CBSE Previous Year Question Paper Class 10 Maths 2019 Delhi

Time Allowed: 3 hours Maximum Marks: 80

General Instructions:

• All questions are compulsory.

• This question paper consists of 30 questions divided into four sections- A, B, C and D.

• Section A contains 6 questions of 1 mark each, Section B contains 6 questions of 2 marks each, Section C contains 10 questions of 3 marks each and Section D contains 8 questions of 4 marks each.

• There is no overall choice. However, an internal choice has been provided in two questions of 1 mark each, two questions of 2 marks each, four questions of 3 marks each and three questions of 4 marks each. You have to attempt only one of the alternatives in all such questions.

• Use of calculators is not permitted.

Section – A

Question 1.
Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, -3) and B is the point (1, 4). [1]
Solution:
Let the coordinates of point A be (x, y) and point O (2, -3) be point the centre, then
By midpoint formula,

The coordinates of point A are (3, -10)

Question 2.
For what values of k, the roots of the equation x² + 4x + k = 0 are real? [1]
OR
Find the value of k for which the roots of the equation 3x² – 10x + k = 0 are reciprocal of each other.
Solution:
The given equation is x² + 4x + k = 0
On comparing the given equation with ax² + bx + c = 0, we get
a = 1, b = 4 and c = k
For real roots, D ≥ 0
or b² – 4ac ≥ 0
or 16 – 4k ≥ 0
or k ≤ 4
For k ≤ 4, equation x² + 4x + k will have real roots.
OR
The given equation is 3x² – 10x + k = 0
On comparing it with ax² + bx + c = 0, we get
a = 3, b = -10, c = k
Let the roots of the equation are α and
Product of the roots =
=
or k = 3

Question 3.
Find A if tan 2A = cot (A – 24°) [1]
OR
Find the value of (sin² 33° + sin² 57°)
Solution:
Given, tan 2A = cot (A – 24°)
or cot (90° – 2A) = cot (A – 24°) [∵ tan θ = cot (90° – θ)]
or 90° – 2A = A – 24°
or 3A = 90° + 24°
or 3A = 114°
A = 38°
OR
sin² 33° + sin² 57°
= sin² 33° + cos² (90° – 57°)
= sin² 33° + cos² 33°
= 1 [∴ sin² θ + cos² θ = 1]

Question 4.
Flow many two digits numbers are divisible by 3? [1]
Solution:
The two-digit numbers divisible by 3 are 12, 15, 18, ……… 99
This is an A.P. in which a = 12, d = 3, a_n = 99
a_n = a + (n – 1) d
99 = 12 + (n – 1) × 3
87 = (n – 1) × 3
or n – 1 = 29
or n = 30
So, there are 30 two-digit numbers divisible by 3.

Question 5.
In Fig., DE || BC, AD = 1 cm and BD = 2 cm. what is the ratio of the ar (ΔABC) to the ar (ΔADE) ? [1]

Solution:
Given, AD = 1 cm, BD = 2 cm
AB = 1 + 2 = 3 cm
Also, DE || BC (Given)
∠ADE = ∠ABC …(i) (corresponding angles)
In ΔABC and ΔADE
∠A = ∠A (common)
∠ABC = ∠ADE [by equation (i)]
ΔABC ~ ΔADE (by AA rule)
Now,

Question 6.
Find a rational number between √2 and √3. [1]
Solution:
As √2 = 1.414 ….
√3 = 1.732…..
So, a rational number between √2 and √3 is 1.5 or we can take any number between 1.414 and 1.732

Section – B

Question 7.
Find the HCF of 1260 and 7344 using Euclid’s algorithm. [2]
OR
Show that every positive odd integer is of the form (4q + 1) or (4q + 3), where q is some integer.
Solution:
Two numbers are 1260 and 7344
Since 7344 > 1260, we apply the Euclid division lemma to 7344 and 1260, we get
7344 = 1260 × 5 + 1044
Also, 1260 = 1044 × 1 + 216
1044 = 216 × 4 + 180
216 = 180 × 1 + 36
180 = 36 × 5 + 0
Now, the remainder is 0, hence our procedure stops here.
H.C.F. of 7344 and 1260 is 36.
OR
Let ‘a’ be any positive odd integer.
We apply the division algorithm with a and b = 4
a = bq + r, where 0 ≤ r < b
or a = 4q + r,
the possible remainders are 0, 1, 2, 3
Then when r = 0, ⇒ a = 4q
r = 1, ⇒ a = 4q + 1
r = 2, ⇒ a = 4q + 2
and when r = 3, ⇒ a = 4q + 3
Since a is odd, a cannot be 4q or 4q + 2
(Since both are divisible by 2)
Therefore, any odd integer is of the form 4q + 1 or 4q + 3.
Hence Proved.

Question 8.
Which term of the A.P. 3, 15, 27, 39, …… will be 120 more than its 21st term? [2]
OR
If S_n, the sum of first nth terms of an A.P. is given by S_n = 3n² – 4n, find the nth term.
Solution:
The given A.P. is 3, 15, 27, 39,…
Here a = 3, d = 12
a₂₁ = a + 20d = 3 + 20 × 12 = 3 + 240 = 243
Now, a_n = a₂₁ + 120 = 243 + 120 = 363
a_n =a + (n – 1) d
363 = 3 + (n – 1) × 12
or 360 = (n – 1) × 12
or n – 1 = 30
n = 31
Hence, the term which is 120 more than its 21st term will be its 31st term.
OR
Given, S_n = 3n² – 4n
We know that
a_n = S_n – S_n-1
= 3n² – 4n – [3 (n – 1)² – 4 (n – 1)]
= 3n² – 4n – [3 (n² – 2n + 1) – 4n + 4]
= 3n² – 4n – (3n² – 6n + 3 – 4n + 4)
= 3n² – 4n – 3n² + 10n – 7
= 6n – 7
So, nth term will be 6n – 7

Question 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 the x-axis. [2]
Solution:
Let the given points be A (1, -3) and B (4, -5) and the line-segment joining by these points is divided by the x-axis, so the coordinates of the point of intersection will be P(x, 0)

Question 10.
A game consists of tossing a coin 3 times and noting the outcome each time. If getting the same result in all the tosses is a success, find the probability of losing the game. [2]
Solution:
When a coin is tossed three times, the set of all possible outcomes is given by,
S = {HHH, HHT, HTH, HTT, TTT, TTH, THT, THH}
Same result on all tosses = HHH, TTT

Question 11.
A die is thrown once. Find the probability of getting a number which
(i) is a prime number
(ii) lies between 2 and 6. [2]
Solution:
In throwing a die Total possible outcomes = 6
i.e., S = {1, 2, 3, 4, 5, 6}
Prime numbers 2, 3, 5

Numbers between 2 and 6 are 3, 4, 5
P (Numbers between 2 and 6) = =

Question 12.
Find c if the system of equations cx + 3y + (3 – c) = 0, 12x + cy – c = 0 has infinitely many solutions? [2]
Solution:
The given equations are
cx + 3y + (3 – c) = 0
and 12x + cy – c = 0
On comparing with equation a₁x + b₁y + c₁ = 0
and equation a₂x + b₂y + c₂ = 0, we get
a₁ = c, b₁ = 3, c₁ = 3 – c
and a₂ = 12, b₂ = c, c₂ = -c
For infinitely many solutions

Section – C

Question 13.
Prove that √2 is an irrational number. [3]
Solution:
Let √2 is a rational number.
So, √2 = a/b where a and b are coprime integers and b ≠ 0
or √2 b = a
Squaring both sides, we get
2b² = a²
Therefore, 2 divides a²
or 2 divides a (from theorem)
Let a = 2c, for some integer c
From equation (i)
2b² = (2c)²
or 2b² = 4c²
or b² = 2c²
It means that 2 divides b² and 2 divides b
Therefore a and b have at least 2 as a common factor.
But this contradicts the fact that a and b are co-prime.
This contradiction is due to our wrong assumption that √2 is rational.
So, we conclude that √2 is irrational.
Hence Proved.

Question 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 to their product. [3]
Solution:
The given quadratic polynomial is x² – (k + 6) x + 2(2k – 1)
Comparing with ax² + bx + c, we get a = 1, b = -(k + 6) and c = 2(2k + 1)
Let the zeroes of the polynomial be α and β
we know that

According to question
Sum of zeroes = 1/2 of their product
α + β = 1/2 αβ
or k + 6 =1/2 × 2(2k – 1) [using equations (i) & (ii)]
or k + 6 = 2k – 1
k = 7

Question 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. [3]
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 present age of father be x years and sum of ages of his two children be y years
According to question
x = 3y …(i)
After 5 years
Father’s age = (x + 5) years
Sum of ages of two children = (y + 5 + 5) years = (y + 10) years
In 2nd case
According to question
x + 5 = 2 (y + 10)
or x + 5 = 2y + 20
or x – 2y = 15
or 3y – 2y = 15 (Using equations (i))
y = 15
Now from equation (i)
x = 3y (Put y = 15)
or x = 3 × 15
x = 45
So, Present age of father = 45 years.
OR
Let the fraction be
According to question =
or 3(x – 2) = y
or 3x – y = 6 …(i)
again, According to question
=
or 2x = y – 1
or 2x – y = -1 …(ii)
On solving equation (i) and (ii), we get
x = 7, y = 15
The required fraction is

Question 16.
Find the point on the y-axis which is equidistant from the points (5, -2) and (-3, 2). [3]
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:
We know that a point on the y-axis is of the form (0, y).
So, let the point P(0, y) be equidistant from A (5, -2) and B (-3, 2)
Then AP = BP
or AP² = BP²
or (5 – 0)² + (-2 – y)² = (-3 – 0)² + (2 – y)²
or 25 + 4 + y² + 4y = 9 + 4 + y² – 4y
8y = -16
y = -2
So, the required point is (0, -2)
OR
The line segment AB is trisected at the points P and Q and P is nearest to A
So, P divides AB in the ratio 1 : 2

P lies on the line 2x – y + k = 0
It will satisfy the equation.
On putting x = 3 and y = -2 in the given equation, we get
2(3) – (-2) + k = 0
6 + 2 + k = 0
k = -8
Hence, k = -8

Question 17.
Prove that (sin θ + cosec θ )² + (cos θ + sec θ)² = 7 + tan² θ + cot² θ. [3]
OR
Prove that (1 + cot A – cosec A) (1 + tan A + sec A) = 2.
Solution:
L.H.S. = (sin θ + cosec θ)² + (cos θ + sec θ)²
= sin² θ + cosec² θ + 2. sin θ. cosec θ + cos² θ + sec² θ + 2 cos θ sec θ.
(∵ (a + b)² = a² + b² + 2ab)

Question 18.
In Fig. PQ is a chord of length 8 cm of a circle of radius 5 cm and centre O. The tangents at P and Q intersect at point T. Find the length of TP. [3]

Solution:
Join OT, let it intersect PQ at the point R
Now, ΔTPQ is an isosceles triangle and TO is the angle bisector of ∠PTQ.
So, OT ⊥ PQ and therefore, OT bisects PQ
PR = RQ = 4 cm

Question 19.
In Fig. ∠ACB = 90° and CD ⊥ AB, prove that CD² = BD × AD. [3]

OR
If P and Q are the points on side CA and CB respectively of ΔABC, right-angled at C, prove that (AQ² + BP²) = (AB² + PQ²).
Solution:
Given, A ΔACB in which ∠ACB = 90° and CD ⊥ AB
To prove : CD² = BD × AD
Proof: In ΔADC and ΔACB
∠A = ∠A (common)
∠ADC = ∠ACB (90° each)
ΔADC ~ ΔACB (By AA rule)…(i)
Similarly,
ΔCDB ~ ΔACB (By AA rule)…(ii)
From equation (i) and (ii)
ΔADC ~ ΔCDB
=
(by the definition of similarity of triangles)
or CD² = AD . BD
or CD² = BD × AD
Hence Proved.
OR
Given, ABC is a right-angled triangle in which ∠C = 90°

To prove : AQ² + BP² = AB² + PQ²
Construction: Join AQ, PB and PQ
Proof: In ΔAQC, ∠C = 90°
AQ² = AC² + CQ² …(i) (Using Pythagoras theorem)
In ΔPBC, ∠C = 90°
BP² = BC² + CP² …(ii) (Using Pythagoras theorem)
Adding equation (i) and (ii)
AQ² + BP² = AC² + CQ² + BC² + CP² = AC² + BC² + CQ² + CP²
or AQ² + BP² = AB² + PQ²
Hence Proved.

Question 20.
Find the area of the shaded region in Fig. if ABCD is a rectangle with sides 8 cm and 6 cm and D is the centre of the circle. [3]
[Take π = 3.14]

Solution:
Given, ABCD is a rectangle with sides AB = 8 cm and BC = 6 cm
In ΔABC
AC² = 8² + 6² (By Pythagoras Theorem)
⇒ AC² = 64 + 36
⇒ AC² = 100
⇒ AC = 10 cm
The diagonal of the rectangle will be the diameter of the circle
radius of the circle = 10/2 = 5 cm
Area of shaded portion = Area of circle – Area of Rectangle
= πr² – l × b
= 3.14 × 5 × 5 – 8 × 6
= 78.50 – 48
= 30.50 cm2
Hence, Area of shaded portion = 30.5 cm²

Question 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? [3]
Solution:
Let b be the width and h be the depth of the canal
b = 6 m and h = 1.5 m
Water is flowing with a speed = 10 km/h = 10,000 m/h
Length of water flowing in 1 hr = 10,000 m
Length (l) of water flowing in 1/2 hr = 5,000 m
Volume of water flowing in 30 min. = l × b × h = 5000 × 6 × 1.5 m³
Let the area irrigated in 30 min (1/2 hr) be x m²
Volume of water required for irrigation = Volume of water flowing in 30 min.
x × 8/100 = 5000 × 6 × 1.5
x = 562500 m² = 56.25 hectares. (∵ 1 hectare = 10⁴ m²)
Hence, the canal will irrigate 56.25 hectares in 30 min.

Question 22.
Find the mode of the following frequency distribution. [3]

Solution:
The given frequency distribution table is

Here, the maximum class frequency is 16
Modal class = 30-40
lower limit (l) of modal class = 30
Class size (h) =10
Frequency (f₁) of the modal class = 16
Frequency (f₀) of preceding class = 10
Frequency (f₂) of succeeding class = 12

Section – D

Question 23.
Two water taps together can fill a tank in 1 hours. The tap with longer diameter takes 2 hours less than the tap with a smaller one to fill the tank separately. Find the time in which each tap can fill the tank separately. [4]
OR
A boat 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 the boat in still water.
Solution:
Let the tap A with longer diameter take x hours and the tap B with smaller diameter take (x + 2) hours to fill the tank.
Portion of tank filled by the tap A in 1 hr. =
and Portion of tank filled by the tap B in 1 hr. =
Portion of the tank filled by both taps in 1 hr. = + =
Time taken by both taps to fill the tank = 1 hrs = 15/8 hrs
Portion of the tank filled by both in 1 hr. = 8/15
According to question,
=
⇒ =
⇒ 15x + 15 = 4x² + 8x
⇒ 4x² – 7x – 15 = 0
⇒ 4x² – 12x + 5x – 15 = 0
⇒ 4x (x – 3) + 5 (x – 3 ) = 0
⇒ (4x + 5)(x – 3) =0
⇒ 4x + 5 = 0 or x – 3 = 0
⇒ x = -5/4 Since, time can not be negative hence, neglected this value is; x = 3
Hence, the time taken with longer diameter tap = 3 hours
and the time taken with smaller diameter tap = 5 hours.
OR
Let the speed of the boat in still water be x km/h and the speed of the stream be y km/h
Then the speed of the boat downstream = (x + y) km/h
and the speed of the boat upstream = (x – y) km/h
We know that,


On solving, we get x = 8 and y = 3
Hence, the speed of the boat in still water = 8 km/h
and the speed of the stream = 3 km/h

Question 24.
If the sum of the first four terms of an A.P. is 40 and that of first 14 terms is 280. Find the sum of the first n terms. [4]
Solution:
Given, S₄ = 40 and S₁₄ = 280
If a be the first term and d be the common difference of an A.P.

Question 25.

Solution:

Question 26.
A man in a boat rowing away from a lighthouse 100 m high takes 2 minutes to change the angle of elevation of the top of the lighthouse from 60° to 30°. Find the speed of the boat in metres per minute. [Use √3 = 1.732] [4]
Solution:
Let AB be the lighthouse C and D be the two positions of the boat, such that,
CD = x m and BC = y m

Question 27.
Construct a ∆ABC in which CA = 6 cm, AB = 5 cm and ∠BAC = 45°. Then construct a triangle whose sides are 3/5 of the corresponding sides of ∆ABC. [4]
Solution:
Steps of Construction are as follows:

1. Draw AB = 5 cm

2. At the point, A draw ∠BAX = 45°

3. From AX cut off AC = 6 cm

4. Join BC, ∆ABC is formed with given data.

5. Draw AY making an acute angle with AB as shown in the figure.
   

6. Draw 5 arcs P₁, P₂, P₃, P₄, and P₅ with equal intervals.

7. Join BP₅.

8. Draw P₃B’ || P₅B meeting AB at B’.

9. From B’, draw B’C’ || BC meeting AC at C’
   ∆AB’C’ ~ ∆ABC
   Hence ∆AB’C’ is the required triangle.

Question 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) [4]
Solution:
Let r and R be the radii of the top and the bottom circular ends of the bucket respectively.
Let h be the height of the bucket.
R = 20 cm and r = 12 cm
Capacity of the bucket = 12308.8 cm³
Volume of bucket (frustum)

Thus, the height of the bucket is 15 cm.
The area of the metal sheet used in making the bucket = CSA of bucket + area of the circular bottom

Area of metal sheet used = π[(R + r)l + r²]
= 3.14 [(20 + 12) × 17 + 12²]
= 3.14 [32 × 17 + 144]
= 3.14 [544 + 144]
= 3.14 × 688
= 2160.32 cm²

Question 29.
Prove that in a right-angle triangle, the square of the hypotenuse is equal to the sum of squares of the other two sides. [4]
Solution:
Given, A ΔABC right angled at B.
To prove : AC² = AB² + BC²

Construction : Draw BD ⊥ AC
Proof: In ΔADB and ΔABC
∠A = ∠A (common)
∠ADB = ∠ABC (90° each)
ΔADB ~ ΔABC (By AA rule)
So, (sides are proportional)
or AB² = AD.AC …(i)
Also, In ΔBDC and ΔABC
∠C = ∠C (common)
∠BDC = ∠ABC (90° each)
ΔBDC ~ ΔABC
So,
or BC² = CD.AC …(ii)
Adding equation (i) and (ii), we get
AB² + BC² = AD.AC + CD.AC
= AC (AD + CD)
= AC × AC
= AC²
or AC² = AB² + BC²
Hence Proved.

Question 30.
If the median of the following frequency distribution is 32.5. Find the values of f₁ and f₂. [4]

OR
The marks obtained by 100 students of a class in an examination are given below.

Draw ‘a less than’ type cumulative frequency curves (ogive). Hence find the median.
Solution:
Median = 32.5




To draw a less than ogive, we mark the upper-class limits of the class intervals on the x-axis and their c.f. on the y-axis by taking a convenient scale.
Here, n = 100 ⇒ n/2 = 50
To get median from graph From 50, we draw a perpendicular to the curve then from that point draw again perpendicular to the x-axis.
The point where this perpendicular meet on the x-axis will be the median.
Median = 29