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E–
Universal Tutorials – X CBSE
Question Paper Universal Tutorials – X CBSE (2018–19) – Maths Page 1 of 3
Full Test 01 (Set 1): Maths Date: ___________
Recommended Time: 3 hours Max. Marks: 80
General Instructions:
1) All questions are compulsory.
2) The question paper consists of 30 questions divided into four sections A, B, C and D.
3) Section A comprises 6 questions of 1 mark each.
Section B comprises 6 questions of 2 marks each.
Section C comprises 10 questions of 3 marks each.
Section D comprises 8 questions of 4 marks each.
4) There is no overall choice. However, an internal choice has been provided in two questions of 1 mark each, two questions of two 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.
5) Use of calculators is not permitted.
Section A: (Question numbers 1 to 6 carry 1 mark each)
1) On applying Euclid’s division lemma to a positive integer a and b = 4. There exist integers q and r such that a = 4q + r. Write the condition which ‘r’ must satisfy. [1]
2) If ax2 + bx + c = 0 has equal roots, what is the value of c? [1]
OR Find the values of k for which the quadratic equation 9x2 – 3kx + k = 0 has equal roots.
3) The nth term of a progression is 3n + 1. Find the 6th term. [1]
4) In fig. if DE || BC, find the value of x. [1]
5) Find the co–ordinates of a point A, where AB is the diameter of a circle whose centre is (2, –3) and B is (1, 4). [1]
6) The ratio of the length of a rod and its shadow is 1 : . Find the angle of elevation of the
sun. [1]
OR If 2 sin = 1, find the value of sec2 – cosec2 .
Section B: (Question numbers 7 to 12 carry 2 marks each)
7) Express 2717 as a product of its prime factors. [2]
OR Determine the value of p and q so that the prime factorization of 2520 is expressible as
23 3P q 7.
8) On comparing the ratios 2
1
a
a,
2
1
b
b and
2
1
c
c, find out whether the following pair of linear
equations are consistent, or inconsistent. 2
3x +
3
5y = 7; 9x – 10y = 14. [2]
9) Is –150 a term of the A.P: 11, 8, 5, 2, …… justify your answer. [2]
OR If 5
4, a, 2 are three consecutive terms of an A.P., then find the value of a.
10) If P(–1, 3), Q(1, –1) and R(5, 1) are the vertices of a triangle PQR, then find the length of the median through the vertex P. [2]
11) Two dice are thrown simultaneously. Find the probability of getting: [2]
a) A total of atleast 10 b) A doublet of even number
12) All kings, queens and aces are removed from a pack of 52 cards. The remaining cards are well shuffled and then a card is drawn from it. Find the probability that the drawn card is:
3
4 cm E
A
B C
D
3 cm
2 cm
x
E– 2
Question Paper Universal Tutorials – X CBSE (2018–19) – Maths Page 2 of 3
i) a black face card ii) a red card [2]
Section C: (Question numbers 13 to 22 carry 3 marks each)
13) Use Euclid’s Division Lemma to show that the cube of any positive integer is either of the form 9m, 9m + 1 or 9m + 8. [3]
14) Find the zeroes of the quadratic polynomial 6x2 – 3 – 7x and verify the relationship between the zeroes and the coefficients of the polynomial. [3]
15) Draw the graphs of the equations x = 3, x = 5 and 2x – y – 4 = 0. Also find the area of the quadrilateral formed by the lines and the x-axis. [3]
16) ABCD is a trapezium in which AB || CD.
he diagonals AC and BD intersect at O. Prove that [3]
i) AOB ~ COD
ii) If OA = 6 cm, OC = 8 cm. Find:COD)( Area
AOB)( Area
OR In Fig. ABC and DBC are on the same base BC.
If AD intersects BC at O, prove that )DBC(Ar
)ABC(Ar
=
DO
AO.
17) The point P(7, b) lies on the line joining A(–5, 2) and B(3, 6). Find the ratio and
hence find b. [3]
OR If A and B are (–2, –2) and (2, –4) respectively, find the coordinate of a point P on line
segment AB such that AP = AB.
18) From a point P, two tangents PA and PB are drawn to a circle with centre O.
If OP = diameter of the circle show that APB is equilateral. [3]
19) If sec + tan = p, show that sec – tan = p
1. Hence, find the values of cos and sin .
[3]
OR Evaluate the following:
75? tan . 60? tan 15? tan
52?cosec 38? cos3
32? sin
58? cos2 .
20) If a wire is bent into the shape of a square, then the area of the square is 81 cm2. When the same wire is bent into a semicircular shape, find the area of the semicircle. [3]
21) A wooden article was made by scooping out of a hemisphere from
each side of a solid cylinder, as shown in fig. If the height of the
cylinder is 10cm, and its base is of radius 3.5cm, find the total surface
area of the article. [3]
OR The radii of the circular ends of a frustum are 33 cm and 27 cm, its slant height is 10 cm. Find its volume and whole surface area.
)(
)(
PBl
APl
7
3
O
A B
C D
E– 3
Question Paper Universal Tutorials – X CBSE (2018–19) – Maths Page 3 of 3
22) The following table shows the marks obtained by 100 students of Class X in a school during a particular academic session. Find the mode of this distribution. [3]
Marks No. of Students
Less than 10 7
Less than 20 21
Less than 30 34
Less than 40 46
Less than 50 66
Less than 60 77
Less than 70 92
Less than 80 100
Section D: (Question numbers 23 to 30 carry 4 marks each)
23) Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an
angle of 60. Justify. [4]
24) If the roots of the quadratic equation (a – b)x2 + (b – c)x + (c – a) = 0, a b are equal, then prove that b + c = 2a. [4]
OR Solve for x :
1x
3x7
3x
1x2 = 5; given that x –3, x 1.
25) Two cars start together in the same direction at the same place. The first goes at uniform speed of 10 km/h. The second goes at a speed of 8 km/h in the first hour and increases
the speed by km each succeeding hour. After how many hours will the second car
overtake the first car if both cars go nonstop? [4]
26) A man in a boat rowing away from a light house 100m high takes 3 minutes to change the
angle of elevation of the top of light house from 60 to 45. Find the speed of the boat in
m/min. ( = 1.73) [4]
27) The ratio of the areas of two similar triangles is equal to the square ratio of their corresponding sides. Prove. [4]
OR) In figure, XY || AC and XY divides triangular region
ABC into two parts of equal areas. Determine AB
AX.
28) Prove that = 1 + tan + cot . [4]
29) A juice seller serves his customers using a glass as shown in Figure.
The inner diameter of the cylindrical glass is 5 cm, but the bottom of the
glass has a hemispherical portion raised which reduces the capacity of
the glass. If the height of the glass is 10 cm, find the apparent capacity
of the glass and its actual capacity. What values does the juice seller
depict by using such a glass? (Use = ) [4]
30) Draw the more than cumulative frequency curve of the following data: [4]
Rent (Rs.) 15-25 25-35 35-45 45-55 55-65 65-75 75-85 85-95
No. of houses 5 10 20 35 30 20 13 7
OR) If the median of the distribution given below is 28.5, find the values of x and y.
Class interval 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 Total
Frequency 5 x 20 15 y 5 60
2
1
3
tan1
cot
cot1
tan
7
22
A
C Y B
X
E–
Universal Tutorials – X CBSE
Model Answer (1st page) Universal Tutorials – X CBSE (2018–19) – Maths Page 4 of 17
(1st few pages form part of question paper)
Full Test 01 (Set 1): Maths (Model Answer) Date: ___________
Recommended Time: 3 hours Max. Marks: 80
Section A: (Question numbers 1 to 6 carry 1 mark each)
1) On applying Euclid’s division lemma to a positive integer a and b = 4. There exist integers q and r such that a = 4q + r. Write the condition which ‘r’ must satisfy. [1]
Ans: 0 r < 4 [1]
2) If ax2 + bx + c = 0 has equal roots, what is the value of c? [1]
Ans: Since the quadratic equation has equal roots, then
b2 – 4ac = 0 4ac = b2 c = a4
b2
. [1]
OR Find the values of k for which the quadratic equation 9x2 – 3kx + k = 0 has equal roots.
Sol: a = 9, b = – 3k, c = k
Since roots of the equation are equal
b2 – 4ac = 0
(–3k)2 – (4 9 k) = 0
9k2 – 36k = 0
k2 – 4k = 0
k(k – 4) = 0
k = 0 or k = 4
Since k = 0 is not possible for the equation
k = 4 [1]
3) The nth term of a progression is 3n + 1. Find the 6th term. [1]
Sol: an = 3n + 1 To find: 6th term, put n = 6
a6 = 3 6 + 1 = 19 a6 = 19. [1]
4) In fig. if DE || BC, find the value of x. [1]
Sol: ADE ABC [AA similarity]
BC
DE
AB
AD
x
4
5
2
2x = 20 x = 10 cm [1]
5) Find the co–ordinates of a point A, where AB is the diameter of a circle whose centre is (2, –3) and B is (1, 4). [1]
Sol: Let A be (x1, y1) Centre is the midpoint of AB
2 = and –3 =
i.e. x1 + 1 = 4 x1 = 3 y1 + 4 = –6 y1 = –10
A is (3, –10) [1]
6) The ratio of the length of a rod and its shadow is 1 : . Find the angle of elevation of the
sun. [1]
Sol: tan = = 30 Angle of elevation of sun is 30. [1]
OR If 2 sin = 1, find the value of sec2 – cosec2 .
Sol: Given, 2 sin = 1
sin = 2
1 = sin 45º
= 45º [½]
Now sec2 – cosec2 = sec2 45º – cosec2 45º [½]
2
1x1
2
41 y
3
3
1
4 cm E
A
B C
D
3 cm
2 cm
x
E– 5
Model Answer Universal Tutorials – X CBSE (2018–19) – Maths Page 5 of 17
= ( 2 )2 – ( 2 )2
= 2 – 2
= 0
Section B: (Question numbers 7 to 12 carry 2 marks each)
7) Express 2717 as a product of its prime factors. [2]
Ans:
[1]
2717 = 11 13 19 [1]
OR Determine the value of p and q so that the prime factorization of 2520 is expressible as
23 3P q 7.
Sol: Prime factorization of 2520 is given by.
2520 = 23 32 5 7 [1]
Given that 2520 = 23 3P q 7,
On comparing both the factorization.
We get, p = 2 and q = 5 [1]
8) On comparing the ratios 2
1
a
a,
2
1
b
b and
2
1
c
c, find out whether the following pair of linear
equations are consistent, or inconsistent. 2
3x +
3
5y = 7; 9x – 10y = 14. [2]
Sol: x + y = 7; 9x – 10y = 14 6
y10x9 = 7,
9x + 10y = 42, and 9x – 10y = 14
9x + 10y – 42 = 0 and 9x – 10y – 14 = 0.
a1 = 9, b1 = 10, c1 = –42 a2 = 9, b2 = –10, c2 = –14. [1]
2
1
a
a =
9
9 = 1,
2
1
b
b =
10
10
= –1
2
1
a
a
2
1
b
b unique solution (consistent) [1]
9) Is –150 a term of the A.P: 11, 8, 5, 2, …… justify your answer. [2]
Sol.: Here, the given A.P. is 11, 8, 5, 2, ……
Then, T2 – T1 = 8 – 11 = –3 d = – 3 Also, a = 11
Let, – 150 be a term of a given A.P. say, nth
Using the General Term Formula, we have Tn = a + (n – 1)d [½]
–150 = 11 + (n – 1) (–3) –150 = 11 – 3n + 3
–150 = –3n + 14 3n = 150 + 14 = 164 n = 3
164 [1]
Which is a rational number. So, our assumption is wrong since n should be a positive integer.
Hence, –150 is not a term of the given A.P. [½]
OR If 5
4, a, 2 are three consecutive terms of an A.P., then find the value of a.
Sol: Since 5
4, a, 2 are the three consecutive terms of an A.P., then
d = a – 5
4 = 2 – a [1]
2a = 2 + 5
4
10a = 10 + 4
10a = 14
2
3
3
5
11 2717 13 247 19
E– 6
Model Answer Universal Tutorials – X CBSE (2018–19) – Maths Page 6 of 17
a = 5
7 [1]
10) If P(–1, 3), Q(1, –1) and R(5, 1) are the vertices of a triangle PQR, then find the length of the median through the vertex P. [2]
Sol: P, Q, R are the vertices of a triangle PQR. PM is the median.
To find the length of median PM.
As M is the mid point of QR
x coordinate of point M = 2
xx 21 = 2
51 =
2
6 = 3
y coordinate of point M = 2
yy 21 = 2
11 =
2
0 = 0
M(x, y) = (3, 0) [1]
Hence, applying the distance formula
PM2 = (–1 –3)2 + (3 – 0)2 = (–4)2 + (3)2 = 16 + 9 = 25.
PM = 25 = 5units. [1]
11) Two dice are thrown simultaneously. Find the probability of getting: [2]
a) A total of atleast 10 b) A doublet of even number
Sol: a) A total of at least 10 = (4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6) = 6
P (a total of at least 10) = 6
1
36
6 [1]
b) A doublet of even numbers = (2, 2) (4, 4), (6, 6) = 3
P (a doublet of even numbers) = 12
1
36
3 [1]
12) All kings, queens and aces are removed from a pack of 52 cards. The remaining cards are well shuffled and then a card is drawn from it. Find the probability that the drawn card is:
i) a black face card ii) a red card [2]
Sol: i) No. of cards removed = 4 kings + 4 queens + 4 aces = 12
No. of cards left = 52 – 12 = 40 = sample space
No. of Face cards: 4 kings + 4 queens + 4 jacks = 12
But 4 kings and 4 queens are removed.
No. of face cards left = 4 jacks.
Out of 4 jacks, only 2 jacks (jacks of club and spade) are black face cards.
P (a black face card) = 20
1
40
2 [1]
ii) Out of 12 cards removed, there are 6 red and 6 black cards (2 each of kings, queens and aces)
Hence no. of red cards left = 26 – 6 = 20 [ Total no. of red cards = 26]
P (red cards) = 2
1
40
20 [1]
Section C: (Question numbers 13 to 22 carry 3 marks each)
13) Use Euclid’s Division Lemma to show that the cube of any positive integer is either of the form 9m, 9m + 1 or 9m + 8. [3]
Sol: Let a is any positive integer and b = 3
By Euclid’s division Algorithm,
a = bq + r where 0 r < b.
a = 3q + r where 0 r < 3. i.e. r = 0, 1, 2
a = 3q or 3q + 1 or 3q + 2 [1]
when a = 3q
a3 = 27q3 = 9(3q3) = 9m where m = 3q3
when a = 3q + 1
a3 = (3q + 1)3 = 27q3 + 27q2 + 9q + 1
= 9 (3q3 + 3q2 + q) + 1 = 9m + 1 where m = 3q3 + 3q2+q [1]
when a = 3q + 2
P(–1,3)
Q(1, –1) M R(5,1)
E– 7
Model Answer Universal Tutorials – X CBSE (2018–19) – Maths Page 7 of 17
a3 = (3q + 2)3 = 27q3+ 3.3q.2 (3q + 2) + 8 = 27q3 + 54q2 + 36q + 8
= 9(3q3 + 6q2 + 4q) + 8 = 9m + 8 where m = 3q3 + 6q2 + 4q [1]
cube of any positive integer is of the from 9m, 9m + 1 or 9m + 8
14) Find the zeroes of the quadratic polynomial 6x2 – 3 – 7x and verify the relationship between the zeroes and the coefficients of the polynomial. [3]
Sol: Let one zero be and the other zero be .
p(x) = 6x2 – 7x – 3 = 6x2 – 9x + 2x – 3 = 3x(2x – 3) + 1(2x – 3) = (3x + 1) (2x – 3)
For p(x) = 0, 3x + 1 = 0 or 2x – 3 = 0
x = 3
1 or x =
2
3 =
3
1, =
2
3 [1]
Sum of zeroes = + = 6
7
2
3
3
1
… (1)
Here a = 6, b = –7, c = –3
Also, sum of zeroes = 6
)7(
a
b=
6
7 … (2) (1) = (2) [½]
Now, product of zeroes = = 2
1
2
3
3
1 … (3)
Also, Product of zeroes = 2
1
6
3
a
c
… (4) [1]
(3) = (4) Hence verified. [½]
15) Draw the graphs of the equations x = 3, x = 5 and 2x – y – 4 = 0. Also find the area of the quadrilateral formed by the lines and the x-axis. [3]
Sol: 2x – y = 4
x 2 3 5
y 0 2 6
Quadrilateral is like trapezium whose parallel sides
are 2 units and 6 units. Distance between parallel
sides is 2 units. So, area of trapezium
= 2
1 (Sum of parallel sides)
(Distance between parallel sides) [Graph 2]
= 2
1 (2 + 6) 2 = 8 sq. units [1]
16) ABCD is a trapezium in which AB || CD.
he diagonals AC and BD intersect at O. Prove that [3]
i) AOB ~ COD
ii) If OA = 6 cm, OC = 8 cm. Find:COD)( Area
AOB)( Area
Sol: ABCD is a trapezium AB || CD
Diagonals AC and BD intersect at O.
OA = 6 cm, OC = 8 cm.
i) In ΔAOB and ΔCOD.
AOB = COD [V.O.A] [fig ½]
OAB = OCD [Alt. angles]
Δ AOB ∼ ΔCOD [By AA similarity] [1]
ii) ΔAOB ∼ ΔCOD
COD)( Area
AOB)( Area
=
16
9
64
36
8
6
OC
OA
OC
OA22
2
2
[1½]
x
y
–1
1 2 3 4
(5, 6)
(3, 2)
1
2
3
4
5
6
(2, 0)
x=5 x=3
2x-y=4
0 5 6
O
A B
C D
O
A B
C D
M
N
E– 8
Model Answer Universal Tutorials – X CBSE (2018–19) – Maths Page 8 of 17
OR In Fig. ABC and DBC are on the same base BC.
If AD intersects BC at O, prove that )DBC(Ar
)ABC(Ar
=
DO
AO.
Sol: Given: Two triangles ABC and DBC in which
i) The base BC is common
ii) AD and BC intersect at O
To prove: )DBC(ar
)ABC(ar
=
DO
AO
Construction:
i) Through A, draw AM BC ii) Through D, draw DN BC [fig ½]
Proof: In triangles AMO and DNO, we have
AMO= DNO [By construction, each 90]
and AOM = DON [Vert. Opp. s]
By ‘AA’ Similarity Criteria
AMO DNO [1]
Their corresponding sides are proportional.
i.e. DO
AO =
DN
AM [CPST] … (1) [½]
Now, )DBC(ar
)ABC(ar
=
DNBC2
1
AMBC2
1
= DN
AM [ (Area of ) =
2
1 Base Altitude]
Hence proved = DO
AO [From (1)] [1]
17) The point P(7, b) lies on the line joining A(–5, 2) and B(3, 6). Find the ratio and
hence find b. [3]
Sol: Let P divide the line AB in the ratio k : 1 [where k = ]
x coordinate of P 7 = 7 =
7(k + 1) = 3k – 5 7k + 7 = 3k – 5 7k – 3k = – 5 – 7 = –12 [½]
4k = –12 k = – k = – 3 …. (i) [1]
As k is negative, point P divides AP, externally as shown where =
Also, y coordinate of P
b = b = = [½]
But k = – 3 [from (i) above]
b = = = = 8 k = – 3 and b = 8 [1]
OR If A and B are (–2, –2) and (2, –4) respectively, find the coordinate of a point P on line
segment AB such that AP = AB.
Sol: AP = AB 7AP = 3AB = 3(AP + PB)
7AP = 3AP + 3PB 7AP – 3AP = 3PB
4AP = 3PB m = 3, n = 4 [1]
)(
)(
PBl
APl
n
m
112
k
xkx
1
)5()3(
k
k
4
12
PB
AP
1
3
112
k
yky
1
2)6(
k
k
1
26
k
k
13
2)3(6
2
218
2
16
7
3
7
3
4
3
PB
AP
A P B
E– 9
Model Answer Universal Tutorials – X CBSE (2018–19) – Maths Page 9 of 17
p(x) = [1]
p(y) = [1]
p(x, y) = .
18) From a point P, two tangents PA and PB are drawn to a circle with centre O.
If OP = diameter of the circle show that APB is equilateral. [3]
Sol: PA and PB are tangents
PA = PB, Also, OA = OB = r
And OAP = OBP = 90º [radius tangent relation]
In OAP and OBP
OA = OB = r
PA = PB [Equal tangents]
PO = PO [Common]
OAP OBP OPA = OPB = [CPCT] [1]
Now in right angle OAP, sin =
sin = [ OP = diameter = 2r given]
sin = = sin 30º = 30º.
APB = + = 2 = 2 30º = 60º. [1]
Also, PA = PB PAB = PBA = x [Angles opposite to equal sides are equal]
In PAB: x + x + 2 = 180º [Angle sum property]
2x + 2 30 = 180 2x = 180 – 60 = 120
x = = 60º PAB = PBA = APB = 60º
APB is an equilateral triangle. [1]
19) If sec + tan = p, show that sec – tan = p
1. Hence, find the values of cos and sin .
[3]
Sol: sec + tan = p (i)
p
1 =
tan sec
) tan (sec
tan sec
1 [½]
p
1 =
tan sec
tan sec
tan sec22
i.e p
1= sec – tan (ii) [½]
Adding (i) and (ii)
We get sec = p2
1p
p
1p
2
1 2
Subtracting eq (ii) from eq (i) we get
tan = p2
1p
p
1p
2
1 2
cos = sec
1 [1]
cos = 1p
2p2
and tan = p2
1p2
p2
1p
cos
sin 2
7
2
7
86
43
)2(4)2(3
nm
nxmx 12
7
20
7
812
43
)2(4)4(3
nm
nymy 12
7
20,
7
2
OP
OA
hypotenuse
opposite
r2
r
2
1
2
120
P
A
B
O
r
r
x
x
E– 10
Model Answer Universal Tutorials – X CBSE (2018–19) – Maths Page 10 of 17
sin = cos . p2
1p2 =
1p
1p
p2
1p
1p
p22
22
2
[1]
OR Evaluate the following:
75? tan . 60? tan 15? tan
52?cosec 38? cos3
32? sin
58? cos2 .
Sol:
75? tan . 60? tan 15? tan
52?cosec 38? cos3
32? sin
58? cos2
=
)5?1 (90? tan . 15? tan 60? tan
)38? (90?cosec 38? cos3
58? (90? sin
58? cos2 [1]
= 13
132
)5?1 cot . 15? tan 3
38?sec 38? cos3
58? cos
58? cos2
= 2 – 1 = 1 [2]
[ tan 60º = , cos sec = 1, tan cot = 1]
20) If a wire is bent into the shape of a square, then the area of the square is 81 cm2. When the same wire is bent into a semicircular shape, find the area of the semicircle. [3]
Sol: Side of square = = 9 cm
Perimeter of square = 4 9 = 36 = Perimeter of semicircle. [1]
r + 2r = 36 or r = 36 r = = 7 cm [1]
Area of semicircle = r2 = 7 7 = 77 cm2 [1]
21) A wooden article was made by scooping out of a hemisphere from
each side of a solid cylinder, as shown in fig. If the height of the
cylinder is 10cm, and its base is of radius 3.5cm, find the total surface
area of the article. [3]
Sol: Total SA (TSA) of the article = curved SA (CSA) of the cylinder
+ CSA of the upper HS + CSA of the lower HS
TSA of article = 2rh + 2r2 + 2r2 [1]
= 2rh + 4r2 = 2r (h + 2r)
= 2 3.5 (10 + 2 3.5)
[1]
= 2 3.5 17 = 374 cm2
TSA of article = 374 cm2 [1]
OR The radii of the circular ends of a frustum are 33 cm and 27 cm, its slant height is 10 cm. Find its volume and whole surface area.
Sol: Given: r1 = 33 cm, r2 = 27 cm, l = 10 cm.
l = l2 = h2 + (r1 – r2)2 (10)2 = h2 + (33 – 27)2
100 = h2 + 62 = h2 + 36. h2 = 100 – 36 = 64 h = = 8 cm.
V = h (r + r1r2 + r ) = h [½]
V = 8 = 8 (62 + 3 33 27)
= 8 (36 + 2673) = 8 2709 = 22,704 cm3 [1]
Volume = 22,704 cm3.
Whole surface area = curved surface area + upper circular area + bottom circular area.
= (r1 + r2) l + r + r = [½]
3
81
2
7
22
36
736
2
1
2
1
7
22
7
22
7
22
2212 rrh
64
3
1 21
22
3
1 212
21 rr3rr
3
1
7
22 2733327332
3
1
7
22
3
1
7
22
3
1
7
22
21
22 2
22121 rrl rr
10
r = 3.5
E– 11
Model Answer Universal Tutorials – X CBSE (2018–19) – Maths Page 11 of 17
= = [600 + 1089 + 729]
= 2418 = 7599.42 cm2. [1]
22) The following table shows the marks obtained by 100 students of Class X in a school during a particular academic session. Find the mode of this distribution. [3]
Marks No. of Students
Less than 10 7
Less than 20 21
Less than 30 34
Less than 40 46
Less than 50 66
Less than 60 77
Less than 70 92
Less than 80 100
Sol: The given data can be written as:
C.I. (Marks) c.f fi. (No. of Students)
0–10 7 7
10–20 21 14 (21 – 7)
20–30 34 13 (34 – 21)
30–40 46 f0 12 (46 – 34)
M.C 40–50 66 f1 20 (66 – 46)
50–60 77 f2 11 (77 – 66)
60–70 92 15 (92 – 77)
70–80 100 8 (100 – 92)
[Table 1½]
Maximum frequency = 20
Modal class (M. C) is 40 – 50
l (lower limit of M.C) = 40, f1 (frequency of M.C) = 20, f0 (frequency of preceding M.C) = 12 f2 (frequency of succeeding M.C)= 11, h (width of M.C) = 10
Mode = l +
201
01
fff2
ff h [½]
Mode = 40 +
10
1112202
1220 = 40 +
17
80 = 40 + 4.7
= 44.7 marks [1]
Section D: (Question numbers 23 to 30 carry 4 marks each)
23) Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an
angle of 60. Justify. [4]
Sol: [Steps of Construction:
Step I: Draw a circle with centre O and radius 5 cm.
Step II: Draw any diameter AOB.
Step III: Draw a radius OC such that BOC = 60
Step IV: At C, we draw CM OC and at A, we draw AN OA.
Step V: Let the two perpendiculars intersect each other at P.
Then, PA and PC are required tangents.]
Justification:
Since OA is the radius, so PA has to be a tangent to the circle.
Similarly, PC is also tangent to the circle. [Const. 3]
APC = 360 – (OAP + OCP + AOC)
= 360 – (90 + 90 + 120) = 360 – 300 = 60
7
22 22)27(33102733
7
22
7
22
A B
C
P
M N
O
600
E– 12
Model Answer Universal Tutorials – X CBSE (2018–19) – Maths Page 12 of 17
Hence, tangents PA and PC are inclined to each other at an angle of 60. [Justification 1]
24) If the roots of the quadratic equation (a – b)x2 + (b – c)x + (c – a) = 0, a b are equal, then prove that b + c = 2a. [4]
Sol: (a – b)x2 + (b – c)x + (c – a) = 0 A = a – b, B = b – c, C = c – a
As the roots are equal; D = 0 i.e. B2 – 4AC = 0 [1]
(b – c)2 – 4(a – b) (c – a) = 0 b2 + c2 – 2bc – 4ac + 4a2 + 4bc – 4ab = 0 [1]
4a2 + b2 + c2 – 4ab + 2bc – 4ac = 0 (2a – b – c)2 = 0 [1]
b + c = 2a [1]
OR Solve for x :
1x
3x7
3x
1x2 = 5; given that x –3, x 1.
Sol:
1x
3x7
3x
1x2 = 5
Let y3x
1x
y
1
1x
3x
Equation: 2y – 5
y
7
5y
7y2 2
2y2 – 7 = 5y [1]
2y2 – 5y – 7 = 0 2y2 + 2y – 7y – 7 = 0
2y (y + 1) – 7 (y + 1) = 0 (y + 1) (2y – 7)
y = – 1 or 2y – 7 = 0 y = –1 or 2y = 7
y = 2
7 [1]
Case – I: When y = – 1,
13x
1x
x – 1 = – 1 (x + 3) x – 1 = – x – 3 x + x = – 3 + 1
2x = – 2 x = 12
2 [1]
Case – II: When y = 2
7
2
7
3x
1x
7 (x + 3) = 2 (x – 1) 7x + 21 = 2x – 2 7x – 2x = – 2 – 21
5x = –23 x = 5
23 x = –1 or
5
23 [1]
25) Two cars start together in the same direction at the same place. The first goes at uniform speed of 10 km/h. The second goes at a speed of 8 km/h in the first hour and increases
the speed by km each succeeding hour. After how many hours will the second car
overtake the first car if both cars go nonstop? [4]
Sol: Let the second car overtake the first car after n hours.
Distance covered by the 1st car in each hour = 10 km.
Distances covered by 1st car in n hours = 10n km. [½]
Increase in speed of the 2nd car per hour = km
Distance covered in 1st hour, 2nd hour, 3rd hour are 8, 8 , 9,…
This is an A.P. with first term, a = 8 and common difference, d = [½]
Distance covered by the 2nd car in n hours is given by Sn = 8 + 8 + 9 + …
Using Sum Formula of an A.P., we have Sn = [2a + (n 1)d]
2
1
2
1
2
1
2
1
2
1
2
n
E– 13
Model Answer Universal Tutorials – X CBSE (2018–19) – Maths Page 13 of 17
Sn = = = km [1]
At the time of overtaking, both cars have travelled the same distance
= 10n n2 + 31n = 40n n2 9n = 0 n(n 9) = 0 [½]
Either n = 0 or n = 9 [1]
Rejecting n = 0, which corresponds to the time, of their we get n = 9
Hence, the second car overtakes the first car after 9 hours [½]
26) A man in a boat rowing away from a light house 100m high takes 3 minutes to change the
angle of elevation of the top of light house from 60 to 45. Find the speed of the boat in
m/min. ( = 1.73) [4]
Sol: Let AB be the light house and C, D be two position of the boat.
In ABD, = cot 45 = 1 BD = 100 [1]
In ABC, = cot 60 = BC = = [1]
CD = BD – BC = 100 – = = m [1]
Time = 3 min
Speed = = 3 = = m/min = 14 m/min. [1]
27) The ratio of the areas of two similar triangles is equal to the square ratio of their corresponding sides. Prove. [4]
Sol: Given: ABC ~ PQR
To Prove:
Construction: Draw AD BC, PS QR.
Proof: ar(ABC) = BC AD
ar(PQR) = QR PS [Area of triangle = base height]
––– (i) [1]
In ADB and PSQ,
B = Q [Given ABC ~ PQR]
ADB = PSQ [By construction, each 90] [1]
ADB PSQ [AA corollary]
[Corresponding sides of similar triangles are proportional]
But [Corresponding sides of ABC PQR]
–– (ii)
[i and ii] [1]
[As ABC PQR]
Hence, = [1]
2
1)1()8(2
2n
n
2
31
2
nn
4
312 nn
4
312 nn
3
AB
BD
100
BD
AB
BC
100
BC
3
1
3
100
3
3100
3
3100
3
173300
3
127
T
D
3
127
3
127
3
1
9
127
9
1
2
2
2
2
2
2)
PR
AC
QR
BC
PQ
AB
PQR)( Area
ABC( Area
2
1
2
1
2
1
PSQR
ADBC
PSQR
ADBC
PQRar
ABCar
2
12
1
)(
)(
PQ
AB
PS
AD
QR
BC
PQ
AB
QR
BC
PS
AD
2
2
)(
)(
QR
BC
QRQR
BCBC
PQRar
ABCar
PR
AC
QR
BC
PQ
AB
PQRar
ABCar
2
2
2
2
2
2
PR
AC
QR
BC
PQ
AB
A
B D C
45 60
100m
A
B CD
P
Q R
S
E– 14
Model Answer Universal Tutorials – X CBSE (2018–19) – Maths Page 14 of 17
E– 15
Model Answer Universal Tutorials – X CBSE (2018–19) – Maths Page 15 of 17
OR) In figure, XY || AC and XY divides triangular region
ABC into two parts of equal areas. Determine AB
AX.
Sol: given XY || AC.
Area (trapezium AXYC) = area ∆BXY, to determine AB
AX
Now, area (Trapezium AXYC) = area (∆BXY)
BXY)(
AXYC) (trapezium
area
area =
1
1
BXY
AXYC) (trapezium
area
area + 1 = 1 + 1 (adding ‘1’ both sides)
BXY
BXY of area AXYC) (trapezium
area
area = 2
)(
)(
BXYarea
BACarea
= 2 … (1) [1]
In ∆BAC and ∆BXY, XY || AC (given)
BXY = BAC (corresponding angles) and BYX = BCA (corresponding angles)
∆BXY ~ ∆BAC (By AA similarity)
)(
)(
BXYarea
BACarea
=
2
2
BX
AB (Areas of 2 similar triangles are proportional to square of corresponding
sides)
From (1) and (2), 2
2
BX
AB = 2 [1]
2
2
AB
BX =
2
1
AB
BX=
2
1 =
2
1 [½]
AB
AXAB =
2
1 [ BX = AB – AX]
AB
AB –
AB
AX =
2
1 1 –
AB
AX =
2
1 [½]
AB
AX = 1 –
2
1 =
2
12 =
22
122
AB
AX =
2
22 [½+½]
28) Prove that = 1 + tan + cot . [4]
Sol: LHS = = + = + [1]
= = [1]
= = [1]
= =
= tan + cot + 1 = RHS [1]
tan1
cot
cot1
tan
tan1
cot
cot1
tan
sincos
cossin
1
cossin
sincos
1
sincossin
cossin
cossincos
sincos
)sin(cossin
cos
)cos(sincos
sin 22
)cos(sinsin
cos
)cos(sincos
sin 22
)cos(sincossin
cossin 33
)cos(sincossin
)coscossin)(sincos(sin 22
cossin
cossincossin 22
cossin
cossin
cossin
cos
cossin
sin 22
A
C Y B
X
A
C Y B
X
E– 16
Model Answer Universal Tutorials – X CBSE (2018–19) – Maths Page 16 of 17
29) A juice seller serves his customers using a glass as shown in Figure.
The inner diameter of the cylindrical glass is 5 cm, but the bottom of the
glass has a hemispherical portion raised which reduces the capacity of
the glass. If the height of the glass is 10 cm, find the apparent capacity
of the glass and its actual capacity. What values does the juice seller
depict by using such a glass? (Use = ) [4]
Sol: Radius of cylinder = radius of HS = r =
Height of cylinder = h = 10 cm
Apparent Vol. of the glass = r2h = 10 = 196.43
V1 = Apparent Vol. of the glass = 196.43 cm3 [1½]
V = Actual volume of the glass = Apparent volume of the glass – volume of the H.S.
V = V1 – r3 = 196.43 – = 196.43 – 32.74 = 163.69 cm3
V1 = apparent volume ≈ 196.43 cm3, V = Actual Volume ≈ 163.69 cm3. [1½] Value show: Dishonesty, as the juice seller is giving less quantity of juice to the customers. [1]
30) Draw the more than cumulative frequency curve of the following data: [4]
Rent (Rs.) 15-25 25-35 35-45 45-55 55-65 65-75 75-85 85-95
No. of houses 5 10 20 35 30 20 13 7
Sol: We prepare the cumulative frequency table by more than method [tab 1+graph 3]
Class Frequency More than lower limit
Cumulative frequency
15 – 25 5 15 140
25 – 35 10 25 135
35 – 45 20 35 125
45 – 55 35 45 105
55 – 65 30 55 70
65 – 75 20 65 40
75 – 85 13 75 20
85 – 95 7 85 7
Coordinates: (15, 140), (25, 135), (35, 125),
(45, 105), (55, 70), (65, 40),
(75, 20) and (85, 7)
OR) If the median of the distribution given below is 28.5, find the values of x and y.
Class interval 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 Total
Frequency 5 x 20 15 y 5 60
7
22
2
5
7
22
2
5
2
5
3
2
3
2
7
22
2
5
2
5
2
5
5 cm
10 cm
10
20
30
40
50
60
15 25 35
45 55 65
Cu
mu
lati
ve
Fre
qu
en
cy
Lower Limits
O 75 8
5
70
80
90
100
110
120
130
140 (15,140)
(25,135)
(35,125)
(45,105)
(55, 70)
(65, 40)
(75, 20)
(85, 7)
95
E– 17
Model Answer Universal Tutorials – X CBSE (2018–19) – Maths Page 17 of 17
Sol:
C.I Frequency C.f.
0 – 10 5 5
10 – 20 x 5 + x = cf
20 – 30 20 = f 25 + x
30 – 40 15 40 + x
40 – 50 y 40 + x + y
50 – 60 5 45 + x + y = n
f1 = 45 + x + y
[Table 1]
Given: 45 + x + y = 60 = n
x + y = 60 – 45 = 15 ……… (1) [½]
Median = 28.5 [given]
By using formula: Median = l + h [½]
Here, median = 28.5 which falls in class interval 20 – 30. Hence median class (M.C) = 20 – 30
l = lower limit = 20, h = width = 30 – 20 = 10
n = 60 (given) = = 30
f = frequency of median class = 20
c.f = cumulative frequency of the preceding median class = 5 + x
Median = l + h 28.5 = 20 + 10
28.5 – 20 = 8.5 = 8.5 2 = 25 – x
17 = 25 – x x = 25 – 17 = 8 [1]
From (1): x + y = 15 8 + y = 15 y = 15 – 8 = 7 x = 8, y = 7. [1]
f
cfn
2
2
n
2
60
f
cfn
220
)5(30 x
2
530 x
2
25 x