CBSE Class 10 Previous Question Paper (2019) >Maths with Solution
SECTION – A
(1) Find the coordinates of a point A, where AB is diameter of a circle whose centre is (2, -3) and B is the point (1,4).
Solution: Let the point A be (x, y)
∴ 1 x 2 + = 2 and 4 y 2 + = –3
⇒ x = 3 and y = –10
∴ Point A is (3, –10)
(2) For what values of K, the roots of the equation x2 + 4x + k = 0 are real?
Solution: Since roots of the equation x2 + 4x + k = 0 are real
⇒ 16 – 4k ≥ 0
⇒ k ≤ 4
Or
Find the value of K for which the roots of the equation 3x2 – 10x + k = 0 are reciprocal of each other.
Solution: Roots of the equation 3x2 – 10x + k = 0 are reciprocal of each other
⇒ Product of roots = 1
⇒ k 3 = 1 ⇒ k = 3
(3) Find A if tan 2A = cot (A – 240)
Solution: tan 2 A = cot (90° – 2A)
∴ 90° – 2A = A – 24°
⇒ A = 38°
Or
Find the value of (sin2 330 + sin2 570)
Solution: sin 33° = cos 57°
∴ sin2 33° + sin2 57° = cos2 57° + sin2 57° = 1
(4) How many two digits numbers are divisible by 3?
Solution: Numbers are 12, 15, 18, …, 99
∴ 99 = 12 + (n – 1) × 3
⇒ n = 30
(5) In Fig, 1, DE ‖ BC, AD = 1 cm and BD = 2 cm. What is the ratio of the ar (△ ABC) to the are (△ ADE)?
Solution: AB = 1 + 2 = 3 cm
∆ABC ~ ∆ADE
∴ ar(ABC)/ar(ADE) = AB2 / AD2 = 9/1
∴ ar(∆ABC) : ar(∆ADE) = 9 : 1
(6) Find a rational number between √2 and √3.
Solution: Any one rational number between 2 (1.41 approx.) and 3 (1.73 approx.)
e.g., 1.5, 1.6, 1.63 etc.
SECTION – B
(7) Find the HCF of 1260 and 7344 using Euclid’s algorithm.
Solution: Using Euclid’s Algorithm
7344 1260 5 1044
1260 1044 1 216
1044 216 4 180
216 180 1 36
180 36 5 0
HCF of 1260 and 7344 is 36.
Or
Show that every positive odd integer is of the form (4q + 1) or (4q + 3), where q is some integer.
Solution: Using Euclid’s Algorithm
a = 4q + r, 0 ≤ r < 4
⇒ a = 4q, a = 4q + 1, a = 4q + 2 and a = 4q + 3.
Now a = 4q and a = 4q + 2 are even numbers.
Therefore when a is odd, it is of the form
a = 4q + 1 or a = 4q + 3 for some integer q.
(8) Which term of the AP 3, 15, 27, 39,…. will be 120 more than its 21st term?
Solution: an = a21 + 120
= (3 + 20 × 12) + 120
= 363
∴ 363 = 3 + (n – 1) × 12
⇒ n = 31
or 31st term is 120 more than a21.
Or
If Sn, the sum of first n terms of an AP is given by Sn = 3n2 – 4n, find the nth term.
Solution: a1 = S1 = 3 – 4 = –1
a2 = S2 – S1 = [3(2)2 – 4(2)] – (–1) = 5
∴ d = a2 – a1 = 6
Hence an = –1 + (n – 1) × 6 = 6n – 7
Alternate method:
Sn = 3n2 – 4n
∴ Sn – 1 = 3(n – 1)2 – 4(n – 1) = 3n2 – 10n + 7
Hence an = Sn – Sn – 1
= (3n2 – 4n) – (3n2 – 10n + 7)
= 6n – 7
(9) Find the ratio in which the segment joining the points (1, -3) and (4, 5) is divided by x – axis? Also find the coordinates of this point on x – axis.
Solution: Let the required point be (a, 0) and required ratio AP : PB = k : 1
∴ a = 4k +1/ k+ 1
0 = 5k – 3/ k+ 1
⇒ k = 3 /5 or required ratio is 3 : 5
Point P is (17/8,0)
(10) A game consists of tossing a coin 3 times and nothing the outcome each time. If getting the same result in all the tosses is a success, find the probability of losing the game.
Solution: Total number of outcomes = 8
Favourable number of outcomes (HHH, TTT) = 2
Prob. (getting success) = 2/8 or 1/4
∴ Prob. (losing the game) = 1-1/4 = 3/4.
(11) A die is thrown once. Find the probability of a getting a number which (i) is a prime number (ii) lies between 2 and 6.
Solution: Total number of outcomes = 6.
(i) Prob. (getting a prime number (2, 3, 5)) = 3/6 or 1/2.
(ii) Prob. (getting a number between 2 and 6 (3, 4, 5)) = 3/6 or 1/2.
(12) Find c if the system of equations cx + 3y + (3 – c) = 0; 12x + cy – c = 0 has infinitely many solutions?
Solution: System of equations has infinitely many solutions
∴ c /12 = 3/c = 3-c/-c
⇒ c2 = 36 ⇒ c = 6 or c = –6 …(1)
Also –3c = 3c – c2 ⇒ c = 6 or c = 0 …(2)
From equations (1) and (2)
c = 6.
SECTION – C
(13) Prove that √2 is an irrational number.
Solution: Let us assume √2 be a rational number and its simplest form be a /b, a and b are coprime positive integers and b ≠ 0.
So √2 = a/b
⇒ a 2 = 2b2
Thus a2 is a multiple of 2
⇒ a is a multiple of 2.
Let a = 2 m for some integer m
∴ b 2 = 2m2
Thus b2 is a multiple of 2
⇒ b is a multiple of 2
Hence 2 is a common factor of a and b.
This contradicts the fact that a and b are coprimes
Hence √2 is an irrational number.
(14) Find the value of K such that the polynomial x2 – (k + 6) x +2 (2k – 1) has sum of its zeros equal to half of their product.
Solution: Sum of zeroes = k + 6
Product of zeroes = 2(2k – 1)
Hence k + 6 = 1/2 x2(2k -1)
⇒ k = 7
(15) A father’s age is three times the sum of the ages of his two children. After 5 years his age will be two times the sum of their ages. Find the present age of the father.
Solution: Let sum of the ages of two children be x yrs and father’s age be y yrs.
∴ y = 3x …(1)
and y + 5 = 2(x + 10) …(2)
Solving equations (1) and (2)
x = 15
and y = 45
Father’s present age is 45 years.
Or
A fraction becomes 1/3 when 2 is subtracted from the numerator and it becomes 1/2 when 1 is subtracted from the denominator. Find the fraction.
Solution: Let the fraction be x /y
∴ x – 2/ y = 1/3 …(1)
and x/ y –1 = 1/ 2 …(2)
Solving (1) and (2) to get x = 7, y = 15.
∴ Required fraction is 7/ 15
(16) Find the point on y – axis which is equidistant from the points (5, -2) and (-3, 2).
Solution: Let the required point on y-axis be (0, b)
∴ (5 – 0)2 + (–2 – b)2 = (–3 – 0)2 + (2 – b)2
⇒ 29 + 4b + b2 = 13 + b2 – 4b
⇒ b = –2
∴ Required point is (0, –2)
Or
The line segment joining the points A(2, 1) and B(5, -8) is trisected at the points P and Q such that P is nearer to A. If P also lies on the line given by 2x – y + k = 0, find the value of k.
Solution: AP : PB = 1 : 2
x = 4+5/3 = 3 and y = 2-8/3 = 2
Thus point P is (3, –2).
Point (3, –2) lies on 2x – y + k = 0
⇒ 6 + 2 + k = 0
⇒ k = –8.
(17) Prove that (sin θ + cosec θ)2 = 7 + tan2 θ + cot2 θ.
Solution: LHS = sin2 θ + cosec2 θ + 2sin θ cosec θ + cos2 θ + sec2 θ + 2cos θ sec θ
= (sin2 θ + cos2 θ) + cosec2 θ + sec2 θ + 2 sin θ/sin θ +2 cos θ/ cos θ.
= 1 + 1 + cot2 θ + 1 + tan2 θ + 2 + 2
= 7 + cot2 θ + tan2 θ = RHS
Or
Prove that (1 + cot A – cosec A) (1 + tan A + sec A) = 2
Solution: Alternate method
LHS = (1+ cOSa/SinA – 1/Sin A) (1+sinA/CosA + 1/Cos A
= (Sin A + Cos A – 1) (Cos A + Sin A +1). 1/CosA SinA
=[(Sin A + CosA)2 -1]X 1/Sin A Cos A
=(1+2 Sin A Cos -1) x 1/SinA Cos A
= 2 = RHS
(18) In Fig, 2, PQ is a chord of length 8 cm of a circle of radius 5 cm and centre O. The tangents at P and Q interest at point T. Find the length of TP.
Solution: Join OT and OQ.
TP = TQ
∴ TM ⊥ PQ and bisects PQ
Hence PM = 4 cm
Therefore OM = √25 – 16 √9 3 cm.
Let TM = x
From ∆PMT, PT2 = x2 + 16
Hence x2 + 16 = x2 + 9 + 6x – 25
⇒ 6x = 32 ⇒ x = 16/ 3
Hence PT2 = 256/9 +16 = 400/9
∴ PT = 20/3 cm.
(19) In Fig. 3, ∠ACB = 900 and CD ⊥ AB, prove that CD2 = BD X AD.
Solution:
∆ACB ~ ∆ADC (AA similarity)
⇒ AC/ BC = AD /CD …(1)
Also ∆ACB ~ ∆CDB (AA similarity)
⇒ AC/ BC = CD/ BD …(2)
Using equations (1) and (2)
AD/ CD = CD /BD
⇒ CD2 = AD × BD
Or
If P and Q are the points on side CA and CB respectively of △ ABC, right angled at C, prove that (AQ2 + BP2) = (AB2 + PQ2)
Solution: Correct Figure
AQ2 = CQ2 + AC2
BP2 = CP2 + BC2
∴ AQ2 + BP2 = (CQ2 + CP2 ) + (AC2 + BC2 )
= PQ2 + AB2.
(20) Find the area of the shaded region in Fig. 4, if ABCD is a rectangle with sides 8 cm and 6 cm and O is the centre of circle. (Take π = 3.14)
Solution: AC = √64 36 10 cm.
∴ Radius of the circle (r) = 5 cm.
Area of shaded region = Area of circle – Ar(ABCD)
= 3.14 × 25 – 6 × 8
= 78.5 – 48
= 30.5 cm2.
(21) Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/hour. How much area will it irrigate in 30 minutes: if 8 cm standing water is needed?
Solution: Length of canal covered in 30 min = 5000 m.
∴ Volume of water flown in 30 min = 6 × 1.5 × 5000 m3
If 8 cm standing water is needed
then area irrigated = 6x 1.5x 5000/.08 = 562500 m2
(22) Find the mode of the following frequency distribution.
Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
Frequency | 8 | 10 | 10 | 16 | 12 | 6 | 7 |
Solution: Modal class is 30-40
∴ Mode = 1 +(f1 – f0/ 2f1 – f0 – f2) x h
= 30+ (16 -10/32-10-12) x 10
SECTION – D
(23) Two water taps together can fill a tank in 1/7/8 hours. The tap with longer diameter takes 2 hours less than the tap with smaller one to fill the tank separately. Find the time in which each tap can fill the tank separately.
Solution: Let the smaller tap fills the tank in x hrs
∴ the larger tap fills the tank in (x – 2) hrs.
Time taken by both the taps together = 15/8 hrs.
Therefore 1/x + 1/x – 2 = 8/15
⇒ 4x2 – 23x + 15 = 0
⇒ (4x – 3) (x – 5) = 0
x ≠ 3/ 4 ∴ x = 5
Smaller and larger taps can fill the tank seperately in 5 hrs and 3 hrs resp.
Or
A boats goes 30 Km upstream and 44 Km downstream in 10 hours. In 13 hours, it can go 40 Km upstream and 55 km downstream. Determine the speed of the stream and that of that boat in still water.
Solution: Let the speed of the boat in still water be x km/hr and speed of the stream be y km/hr.
Given 30/ x –y + 44 / x + y …(i)
and 40/x-y +55/x + y …(ii)
Solving (i) and (ii) to get
x + y = 11 ..(iii)
and x – y = 5 …(iv)
Solving (iii) and (iv) to get x = 8, y = 3.
Speed of boat = 8 km/hr & speed of stream = 3 km/hr.
(24) If the sum of first four terms of an AP is 40 and that of first 14 terms is 280. Find the sum of its first n terms.
Solution: S4 = 40 ⇒ 2(2a + 3d) = 40 ⇒ 2a + 3d = 20
S14 = 280 ⇒ 7(2a + 13d) = 280 ⇒ 2a + 13d = 40
Solving to get d = 2
and a = 7
∴ Sn = n/2 [14+(n – 1) x 2]
= n(n + 6) or (n2 + 6n)
(25) Prove that sin A – Cos A +1/Sin A + Cos A – 1 = 1/Sec A – tan A
Solution: LHS = Sin A – Cos A +1/ Sin A + Cos A – 1
Dividing num. & deno. by Cos A
= tan A -1 + Sec A/ tan A + 1 – Sec A
= tan A – 1 + Sec A/(tan A – Sec A) + (Sec2 A – tan2 A)
= tan A – 1 + Sec A/(tan A – Sec A ) (1 – Sec A – tan – A)
= -1/tan A – Sec A = 1/Sec A – tan A = RHS
(26) A man in a boat rowing away from a light house 100 m high takes 2 minutes to change the angle of elevation of the top of the light house from 600 to 300. Find the speed of that boat in metres per minute.[ Use √3 = 1.732]
Solution: Correct Figure
Let the speed of the boat be y m/min
∴ CD = 2y
tan 600 = √3 = 100/x ⇒ x = 100/√3
tan 300 = 1/√3 = 100/x + 2y ⇒ x + 2y = 100 √3
∴ y = 100 √3/3 = 27.73
or speed of boat = 57.73 m/min.
Or
Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 600 and 300 respectively. Find the height of the poles and the distances of the point from the poles.
Solutions: Correct Figure
Let BC = x so AB = 80 – x
where AC is the road.
tan 600 = √3 = h/x ⇒ h = x √3
and tan 300 = 1/√3 = h / 80 – x ⇒ h √3 = 80 – x
Solving equation to get
X = 20, h = 20 √3
∴ AB = 60 m, BC = 20 m and h = 20√ 3 m.
(27) Construct a △ ABC in which CA = 6 cm, AB = 5 cm and ∠ BAC = 450. Then construct a triangle whose sides are 3/5 of the corresponding sides of △ ABC.
Solution: Correct construction of ∆ABC.
Correct construction of triangle similar to triangle ABC.
(28) 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 of circular ends of the bucket are 20 cm and 12 cm respectively. Find the height of the bucket and also the area of the metal sheet used in making it (Use π = 3.14)
Solution: Volume of the bucket = 12308.8 cm3
Let r1 = 20 cm, r2 = 12 cm
∴ V = πh/3 (r21 + r22 + r1 r2)
∴ 12308.8 = 3.14 x h / 3 ( 400 + 144 + 240)
⇒ h = 12308.8x 3/3.14 x 784 = 15 cm
Now l2 = h2 + (r1 – r2) 2 = 225 + 64 = 289
⇒ l = 17 cm.
Surface area of metal sheet used = π 22 + πl (r1 + r2)
= 3.14 (144 + 17 x 32)
= 2160.32 cm2.
(29) Prove that in a right angle triangle, the square of the hypotenuse is equal the sum of squares of the other two sides.
Solution: Correct given, to prove, figure and construction
Correct proof.
(30) If the median of the following frequency distribution is 32.5. Find the values of f1 and f2.
Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | Total |
Frequency | F1 | 5 | 9 | 12 | F2 | 3 | 2 | 40 |
Solution:
Class | Frequency | Cumulative freq. |
0-10 | f 1 | f 1 |
10-20 | 5 | 5 + f1 |
20-30 | 9 | 14 + f1 |
30-40 | 12 | 26 + f1 |
40-50 | f 2 | 26 + f1 + f2 |
50-60 | 3 | 29 + f1 + f2 |
60-70 | 2 | 31 + f1 + f2 |
40 |
Median = 32.5 ⇒ median class is 30-40.
Now 32.5 = 30+ 10/12(20 – 14 –f1)
⇒ f 1 = 3
Also 31 + f1 + f2 = 40
⇒ f2 = 6
Or
The marks obtained by 100 students of a class in an examination are given below.
Marks | No. Of students |
0-5 | 2 |
5-10 | 5 |
10-15 | 6 |
15-20 | 8 |
20-25 | 10 |
25-30 | 25 |
30-35 | 20 |
35-40 | 18 |
40-45 | 4 |
45-50 | 2 |
Draw ‘a less than’ type cumulative frequency curves (ogive). Hence find median.
Solution: Less than type distribution is as follows
Mark | No. of students |
Less than 5 | 2 |
Less than 10 | 7 |
Less than 15 | 13 |
Less than 20 | 21 |
Less than 25 | 31 |
Less than 30 | 56 |
Less than 35 | 76 |
Less than 40 | 94 |
Less than 45 | 98 |
Less than 50 | 100 |
Plotting of points (5, 2), (10, 7) (15, 13), (20, 21), (25, 31), (30, 56),
(35, 76), (40, 94), (45, 98), (50, 100)
Joining to get the curve
Getting median from graph (approx. 29)
Others Set | Solution |
30 / 1 / 2 | Click here |
30 / 1 / 3 | Click here |