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Mathematics
Class – IX
Time Allowed: 3 hours Maximum Marks: 80
General Instructions:
1. This question paper contains two parts A and B
2. Both Part A and Part B have internal choices
Part – A:
1. It consists of two section I and II
2. Section I has 22 questions. Internal choices provided in 7 questions.
3. Section II has Q.23 and 24 two case study-based questions. Each case study has 5 case –based sub parts. An examinee is to attempt any 4 out of 5 sub-parts.
Part – B:
1. Question No 25 to 31 are Very Short Answer type questions of 2 marks each
2. Question No 32 to 38 are Short Answer type questions of 3 marks each
3. Question No 39 to 41 are Long Answer type questions of 5 marks each.
4. Internal choice is provided in 2 questions of 2 marks, 2 questions of 3 marks and 1 question of 5 marks.
Part – A
Section – I
1. The product of any two irrational numbers can be __________ or ___________.
View AnswerAns. rational, irrational
OR
Write a real number which has terminating decimal expansion.
View AnswerAns. 51/2 = 25.5
2. What is the degree of the polynomial √2?
View AnswerAns. √2 = √2x0
Because exponent of x is 0, so required degree is 0
3. Write the equation x = 7, in two variables, in the other form.
View AnswerAns. Co-efficient of y in the given equation x = 7 is 0. Hence, the required equation is 1.x + 0,y = 7
4. Name the quadrant where the point (-3, 5) lies.
View AnswerAns. In point (-3, 5), x coordinate is negative and y coordinate is positive. So, the point lies in the second quadrant
5. If AB = QR, BC = PR and CA = PQ, then ____________
View AnswerAns. AB = QR & BC = PR => ∠B = ∠R …….. (i)
BC = PR & CA = PQ => ∠C = ∠P ……… (ii)
CA = PQ & AB = QR => ∠A = ∠Q ………(iii)
From equation (i), (ii) and (iii), we get
ΔCBA ≅ ΔPRQ
6. Three angles of a quadrilateral are 75°, 90° and 75°. Find the fourth angle.
View AnswerAns. We know that the sum of angles in a quadrilateral = 360
Therefore,
Fourth angle of the quadrilateral
= 360 – (75 + 90 + 75)
= 360 – 240
= 120
OR
Name the two quadrilaterals whose diagonals meet at right angles.
View AnswerAns. Square and rhombus
7. In triangles ABC and DEF, AB = FD and ∠A = ∠D. Write the condition so that the two triangles will become congruent by SAS axiom.
View AnswerAns. Given in ΔABC and ΔDEF, AB = FD, ∠A = ∠D
We know that, two triangles will be congruent by SAS axiom if two sides and the included angle between them two triangles are equal.
Clearly, required condition is AC = DE
8. A diagonal of a rectangle is inclined to one side of the rectangle at 25. Find the acute angle between the diagonals.
View AnswerAns. ABCD is a rectangle in which diagonal AC is inclined to one side, AB of the rectangle at an angle of 25.
Now AC = BD (Diagonals of a rectangle are equal)
1/2AC = 1/2BD
Or OA = OB
In ΔAOB, we have
OA = OB
∠OBA = ∠BAO= 25
By angle sum property, we have ∠OBA + ∠AOB + ∠BAO = 180
25 + 25 + ∠AOB = 180
∠AOB = 180 – 50 = 130
We know that, ∠AOB and ∠AOD form linear pair
∠AOB + ∠AOD = 180
130 + ∠AOD = 180
∠AOD = 180 – 130 = 50
Therefore, the acute angle between the diagonal is 50
9. In the given figure, if OA = 5 cm, AB = 8 cm and OD is perpendicular to AB, then find the length of CD.
Ans. The perpendicular drawn from the centre to a chord bisect the chord
AC = ½ x AB = ½ x 8 = 4 cm
OC = √OA2 – AC2
= √52 – 42
√25 – 16
√9 = 3 cm
OC = 3 cm
Now, CD = OD – OC
= 5 cm – 3 cm = 2 cm
10. The perimeter of an equilateral triangle is 60 m. What is the area of the triangle?
View AnswerAns. Perimeter of equilateral triangle = 60 m
3a = 60 m
a = 20 m
Its area √3/4 a2 = √3/4 x 20 x 20 = 100√3 m2
OR
ΔABC is an isosceles right angled triangle in which ∠A = 90. Calculate ∠B.
View AnswerAns. AB = AC
∠C = ∠B
∠A + ∠B + ∠C = 180 (Angle sum property)
Or 90 + ∠B + ∠= 180
Or 2∠B = 90
∠B = 45
11. Find the class mark of the class 125 – 145
View AnswerAns. Class mark = (125 + 145)/2 = 270/2 = 135
12. Find the value of x for which
Ans.
13. Rationalize the denominator of 2/3√3
View AnswerAns.
14. From the figure given below find the co-ordinates of point Q
Ans. The co-ordinate of point Q = (-3, -3.5)
15. In which quadrants the point P(2, -3) and Q(-3, 2) lie?
View AnswerAns. P(2, -3) and Q(-3, 2) lie in IV and II quadrants
OR
If (a, b) = (0, -2). Find the value of a and b
View AnswerAns. (a, b) = (0, -2)
a = 0, b = -2
16. Compute the curved surface area of a hemisphere whose diameter is 21 cm.
View AnswerAns. Given, diameter of hemisphere = 21 cm
Curved surface area = 2πr2
= 2 x 22/7 x 21/2 x 21/2
= 693
OR
If perimeter of an equilateral triangle is 120 m, then find its area.
View AnswerAns. Perimeter of equilateral triangle = 120
Therefore, 3a = 120 (where a is the side of triangle)
=> a = 40
Area = √3/4 a2
= √3/4 x 40 x 40
= 400√3 m2
17. In quadrilateral ABCD, ∠A + ∠D = 180. What special name can be given to this quadrilateral?
View AnswerAns. In quadrilateral ABCD, ∠A + ∠D = 180, that is, the sum of two consecutive angles is 180. So, pair of opposite side AB and CD are parallel.
Therefore, quadrilateral ABCD is trapezium
Hence, special name which can be given to this quadrilateral ABCD is trapezium.
18. How many number of dimensions a solid has?
View AnswerAns. A solid has shape, size and position and can be moved from one place to another. So, a solid has three dimensions
OR
In how many chapters, Euclid divided his famous treatise ‘The Elements’.
View AnswerAns. Euclid divided his famous treatise “The Elements” into 13 chapters
19. Find the point where equation 7x + 11y = 19 intersects Y-axis.
View AnswerAns. As the line intersect y-axis, put x = 0 in the given equation, we get
7(0) + 11y = 19
y = 19/11
The required point is (0, 19/11)
OR
If (4, 0) is a solution of the linear equation 4x + 6y = k, then find value of k
View AnswerAns. Put x = 4 and y = 0 in the given equation 4x + 6y = k, we get
4(4) + 6(0) = k
k = 16
20. Find value of 3942 – 3492
View AnswerAns. 3942 – 3492 = (394 – 349)(394 + 349) [a2 – b2= (a – b)(a+ b)]
= 45 x 743
= 33,435
21. In ΔABC, BC = AB and ∠B = 70, then find ∠A.
View AnswerAns. In ΔABC, BC = AB, B = 70
∠C = ∠A (angles opposite to equal sides are equal)
Now ∠A + ∠B + ∠C = 180 (sum of all angles of triangle)
∠A + 70 + ∠A = 180
2∠A = 110
∠A = 55
22. Find value of √20 x √15
View AnswerAns. We have,
√20 x √15 = √(20 x 15)
= √(4 x 5 x 5 x 3)
= √(2 x 2 x 5 x 5 x 3)
= 10√3
Section – II
23. The National Service Scheme (NSS) is an Indian government sponsored public service program that aimed at developing student’s personality through community service
The volunteers of NSS erected a conical tent made of tarpaulin in a flood effected area. If vertical height and diameter of the conical tent are 3 m and 8 m, respectively and width of tarpaulin is 250 cm, then answer the following:
A) What is the slant height of the conical tent?
i) 3m
ii) 6m
iii) 5m
iv) 7m
View AnswerAns. iii) 5m
Vertical height = 3 m
Diameter = 8 m
So, radius = 4 m
Since, l2 = h2 + r2
l2 = (3)2 + (4)2 = 9 + 16 = 25
l = √25 = 5 m
B) What is the formula to calculate the curved surface of a cone?
i) πr2l
ii) πr
iii) πrl
iv) 2 πrl
View AnswerAns. iii) πrl
C) What is the area of the tarpaulin used?
i) 62.8 m2
ii) 66m2
iii) 64.2 m2
iv) 68m2
View AnswerAns. i) 62.8 m2
Area of the tarpaulin used = Curved surface area of a conical tent
= πrl
= 3.14 x 4 x 5
= 62.8 m2
D) What is the length of the tarpaulin used?
i) 25 m
ii) 15 m
iii) 20 m
iii) 25.12 m
View AnswerAns. iii) 25.12 m
Width of tarpaulin used = 250 cm = 2.5 m
Length of the tarpaulin used
= Area of the tarpaulin/Width of tarpaulin
= 62.8/2.5
= 25.12 m
E) What is the length of the tarpaulin used, if assumed that 10% extra material is required for stitching margin and wastage in cutting ? (Take π= 3.14)
i) 26.45 m
ii) 27.63 m
iii) 34.56 m
iv) 43.23 m
View AnswerAns. ii) 27.63 m
Area of the tarpaulin used = 62.8 + 10% of 62.8
= 62.8 + 6.28
= 69.008
Length of the tarpaulin used = Area of the tarpaulin/Width of tarpaulin
= 69.08/2.5
= 27.63 m
24. Isosceles triangles were used to construct a bridge in which the base (unequal side) of an isosceles triangle is 4 cm and its perimeter is 20 cm.
A) What is the length of equal sides?
i) 2 cm
ii) 3 cm
iii) 8 cm
iv) 10 cm
View AnswerAns. iii) 8 cm
Let x cm be the length of equal sides of the isosceles triangle
So, x + x + 4 = 20
2x + 4 = 20
2x =16
x = 16/2 = 8 cm
B) What is the Heron’s formula for a triangle?
i) √[s(s + a)(s – b)(s – c)]
ii) √[s(s + a)(s + b)(s + c)]
iii) √[s(s – a)(s – b)(s – c)]
iv) √[s(s . a)(s . b)(s . c)] View Answer
Ans. iii) √[s(s – a)(s – b)(s – c)]
C) What is the semi perimeter of the highlighted triangle?
i) 30 cm
ii) 40 cm
iii) 10 cm
iv) 50 cm
View AnswerAns. iii) 10 cm
Required semi-perimeter = Perimeter/2 = 20/2 = 10 cm
D) What is the area of highlighted triangle?
i) 4√15 cm2
ii) 4 cm2
iii) √15 cm2
iv) 20 cm2
View AnswerAns. i) 4√15 cm2
Since, semi perimeter
s = 10 cm
Thus, area of the triangle
√[s(s – a)(s – b)(s – c)] =√10(10 – 8)(10 – 8)(10 – 4)
= √10 x 2 x 2 x 6
= 4√15 cm2
E) If the sides of a triangle are in the ratio 3 : 5: 7 and its perimeter is 300 m. Find its area.
i) 100√2 m2
ii) 500√2m2
iii) 1500√3 m2
iv) 200√3 m2
View AnswerAns. iii) 1500√3 m2
Let the sides of a triangle are
a = 3x, b = 5x, c = 7x
the a + b + c = 300
15x = 300
x = 20 m
So a = 60, b = 100, c = 140
s = (a + b + c)/2
= 300/2= 150
Area of triangle = √[s(s – a)(s – b)(s – c)]
= √[150(150 – 60)(150 – 100)(150 – 140)]
= √150 x 90 x 50 x 10
= 1500√3 m2
Part – B
25. If y = 2 and y = 0 are the zeroes of the polynomial f(y) = 2y3 – 5y2 + ay + y, find the value of a and b
View AnswerAns. Given f(y) = 2y3 – 5y2 + ay + b
f(2) = 2(2)3 – 5(2)2 + a(2) + b = 0
16 – 20 + 2a + b = 0
2a + b = 4 ………………… (i)
f(0) = 0
From equ (i)
2a + 0 = 4
a = 2, b = 0
26. ABCD is a cyclic quadrilateral in which AB || CD. If ∠D = 70, find all the remaining angles
View AnswerAns. Since sum of the opposite pairs of angles in a cyclic quadrilateral is 180
Hence ∠B + ∠D = 180
Or ∠B = 180 – 70 = 110
Again AB ||CD and AD is its transversal, so
∠A + ∠D = 180 (consecutive interior angles)
∠A + 70 = 180
Or ∠A = 180 – 70 = 110
And ∠A + ∠C = 180
Or 110 + ∠C = 180
∠C = 180 – 110 = 70
OR
ABCD is a cyclic quadrilateral in which AC and BD are its diagonals. If ∠DBC = 55 and ∠BAC = 45, find ∠BCD
View AnswerAns.
∠BAC = ∠BDC = 45 (angles in the same segment)
In ΔDBC,
∠DBC + ∠BCD + ∠CDB = 180 (Angle sum property)
Or 55 + ∠BCD + 45 = 180
Or ∠BCD = 80
27. By actual division, find the quotient and remainder when p(x) = 2x3 + 3x2 – 9x + 4 is divided by 2x – 1
View AnswerAns.
28. In the given figure, AB and CD are two chords of a circle with centre O at a distance of 6 cm and 8 cm from O. if the radius of the circle is 10 cm, find the length of chords.
Ans. Join OA and OC
Since, perpendicular from centre bisects the chord
AP = BP = 1/2AB
CQ = QD = 1/2CD
In ΔOAP, by Pythagoras theorem,
AP2 = OA2 – OP2
= 102 – 62 = 64
AP = 8 cm or AB = 16 cm
In ΔOQC
CQ2 = OC2 – OQ2
= 102 – 82
= 36
Or CQ = 6 cm or CD = 12 cm
29. The radius and slant height of a cone are in the ratio 4: 7. If its curved surface area is 792cm2, find its radius.
View AnswerAns. Let the radius of a cone r = 4x
And slant height h = 7x
CSA = 792 cm2
πrl = 792
22/7 x 4x x 7x = 792
x2 = (792 x 7)/(22x 4 x 7)
= 9
x = 3 cm
radius = 4 x 3 = 12 cm
OR
How much ice-cream can be put into a cone with base radius 3.5 cm and height 12 cm?
View AnswerAns. r = 3.5 cm, h = 12 cm
Therefore, Amount of ice cream = 1/3πr2h
1/3 x 22/7 x 3.5 x 3.5 x 12 = 154 cm2
30. Below is the marks of 35 students in mathematics test (out of 10). Average these marks in tabular form using tally mark.
Ans.
31. Rainwater harvesting system is a technology that collects and stores rainwater for human use.
Anup decided to do rainwater harvesting. He collected rainwater in the underground tank at the rate of 30 cm3/s.
i) What will be the equation formed if volume of water collected in x seconds is taken as y cm3 ?
ii) How much time will it take to collect water in 900 cm3 ?
View AnswerAns. (i) y = 30x
(ii) 30s
Since y = 30x
If y = 900, then 900 = 30x
X = 900/30 = 30
Required time is 30 s
32.
Ans.
OR
If (25)x = (125)y then x : y equals?
View AnswerAns. (25)x = (125)y
x/y = 125/25
x/y = 5/1
x : y= 5 : 1
33. In the given figure, AB || CD and CD || EF, If y : z = 4 : 5, then find x, y and z
Ans. As AB || CD
∠x + ∠y = 180 (Co-interior angles) …….. (i)
And as CD||EF
AB || EF
So, ∠x = ∠z (alternate angle) ……… (ii)
From eqn (i) and (ii), we get
∠y + ∠z = 180 ………. (iii)
Let y = 4p and z = 5p where p is common ratio
Then from eqn (iii)
4p + 5p = 180
p = 20
∠y = 4 x 20 = 80
∠z = 5 x 20 = 100
∠x = 180 – ∠y
= 180 – 80
= 100
34. Write the equation 4x = 6(1 – y) + 3x in the form ax + by = c and also find the co-ordinate of the points where its graph cuts the two axes?
View AnswerAns. 4x = 6(1 – y) + 3x
4x = 6 – 6y + 3x
x = 6 – 6y
x + 6y = 6
OR
ABCD is a rectangle. Write the equation of its sides. Also, find its area.
Ans. Equation of the sides are,
AB: Y = 0
BC: X = -1
CD: Y = -4
DA: X = -4
Area = 4 x 3 = 12 sq units
35. For what value of p; x = 2, y = 3 is a solution of (p +1)x – (2p + 3)y – 1 = 0 and write the equation.
View AnswerAns. Given equation is (p +1)x – (2p + 3)y – 1 = 0 ……… (i)
If x = 2, y = 3 is the solution of the equation (i)
Then (p +1)2 – (2p + 3)3 – 1 = 0
2p + 2 – 6p – 9 – 1 = 0
-4p – 8 = 0
p = -2
Put the value of p in equation (i), then
-x + y – 1 = 0
Or x – y + 1 = 0, is the required equation.
36. The angles A, B, C and D of a quadrilateral ABCD are in the ratio 2: 4: 5: 7. Find the measure of these angles. What type of quadrilateral is it? Give reasons.
Ans. Let the measures of the angles be 2x, 4x, 5x and 7x
2x + 4x + 5x + 7x = 360 (Angle sum property)
18x = 360
x = 20
∠A = 40
∠B = 80
∠C = 100
∠D = 140
As ∠A + ∠D = 180 and ∠B + ∠C = 180
Or CD || AB
Hence ABCD is a trapezium
37. In the given figure, PQ = QR = RS and ∠PQR = 128. Find ∠PTQ, ∠PTS and ∠ROS
Ans. Given PQ = QR = RS, ∠PQR = 128
∠1 = ∠2 = (180 – 128)/2 = 52/2 = 26°
∠PTQ = ∠QRP = 26 (Angle in same segment)
PQRT is cyclic quadrilateral
∠PQR + ∠RTP = 180
128 + ∠4 + ∠QTP = 180
∠QTP = 26
Since ∠PTP = ∠QTP + ∠4 + ∠3
∠PTS = 26 + ∠4 + ∠3(∠4 = ∠1 = 26)
= 26 + 26 + 26
= 78
∠ROS = 2∠RTS (Angle subtended by an arc at the centre is twice at circumference)
= 2 x 26 = 52
38. There are two cones. The ratio of their radii are 4: 1. Also, the slant height of the second cone is twice that of the former. Find the relationship between their curved surface area.
View AnswerAns. Let r1 and l1 be the radius and slant height of first cone
Let r2 and l2 be the radius and slant height of second cone
Curved surface area of first cone (CSA1) = πr1l1 and curved surface area of second cone (CSA2) = πr2l2
According to the question,
R1 : r2 = 4 : 1
Or r1/r2 = 4/1 and l2 = 2l1 or l1/l2 = ½
CSA1/ CSA2 = πr1l1/ πr2l2 = (r1/r2)(l1/l2)
= (4/1)(1/2)
= 2/1
39. If x + 1/x = √3, evaluate x3 + 1/x3
View AnswerAns.
40. In the adjoining figure, AB || CD and l is transversal. Find values of x and y
Ans. ∠EFD = 60
Or 25 + y = 60
Or y = 35
∠BEF + ∠EFD = 180 (Sum of co-interior angles on same side of transversal is supplementary)
Or ∠BEF + 60 = 180
∠BEF = 120
Or ∠PEF + 40 = 120
∠PEF = 80
Now in ΔPEF
OR x + 80 + 25 = 180 (Angle sum property)
Or x = 75
OR
A transversal intersects two parallel lines. Prove that the bisectors of any pair of corresponding angles so formed are parallel.
View AnswerAns. Given l ||m line t is a transversal intersecting them at P and Q respectively
To prove : PR || QS
Proof: ∠5 = ∠6 (Corresponding angles and l || m)
Or ½ ∠5 = ½ ∠6
Or ∠1 = ∠3
Therefore, PR || QS
41. Two sides AB and BC and median AM of one triangle ΔABC are respectively equal to sides PQ and QR and median PN of ΔPQR. Show that
i) ΔABM ≅ ΔPQN
ii) ΔABC ≅ ΔPQR
View AnswerAns.
Given: AB = PQ
BC = QR
AM = PN
To prove: ΔABM ≅ ΔPQN
BC = QR
1/2BC = 1/2QR
BM = QN …….(1)
(M and N are the mid-points of sides BC & QR, respectively)
Now, in ΔABM & ΔPQN
AB = PQ (given)
AM = PN
BM = QN (from eqn (1))
ΔABM ≅ ΔPQN (By SSS rule)
∠1 = ∠2 ……. (2) CPCT
ii) Now in ΔABC & ΔPQR,
AB = PQ (Given)
∠1 = ∠2 (From eqn (2)
BC = QR (given)
ΔABC ≅ ΔPQR (By SAS rule)