30/3/1 2019 Class 10 Maths Question Paper Solution
MATHEMATICS
SECTION A
1) Write the discriminant of the quadratic equation (x + 5)2 = 2 (5x – 3).
Ans: (x + 5)2 = 2(5x – 3)=> x2 + 31 = 10
D=-124
2) Find after how many places of decimal the decimal form of the number 27/23.54.32 will terminate.
Ans: 27/23.54.32=3/23.54
It will terminate after 4 decimal places
OR
Express 429 as a product of its prime factors.
Ans: 429 = 3 × 11 × 13
3) Find the sum of first 10 multiples of 6.
Ans: S10=10/2[2× 6 +9× 6]
=330
4) Find the value(s) of x, if the distance between the points A(0, 0) and B(x, – 4) is 5 units.
Ans: AB = 5
=>√(x-0)2+(-4-0)2=5
x2+16=25
x = ±3
5) Two concentric circles of radii a and b (a > b) are given. Find the length of the chord of the larger circle which touches the smaller circle.
Ans: Length of chord =2√a2-b2
6) In Figure 1, PS = 3 cm, QS = 4 cm, Ð PRQ = q, Ð PSQ = 90°, PQ ⊥RQ and RQ = 9 cm. Evaluate tan q.
Ans: PQ = 5 cm
tan θ=PQ/PR=5/9
or
If tan a =5/12. find the value of sec a.
sec α =√1+tan2 α
=√1+25/144=13/12
SECTION B
7) Points A(3, 1), B(5, 1), C(a, b) and D(4, 3) are vertices of a parallelogram ABCD. Find the values of a and b.
Ans: Diagonals of parallelogram bisect each other
∴ (3+a/2,1+b/2)=(5+4/2,1+3/2)
3 + a = 9, 1 + b = 4
So a = 6, b = 3
OR
Points P and Q trisect the line segment joining the points A(– 2, 0) and B(0, 8) such that P is near to A. Find the coordinates of points P and Q.
Ans:
P divides AB in the ratio 1 : 2
∴ Coordinates of P are(0-4/3,8+0/2)=(-4/3,8/3)
Q divides AB in the ratio 2 : 1
∴ Coordinates of Q are (0-2/3,16+0/3)=(-2/3,16/3)
8) Solve the following pair of linear equations :
3x – 5y = 4
2y + 7 = 9x
Ans:
3x – 5y = 4
9x – 2y = 7
9x – 15y = 12
9x – 2y = 7
– + –
—————————–
–13y = 5 => y = –5/13
From (1), x = 9/13 ∴ solution is (9/13,-5/13)
9) If HCF of 65 and 117 is expressible in the form 65n – 117, then find the value of n.
Ans: HCF (65, 117) = 13
13 = 65n – 117
Solving, we get, n = 2
OR
On a morning walk, three persons step out together and their steps measure 30 cm, 36 cm and 40 cm respectively. What is the minimum distance each should walk so that each can cover the same distance in complete steps ?
Ans: Required minimum distance = LCM (30, 36, 40)
30 = 2 × 3 × 5 = 23 × 32 × 5
36 = 22 × 32 = 360 cm
40 = 23 × 5
10) A die is thrown once. Find the probability of getting (i) a composite number, (ii) a prime number.
Ans: Composite numbers on a die are 4 and 6
∴ P (composite number) =2/6 or 1/3
Prime numbers are 2, 3 and 5
∴ P(prime number) =3/6 or 1/2
11) Using completing the square method, show that the equation x2 – 8x + 18 = 0 has no solution.
Ans: x 2 – 8x + 18 = 0
x 2 – 8x + 16 + 2 = 0
(x-4)2=-2
Square of a number can’t be negative
∴ The equation has no solution.
12) Cards numbered 7 to 40 were put in a box. Poonam selects a card at random. What is the probability that Poonam selects a card which is a multiple of 7 ?
Ans: Total number of possible outcomes = 34
Favourable number of outcomes is (7, 14, 21, 28 and 35) = 5
P(multiple of 7) =5/34
SECTION C
13) The perpendicular from A on side BC of a D ABC meets BC at D such that DB = 3CD. Prove that 2AB2 = 2AC2 + BC2
Ans:
AB2 = AD2 + BD2
AC2 = AD2 + CD2
AB2 – AC2 = BD2 – CD2
= (3CD)2 – CD2
= 8 CD2
=8 ×(1/4BC)2
=>2AB2 – 2AC2 = BC2
or
2AB2 = 2AC2 + BC2
OR
AD and PM are medians of triangles ABC and PQR respectively where D ABC ~ D PQR. Prove that AB/PQ=AD/PM
Ans:
Correct Figure
∆ABC ~ ∆PQR
∴ AB/ PQ = BC /QR=AC/PR
AB /PQ = 2BD/2QM or BD /QM
Also ∠B = ∠Q
∴ ∆ABD ~ ∆PQM
So AB/ PQ = AD /PM
14) Check whether g(x) is a factor of p(x) by dividing polynomial p(x) by polynomial g(x),
where p(x) = x5 – 4x3 + x2 + 3x + 1, g(x) = x3 – 3x + 1
Ans:
Since remainder ≠ 0 ∴ g(x) is not a factor of p(x)
15) Find the area of the triangle formed by joining the mid-points of the sides of the triangle ABC, whose vertices are A(0, – 1), B(2, 1) and C(0, 3).
Ans:
Coordinates of mid points are
D(1, 2)
E (1, 0)
F(0 ,1)
Area of ∆DEF =1/2[1(0-1)+1(1-2)+0]
16) Draw the graph of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Using this graph, find the values of x and y which satisfy both the equations.
Ans:
= 1/2 (–2) 1sq. unit
Correct graph
Solution is
x = 2, y = 3
17) Prove that √3 is an irrational number.
Ans: Let us assume that √3 be a rational number
√3=p/q where p and q are co-primes and q ≠ 0
=>p2 = 3q2
∴ 3 divides p2
i.e., 3 divides p also
Let p = 3m, for some integer m
From (1), 9m2 = 3q2
=>q2 = 3m2
∴ 3 divides q2 i.e., 3 divides q also
From (2) and (3), we get that 3 divides p and q both which is a contradiction to the fact that p and q are co-primes
Hence our assumption is wrong
∴ √3 is irrational
OR
Find the largest number which on dividing 1251, 9377 and 15628 leaves remainders 1, 2 and 3 respectively.
Ans: 1251 – 1 = 1250, 9377 – 2 = 9375, 15628 – 3 = 15625
Required largest number = HCF (1250, 9375, 15625)
1250=2×54
9375=3×54
6250=2×55
HCF (1250, 9375, 15625) = 54 = 625
18) A, B and C are interior angles of a triangle ABC. Show that
Ans: A, B, C are interior angles of ∆ABC
∴ A + B + C = 180°
(i) sin(B+C/2)=cos(A/2)
=> sin(B+C/2)=sin(180°-A/2)
=sin(90°-A/2)
=cosA/2
(ii) If Ð A = 90°, then find the value of than (B+C/2)
Ans: tan(B+C/2)=tan(90°/2)
=tan 45°
=1
OR
If tan (A + B) = 1 and tan (A – B) =1/√3, 0°<A+B<90°,A>B, then find the values of A and B.
Ans: tan (A + B) = 1 ∴ A + B = 45°
tan (A – B) =1/√3∴ A – B = 30°
Solving, we get ∠A =37(1°/2 or 37.5°
∠B = 7(1°/2) or 7.5°
19) In Figure 2, PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T. Find the length TP.
Ans:
Let TR be x cm and TP be y cm
OT is ⊥ bisector of PQ
So PR = 4 cm
ln ∆OPR, OP2 = PR2 + OR2
∴ OR = 3 cm
ln ∆PRT, y2 = x2 + 42
ln ∆OPT, (x + 3)2 = 52 + y2
∴ (x + 3)2 = 52 + x2 + 16
Solving we get x=16/3cm
From (1), y2=256/9+16=400/9
So y=20/3cm
OR
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Ans:
∆ROC ≅ ∆QOC
∴ ∠1 = ∠2
Similarly∠ 4=∠3
∠5 =∠6
∠8=∠7
∠ROQ + ∠QOP + ∠POS + ∠SOR = 360°
∴ 2∠1 + 2∠4 + 2∠5 + 2∠8 = 360
=>∠1 + ∠4 + ∠5 + ∠8 = 180°
So, ∠DOC + ∠AOB = 180°
and ∠AOD + ∠BOC = 180°.
20) Water in a canal, 6 m wide and 1·5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes if 8 cm of standing water is needed ?
Ans: Volume of water flowing through canal in 30 minutes
=5000 × 6 × 1.5= 45000 m3
Area = 45000÷8/ 100
=562500 m2
21) A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.
Number of days : | 0 – 6 | 6 – 12 | 12 – 18 | 18 – 24 | 24 – 30 | 30 – 36 | 36 – 42 |
Number of students : | 10 | 11 | 7 | 4 | 4 | 3 | 1 |
Ans:
Number of days | Number of students (fi) | xi | f i xi |
0-6 | 10 | 3 | 30 |
6-12 | 11 | 9 | 99 |
12-18 | 7 | 15 | 105 |
18-24 | 4 | 21 | 84 |
24-30 | 4 | 27 | 108 |
30-36 | 3 | 33 | 99 |
36-42 | 1 | 39 | 39 |
Total | 40 | 564 |
x̄=Σfixi/Σfi=564/40
=14.1
22) A car has two wipers which do not overlap. Each wiper has a blade of length 21 cm sweeping through an angle 120°. Find the total area cleaned
at each sweep of the blades. (Take π=22/7)
Ans: Total area cleaned = 2 × Area of sector
=2×πr2θ/260°
=2× 22/7 × 21× 21×120°/360°
=924cm2
SECTION D
23) A pole has to be erected at a point on the boundary of a circular park of diameter 13 m in such a way that the difference of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 m. Is it possible to do so ? If yes, at what distances from the two gates should the pole be erected ?
Ans:
Correct Figure
PB-PA=7m
Let AP be x m ∴ PB = (x + 7) m
AB2 = PB2 + AB2
∴ 132 = (x + 7)2 + x2
x 2 + 7x – 60 = 0
= (x + 12) (x – 5) = 0
∴ x = 5, –12 Rejected
∴ Situation is possible
∴ Distance of pole from gate A = 5 m
and distance of pole from gate B = 12 m.
24) If m times the mth term of an Arithmetic Progression is equal to n times its nth term and m ¹ n, show that the (m + n)th term of the A.P. is zero.
Ans: mam = nan
=>ma + m(m – 1)d = na + n(n – 1)d
=>(m – n)a + (m2 – m – n2 + n)d = 0
(m – n)a + [(m – n) (m + n) – (m – n)d] = 0
Dividing by (m – n)
So, a + (m + n – 1)d = 0
or
am + n = 0
OR
The sum of the first three numbers in an Arithmetic Progression is 18. If the product of the first and the third term is 5 times the common difference, find the three numbers.
Ans: Let first three terms be a –d, a and a + d
a – d + a + a + d = 18
So a = 6
(a – d) (a + d) = 5d
=>62 – d2 = 5d
or
d 2 + 5d – 36 = 0
(d + 9) (d – 4) = 0
so, d=–9 or 4
For d = –9 three numbers are 15, 6 and –3
For d = 4 three numbers are 2, 6 and 10
25) Construct a triangle ABC with side BC = 6 cm, AB = 5 cm and Ð ABC = 60°. Then construct another triangle whose sides are 3/4 of the corresponding sides of the triangle ABC.
Ans: Correct construction of ∆ABC
Correct construction of triangle similar to ∆ABC
26) In Figure 3, a decorative block is shown which is made of two solids, a cube and a hemisphere. The base of the block is a cube with edge 6 cm and the hemisphere fixed on the top has a diameter of 4·2 cm. Find
(a) the total surface area of the block.
=>Total surface area of block
=TSA of cube + CSA of hemisphere – Base area of hemisphere
=6a2 + 2πr 2 – πr 2
=6a2 + πr 2
=(6× 62+22/7×2.1)cm2
=(216+13.86)cm2
=229.86 cm2
=229.86cm2
(b) the volume of the block formed.( (Take π=22/7)
=>Volume of block
=63+2/3×22/7×(2.1)3
=(216+19.40) cm3
=235.40 cm3
OR
A bucket open at the top is in the form of a frustum of a cone with a capacity of 12308·8 cm3 . The radii of the top and bottom circular ends are 20 cm and 12 cm respectively. Find the height of the bucket and the area of metal sheet used in making the bucket. (Use p = 3·14)
Ans: Volume of frustum = 12308.8 cm3
∴1/3πh(r21+ r22+r1r2)= 12308.8
=>1/3×3.14×h(202+20×12)= 12308.8
h=12308.8×3/784×3.14
h=15cm
1=√152+(20-12)2=17cm
Area of metal sheet used = πl(r1+r2)+ πr22
= 3.14[17 × 32 + 122 ]
= 3.14 × 688 cm2
=2160.32 cm2
27) 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: Correct figure, given, to prove and construction
Correct proof.
OR
Prove that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Correct figure, given, to prove and construction
Correct proof.
28) If 1 + sin2 q = 3 sin q cos q, then prove that tan q = 1 or tan q = 1/2
Ans: 1 + sin2 θ = 3sin θ cos θ
Dividing by cos2 θ
sec2 θ + tan2 θ = 3tan θ
=>1 + tan2 θ + tan2 θ = 3tan θ
=>2 tan2 θ – 3tan θ + 1 = 0
(tan θ – 1) (2tan θ – 1) = 0
So tan θ =1 or 1/2
29) Change the following distribution to a ‘more than type’ distribution. Hence draw the ‘more than type’ ogive for this distribution.
Class interval : | 20 -30 | 30 -40 | 40 -50 | 50 -60 | 60 -70 | 70 -80 | 80 -90 |
Frequency : | 10 | 8 | 12 | 24 | 6 | 25 | 15 |
Ans:
Class interval | Cumulative Frequency |
More than or equal to 20 | 100 |
More than or equal to 30 | 90 |
More than or equal to 40 | 82 |
More than or equal to 50 | 70 |
More than or equal to 60 | 46 |
More than or equal to 70 | 40 |
More than or equal to 80 | 15 |
Plotting of points (20, 100), (30, 90), (40, 82), (50, 70), (60, 46), (70, 40) and (80, 15)
Joining the points to get a curve
30) The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun’s altitude is 30° than when it was 60°. Find the height of the tower. (Given √3=1.732)
Ans:
Let AB = h be the height of tower
ln ∆ABC, h /x =tan 60°
h = x √3
ln ∆ABD, h /x+ 40 = tan 30°
=> h √3 = x + 40
3x = x + 40
∴ x = 20
So, height of tower = h =20 √3 = m
=20 × 1.732 m
=34.64 m
CBSE Class 10 Previous Question Paper 2019 Solution
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30/3/2 |
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30/3/3 |