Class 10

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Problem Set 1 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 6 Statistics.

10th Standard Maths 2 Problem Set 1 Chapter 1 Similarity Textbook Answers Maharashtra Board

Class 10 Maths Part 2 Problem Set 1 Chapter 1 Similarity Questions With Answers Maharashtra Board

Question 1.
Select the appropriate alternative.
i. In ∆ABC and ∆PQR, in a one to one correspondence \(\frac { AB }{ QR } \) = \(\frac { BC }{ PR } \) = \(\frac { CA }{ PQ } \), then

(A) ∆PQR – ∆ABC
(B) ∆PQR – ∆CAB
(C) ∆CBA – ∆PQR
(D) ∆BCA – ∆PQR
Answer:
(B)

ii. If in ∆DEF and ∆PQR, ∠D ≅ ∠Q, ∠R ≅ ∠E, then which of the following statements is false?

(A) \(\frac { EF }{ PR } \) = \(\frac { DF }{ PQ } \)
(B) \(\frac { DE }{ PQ } \) = \(\frac { EF }{ RP } \)
(C) \(\frac { DE }{ QR } \) = \(\frac { DF }{ PQ } \)
(D) \(\frac { EF }{ RP } \) = \(\frac { DE }{ QR } \)
Answer:
∆DEF ~ ∆QRP … [AA test of similarity]
∴ \(\frac { DE }{ QR } \) = \(\frac { EF }{ RP } \) = \(\frac { DF }{ PQ } \) …[Corresponding sides of similar triangles]
(B)

iii. In ∆ABC and ∆DEF, ∠B = ∠E, ∠F = ∠C and AB = 3 DE, then which of the statements regarding the two triangles is true?

(A) The triangles are not congruent and not similar.
(B) The triangles are similar but not congruent.
(C) The triangles are congruent and similar.
(D) None of the statements above is true.
Answer:
(B)

iv. ∆ABC and ∆DEF are equilateral triangles, A(∆ABC) : A(∆DEF) = 1 : 2. If AB = 4, then what is length of DE?

(A) 2√2
(B) 4
(C) 8
(D) 4√2
Answer:
Refer Q. 6 Practice Set 1.4
(D)

v. In the adjoining figure, seg XY || seg BC, then which of the following statements is true?

(A) \(\frac { AB }{ AC } \) = \(\frac { AX }{ AY } \)
(B) \(\frac { AX }{ XB } \) = \(\frac { AY }{ AC } \)
(C) \(\frac { AX }{ YC } \) = \(\frac { AY }{ XB } \)
(D) \(\frac { AB }{ YC } \) = \(\frac { AC }{ XB } \)
Answer:
∆ABC ~ ∆AXY … [AA test of similarity]
∴ \(\frac { AB }{ AX } \) = \(\frac { AC }{ AY } \) …[Corresponding sides of similar triangles]
∴ \(\frac { AB }{ AC } \) = \(\frac { AX }{ AY } \) …[Altemendo]
(A)

Question 2.
In ∆ABC, B-D-C and BD = 7, BC = 20, then find following ratios.

Solution:

Draw AE ⊥ BC, B – E – C.
BC = BD + DC [B – D – C]
∴ 20 = 7 + DC
∴ DC = 20 – 7 = 13

i. ∆ABD and ∆ADC have same height AE.
\(\frac{\mathrm{A}(\Delta \mathrm{ABD})}{\mathrm{A}(\Delta \mathrm{ADC})}=\frac{\mathrm{BD}}{\mathrm{DC}}\) [Triangles having equal height]
∴ \(\frac{A(\Delta A B D)}{A(\Delta A D C)}=\frac{7}{13}\)

ii. ∆ABD and ∆ABC have same height AE.
\(\frac{\mathrm{A}(\Delta \mathrm{ABD})}{\mathrm{A}(\Delta \mathrm{ABC})}=\frac{\mathrm{BD}}{\mathrm{BC}}\) [Triangles having equal height]
∴ \(\frac{A(\Delta A B D)}{A(\Delta A B C)}=\frac{7}{20}\)

iii. ∆ADC and ∆ABC have same height AE.
\(\frac{A(\Delta A D C)}{A(\Delta A B C)}=\frac{D C}{B C}\) [Triangles having equal height]
∴ \(\frac{A(\Delta A D C)}{A(\Delta A B C)}=\frac{13}{20}\)

Question 3.
Ratio of areas of two triangles with equal heights is 2 : 3. If base of the smaller triangle is 6 cm, then what is the corresponding base of the bigger triangle?
Solution:
Let A₁ and A₂ be the areas of two triangles. Let b₁ and b₂ be their corresponding bases.
A₁ : A₂ = 2 : 3

∴ The corresponding base of the bigger triangle is 9 cm.

Question 4.
In the adjoining figure, ∠ABC = ∠DCB = 90°, AB = 6, DC = 8, then \(\frac{\mathbf{A}(\Delta \mathbf{A} \mathbf{B} \mathbf{C})}{\mathbf{A}(\mathbf{\Delta D C B})}=?\)

Solution:
∆ABC and ∆DCB have same base BC.

Question 5.
In the adjoining figure, PM = 10 cm, A(∆PQS) = 100 sq. cm,
A(∆QRS) = 110 sq. cm, then find NR.

Solution:

∴ NR = 11 cm

Question 6.
∆MNT ~ ∆QRS. Length of altitude drawn from point T is 5 and length of altitude drawn from point S is 9. Find the ratio \(\frac{A(\Delta M N T)}{A(\Delta Q R S)}\)

Solution:
∆MNT- ∆QRS [Given]
∴ ∠M ≅ ∠Q (i) [Corresponding angles of similar triangles]
In ∆MLT and ∆QPS,
∠M ≅ ∠Q [From (i)]
∠MLT ≅ ∠QPS [Each angle is of measure 90°]
∴ ∆MLT ~ ∆QPS [AA test of similarity]

Question 7.
In the adjoining figure, A – D – C and B – E – C. seg DE || side AB. If AD = 5, DC = 3, BC = 6.4, then find BE.

Solution:
In ∆ABC,
seg DE || side AB [Given]
∴ \(\frac { DC }{ AD } \) = \(\frac { EC }{ BE } \) [Basic proportionality theorem]
∴ \(\frac { 3 }{ 4 } \) = \(\frac { 6.4-x }{ x } \)
∴ 3x = 5 (6.4 – x)
∴ 3x = 32 – 5x
∴ 8x = 32
∴ x = \(\frac { 32 }{ 8 } \) =4
∴ BE = 4 units

Question 8.
In the adjoining figure, seg PA, seg QB, seg RC and seg SD are perpendicular to line AD. AB = 60, BC = 70, CD = 80, PS = 280, then find PQ, QR and RS.

Solution:
seg PA, seg QB, seg RC and seg SD are perpendicular to line AD. [Given]
∴ seg PA || seg QB || seg RC || seg SD (i) [Lines perpendicular to the same line are parallel to each other]
Let the value of PQ be x and that of QR be y.
PS = PQ + QS [P – Q – S]
∴ 280 – x + QS
∴ QS = 280 – x (ii)
Now, seg PA || seg QB || seg SD [From (i)]
∴ \(\frac { AB }{ BD } \) = \(\frac { PQ }{ QS } \) [Property of three parallel lines and their transversals]
∴\(\frac { AB }{ BC+CD } \) = \(\frac { PQ }{ QS } \) [B – C – D]
∴ \(\frac { 60 }{ 70+80 } \) = \(\frac { x }{ 280-x } \)
∴ \(\frac { 60 }{ 150 } \) = \(\frac { x }{ 280-x } \)
∴ \(\frac { 2 }{ 5 } \) = \(\frac { x }{ 280-x } \)
∴ 5x = 2 (280 – x)
∴ 5x = 560 – 2x
∴ 7x = 560
∴ x = \(\frac { 560 }{ 7 } \) = 80
∴ PQ = 80 units
QS = 280 – x [From (ii)]
= 280 – 80
= 200 units
But, QS = QR + RS [Q – R – S]
∴ 200 = y + RS
∴ RS = 200 – y (ii)
Now, seg QB || seg RC || seg SD [From (i)]
∴\(\frac { BC }{ CD } \) = \(\frac { QR }{ RS } \) [Property of three parallel lines and their transversals]
∴ \(\frac { 70 }{ 80 } \) = \(\frac { y }{ 200-y } \)
∴ \(\frac { 7 }{ 8 } \) = \(\frac { y }{ 200-y } \)
∴ 8y = 7(200 – y)
∴ 8y = 1400 – 7y
∴ 15y = 1400
∴ y = \(\frac { 1400 }{ 15 } \) = \(\frac { 280 }{ 3 } \)
∴ QR = \(\frac { 280 }{ 3 } \) units
RS = 200 – 7 [From (iii)]
= 200 – \(\frac { 280 }{ 3 } \)
= \(\frac{200 \times 3-280}{3}\)
= \(\frac { 600-280 }{ 3 } \)
∴ RS = \(\frac { 320 }{ 3 } \) units

Question 9.
In ∆PQR, seg PM is a median. Angle bisectors of ∠PMQ and ∠PMR intersect side PQ and side PR in points X and Y respectively. Prove that XY || QR
Complete the proof by filling in the boxes.

Solution:
Proof:
In ∆PMQ, ray MX is bisector of ∠PMQ.

Question 10.
In the adjoining figure, bisectors of ∠B and ∠C of ∆ABC intersect each other in point X. Line AX intersects side BC in point Y.
AB = 5, AC = 4, BC = 6, then find \(\frac { AX }{ XY } \).

Solution:
Let the value of BY be x.
BC = BY + YC [B – Y – C]
∴ 6 = x + YC
∴ YC = 6 – x
in ∆BAY, ray BX bisects ∠B. [Given]
∴ \(\frac { AB }{ BY } \) = \(\frac { AX }{ XY } \) (i) [Property of angle bisector of a triangle]
Also, in ∆CAY, ray CX bisects ∠C. [Given]

Question 11.
In ꠸ABCD, seg AD || seg BC. Diagonal AC and diagonal BD intersect each other in point P. Then show that \(\frac { AP }{ PD } \) = \(\frac { PC }{ BP } \)

Solution:
proof:
seg AD || seg BC and BD is their transversal. [Given]
∴ ∠DBC ≅ ∠BDA [Alternate angles]
∴ ∠PBC ≅ ∠PDA (i) [D – P – B]
In ∆PBC and ∆PDA,
∠PBC ≅ ∠PDA [From (i)]
∠BPC ≅ ∠DPA [Vertically opposite angles]
∴ ∆PBC ~ ∆PDA [AA test of similarity]
∴ \(\frac { BP }{ PD } \) = \(\frac { PC }{ AP } \) [Corresponding sides of similar triangles]
∴ \(\frac { AP }{ PD } \) = \(\frac { PC }{ BP } \) [By altemendo]

Question 12.
In the adjoining figure, XY || seg AC. If 2 AX = 3 BX and XY = 9, complete the activity to find the value of AC.

Solution:
2 AX = 3 BX [Given]

Question 13.
In the adjoining figure, the vertices of square DEFG are on the sides of ∆ABC. If ∠A = 90°, then prove that DE² = BD × EC.
(Hint: Show that ∆GBD is similar to ∆ CFE. Use GD = FE = DE.)

Solution:
proof:
꠸DEFG is a square.
∴ DE = EF = GF = GD (i) [Sides of a square]
∠GDE = ∠DEF = 90° [Angles of a square]
∴ seg GD ⊥ side BC, seg FE ⊥ side BC (ii)
In ∆BAC and ∆BDG,
∠BAC ≅ ∠BDG [From (ii), each angle is of measure 90°]
∠ABC ≅ ∠DBG [Common angle]
∴ ∆BAC – ∆BDG (iii) [AA test of similarity]
In ∆BAC and ∆FEC,
∠BAC ≅ ∠FEC [From (ii), each angle is measure 90°]
∠ACB ≅ ∠ECF [Common angle]
∴ ∆BAC – ∆FEC (iv) [AA test of similarity]
∴ ∆BDG – ∆FEC [From (iii) and (iv)]
∴ \(\frac { BD }{ EF } \) = \(\frac { GD }{ EC } \) (v) [Corresponding sides of similar triangles]
∴ \(\frac { BD }{ DE } \) = \(\frac { DE }{ EC } \) [From (i) and (v)]
∴ DE² = BD × EC

Class 10 Maths Digest

• Practice Set 1.1 Class 10 Answers
• Practice Set 1.2 Class 10 Answers
• Practice Set 1.3 Class 10 Answers
• Practice Set 1.4 Class 10 Answers
• Problem Set 1 Class 10 Answers

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 1.4 Algebra 10th Class Maths Part 2 Answers Solutions Chapter 1 Similarity.

10th Standard Maths 2 Practice Set 1.4 Chapter 1 Similarity Textbook Answers Maharashtra Board

Class 10 Maths Part 2 Practice Set 1.4 Chapter 1 Similarity Questions With Answers Maharashtra Board

Question 1.
The ratio of corresponding sides of similar triangles is 3 : 5, then find the ratio of their areas.
Solution:
Let the corresponding sides of similar triangles be S₁ and S₂.
Let A₁ and A₂ be their corresponding areas.

∴ Ratio of areas of similar triangles = 9 : 25

Question 2.
If ∆ABC ~ ∆PQR and AB : PQ = 2:3, then fill in the blanks.
Solution:

Question 3.
If ∆ABC ~ ∆PQR, A(∆ABC) = 80, A(∆PQR) = 125, then fill in the blanks.
Solution:

Question 4.
∆LMN ~ ∆PQR, 9 × A(∆PQR) = 16 × A(∆LMN). If QR = 20, then find MN.
Solution:
9 × A(∆PQR) = 16 × A(∆LMN) [Given]
∴ \(\frac{\mathrm{A}(\Delta \mathrm{LMN})}{\mathrm{A}(\Delta \mathrm{PQR})}=\frac{9}{16}\) (i)
Now, ∆LMN ~ ∆PQR [Given]
∴ \(\frac{\mathrm{A}(\Delta \mathrm{LMN})}{\mathrm{A}(\Delta \mathrm{PQR})}=\frac{\mathrm{MN}^{2}}{\mathrm{QR}^{2}}\) (ii) [Theorem of areas of similar triangles]
∴ \(\frac{\mathrm{MN}^{2}}{\mathrm{QR}^{2}}=\frac{9}{16}\) [From (i) and (ii)]
∴ \(\frac{M N}{Q R}=\frac{3}{4}\) [Taking square root of both sides]
∴ \(\frac{\mathrm{MN}}{20}=\frac{3}{4}\)
∴ MN = \(\frac{20 \times 3}{4}\)
∴ MN = 15 units

Question 5.
Areas of two similar triangles are 225 sq. cm. and 81 sq. cm. If a side of the smaller triangle is 12 cm, then find corresponding side of the bigger triangle.
Solution:
Let the areas of two similar triangles be A₁ and A₂.
A₁ = 225 sq. cm. A₂ = 81 sq. cm.
Let the corresponding sides of triangles be S₁ and S₂ respectively.
S₁ = 12 cm

∴ The length of the corresponding side of the bigger triangle is 20 cm.

Question 6.
∆ABC and ∆DEF are equilateral triangles. If A(∆ABC): A(∆DEF) = 1:2 and AB = 4, find DE.
Solution:
In ∆ABC and ∆DEF,

Question 7.
In the adjoining figure, seg PQ || seg DE, A(∆PQF) = 20 sq. units, PF = 2 DP, then find A (꠸ DPQE) by completing the following activity.
Solution:
A(∆PQF) = 20 sq.units, PF = 2 DP, [Given]
Let us assume DP = x.
∴ PF = 2x

Class 10 Maths Digest

• Practice Set 1.1 Class 10 Answers
• Practice Set 1.2 Class 10 Answers
• Practice Set 1.3 Class 10 Answers
• Practice Set 1.4 Class 10 Answers
• Problem Set 1 Class 10 Answers

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Problem Set 6 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 6 Statistics.

10th Standard Maths 1 Problem Set 6.1 Chapter 6 Statistics Textbook Answers Maharashtra Board

Class 10 Maths Part 1 Problem Set 6.1 Chapter 6 Statistics Questions With Answers Maharashtra Board

10th Geometry Problem Set 6 Question 1.
Find the correct answer from the alternatives given.

i. The persons of O – blood group are 40%. The classification of persons based on blood groups is to be shown by a pie diagram. What should be the measures of angle for the persons of O – blood group?
(A) 114°
(B) 140°
(C) 104°
(D) 144°
Answer:
Measure of the central angle = \(\frac { 40 }{ 100 } \) × 360° = 144°
(D)

ii. Different expenditures incurred on the construction of a building were shown by a pie diagram. The expenditure of ₹ 45,000 on cement was shown by a sector of central angle of 75°. What was the total expenditure of the construction?
(A) 2,16,000
(B) 3,60,000
(C) 4,50,000
(D) 7,50,000
Answer:
Measure of the central angle = \(\frac{\text { Expenditure of cement }}{\text { Total expenditure }} \times 360^{\circ}\)

(A)

iii. Cumulative frequencies in a grouped frequency table are useful to find.
(A) Mean
(B) Median
(C) Mode
(D) All of these
Answer:
(B)

iv. The formula to find mean from a grouped frequency table is \(\overline{\mathrm{X}}=\mathrm{A}+\frac{\sum f_{i} u_{i}}{\sum f_{i}} \times g\)
in the formula u_i = _________.

Answer:

(C)

v.

The median of the distances covered per litre shown in the above data is in the group
(A) 12 – 14
(B) 14 – 16
(C) 16 – 18
(D) 18 – 20
Answer:
(C)

vi.

The above data is to be shown by a frequency polygon. The coordinates of the points to show number of students in the class 4 – 6 are.
(A) (4, 8)
(B) (3,5)
(C) (5,8)
(D) (8,4)
Answer:
Class mark = 5
Frequency = 8
∴ Co-ordinates of the point = (5, 8)
(C)

Statistics Problem Set 6 Question 2.
The following table shows the income of farmers in a grape season. Find the mean of their income.

Solution:

∴ The mean of the income of the farmers is ₹ 52,500.

Statistics Problem Set Question 3.
The loans sanctioned by a bank for construction of farm ponds are shown in the following table. Find the mean of the loans.

Solution:

∴ The mean of the loans given by the bank is ₹ 65,400.

Question 4.
The weekly wages of 120 workers in a factory are shown in the following frequency distribution table. Find the mean of the weekly wages.

Solution:

∴ The mean of the weekly wages of the workers is ₹ 4250.

Problem Set 6 Algebra Class 9 Question 5.
The following frequency distribution table shows the amount of aid given to 50 flood affected families. Find the mean of the amount of aid.

Solution:

∴ The mean of the amount of aid given to families is ₹ 72,400.
[Note: The above problems are solved using direct method. Students can solve these problems by using other method.]

Problem Set 6 Algebra Class 10 Question 6.
The distances covered by 250 public transport buses in a day is shown in the following frequency distribution table. Find the median of the distances.

Solution:

Cumulative frequency which is just greater than (or equal) to 125 is 180.
∴ The median class is 220 – 230.
Now, L = 220, f = 80, cf = 100, h = 10

∴ The median of the distances is 223.13 km (approx.).

Algebra 10th Class Problem Set 6 Question 7.
The prices of different articles and demand for them is shown in the following frequency distribution table. Find the median of the prices.

Solution:

Cumulative frequency which is just greater than (or equal) to 200 is 240.
∴ The median class is 20 – 40.
Now,L = 20, f = 100,cf = 140, h = 20

∴ The median of the prices of different articles is ₹ 32.

10th Algebra Problem Set 6 Question 8.
The following frequency table shows the demand for a sweet and the number of customers. Find the mode of demand of sweet.

Solution:


∴ The mode of the demand of sweet is 397.06 grams.

Question 9.
Draw a histogram for the following frequency distribution.

Solution:

Question 10.
In a handloom factory different workers take different periods of time to weave a saree. The number of workers and their required periods are given below. Present the information by a frequency polygon.

Solution:

Problem Set 6 Question 11.
The time required for students to do a science experiment and the number of students is shown in the following grouped frequency distribution table. Show the information by a histogram and also by a frequency polygon.

Solution:

Question 12.
Draw a frequency polygon for the following grouped frequency distribution table.

Solution:

Question 13.
The following table shows the average rainfall in 150 towns. Show the information by a frequency polygon.

Solution:

Question 14.
Observe the given pie diagram. It shows the percentages of number of vehicles passing a signal in a town between 8 am and 10 am.
i. Find the central angle for each type of vehicle.
ii. If the number of two-wheelers is 1200, find the number of all vehicles.

Solution:

∴ The total number of vehicles is 3000.

Problem Set 6 Geometry Class 10 Question 15.
The following table shows causes of noise pollution. Show it by a pie diagram.

Solution:

Question 16.
A survey of students was made to know which game they like. The data obtained in the survey is
presented in the given pie diagram. If the total number of students are 1000,
i. how many students like cricket?
ii. how many students like football?
iii. how many students prefer other games?

Solution:

∴ 225 students like cricket.

ii. Central angle for football (θ) = 63°

∴ 175 students like football.

iii. Central angle for other games (θ) = 72°

∴ 200 students like other games.

Question 17.
Medical check up of 180 women was conducted in a health centre in a village. 50 of them were short of hemoglobin, 10 suffered from cataract and 25 had respiratory disorders. The remaining women were healthy. Show the information by a pie diagram.
Solution:
Total number of women = 180

Question 18.
On an environment day, students in a school planted 120 trees under plantation project. The information regarding the project is shown in the following table. Show it by a pie diagram.

Solution:
Total number of trees planted = 120

Maharashtra Board Class 10 Maths Solutions

Class 10 Maths Digest

• Practice Set 6.1 Class 10 Answers
• Practice Set 6.2 Class 10 Answers
• Practice Set 6.3 Class 10 Answers
• Practice Set 6.4 Class 10 Answers
• Practice Set 6.5 Class 10 Answers
• Practice Set 6.6 Class 10 Answers
• Problem Set 6 Class 10 Answers

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 6.6 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 6 Statistics.

10th Standard Maths 1 Practice Set 6.6 Chapter 6 Statistics Textbook Answers Maharashtra Board

Class 10 Maths Part 1 Practice Set 6.6 Chapter 6 Statistics Questions With Answers Maharashtra Board

Question 1.
The age group and number of persons, who donated blood in a blood donation camp is given below.
Draw a pie diagram from it.

Solution:
Total number of persons = 80 + 60 + 35 + 25 = 200

Question 2.
The marks obtained by a student in different subjects are shown. Draw a pie diagram showing the information.

Solution:
Total marks obtained = 50 + 70 + 80 + 90 + 60 + 50 = 400

Question 3.
In a tree plantation programme, the number of trees planted by students of different classes is given in the following table. Draw a pie diagram showing the information.

Solution:
Total number of trees planted = 40 + 50 + 75 + 50 + 70 + 75 = 360

Question 4.
The following table shows the percentages of demands for different fruits registered with a fruit vendor. Show the information by a pie diagram.

Solution:
Total percentage = 30 + 15 + 25 + 20 + 10 = 100%

Question 5.
The pie diagram in the given figure shows the proportions of different workers in a town. Answer the following questions with its help.
i. If the total workers is 10,000, how many of them are in the field of construction?
ii. How many workers are working in the administration?
iii. What is the percentage of workers in production?

Solution:

∴ There are 2000 workers working in the field of construction.

∴ There are 1000 workers working in the administration.

∴ 25% of workers are working in the production field.

Question 6.
The annual investments of a family are shown in the given pie diagram. Answer the following questions based on it.
i. If the investment in shares is ? 2000, find the total investment.
ii. How much amount is deposited in bank?
iii. How much more money is invested in immovable property than in mutual fund?
iv. How much amount is invested in post?

Solution:

The total investment is ₹ 12000.

ii. Central angle for deposit in bank (θ) = 90°

∴ The amount deposited in bank is ₹ 3000.

iii. Difference in central angle for immovable property and mutual fund (θ) = 120° – 60° = 60°

∴ ₹ 2000 more is invested in immovable property than in mutual fund.

iv. Central angle for post (θ) = 30°

∴ The amount invested in post is ₹ 1000.

Class 10 Maths Digest

• Practice Set 6.1 Class 10 Answers
• Practice Set 6.2 Class 10 Answers
• Practice Set 6.3 Class 10 Answers
• Practice Set 6.4 Class 10 Answers
• Practice Set 6.5 Class 10 Answers
• Practice Set 6.6 Class 10 Answers
• Problem Set 6 Class 10 Answers

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 6.5 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 6 Statistics.

10th Standard Maths 1 Practice Set 6.5 Chapter 6 Statistics Textbook Answers Maharashtra Board

Class 10 Maths Part 1 Practice Set 6.5 Chapter 6 Statistics Questions With Answers Maharashtra Board

Question 1.
Observe the following frequency polygon and write the answers of the questions below it.
i. Which class has the maximum number of students?
ii. Write the classes having zero frequency.
iii. What is the class mark of the class, having frequency of 50 students?
iv. Write the lower and upper class limits of the class whose class mark is 85.
v. How many students are in the class 80 – 90?

Solution:
i. The class 60 – 70 has the maximum number of students.
ii. The classes 20 – 30 and 90 – 100 have frequency zero.
iii. The class mark of the class having 50 students is 55.
iv. The lower and upper class limits of the class having class mark 85 are 80 and 90 respectively.
v. There are 15 students in the class 80 – 90.

Question 2.
Show the following data by a frequency polygon.

Solution:

Question 3.
The following table shows the classification of percentages of marks of students and the number of students. Draw a frequency polygon from the table.

Solution:

Class 10 Maths Digest

• Practice Set 6.1 Class 10 Answers
• Practice Set 6.2 Class 10 Answers
• Practice Set 6.3 Class 10 Answers
• Practice Set 6.4 Class 10 Answers
• Practice Set 6.5 Class 10 Answers
• Practice Set 6.6 Class 10 Answers
• Problem Set 6 Class 10 Answers

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 6.4 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 6 Statistics.

10th Standard Maths 1 Practice Set 6.4 Chapter 6 Statistics Textbook Answers Maharashtra Board

Class 10 Maths Part 1 Practice Set 6.4 Chapter 6 Statistics Questions With Answers Maharashtra Board

Question 1.
Draw a histogram of the following data.

Solution:

Question 2.
The table below shows the yield of jowar per acre. Show the data by histogram.

Solution:

Question 3.
In the following table, the investment made by 210 families is shown. Present it in the form of a histogram.

Solution:

Question 4.
Time allotted for the preparation of an examination by some students is shown in the table. Draw a histogram to show the information.

Solution:

Class 10 Maths Digest

• Practice Set 6.1 Class 10 Answers
• Practice Set 6.2 Class 10 Answers
• Practice Set 6.3 Class 10 Answers
• Practice Set 6.4 Class 10 Answers
• Practice Set 6.5 Class 10 Answers
• Practice Set 6.6 Class 10 Answers
• Problem Set 6 Class 10 Answers

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 6.3 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 6 Statistics.

10th Standard Maths 1 Practice Set 6.3 Chapter 6 Statistics Textbook Answers Maharashtra Board

Class 10 Maths Part 1 Practice Set 6.3 Chapter 6 Statistics Questions With Answers Maharashtra Board

Question 1.
The following table shows the information regarding the milk collected from farmers on a milk collection centre and the content of fat in the milk, measured by a lactometer. Find the mode of fat content.

Solution:

Here, the maximum frequency is 80.
∴ The modal class is 4 – 5.
L = lower class limit of the modal class = 4
h = class interval of the modal class = 1
f₁ = frequency of the modal class = 80
f₀ = frequency of the class preceding the modal class = 70
f₂ = frequency of the class succeeding the modal class = 60

∴ The mode of the fat content is 4.33%.

Question 2.
Electricity used by some families is shown in the following table. Find the mode of use of electricity.

Solution:

Here, the maximum frequency is 100.
∴ The modal class is 60 – 80.
L = lower class limit of the modal class = 60
h = class interval of the modal class = 20
f₁ = frequency of the modal class = 100
f₀ = frequency of the class preceding the modal class = 70
f₂ = frequency of the class succeeding the modal class = 80

∴ The mode of use of electricity is 72 units.

Question 3.
Grouped frequency distribution of supply of milk to hotels and the number of hotels is given in the following table. Find the mode of the supply of milk.

Solution:

Here, the maximum frequency is 35.
∴ The modal class is 9 – 11.
L = lower class limit of the modal class = 9
h = class interval of the modal class = 2
f₁ = frequency of the modal class = 35
f₀ = frequency of the class preceding the modal class = 20
f₂ = frequency of the class succeeding the modal class = 18

∴ The mode of the supply of milk is 9.94 litres (approx.).

Question 4.
The following frequency distribution table gives the ages of 200 patients treated in a hospital in a week. Find the mode of ages of the patients.

Solution:

Here, the maximum frequency is 50.
The modal class is 9.5 – 14.5.
L = lower class limit of the modal class = 9.5
h = class interval of the modal class = 5
f₁ = frequency of the modal class = 50
f₀ = frequency of the class preceding the modal class = 32
f₂ = frequency of the class succeeding the modal class = 36

∴ The mode of the ages of the patients is 12.31 years (approx.).

Class 10 Maths Digest

• Practice Set 6.1 Class 10 Answers
• Practice Set 6.2 Class 10 Answers
• Practice Set 6.3 Class 10 Answers
• Practice Set 6.4 Class 10 Answers
• Practice Set 6.5 Class 10 Answers
• Practice Set 6.6 Class 10 Answers
• Problem Set 6 Class 10 Answers

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 6.2 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 6 Statistics.

10th Standard Maths 1 Practice Set 6.2 Chapter 6 Statistics Textbook Answers Maharashtra Board

Class 10 Maths Part 1 Practice Set 6.2 Chapter 6 Statistics Questions With Answers Maharashtra Board

Statistics Practice Set 6.2 Question 1.
The following table shows classification of number of workers and the number of hours they work in a software company. Find the median of the number of hours they work.

Solution:

Cumulative frequency which is just greater than (or equal) to 500 is 650.
∴ The median class is 10 – 12.
Now, L = 10, f = 500, cf = 150, h = 2

∴ The median of the number of hours the workers work is 11.4 hours.

10th Class Algebra Practice Set 6.2 Question 2.
The frequency distribution table shows the number of mango trees in a grove and their yield of mangoes. Find the median of data.

Solution:

Here, total frequency = ∑f_i = N = 250
∴ \(\frac { N }{ 2 } \) = \(\frac { 250 }{ 2 } \) = 125
Cumulative frequency which is just greater than (or equal) to 125 is 153.
∴ The median class is 150 – 200.
Now, L = 150, f = 90, cf = 63, h = 50

∴ The median of the given data is 184 mangoes (approx).

Statistics Class 10 Practice Set 6.2 Question 3.
The following table shows the classification of number of vehicles and their speeds on Mumbai-Pune express way. Find the median of the data.

Solution:

Here, total frequency = ∑f_i = N = 200
∴ \(\frac { N }{ 2 } \) = \(\frac { 200 }{ 2 } \) = 100
Cumulative frequency which is just greater than (or equal) to 100 is 184.
∴ The median class is 74.5 – 79.5.
Now, L = 74.5, f = 85, cf = 99, h = 5

∴ The median of the given data is 75 km/hr (approx.).

Practice Set 6.2 Geometry Class 10 Question 4.
The production of electric bulbs in different factories is shown in the following table. Find the median of the productions.

Solution:

Cumulative frequency which is just greater than (or equal) to 52.5 is 67.
∴ The median class is 50 – 60.
Now, L = 50, f = 20, cf = 47, h = 10

∴ The median of the productions is 52750 bulbs (approx.).

Practice Set 6.2 Question 1.
If the number of scores is odd, then the (\(\frac { n+1 }{ 2 } \))^th score is the median of the data. That is, the number of scores below as well as above \({ K }_{ \frac { n+1 }{ 2 } }\) is \(\frac { n-1 }{ 2 } \) Verify the fact by taking n = 2m + I. (Textbk pg. no. 139)
Solution:

The sequence of the terms of scores is 1,2, 3, …….., m, m + 1, m + 2, …, 2m + 1
Thus, we have to prove that m + 1 is the middle term if the number of scores is 2m + 1
i.e. to prove
number of terms from 1 to m = number of terms from m + 2 to 2m + 1 …(i)
Consider the L.H.S. of equation (i)
The sequence is an A.P. with a = 1,d = 1, t_n1 = m
t_n1 = a + (n₁ – 1) d
∴ m = 1 + (n₁ – 1)1
∴ m = 1 + n₁ – 1
∴ m = n₁
Consider the R.H.S. of equation (ii)
The sequence is an A.P. with a = m + 2, d = 1, t_n2 = 2m + 1
t_n2 = a + (n₂ – 1)d
∴ 2m + 1 = m + 2 + (n₂ – 1)1
∴ 2m + 1 = m + n₂ + 1
∴ m = n₂
∴ number of terms from 1 to m = number of terms from m + 2 to 2m + 1 = m = \(\frac { n-1 }{ 2 } \)
∴ m + 1 is the middle term if the number of scores is 2m + 1.

Question 2.
If the number of the scores is even, then the mean of the middle two terms is the median. This is because the number of terms below \({ K }_{ \frac { n }{ 2 } }\) and above \({ K }_{ \frac { n+2 }{ 2 } }\) is equal, which is \(\frac { n-2 }{ 2 } \). Verify this by taking n = 2m. (Textbook pg. no. 139)
Solution:

The sequence of the terms of scores is 1, 2, 3 … m – 1, m, m + 1, m + 2,…., 2m
Thus, we have to prove that m and m + 1 are the middlemost terms if the number of scores is 2m.
i.e. to prove
number of terms from 1 to m – 1 = number of terms from m + 2 to 2m …(i)
Consider the L.H.S. of equation (i)
The sequence is an A.P. with a = 1, d = 1, t_n1 = m – 1
t_n1 = a + (n₁ – 1)d
∴ m – 1 = 1 + (n₁ – 1)1
∴m – 1 = 1 + n₁ – 1
∴ n₁ = m – 1
Consider the R.H.S. of equation (i)
The sequence is an A.P. with a = m + 2, d = 1, t_n2= 2m
t_n2= a + (n₂ – 1) d
∴ 2m = m + 2 + (n₂ – 1)1
∴ 2m = m + 2 + n₂ – 1
∴ n₂ = m – 1
∴ number of terms from 1 to m – 1 = number of terms from m + 2 to 2m = m – 1 = \(\frac { n-2 }{ 2 } \)
∴ m and m + 1 are the middlemost terms if the number of scores is 2m.

Class 10 Maths Digest

• Practice Set 6.1 Class 10 Answers
• Practice Set 6.2 Class 10 Answers
• Practice Set 6.3 Class 10 Answers
• Practice Set 6.4 Class 10 Answers
• Practice Set 6.5 Class 10 Answers
• Practice Set 6.6 Class 10 Answers
• Problem Set 6 Class 10 Answers

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Problem Set 5 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 5 Probability.

10th Standard Maths 1 Problem Set 5 Chapter 5 Probability Textbook Answers Maharashtra Board

Class 10 Maths Part 1 Problem Set 5 Chapter 5 Probability Questions With Answers Maharashtra Board

Question 1.
Choose the correct alternative answer for each of the following questions.

i. Which number cannot represent a probability?
(A) \(\frac { 2 }{ 3 } \)
(B) 1.5
(C) 15%
(D) 0.7
Answer:
The probability of any 0 to 1 or 0% to 100%. event is from
(B)

ii. A die is rolled. What is the probability that the number appearing on upper face is less than 3?
(A) \(\frac { 1 }{ 6 } \)
(B) \(\frac { 1 }{ 3 } \)
(C) \(\frac { 1 }{ 2 } \)
(D) 0
Answer:
(B)

iii. What is the probability of the event that a number chosen from 1 to 100 is a prime number?
(A) \(\frac { 1 }{ 5 } \)
(B) \(\frac { 6 }{ 25 } \)
(C) \(\frac { 1 }{ 4 } \)
(D) \(\frac { 13 }{ 50 } \)
Answer:
n(S) = 100
Let A be the event that the number chosen is a prime number.
∴ A = {2, 3, 5. , 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}
∴ n(A) = 25
∴ P(A) = \(\frac { n(A) }{ n(S) } \) = \(\frac { 25 }{ 100 } \) = \(\frac { 1 }{ 4 } \)
(C)

iv. There are 40 cards in a bag. Each bears a number from 1 to 40. One card is drawn at random. What is the probability that the card bears a number which is a multiple of 5?
(A) \(\frac { 1 }{ 5 } \)
(B) \(\frac { 3 }{ 5 } \)
(C) \(\frac { 4 }{ 5 } \)
(D) \(\frac { 1 }{ 3 } \)
Answer:
(A)

v. If n(A) = 2, P(A) = \(\frac { 1 }{ 5 } \), then n(S) = ?
(A) 10
(B) \(\frac { 5 }{ 2 } \)
(C) \(\frac { 2 }{ 5 } \)
(D) \(\frac { 1 }{ 3 } \)
Answer:
(A)

Question 2.
Basketball players John, Vasim, Akash were practising the ball drop in the basket. The probabilities of success for John, Vasim and Akash are \(\frac { 4 }{ 5 } \), 0.83 and 58% respectively. Who had the greatest probability of success ?
Solution:
The probability that the ball is dropped in the basket by John = \(\frac { 4 }{ 5 } \) = 0.80
The probability that the ball is dropped in the basket by Vasim = 0.83
The probability that the ball is dropped in the basket by Akash = 58% = \(\frac { 58 }{ 100 } \) = 0.58
0.83 > 0.80 > 0.58
∴ Vasim has the greatest probability of success.

Question 3.
In a hockey team there are 6 defenders , 4 offenders and 1 goalie. Out of these, one player is to be selected randomly as a captain. Find the probability of the selection that:
i. The goalie will be selected.
ii. A defender will be selected.
Solution:
Total number of players in the hockey team
= 6 + 4 + 1 = 11
∴ n(S) = 11

i. Let A be the event that the captain selected will be a goalie.
There is only one goalie in the hockey team.

ii. Let B be the event that the captain selected will be a defender.
There are 6 defenders in the hockey team.

Question 4.
Joseph kept 26 cards in a cap, bearing one English alphabet on each card. One card is drawn at random. What is the probability that the card drawn is a vowel card ?
Solution:
Each card bears an English alphabet.
∴ n(S) = 26
Let A be the event that the card drawn is a vowel card.
There are 5 vowels in English alphabets.
∴ A = {a, e, i, o, u}
∴ n(A) = 5

∴ The probability that the card drawn is a vowel card is \(\frac { 5 }{ 26 } \).

Question 5.
A balloon vendor has 2 red, 3 blue and 4 green balloons. He wants to choose one of them at random to give it to Pranali. What is the probability of the event that Pranali gets,
i. a red balloon.
ii. a blue balloon,
iii. a green balloon.
Solution:
Let the 2 red balloon be R₁, R₂,
3 blue balloons be B₁, B₂, B₃, and
4 green balloons be G₁, G₂, G₃, G₄.
∴ Sample space
S = {R₁, R₂, B₁, B₂, B₃, G₁, G₂, G₃, G₄}
∴ n(S) = 9

i. Let A be the event that Pranali gets a red balloon.

∴ The probability that Pranali gets a red balloon is \(\frac { 2 }{ 9 } \)

ii. Let B be the event that Pranali gets a blue balloon.

∴ The probability that Pranali gets a blue balloon is \(\frac { 1 }{ 3 } \).

iii. Let C be the event that Pranali gets a green balloon.

∴ The probability that Pranali gets a green balloon is \(\frac { 4 }{ 9 } \).

Question 6.
A box contains 5 red, 8 blue and 3 green pens. Rutuja wants to pick a pen at random. What is the probability that the pen is blue?
Solution:
Let 5 red pens be R₁, R₂, R₃, R₄, R₅.
8 blue pens be B₁, B₂, B₃, B₄, B₅, B₆, B₇, B₈. and
3 green pens be G₁, G₂, G₃.
∴ Sample space
S = {R₁, R₂, R₃, R₄, R₅, B₁, B₂, B₃, B₄, B₅, B₆, B₇, B₈, G₁, G₂, G₃}
∴ n(S) = 16
Let A be the event that Rutuja picks a blue pen.
∴ A = {B₁, B₂, B₃, B₄, B₅, B₆, B₇, B₈}

∴ The probability that Rutuja picks a blue pen is \(\frac { 1 }{ 2 } \).

Question 7.
Six faces of a die are as shown below.

If the die is rolled once, find the probability of
i. ‘A’ appears on upper face.
ii. ‘D’ appears on upper face.
Solution:
Sample space
S = {A, B, C, D, E, A}
∴ n (S) = 6
i. Let R be the event that ‘A’ appears on the upper face.
∴ R = {A, A}
∴ n(R) = 2

ii. Let Q be the event that ‘D’ appears on the upper face.
Total number of faces having ‘D’ on it = 1
Q = {D}
∴ n(Q) = 1

Question 8.
A box contains 30 tickets, bearing only one number from 1 to 30 on each. If one ticket is drawn at random, find the probability of an event that the ticket drawn bears
i. an odd number.
ii. a complete square number.
Solution:
Sample space,
S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30}
∴ n(S) = 30

i. Let A be the event that the ticket drawn bears an odd number.
∴ A = {1,3,5,7,9,11,13,15,17,19,21, 23,25,27,29}
∴ n(A) =15

ii. Let B be the event that the ticket drawn bears a complete square number.
∴ B = {1,4,9,16,25}
∴ n(B) = 5

Question 9.
Length and breadth of a rectangular garden are 77 m and 50 m. There is a circular lake in the garden having diameter 14 m. Due to wind, a towel from a terrace on a nearby building fell into the garden. Then find the probability of the event that it fell in the lake.

Solution:
Area of the rectangular garden
= length × breadth
= 77 × 50
∴ Area of the rectangular garden = 3850 sq.m
Radius of the lake = \(\frac { 14 }{ 2 } \) = 7 m

∴ The probability of the event that the towel tell in the lake is \(\frac { 1 }{ 25 } \).

Question 10.
In a game of chance, a spinning arrow comes to rest at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8. All these are equally likely outcomes. Find the probability that it will rest at
i. 8.
ii. an odd number.
iii. a number greater than 2.
iv. a number less than 9.

Solution:
Sample space (S) = {1,2, 3, 4, 5, 6, 7, 8}
∴ n(S) = 8
i. Let A be the event that the spinning arrow comes to rest at 8.

ii. Let B be the event that the spinning arrow comes to rest at an odd number.

iii. Let C be the event that the spinning arrow comes to rest at a number greater than 2.

iv. Let D be the event that the spinning arrow comes to rest at a number less than 9.
∴ D = {1,2, 3, 4, 5, 6, 7, 8}

Question 11.
There are six cards in a box, each bearing a number from 0 to 5. Find the probability of each of the following events, that a card drawn shows,
i. a natural number.
ii. a number less than 1.
iii. a whole number.
iv. a number greater than 5.
Solution:
Sample space (S) = {0, 1, 2, 3, 4, 5}
∴ n(S) = 6

i. Let A be the event that the card drawn shows a natural number.
∴ A = {1,2,3,4,5}
∴ n(A) = 5

ii. Let B be the event that the card drawn shows a number less than 1.
∴ B = {0}
∴ n(B) = 1

iii. Let C be the event that the card drawn shows a whole number.
∴ C = {0,1, 2, 3, 4, 5}
∴ n(C) = 6

iv. Let D be the event that the card drawn shows a number greater than 5.
Here, the greatest number is 5.
∴ Event D is an impossible event.
∴ D = { }
∴ n(D) = 0

Question 12.
A bag contains 3 red, 3 white and 3 green balls. One ball is taken out of the bag at random. What is the probability that the ball drawn is:
i. red.
ii. not red.
iii. either red or white.
Solution:
Let the three red balls be R₁, R₂, R₃, three white balls be W₁, W₂, W₃ and three green balls be G₁, G₂, G₃.
∴ Sample space,
S = {R₁, R₂, R₃, W₁, W₂, W₃, G₁, G₂, G₃}
∴ n(S) = 9

i. Let A be the event that the ball drawn is red.
∴ A = {R₁, R₂, R₃}
∴ n(A) = 3

ii. Let B be the event that the ball drawn is not red.
B = {W₁,W₂,W₃,G₁,G₂,G₃}
∴ n(B) = 6

iii. Let C be the event that the ball drawn is red or white.
∴ C = {R₁, R₂, R₃, W₁, W₂, W₃}
∴ n(C) = 6

Question 13.
Each card bears one letter from the word ‘mathematics’. The cards are placed on a table upside down. Find the probability that a card drawn bears the letter ‘m’.
Solution:
Sample space
= {m, a, t, h, e, m, a, t, i, c, s}
∴ n(S) = 11
Let A be the event that the card drawn bears the letter ‘m’
∴ A = {m, m}
∴ n(A) = 2

∴ The probability that a card drawn bears letter ‘m’ is \(\frac { 2 }{ 11 } \).

Question 14.
Out of 200 students from a school, 135 like Kabaddi and the remaining students do not like the game. If one student is selected at random from all the students, find the probability that the student selected dosen’t like Kabaddi.
Solution:
Total number of students in the school = 200
∴ n(S) = 200
Number of students who like Kabaddi = 135
∴ Number of students who do not like Kabaddi
= 200 – 135 = 65
Let A be the event that the student selected does not like Kabaddi.
∴ n(A) = 65

∴ The probability that the student selected doesn’t like kabaddi is \(\frac { 13 }{ 40 } \).

Question 15.
A two digit number is to be formed from the digits 0, 1, 2, 3, 4. Repetition of the digits is allowed. Find the probability that the number so formed is a:
i. prime number.
ii. multiple of 4.
iii multiple of 11.
Solution:
Sample space
(S) = {10, 11, 12, 13, 14,
20, 21, 22, 23, 24,
30, 31, 32, 33, 34,
40, 41, 42, 43, 44}
∴ n(S) = 20

i. Let A be the event that the number so formed is a prime number.
∴ A = {11,13,23,31,41,43}
∴ n(A) = 6

ii. Let B be the event that the number so formed is a multiple of 4.
∴ B = {12,20,24,32,40,44}
∴ n(B) = 6

iii. Let C be the event that the number so formed is a multiple of 11.
∴ C = {11,22,33,44}
∴ n(C) = 4

Question 16.
The faces of a die bear numbers 0,1, 2, 3,4, 5. If the die is rolled twice, then find the probability that the product of digits on the upper face is zero.
Solution:
Sample space,
S = {(0, 0), (0,1), (0,2),
(1,0), (1,1), (1,2),
(2,0), (2,1), (2,2),
(3.0), (3,1), (3,2),
(4.0), (4,1), (4,2),
(5.0), (5,1), (5,2),
∴ n(S) = 36
Let A be the event that the product of digits on the upper face is zero.
∴ A = {(0, 0), (0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (1,0), (2, 0), (3,0), (4, 0), (5,0)}
∴ n(A) = 11

∴ The probability that the product of the digits on the upper face is zero is \(\frac { 11 }{ 36 } \).

Class 10 Maths Digest

• Practice Set 5.1 Class 10 Answers
• Practice Set 5.2 Class 10 Answers
• Practice Set 5.3 Class 10 Answers
• Practice Set 5.4 Class 10 Answers
• Problem Set 5 Class 10 Answers

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 6.1 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 6 Statistics.

10th Standard Maths 1 Practice Set 6.1 Chapter 6 Statistics Textbook Answers Maharashtra Board

Class 10 Maths Part 1 Practice Set 6.1 Chapter 6 Statistics Questions With Answers Maharashtra Board

Question 1.
The following table shows the number of students and the time they utilized daily for their studies. Find the mean time spent by students for their studies by direct method.

Solution:

∴ The mean of the time spent by the students for their studies is 4.36 hours.

Question 2.
In the following table, the toll paid by drivers and the number of vehicles is shown. Find the mean of the toll by ‘assumed mean’ method.

Solution:
Let us take the assumed mean (A) = 550

∴ The mean of the toll paid by the drivers is ₹ 521.43.

Question 3.
A milk centre sold milk to 50 customers. The table below gives the number of customers and the milk they purchased. Find the mean of the milk sold by direct method.

Solution:

∴ The mean of the milk sold is 2.82 litres.

Question 4.
A frequency distribution table for the production of oranges of some farm owners is given below. Find the mean production of oranges by ‘assumed mean’ method.

Solution:
Let us take the assumed mean (A) = 37.5

∴ The mean of the production of oranges is ₹ 35310.

Question 5.
A frequency distribution of funds collected by 120 workers in a company for the drought affected people are given in the following table. Find the mean of he funds by ‘step deviation’ method.

Solution:
Here, we take A = 1250 and g = 500

∴ The mean of the funds collected is ₹ 987.5.

Question 6.
The following table gives the information of frequency distribution of weekly wages of 150 workers of a company. Find the mean of the weekly wages by ‘step deviation’ method.

Solution:
Here, we take A = 2500 and g = 1000.

∴ The mean of the weekly wages is ₹ 3070.

Question 1.
The daily sale of 100 vegetable vendors is given in the following table. Find the mean of the sale by direct method. (Textbook pg. no. 133 and 134)

Solution:

The mean of the sale is 2150.

Question 2.
The amount invested in health insurance by 100 families is given in the following frequency table. Find the mean of investments using direct method and assumed mean method. Check whether the mean found by the two methods is the same as calculated by step deviation method (Ans: ₹ 2140). (Textbook pg. no. 135 and 136)

Solution:

∴ The mean of investments in health insurance is ₹ 2140.
Assumed mean method:

∴ The mean of investments in health insurance is ₹ 2140.
∴ Mean found by direct method and assumed mean method is the same as calculated by step deviation method.

Question 3.
The following table shows the funds collected by 50 students for flood affected people. Find the mean of the funds.

If the number of scores in two consecutive classes is very low, it is convenient to club them. So, in the above example, we club the classes 0 – 500, 500 – 1000 and 2000 – 2500, 2500 – 3000. Now the new table is as follows

i. Solve by direct method.
ii. Verily that the mean calculated by assumed mean method is the same.
iii. Find the mean in the above example by taking A = 1750. (Textbook pg. no. 137)
Solution:
i. Direct method:

∴ The mean of the funds is ₹ 1390.

ii. Assumed mean method:
Here, A = 1250

∴ The mean calculated by assumed mean method is the same.

iii. Step deviation method:
Here, we take A = 1750 and g = 250

∴ The mean of the funds is ₹ 1390.

Class 10 Maths Digest

• Practice Set 6.1 Class 10 Answers
• Practice Set 6.2 Class 10 Answers
• Practice Set 6.3 Class 10 Answers
• Practice Set 6.4 Class 10 Answers
• Practice Set 6.5 Class 10 Answers
• Practice Set 6.6 Class 10 Answers
• Problem Set 6 Class 10 Answers