30/5/1 2019 Class 10 Maths Question Paper Solution
MATHEMATICS
SECTION A
1) The HCF of two numbers a and b is 5 and their LCM is 200. Find the product ab.
Ans: a.b = 1000
2) Find the value of k for which x = 2 is a solution of the equation kx2 + 2x – 3 = 0.
Ans: k(2)2 + 2(2) – 3 = 0
k = -1 /4
OR
Find the value/s of k for which the quadratic equation 3x2 + kx + 3 = 0 has real and equal roots.
Ans: For real and equal roots
k 2 – 4 × 3 × 3 = 0
k = ±6
3) If in an A.P., a = 15, d = – 3 and an = 0, then find the value of n.
Ans: 15 + (n – 1)(–3) = 0
n = 6
4) If sin x + cos y = 1; x = 30° and y is an acute angle, find the value of y.
Ans: sin 30° + cos y = 1
cos y = 1 /2
=> y = 60°
OR
Find the value of (cos 48° – sin 42°).
Ans: cos 48° – sin 42°
= cos 48° – cos (90° – 42°)
= 0
5) The area of two similar triangles are 25 sq. cm and 121 sq. cm. Find the ratio of their corresponding sides.
Ans: 5 : 11
6) Find the value of ‘a’ so that the point (3, a) lies on the line represented by 2x – 3y = 5.
Ans: 6 – 3a = 5
a=1/3
SECTION B
7) If Sn, the sum of the first n terms of an A.P. is given by Sn = 2n2 + n, then find its nth term.
a1 = S1 = 2(1)2 + 1 = 3
a 2 = S2 – S1 = 10 – 3 = 7
AP 3, 7 …, => d = 4
a n = 3 + (n – 1)4 = (4n – 1)
OR
If the 17th term of an A.P. exceeds its 10th term by 7, find the common difference.
Ans: a17 = a10 + 7
a + 16d = a + 9d + 7
d = 1
8) The mid-point of the line segment joining A(2a, 4) and B(–2, 3b) is (1, 2a + 1). Find the values of a and b.
Ans: 2a-2/2=1
=> a=2
4+3b/2=2a+1
9) A child has a die whose 6 faces show the letters given below :
A B C A A B
The die is thrown once. What is the probability of getting (i) A (ii) B ?
Ans: i) P (getting A) = 3/6 or 1/2
(ii) P (getting B) = 2 /6 or 1/3
10) Find the HCF of 612 and 1314 using prime factorisation.
Ans: 612 = 22 × 32 × 17
1314 = 2 × 32 × 73
HCF (612, 1314) = 2 × 32 = 18
OR
Show that any positive odd integer is of the form 6m + 1 or 6m + 3 or 6m + 5, where m is some integer.
Let a be any +ve integer
and b = 6
=> a = 6m + r 0 ≤ r < 6, for any +ve integer m
Possible forms of ‘a’ are
6m, 6m + 1, 6m + 2, 6m + 3, 6m + 4, 6m + 5
Out of which 6m, 6m + 2 and 6m + 4 are even.
Hence, any +ve odd integer can be 6m + 1, 6m + 3 or 6m + 5
11) Cards marked with numbers 5 to 50 (one number on one card) are placed in a box and mixed thoroughly. One card is drawn at random from the box. Find the probability that the number on the card taken out is (i) a prime number less than 10, (ii) a number which is a perfect square.
Ans: Total cards = 46
(i) P [Prime number less than 10(5, 7)] = 2/ 46 or 1/23
(ii) P [A number which is perfect square (9, 16, 25, 36, 49)] = 5/ 46
12) For what value of k, does the system of linear equations
2x + 3y = 7
(k – 1) x + (k + 2) y = 3k
have an infinite number of solutions ?
Ans: For infinitely many solutions
2/k-1=3/k+2=7/3k
2k+4=3k-3;
k=7
9k=7k+14
k=7
Hence k=7
SECTION C
13) Prove that √5 is an irrational number.
Ans: Let √5 be rational.
√5 = a /b ,b ≠ 0. a, b are positive integers, HCF (a, b) = 1
On squaring
5 =a2/b2
b2=a2/5
=>5 divides a2
=>5 divides a also.
a = 5m, for some +ve integer m.
b 2=25m2/5
b 2 = 5m2
=> 5 divides b2
=> 5 divides b also
=> 5 divides a and b both.
Which is the contradiction to the fact that HCF (a, b) = 1
Hence our assumption is wrong.
√5 is a irrational
14) Find all the zeroes of the polynomial x 4 + x 3 – 14x2 – 2x + 24, if two of its
zeroes are √2 and –√2 .
Ans: Given √2 and – √2 are zeroes of given polynomial.
∴ (x -√2) − and (x √2) + are two factors i.e. x2 – 2 is a factor
x 2 + x – 12 = x2 + 4x – 3x – 12
=(x + 4) (x – 3)
∴ –4, 3 are the zeroes.
Hence, all zeroes are –4, 3, √2,- √ 2
15) Point P divides the line segment joining the points A(2, 1) and B(5, –8) such that AP/AB=1/3. If P lies on the line 2x – y + k = 0, find the value of k.
Ans:
AP/AB =1/3 =>AP /PB=1/ 2
Coordinates of P are (5+4/3, -8+2/3)=(3, -2)
Now, P lies on 2x – y + k = 0
∴ 2(3) – (–2) + k = 0
=>k = –8
OR
For what value of p, are the points (2, 1), (p, –1) and (–1, 3) collinear ?\
Ans: Three points are collinear => area of ∆ formed by these points is zero.
∴ 1/2[2(-1-3)+p(3-1)-(1+1)]=0
-8+2p-2=0
p=5
16) Prove that :
tan θ/1-tan θ-cot θ/1-cot θ=cos θ +sin θ/cos θ –sin θ
LHS= tan/1- tan -cot /1-cot
= sin θ/cos θ/1-sin θ/cos θ-cos θ/sin θ/1-cos θ/sin θ
=sin θ/cos θ-sin θ+cos θ/cos θ-sin θ
=sin θ+cos θ/cos θ-sin θ= RHS
OR
If cos q + sin q = √2 cos q, show that cos q – sin q = √2 sin q.
Ans: sin θ =( √2-1) cos θ
(√2+1) sin θ =( √2-1)( √2+1) cos θ
(√2+1) sin θ = cos θ
=> √2 sin θ =cos θ –sin θ
Alternate method
cos θ + sin θ = 2 cos θ
On squaring
cos2 θ + sin2 θ + 2 cos θ sin θ = 2 cos2 θ
sin2 θ + 2 cos θ sin θ = cos2 θ
2 cos θ sin θ = cos2 θ – sin2 θ
2 cos θ sin θ = (cos θ – sin θ) (cos θ + sin θ)
2 cos θ sin θ = (cos sin )( 2 cos )
2 sin cos sin θ = θ − θ
17) A part of monthly hostel charges in a college hostel are fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 25 days, he has to pay ₹ 4,500, whereas a student B who takes food for 30 days, has to pay ₹ 5,200. Find the fixed charges per month and the cost of food per day.
Ans: Let the fixed charges per student = ₹ x
Cost of food per day per student = ₹ y
x + 25y = 4500
x + 30y = 5200
On solving 5y = 700
∴ y = 140
x = 1000
∴ Fixed charges = ₹1000 & cost of food per day ₹140
18) In D ABC, Ð B = 90° and D is the mid-point of BC. Prove that AC2 = AD2 + 3CD2 .
Ans:
Correct Figure
∆ABC is right angled at B
∴ AC2 = AB2 + BC2
AC2 = AB2 + (2CD)2
AC2 – 4CD2 = AB2
∆ABD is right angled at B,
∴ AD2 – BD2 = AB2
By (1) & (2) AC2 – 4CD2 = AD2 – BD2
AC2 = AD2 – CD2 + 4CD2 = AD2 + 3CD2 (∵ BD = CD)
OR
In Figure 1, E is a point on CB produced of an isosceles ∆ ABC, with side AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ∆ ABD ~ ∆ ECF.
Ans: AB = AC => ∠ C = ∠ B
In ∆ ABD & ∆ ECF
∠ADB = ∠EFC (each 90°)
∠ABD = ∠ECF (by (1))
By AA similarity
∆ABD ~ ∆ECF
19) Prove that the parallelogram circumscribing a circle is a rhombus.
Ans:
Correct Figure
Let parallelogram ABCD circumscribes a circle
AP= AS
PB =BQ
DR= DS
CR =CQ
tangents from an external point to a circle.
AP + PB + DR + RC = AS + BQ + DS + CQ
AB + DC = AD + BC
AB + AB = AD + AD (opp. sides equal)
2AB = 2AD
=> AB = AD
=> ABCD is a rhombus
20) In Figure 2, three sectors of a circle of radius 7 cm, making angles of 60°, 80° and 40° at the centre are shaded. Find the the area of the shaded region.
Ans: Area of shaded region = 80°/360°π (7)2+40°/360°π (7)260°/360°π (7)2
= 22/7×7×7[180°/360°]
= 77cm2
21) The following table gives the number of participants in a yoga camp :
Age (in years) : | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 | 60 – 70 |
No. of Participants : | 8 | 40 | 58 | 90 | 83 |
Find the modal age of the participants.
Ans: Modal class: 50 – 60
mode= 50+(90-58/180-58-83) × 10
= 50+ 32/39×10
=58.2
∴ Modal age = 58.2 years.
22) A juice seller was serving his customers using glasses as shown in Figure 3. The inner diameter of the cylindrical glass was 5 cm but bottom of the glass had a hemispherical raised portion which reduced the capacity of the glass. If the height of a glass was 10 cm, find the apparent and actual capacity of the glass. (Use p = 3·14)
Ans: Apparent capacity = πr 2 h
= 3.14×5/2×5/2×10
=196.25 cm2
OR
A girl empties a cylindrical bucket full of sand, of base radius 18 cm and height 32 cm on the floor to form a conical heap of sand. If the height of this conical heap is 24 cm, then find its slant height correct to one place of decimal.
Ans: π(18)2×32=1/3 πr2×24
r2=(18)2×4
r=36 cm
l 2 = (36)2 + (24)2
l 2 = 1872
l = 43.2 cm
SECTION D
23) A train travels 360 km at a uniform speed. If the speed had been 5 km/hr more, it would have taken 1 hr less for the same journey. Find the speed of the train.
Ans: Let speed of train be x km/h
360/x-360/x+5=1
360[x+5-x/x(x+5)]=1
x2+5x-1800=0
(x + 45) (x – 40) =0
x = –45, x = 40
(Rejected)
Hence, speed of train = 40 km/h
OR
Solve for x :
1/a+b+x=1/a+1/b+1/x;a¹b¹0,x¹-(a+b)
Ans: 1/a+b+x-1/x=1/a+1/b
x-a-b-x/x(a+b+x)=b+a/ab
-ab= x2+(a+b)x
x2+(a+b)x+ab=0
(x+a)(x+b)=0
x+-a, x=-b
24) If the sum of the first p terms of an A.P. is q and the sum of the first q terms is p; then show that the sum of the first (p + q) terms is { – (p + q)}.
Ans: p/2 (2a+ (p- 1) d=q
2a+ (p- 1)d=2q/p
q/2[(2a+(q-1)d]=p
2a+(q-1)d=2p/q
On solving (1) and (2) for a and d
d = -2(p+ q)/pq
a=q2+p2-p+pq-q/pq
Sp+q=p+q/2(2a+(p+q-1)d)
=p+q/2[2(q2+p2-p+pq-q/pq)+(p+q-1)(2(p+q)/pq)]
=(p+q)[a2+p2-p+pq-q-p2-q2-2pq+p+q/pq]
=(p+q) ×-pq/pq=-(p+q)
25) In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then prove that the angle opposite to the first side is a right angle.
Ans: For Correct Given, To Prove, Construction, Figure
For Correct Proof
26) Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 3/4 times the corresponding sides of the isosceles triangle.
Ans: For Correct Construction of triangle
For construction of similar triangle
27) A boy standing on a horizontal plane finds a bird flying at a distance of 100 m from him at an elevation of 30°. A girl standing on the roof of a 20 m high building, finds the elevation of the same bird to be 45°. The boy and the girl are on the opposite sides of the bird. Find the distance of the bird from the girl. (Given √2 = 1·414)
Ans:
Correct Figure
In ∆ABC
sin 30° =BC/100
=>BC = 50 m
CF = 50 – 20 = 30 m
In ∆CFE
sin 45°=30/CE
CE=30√2
=30×1.414
=42.42 m
OR
The angle of elevation of an aeroplane from a point A on the ground is 60°. After a flight of 30 seconds, the angle of elevation changes to 30°. If the plane is flying at a constant height of 3600 √3 metres, find the speed of the aeroplane.
Ans:
Correct Figure
In ∆ABC, tan 60°=3600 √3/x
x = 3600
In ∆ADE, tan 30° =3600 √3/x+y
3600 + y = 3600 × 3
y = 7200
Speed = 7200/30=240 m/s
28) Find the values of frequencies x and y in the following frequency distribution table, if N = 100 and median is 32.
Class : | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 | Total |
Frequency : | 10 | x | 25 | 30 | y | 10 | 100 |
Ans:
Marks |
fi |
cf |
0–10 | 10 | 10 |
10–20 | x | 10 + x |
20–30 | 25 | 35 + x |
30–40 | 30 | 65 + x |
40–50 | y | 65 + x + y |
50–60 | 10 | 75 + x + y |
Total | 100 |
Median class = 30 – 40
75 + x + y = 100
x + y = 25
32 =30+(50-35-x/30) ×10
2=15-x/3
x=9
y=16
OR
For the following frequency distribution, draw a cumulative frequency curve (ogive) of ‘more than type’ and hence obtain the median value.
Class : | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 | 60 – 70 |
Frequency : | 5 | 15 | 20 | 23 | 17 | 11 | 9 |
Ans:
Class |
cf |
More than or equal to 0 | 100 |
More than or equal to 10 | 95 |
More than or equal to 20 | 80 |
More than or equal to 30 | 60 |
More than or equal to 40 | 37 |
More than or equal to 50 | 20 |
More than or equal to 60 | 9 |
Plotting of points (0, 100), (10, 95), (20, 80), (30, 60), (40, 37), (50, 20) and (60, 9)
Joining the points to get curve
Median = 35 (approx.)
29) Prove that :
(1+cot θ +tan θ )(sin θ – cos θ )/(sec3 θ -cosec3 θ )=sin2 θ cos2 θ
Ans: LHS= (1+cot θ +tan θ)(sin θ – cos θ)/sec3 θ – cosec3 θ
=(1+cos θ /sin θ + sin θ / cos θ)(sin θ – cos θ )/1/cos3 θ –sin3 θ
=(cos θ sin θ +cos2 θ+ sin2 θ)(sin θ- cos θ)/cos θ sin θ/sin3 θ- cos3 θ/cos3 θ sin3 θ
=sin3 θ-soc3 θ/sin θ cos θ × cos3 θ sin3 θ/sin3 θ-cos3 θ
=cos2 θ sin2 θ=RHS
30) An open metallic bucket is in the shape of a frustum of a cone. If the diameters of the two circular ends of the bucket are 45 cm and 25 cm and the vertical height of the bucket is 24 cm, find the area of the metallic sheet used to make the bucket. Also find the volume of the water it can hold. (Use p=22/7)
Ans: l2 = (24)2+(45/2-25/2)2
I2= 576 + 100 = 676
l = 26 cm.
TSA =22/7×26(25/2+45/2)+22/7×25/2×25/2
=2860 + 491.07
= 3351.07 cm2
Volume =1/3×22/7×24(625/4+2025/4+1125/4)
=1/3×22/7×24×3775/4
=166100 /7(cm3)
or 23728.57 cm3
CBSE Class 10 Previous Question Paper 2019 Solution
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