Class 9

A mathematics teacher ask students to collect marks of Mathematics in . . .

Case Study based question on Statistics

A mathematics teacher ask students to collect marks of Mathematics in Half Yearly Exams. She instructed all the students to prepare a frequency distribution table using the data collected. Ram collected the following marks (out of 50) obtained in Mathematics by 60 students of class IX.

21, 10, 30, 22, 33, 5, 37, 12, 25, 42, 15, 39, 26, 32, 18, 27, 28, 19, 29, 35, 31, 24, 36, 18, 20, 38, 22, 44, 16, 24, 10, 27, 39, 28, 49, 29, 32, 23, 31, 21, 34, 22, 23, 36, 24, 36, 33, 47, 48, 50, 39, 20, 7, 16, 36, 47, 30, 22, 17

(i) How many students scored more than 20 but less than 30 marks? [1 Marks]

(ii) How many students scored less than 20 marks? [1 Marks]

(iii) How many students scored more than 60% marks? [2 Marks]

What is the class size and class marks of the class interval 30-40? [2 Marks]

Detailed Solution :

(i) The marks more than 20 but less than 30 (i.e., 21–29) are:

{21, 22, 25, 26, 27, 28, 29, 24, 22, 24, 27, 28, 29, 23, 21, 22, 23, 24, 22} = 19

So, there are 19 students who scored more than 20 but less than 30 marks.

(ii) Number of students scored less than 20 marks (i.e. 1 to 19) are :

{10, 5, 12, 15, 18, 19, 18, 16, 10, 7, 16, 17} = 12

So, there are 12 students who scored less than 20 marks.

(iii) No. of students who scored more than 60% marks i.e.
60% of 50
=(60×50)/100
= 3000/100
= 30 Marks

{33, 37, 42, 39, 32, 35, 31, 36, 38, 44, 39, 49, 32, 31, 34, 36, 36, 33, 47, 48, 50, 39, 36, 45, 47} = 25

So, the number of students who scored more than 60% marks is 25.

Class size
= UL-LL
= 40-30
= 10

Class Interval : 30-40?

Class Marks (xi)
= (LL+UL )/ 2
= (30+40)/2
= 70/2
= 35

One day Hemesh was going on a road trip with his father. They started . . .

Case Study Based Question on Lines and Angles

One day Hemesh was going on a road trip with his father. They started their journey on a straight road. Later walking some distance they reach a crossroad where one straight road intersects another at 45° as shown in figure.

Based on above information, answer the following questions :

(i) Find the measure of ∠BOC

(ii) Find the measure of ∠BOD

(iii) Find the reflex ∠BOC

Vertically opposite angles have common arms, Is it true?

Detailed Solution :

(i) ∠AOC + ∠BOC = 180°
45°+∠BOC = 180°
∠BOC = 180°-45°

∠BOC = 135°

(ii) ∠BOD=∠AOC [Vertically opposite angles]

∠BOD=45° [∵∠AOC=45°]

(iii) Reflex ∠BOC
= 360°-∠BOC
=360°-135°
= 225°

Hence, Reflex ∠BOC=225°

False, vertically opposite angles do not have common arms. Adjacent angles and linear pairs of angles have common arms.

A grinding mill manufactures spherical steel balls of various radii . . .

Case Study Based Question on Surface Area and Volume

A grinding mill manufactures spherical steel balls of various radii, ie. 5mm, 7mm, 10mm, 14mm, 16mm and so on. One day the mill manufactured 10 balls of 7mm radius and 20 balls of 3.5mm radius. The mill was having 18,000 mm3 steel. This 18,000 mm3 steel was melted and 10 balls of 7mm radius and 20

balls of 3.5 mm radius were made and the remaining steel was stored for future use.

Answer the following questions :

i) Find the volume of 10 balls each of radius 7 mm.

ii) Find the volume of 20 balls each of radius 3.5 mm.

iii) How much steel was kept for future use?

Find the surface area of one ball of radius 7 mm.

Detailed Solution :

(i) Volume of 10 balls each of radius 7 mm
= 10×Volume of each ball of radius 7 mm
= 10×(4/3)πr³
= 10×(4/3)(22/7)(7)³
= 10×(4/3)22×49
= (40×22×49)/3
= 43120/3
= 14373.33 mm³ (approx)

(ii) Volume of 20 balls each of radius 3.5 mm
r=3.5 mm
r= 3.5/10 = 7/2 mm

Required Volume = 20×Volume of each ball of radius 7/2 mm
= 20×(4/3)πr³
= 20×(4/3)(22/7)(7/2)³
= 10×(1/3)(22)(49)
= (10×22×49)/3
= 10780/3
= 3593.33 mm³ (approx)

(iii) Total amount of steel = 18000 mm³
Amount of steel used = 14373.33 + 3593.33

= 17966.66 mm³

Amount of steel kept for future use
= 18000-17966.66
= 33.34 mm³ (approx)

OR

Surface area of one ball of radius 7 mm
= 4πr²
= 4(22/7)(7)²
= 4×22×7
= 28×22
= 616 mm²

MCQ Questions of Quadrilaterals Class IX Maths

1.) Three angles of a quadrilateral are 75º, 90º and 75º. The fourth angle is

(A) 90º
(B) 95º
(C) 105º
(D) 120º

2. A diagonal of a rectangle is inclined to one side of the rectangle at 25º. The acute angle between the diagonals is

(A) 55º
(B) 50º
(C) 40º
(D) 25º

3. ABCD is a rhombus such that ∠ACB = 40º. Then ∠ADB is

(A) 40º
(B) 45º
(C) 50º
(D) 60º

Hint : Diagonal of Rhombus bisect each other at 90º.

4. The quadrilateral formed by joining the mid-points of the sides of a quadrilateral

PQRS, taken in order, is a rectangle, if

(A) PQRS is a rectangle

(B) PQRS is a parallelogram

(D) diagonals of PQRS are equal.

5. The quadrilateral formed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rhombus, if

(A) PQRS is a rhombus

(B) PQRS is a parallelogram

(D) diagonals of PQRS are equal.

6. If angles A, B, C and D of the quadrilateral ABCD, taken in order, are in the ratio 3:7:6:4, then ABCD is a

(A) rhombus
(B) parallelogram

Hint : Two pair of adjacent agnles are supplementary.

7. If bisectors of ∠A and ∠B of a quadrilateral ABCD intersect each other at P, of ∠B and ∠C at Q, of ∠C and ∠D at R and of ∠D and ∠A at S, then PQRS is a

(A) rectangle
(B) rhombus
(C) parallelogram

(D) quadrilateral whose opposite angles are supplementary

8. If APB and CQD are two parallel lines, then the bisectors of the angles APQ, BPQ, CQP and PQD form

(A) a square
(B) a rhombus

If two parallel lines are intersected intersect by a transversal, then the bisectors of the two pairs of interior angles enclose

(a) a square

(b) a parallelogram

(d) a trapezium

9. The figure obtained by joining the mid-points of the sides of a rhombus, taken in order, is
(A) a rhombus
(B) a rectangle
(C) a square
(D) any parallelogram

Hint : Check Question No. 4.

10. D and E are the mid-points of the sides AB and AC of ∆ABC and O is any point on side BC. O is joined to A. If P and Q are the mid-points of OB and OC respectively, then DEQP is
(A) a square
(B) a rectangle
(C) a rhombus
(D) a parallelogram

Hint : If one pair of opposite sides of a quadrilateral in equal and parallel then it is a _________.

11. The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only if,

(A) ABCD is a rhombus

(B) diagonals of ABCD are equal

(D) diagonals of ABCD are perpendicular.

Hint : Check Question No. 5.

12. The diagonals AC and BD of a parallelogram ABCD intersect each other at the

point O. If ∠DAC = 32º and ∠AOB = 70º, then ∠DBC is equal to

(A) 24º
(B) 86º
(C) 38º
(D) 32º

Hint : Use the concept of alternate interior angles, linear pair and angle sum property.

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

(A) opposite sides are equal

(B) opposite angles are equal

(D) diagonals bisect each other.

Answer Key

1. (D) 120º

2. (B) 50º

3. (C) 50º

4. (C) diagonals of PQRS are perpendicular

5. (D) diagonals of PQRS are equal.

6. (C) trapezium

7. (D) quadrilateral whose opposite angles are supplementary

8. (C) a rectangle or (c) a rectangle

9. (B) a rectangle
10. (D) a parallelogram

11. (C) diagonals of ABCD are equal and perpendicular

12. (C) 38º

13. (C) opposite angles are bisected by the diagonals

Bisectors of angles B and C of a triangle ABC intersect . . .

Question : Bisectors of angles B and C of a triangle ABC intersect each other at the point O. Prove that ∠BOC=90°+∠A/2

OR

In a ΔABC, if the bisectors of ∠B and ∠C meet at point O, then ∠BOC is equal to 90°+∠A/2

Solution :

In ΔABC

∠A+∠B+∠C=180°(Angle Sum Property)
Multiply both sides by ½
½×∠A+½×∠B+½×∠C=½×180°
∠A/2+∠B/2+∠C/2 =90°
∠A/2+∠B/2+∠C/2 =90°
∠A/2+∠1+∠2 =90°
∠1+∠2 =90°-∠A/2— (1)

In ΔOBC

∠BOC+∠1+∠2=180° (ASP)
∠BOC+90°-∠A/2=180° [From eq. (1)]
∠BOC=180°-90°+∠A/2

∠BOC=90°+∠A/2

For a particular year, following in the distribution of ages . . .

Case Study Based Question on Statistics
For a particular year, following in the distribution of ages (in years) of primary school teachers in a district :

Age (in years)	15-20	20-25	25-30	30-35	35-40	40-45	45-50
No. of Teachers	10	30	50	50	30	6	4

(a) Determine the class limits of the fourth class interval.

(b) Find the class marks of the class 45-50. Also determine the class size.

(d) How many primary school teachers are of age more than or equal to 35 years.

Doubt by Veer

Solution :

(a) Fourth Class Interval is 30-35
Lower class limit is 30
Upper Class limit is 35

(b) Class Marks & Class size of the class 45-50

Class Marks (xi)
= [LL+UL]/ 2
= [45+50]/2
= 95/2
= 47.5

Class Size (h)
= UL-LL
= 50-45
= 5

Upper limit of fifth class interval (35–40)
= 40

Sum = 15+40
= 55

(d) No. of primary school teachers which are of age more than or equal to 35 years
= 30+6+4
= 40

An angle is 24° less than its complementary angle. . .

Question : An angle is 24° less than its complementary angle. The angles measure is :

(a) 36°

(b) 33°

(d) 57°

Doubt by Veer

Solution :

Let the angle be x°
Compelent of x° = (90°-x°)

ATQ
(90°-x°)-x°=24°
90°-x°-x°=24°
90°-2x°=24°
90°-24°=2x°
66°=2x°
66°/2=x°
33°=x°
x°=33°

Hence, the correct option is (b) 33°.

If the drops of water are spherical and their diameter is 0.1 cm . . .

Question : If the drops of water are spherical and their diameter is 0.1 cm, 32000 such drops fills a conical glass up to its brim, with diameter equal to height, find the height of the glass.

Doubt by Veer

Solution :

Diameter of spherical drop (d) = 0.1 cm

Radius of spherical drop (r) = 0.1/2
=1/20 cm

For Conical Glass
D=H
2R=H
R=H/2

Volume of 32000 raindrops = volume of conical glass

32000×(4/3)πr³ = (1/3)πR²H 32000×4r³ = R²H 32000×4(1/20)³ = (H/2)²H 32000×4(1/8000) = (H²/4)H 4×4 = H³/4 4×4×4=H³ H³=(4)³ H=4 cm

Hence, the height of the glass is 4 cm

191125

Question : In the given figure, O is the centre of a circle prove that ∠x+∠ y=∠ z.

Doubt by Kashvi

Solution : Coming Soon

MCQs of Circles

MCQ Questions of Circles

1. AD is a diameter of a circle and AB is a chord. If AD=34 cm, AB=30 cm, the distance of AB from the centre of the circle is :

(A) 17 cm
(B) 15 cm
(C) 4 cm
(D) 8 cm

2. In given figure, if OA=5 cm, AB=8 cm and OD is perpendicular to AB, then CD is equal to:

(A) 2 cm
(B) 3 cm

3. If AB=12 cm, BC=16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is:

(A) 6 cm
(B) 8 cm

4. In given figure, if ∠ABC=20º, then ∠AOC is equal to:

(A) 20º
(B) 40º
(C) 60º
(D) 10º

5. In given figure, if AOB is a diameter of the circle and AC=BC, then ∠CAB is equal to:

(A) 30º
(B) 60º

6. In given figure, if ∠OAB=40º, then ∠ACB is equal to :

(A) 50º
(B) 40º
(C) 60º
(D) 70°

7. In given figure, if ∠DAB=60º, ∠ABD=50º, then ∠ACB is equal to:

(A) 60º
(B) 50º
(C) 70º
(D) 80º

8. ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and

∠ADC=140º, then ∠BAC is equal to:

(A) 80º
(B) 50º

9. In given figure, BC is a diameter of the circle and ∠BAO=60º. Then ∠ADC is equal to :

(A) 30º
(B) 45º

10. In given figure, ∠AOB=90º and ∠ABC=30º, then ∠CAO is equal to:

(A) 30º
(B) 45º
(C) 90º
(D) 60º

11. In given figure, two congruent circles have centres O and O′. Arc AXB subtends an angle of 75º at the centre O and arc A′ Y B′ subtends an angle of 25º at the centre O′. Then the ratio of arcs A X B and A′ Y B′ is:

(A) 2 : 1
(B) 1 : 2
(C) 3 : 1
(D) 1 : 3

12. In given figure, AB and CD are two equal chords of a circle with centre O. OP and OQ are perpendiculars on chords AB and CD, respectively. If ∠POQ=150º, then ∠APQ is equal to

(A) 30º
(B) 75º

ANSWER KEY

1. (D)

2. (A)

3. (C)

4. (B)

5. (D)

6. (A)

7. (C)

8. (B)

9. (C)

10. (D)

11. (C)
12. (B)

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