place are its coordinates and the distance formula is used to find the distance between two places. The distance between two parallel lines is approximately 150 km. A family from Uttar Pradesh planned a round trip from Lucknow (L) to Puri (P) via Bhuj (B) and Nashik (N) as shown in the given figure below.
Ans: (1) LB = √ (2 − 1 ) 2 + (2 − 1 ) 2 ⇒ LB = √ (0 − 5) 2 + (7 − 10)
LB = √ (5) 2 + (3) 2 ⇒ LB = √25 + 9 LB = √34
Hence the distance is 150 √34 km
(2) Coordinate of Kota (K) is ( 3 x 5 + 2 x 0 3 + 2 , 3 x 7 + 2 x 10 3 + 2 )
= ( 15+0 5 , 21+20 5 )= (3, 41 5 )
(3) L(5, 10), N(2,6), P(8,6)
LN = √ (2 − 5) 2 + (6 − 10) 2 = √ (3) 2 + (4) 2 = √9 + 16 = √25 = 5
NP = √ (8 − 2) 2 + (6 − 6) 2 = √ (4) 2 + (0) 2 = 4
PL = √ (8 − 5) 2 + (6 − 10) 2 = √ (3) 2 + (4) 2 ⇒ LB = √9 + 16 = √25 = 5
as LN = PL ≠ NP, so ∆ LNP is an isosceles triangle.
[OR]
Find a place (point) on the longitude (y-axis) which is equidistant from the points Lucknow (L) and Puri (P).
Ans: Let A (0, b ) be a point on the y – axis then AL = AP
⇒ √ (5 − 0) 2 + (10 − b) 2 = √ (8 − 0) 2 + (6 − b) 2
⇒ (5) 2 + (10 − b) 2 = (8) 2 + (6 − b) 2
⇒ 25 + 100 − 20 + b 2 = 64 + 36 − 12 + b 2 ⇒ 8b = 25 ⇒ b = 25/ 8
So, the coordinate on y axis is (0, 25 /8 )
(38) Case Study – 3
Lakshaman Jhula is located 5 kilometers north-east of the city of Rishikesh in the Indian state of Uttarakhand. The bridge connects the villages of Tapovan to Jonk. Tapovan is in Tehri Garhwal district, on the west bank of the river, while Jonk is in Pauri Garhwal district, on the east bank. Lakshman Jhula is a pedestrian bridge also used by motorbikes. It is a landmark of Rishikesh.
A group of Class X students visited Rishikesh in Uttarakhand on a trip. They observed from a point (P) on a river bridge that the angles of depression of opposite banks of the river are 60° and 30° respectively. The height of the bridge is about 18 meters from the river.
Based on the above information answer the following questions.
(1) Find the distance PA.
Ans: sin 60° = PC/ PA
⇒ √3 2 = 18/ PA ⇒ PA = 12 √3 m
(2) Find the distance PB
Ans: sin 30° = PC/ PB
⇒ 1 2 = 18/ PB ⇒ PB = 36 m
(3) Find the width AB of the river
Ans: tan 60° = PC/ AC ⇒ √3 = 18 AC ⇒ AC = 6 √3 m
tan 30° = PC/ CB ⇒ 1 √3 = 18/ CB ⇒ CB = 18 √3 m
Width AB = AC + CB = 6 √3 + 18 √3 = 24 √3 m
[OR]
Find the height BQ if the angle of the elevation from P to Q be 30°.
Ans: RB = PC =18 m & PR = CB = 18 √3 m
tan 30° = QR /PR ⇒ 1 √3 = QR/ 18 √3 ⇒ QR = 18 m
QB = QR + RB = 18 + 18 = 36m. Hence height BQ is 36m
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