30/4/2 2019 Class 10 Maths Question Paper Solution
MATHEMATICS
SECTION A
1) For what values of k does the quadratic equation 4x2 – 12x – k = 0 have no real roots ?
Ans: Disc. = 144 – 4 × 4 × (–k) < 0
16k < –144
k < –9
2) Find the distance between the points (a, b) and (– a, – b).
Ans: Required distance = √(-a-a)2+(-b-b)2
= √4(a2+b2) or 2 √a2+b2
3) Find a rational number between √2 and √7 .
Ans: Here 1.41 < x < 2.6
Any rational number lying between 1.4 … & 2.6 …
(variable answer)
OR
Write the number of zeroes in the end of a number whose prime factorization is 2 2 ´ 5 3 ´ 3 2 ´ 17.
Ans: 2 2 × 52 × 5 × 32 × 17 = (10)2 × 5 × 32 × 17
∴ No. of zeroes in the end of the number = Two
4) Let D ABC ~ D DEF and their areas be respectively, 64 cm2 and 121 cm2 . If EF = 15·4 cm, find BC.
Ans: Here BC/EF=8/11
∴ BC = 8/11 × 15.4= 11.2 cm
5) Evaluate :
tan 65°/cot 25°
Ans: tan 65°/cot 25°= tan(90°-25°)/cot 25°
= cot 25°/ cot 25°=1
OR
Express (sin 67° + cos 75°) in terms of trigonometric ratios of the angle between 0° and 45°.
Ans: sin 67° + cos 75° = sin (90° – 23°) + cos (90° – 15°)
=cos 23° + sin 15°
6) Find the number of terms in the A.P. : 18, 15 (2 /1) , 13, …, – 47.
Ans: Here – 47= 18+(n-1)(-5/2)
=> n= 27
SECTION B
7) A bag contains 15 balls, out of which some are white and the others are black. If the probability of drawing a black ball at random from the bag is 2/3 , then find how many white balls are there in the bag.
Ans: Let the number of white balls = x
∴ The number of black balls = 15 – x
P(Black) = 2 /3
=> 15-x/15=2/3
=> 45- 3x= 30
=> x=5
Hence number of white balls = 5.
8) A card is drawn at random from a pack of 52 playing cards. Find the probability of drawing a card which is neither a spade nor a king.
Ans: No. of spade cards + 3 other kings = 13 + 3 = 16
∴ Cards which are neither spade nor kings = 52 – 16 = 36
Hence P (neither spade nor king) = 36/52 or 9/13
9) Find the solution of the pair of equations :
3/x+8/y=-1;1/x-2/y=2, x, y ≠0
Ans: 3/x+8/y=-1…(i)
1/x-2/y=2…(ii)
Multiply (ii) by 3 and subtract from (i), we get
14/y=-7=>y=-2
Substitute this value of y = –2 in (i), we get x = 1
Hence, x = 1, y = –2
OR
Find the value(s) of k for which the pair of equations has a unique solution. kx+ 2y =3
3x + 6y= 10
has a unique solution.
Ans: For unique solution a1/a2≠b1/b2=>k/3≠2/6
=> k≠ 1
The pair of equations have unique solution for all real values of k except 1.
10) How many multiples of 4 lie between 10 and 205 ?
Ans: 12, 16, 20, …, 204
Let the number of multiples be n.
∴ tn=12+(n-1) × 4=204
=>n=49
OR
Determine the A.P. whose third term is 16 and 7th term exceeds the 5 th term by 12.
Ans: Here t3 = 16 and t7 = t5 + 12
=>a + 2d = 16 (i) and a + 6d = a + 4d + 12 (ii)
From (ii), d = 6
From (i), a = 4
∴ A.P. is 4, 10, 16, …
11) Use Euclid’s division algorithm to find the HCF of 255 and 867.
Ans: 867= 3× 255+ 102
255= 2× 102+ 51
102= 2× 51+ 0
∴ HCF = 51
12) The point R divides the line segment AB, where A(– 4, 0) and B(0, 6) such that AR = 3/4 AB . Find the coordinates of R.
Ans:
AR/ AB =3/4=> AR/RB=3/1
R=(3 × 0+ 1(-4)/4,3× 6+1× 0/4), i.e.(-1,9/2)
SECTION C
13) Prove that :
(sin q + 1 + cos q) (sin q – 1 + cos q) . sec q cosec q = 2
Ans: LHS = (sin θ + cos θ + 1)(sin θ + cos θ – 1) sec θ cosec θ
= [(sin θ + cos θ) 2 – 1] sec θ cosec θ
= 2 sin θ cos θ sec θ cosec θ
= 2 = RHS
OR
Prove that :
√ sec θ -1/ sec θ +1 + √sec θ +1/sec θ-1= 2 cosec θ
Ans: LHS =√sec θ-1/sec θ+ 1+√sec θ +1/sec θ-1 =(sec θ-1) +(sec θ+1)/ √sec2 θ-1
=2 sec θ/tan θ
= 2/sin θ=2 cosec θ= RHS
14) In what ratio does the point P(– 4, y) divide the line segment joining the points A(– 6, 10) and B(3, – 8) ? Hence find the value of y.
Ans:
Let point P divides the line segment AB in the ratio k : 1
∴ 3k-6/k+1=-4
=> 3k – 6=-4k-4
=> 7k= 2 i.e..k==2/7 ∴ Ratio is 2 : 7
Again 2×(-8)+7×10/2+7=y=>y=6
Hence y = 6
OR
Find the value of p for which the points (– 5, 1), (1, p) and (4, – 2) are collinear.
Ans: The points are collinear if the area of triangle formed is zero
i.e., –5(p + 2) + 1(–2 – 1) + 4(1 – p) = 0
–5p – 10 – 3 + 4 – 4p = 0
–9p = 9
p = –1
15) ABC is a right triangle in which ∠ B = 90°. If AB = 8 cm and BC = 6 cm, find the diameter of the circle inscribed in the triangle.
Ans:
AC =√AB2+ BC2=√64+36=10cm
Area of △ABC = 1/ 2 ×8× 6= 24 cm2
Let r be the radius of inscribed circle.
ar(△ABC) = ar(AOB) + ar(△BOC) + ar(△AOC)
=1/2× 8r+1/2× 6r+1/2× 10r
=1/2r(8+6+10)=12r
12r=24=>r=2cm
∴ Diameter = 4 cm
16) In Figure 1, BL and CM are medians of a △ ABC right-angled at A. Prove that 4 (BL2 + CM2 ) = 5 BC2 .
Ans:
In right angled triangle CAM,
CM2= CA2+ AM2…(i)
Similarly, BC2= AC2 +AB2…(ii)
and
BL2 =AL2 +AB2…(iii)
Now 4(BL2 + CM2 ) = 4(AL2 + AB2 + AC2 + AM2 )
But AL = LC = 1 /2 AC and AM=MB=1/2 AB
∴ 4(BL2 + CM2 ) = 4 (AC2/4+AB2+AC2+AB2/4)
= 4(5/4 AB2+5/4 AC2)
= 5(AB2 + AC2 ) = 5BC2
OR
Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
Ans:
Let ABCD be rhombus and its diagonals intersect at O.
In △AOB, AB2 = AO2 + OB2
= (AC/2)2 +(BD/2)2
=1/4(AC2 +BD2)
=> 4AB2 =AC2 + BD2
=> AB2 +BC2+CD2+AD2=AC2+BD2 (ABCD being rhombus)
17) In Figure 2, two concentric circles with centre O, have radii 21 cm and 42 cm. If ∠ AOB = 60°, find the area of the shaded region.
Ans: Area of shaded region
= [π(42)2– π(21)2]300°/360°
=22/7×63×21×5/6
=3465 cm2
18) Calculate the mode of the following distribution :
Class : | 10 – 15 | 15 – 20 | 20 – 25 | 25 – 30 | 30 – 35 |
Frequency : | 4 | 7 | 20 | 8 | 1 |
Ans: Here the modal class is 20 – 25
Mode = 20 + 20-7/40-7-8×5
=20+13/25 ×5= 22.6
Hence mode = 22.6
19) A cone of height 24 cm and radius of base 6 cm is made up of modelling clay. A child reshapes it in the form of a sphere. Find the radius of the sphere and hence find the surface area of this sphere.
Ans: Volume of cone =1/3 πr2h=1/3 π(6)2×24 cm2
Let the radius of the sphere be R cm
∴ 4/3πR3=1/3π×36×24
=>R3 = 6 × 6 × 6
=>R = 6 cm
Surface area = 4πR 2 = 144π cm2
OR
A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in his field which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/hr, in how much time will the tank be filled ?
Ans: Water required to fill the tank = π(5)2 × 2 = 50π m3
Water flown in 1 hour = π(1/10)2× 3000 m3
= 30π m3
Time taken to fill 30π m3 = 60 minutes
Time taken to fill 50π m3=60/30×50=100 minutes
20) Prove that 2 + 3 √3 is an irrational number when it is given that 3 is an irrational number.
Ans: Let 2+ 3 √3 = a where a is a rational number
Then √3=a-2/3
Which is contradiction as LHS in irrational and
RHS is rational
∴ 2+ 3√ 3 + is irrational
21) Sum of the areas of two squares is 157 m2 . If the sum of their perimeters is 68 m, find the sides of the two squares.
Ans: Let x and y be length of the sides of two squares.
∴ x 2 + y2 = 157 and 4(x + y) = 68 =>x + y = 17
∴ x 2 + (17 – x)2 = 157
x 2 + 289 + x2 – 34x – 157 = 0
or
x 2 – 17x + 66 = 0
(x – 6) (x – 11) = 0
∴ x = 6 or 11
∴ y = 11 or 6
Hence length of sides of squares are 6 m and 11 m.
22) Find the quadratic polynomial, sum and product of whose zeroes are –1 and – 20 respectively. Also find the zeroes of the polynomial so obtained.
Ans: If α, β are zeroes of the polynomial, then
α + β = –1, αβ = –20
∴ Polynomial is (x2+x-20)
(x + 5) (x – 4)
∴ Zeroes of the polynomial are 4 and –5
SECTION D
23) A plane left 30 minutes later than the scheduled time and in order to reach its destination 1500 km away on time, it has to increase its speed by 250 km/hr from its usual speed. Find the usual speed of the plane.
Let x km/hr be the usual speed of the plane
∴ 1500/x-1500/x+250=1/2
=>x2+250x-750000=0
=>x = –1000 or 750
∴ Speed of the plane = 750 km/h
OR
Find the dimensions of a rectangular park whose perimeter is 60 m and area 200 m2 .
Ans: Let l be the length and b be the breadth of the park
∴ 2(l + b) = 60 ⇒ l + b = 30 and l × b = 200
l(30 – l) = 200
=>l 2 – 30l + 200 = 0
=>(l – 20) (l – 10) = 10
=>l = 20 or 10
Hence length = 20 m, breadth = 10 m
24) Find the value of x, when in the A.P. given below
2 + 6 + 10 + … + x = 1800.
Ans: Let x be the nth term
∴ t n = x = 2 + (n – 1)4 i.e. x = 4n – 2
Also Sn = 1800=n/2 {4 (n – 1)4}
i.e. 4n/2= 1800
n 2 = 900 => n = 30
∴ x = 30 × 4 – 2 = 118
25) If sec q + tan q = m, show that m2-1/m2+1= sin θ
Ans: sec θ + tan θ = m
We know that sec2 θ – tan2 θ = 1
sec θ – tan θ =1/m
From (i) and (ii), 2 sec θ = m +1/m and 2 tan θ =m-1/m
Now sin θ =2 tan θ/2 sec θ=m-1/m/m+1/m=m2-1/m2+1
26) In D ABC (Figure 3), AD ⊥ BC. Prove that
AC2 = AB2 + BC2 – 2BC ´ BD
Ans: In ∆ABD, AB2 = AD2 + BD2 => AD2 = AB2 – BD2
In ∆ADC, AC2 = AD2 + CD2
= AB2 – BD2 + (BC – BD)2
= AB2 – BD2 + BC2 + BD2 – 2BC × BD
= AB2 + BC2 – 2BC × BD
27) A moving boat is observed from the top of a 150 m high cliff moving away from the cliff. The angle of depression of the boat changes from 60° to 45° in 2 minutes. Find the speed of the boat in m/min.
Ans:
Correct Figure
150/BC= tan 60°=√3
=> BC=150/√3=50√3M
Also AB/BD=tan 45°=1=>AB=BD=150M
Now CD = BD – BC = (150 -50√ 3) m
Distance travelled in 2 minutes = (150- 50 √3) m
∴ Distance travelled in 1 minute = (75- 25√ 3) m
or 75 – 25(1.732) = 75 – 43.3 = 31.7 m/minute
Hence speed of boat is (75 -25√ 3) m − /minutes or 31.7 m/minutes
OR
There are two poles, one each on either bank of a river just opposite to each other. One pole is 60 m high. From the top of this pole, the angle of depression of the top and foot of the other pole are 30° and 60° respectively. Find the width of the river and height of the other pole.
Ans:
Correct Figure
In ∆ABC,AB/AC=tan 60°
60/AC=√3
AC=20√3M
In ∆BED, 60-/DE=tan 30°=1/√3
i.e. 60-y/20√3=1/√3=> 60-y= 20 i.e y=40m
Hence width of river = 20√3 m and
height of other pole=40m
28) Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are 3/5 of the corresponding sides of the first triangle.
Ans: Correct Construction of triangle
Correct Construction of similar triangle
29) Calculate the mean of the following frequency distribution :
Class : | 10 – 30 | 30 – 50 | 50 – 70 | 70 – 90 | 90 – 110 | 110 – 130 |
Frequency : | 5 | 8 | 12 | 20 | 3 | 2 |
Classes |
Class mark (X) | Frequency (fi ) |
fi xi |
10-30 | 20 | 5 | 100 |
30-50 | 40 | 8 | 320 |
50-70 | 60 | 12 | 720 |
70-90 | 80 | 20 | 1600 |
90-110 | 100 | 3 | 300 |
110-130 | 120 | 2 | 240 |
Mean= Σfixi/Σfi
=3280/50
=65.6
Alternate methods by assuming mean are acceptable.
OR
Change the distribution to a ‘more than type’ distribution, and draw its ogive.
Production yield (kg/hectare) : | 40 – 45 | 45 – 50 | 50 – 55 | 55 – 60 | 60 – 65 | 65 – 70 |
Number of farms : | 4 | 6 | 16 | 20 | 30 | 24 |
More than or equal to 65 | 24 |
More than or equal to 60 | 54 |
More than or equal to 55 | 74 |
More than or equal to 50 | 90 |
More than or equal to 45 | 96 |
More than or equal to 40 | 100 |
Plotting graph of (40, 100), (45, 96), (50, 90), (55, 74), (60, 54)
and (65, 24) and joining the points
30) A container opened at the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm respectively. Find the cost of milk which can completely fill the container, at the rate of ₹50 per litre. Also find the cost of metal sheet used to make the container, if it costs ₹ 10 per 100 cm2 . (Take p = 3·14)
Ans: Volume of the container =π/3(h)(r21+r22+r1r2)
=3.14/3×16(202+82+20×8)
=3.14×16×208=10450 cm3
=10.45 litres
Cost of milk = 10.45 × 50 = ₹ 522.50
Slant height of frustum =√162+122=20cm
Surface area = π[(r1+r2)1 +r22]
= 3.14[(8+20) 20+82]
= 3.14 × 624 = 1959.36 cm2
∴ Cost of metal used =10/100×1959.36=₹ 195.93
CBSE Class 10 Previous Question Paper 2019 Solution
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