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Grade IX

Mathematics Mock Test

Math Mock Test Grade - IX

Time allowed: 3 Hours Maximum Marks: 80 General Instructions: (i) All questions are compulsory. (ii) The question paper consists of 40 questions divided into 4 sections A, B, C and D. (iii) Section A comprises of 20 questions of 1 mark each. Section B comprises of 6 questions of 2 marks each. Section C comprises of 8 questions of 3 marks each. Section D comprises of 6 questions of 4 marks each. (iv) 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, three 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. (vi) Use of calculators is not permitted.

Section A

Q 1- 10 are multiple choice questions. Select the most appropriate answer from the given options.

1. Which of the following statements is true? Between any two irrational numbers

A. there are infinitely many integers B. there is no rational number C. there is exactly one rational number D. there are infinitely many rational and irrational numbers

2. If a transversal intersects two parallel lines, then

Which of the following statements is not correct? A. corresponding angles are equal and conversely B. alternate interior angles are equal and conversely C. alternate exterior angles are equal and conversely D. interior angles on the same side of the transversal are equal and conversely

3. Which of the following is not a criterion for congruence of triangles?

A. RHS criterion B. AAS criterion C. SSS criterion D. AAA criterion

4. Which of the following is not true for a parallelogram?

A. Opposite angles are equal B. Opposite sides are equal C. The diagonals are equal D. The diagonals bisect each other

5. Which of the following statements is true

Statement-I: Two congruent figures always have equal areas. Statement-II: The converse of the above statement is always true. A. Statement-I is true B. Statement-II is true C. Both the statements are true D. Both the statements are not true

6. If two chords of a circle are equal, then which of the following will also be equal?

A. The corresponding arc B. The angle subtended at the centre C. The perpendicular distance from the center to the chord D. All of the above

7. The locus of all the points whose abscissa is same as the ordinate lie on

A. the line B. the curve C. Both (A) and (B) D. Neither (A) nor (B)

8. Let be a zero polynomial. Its zeroes are

A. 0 only B. 1 only C. any real number D. not defined

9. Three spheres of same radius are inserted inside a right circular cylinder as shown in the figure. What is the capacity of the cylinder for r = 1?

A. 8 B. 6 C. 4 D. 2

10. Which of the following methods does not give the graphical representation of data?

A. Tally marks B. Bar graphs C. Histograms D. Frequency polygons

(11 – 15) Fill in the blanks: 11. If the angles of linear pair made by a transversal intersecting a straight line are in the ratio 3 : 5, then the difference between the two angles is _____. 12. A point equidistant from two given points lies on the _____ of the line-segment joining the two points.

OR In an isosceles triangle ABC, right-angled at B, the longest side of the triangle is _____. 13. A quadrilateral is said to be a _____, if it satisfies all the properties mentioned below.

1. its opposite angles are equal 2. its opposite sides are equal 3. its diagonals bisect each other 4. a pair of opposite sides is equal and parallel

14. Two coins are tossed 500 times and the outcomes are recorded as below:

Number of heads 2 1 0

Frequency 150 240 110

Based on this information, the probability of getting two tails is _____. 15. If two triangles have equal bases and equal areas, then their corresponding _____ will also be equal. (16 – 20) Answer the following:

16. Find the value of 17. If point (k, −k) lies on the line x + y + c = 0, then what can be said about c? 18. If the semi-perimeter of ΔABC is equal to that of ΔADC, then what is the relation between a and b?

19. What is the mean of the multiples of number 4 that are less than 40?

OR What is the mean of the first five numbers that are multiples of both 2 and 5?

20.

Check whether the given statement is true or false with respect to the given triangle. “The length of AC cannot be less than 14.9 cm.”

Section B

21. If a and b are rational numbers and , find the value of a and b.

OR

If the sides containing the right angle of a right triangle are 2 units and 4 units long, then is the length of the hypotenuse a rational or an irrational number?

22. Determine the solution of the equation mx + ny = p, when .

23. (a) Express a in terms of d and e.

(b) Express a in terms of c.

24. Prove that hypotenuse is the longest side in a right triangle.

25. The sum of the lengths of two sides of a triangle is 48 cm. If the perimeter of the triangle is 70 cm, then what is the difference between the semi-perimeter and the third side of the triangle?

OR

What is the area of the triangle having sides of length 7 cm, 8 cm and 9 cm?

26. Two cubes each of volume 729 cm3 are joined end to end. What is the total surface area of the resultant shape?

Section C

27. Using the factor theorem, show that (x − l) is a factor of the polynomial x2 (l − m) + l2 (m − x) + m2 (x − l).

28. (2k, 2k − 1) is a solution of the equation 3x − 2y = a, and (a, −3) is a solution of the equation x + y = 3. What is the value of k?

OR

Plot the graph of the linear equation 2x + y = 1.

29. The coordinates of the points P and Q are (2, 1) and (0, −3) respectively. Draw a line passing through the points P and Q on the Cartesian plane and determine the coordinates of three different points other than P and Q, which lie on the line PQ.

OR

Draw a ΔABC whose vertices have coordinates A (−1, 1), B (−1, 5) and C (4, 4). What is the area of ΔABC?

30. If 0 and −1 are the zeroes of the polynomial 2x3 + x2 + ax + b, then find the value of the polynomial x3 + 3x2 + 3x + 2 at x = a + b.

31. In the given figure, ABCD is a parallelogram where P is point on side BC. A line is drawn through C and parallel to AP such that it intersects the extended line BA at Q and then DQ is joined.

Show that: ar (ΔADQ) = ar (PCQ)

32. From the given figure, find ∠AOC.

OR The given figure shows a circle with centre O where AB and EF are its two equal chords. AO and FO are extended to the points D and C respectively on the circle.

Show that: arc BC = arc DE

33. The area of an equilateral triangle is cm2. Determine the lengths of its sides using Heron’s formula? 34. The given table shows the number of times different numbers appear on a dice in 1000 throws.

Number on dice 1 2 3 4 5 6

Number of times it appears 321 128 111 137 217 86

What is the probability that an odd number appeared on the dice?

Section D

35. Find three irrational numbers between and . 36. If the volume of a cylinder is (3x2 − 12x + 12)π cubic units, where x > 2, and its height is 3 units, then what is the radius of the cylinder? 37. In the given figure, ABCD is a parallelogram. The points L and M trisect the diagonal AC.

Show that X and Y are the mid-points of sides CD and AB respectively.

OR

In the given figure, ABCD is a quadrilateral where AD = BC. When BD is joined, ar (ΔABD) = ar (ΔBCD). Show that ABCD is a parallelogram.

38. Construct an isosceles right-angled triangle ABC whose perimeter is 10 cm.

OR

Construct a ΔABC in which BC = 7 cm, ∠B = 60°, and AB + AC = 12 cm.

39. A cylindrical container can hold a maximum of 3 L 80 mL milk. A new container is made by decreasing the base diameter of the original container by 20% and increasing its height by 25%. Find the maximum amount of milk that the new container can hold.

OR

The volume and total surface area of a hollow metallic cylinder, whose height is 18 cm, are 3564 cm3and 2772 cm2. Find the thickness of the metallic cylinder.

40. The results of a random survey of the number of school dropouts of various age groups of a certain locality are displayed in the given frequency table.

Age (years) Number of school dropouts

5 − 8 12

8 − 10 10

10 − 14 14

14 − 17 9

Represent the given data in the form of a histogram.

Maths Mock Test Solutions

1. We know that are two irrational numbers.

Option-I is wrong as there is no integer between .

Option-II and III are wrong as there lies infinitely many rational numbers between .

Option-IV is correct as there are infinitely many rational and irrational numbers between .

Hence, the correct answer is option D.

2. If a transversal intersects two parallel lines, then the interior angles on the same side of the transversal are supplementary and conversely. Hence, the correct answer is option D.

3. There are 5 criteria for proving congruence between triangles which are SSS, ASA, SAS, AAS and RHS. Hence, the correct answer is option D.

4. By the definition of parallelograms, we know that in a parallelogram

(i) opposite angles are equal

(ii) opposite sides are equal

(iii) diagonals bisect each other.

The two diagonals are not equal in a parallelogram.

Hence, the correct answer is option C.

5. We know that if two figures are congruent then they will have equal areas. So, statement-I is true.

But, two figures if they have equal areas then they may or may not be congruent. Statement-II is false.

Hence, the correct answer is option A.

6. If two chords of a circle are equal, then all of the above-mentioned quantities will be equal.

Hence, the correct answer is option D.

7. The locus of all the points whose abscissa is same as the ordinate i.e. lie on the line .

Hence, the correct answer is option A.

8.

This implies that is a zero polynomial.

Its zero can be obtained by equating to zero.

i.e.

Now the zero of this polynomial is the value of x for which .

The above equation is true for any real number.

Hence, the correct answer is option C.

9. The radius of the base of cylinder = r

The height of the cylinder = 6r

Therefore, the capacity of the cylinder =

For r = 1, the capacity is .

Hence, the correct answer is option B.

10. Tally marks are used to denote the frequency of a particular class.

They are not used to give a graphical representation of data.

Hence, the correct answer is option A.

11. Let the two angles be

Now,

Therefore, the two angles are and the difference between them is .

12. A point equidistant from two given points lies on the perpendicular bisector of the line-segment joining the two points.

OR

Since the given triangle is right-angled at B, therefore, AB = BC.

Using the angle sum property of triangles, we get

Since

Hence, the longest side of the triangle is AC.

13. A quadrilateral is said to be a parallelogram, if it satisfies all the properties mentioned below.

1. its opposite angles are equal

2. its opposite sides are equal

3. its diagonals bisect each other

4. a pair of opposite sides is equal and parallel

14. When a coin is tossed, the only two possible outcomes are Head and Tail.

So, the probability of getting 2 tails = the probability of getting no head =

.

15. If two triangles have equal bases and equal areas, then their corresponding altitude will also be equal.

16. 17. It is given that the point (k, −k) lies on the line x + y + c = 0.

∴ k + (−k) + c = 0

⇒ 0 + c = 0

⇒ c = 0

Thus, the value of c must be zero.

18. It is given that the semi-perimeter of ΔABC is equal to that of ΔADC.

⇒ a + 11.5 cm + AC = 17.5 cm + b + AC

⇒ a = 6 cm + b

∴ a − b = 6 cm

This is the required relation between a and b.

19. Multiples of 4 that are less than 40 are as follows:

4, 8, 12, 16, 20, 24, 28, 32, 36

Thus, the mean of the multiples of number 4 that are less than 40 is 20.

OR

The first five numbers that are multiples of both 2 and 5 are:

10, 20, 30, 40, 50

Thus, the mean of the first five multiples of 2 and 5 both is 30.

20. The sum of two sides of a triangle is always greater than the third side of the triangle.

∴ AB + BC > AC

⇒ 8.4 cm + 6.5 cm > AC

⇒ AC < 14.9 cm

∴The length of AC will be less than 14.9 cm.

Thus, the given statement is false.

21.

Rationalise the denominator.

Compare both sides, we get;

OR

On applying Pythagoras theorem, we obtain

AC2 = AB2 + BC2

⇒ AC2 = (2 units)2 + (4 units)2

⇒ AC2 = 4 units + 16 units

⇒ AC2 = 20 units

⇒ AC =

⇒ AC =

Since 2 is a rational number and is an irrational number, is an irrational number.

Thus, the length of the hypotenuse is an irrational number.

22. The given equation is mx + ny = p.

At , the solution of the equation is:

Thus, the solution of the given equation at is .

23. (a) We have,

∠AOD + ∠DOB = 180° (Linear pair)

⇒ ∠AOE + ∠EOD + ∠DOB = 180°

⇒ 42° + a + b = 180°

⇒ a = 180° − 42° − b

⇒ a = 138° − b

⇒ a = 138° − (d + e) [ b = e + d as they are a pair of vertically opposite angles]

⇒ a = 138° − d − e

(b) ∠AOD = ∠BOC (Vertically opposite angles)

⇒ ∠AOE + ∠EOD = ∠BOC

⇒ 42° + a = c

⇒ a = c − 42°

24. Let ΔABC be right-angled at B.

i.e., ∠B = 90° ... (1)

The sum of the measures of all the angles of a triangle is 180°.

∴∠A + ∠B + ∠C = 180°

⇒ ∠A + 90° + ∠C = 180°

⇒ ∠A + ∠C = 90° ... (2)

From equations (1) and (2):

∠A + ∠C = ∠B

⇒ ∠A < ∠B and ∠C < ∠B

Thus, in a right triangle, the right angle is the greatest angle.

It is also known that in a triangle, the side opposite the larger angle is longer.

In a right triangle, the side opposite the right angle is the hypotenuse.

Hence, the hypotenuse is the longest side in a right triangle.

25. Let the sides of the triangle be a, b and c.

It is given that the sum of the lengths of two sides of the triangle is 48 cm.

Let a + b = 48 cm.

It is also given that the perimeter is 70 cm.

∴ Semi-perimeter of the triangle,

Thus, difference between semi-perimeter and third side = 35 cm − 22 cm = 13 cm.

OR

The lengths of the sides of the triangle are given as a =7 cm, b = 8 cm and c = 9 cm.

∴Semi-perimeter (s) of the triangle

Thus, the area of the triangle [Heron’s formula]

26. It is given that the volume of each cube is 729 cm3.

It is known that the volume of a cube is given by l3, where l is the edge of the cube.

Thus, the edge of each cube is 9 cm.

By joining two cubes each of edge 9 cm, a cuboid of dimensions 18 cm × 9 cm × 9 cm will be obtained.

It is known that the surface area of a cuboid is given by 2(lb + bh + hl), where l, b, and h are length, breadth, and height of the cuboid respectively.

∴ Surface area of the cuboid

Thus, the surface area of the resultant cuboid is 810 cm2.

27. Let p(x) = x2 (l − m) + l2 (m − x) + m2 (x − l)

According to the factor theorem, if p(x) is a polynomial of degree n ≥ 1 and h(x) is a linear polynomial, then h(x) is a factor of p(x) if p(x) = 0 at zero of h(x).

The zero of (x − l) is l.

Thus, (x − l) is a factor of the polynomial x2 (l − m) + l2 (m − x) + m2 (x − l).

28. The given equations are:

3x − 2y = a ... (1)

x + y = 3 ... (2)

(a, − 3) is a solution of equation (2).

∴ a + (−3) = 3

⇒ a = 6

Substituting the value of a in equation (1),

3x − 2y = 6 ... (3)

(2k, 2k − 1) is a solution of equation (1).

∴ 3(2k) − 2 (2k − 1) = 6

⇒ 6k − 4k + 2 = 6

⇒ 2k = 4

⇒ k = 2

Thus, the value of k is 2.

OR

The given equation is 2x + y = 1.

In order to obtain the graph of the given linear equation, some ordered pairs satisfying the equation are determined.

Now, 2x + y = 1

⇒ y = 1 − 2x

Some solutions of the given equation have been shown in the table as:

x −2 −1 0

2

y = 1 − 2x 5 3 1 0 −3

On plotting and joining the points (−2, 5), (−1, 3), (0, 1), and (2, −3) on the coordinate plane, a straight line representing the given equation is obtained.

This can be represented as:

29. The point P (2, 1) can be located by moving 2 units on the positive x-axis from the origin (0, 0), and then 1 unit along the positive y-axis. The point Q (0, −3) can be located by moving 3

units on the negative y-axis. The line passing through P and Q is obtained by joining P and Q and extending in both directions. This can be done as:

It is seen that, (3, 3), (1, −1) and (5, 7) are three different points lying on the line other than P and Q.

OR

The points A (−1, 1), B (−1, 5) and C (4, 4) can be plotted on the coordinate plane as:

Draw CM ⊥ AB.

It is seen that, AB = 4 units and CM = 5 units.

Area of ΔABC = 10 square units

Thus, the area of ΔABC is 10 square units.

30. Let p(x) = 2x3 + x2 + ax + b

Given, 0 is a zero of p(x), therefore, p(0) = 0.

p(0) = 0

⇒ 2 (0)3 + (0)2 + a (0) + b = 0

⇒ b = 0

Given, −1 is a zero of p(x), therefore, p(−1) = 0.

p(−1) = 0

⇒ 2 (−1)3 + (−1)2 + a (−1) + b = 0

⇒ −2 + 1 − a + 0 = 0 [b = 0]

⇒ a = −1

∴ a + b = −1 + 0 = −1

Let q(x) = x3 + 3x2 + 3x + 2

∴ q(−1) = (−1)3 + 3(−1)2 + 3(−1) + 2 = −1 + 3 − 3 + 2 = 1

Thus, the value of the polynomial x3 + 3x2 + 3x + 2 at x = a + b is 1.

31. Join AC.

It is known that triangles on the same base (or equal bases) and between the same parallels are equal in area.

It is observed that ΔADQ and ΔACQ lie on the same base AQ and between the same parallels QB and CD.

∴ ar (ΔADQ) = ar (ΔACQ) … (1)

Also, ΔAQC and ΔQPC lie on the same base QC and between the same parallels QC and AP.

∴ ar (ΔAQC) = ar (ΔQPC) … (2)

From equations (1) and (2):

ar (ΔADQ) = ar (ΔQPC)

Hence, the result is proved.

32. In the given figure, ∠BAO = 21°and ∠BCO = 17°

Since OA and OB are radii of circle, OA = OB

∴ ∠OBA = ∠BAO

⇒ ∠OBA = 21°

Also, since OB and OC are radii of circle,

∴ ∠OBC = ∠BCO

⇒ ∠OBC = 17°

Now, ∠ABC = ∠OBA + ∠OBC

= 21°+ 17°= 38°

We know that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

∴ ∠AOC = 2 ∠ABC

⇒ ∠AOC = 2 ×38°

⇒ ∠AOC = 76°

OR

Comparing ΔABD and ΔCEF:

∠ABD = ∠CEF (Angle in a semi-circle is a right angle)

AD = CF (Diameter of the same circle)

AB = EF (Given)

∴ ΔABD ≅ ΔFEC (RHS congruence criterion)

The corresponding parts of congruent triangles are equal.

BD = CE

⇒ arc BD = arc CE

⇒ arc BD − arc CD = arc CE − arc CD

⇒ arc BC = arc DE

Hence, the result is proved.

33. Let the length of each side of the equilateral triangle be a.

∴ Semi-perimeter, s =

Thus, the length of each side of the equilateral triangle is .

34. Number of throws of dice = 1000

Number of times odd numbers appeared = 321 + 111 + 217 = 649

∴ Required probability

35.

A number that has a non-terminating and non-recurring decimal expansion is an irrational

number. Therefore, to find three irrational number between and , it is required to search for three numbers that are non-terminating and non−recurring and also lying

between the numbers and 4.898.

Thus, the three irrational numbers between and are

4.46101101110..., 4.67484884888...and 4.7214345414...

36. It is given that the volume of the cylinder is (3x2 − 12x + 12)π cubic units.

(3x2 − 12x + 12)π cubic units

= 3 (x2 − 4x + 4)π cubic units

= 3 (x2 − 2 × x × 2 + 22)π cubic units

= π (x − 2)2 × 3 cubic units

It is known that the volume of a cylinder with base radius r and height h is given by πr2h.

It is also given that the height of the cylinder is 3 units.

Therefore, on comparing π (x − 2)2 × 3 with πr2h, it is obtained that r = x − 2 and h = 3.

Thus, the radius of the cylinder is (x − 2) units.

37. It is given that, L and M trisect the diagonal AC.

∴ AL = LM = MC …(1)

It is known that, the opposite sides of a parallelogram are equal and parallel.

∴ AD = BC and AD || BC

In ΔALD and ΔCMB:

AD = BC (Given)

∠DAL = ∠BCM (Alternate interior angles)

AL = MC [{From (1)]

∴ ΔALD ≅ ΔCMB [By SAS congruence rule]

It is known that, the corresponding parts of congruent triangles are equal.

∴ ∠ALD = ∠BMC

⇒ 180° − ∠ALD = 180° − ∠BMC

⇒ ∠DLM = ∠BML

However, these are alternate interior angles on DY and BX.

∴ DY || BX

According to the converse of mid-point theorem, the line drawn through the mid-point of one side of a triangle and parallel to another side bisects the third side.

In ΔDCL,

M is the mid-point of side CL and DL || XM.

∴ X is the mid-point of CD.

Similarly, in ΔABM,

Y is the mid-point of AB.

Thus, X and Y are mid-points of AB and CD respectively.

OR

Draw perpendiculars AL and CM to BD.

It is given that: ar (ΔABD) = ar (ΔBCD)

⇒ × AL ×BD = ×CM ×BD

⇒ AL = CM …(1)

Comparing ΔALD and ΔBCM:

∠ALD = ∠CMB [Each is a right angle]

AD = BC [Given]

AL = CM [Using (1)]

∴ ΔALD ≅ ΔCMB [RHS congruence rule]

Corresponding parts of congruent triangles are equal.

∴ ∠ADL = ∠MBC

These angles are the alternate interior angles with respect to the lines AD and BC.

∴ AD||BC

If a pair of opposite sides is parallel and equal, then the quadrilateral is a parallelogram.

In quadrilateral ABCD, AD||BC and AD = BC.

Thus, ABCD is a parallelogram.

38. The angles of an isosceles right-angled triangle are 90°, 45° and 45°.

In ΔABC, let ∠A = 90°, ∠B = ∠C = 45°.

It is given that the perimeter of ΔABC is 10 cm.

The steps of construction are as follows:

(i) Draw a line segment XY of length of 10 cm.

(ii) At points X and Y, draw angles ∠LXY and ∠MYX of measures 45° each.

(iii) Draw the bisectors of ∠LXY and ∠MYX such that they intersect each other at A.

(iv) Draw the perpendicular bisectors of AX and AY to intersect XY at points B and C respectively.

(v) Join AB and AC.

ΔABC is the required triangle.

OR

The required triangle can be constructed as follows.

1. Draw a line segment BC of length 7 cm.

2. Draw a ray BX making an angle of 60° with BC.

3. With B as centre and radius equal to 12 cm, draw an arc which cuts BX at D.

4. Join DC and draw its perpendicular bisector which intersects BD at point A.

5. Join AC. Thus, ΔABC is the required triangle.

39. Let d and h be the respective diameter and height of the original container.

∴ Radius of the original container,

Capacity of the original container = V = πr2h = 3L 80 mL = 3080 mL

The diameter of the original container is decreased by 20%.

∴ Diameter ( ) of the new container

∴ Radius ( ) of the new container

The height of the original container is also increased by 25%.

∴ Height ( ) of the new container

Then, volume of the new container

∴ Capacity of the new container is given by,

OR

Let r2and r1be the external and internal radii of the cylinder. Its height, h= 18 cm

Now, the volume of the cylinder

It is given that total surface area of the metallic cylinder is 2772 cm2.

Dividing equation (1) by (2), we obtain

Therefore, the thickness of the metallic cylinder is 3 cm.

40. It is observed that intervals are not equally spaced in the given data.

The areas of rectangles in a histogram are proportional to their corresponding frequencies. Since the intervals are not equally spaced in this case, the width of the rectangles will vary, and the bar graph will not be correct.

In order to draw the histogram correctly, the new lengths of the rectangles proportional to the minimum class size, i.e., 2 (= 10 − 8) need to the calculated.

This can be done as:

Age (years) Frequency Width of the class Length of the rectangle

5 − 8 12 3

8 − 10 10 2

10 − 14 14 4

14 − 17 9 3

Now, the required histogram can be constructed by taking the new lengths of the rectangles along the vertical axis, and the age along the horizontal axis.

This can be drawn as: