430/4/1 2020 Class 10 Maths Basic Question Paper Solution
MATHEMATICS
(BASIC)
SECTION A
1.) Given that HCF (156, 78) = 78, LCM (156, 78) is
(A) 156 (B) 78 (C) 156 × 78 (D) 156 × 2
Ans: (A) 156
2) Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio
(A) 4 : 9 (B) 2 : 3 (C) 81 : 16 (D) 16 : 81
Ans: (D) 16 : 81
3) The distance between the points (– 1, – 3) and (5, – 2) is
(A) √61 units (B) √37 units (C) 5 units (D) √17 units
Ans: (B) √37 units
5) Number of zeroes of the polynomial p(x) shown in Figure-1, are
(A) 3 (B) 2 (C) 1 (D) 0
Ans: (C) 1
6) A dice is thrown once. The probability of getting an odd number is
(A) 1 (B) 1/ 2 (C) 4/ 6 (D) 2/ 6
Ans: (B) 1/ 2
7) The value of k for which the equations 3x – y + 8 = 0 and 6x + ky = – 16 represent coincident lines, is
(A) -1/ 2 (B) 1 /2 (C) 2 (D) – 2
Ans: (D) –2
8) If sin A = cos A, 0 ≤ A ≤ 90°, then the angle A is equal to
(A) 30° (B) 60° (C) 0° (D) 45°
Ans: (D) 45°
9) The second term from the end of the A.P. 5, 8, 11, …, 47 is
(A) 50 (B) 45 (C) 44 (D) 41
Ans: (C) 44
10) Total surface area of a solid hemisphere is
(A) 3πr 2 (B) 2πr 2 (C) 4πr 2 (D) 2/ 3π r3
Ans: (A) 3πr 2
Fill in the blanks in question numbers 11 to 15.
11) The roots of the equation, x2 + bx + c = 0 are equal if _______.
Ans: b2 = 4c
12) The mid-point of the line segment joining the points (– 3, – 3) and (– 3, 3) is _______.
Ans: (–3, 0)
13) The lengths of the tangents drawn from an external point to a circle are _______.
Ans: Equal
14) For a given distribution with 100 observations, the ‘less than’ ogive and ‘more than’ ogive intersect at (58, 50). The median of the distribution is _______.
Ans: 58
15) In the quadratic polynomial t2 – 16, sum of the zeroes is _______.
Ans: 0
Answer the following question numbers 16 to 20.
16) Write the 26th term of the A.P. 7,4,1, –2, …
Ans: d = –3
a26 = –68
17) Find the coordinates of the point on x-axis which divides the line segment joining the points (2, 3) and (5, – 6) in the ratio 1 : 2.
Ans: Let the point on x-axis be (x, 0)
∴ Required point is (3, 0)
18) If cosec θ = 5 /4, find the value of cot θ
Ans: cot2 θ=25/16-1=9/16
cot θ=3/4
OR
Find the value of sin 42° – cos 48 °.
Ans: sin 42° = cos (90° – 42°) = cos 48°
∴ sin 42° – cos 48° = 0
19) The angle of elevation of the top of the tower AB from a point C on the ground, which is 60 m away from the foot of the tower, is 30°, as shown in Figure-2. Find the height of the tower.
Ans: tan 30° = AB/ 60°
⇒ AB = 60/ √3 or 20√3m
20) In Figure-3, find the length of the tangent PQ drawn from the point P to a circle with centre at O, given that OP = 12 cm and OQ = 5 cm.
Ans: PQ2 = 144 – 25 = 119
∴ PQ = √119 units
SECTION B
Question numbers 21 to 26 carry 2 marks each.
21) A cylindrical bucket, 32 cm high and with radius of base 14 cm, is filled completely with sand.
Find the volume of the and. (Use π = 22 /7 )
Ans: Volume of sand = 22/7×14×14×32
= 19712 cm3
22) In Figure-4, ΔABC and ΔXYZ are shown. If AB = 3.8 cm, AC = 3√ 3 cm, BC = 6 cm, XY = 6 √3 cm, XZ = 7.6 cm, YZ = 12 cm and ∠A = 65°, ∠B = 70°, then find the value of ∠Y.
Ans: ΔABC ~ ΔXZY
=>∠Y = ∠C = 180° – (∠A + ∠B)
=>∠Y = 45°
OR
If the areas of two similar triangles are equal, show that they are congruent
Ans: Let ΔABC ~ ΔPQR
∴ ar(ABC)/ar(PQR)=1=AB2/PQ2=AC2/PR2
=> AB=PQ, BC=QR, AC=PR
∴ Δ ≅Δ ABC PQR (SSS congrence rule)
23) If sec 2A = cosec (A – 30°), 0° < 2A < 90°, then find the value of ∠A.
Ans: sec 2A = cosec (A – 30°)
=> cosec (90° – 2A) = A – 30°
=> 90° – 2A = A – 30°
=> ∠A = 40°
24) Show that every positive even integer is of the form 2q and that every positive odd integer is of the form 2q + 1, where q is some integer
Ans: Let ‘a’ be any positive integer and b = 2
Using Euclid’s Division lemma
a = 2q + r, r = 0, 1
∴ a = 2q or 2q + 1
Because 2q is an even integer and every positive ineger is either even or odd.
∴ a = 2q + 1 is an odd positive integer.
25) How many two-digit numbers are divisible by 6?
Ans: Two digit numbers divisible by 6 are 12, 18, 24,…96
96 = 12 + (n – 1) × 6
=>n = 15
OR
In an A.P. it is given that common difference is 5 and sum of its first ten terms is 75. Find the first term of the A.P.
Ans: S10 = 75 and n = 10
10/2(2a+9×5)=75
=>a=-15
26) The following table shows the ages of the patients admitted in a hospital during a year:
Age (in years): | 5 – 15 | 15 – 25 | 25 – 35 | 35 – 45 | 45 – 55 | 55 – 65 |
Number of patients: | 60 | 110 | 210 | 230 | 150 | 50 |
Find the mode of the distribution.
Ans: Modal class is 35 – 45
∴ Mode = 35+230-210/460-210-150× 10
= 37
SECTION C
Question numbers 27 to 34 carry 3 marks each.
27) Seema has a 10 m × 10 m kitchen garden attached to her kitchen. She divides it into a 10 × 10 grid and wants to grow some vegetables and herbs used in the kitchen. She puts some soil and manure in that and sows a green chilly plant at A, a coriander plant at B and a tomato plant at C. Her friend Kusum visited the garden and praised the plants grown there. She pointed out that they seem to be in a straight line. See the below diagram carefully and answer the following questions:
(i) Write the coordinates of the points A, B and C taking the 10 × 10 grid as coordinate axes.
(ii) By distance formula or some other formula, check whether the points are collinear.
Ans: (i) Coordinates of A, B and C are
(2, 2), (5, 4), (7, 6)
(ii) Area of triangle = 1/ 2[2(4 -6) +5(6- 2)+ 7(2-4)]
= 1 ≠0
∴ Points are not collinear
28) In Figure-5, a circle is inscribed in a ΔABC touching BC, CA and AB at P, Q and R respectively. If AB = 10 cm, AQ = 7 cm, CQ = 5 cm, find the length of BC.
Ans: AR = AQ = 7 cm
BR = AB – AR = 10 – 7 = 3 cm
BC = BP + PC
= 3 + 5 = 8 cm
OR
In Figure-6, two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that ∠PTQ = 2 ∠OPQ.
Ans: Let ∠PTQ = θ
Q TP = TQ
∴ ∠TPQ = ∠TQP =1/2(180°- θ)
=90 ° -1/2 θ
∠TPO = 90°
∴ ∠OPQ = (90°-1/2 θ)
= 1/2 θ=1/2∠PTQ
=> ∠PTQ = 2 ∠OPQ
29) Prove that √2 is an irrational number.
Ans: Let us assume √2 is a rational number.
∴ √2=p/q,q≠0, HCF(p,q)=1, p&q are integers
Squaring both sides
2=p2/q2=>p2=2q2
=>p=2m
Using equations (i) and (ii)
4m2 = 2q2
=>q = 2m
Using equation (ii) and (iii) we get p and q both are multiples of 2 which contradicts the assumption.
Hence √2 is an irrational number
30) Prove that:
(cosec θ – cot θ) 2 = 1 -cos θ /1 +cos θ
Ans: LHS = (cosec θ – cot θ) 2
= (1/sin θ -cos θ /sin θ)2
=(1-cos θ)/sin2 θ
=(1-cos θ)2/(1-cos θ)(1+cos θ)
=1-cos θ/1+cos θ=RHS
31) 5 pencils and 7 pens together cost ₹ 250 whereas 7 pencils and 5 pens together cost ₹ 302. Find the cost of one pencil and that of a pen.
Ans: Let the cost of 1 pencil be ₹ x and cost of 1 pen be ₹ y.
5x + 7y = 250
7x + 5y = 302
Solving (i) and (ii)
x = 36 and y = 10
Hence, cost of 1 pencil = ₹36
cost of 1 pen = ₹ 10
OR
Solve the following pair of equations using cross-multiplication method:
x – 3y – 7 = 0
3x – 5y – 15 = 0
Ans: x/( -3) (- 15)- (- 5) (- 7) =y/( -7)3 -(- 15) =1/-5- 3( -3)
=>x/10=y/-6=1/4
=>x=5/2
=>y=-3/2
32) One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting
(i) a king of red colour.
(ii) the queen of diamonds.
(iii) an ace
Ans: Toal number of possible outcomes = 52
(i) P(Card is a king of Red colour) = 2/52 or1/26
(ii) P(card is a queen of diamonds) = 1 /52
(iii) P(Card is an ace) = 4 /52 or 1/13
OR
A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears
(i) a two-digit number.
(ii) a perfect square number.
(iii) a prime, number less than 15
Ans: Total number of possible outcomes = 90
(i) P(a two digit number) = 81 /90 or 9/10
(ii) P(a perfect square number) = 9/ 90 or 1/10
(iii) P(a prime number less than 15) = 6 /90 or 1 /15
33) In Figure-7, ABCD is a square of side 14 cm. From each corner of the square, a quadrant of a circle of radius 3.5 cm is cut and also a circle of radius 4 cm is cut as shown in the figure. Find the area of the remaining (shaded) portion of the square.
Ans: Area of the square = 14 × 14 = 196 cm2
Area of middle circle = 22/7×4×4=50.28 cm2
Hence, Area of Remaining part of square.
= 196 – (50.28 + 38.5)
= 107.22 cm2
34) Draw a circle of radius 3 cm. Take a point P outside the circle at a distance of 7 cm from the centre O of the circle and draw two tangents to the circle.
Ans: Drawing a circle and locating point P.
Constructing tangents from P.
Question numbers 35 to 40 marks each.
35) In a right-angled triangle, prove that the square of the hypotenuse is equal to the sum of the squares of the remaining two sides.
Ans: For correct given, to prove, figure and construction.
Correct proof
36) Divide polynomial – x3 + 3x2 – 3x – 3x + 5 by the polynomial x2 + x – 1 and verify the division algorithm.
Ans: On dividing –x3 + 3x2 – 3x + 5 by x2 + x – 1
We get quotient = –x + 4
and Remainder = –8x + 9
Verification
(x2 + x – 1) (–x + 4) + (9 – 8x)
=-x3-x2+x+4x2+4x-4+9-8x
= -x3+3x2-3x+5
OR
Find other zeroes of the polynomial
p(x) = 2x4 – 3x3 – 3x2 + 6x – 2
if two of its zeroes are √2 and – √2 .
Ans: Two factors of p(x) are(x-√2) and (x+√2)
g(x) = (x+√2) and (x-√2)
= x2 – 2
Now, 2x4-3x3-3x2+6x-2/x2-2=2x2-3x+1
Also, 2x2 – 3x + 1 = (2x – 1) (x – 1)
Other zeroes are 1/ 2,1
37) From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower, fixed at the top of a 20 m high building, are 45° and 60° respectively. Find the height of the tower. (Use √ 3 = 1.73)
Ans:
tan 45° =20/x
=>x=20m
Now, tan 60° =20+h/x
=>20√3-20=h
⇒ h = 20(√3- 1)
= 20 × 0.73
= 14.6 m
38) A bucket is in the form of a frustum of a cone of height 30 cm with the radii of its lower and upper circular ends as 10 cm and 20 cm respectively. Find the capacity of the bucket. (Use π = 3.14)
Ans: Capacity of bucket =1/3πh(r21+r22+r1r2)
=1/3×3.14×30(100+400+200)
=21980 cm3
OR
Water in a canal 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/hr. How much area will it irrigate in 30 minutes if 4 cm of standing water is needed?
Ans: Length of canal covered by water in 30 min. = 5000 m
Volume of water flown in 30 min. = 6 × 1.5 × 5000 m3
Hence, 6 × 1.5 × 5000 = (Area of field) × 4/ 100
∴ Area of field = 1125000 m2
39) Draw a ‘more than’ ogive for the following distribution:
Weigth (in kg): | 40 – 44 | 44 – 48 | 48 – 52 | 52 – 56 | 56 – 60 | 60 – 64 | 64 – 68 |
Number of Students: | 4 | 10 | 30 | 24 | 18 | 12 | 2 |
Ans: Points to be plotted for more than type ogive
(40, 100), (44, 96), (48, 86), (52, 56), (56, 32), (60, 14), (64, 2)
For drawing correct ogive
40) A train travels 360 km at a uniform speed. If the speed had been 5 km/hr more, it would have taken 1 hour less for the same journey. Find the original speed of the train.
Ans: Let original speed of the train be x km/h.
∴ 360 /x-360/x+5= 1
=>x2 + 5x – 1800 = 0
=>(x + 45) (x – 40) = 0
=>x = 40
∴ Speed of the train is 40 km/h
OR
Sum of the areas of two squares is 468 m2 . If the difference of their parameters is 24 m, find the sides of the two squares.
Ans: Let the side of squares be x m, y m (x > y)
x2 + y2 = 468
and 4(x – y) = 24
Simplify (i) and (ii) to get
x2 –6x – 216 = 0
=>(x – 18) (x + 12) = 0
=>x = 18
and y = 12
∴ Sides of square are 18 m and 12 m.
CBSE Class 10 Previous Question Paper 2020 Solution
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430/4/2 | |