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RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency VSAQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency VSAQS

Other Exercises

• RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.1
• RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.2
• RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.3
• RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.4
• RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency VSAQS
• RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency MCQS

Question 1.
If the ratio of mode and median of a certain data is 6 : 5, then find the ratio of its mean and median.
Solution:
We know that
Mode = 3 median – 2 mean…(i)
and \(\frac { mode }{ median } \) = \(\frac { 6 }{ 5 } \)
Mode = \(\frac { 6 }{ 5 } \)median
∴From (i), \(\frac { 6 }{ 5 } \) median = 3 median – 2 mean
=> 2 mean = 3 median – \(\frac { 6 }{ 5 } \)median
2 mean = \(\frac { 15-6 }{ 5 } \)median = \(\frac { 9 }{ 5 } \)median
\(\frac { mean }{ median } \) = \(\frac { 9 }{ 5X2 } \) = \(\frac { 9 }{ 10 } \)
∴Ratio = 9:10

Question 2.
If the mean of x + 2, 2x + 3, 3x + 4, 4x + 5 is x + 2, find x.
Solution:
Mean of x + 2, 2x + 3, 3x + 4, 4x + 5 = x + 2
=> \(\frac { x + 2+2x + 3+3x + 4+4x + 5 }{ 4 } \) = x + 2
=> 10x + 14 = 4x + 8
=> 10x – 4x = 8 – 14
=> 6x= – 6
∴ x = – 1

Question 3.
If the median of scores ,\(\frac { x }{ 2 } \), \(\frac { x }{ 3 } \), \(\frac { x }{ 4 } \), \(\frac { x }{ 5} \) and \(\frac { x }{ 6 } \) (where x > 0) is 6, then find the value \(\frac { x }{ 6 } \)
Solution:
\(\frac { x }{ 2 } \), \(\frac { x }{ 3 } \), \(\frac { x }{ 4 } \), \(\frac { x }{ 5} \), \(\frac { x }{ 6 } \)
Here n = 5
Median = \(\frac { n+1 }{ 2 } \) th term = \(\frac { 5+1 }{ 2 } \) th
\(\frac { 6 }{ 2 } \) = 3rd term = \(\frac { x }{ 4 } \)
\(\frac { x }{ 4 } \) = 6 => x = 24
\(\frac { x }{ 6 } \) = \(\frac { 24 }{ 6 } \) = 4
∴Hence = \(\frac { x }{ 6 } \) = 4

Question 4.
If the mean of 2, 4, 6, 8, x, y is 5, then find the value of x + y.
Solution:
Mean of 2, 4, 6, 8, x, y is 5
\(\frac { 2+4+6+8+x+y }{ 6 } \) = 5
\(\frac { 20+x+y }{ 6 } \) = 5
=> 20 + (x +y) = 30
=> x + y = 30 – 20 = 10
∴x + y = 10

Question 5.
If the mode of scores 3, 4, 3, 5, 4, 6, 6, x is 4, find the value of x.
Solution:
Mode of 3, 4, 3, 5, 4, 6, 6, x is 4
∴ 4 comes in maximum times
But here ,
3 2
4 2
5 1
6 2
3, 4 and 6 are equal in number
∴ x must be 4 so that it becomes in maximum times

Question 6.
If the median of 33, 28, 20. 25, 34, x is 29. find the maximum possible value of x.
Solution:
Median of 33, 28, 20, 25, 34, x is 29
Now arranging in ascending order 20, 25, 28, x, 33, 34
Here n = 6
Median = \(\frac { 1 }{ 2 } \left[ \frac { 6 }{ 2 } th\quad term+\left( \frac { 6 }{ 2 } +1 \right) th\quad term \right] \)
29 = \(\frac { 1 }{ 2 } \) [3rd term + 4th term]
29 = \(\frac { 1 }{ 2 } \) [28+x]
58 = 28 + x
=> x = 58 – 28 = 30
∴Possible value of x = 30

Question 7.
If the median of the scores 1, 2, x, 4, 5 (where 1 <2 <x <4 <5) is 3, then find the mean of the scores.
Solution:
Scores are 1, 2, x, 4, 5 and median 3
Here n = 5 which is odd
Median = \(\frac { n+1 }{ 2 } \) th term = \(\frac { 5+1 }{ 2 } \) = \(\frac { 6 }{ 2 } \) th
=> 3 = 3rd term = x
=> 3 = x
∴ x = 3
Mean of the score = \(\frac { 1+2+3+4+5 }{ 5 } \) = 3

Question 8.
If the ratio of mean and median of a certain data is 2 : 3, then find the ratio of its mode and mean.
Solution:
We know that mode = 3 median – 2 mean

\(\frac { mode }{ mean } \) = \(\frac { 5 }{ 2 } \)
Ratio in mode and mean = 5 : 2

Question 9.
The arithmetic mean and mode of a data are 24 and 12 respectively, then find the median of the data.
Solution:
Mean = 24
Mode = 12
We know that mode = 3 median – 2 mean
12 = 3 median – 2 x 24
12 = 3 median – 48
3 median 12 + 48 = 60
Median = \(\frac { 60 }{ 3 } \) = 20

Question 10.
If the difference of mode and median of a data is 24, then find the difference of median and mean.
Solution:
Mode – Median = 24
Mode = 24 + median
But mode = 3 median – 2 mean
3 median – 2 mean = 24 + median
3 median – median – 2 mean = 24
=> 2 median – 2 mean = 24
=> Median – Mean = 12 (Dividing by 2)

Hope given RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency VSAQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry Ex 11.2

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry Ex 11.2

Other Exercises

• RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry Ex 11.1
• RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry Ex 11.2
• RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry VSAQS
• RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS

Question 1.
The exterior angles obtained on producing the base of a triangle both ways are 104° and 136°. Find all the angles of the triangle.
Solution:
In ∆ABC, base BC is produced both ways to D and E respectivley forming ∠ABE = 104° and ∠ACD = 136°

Question 2.
In the figure, the sides BC, CA and AB of a ∆ABC have been produced to D, E and F respectively. If ∠ACD = 105° and ∠EAF = 45°, find all the angles of the ∆ABC.

Solution:
In ∆ABC, sides BC, CA and BA are produced to D, E and F respectively.
∠ACD = 105° and ∠EAF = 45°
∠ACD + ∠ACB = 180° (Linear pair)
⇒ 105° + ∠ACB = 180°
⇒ ∠ACB = 180°- 105° = 75°
∠BAC = ∠EAF (Vertically opposite angles)
= 45°
But ∠BAC + ∠ABC + ∠ACB = 180°
⇒ 45° + ∠ABC + 75° = 180°
⇒ 120° +∠ABC = 180°
⇒ ∠ABC = 180°- 120°
∴ ∠ABC = 60°
Hence ∠ABC = 60°, ∠BCA = 75°
and ∠BAC = 45°

Question 3.
Compute the value of x in each of the following figures:

Solution:
(i) In ∆ABC, sides BC and CA are produced to D and E respectively


(ii) In ∆ABC, side BC is produced to either side to D and E respectively
∠ABE = 120° and ∠ACD =110°
∵ ∠ABE + ∠ABC = 180° (Linear pair)

(iii) In the figure, BA || DC

Question 4.
In the figure, AC ⊥ CE and ∠A: ∠B : ∠C = 3:2:1, find the value of ∠ECD.

Solution:
In ∆ABC, ∠A : ∠B : ∠C = 3 : 2 : 1
BC is produced to D and CE ⊥ AC
∵ ∠A + ∠B + ∠C = 180° (Sum of angles of a triangles)
Let∠A = 3x, then ∠B = 2x and ∠C = x
∴ 3x + 2x + x = 180° ⇒ 6x = 180°
⇒ x = \(\frac { { 180 }^{ \circ } }{ 6 }\)  = 30°
∴ ∠A = 3x = 3 x 30° = 90°
∠B = 2x = 2 x 30° = 60°
∠C = x = 30°
In ∆ABC,
Ext. ∠ACD = ∠A + ∠B
⇒ 90° + ∠ECD = 90° + 60° = 150°
∴ ∠ECD = 150°-90° = 60°

Question 5.
In the figure, AB || DE, find ∠ACD.

Solution:
In the figure, AB || DE
AE and BD intersect each other at C ∠BAC = 30° and ∠CDE = 40°
∵ AB || DE
∴ ∠ABC = ∠CDE (Alternate angles)

⇒ ∠ABC = 40°
In ∆ABC, BC is produced
Ext. ∠ACD = Int. ∠A + ∠B
= 30° + 40° = 70°

Question 6.
Which of the following statements are true (T) and which are false (F):
(i) Sum of the three angles of a triangle is 180°.
(ii) A triangle can have two right angles.
(iii) All the angles of a triangle can be less than 60°.
(iv) All the angles of a triangle can be greater than 60°.
(v) All the angles of a triangle can be equal to 60°.
(vi) A triangle can have two obtuse angles.
(vii) A triangle can have at most one obtuse angles.
(viii) If one angle of a triangle is obtuse, then it cannot be a right angled triangle.
(ix) An exterior angle of a triangle is less than either of its interior opposite angles.
(x) An exterior angle of a triangle is equal to the sum of the two interior opposite angles.
(xi) An exterior angle of a triangle is greater than the opposite interior angles.
Solution:
(i) True.
(ii) False. A right triangle has only one right angle.
(iii) False. In this, the sum of three angles will be less than 180° which is not true.
(iv) False. In this, the sum of three angles will be more than 180° which is not true.
(v) True. As sum of three angles will be 180° which is true.
(vi) False. A triangle has only one obtuse angle.
(vii) True.
(viii)True.
(ix) False. Exterior angle of a triangle is always greater than its each interior opposite angles.
(x) True.
(xi) True.

Question 7.
Fill in the blanks to make the following statements true:
(i) Sum of the angles of a triangle is ………
(ii) An exterior angle of a triangle is equal to the two …….. opposite angles.
(iii) An exterior angle of a triangle is always …….. than either of the interior opposite angles.
(iv) A triangle cannot have more than ………. right angles.
(v) A triangles cannot have more than ……… obtuse angles.
Solution:
(i) Sum of the angles of a triangle is 180°.
(ii) An exterior angle of a triangle is equal to the two interior opposite angles.
(iii) An exterior angle of a triangle is always greater than either of the interior opposite angles.
(iv) A triangle cannot have more than one right angles.
(v) A triangles cannot have more than one obtuse angles.

Question 8.
In a ∆ABC, the internal bisectors of ∠B and ∠C meet at P and the external bisectors of ∠B and ∠C meet at Q. Prove that ∠BPC + ∠BQC = 180°.
Solution:
Given : In ∆ABC, sides AB and AC are produced to D and E respectively. Bisectors of interior ∠B and ∠C meet at P and bisectors of exterior angles B and C meet at Q.

To prove : ∠BPC + ∠BQC = 180°
Proof : ∵ PB and PC are the internal bisectors of ∠B and ∠C
∠BPC = 90°+ \(\frac { 1 }{ 2 }\) ∠A …(i)
Similarly, QB and QC are the bisectors of exterior angles B and C
∴ ∠BQC = 90° + \(\frac { 1 }{ 2 }\) ∠A …(ii)
Adding (i) and (ii),
∠BPC + ∠BQC = 90° + \(\frac { 1 }{ 2 }\) ∠A + 90° – \(\frac { 1 }{ 2 }\) ∠A
= 90° + 90° = 180°
Hence ∠BPC + ∠BQC = 180°

Question 9.
In the figure, compute the value of x.

Solution:
In the figure,
∠ABC = 45°, ∠BAD = 35° and ∠BCD = 50° Join BD and produce it E

Question 10.
In the figure, AB divides ∠D AC in the ratio 1 : 3 and AB = DB. Determine the value of x.

Solution:
In the figure AB = DB

Question 11.
ABC is a triangle. The bisector of the exterior angle at B and the bisector of ∠C intersect each other at D. Prove that ∠D = \(\frac { 1 }{ 2 }\) ∠A.
Solution:
Given : In ∠ABC, CB is produced to E bisectors of ext. ∠ABE and into ∠ACB meet at D.

Question 12.
In the figure, AM ⊥ BC and AN is the bisector of ∠A. If ∠B = 65° and ∠C = 33°, find ∠MAN.

Solution:

Question 13.
In a AABC, AD bisects ∠A and ∠C > ∠B. Prove that ∠ADB > ∠ADC.
Solution:
Given : In ∆ABC,
∠C > ∠B and AD is the bisector of ∠A

To prove : ∠ADB > ∠ADC
Proof: In ∆ABC, AD is the bisector of ∠A
∴ ∠1 = ∠2
In ∆ADC,
Ext. ∠ADB = ∠l+ ∠C
⇒ ∠C = ∠ADB – ∠1 …(i)
Similarly, in ∆ABD,
Ext. ∠ADC = ∠2 + ∠B
⇒ ∠B = ∠ADC – ∠2 …(ii)
From (i) and (ii)
∵ ∠C > ∠B (Given)
∴ (∠ADB – ∠1) > (∠ADC – ∠2)
But ∠1 = ∠2
∴ ∠ADB > ∠ADC

Question 14.
In ∆ABC, BD ⊥ AC and CE ⊥ AB. If BD and CE intersect at O, prove that ∠BOC = 180°-∠A.
Solution:
Given : In ∆ABC, BD ⊥ AC and CE⊥ AB BD and CE intersect each other at O

To prove : ∠BOC = 180° – ∠A
Proof: In quadrilateral ADOE
∠A + ∠D + ∠DOE + ∠E = 360° (Sum of angles of quadrilateral)
⇒ ∠A + 90° + ∠DOE + 90° = 360°
∠A + ∠DOE = 360° – 90° – 90° = 180°
But ∠BOC = ∠DOE (Vertically opposite angles)
⇒ ∠A + ∠BOC = 180°
∴ ∠BOC = 180° – ∠A

Question 15.
In the figure, AE bisects ∠CAD and ∠B = ∠C. Prove that AE || BC.

Solution:
Given : In AABC, BA is produced and AE is the bisector of ∠CAD
∠B = ∠C

To prove : AE || BC
Proof: In ∆ABC, BA is produced
∴ Ext. ∠CAD = ∠B + ∠C
⇒ 2∠EAC = ∠C + ∠C (∵ AE is the bisector of ∠CAE) (∵ ∠B = ∠C)
⇒ 2∠EAC = 2∠C
⇒ ∠EAC = ∠C
But there are alternate angles
∴ AE || BC

Hope given RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry Ex 11.2 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 9 Solutions Chapter 25 Probability Ex 25.1

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 25 Probability Ex 25.1

Other Exercises

• RD Sharma Class 9 Solutions Chapter 25 Probability Ex 25.1
• RD Sharma Class 9 Solutions Chapter 25 Probability VSAQS
• RD Sharma Class 9 Solutions Chapter 25 Probability MCQS

Question 1.
A coin is tossed 1000 times with the following frequencies
Head : 455, Tail : 545.
Compute the probability for each event.
Solution:
Total number of events (m) 1000
(i) Possible events (m) 455
∴Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 455 }{ 1000 } \)
= \(\frac { 91 }{ 200 } \) = 0.455
(ii) Possible events (m) = 545
∴Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 545 }{ 1000 } \) = \(\frac { 109 }{ 200 } \) = 0.545

Question 2.
Two coins are tossed simultaneously 500 times with the following frequencies of different
outcomes:
Two heads : 95 times
One tail : 290 times
No head: 115 times
Find the probability of occurrence of each of these events.
Solution:
Two coins are tossed together simultaneously 500 times
∴ Total outcomes (n) 500
(i) 2 heads coming (m) = 95 times
∴Probability P(A) = \(\frac { m }{ n } \)
= \(\frac { No. of possible events }{ Total number of events } \)
= \(\frac { 95 }{ 500 } \) = \(\frac { 19 }{ 100 } \) = 0.19
(ii) One tail (m) = 290 times
∴Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 290 }{ 500 } \) = \(\frac { 580 }{ 1000 } \) = \(\frac { 58 }{ 100 } \) = 0.58
(iii) No head (m) = 115 times
∴Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 115 }{ 500 } \) = \(\frac { 23 }{ 100 } \) = 0.23

Question 3.
Three coins are tossed simultaneously 1oo times with the following frequencies of different outcomes:

If the three coins are simultaneously tossed again, compute the probability of:
(i) 2 heads coming up.
(ii) 3 heads coming up.
(iii) at least one head coming up.
(iv) getting more heads than tails.
(v) getting more tails than heads.
Solution:
Three coins are tossed simultaneously 100 times
Total out comes (n) = 100

(i) Probability of 2 heads coming up (m) = 36
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 36 }{ 100 } \) = 0.36
(ii) Probability of 3 heads (m) = 12
ProbabilityP(A)= \(\frac { m }{ n } \) = \(\frac { 12 }{ 100 } \) = 0.12
(iii) Probability of at least one head coming up (m) = 38 + 36 + 12 = 86
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 86 }{ 100 } \) = 0.86
(iv) Probability of getting more heads than tails (m) = 36 + 12 = 48
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 48 }{ 100 } \) = 0.48
(v) Getting more tails than heads (m) = 14 + 38 = 52
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 52 }{ 100 } \) = 0.52

Question 4.
1500 families with 2 children were selected randomly and the following data were recorded:

If a family is chosen at random, compute the probability that it has:
(i) No girl
(ii) 1 girl
(iii) 2 girls
(iv) at most one girl
(v) more girls than boys
Solution:
Total number of families (n) = 1500

(i) Probability of a family having no girls (m) = 211
∴Probability P(A)= \(\frac { m }{ n } \) = \(\frac { 211 }{ 1500 } \) = 0.1406
(ii) Probability of a family having one girl (in) = 814
∴Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 814 }{ 1500 } \) = 0.5426
(iii) Probability of a family having 2 girls (m) = 475
∴Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 475 }{ 1500 } \) = 0.3166
(iv) Probability of a family having at the most one girls
∴m = 814 + 211 = 1025
∴Probability P(A) =\(\frac { m }{ n } \) = \(\frac { 1025 }{ 1500 } \) = 0.6833
(v) Probability of a family having more girls than boys (m) = 475
∴Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 475 }{ 1500 } \) = 0.3166

Question 5.
In a cricket match, a batsman hits a boundary 6 times out of 30 balls he plays. Find the probability that on a ball played:
(i) he hits boundary
(ii) he does not hit a boundary.
Solution:
Total balls played (n) 30
No. of boundaries = 6
(i) When the batsman hits the boundary = 6
∴m = 6
∴Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 6 }{ 30 } \) = \(\frac { 1 }{ 5 } \) = 0.2
(ii) When the batsman does not hit the boundary (m) = 30 – 6 = 24
∴Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 24 }{ 30 } \) = \(\frac { 4 }{ 5 } \) = 0.8

Question 6.
The percentage of marks obtained by a student in monthly unit tests are given below:

Find the probability that the student gets:
(i) more than 70% marks
(ii) less than 70% marks
(iii) a distinction.
Solution:
Percentage of marks obtain in

(i) Probability of getting more than 70% marks (m) = In unit test II, III, V = 3
Total unit test (n) = 5
∴Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 3 }{ 5 } \) = 0.6
(ii) Getting less then 70% marks = units test I and IV
∴m = 2
Total unit test (n) = 5
∴Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 2 }{ 5 } \) = 0.4
(iii) Getting a distinction = In test V (76 of marks)
∴m = 1
Total unit test (n) = 5
∴Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 1 }{ 5 } \) = 0.2

Question 7.
To know the opinion of the students about Mathematics, a survey of 200 students was conducted. The data is recorded in the following table:

Find the probability that a student chosen at random
(i) likes Mathematics
(ii) does not like it.
Solution:
Total number of students (n) = 200

(i) Probability of students who like mathematics (m) = 135
∴Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 135 }{ 200 } \) = 0.675
(ii) Probability of students who dislike mathematics (m) = 65
∴Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 65 }{ 200 } \) = 0.325

Question 8.
The blood groups of 30 students of class IX are recorded as follows:
A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O,
A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O,
A student is selected at random from the class from blood donation. Find the probability that the blood group of the student chosen is:
(i) A (ii) B (iii) AB (iv) O
Solution:
Total number of students of IX class = 30
No. of students of different blood groups
A AB B O
9 3 6 12

Question 9.
Eleven bags of wheat flour, each marked 5 kg, actually contained the following weights of flour
(in kg):
4.97, 5.05, 5.08, 5.03, 5.00, 5.06, 5.08, 4.98, 5.04, 5.07, 5.00
Find the probability that any of these bags chosen at random contains more than 5 kg of flour.
Solution:
Number of total bags (n) = 11
No. of bags having weight more than 5 kg (m) = 7
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 7 }{ 11 } \)

Question 10.
Following table shows the birth month of 40 students of class IX.

Find the probability that a student was born in August.
Solution:
Total number of students (n) = 40

Number of students who born in Aug. (m) = 6
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 6 }{ 40 } \) = \(\frac { 3 }{ 20 } \)

Question 11.
Given below is the frequency distribution table regarding the concentration of sulphur dioxide in the air in parts per million of a certain city for 30 days.

Find the probability of concentration of sulphur dioxide in the interval 0.12 – 0.16 on any of these days.
Solution:
Total number of days (n) = 30

Probability of cone, of S02 of the interval 0.12-0.16 (m) = 2
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 2 }{ 30 } \) = \(\frac { 1 }{ 15 } \)

Question 12.
A company selected 2400 families at random and survey them to determine a relationship between income level and the number of vehicles in a home. The information gathered is listed in the table below:

If a family is chosen, find the probability that the family is:
(i)earning Rs 10000-13000 per month and owning exactly 2 vehicles.
(ii)earning Rs 16000 or more per month and owning exactly I vehicle.
(iii)earning less than Rs 7000 per month and does not own any vehicle.
(iv)earning Rs 13000-16000 per month and owning more than 2 vehicle.
(v)owning not more than 1 vehicle.
(vi)owning at least one vehicle.
Solution:
Total number of families (n) = 2400

(i) Number of families earning income Rs 10000-13000 and owning exactly 2 vehicles (m) = 29
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 29 }{ 2400 } \)
(ii) Number of families earning income Rs 16000 or more having one vehicle (m) = 579
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 579 }{ 2400 } \)
(iii) Number of families earning income less than Rs 7000 having no own vehicle (m) = 10
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 10 }{ 2400 } \) = \(\frac { 1 }{ 240 } \)
(iv) Number of families having X13000 to X16000 having more than two vehicles (m) = 25
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 25 }{ 2400 } \) = \(\frac { 1 }{ 96 } \)
(v) Number of families owning not more than one vehicle (m)
= 10 + 1 + 2 + 1 + 160 + 305 + 533 + 469 + 579 = 2062
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 2062 }{ 2400 } \) = \(\frac { 1031 }{ 1200 } \)
(vi) Number of families owning at least one vechile (m) = 2048 + 192 + 110 = 2356
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 2356 }{ 2400 } \) = \(\frac { 589 }{ 600 } \)

Question 13.
The following table gives the life time of 400 neon lamps:

A bulb is selected at random. Find the probability that the life time of the selected bulb is: (i) less than 400 (ii) between 300 to 800 hours (iii) at least 700 hours.
Solution:
Total number of neon lamps (n) = 400

A bulb is chosen:
(i)No. of bulbs having life time less than 400 hours (m) = 14
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 14 }{ 400 } \) = \(\frac { 7 }{ 200 } \)
(ii)No. of bulbs having life time between 300 to 800 hours (m) = 14 + 56 + 60 + 86 + 74 = 290
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 290 }{ 400 } \) = \(\frac { 29 }{ 40 } \)
(iii)No. of bulbs having life time at least 700 hours (m) = 74 + 62 + 48 = 184
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 184 }{ 400 } \) = \(\frac { 23 }{ 50 } \)

Question 14.
Given below is the frequency distribution of wages (in Rs) of 30 workers in a certain factory:

A worker is selected at random. Find the probability that his wages are:
(i) less than Rs 150
(ii) at least Rs 210
(iii) more than or equal to 150 but less than Rs 210.
Solution:
Number of total workers (n) = 30

A worker is selected.
(i)No. of workers having less than Rs 150 (m) = 3 + 4 = 7
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 7 }{ 30 } \)
(ii)No. of workers having at least Rs 210 (m) = 4 + 3 = 7
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 7 }{ 30 } \)
(iii)No. of workers having more than or equal to Rs 150 but less than Rs 210 = 5 + 6 + 5 = 16
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 16 }{ 30 } \) = \(\frac { 8 }{ 15 } \)

Hope given RD Sharma Class 9 Solutions Chapter 25 Probability Ex 25.1 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 9 Solutions Chapter 25 Probability VSAQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 25 Probability VSAQS

Other Exercises

• RD Sharma Class 9 Solutions Chapter 25 Probability Ex 25.1
• RD Sharma Class 9 Solutions Chapter 25 Probability VSAQS
• RD Sharma Class 9 Solutions Chapter 25 Probability MCQS

Question 1.
Define a trial.
Solution:
When we perform an experiment, it is called a trial of the experiment.

Question 2.
Define an elementary event.
Solution:
An outcome of a trial of an experiment is called an elementary event.

Question 3.
Define an event.
Solution:
An event association to a random experiment is said to occur in a trial.

Question 4.
Define probability of an event.
Solution:
In n trials of a random experiment if an event A happens m times, then probability of happening
of A is given by P(A) = \(\frac { m }{ n } \)

Question 5.
A bag contains 4 white balls and some red balls. If the probability of drawing a white ball from the bag is \(\frac { 2 }{ 5 } \), find the number of red balls in the bag
Solution:
No. of white balls = 4
Let number of red balls = x
Then total number of balls (n) = 4 white + x red = (4 + x) balls

Question 6.
A die is thrown 100 times. If the probability of getting an even number is \(\frac { 2 }{ 5 } \). How many times an odd number is obtained?
Solution:
Total number of a die is thrown = 100
Let an even number comes x times, then probability of an even number = \(\frac { x }{ 100 } \)

Question 7.
Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes

Find the probability of getting at most two heads.
Solution:
Total number of three coins are tossed (n) = 200
Getting at the most 2 heads (m) = 72 + 77 + 28 = 177
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 177 }{ 200 } \)

Question 8.
In the Q. No. 7, what is the probability of getting at least two heads?
Solution:
Total number of possible events = 200
No. of events getting at the least = 2 heads (m) = 23 + 72 = 95
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 95 }{ 200 } \) = \(\frac { 19 }{ 40 } \)

Hope given RD Sharma Class 9 Solutions Chapter 25 Probability VSAQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.