Class-ix mathematics formative Assignment

SECTION A (4 x 1M= 4M)

1. In the given figure fig. 9.1, ABCD is a parallelogram and AEB is a triangle. If Ar (ABCD) = 172.6cm²,

 then Ar ( ABE) =_________  (a) 86.3cm2 (b) 172.6cm2 (c) 345.2cm2 (d) cannot be determined.

2. In the given figure fig. 9.2, chord AB at a distance of 9cm from the center O of the given circle. If radius of the circle is 41cm, length of chord AB is:         

(a) 40cm (b) 80cm (c) 50cm (d) cannot be determined

3. In fig. 9.3 ABCD is a parallelogram, AP ^ DC and CQ ^ AD. If AB = 10cm, AD = 8cm and AP = 8cm then 

CQ = ----                                           

(a) 16cm (b) 10cm (c) 8 cm (d) none of these.

4. In fig. 9.4, If O is the center of the circle and AB = CD, then OL : OM = __ (a) 2:1 (b) 1:2 (c) 1:1 (d) none of these.

                                                                 

 SECTION B (3 x 2M = 6M)

5. in fig. 9.5, A and B are points on sides PQ and QR of parallelogram PQRS. Show that Ar (ARS) = Ar(PBS)                                                                                                                              

6. In fig. 9.6, AB and BC are two chords of a circle with centre O such that <ABO = <CBO.  Show that AB = BC

7. Construct the angle of measurement 22½ ° 

                                                                                

SECTION C (2 x 3M = 6M)      

                                                  

8. A line l parallel to side BC of ABC meets AB at X and AC at Y. Also lines parallel to AB through C and AC through B meets line l at E and F respectively. 

Then prove that Ar (ABF) = Ar (ACE).

9. In fig. 9.7, O is the centre of the circle. Determine ÐDAC, ÐACB, ÐADE.      

   

 SECTION D (1 x 4M = 4M)

10. Prove 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.                                                                                                      

                                           (OR),

In fig. 9.8, P is a point in the interior of a parallelogram ABCD. Show that 

1) Ar (APB) + Ar (PCD) = ½ Ar(ABCD)       

2) Ar(APD) + Ar(PBC) = Ar(APB) + Ar(PCD)

9th ncert solution Ch area of parallelogram and triangle optional exercise

1.Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.

Solution. IIgm ABCD and rectangle ABEF are between the same parallels AB and CF.

 AB = EF (For rectangle) and AB = CD (For parallelogram)

∴ CD = EF  ⇒ AB + CD = AB + EF ... (1)

Of all the line segments that can be drawn to a given line from a point not lying on it, the perpendicular line segment is the shortest.

∴ AF < AD

similarly we write , BE < BC          ∴ AF + BE < AD + BC ... (2)

From equations (1) and (2), we get

AB + EF + AF + BE < AD + BC + AB + CD
Perimeter of rectangle ABEF < Perimeter of parallelogram ABCD

NCERT Solutions IX Area of parallelograms and triangles

Q.1 A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented.
Ans: Let quadrilateral ABCD be original shape of field. We need to join diagonal BD and have to draw a line parallel to BD through point A. that meet the extended side CD at point E.

Now we join BE and AD. that intersect each other at O. 

In this way part ∆AOB can be cut from the original field to make new shape of field will be ∆ BCE.


Now we have to prove that the area of ∆AOB (portion that was cut so as to construct Health Centre) is equal to the area of the ∆DEO (portion added to the field so as to make the area of new field so formed equal to the area of original field) 

∆DEB and ∆DAB lie on same base BD and are between same parallels BD and AE.
∴ Area (∆DEB) = area (∆DAB)
⇒ Area (∆DEB) – area (∆DOB) = area (∆DAB) – area (∆DOB)
⇒ Area (∆DEO) = area (∆AOB)

2. In the given figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. show that   (i) ar (ACB) = ar (ACF)  (ii) ar (AEDF) = ar (ABCDE) 

  

Solution : (i)  ∆ACB and ∆ ACF are on the same base AC and are between  the same parallels AC and BF   ∴ area (∆ACB) = area (∆ACF) 

(ii) Area (∆ACB) = area (∆ACF)   

⇒ Area (∆ACB) + area (ACDE) = area (ACF) + area (ACDE) 

⇒ Area (ABCDE) = area (AEDF)

3.  In the given figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that:  

 (i) ar (DOC) = ar (AOB)  (ii) ar (DCB) = ar (ACB)   (iii) DA || CB or ABCD is a parallelogram.

Solution: We have to draw DN ⊥ AC and BM ⊥ AC 

(i) In ∆DON and ∆BOM

   ∠DNO = ∠BMO   (By construction) 

∠DON = ∠BOM   (Vertically opposite angles) 

OD = OB   (Given) 

By A–A–S congruence rule 

∆DON ≅ ∆BOM 

∴ DN = BM      ... (1) 

We know that congruent triangles have equal areas. 

∴ Area (∆DON) = area (∆BOM)    ... (2) 

In ∆DNC and ∆BMA 

∠DNC = ∠BMA  

CD = AB   

DN = BM   

∴ ∆DNC ≅ ∆BMA   (RHS congruency) 

⇒ area (∆DNC) = area (∆BMA)   ... (3) 

On adding equation (2) and (3), we have 

Area (∆DON) + area (∆DNC) = area (∆BOM) + area (∆BMA) 

So, area (∆DOC) = area (∆AOB) 

(ii) We have 

Area (∆DOC) = area (∆AOB) 

   ⇒ Area (∆DOC) + area (∆OCB) = area (∆AOB) + area (∆OCB) 

      (Adding area (∆OCB) to both sides) 

   ⇒ Area (∆DCB) = area (∆ACB) 

(iii) Area (∆DCB) = area (∆ACB) 

Now if two triangles are having same base and equal areas, these will be between same parallels 

∴ DA || CB      ... (4) 

For quadrilateral ABCD, we have one pair of opposite sides are equal 

(AB = CD) and other pair of opposite sides are parallel (DA || CB).  

Therefore, ABCD is parallelogram 

VIII MATHS ASSIGNMENT SA-2 unit Exponents and Radicals, Polynomials, Compound Interest

Assignment S.A.–2 Class:X Quadratic Equations

An equation of the form ax^2 + bx + c = 0, where a, b, c are real numbers but a ≠ 0. is called  a quadratic equation.

A quadratic equation in which terms  are arranged in descending order of their degrees  is called the standard form of the equation. That is, ax^2 + bx + c = 0, a ≠ 0

The roots of the quadratic equation: a real number α is called a root of the quadratic equation if  a α2 + bα + c = 0

Class 10  Quadratic Equation Solved CBSE Test Papers

10th Quadratic Equations Solved CBSE Exam Paper
10th Quadratic Equations Solved CBSE Test Paper -1
10th Quadratic Equations Solved CBSE Test Paper -2
Class 10 Quadratic Equation Solved Practice Paper-1
10th Quadratic Equations Solved CBSE Test Paper -3
Class 10 Quadratic Equation Solved Question Paper-2
Class 10 Quadratic Equation Solved Practice Paper-3
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Class 9th (SA-II) MATHEMATICS Assignment-2011for FA-3

JSUNIL TUTORIAL,SAMASTIPUR

Class 9^th (SA-II)

^{MATHEMATICS Assignment-I (For month of November)}     

Q1.  Determine the point on the graph of the linear equation x+y=6, whose ordinate is twice its abscissa.

Q2.  How many solution(s) of the equation 3x+2=2x-3 are there on the

          i) Number Line            ii) Cartesian plane

Q3.  Draw the graph of the equation represented by the straight line which is parallel to the x-axis and 3 units above it.

Q4. Find the solutions of the linear equation x+2y=8, which represents a point on

           i) x axis                            ii) y-axis

Q5.  For what values of c, the linear equation 2x+cy=8 has equal values of x and y as its solution.

Q6. Give the geometrical interpretations of 5x+3=3x-7 as an equation

           i) in one variable              ii) In two variables

Q7. Draw the graph of the equation 3x+4y=6. At what points, the graph cut the x-axis and the y-axis.

Q8. At what point does the graph of equation 2x+3y=9 meet a line which is parallel to y -axis at a distance 4 units from the origin and on the right side of the y-axis.

Q9.  P is the mid point of side BC of parallelogram ABCD such that AP bisects angle A.

Prove that AD =2CD.

Q10. Prove that bisector of any two consecutive angles of parallelogram intersect at right angles.

Q11. E and F are respectively the midpoints of non parallel sides AD and BC of trapezium. Prove that EF is parallel to AB and EF=1/2(AB+CD).

Q12.  ABCD is a rectangle in which diagonal BD bisects angle B. Show that ABCD is a Square.

Q13.  Diagonals of Quadrilateral ABCD bisect each other. If angle A = 35 degree, determine angle B.

Q14. The bisectors of angle B and angle D of quadrilateral ABCD meet CD and AB, produced at point P and Q respectively. Prove that angle P+angle Q= ½(angle ABC+ angle ADC).

Q15. In parallelogram ABCD, AB=10cm, AD= 6cm. The bisector of angle A meets DC in A. AE and BC produced meet at F. Find the length of CF.

Q16. Evaluate: (5x+1) (x+3)-8= 5(x+1) (x+2).

IX Chapter-15 probability Points to Remember

IX Chapter-15 : PROBABILITY - NCERT solution - Key points CCE Test papers Solved Theorem and Illustrated examples
1. An activity which gives a result is called an experiment.
2. An experiment which can be repeated a number of times under the same set of conditions, and the outcomes are not predictable is called a Random Experiment.
3. Performing an experiment is called a trial.
4. Any outcome of an experiment is known as an event.
5. In n trials of a random experiments, if an event E happens m times, then the probability of happening of E is given by,
P (E)= Number of outcomes favour to E /Total number of possible outcomes
6. For any event E, which is associated to an experiment, we have 0 lessthan and equal P(E) greater and equalto 1.
7. If E1, E2, E3, ....., En are n elemantary events associated to a random experiment, then
P(E1) + P(E2) + P(E3) + ........... + P(En) = 1
PRACTICE EXERCISE
1. A number is chosen from 1 to 20. Find the probability that the number chosen is :
(i) a prime number (ii) a composite number
(iii) a square number (iv) an odd number
(v) an even number (vi) number between 7 and 14
2. A bag contains 9 red and 6 blue balls. Find the probability that a ball drawn from a bag at random is (i) Red ball (ii) blue ball
3. In a sample of 500 items, 120 are found to be defective. Find the probability that the item selected at random is (i) defective (ii) non-defective
4. In a school of 1800 students, there are 875 girls. Find the probability that a student chosen at random is (i) a boy (ii) a girl
5. In a cricket match, a batsman hit a boundary 12 times out 45 balls he plays. Find the probability that hedid not hit a boundary.
6. A coin is tossed 700 times and we get:
head : 385 times; tail : 315 times.
When a coin is tossed at random,what is the probability of getting : (i) a head? (ii) a tail?
7. Two coins are tossed 600 times and we get two heads : 138 times, one head : 192 times ; no head : 270times.
When two coins are tossed at random, what is the probability of getting :
(i) 2 heads? (ii) 1 head? (iii) no head?
8. Three coins are tossed 250 times and we get:
3 heads : 46 times; 2 heads : 56 times; 1 head : 70 times; 0 head : 78 times.
When three coins are tossed at random, what is the probability of getting :(i) 3 heads ? (ii) 2 heads? (iii) atleast 2 heads? (iv) atmost 2 heads?

Class IX Mathematics Ch 15. Probability NCERT Solutions by topperlearning.

IXQuadrilateral CCE Test Papers

CBSE ADDA: 9th Quadrilateral CCE Test Papers

Prove that followings:

1. A diagonal of a parallelogram divides it into two congruent triangles.
2. In a parallelogram, opposite sides and angle are equal.
3. If each pair of opposite sides of quadrilateral is equal, then it is a parallelogram.
4. If in a quadrilateral, each pair of opposite angles is equal, then it is a parallelogram.
5. The diagonals of a parallelogram bisect each other.
6. If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.
7. A quadrilateral is a parallelogram if a pair of opposite sides is equal and parallel.
8. The line drawn through the mid-point of one side of a triangle, parallel to another side bisects the third side.
9. The line segment joining the mid- points of the two sides of a triangle is parallel to the third side.
10. Show that each angle of a rectangle is a right angle.
11. Show that the diagonal of a rhombus are perpendicular to each other.
12. Show that the bisectors of the angles of a parallelogram form a rectangle.
13. ABCD is a parallelogram (||gm) in which P and Q are mid-points of opposite side AB and CD. If AQ intersects DP at S and BQ intersects CO at R, show that (i) APCQ is ||gm(ii)DPBQ is ||gm(iii) PSQR is ||gm
14. In Triangle ABC, D, E and Fare respectively the mid points of sides AB, BC and CA. Show that triangle ABC is divided into four congruent triangle by joining D, E and F
15. If the diagonal of a parallelogram are equal, then show that it is a rectangle.
16. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
17. Show that the diagonals of a square are equal and bisect each other at right angles.
18. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
19. In Δ ABC and Δ DEF, AB=DE, AB||DE, BC=EF and BC||EF. Vertices A, B and C are joined to vertices D, E and F respectively. Show that:(i) Quadrilateral ABCD is a parallelogram.(ii) Quadrilateral BEFC is a parallelogram.(iii) AD||CF and AD=CF(iv) Quadrilaterals ACFD is a parallelogram(v) AC=DF(vi) Δ ABC @ Δ DEF.
20. ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA. AC is a diagonal. Show that:(i) SR||AC and SR =1/2 AC(ii) PQ=SR(iii) PQRS is a parallelogram.
21. ABCD is a rhombus and P, Q, R and S are the mid- point of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.
22. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.
23. Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.
24. ABC is a triangle right angle at C. A line through the mid-points M of hypotenuse AM and parallel to BC intersects AC at D. Show that(i) D is the mid –point of AC(ii) MD ┴ AC(iii) CM=MA=1/2 AB.

Activity related to Area of parallelogram and triangle for class 9th

Activity related to to find a relation, between the areas of two parallelograms on the same base and between the same parallels.

Let us take a graph sheet and draw two parallelograms ABCD and PQCD on it as shown in Fig

Now find the areas of these two parallelograms by counting the number of complete squares enclosed by the figure, the number of squares a having more than half their parts enclosed by the figure and the number of squares having half their parts enclosed by the figure. The squares whose less than half parts are enclosed by the figure are ignored.
You will find that areas of both the parallelograms are (approximately) 15 Sq.cm

Hence, you to conclude that parallelograms on the same base and between the same parallels are equal in area

Theorem : Parallelograms on the same base and between the same parallels are equal in area.

Proof : Two parallelograms ABCD and EFCD, on the same base DC and between the same parallels

AF and DC are given (see Fig).
We need to prove that ar (ABCD) = ar (EFCD).
In Δ ADE and Δ BCF,
∠ DAE = ∠ CBF (Corresponding angles from AD || BC and transversal AF) (1)
∠ AED = ∠ BFC (Corresponding angles from ED || FC and transversal AF) (2)
Therefore, ∠ ADE = ∠ BCF (Angle sum property of a triangle) (3)
Also, AD = BC (Opposite sides of the parallelogram ABCD) (4)
So, Δ ADE ≅ Δ BCF [By ASA rule, using (1), (3), and (4)]
Therefore, ar (ADE) = ar (BCF) (Congruent figures have equal areas) (5)
Now,adding ar (EDCB) both the sides
 ar (ADE) + ar (EDCB) = ar (BCF)+ ar (EDCB)
                 ar (ABCD) = ar (EFCD)
So, parallelograms ABCD and EFCD are equal in area.

Problem: If a triangle and a parallelogram are on the same base and between the same parallels, then prove that the area of the triangle is equal to half the area of the parallelogram.
Solution : Let Δ ABP and parallelogram ABCD be on the same base AB and between the same parallels
AB and PC (see Fig).

To prove :  ar (PAB) =1/2 ar (ABCD)
Construction: Draw BQ || AP to obtain another parallelogram ABQP. So that parallelograms ABQP
and ABCD will be on the same base AB and between the same parallels AB and PC.
Therefore, ar (ABQP) = ar (ABCD) (By Theorem that parallelograms on the same base and between the same parallels are equal in area) --------(1)

But Δ PAB ≅ Δ BQP (As diagonal PB divides parallelogram ABQP into two congruent triangles.)

So, ar (PAB) = ar (BQP) -------------(2)

Therefore, ar (PAB) =1/2ar (ABQP) [From (2)]--------------------- (3)

This gives ar (PAB) =1/2 ar (ABCD) [From (1) and (3)]

Theorem :Two triangles on the same base (or equal bases) and between the same parallels are equal in area.

Let two triangles ABC and PBC on the same base BC and between the same parallels BC and AP

Construction : Draw CD || BA and CR || BP such that D and R lie on line AP(see Fig) to get two parallelograms PBCR and ABCD on the same base BC and between the same parallels BC and AR.

Now, Parallelograms PBCR and ABCD on the same base BC and between the same parallels BC and AR. Therefore, ar (ABCD) = ar (PBCR) ----------(1)

Now Δ ABC ≅ Δ CDA and Δ PBC ≅ Δ CRP ( The diagonals of a parallelogram divides it  into two congruent triangles)

So, ar(Δ ABC) = ar(Δ CDA) and ---------(ii)

ar (ABCD) = ar(Δ ABC) + ar(Δ CDA) --------------(iii)

ar (PBCR)  = ar(Δ PBC) = ar(ΔCRP) ----------------(iv)

From (1),(ii),(iii) and (iv) , we have 

So, ar (ABC) =1/2 ar (ABCD)

and ar (PBC) =1/2 ar (PBCR)

Therefore, ar (ABC) = ar (PBC)

Problem: A farmer was having a field in the form of a parallelogram PQRS. She took any point A

on RS and joined it to points P and Q. In how many parts the fields is divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it?

When she took any point A on RS and joined it to points P and Q

A divides the field into three parts. These parts are triangular in shape –> ∆PSA, ∆PAQ and ∆QRA 

So, Area of   IIgm  PQRS =  Area of ∆PSA + Area of ∆PAQ + Area of ∆QRA  ... (1)  

∴ Area (∆PAQ) = 1/2 area (PQRS)      ... (2) [a parallelogram and triangle are on the same base and 

between the same parallels, the area of triangle is half the area of the parallelogram]

From equations (1) and (2), we get

Area (∆PSA) + area (∆QRA) + 1/2 area (PQRS) = area (PQRS)   

Area (∆PSA) + area (∆QRA)=  area (PQRS) -1/2 area (PQRS) 

Area (∆PSA) + area (∆QRA)=  1/2 area (PQRS) 

Area (∆PSA) + area (∆QRA)= Area (∆PAQ)

Clearly, farmer must sow wheat in triangular part PAQ and pulses in other two triangular parts PSA and QRA or wheat in triangular part PSA and QRA and pulses in triangular parts PAQ.

Class-IX. Math. Chapter : Area of Parallelogram and Triangles. NCERT Solutions Topperlearning Download
9th Area of Parallelogram and Triangle Test paper

10th probability solved examples

ILLUSTRATIVE EXAMPLESExample 1. An unbiased die is thrown. What is the probability of getting :
(i) an odd number (ii) a multiple of 3 (iii) a perfect square number (iv) a number less than 4.

Solution.An unbiased die is thrown we may get 1, 2, 3,4,5,6So, total number of all possible outcomes= 6(i) Favourable outcomes for an odd number are 1, 3, 5.So, no. of favourable outcomes = 3P (an odd number) = No. of favourable outcomes/ Total no. of possible outcome= 3/6=1/2(ii) favourable outcomes for a multiple of 3 are 3 and 6.So, no. of favourable outcomes = 2P (a multiple of 3) = 2/6= 1/3
(iii) favourable outcomes for ) a perfect square number are 1 and 4.So, no. of favourable outcomes = 2 P (a perfect square number) =2/6= 1/3
(iv) favourable outcomes for a number less than 4. are 1, 2 and 3.So, no. of favourable outcomes = 3P (a number less than 4) = 3/6 =1/2

2. Three unbiased coins are tossed together. Find the probability of getting : (i) all heads (ii) two heads (iii) one head (iv) at least two heads

Solution. When three unbiased coins are tossed together, possible outcomes areHHH, HHT, HTH, HTT, THH, THT, TTH and TTT. So, total no. of possible outcomes = 8(i) favourable outcome = HHHSo, No. of favourable outcome = 1P (all heads) =no. of favourable outcomes/Total no. possible outcomes= 1/8(ii) favourable outcomes are HHT, THH and HTH.So, no. of favourable outcomes = 3P (two heads) = 3/8(iii) favourable outcomes are HTT, THT and TTH. So, no. of favourable outcomes = 3 P (one head) = 3/8(iv) favourable outcomes are HHH, HHT, HTH and THH.So, no. of favourable outcomes = 4P (at least two heads) 4/8 =1/2

Example 3. Find the probability that a leap year selected at random will contain 53 Sundays.

Solution. In a leap year, there are 366 days. But 366 days = 52 weeks + 2 days.Thus, a leap year has always 52 sundays.The remaining 2 days can be :(i) Sunday and Monday(ii) Monday and Tuesday(iii) Tuesday and Wednesday(iv) Wednesday and Thursday(v) Thursday and Friday(vi) Friday and Saturday(vii) Saturday and SundayClearly, there are seven elementary events associated with this random experiment.Let E be the event that a leap year has 53 sundays. Clearly, the event E will happer if the last twodays of the leap year are either Sunday and Monday or Saturday and Sunday.Favourable no. of elementary events = 2Hence, required probability = 2/7

Example 4. One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn.

Find the probability that the card drawn is :(i) an ace (ii) either red or king (iii) a face card (iv) a red face cardSolution. here, total no. of possible outcomes = 52.(i) There are 4 ace cards in a pack of 52 cards. One ace can be chosen in 4 ways.So, favourable no. of outcomes = 4P (an ace) no. = of favourable outcomes/Total no. of possible outcomes= 4/52=1/13
(ii) There are 26 red cards, including 2 red kings. Also, there are 4 kings, two red and two black.card drawn will be a red card or a king if it is any one of 28 cards (26 red cards and 2 black kings)So, favourable no. of outcomes = 28 P(either red or king) =28/52= 7/13
(iii) Kings, queens and jacks are the face cards.So, favourable no. of outcomes = 3 × 4 = 12 P(a face card) = 12/52 =3/13
(iv) There are 6 red face cards, 3 each from diamonds and hearts.So, favourable no. of outcomes = 6 P(a red face card) = 6/52= 3/26

Example 5. Cards marked with the numbers 2 to 101 are placed in a box and mixed thoroughly. One card is drawn from this box. Find the probability that the number of the card is :
(i) an even number (ii) a number less than 14(iii) a number which is a perfect square (iv) a prime number less than 20.

Solution.From 2 to 101, these are (101–2) + 1 = 100 numbers.So, total no. of possible outcomes = 100.(i) From 2 to 101, the even numbers are 2, 4, 6, ...., 100 which are 50 in number.So, number of favourable outcomes = 50 P(an even number) = no. of favourable outcomes/Total no. of possible outcomes = 50/100 = 1/2(ii) From 2 to 101, the numbers less than 14 are 2, 3, ...., 13 which are 12 in number.So, no. of favourable outcomes = 12 P(a number less than 14) = 12/100 = 3/25(iii) From 2 to 101, the perfect squares are 4, 9, 16, ..... 100, which are 9 in number. So, no. of favourable outcomes = 9 P (a number which is a perfect square) = 9/100(iv) From 2 to 101, the prime numbers less than 20 are 2, 3, 5, 7, 11, 13, 17 and 19 which are 8 innumber.So, no. of favourable outcomes = 8 P (a prime no. less than 20) = 8/100=2/25

Example 6. A bag contains 3 red balls and 5 black balls.A ball is drawn at random from a bag. What is the
probability that the ball drawn is : (i) red (ii) not red

Solution. Total number of balls = 3 + 5 = 8(i) P (red ball) = no. of red balls /Total no.of balls=3/8(ii) P (not red ball) = 1 – P(red ball) =1 – 3/8 =5/8

Example 7. A jar contains 24 marbles, some are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green is 2/3 . Find the number of blue marbles in the jar.

Solution. Total number of elementary events = 24.Let there be x green marbles.P (green marbles is drawn) = x/24but, P(green marbles is drawn)= 2/3 (given)So, x / 24 = 2/3 x=24x2/3 x = 16 Number of green marbles = 16Number of blue marbles = 24 – 16 = 8 Ans.

Example 8. Suppose you drop a die at random on the rectangular region shown in the figure. What is the probability that it will land inside the circle with diameter 1 m?

Solution. Total area of rectangular region = 3 m × 2 m = 6 m2Area of the circle = p r² = p (1/2)² m² = p/ 4 m²P (die to land inside the circle) = p/ 4 m² ÷ 6 = p/24

Example 9. A bag contains 12 balls out of which x are white.

(i) If one ball is drawn at random, what is the probability that it will be a white ball?

(ii) If 6 more white balls are put in the bag, the probability of drawing a white ball will be
double than that in (i). Find x.

Solution. (i) Total number elementary events = 12.There are x white balls out of which one can be chosen in x ways.So, favourable number of elementary events = x
p1 = P (white ball) = no. of favourable outcomes / Total no. of possible outcomes = x /12
(ii) If 6 more white balls are put in the bag, then total number of balls in the bag =12 + 6 = 18and, no. of white balls in the bag = (x + 6)P₂ = P (getting a white ball) = ( x + 6 )/ 18It is given that, p₂ = 2 p₁
( x + 6 )/ 18 = 2 ( x / 2)x = 3

Example 10. Two dice are thrown simultaneously. Find the probability of getting :

(i) a doublet i.e. same number on both dice. (ii) the sum as a prime number.
Solution. Possible outcomes associated to the random experiment of throwing two dice are :

(1, 1), (1, 2), ......., (1, 6) (2, 1), (2, 2), ......., (2, 6)..............................................................................(6,1) , (6, 2), ......., (6, 6)

Total number of possible outcomes = 6 × 6 = 36

(i) The favourable outcomes are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6).

Total no. of favourable outcomes = 6

So, P(a doublet) = no. of favourable outcomes/Total no. of possible outcomes = 6/36= 1/6

(ii) Here, favourable sum (as a prime number) are 2, 3, 5, 7 and 11.

So, favourable outcomes are (1, 1), (1, 2), (2, 1), (1, 4), (4, 1), (2, 3), (3, 2), (1, 6), (6, 1), (2, 5), (5, 2),(3, 4), (4,3), (6, 5) and (5, 6).

no. of favourable outcomes = 15

P (the sum as a prime number) = 15/36 = 5/12

10th Maths SA-2 Chapter wise Test Papers Links
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