30/2/1 2019 Class 10 Maths Question Paper Solution
MATHEMATICS
SECTION A
1) If HCF (336, 54) = 6, find LCM (336, 54).
Ans: LCM (336, 54) =336×54/6
=336 × 9 = 3024
2) Find the nature of roots of the quadratic equation 2x2– 4x + 3 = 0.
Ans:3-a/3a-1/a=3-a-3/3a=-1/3
3) Find the common difference of the Arithmetic Progression (A.P.)
1/a,3-a/3a,3-2a/3a,…(a≠0)
Ans: 2x2 – 4x + 3 = 0 à D = 16 – 24 = –8
∴ Equation has NO real roots
4) Evaluate :
sin260°+2 tan 45°– cos230°
Ans: sin2 60° + 2 tan 45° – cos2 30 =(√3/2)2+2(1)-( √3/2)2
=2
OR
If sin A=3/4, calculate sec A.
Ans: sin A = 3/4 =>cos A=√1-9/16=c7/4
sec A =4/√7
5) Write the coordinates of a point P on x-axis which is equidistant from the points A(-2,0) and B(6,0)
Ans: Point on x-axis is (2, 0)
6) In Figure 1, ABC is an isosceles triangle right angled at C with AC=4cm. Find the length of AB.
Ans: ∆ABC: Isosceles ∆ => AC = BC = 4 cm.
AB =√42+ 4 2=4√2 cm
OR
In Figure 2, DE∥ BC. Find the length of side AD, given that AE=1.8 cm, BD=7.2cm and CE=5.4cm
Ans: ΑD /BD = AE /CE=> 1.8 /5.4
∴ AD = 7.2× 1.8/5.4= 2.4 cm.
SECTION B
7) Write the smallest number which is divisible by both 306 and 657.
Ans: Smallest number divisible by 306 and 657 = LCM (306, 657)
LCM (306, 657) = 22338
8) Find a relation between x and y if the points A(x, y), B(– 4, 6) and C(– 2, 3) are collinear.
Ans: A, B, C are collinear ⇒ ar. (∆ABC) = 0
∴ 1/2 [x(6 – 3) – 4(3 – y) – 2(y – 6)] 2 = 0
=> 3x + 2y = 0
OR
Find the area of a triangle whose vertices are given as (1, – 1) (– 4, 6) and (– 3, – 5).
Ans: Area of triangle = 1/2 [1(6 +5) – 4(–5 +1) – 3(–1 – 6)]
=1/2[11+16+21]=48/2=24 sq.units.
9) The probability of selecting a blue marble at random from a jar that contains only blue, black and green marbles is 5 /1 . The probability of selecting a black marble at random from the same jar is 4/ 1 . If the jar contains 11 green marbles, find the total number of marbles in the jar.
Ans: P(blue marble) =1/5, P(black marble)=1/4
∴ P(green marble) =1-(1 /5+1/4) =11/20
Let total number of marbles be x
then11/20× x=11=> x=20
10) Find the value(s) of k so that the pair of equations x+2y=5 and 3x+ky+15=0 has a unique solution.
Ans: For unique solution1/3≠2/k
=> k≠6
11) The larger of two supplementary angles exceeds the smaller by 18°. Find the angles.
Ans: Let larger angle be x°
Let larger angle be x°
∴ (x) – (180 – x) = 18
2x = 180 + 18 = 198 => x = 99
∴ The two angles are 99°, 81°
OR
Sumit is 3 times as old as his son. Five years later, he shall be two and a half times as old as his son. How old is Sumit at present ?
Ans: Let Son’s present age be x years
Then Sumit’s present age = 3x years.
∴ 5 Years later, we have, 3x + 5 = 5/2 (x 5)
6x + 10 = 5x + 25 => x = 15
∴ Sumit’s present age = 45 years
12) Find the mode of the following frequency distribution :
Class Interval : | 25 – 30 | 30 – 35 | 35 – 40 | 40 – 45 | 45 – 50 | 50 – 55 |
Frequency : | 25 | 34 | 50 | 42 | 38 | 14 |
Ans: Maximum frequency = 50, class (modal) = 35 – 40.
Mode = L+(f1-f0/2f1-f0-f2) ×h
=35+50-34/100-34-42×5
=35+16/24×5=38.33
SECTION C
13) Prove that 2 + 5 √3 is an irrational number, given that √3 is an irrational number.
Ans: Let 2+ 5√ 3 = a, where ‘a’ is a rational number.
than √3=a-2/5
Which is a contradiction as LHS is irrational and RHS is rational
∴ 2+5√3 can not be rational
Hence 2+ 5√ 3 is irrational.
Alternate method:
Let 2+ 5√ 3 + be rational
∴ 2+ 5√ 3 = p/q , p, q are integers, q ≠ 0
=>√3=(p/q-2) ÷5=p-2q/5q
LHS is irrational and RHS is rational
which is a contradiction.
∴ 2 +5√ 3 is irrational.
OR
Using Euclid’s Algorithm, find the HCF of 2048 and 960.
Ans: 2048 = 960 × 2 + 128
960 = 128 × 7 + 64
128 = 64 × 2 + 0
∴ HCF (2048, 960) = 64
14) Two right triangles ABC and DBC are drawn on the same hypotenuse BC and on the same side of BC. If AC and BD intersect at P, prove that AP ´ PC = BP ´ DP.
Ans:
Correct Figure
∆APB ~ ∆DPC [AA similarity]
AP /DP = BP/ PC
=>AP × PC = BP × DP
OR
Diagonals of a trapezium PQRS intersect each other at the point O, PQ RS and PQ = 3RS. Find the ratio of the areas of triangles POQ and ROS.
Ans:
Correct Figure
In ∆POQ and ∆ROS
∠p=∠R
∠Q=∠S
alt. ∠s
∴ ∆POQ ~ ∆ROS [AA similarity]
∴ar (△ POQ) /ar (△ROS)=( PQ/RS)2
=(3/1)2=9/1
∴ ar(∆POQ) : ar(∆ROS) = 9 : 1
15) In Figure 3, PQ and RS are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting PQ at A and RS at B. Prove that Ð AOB = 90°.
Ans:
∆AOD ≅ AOC [SAS]
=> ∠1 = ∠2
Similarly ∠4 = ∠3
=> ∠1 + ∠4 = ∠2+∠3=1/2(180°)
=>∠2 + ∠3 = 90° or ∠AOB = 90°
Alternate method:
Correct Figure
∆OAD ≅ ∆AOC [SAS]
=> ∠1 = ∠2
Similarly ∠4 = ∠3
But ∠1 + ∠2 + ∠3 + ∠4 = 180°
[∵ PQ || RS]
=>∠2 + ∠3 = ∠1 + ∠4=1/2(180°)=90°
∴ In ∆AOB, ∠AOB = 180° – (∠2 + ∠3) = 90°
16) Find the ratio in which the line x – 3y = 0 divides the line segment joining the points (– 2, – 5) and (6, 3). Find the coordinates of the point of intersection.
Ans:
Let the line x – 3y = 0 intersect the segment
joining A(–2, –5) and B(6, 3) in the ratio k : 1
∴ Coordinates of P are(6k-2/k+1,3k-5/k+1)
∴ Ratio is 13 : 3
=>Coordinates of P are (9/2,3/2)
17) Evaluate :
(3 sin 43°/cos 47°)2-cos 37° cosec 53°/tan 5° tan 25° tan 45° tan 65° tan 85°
Ans: (3 sin 43°/cos 47°)2-cos 37° cosec 53°/tan 5° tan 25° tan 45° tan 65° tan 85°
= (3 sin 43°/cos (90°- 43°)-cos 37° cosec ( 90°-37°)/tan 5° tan 25° (i) tan (90°-25°) tan(90°-5°)
= (3 sin 43°/sin 43°)2-cos 37°.sec 37°/tan 5°. tan 25°(1) cot 25° cot 5°
=9-1/1=8
18) In Figure 4, a square OABC is inscribed in a quadrant OPBQ. If OA = 15 cm, find the area of the shaded region. (Use p = 3·14)
Ans:
Radius of quadrant= OB=√152+152=15√2 cm
Shaded area= Area of quadrant- Area of square
=1/4(3.14)[(15√2)2-(15)2]
=(15)2(1.57-1)=128.25 cm2
OR
In Figure 5, ABCD is a square with side 2√ 2 cm and inscribed in a circle. Find the area of the shaded region. (Use p = 3·14)
BD√(2√2)2+(2√2)2=√16=4 cm
∴ Radius of circle = 2 cm
∴ Shaded area = Area of circle – Area of square
=3.14× 22 -(2√ 2)2
=12.56-8=4.56 cm2
19) A solid is in the form of a cylinder with hemispherical ends. The total height of the solid is 20 cm and the diameter of the cylinder is 7 cm. Find
the total volume of the solid. (Use p = 22/7)
Ans:
Height of cylinder = 20 – 7 = 13 cm.
Total volume = π(7/2)2.13+4/3 π(7/2)3cm3
=22/7 ×49/4(13+4/3.7/2)cm3
=77 ×53/6=680.17 cm3
20) The marks obtained by 100 students in an examination are given below :
Marks : | 30 – 35 | 35 – 40 | 40 – 45 | 45 – 50 | 50 – 55 | 55 – 60 | 60 – 65 |
Number of Students : | 14 | 16 | 28 | 23 | 18 | 8 | 3 |
Ans:
xi : | 32.5 | 37.5 | 42.5 | 47.5 | 52.5 | 57.5 | 62.5 | ||||
f i : | 14 | 16 | 28 | 23 | 18 | 8 | .3 | Σf i = 110 | |||
ui : | –3 | –2 | –1 | 0 | 1 | 2 | 3 | ||||
fi ui : | –42 | –32 | –28 | 0 | 18 | 16 | 9, | Σf i ui = – 59 |
Mean =47.5- 59× 5/110=47.5-2.68=44.82
Note: If N is taken as 100, Ans. 44.55
21) For what value of k, is the polynomial
f(x) = 3x4 – 9x3 + x2 + 15x + k,
completely divisible by 3x2 – 5 ?
Ans:
OR
Find the zeroes of the quadratic polynomial 7y2-11/3(y)-2/3 and verify the relationship between the zeroes and the coefficients.
p(y)=7y2-11/3y-2/3=1/3(21y2-11y-2)
=1/3[(7y+1)(3y-2)]
∴ Zeroes are 2/3, –1/7
Sum of zeroes =2/3-1/7=11/21
-b/a=11/21∴ sum of zeroes =-b/a
Product of zeroes =(2/3)(-1/7)=-2/21
c/a=-2/3(1/7)=-2/21∴ Product =c/a
22) Write all the values of p for which the quadratic equation x2 + px + 16 = 0 has equal roots. Find the roots of the equation so obtained.
Ans:x2+px+16= 0 have equal roots if D =p2-4(16)(1)=0
p2=64=>p= ±8
∴ x2 ±8x+16=0=>(x ±4)2=0
x ±4=0
∴ Roots are x = –4 and x = 4
SECTION D
23) If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then prove that the other two sides are divided in the same ratio.
Ans: For correct, given, to prove, construction and figure
For correct proof.
24) Amit, standing on a horizontal plane, finds a bird flying at a distance of 200 m from him at an elevation of 30°. Deepak standing on the roof of a 50 m high building, finds the angle of elevation of the same bird to be 45°. Amit and Deepak are on opposite sides of the bird. Find the distance of the bird from Deepak.
Ans:
Correct Figure
In ∆APQ
PQ/ AP =in 30°=1/2
PQ=(200)(1/2)=100m
PR=100-50=50m
PR=100-50=50m
In ∆PRD, PR/ PD= sin 45°=1/√2
PD = (PR)( √2)=50√2m
25) A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately 8 gm mass. (Use p = 3·14)
Ans:
Total volume = 3.14 (12)2 (220) + 3.14(8)2 (60) cm3
= 99475.2 + 12057.6 = 111532.8 cm3
Mass= 111532.8×8/1000kg
=892.262kg
26) Construct an equilateral D ABC with each side 5 cm. Then construct another triangle whose sides are 2/3 times the corresponding sides of
D ABC.
Ans: Constructing an equilateral triangle of side 5 cm
Constructing another similar ∆ with scale factor 2/ 3
OR
Draw two concentric circles of radii 2 cm and 5 cm. Take a point P on the outer circle and construct a pair of tangents PA and PB to the smaller circle. Measure PA.
Ans: Constructing two concentric circle of radii 2 cm and 5 cm
Drawing two tangents PA and PB
PA = 4.5 cm
27) Change the following data into ‘less than type’ distribution and draw its ogive :
Class Interval : | 30 – 40 | 40 – 50 | 50 – 60 | 60 – 70 | 70 – 80 | 80 – 90 | 90 – 100 |
Frequency : | 7 | 5 | 8 | 10 | 6 | 6 | 8 |
Ans:
Less than 40 | Less than 50 | Less than 60 | Less than 70 | Less than 80 | Less than 90 | Less than 100 |
7 | 12 | 20 | 30 | 36 | 42 | 52 |
Plotting of points (40, 7), (50, 12), (60, 20), (70, 30), (80, 36), (90, 42) and (100, 50)
Joining the points to get the curve
28) Prove that :
tanq/1-cotq +cot/1-tanq =1+secq cosecq
Ans: LHS=tan θ/1-(1/tan θ)+1/tan θ/1-tan θ=tan2 /tan θ(tanθ-1)
=tan3 θ-1/tan θ(tan θ-1)=(tan θ-1)(tan2 θ+tan θ+1)/tan θ(tan θ-1)
=tan θ+1+cot θ=1+sin θ/cos θ+cos θ/sin θ=1+sin2 θ+cos2 θ/sin θcos θ
=1+1/sin θcos θ=1+cosec θsec θ=RHS
OR
Prove that :
sinq/cotq +cosecq =2+sinq /cotq –cosec q
Ans: consider
sin θ/cosec θ+cot θ-sin θ/cot θ-cosec θ=sin θ/cosec θ+cot θ+sin θ/cosec θ-cot θ
=sin θ[cosec θ-cot θ+cosec θ+cot θ]/cosec2 θ-cot2 θ=sin θ(2 cosec θ)/1=2
Hence(sin θ/cosec θ+cot θ=2+sin θ/cot θ-cosec θ
29) Which term of the Arithmetic Progression –7, –12, –17, –22, … will be –82 ? Is –100 any term of the A.P. ? Give reason for your answer.
Ans: Let -82=aa ∴ –82 = –7 + (n – 1) (–5)
=> 15 = n – 1 or n = 16
Again –100 = am = –7 + (m – 1) (–5)
=> (m – 1)(–5) = –93
m-1=93/5 or m=93/5+1 ∉N
∴ –100 is not a term of the AP
OR
How many terms of the Arithmetic Progression 45, 39, 33, … must be taken so that their sum is 180 ? Explain the double answer.
Ans: Sn=180=n/2.[90+(n-1)(-6)]
360=90n-6n2+6n=>6n2-96n+360=0
=>6[(n-6)(n-10)]=0=>n=6.n=6 or n=10
Sum of a7 , a8 , a9 , a10 = 0 ∴ n = 6 or n = 10
30) In a class test, the sum of Arun’s marks in Hindi and English is 30. Had he got 2 marks more in Hindi and 3 marks less in English, the product of the marks would have been 210. Find his marks in the two subjects.
Ans: Let marks in Hindi be x
Then marks in Eng = 30 – x
∴ (x + 2) (30 – x – 3) = 210
=>x2-25x+156=0 or(x-13)(x-12)=0
=> x = 13 or x = 12
∴ 30 – 13 = 17 or 30 – 12 = 18
∴ Marks in Hindi & English are
(13, 17) or (12, 18)
CBSE Class 10 Previous Question Paper 2019 Solution
Others Set |
Solution |
30/2/2 | |