430/5/1 2020 Class 10 Maths Basic Question Paper Solution
MATHEMATICS
(BASIC)
SECTION A
Question numbers 1 to 20 carry 1 mark each.
1) If a pair of linear equations is consistent, then the lines represented by them are
(A) parallel (B) intersecting or coincident (C) always coincident (D) always intersecting
Ans: (B) Intersecting or coincident.
2) The distance between the points (3, – 2) and (– 3, 2) is
(A) √52 units (B) 4 √10 units (C) 2√ 10 units (D) 40 units
Ans: (A) √52 units
3) 8 cot2 A – 8 cosec2 A is equal to
(A) 8 (B) 1 /8 (C) – 8 (D)- 1/ 8
Ans: (C) –8
4) The total surface area of a frustum-shaped glass tumbler is (r1> r2)
Ans: (B) πl (r1 + r2) + πr 2 2
5) 120 can be expressed as a product of its prime factors as
(A) 5 × 8 × 3 (B) 15 × 23 (C) 10 × 22 × 3 (D) 5 × 23 × 3
Ans: (D) 5 × 23 × 3
6) The discriminant of the quadratic equation 4x2 – 6x + 3 = 0 is
(A) 12 (B) 84 (C) 2√ 3 (D) – 12
Ans: (D) –12
7) If (3, –6) is the mid-point of the line segment joining (0, 0) and (x, y), then the point (x, y) is
Ans: (A) (–3, 6) (B) (6, –6) (C) (6, –12) (D) (3/ 2,-3)
8) In the given circle in Figure-1, number of tangents parallel to tangent PQ is
(A) 0 (B) many (C) 2 (D) 1
Ans: (D) 1
9) For the following frequency distribution:
Class: | 0 – 5 | 5 – 10 | 10 – 15 | 15 – 20 | 20 – 25 |
Frequency | 8 | 10 | 19 | 25 | 8 |
The upper limit of median class is
(A) 15 (B) 10 (C) 20 (D) 25
Ans: (A) 15
10) The probability of an impossible event is
(A) 1 (B) 1/ 2 (C) not defined (D) 0
Ans: (D) 0
Fill in the blanks in question numbers 11 to 15.
11) A line intersecting a circle in two points is called a _______.
Ans: Secant
12) If 2 is a zero of the polynomial ax2 – 2x, then the value of ‘a’ is _______.
Ans: 1
13) All squares are _______ (congruent/similar).
Ans: Similar
14) If the radii of two spheres are in the ratio 2 : 3, then the ratio of their respective volumes is _______.
Ans: 8/27 or 8 : 27
15) If ar (Δ PQR) is zero, then the points P, Q and R are _______.
Ans: Collinear
Answer the following question numbers 16 to 20:
16) In Figure-2, the angle of elevation of the top of a tower AC from a point B on the ground is 60 °. If the height of the tower is 20 m, find the distance of the point from the foot of the tower.
Ans: AC /AB = tan 60 °
20 /AB= √3
AB=20 √3/3 or 20/√3
17) Evaluate:
tan 40° × tan 50°
Ans: tan 40° × cot 40°
=1
OR
If cos A = sin 42°, then find the value of A.
Ans: cos A = sin (90° – 48°)
= cos 48°
=>A=48°
18) A coin is tossed twice. Find the probability of getting head both the times.
Ans: Total outcomes = 4
P(getting head both the times) = 1/ 4
19) Find the height of a cone of radius 5 cm and slant height 13 cm.
Ans: h =√(13)2-(5)2
h=12cm
20) Find the value of x so that – 6, x, 8 are in A.P.
Ans: x + 6 = 8 – x
x=1
OR
Find the 11th term of the A.P. – 27, – 22, –17, –12, … .
Ans: a = –27, d = 5
a11 = –27 + 50 = 23
SECTION B
Question numbers 21 to 26 carry 2 marks each.
21) Find the roots of the quadratic equation
3×2-4√3x+4=0
Ans: 3x 2 -2√3x-2√3x+4=0
(√3x-2)( √3x-2)=0
√3x-2=0=>x=2√3
22) Check whether 6n can end with the digit ‘0’ (zero) for any natural number n.
Ans: 6n = (2 × 3)n = 2n × 3n
It is not in form of 2n × 5m
6n can’t end with digit ‘0’
OR
Find the LCM of 150 and 200.
Ans: 150 = 2 × 3 × 52
200 = 23 × 52
LCM = 23 × 52 × 3
=600
23) If tan (A + B) = √3 and tan (A – B) = 1 /√3, 0 < A + B ≤ 90°, A > B, then find the value of A and B.
Ans: A + B = 60°
A-b=30°
From (i) and (ii)
A=45°
B=15°
24) In Figure-3, ΔABC and ΔXYZ are shown. If AB = 3 cm BC = 6 cm, AC = 2 √3 cm, ∠A = 80°, ∠B = 60º, XY = 4 √3 cm YZ = 12 cm and XZ = 6 cm, then find the value of ∠Y
Ans: AB/XZ=BC/YZ=AC/XY=1/2
∴ ΔABC ~ ΔXZY
∠C = ∠Y = 40°
25) 4 defective bulbs are accidentally mixed with 98 good ones. It is not possible to just look at the bulb and tell whether it is defective or not. One bulb is taken out at random from this lot. Determine the probability that the bulb taken out is a good one.
Ans: Total outcomes = 14 + 98 = 112
P(good bulb) = 98/112 or 7/8
26) Find the mean for the following distribution:
Ans:
Classes: | 5 – 15 | 15 – 35 | 25 – 35 | 35 – 45 |
Frequency: | 2 | 4 | 3 | 1 |
Ans:
Classes |
Freq. | Mid value = x |
f × x |
5-15 | 2 | 10 | 20 |
15-25 | 4 | 20 | 80 |
25-35 | 3 | 30 | 90 |
35-45 | 1 | 40 | 40 |
Σf = 10 | Σfx = 230 |
OR
The following distribution shows the transport expenditure of 100 employees:
Expenditure (in (`): |
200 – 400 | 400 – 600 | 600 – 800 | 800 – 1000 |
1000 – 1200 |
Number of employees: | 21 | 25 | 19 | 23 | 12 |
Find the mode of the distribution.
Ans: Modal class = 400 – 600
Mode =1+[f1-f0/2f1-f0-f2] ×h
=400+[25-21/50-21-19] ×200
=400 + 80 = 480
SECTION C
Question numbers 27 to 34 carry 3 marks each.
27) A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC.
Ans:
Proof: AP =AS
BP= BQ
CR=CQ
DR =DS
∴ By adding,
AP + BP + CR + DR = AS + BQ + CQ + DS
AB + CD = AD + BC
28) The difference between two numbers is 26 and the larger number exceeds thrice of the smaller number by 4. Find the numbers.
Ans: Let larger No. = x
Let smaller No = y
x – y = 26
x – 3y = 4
By solving (i) & (ii), we get
x = 37
y = 11
OR
Solve for x and y:
2/x+3/y=13 and 5/x-4/y=-2
Ans: Let 1/x=p& 1/y=q
2p+3q=13
5p-3q=-2
By solving (i) & (ii), we get
p = 2, q = 3
1/x=2, 1/y=3
x=1/2 y=1/3
29) Prove that √ 3 is an irrational number
Ans: Let √3 is rational
√3=a/b(where a & b are +ve integers & co-prime, b ≠ 0)
a2 = 3b2
3 divides a2
3 divides a also
Let a = 3c & put in (i)
(3c)2 = 3(b)2
3c2 = b2
=> 3 divides b2
3 divides b also
3 divides a and b both
This contradicts our assumption
Therefore, √3 is irrational no
30) Krishna has an apple orchard which has a 10 m × 10 m sized kitchen garden attached to it. She divides it into a 10 × 10 grid and puts soil and manure into it. She grows a lemon plant at A, a coriander plant at B, an onion plant at C and a tomato plant at D. Her husband Ram praised her kitchen garden and points out that on joining A, B, C and D they may form a parallelogram. Look at the below figure carefully and answer the following questions:
(i) Write the coordinates of the points A, B, C and D, using the 10 × 10 grid as coordinate axes.
Ans: (i) Coordinates are A(2, 2), B(5, 4), C(7, 7), D(4, 5)
(ii)AB= √(5-2)2+(4-2)2= √13
BC= √13
CD= √13
DA= √13
AB =BC =CD= DA
ABCD is a parallelogram
31) If the sum of the first 14 terms of an A.P. is 1050 and its first term is 10 then find the 21st term of the A.P.
Ans: n/2[2a+(n-1)d]=1050
7[20+13d]=1050
∴ d = 10
a21 = a + 20d = 10 + 20 × 10 = 210
32) Construct a triangle with its sides 4 cm, 5 cm and 6 cm. Then construct a triangle similar to it whose sides are 2 /3 of the corresponding sides of the first triangle.
Ans: For correct construction of Δ
For construction of similar Δ
OR
Draw a circle of radius 2.5 cm. Take a point P at a distance of 8 cm from its centre. Construct a pair of tangents from the point P to the circle.
Ans: For draw the correct circle & exterior pt.
For construction of the pair of tangents
33) Prove that:
(cosec A – sin A) (sec A – cos A) =1/tan A+ cot A
Ans: LHS=(1/sin A- sin A)(1/ cos A- cos A)
=1-sin2 A/sin A×1- cos2 A/ cos A
=cos2 A/sin A ×sin2 A/cos A
=cos A . sin A
RHS=1/sin A/cos A+cos A/sin A
=1/sin2 A+ cos2 A/sin A. cos A
= sin A ⋅ cos A
∴ LHS = RHS
34) In Figure-4, AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, then find the area of the shaded region.
Ans: Area of smaller circle = πr2=22/7×7/2×7/2=77/2= 38.5 cm2
Area of Big semi-circle =1/2× 22/7× 7×7 =77cm2
Area of ΔABC =1/2× 14× 7=49cm2
Area of shaded portion = ar. of smaller circle + ar. of big semicircle – ar. of ΔABC
= 38.5 + 77 – 49 = 66.5 cm2
OR
In Figure-5, ABCD is a square with side 7 cm. A circle is drawn circumscribing the square. Find the area of the shaded region.
Ans: Area of sqaure ABCD = a2 = 72 = 49 cm2
Diagonal of square =√2a=7√2cm
∴ Radius of circle =7√2/2cm
Area of circle =22/7× (7√2/2)=77cm2
Area of shaded, portion = 77 – 49 = 28 cm2
SECTION D
Question numbers 35 to 40 carry 4 marks each.
35) Find other zeroes of the polynomial
p(x) =3x4 – 4x3 – 10x2 + 8x + 8,
if two of its zeroes are √2 and – √2 .
Ans: (x-√2)&(x+ √2) are two factors
i.e. x2 – 2 is a factor
3x2 – 4x – 4 = (3x + 2) (x – 2)
–2/3, 2 are other two zeroes.
OR
Divide the polynomial g(x) = x3 – 3x2 + x + 2 by the polynomial x2 – 2x + 1 and verify the division algorithm.
Ans:
Verify,
P(x) = q(x) × g(x) + r(x)
= (x – 1) (x2 – 2x + 1) + (–2x + 3)
= x3 – 3x2 + 3x – 1 – 2x + 3
= x3 – 3x2 + x + 2
36) From the top of a 75 m high lighthouse from the sea level, the angles of depression of two ships are 30° and 45°. If the ships are on the opposite sides of the lighthouse, then find the distance between the two ships.
Ans:
Correct figure
In ΔACD
75/x=1/√3
∴ x = 75√ 3
In ΔBCD
75/ 1= y
∴ y = 75
∴ Distance b/w two ships i.e. AB = x + y
= 75 √3+ 75
= 75(√ 3+ 1)
37) 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.
Ans: For correct given, To prove, Construction, Figure 4× 1 2 =2 For correct proof
OR
If Figure-6, in an equilateral triangle ABC, AD ⊥ BC, BE ⊥ AC and CF ⊥ AB. Prove that 4(AD2 + BE2 + CF2 ) = 9 AB2 .
Ans: Proof
In ΔABD, AD2=AB2-BD2
In ΔBCE BE2=BC2-CE2
In Δ ACF CF2=AC2– AF2
AD2 + BE2 + CF2 = AB2 + BC2 + AC2 – BD2 – CE2 – AF2
= 3AB2-(BC/2)2-(AC/2)2-(AB/2)2
= 3AB2-3/4 AB2
=9/4 AB2
4(AD2 + BE2 + CF2 ) = 9AB2
38) A container open at the top and made up of a metal sheet, is in the form of a frustum of a cone of height 14 cm with radii of its lower and upper circular ends as 8 cm and 20 cm, respectively. Find the capacity of the container.
Ans: Vol. of container =1/3 πh(r21+r22+r1r2)
=1/3× 22/7× 14[(8)2 +(20)2 + 8 ×20]
= 1/3× 22/7× 14[64+400+160]
= 9152 cm3
39) Two water taps together can fill a tank in9(3/8) hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Ans: Let smaller diameter tap takes x hours to fill the tank
Then, time taken by larger diameter tap to fill the tank = (x – 10) hr
ATQ
1/x+1/x-10=8/75
8x2 – 230x + 750 = 0
(8x – 30) (x – 25) = 0
x=15/4 and x=25
Rejected x = 15/4
Hence, time taken by smaller diameter tap = 25 hrs
Time taken by larger diameter tap = 25 – 10 = 15 hrs
OR
A rectangular park is to be designed whose breadth is 3 m less than its length. Its area is to be 4 square metres more than the area of a park that has already been made in the shape of an isosceles triangle with its base as the breadth of the Rectangular park and of altitude 12 m. Find the length and breadth of the park.
Ans:
Let length of rectangle = x
∴ Breadth = x – 3
- of reactangle = x(x – 3)
= x2 – 3x
Area of Isosceles ΔADE
= 1/2(x-3) × 12
= 6x – 18
ATQ
x2 – 3x = 6x – 18 + 4
x 2 – 9x + 14 = 0
(x – 7) (x – 2) = 0
x = 7, x = 2 Rejected
∴ Length of rectangle = 7 cm
Breadth of rectangle = 4 cm
40) Draw a ‘less than’ ogive for the following frequency distribution:
Classes: | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 | 60 – 70 | 70 – 80 |
Frequency: | 7 | 14 | 13 | 12 | 20 | 11 | 15 | 8 |
Ans: getting the pts (10, 7), (20, 21)
(30, 34), (40, 46), (50, 66)
(60, 77), (70, 92), (80, 100)
Plotting and Joining the pts to get the correct ogive
CBSE Class 10 Previous Question Paper 2020 Solution
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