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9th Standard Maths 1 Practice Set 5.2 Chapter 5 Linear Equations in Two Variables Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 5.2 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 5 Linear Equations in Two Variables.

Class 9 Maths Part 1 Practice Set 5.2 Chapter 5 Linear Equations in Two Variables Questions With Answers Maharashtra Board

Question 1.
In an envelope there are some ₹5 notes and some ₹10 notes. Total amount of these notes together is ₹350. Number of ₹5 notes are less by 10 than twice the number of ₹10 notes. Then find the number of ₹5 and ₹10 notes.
Solution:
Let the number of ₹5 notes be ‘x’ and the number of ₹10 notes be ‘y’
Total amount of x notes of ₹ 5 = ₹ 5x
Total amount ofy notes of ₹ 10 = ₹ 10y
∴ Total amount = 5x + 10y
According to the first condition,
total amount of the notes together is ₹350.
∴ 5x + 10y = 350 …(i)
According to the second condition,
Number of ₹ 5 notes are less by 10 than twice the number of ₹ 10 notes.
∴ x = 2y – 10
∴ x – 2y = -10 …..(ii)
Multiplying equation (ii) by 5,
5x – 10y = -50 …(iii)
Adding equations (i) and (iii),
5x + 10y =350
+ 5x – 10y = -50
10x =300
∴ x = \(\frac { 300 }{ 10 }\)
∴ x = 30
Substituting x = 30 in equation (ii),
x – 2y = -10
30 – 2y = -10
∴ 30 + 10 = 2y
∴ 40 = 2y
∴ y = \(\frac { 40 }{ 2 }\)
∴ y = 20
There are 30 notes of ₹ 5 and 20 notes of ₹ 10 in the envelope.

Question 2.
The denominator of a fraction is 1 less than twice its numerator. If 1 is added to numerator and denominator respectively, the ratio of numerator to denominator is 3 : 5. Find the fraction.
Solution:
Let the numerator of the fraction be ‘x’ and its denominator be ‘y’.
Then, the required fraction is \(\frac { x }{ y }\) .
According to the first condition,
the denominator is 1 less than twice its numerator.
∴ y = 2x – 1
∴ 2x – y = 1 …(i)
According to the second condition,
if 1 is added to the numerator and the denominator, the ratio of numerator to denominator is 3 : 5.
∴ \(\frac { x+1 }{ y+1 }\) = \(\frac { 3 }{ 5 }\)
∴ y + 1 = 5
∴ 5(x + 1) = 3(y + 1)
∴ 5x + 5 = 3y + 3
∴ 5x – 3y = 3 – 5
∴ 5x – 3y = -2 ……(ii)
Multiplying equation (i) by 3,
6x – 3y = 3 …(iii)
Subtracting equation (ii) from (iii),

Substituting x = 5 in equation (i),
∴ 2x – y = 1
∴ 2(5) – y = 1
∴ 10 – y = 1
∴ y= 10 – 1 =9
∴ The required fraction is \(\frac { 5 }{ 9 }\).

Question 3.
The sum of ages of Priyanka and Deepika is 34 years. Priyanka is elder to Deepika by 6 years. Then find their present ages.
Solution:
Let the present age of Priyanka be ‘x’ years and that of Deepika be ‘y’ years.
According to the first condition,
Priyanka’s age + Deepika’s age = 34 years
∴ x + y = 34 …(i)
According to the second condition,
Priyanka is elder to Deepika by 6 years.
∴ x =y + 6
∴ x – y = 6 …..(ii)
Adding equations (i) and (ii),

∴ x = 20
Substituting x = 20 in equation (i),
x + y = 34
∴ 20 + y = 34
∴ y = 34 -20= 14
∴ The present age of Priyanka is 20 years and that of Deepika is 14 years.

Question 4.
The total number of lions and peacocks in a certain zoo is 50. The total number of their legs is 140. Then find the number of lions and peacocks in the zoo.
Solution:
Let the number of lions in the zoo be ‘x’ and the number of peacocks be ‘y’.
According to the first condition,
the total number of lions and peacocks is 50.
∴ x + y = 50 …(i)
Lion has 4 legs and Peacock has 2 legs.
According to the second condition,
the total number of their legs is 140.
∴ 4x + 2y = 140
Dividing both sides by 2,
2x + y = 70 …(ii)
Subtracting equation (i) from (ii),

Substituting x = 20 in equation (i),
x + y = 50
∴ 20 + y = 50
∴ y = 50 – 20 = 30
∴ The number of lions and peacocks in the zoo are 20 and 30 respectively.

Question 5.
Sanjay gets fixed monthly income. Every year there is a certain increment in his salary. After 4 years, his monthly salary was ₹ 4500 and after 10 years his monthly salary became ₹ 5400, then find his original salary and yearly increment.
Solution:
Let the original salary of Sanjay be ₹ ‘x’ and his yearly increment be ₹ ‘y’.
According to the first condition, after 4 years his monthly salary was ₹ 4500
∴ x + 4y = 4500 …..(i)
According to the second condition,
after 10 years his monthly salary became ₹ 5400
∴ x + 10y = 5400 …(ii)
Subtracting equation (i) from (ii),

∴ y = 150
Substituting y = 150 in equation (i),
x + 4y = 4500
∴ x +4(150) = 4500
∴ x + 600 = 4500
∴ x = 4500 – 600 = 3900
∴ The original salary of Sanjay is ₹ 3900 and his yearly increment is ₹ 150.

Question 6.
The price of 3 chairs and 2 tables is ₹ 4500 and price of 5 chairs and 3 tables is ₹ 7000, then find the price of 2 chairs and 2 tables.
Solution:
Let the price of one chair be ₹ ‘x’ and that of one table be ₹ ‘y’.
According to the first condition,
the price of 3 chairs and 2 tables is ₹ 4500
∴ 3x + 2y = 4500 ,..(i)
According to the second condition, the price of 5 chairs and 3 tables is ? 7000
∴ 5x + 3y = 7000 …(ii)
Multiplying equation (i) by 3,
9x + 6y = 13500 ….(iii)
Multiplying equation (ii) by 2,
10x + 6y= 14000 …(iv)
Subtracting equation (iii) from (iv),

Substituting x = 500 in equation (i),
3x + 2y = 4500
∴ 3(500)+ 2y = 4500
∴ 1500 + 2y = 4500
∴ 2y = 4500- 1500
∴ 2y = 3000
∴ y = \(\frac { 3000 }{ 2 }\)
∴ y = 1500
∴ Price of 2 chairs and 2 tables = 2x + 2y
= 2(500)+ 2(1500)
= 1000 + 3000 = ₹ 4000
∴ The price of 2 chairs and 2 tables is ₹ 4000.

Question 7.
The sum of the digits in a two-digit number is 9. The number obtained by interchanging the digits exceeds the original number by 27. Find the two-digit number.
Solution:
Let the digit in unit’s place be ‘x’ and the digit in ten’s place be ‘y’.

According to the first condition.
the sum of the digits in a two-digit number is 9
x + y = 9 …(i)
According to the second condition,
the number obtained by interchanging the digits exceeds the original number by 27
∴ 10x + y = 10y + x + 27
∴ 10x – x + y – 10y = 27
∴ 9x – 9y = 27
Dividing both sides by 9,
x – y = 3 …….(ii)
Adding equations (i) and (ii),

∴ x = 6
Substituting x = 6 in equation (i),
x + y = 9
∴ 6 + y = 9
∴ y = 9 – 6 = 3
∴ Original number = 10y + x = 10(3)+ 6
= 30 + 6 = 36
∴ The two digit number is 36.

Question 8.
In ∆ABC, the measure of ∠A is equal to the sum of the measures of ∠B and ∠C. Also the ratio of measures of ∠B and ∠C is 4 : 5. Then find the measures of angles of the triangle.
Solution:
Let the measure of ∠B be ‘x°’ and that of ∠C be ‘y°’.
According to the first condition,
m∠A = m∠B + m∠C
∴ m∠A = x° + y°
In AABC,
m∠A + m∠B + m∠C = 180° …[Sum of the measures of the angles of a triangle is 180°]
∴ x + y + x + y = 180 ,
∴ 2x + 2y = 180
Dividing both sides by 2,
x + y = 90 …(i)
According to the second condition,
the ratio of the measures of ∠B and ∠C is 4 : 5.
∴ \(\frac { x }{ y }\) = \(\frac { 4 }{ 5 }\)
∴ 5x = 4y
∴ 5x – 4y = 0 …….(ii)
Multiplying equation (i) by 4,
4x + 4y = 360 …(iii)
Adding equations (ii) and (iii),

∴ x = 40
Substituting x = 40 in equation (i),
x + y = 90
∴ 40 + y = 90
∴ y = 90 – 40
∴ y = 50
∴ m∠A = x° + y° = 40° + 50° = 90°
∴ The measures of ∠A, ∠B and ∠C are 90°, 40°, and 50° respectively.

Question 9.
Divide a rope of length 560 cm into 2 parts such that twice the length of the smaller part is equal to \(\frac { 1 }{ 3 }\) of the larger part. Then find the length of the larger part.
Solution:
Let the length of the smaller part of the rope be ‘x’ cm and that of the larger part be ‘y’ cm.
According to the first condition,
total length of the rope is 560 cm.
∴ x + y = 560 …(i)
Twice the length of the smaller part = 2x
\(\frac { 1 }{ 3 }\)rd length of the larger part = \(\frac { 1 }{ 3 }\)y
According to the second condition,
2x = \(\frac { 1 }{ 3 }\) 3
∴ 6x = y
∴ 6x – y = 0 ……(ii)
Adding equations (i) and (ii),

∴ x = 80
Substituting x = 80 in equation (ii),
6x – y = 0
∴ 6(80) – y = 0
∴ 480 – y = 0
∴ y = 480
∴ The length of the larger part of the rope is 480 cm.

Question 10.
In a competitive examination, there were 60 questions. The correct answer would carry 2 marks, and for incorrect answer 1 mark would be subtracted. Yashwant had attempted all the questions and he got total 90 marks. Then how many questions he got wrong?
Solution:
Let us suppose that Yashwant got ‘x’ questions right and ‘y’ questions wrong.
According to the first condition, total number of questions in the examination are 60.
∴ x + y = 60 …(i)
Yashwant got 2 marks for each correct answer and 1 mark was deducted for each wrong answer.
∴ He got 2x – y marks.
According to the second condition,
he got 90 marks.
2x – y = 90 … (ii)
Adding equations (i) and (ii),

∴ x = 50
Substituting x = 50 in equation (i),
50 + y = 60
∴ y = 60 – 50 = 10
∴ Yashwant got 10 questions wrong.

Maharashtra Board Class 9 Maths Chapter 5 Linear Equations in Two Variables Practice Set 5.2 Intext Questions and Activities

Question 1.
The population of a certain town was 50,000. In a year, male population was increased by 5% and female population was increased by 3%. Now the population became 52020. Then what was the number of males and females in the previous year? (Textbook pg. no. 89)
Solution:
Step 1: Read the given word problem carefully and try to understand it.

Step 2: Make assumptions using two variables x and y.
Let the number of males in previous year be
‘x’ and the number of females be ‘y’.

Step 3: From the given information, form mathematical statements using the above variables.
According to the first condition,
the total population of town was 50,000.
∴ x + y = 50000 …(i)
Male population increased by 5%.
∴ Number of males = x + 5% of x , 5

Female population increased by 3%.
∴ Number of females = y + 3% of y

According to the second condition,
in a year population became 52020
∴ \(\frac{105}{100} x+\frac{103}{100} y=52020\)
∴ 105 x + 103 y = 5202000 …(ii)
Multiplying equation (i) by 103,
103 x + 103 y = 5150000 …(iii)

Step 4: Here, we use elimination method.
Subtracting equation (iii) from (ii),

∴ x = 26000
Substituting x = 26000 in equation (i),
∴ 26000 + y = 50000
∴ y = 50000 – 26000
∴ y = 24000
∴ Number of males = x = 26000
∴ Number of females = y = 24000

Step 5: Write the answer.
The number of males and females in the previous year were 26,000 and 24,000 respectively.

Step 6: Verify your result using smart check.

9th Class Maths Solution Maharashtra Board 

• Linear Equations in Two Variables Practice Set 5.1 Class 9 Maths Solutions
• Linear Equations in Two Variables Practice Set 5.2 Class 9 Maths Solutions
• Linear Equations in Two Variables Problem Set 5 Class 9 Maths Solutions

9th Standard Maths 1 Practice Set 6.2 Chapter 6 Financial Planning Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 6.2 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 6 Financial Planning.

Class 9 Maths Part 1 Practice Set 6.2 Chapter 6 Financial Planning Questions With Answers Maharashtra Board

Question 1.
Observe the table given below. Check and decide, whether the individuals have to pay income tax.

Solution:
i. Miss Nikita’s age = 27 years < 60 years
Miss Nikita’s income = ₹ 2,34,000
Miss Nikita’s income is below the basic
exemption limit of ₹ 2,50,000.
∴ Miss Nikita will not have to pay income tax.

ii. Mr. Kulkarni’s age 36 years < 60 years
Mr. Kulkarni’s income = ₹3,27,000
Mr. Kulkarni’s income is above the basic exemption Limit of ₹2,50,000.
∴ Mr. Kulkarni will have to pay income tax.

iii. Miss Mehta’s age = 44 years < 60 years Miss Mehta’s income = ₹5.82,000
Miss Mehta’s income is above the basic exemption limit of ₹2,50,000.
∴ Miss Mehta will have to pay income tax.

iv. Mr. Bajaj’s age = 64 years (Age 60 to 80 years)
Mr. Bajaj’s income = ₹8,40,000
Mr. Bajaj’s income is above the basic exemption Limit of ₹3,00,000.
∴ Mr. Bajaj will have to pay income tax.

v. Mr. Desilva’s age = 81 years > 80 years
Mr. Desilva’s income = ₹4,50,000
Mr. Desilva’s income is below the basic exemption limit of ₹ 5,00.000.
∴ Mr. Desilva will not have to pay income tax.

Question 2.
Mr. Kartarsingh (age 48 years) works in a private company. His monthly income after deduction of allowances is ₹ 42,000 and every month he contributes ₹ 3000 to GPF. He has also bought ₹ 15,000 worth of NSC (National Savings Certificate) and donated ₹ 12,000 to the PM’s Relief Fund. Compute his income tax.
Solution:
Mr. Kartarsingh’s monthly income = ₹ 42,000
Mr. Kartarsingh’s yearly income = 42,000 x 12 = ₹ 5,04,000

Mr. Kartarsingh’s investment
= GPF + NSC
= (3000 x 12)+ 15,000
= 36,000 + 15,000
= ₹ 51,000

Donation to PM’s relief fund = ₹ 12, 000
∴ Taxable income
= yearly income – (investment + donation)
= 5,04,000 – (51,000 + 12,000)
= 5,04,000 – 63,000 = ₹ 4,41,000
Mr. Kartarsingh income falls in the slab 2,50,001 to 5,00,000.
∴ Income tax = 5% of (Taxable income – 250000) = 5% of (4,41,000 – 2,50,000)
= \(\frac { 5 }{ 100 }\) x 1,91,000 100
= ₹ 9550
Education cess = 2% of income tax
= \(\frac { 2 }{ 100 }\) x 9550
= 191
Secondary and Higher Education cess = 1% of income tax
= \(\frac { 1 }{ 100 }\) x 9550 100
= 95.50
Total income tax = Income tax + Education cess + Secondary and higher education cess
= 9550 + 191 + 95.50
= ₹ 9836.50
∴ Mr. Kartarsingh’s income tax is ₹ 9836.50

Maharashtra Board Class 9 Maths Chapter 6 Financial Planning Practice Set 6.2 Intext Questions and Activities

Question 1.
Use Table I given above and write the appropriate amount/figure in the boxes for the example given below. (Textbook pg. no. 102)
Mr. Mehta’s annual income is ₹4,50,000
i. If he does not have any savings by which he can claim deductions from his income, to which slab does his taxable income belong ? ______
ii, What is the amount on which he will have to pay income tax and at what percent rate? on ₹ _______
percentage _______
iii. On what amount will the cess be levied? _______
Answer:
1. ₹2,50,001 to ₹5,00,000
ii. 5% of (4,50,000 – 2,50,000)
i.e. 5% of ₹2,00,000
iii. income tax = 5% of 2,00,000
= \(\frac { 5 }{ 100 }\) x 2,00,000
= ₹10,000
∴ Education cess and Secondary and higher education cess will be levied on the income tax i.e., on ₹10,000.

Question 2.
Use table lito carry out the following activity.
Mr. Pandit is 75 years old. Last year his annual income was ₹ 13,25,000. How much is his taxable income? How much tax does he have to pay? (Textbook pg. no. 103)
Solution:
Mr. Pandit’s age = 75 years (Age 60 to 80 years)
Mr. Pandit’s income is more than 10,00,000.
According to the table,
Income tax = ₹ 1,10,000 + 30 % of (taxable income – 10,00,000)
Taxable income – 10,00,000 = 13,25,000 – 10,00,000 = 3,25,000
In addition, on ₹ 3,25,000 rupees he has to pay 30% income tax.
3,25,000 x \(\frac { 30 }{ 100 }\) = ₹ 97500
Therefore, his total income tax amounts to 1,10,000 + 97,500 ₹ 207500
Besides this, education cess willi be 2% of income tax 207500 x \(\frac { 2 }{ 100 }\) = ₹ 4150.
A secondary and higher education cess at 1% of income tax = 207500 x \(\frac { 1 }{ 100 }\) = ₹ 2075.
∴ Total income tax = Income tax + education cess + secondary and higher education cess
= 207500 + 4150 + 2075
= ₹2,13,725

Maharashtra Board 9th Maths Solution 

• Financial Planning Practice Set 6.1 Class 9 Maths Solutions
• Financial Planning Practice Set 6.2 Class 9 Maths Solutions
• Financial Planning Problem Set 6 Class 9 Maths Solutions

10th Standard Maths 1 Practice Set 3.4 Chapter 3 Arithmetic Progression Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 3.4 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 3 Arithmetic Progression.

Class 10 Maths Part 1 Practice Set 3.4 Chapter 3 Arithmetic Progression Questions With Answers Maharashtra Board

Question 1.
On 1^st Jan 2016, Sanika decides to save ₹ 10, ₹ 11 on second day, ₹ 12 on third day. If she decides to save like this, then on 31st Dec 2016 what would be her total saving?
Solution:

∴ Sanika’s total saving on 31^st December 2016 would be ₹ 70455.

Question 2.
A man borrows ₹ 8000 and agrees to repay with a total interest of ₹ 1360 in 12 monthly instalments. Each instalment being less than the preceding one by ₹ 40. Find the amount of the first and last instalment.
Solution:
i. The instalments are in A.P.
Amount repaid in 12 instalments (S12)
= Amount borrowed + total interest
= 8000 + 1360
∴ S₁₂ = 9360
Number of instalments (n) = 12
Each instalment is less than the preceding one by ₹ 40.
∴ d = -40

∴ Amount of the first instalment is ₹ 1000 and that of the last instalment is ₹ 560.

Question 3.
Sachin invested in a national saving certificate scheme. In the first year he invested ₹ 5000, in the second year ₹ 7000, in the third year ₹ 9000 and so on. Find the total amount that he invested in 12 years.
Solution:
i. Amount invested by Sachin in each year are as follows:
5000, 7000, 9000, …
The above sequence is an A.P.
∴ a = 5000, d = 7000 – 5000 = 2000, n = 12

∴ The total amount invested by Sachin in 12 years is ₹ 1,92,000.

Question 4.
There is an auditorium with 27 rows of seats. There are 20 seats in the first row, 22 seats in the second row, 24 seats in the third row and so on. Find the number of seats in the 15th row and also find how many total seats are there in the auditorium?
Solution:
i. The number of seats arranged row-wise are as follows:
20, 22, 24,
The above sequence is an A.P.
∴ a = 20, d = 22 – 20 = 2, n = 27

ii. t_n = a + (n – 1)d
∴ t₁₅ = 20 + (15 – 1)2
= 20 + 14 × 2
= 20 + 28
∴ t₁₅ = 48
∴ The number of seats in the 15^th row is 48.

∴ Total seats in the auditorium are 1242.

Question 5.
Kargil’s temperature was recorded in a week from Monday to Saturday. All readings were in A.P. The sum of temperatures of Monday and Saturday was 5°C more than sum of temperatures of Tuesday and Saturday. If temperature of Wednesday was -30° Celsius then find the temperature on the other five days.
Solution:
Let the temperatures from Monday to Saturday in A.P. be
a, a + d, a + 2d, a + 3d, a + 4d, a + 5d.
According to the first condition,
(a) + (a + 5d) = (a + d) + (a + 5d) + 5°
∴ d = -5°
According to the second condition,
a + 2d = -30°
∴ a + 2(-5°) = -30°
∴ a – 10° = -30°
∴ a = -30° + 10° = -20°
∴ a + d = -20° – 5° = – 25°
a + 3d = -20° + 3(- 5°) = -20° – 15° = -35°
a + 4d = -20° + 4(-5°) = -20° – 20° = -40°
a + 5d = -20° + 5(-5°) = -20° – 25° = -45°
∴ The temperatures on the other five days are
-20°C, -25° C, -35° C, -40° C and -45° C.

Question 6.
On the world environment day tree plantation programme was arranged on a land which is triangular in shape. Trees are planted such that in the first row there is one tree, in the second row there are two trees, in the third row three trees and so on. Find the total number of trees in the 25 rows.
Solution:
i. The number of frees planted row-wise are as follows:
1,2,3,…
The above sequence is an A.P.
∴ a = 1, d = 2 – 1 = 1,n = 25

∴ The total number of trees in 25 rows are 325.

Maharashtra State Board Class 10 Maths Solutions Part 1

• Arithmetic Progression Practice Set 3.1 Class 10 Maths Solutions
• Arithmetic Progression Practice Set 3.2 Class 10 Maths Solutions
• Arithmetic Progression Practice Set 3.3 Class 10 Maths Solutions
• Arithmetic Progression Practice Set 3.4 Class 10 Maths Solutions
• Arithmetic Progression Problem Set 3 Class 10 Maths Solutions
• Financial Planning Practice Set 4.1 Class 10 Maths Solutions
• Financial Planning Practice Set 4.2 Class 10 Maths Solutions
• Financial Planning Practice Set 4.3 Class 10 Maths Solutions
• Financial Planning Practice Set 4.4 Class 10 Maths Solutions
• Financial Planning Problem Set 4A Class 10 Maths Solutions
• Financial Planning Problem Set 4B Class 10 Maths Solutions

9th Standard Maths 1 Practice Set 5.1 Chapter 5 Linear Equations in Two Variables Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 5.1 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 5 Linear Equations in Two Variables.

Class 9 Maths Part 1 Practice Set 5.1 Chapter 5 Linear Equations in Two Variables Questions With Answers Maharashtra Board

Question 1.
By using variables x and y form any five linear equations in two variables.
Answer:
The general form of a linear equation in two variables x and y is ax + by + c = 0,
where a, b, c are real numbers and a ≠ 0, b ≠ 0.
Five linear equations in two variables are as follows:
i. 3x + 4y – 12 = 0
ii. 3x – 4y + 12 = 0
iii. 5x + 5y – 6 = 0
iv. 7x + 12y – 11 = 0
v. x – y + 5 = 0

Question 2.
Write five solutions of the equation x + y = 1.
Answer:
i. x = 1, y = 6
ii. x = -1, y = 8
iii. x = 5, y = 2
iv. x = 0, y = 7
v. x = 10, y = -3

Question 3.
Solve the following sets of simultaneous equations.
i. x + y = 4 ; 2x – 5y = 1
ii. 2x + y = 5 ; 3x – y = 5
iii. 3x – 5y = 16; x – 3y= 8
iv. 2y – x = 0; 10x + 15y = 105
v. 2x + 3y + 4 = 0; x – 5y = 11
vi. 2x – 7y = 7; 3x + y = 22
Solution:
i. Substitution Method:
x + y = 4
∴ x = 4 – y …(i)
2x – 5y = 1 ……(ii)
Substituting x = 4 – y in equation (ii),
2(4 – y) – 5y = 1
∴ 8 – 2y – 5y = 1
∴ 8 – 7y = 1
∴ 8 – 1 = 7y
∴ 7 = 7y
∴ y = \(\frac { 7 }{ 7 }\)
∴ y = 1
Substituting y = 1 in equation (i),
x = 4 – 1 = 3
∴ (3,1) is the solution of the given equations.

Alternate method:
Elimination Method:
x + y = 4 …(i)
2x – 5y = 1 ……(ii)
Multiplying equation (i) by 5,
5x + 5y = 20 … (iii)
Adding equations (ii) and (iii),
2x – 5y = 1
+ 5x + 5y = 20
7 = 21
∴ x = \(\frac { 21 }{ 7 }\)
∴ x = 3
Substituting x = 3 in equation (i),
3 + y = 4
∴ y = 4 – 3 = 1
(3,1) is the solution of the given equations.

ii. 2x + y = 5 …(i)
3x – y = 5 …(ii)
Adding equations (i) and (ii),
2x + y = 5
+ 3x – y = 5
5x = 10
∴ x = \(\frac { 10 }{ 5 }\)
∴ x = 2
Substituting x = 2 in equation (i),
2(2) + y = 5
4 + y = 5
∴ y = 5 – 4 = 1
∴ (2, 1) is the solution of the given equations.

iii. 3x – 5y = 16 …(i)
x – 3y = 8
∴x = 8 + 3y …..(ii)
Substituting x = 8 + 3y in equation (i),
3(8 + 3y) – 5y = 16
24 + 9y- 5y = 16
∴4y= 16 – 24
∴ 4y = -8
∴ y = \(\frac { -8 }{ 4 }\)
y = -2
Substituting y = -2 in equation (ii),
x = 8 + 3 (-2)
∴ x = 8 – 6 = 2
∴ (2, -2) is the solution of the given equations.

iv. 2y – x = 0
∴ x = 2y …(i)
10x + 15y = 105 …(ii)
Substituting x = 2y in equation (ii),
10(2y) + 15y = 105
∴ 20y + 15y = 105
∴ 35y = 105
∴ y = \(\frac { 105 }{ 35 }\)
∴ y = 3
Substituting y = 3 in equation (i),
x = 2y
∴ x = 2(3) = 6
∴ (6, 3) is the solution of the given equations.

v. 2x + 3y + 4 = 0 …(i)
x – 5y = 11
∴x = 11 + 5y …(ii)
Substituting x = 11 + 5y in equation (i),
2(11 +5y) + 3y + 4 = 0
∴ 22 + 10y + 3y + 4 = 0
∴ 13y + 26 = 0
∴ 13y = -26
∴y = \(\frac { -26 }{ 13 }\)
∴ y = -2
Substituting y = -2 in equation (ii),
x = 11 + 5y
∴ x = 11 + 5(-2)
∴ x = 11 – 10 = 1
∴ (1, -2) is the solution of the given equations.

vi. 2x – 7y = 7 …(i)
3x + y = 22
∴ y = 22 – 3x ……(ii)
Substituting y = 22 – 3x in equation (i),
2x – 7(22 – 3x) = 7
∴ 2x – 154 + 21x = 7
∴ 23x = 7 + 154
∴ 23x = 161
∴ x = \(\frac { 161 }{ 23 }\)
∴ x = 7
Substituting x = 7 in equation (ii),
y = 22 – 3x
∴ y = 22 – 3(7)
∴ 7 = 22 -21= 1
∴ (7, 1) is the solution of the given equations.

Question 1.
Solve the following equations. (Textbook pg. no. 80)
i. m + 3 = 5
ii. 3y + 8 = 22
iii. \(\frac { x }{ 3 }\) = 2
iv. 2p = p + \(\frac { 4 }{ 9 }\)
Solution:
i. m + 3 = 5
m = 5 – 3
∴m = 2

ii. 3y + 8 = 22
∴ 3y = 22 – 8
∴ 3y = 14
∴ y = \(\frac { 14 }{ 9 }\)

iii. \(\frac { x }{ 3 }\) = 2
∴ x = 2 × 3
∴ x = 6

iv. 2p = p + \(\frac { 4 }{ 9 }\)
∴ 2p – p = \(\frac { 4 }{ 9 }\)
∴ p = \(\frac { 4 }{ 9 }\)

Question 2.
Which number should be added to 5 to obtain 14? (Textbook pg. no. 80)
Solution:
x + 5 = 14
∴ x = 14 – 5
x = 9
∴ 9 + 5 = 14

Question 3.
Which number should be subtracted from 8 to obtain 2? (Textbook pg. no. 80)
Solution:
8 – y = 2
∴ y = 8 – 2
∴ y = 6
∴ 8 – 6 = 2

Question 4.
x + y = 5 and 2x + 2y= 10 are two equations in two variables. Find live different solutions of x + y = 5, verify whether same solutions satisfy the equation 2x + 2y = 10 also. Observe both equations. Find the condition where two equations in two variables have all solutions in common. (Textbook pg. no. 82)
Solution:
Five solutions of x + y = 5 are given below:
(1,4), (2, 3), (3, 2), (4,1), (0, 5)
The above solutions also satisfy the equation 2x + 2y = 10.
∴ x + y = 5 …[Dividing both sides by 2]
∴ If the two equations are the same, then the two equations in two variables have all solutions common.

Question 5.
3x – 4y – 15 = 0 and y + x + 2 = 0. Can these equations be solved by eliminating x ? Is the solution same? (Textbook pg. no. 84)
Solution:
3x – 4y – 15 = 0
∴ 3x – 4y = 15 …(i)
y + x + 2 = 0
∴ x + y = -2 ……(ii)
Multiplying equation (ii) by 3,
3x + 3y = -6 …(iii)
Subtracting equation (iii) from (i),

∴ y = -3
Substituting y = -3 in equation (ii),
∴ x – 3 = -2
∴ x = – 2 + 3
∴ x = 1
∴ (x, y) = ( 1, -3)
Yes, the given equations can be solved by eliminating x. Also, the solution will remain the same.

9th Class Maths Solution Maharashtra Board 

• Linear Equations in Two Variables Practice Set 5.1 Class 9 Maths Solutions
• Linear Equations in Two Variables Practice Set 5.2 Class 9 Maths Solutions
• Linear Equations in Two Variables Problem Set 5 Class 9 Maths Solutions

9th Standard Maths 1 Practice Set 4.3 Chapter 4 Ratio and Proportion Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 4.3 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 4 Ratio and Proportion.

Class 9 Maths Part 1 Practice Set 4.3 Chapter 4 Ratio and Proportion Questions With Answers Maharashtra Board

Ratio and Proportion Practice Set 4.3 Question 1.
If \(\frac { a }{ b }\) = \(\frac { 7 }{ 3 }\), then find the aIues of the following ratios.

Solution:

Ratio and Proportion Class 9 Practice Set 4.3 Question 2.
If \(\frac{15 a^{2}+4 b^{2}}{15 a^{2}-4 b^{2}}=\frac{47}{7}\), then find the value of the following ratios.

Solution:

Practice Set 4.3 Algebra 9th Question 3.
If \(\frac{3 a+7 b}{3 a-7 b}=\frac{4}{3}\)then find the value of the ratio \(\frac{3 a^{2}-7 b^{2}}{3 a^{2}+7 b^{2}}\).
Solution:

Class 9 Maths Chapter 4 Ratio And Proportion Practice Set 4.3 Question 4.
Solve the following equations.

Solution:

This equation is true for x = 0
∴ x = 0 is one of the solutions.
If x ≠ 0, then x² ≠ 0
∴ \(\frac { 1 }{ 12x – 20 }\) = \(\frac { 1 }{ 8x + 12 }\) … [Dividing both sides by x²]
∴ 8x + 12 = 12x – 20
∴ 12 + 20 = 12x – 8x
∴ 32 = 4x
∴ x = 8
∴ x = 0 or x = 8 are the solutions of the given equation.

∴ 21(x – 5) = 4(2x + 3)
∴ 21x – 105 = 8x + 12
∴ 21x – 8x = 12 + 105
∴ 13x = 117
∴ x = 9
∴ x = 9 ¡s the solution of the given equation.

∴ 9(4x + 1) = 25(x + 3)
∴36x + 925x + 75
∴ 36x – 25 = 75 – 9
∴11x = 66
∴ x = 6
∴ x = 6 is the solution of the given equation.


∴ 4(3x – 4) = 5(x + 1)
∴ 12x – 16 = 5x + 5
∴ 12x – 5x = 5 + 16
∴ 7x = 21
∴ x = 3
∴ x = 3 ¡s the solution of the given equation.

Class 9 Maths Solution Maharashtra Board 

• Ratio and Proportion Practice Set 4.1 Class 9 Maths Solutions
• Ratio and Proportion Practice Set 4.2 Class 9 Maths Solutions
• Ratio and Proportion Practice Set 4.3 Class 9 Maths Solutions
• Ratio and Proportion Practice Set 4.4 Class 9 Maths Solutions
• Ratio and Proportion Practice Set 4.5 Class 9 Maths Solutions
• Ratio and Proportion Problem Set 4 Class 9 Maths Solutions

9th Standard Maths 1 Practice Set 4.5 Chapter 4 Ratio and Proportion Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 4.5 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 4 Ratio and Proportion.

Class 9 Maths Part 1 Practice Set 4.5 Chapter 4 Ratio and Proportion Questions With Answers Maharashtra Board

Question 1.
Which number should be subtracted from 12, 16 and 21 so that resultant numbers are in continued proportion?
Solution:
Let the number to be subtracted be x.
∴ (12 – x), (16 – x) and (21 – x) are in continued proportion.

∴ 84 – 4x = 80 – 5x
∴ 5x – 4x = 80 – 84
∴ x = -4
∴ -4 should be subtracted from 12,16 and 21 so that the resultant numbers in continued proportion.

Question 2.
If (28 – x) is the mean proportional of (23 – x) and (19 – x), then find the value ofx.
Solution:
(28 – x) is the mean proportional of (23 – x) and (19-x). …[Given]

∴ -5(19 – x) = 9(28 – x)
∴ -95 + 5x = 252 – 9x
∴ 5x + 9x = 252 + 95
∴ 14x = 347
∴ x = \(\frac { 347 }{ 14 }\)

Question 3.
Three numbers are in continued proportion, whose mean proportional is 12 and the sum of the remaining two numbers is 26, then find these numbers.
Solution:
Let the first number be x.
∴ Third number = 26 – x
12 is the mean proportional of x and (26 – x).
∴ \(\frac { x }{ 12 }\) = \(\frac { 12 }{ 26 – x }\)
∴ x(26 – x) = 12 x 12
∴ 26x – x² = 144
∴ x² – 26x + 144 = 0
∴ x² – 18x – 8x + 144 = 0
∴ x(x – 18) – 8(x – 18) = 0
∴ (x – 18) (x – 8) = 0
∴ x = 18 or x = 8
∴ Third number = 26 – x = 26 – 18 = 8 or 26 – x = 26 – 8 = 18
∴ The numbers are 18, 12, 8 or 8, 12, 18.

Question 4.
If (a + b + c)(a – b + c) = a² + b² + c², show that a, b, c are in continued proportion.
Solution:
(a + b + c)(a – b + c) = a² + b² + c² …[Given]
∴ a(a – b + c) + b(a – b + c) + c(a – b + c) = a2 + b2 + c2
∴ a² – ab + ac + ab – b² + be + ac – be + c² = a² + b² + c²
∴ a² + 2ac – b² + c² = a² + b² + c²
∴ 2ac – b² = b²
∴ 2ac = 2b²
∴ ac = b²
∴ b² = ac
∴ a, b, c are in continued proportion.

Question 5.
If \(\frac { a }{ b }\) = \(\frac { b }{ c }\) and a, b, c > 0, then show that,
i. (a + b + c)(b – c) = ab – c²
ii. (a² + b²)(b² + c²) = (ab + be)²
iii. \(\frac{a^{2}+b^{2}}{a b}=\frac{a+c}{b}\)
Solution:
Let \(\frac { a }{ b }\) = \(\frac { b }{ c }\) = k
∴ b = ck
∴ a = bk =(ck)k
∴ a = ck² …(ii)

i. (a + b + c)(b – c) = ab – c²
L.H.S = (a + b + c) (b – c)
= [ck² + ck + c] [ck – c] … [From (i) and (ii)]
= c(k² + k + 1) c (k – 1)
= c² (k² + k + 1) (k – 1)
R.H.S = ab – c²
= (ck²) (ck) – c² … [From (i) and (ii)]
= c²k³ – c²
= c²(k³ – 1)
= c² (k – 1) (k² + k + 1) … [a³ – b³ = (a – b) (a² + ab + b²]
∴ L.H.S = R.H.S
∴ (a + b + c) (b – c) = ab – c²

ii. (a² + b²)(b² + c²) = (ab + bc)²
b = ck; a = ck²
L.H.S = (a² + b²) (b² + c²)
= [(ck²) + (ck)²] [(ck)² + c²] … [From (i) and (ii)]
= [c²k⁴ + c²k²] [c²k² + c²]
= c²k² (k² + 1) c² (k² + 1)
= c4k² (k² + 1)²
R.H.S = (ab + bc)²
= [(ck²) (ck) + (ck)c]² …[From (i) and (ii)]
= [c²k³ + c²k]²
= [c²k (k² + 1)]2 = c⁴(k² + 1)²
∴ L.H.S = R.H.S
∴ (a² + b²) (b² + c²) = (ab + bc)²

iii. \(\frac{a^{2}+b^{2}}{a b}=\frac{a+c}{b}\)

9th Standard Algebra Practice Set 4.5 Question 6. Find mean proportional of \(\frac{x+y}{x-y}, \frac{x^{2}-y^{2}}{x^{2} y^{2}}\).
Solution:
Let a be the mean proportional of \(\frac{x+y}{x-y}\) and \(\frac{x^{2}-y^{2}}{x^{2} y^{2}}\)

Class 9 Maths Solution Maharashtra Board 

• Ratio and Proportion Practice Set 4.1 Class 9 Maths Solutions
• Ratio and Proportion Practice Set 4.2 Class 9 Maths Solutions
• Ratio and Proportion Practice Set 4.3 Class 9 Maths Solutions
• Ratio and Proportion Practice Set 4.4 Class 9 Maths Solutions
• Ratio and Proportion Practice Set 4.5 Class 9 Maths Solutions
• Ratio and Proportion Problem Set 4 Class 9 Maths Solutions

9th Standard Maths 1 Problem Set 4 Chapter 4 Ratio and Proportion Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Problem Set 4 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 4 Ratio and Proportion.

Class 9 Maths Part 1 Problem Set 4 Chapter 4 Ratio and Proportion Questions With Answers Maharashtra Board

Question 1.
Select the appropriate alternative answer for the following questions.

i . If 6 : 5 = y : 20, then what will be the value of y?
(A) 15
(B) 24
(C) 18
(D) 22.5
Answer:
(B) 24

ii. What is the ratio of 1 mm to 1 cm ?
(A) 1 : 100
(B) 10: 1
(C) 1 : 10
(D) 100: 1
Answer:
(C) 1 : 10

iii. The ages of Jatin, Nitin and Mohasin are 16, 24 and 36 years respectively. What is the ratio of Nitin’s age to Mohasin’s age ?
(A) 3 : 2
(B) 2 : 3
(C) 4 : 3
(D) 3 : 4
Answer:
(B) 2 : 3

iv. 24 bananas were distributed between Shubham and Anil in the ratio 3 : 5, then how many bananas did Shubham get?
(A) 8
(B) 15
(C) 12
(D) 9
Answer:
(D) 9

v. What is the mean proportional of 4 and 25?
(A) 6
(B) 8
(C) 10
(D) 12
Answer:
(C) 10

Hints:
i . \(\frac{6}{5}=\frac{y}{20}\)
∴ \(\quad y=\frac{6 \times 20}{5}=24\)

ii. \(\frac{1 \mathrm{mm}}{1 \mathrm{cm}}=\frac{1 \mathrm{mm}}{10 \mathrm{mm}}=\frac{1}{10}=1 : 10\)

iii. \(\frac{\text { Age of Nitin }}{\text { Age of Mohasin }}=\frac{24}{36}=\frac{12 \times 2}{12 \times 3}\)
\(\frac { 2 }{ 3 }\) = 2 : 3

iv. 3x + 5x = 24
∴ 8x = 24
∴ x = 3
∴ Number of bananas with Shubham = 3x = 9

v. x² = 4 x 25
∴ x² = 100
∴ x = 10

Question 2.
For the following numbers write the ratio of first number to second number in the reduced form. [1 Mark each]
i. 21,48
ii. 36,90
iii. 65,117
iv. 138,161
v. 114,133
Solution:
i. 21,48
\(\text { Ratio }=\frac{21}{48}=\frac{3 \times 7}{3 \times 16}=\frac{7}{16}=7 : 16\)
ii. 36,90
\(\text { Ratio }=\frac{36}{90}=\frac{18 \times 2}{18 \times 5}=\frac{2}{5}=2 : 5\)
iii. 65,117
\(\text { Ratio }=\frac{65}{117}=\frac{13 \times 5}{13 \times 9}=\frac{5}{9}=5 : 9\)
iv. 138,161
\(\text { Ratio }=\frac{138}{161}=\frac{23 \times 6}{23 \times 7}=\frac{6}{7}=6 : 7\)
v. 114,133
\(\text { Ratio }=\frac{114}{113}=\frac{19 \times 6}{19 \times 7}=\frac{6}{7}=6 : 7\)

Question 3.
Write the following ratios in the reduced form.
i. Radius to the diameter of a circle.
ii. The ratio of diagonal to the length of a rectangle, having length 4 cm and breadth 3 cm.
iii. The ratio of numbers denoting perimeter to area of a square, having side 4 cm.
Solution:
i. Radius to the diameter of a circle.
Let radius of the circle be r
then, diameter = 2 x radius = 2r
Ratio of radius to diameter of circle

∴ Ratio of radius to diameter of circle is 1 : 2.

ii. The ratio of diagonal to the length of a rectangle, having length 4 cm and breadth 3 cm.

Let □ ABCD be a rectangle.
In ∆ABC, ∠B = 90°
AC² = AB² + BC² … [Pythagoras theorem]
= 4² + 3² = 16 + 9
∴ AC² = 25
AC = 5 … [Taking square root on both side]
The ratio of diagonal to the length of a rectangle = \(\frac { AC }{ AB }\)
= \(\frac { 5 }{ 4 }\)
= 5 : 4
∴ The ratio of diagonal to the length of a rectangle is 5 : 4

iii. The ratio of numbers denoting perimeter to area of a square, having side 4 cm. side of square = 4cm
Perimeter of square = 4 x side = 4 x 4 = 16 cm
Area of square = (side)² = (4)² = 16 cm²
Ratio of numbers denoting perimeter to area of square

∴ The ratio of numbers denoting perimeter to area of a square is 1 : 1.

Question 4.
Check whether the following numbers are in continued proportion.
i. 2, 4, 8
ii. 1, 2, 3
iii. 9, 12, 16
iv. 3, 5, 8
Solution:
If a, b, c are in continued proportion then b² = ac.
i. 2, 4, 8
Let, a = 2, b = 4 and c = 8
Here, b² = 4² = 16
ac = 2 x 8 = 16
∴ b² = ac
∴ 2, 4,8 are in continued proportion.

ii. 1, 2, 3
Let, a = 1, b = 2 and c = 3
Here, b² = 2² = 4
ac = 1 x 3 = 3
∴ b² ≠ ac
∴ 1, 2,3 are not in continued proportion.

iii. 9, 12, 16
Let, a = 9, b = 12 and c = 16
Here, b² = 12² = 144
ac = 9 x 16 = 144
∴ b² = ac
∴ 9, 12, 16 are in continued proportion.

iv. 3, 5, 8
Let, a = 3, b = 5 and c = 8
Here, b² = 5² = 25
ac = 3 x 8 = 24
∴ b² ≠ ac
∴ 3, 5, 8 are not in continued proportion.

Question 5.
a, b, c are in continued proportion. If a = 3 and c = 27, then find b.
Solution:
a, b, c are in continued proportion. …[Given]
∴ b² = ac
∴ b² = 3 x 27 …[∵ a = 3 and c = 27]
∴ b² = 81
∴ b = 9 …[Taking square root of both sides]

Question 6.
Convert the following ratios into percentages.

Solution:
i. Let 37 : 500 = x%

∴ 37 : 500 = 7.4%

Question 7.
Write the ratio of first quantity to second quantity in the reduced form.
i. 1024 MB, 1.2 GB [(1024 MB = 1GB)]
ii. ₹ 17, ₹ 25 and 60 paise
iii. 5 dozen, 120 units
iv. 4 sq.m, 800 sq.cm
v. 1.5 kg, 2500 gm
Solution:
i. 1024 MB, 1.2 GB
1024 MB = 1 GB

ii. ₹ 17, ₹ 25 and 60 paise
₹ 17 = 17 x 100 paise = 1700 paise
₹ 25 and 60 paise = (25x 100) paise + 60 paise
= (2500 + 60) paise
= 2560 paise

iii. 5 dozen, 120 units
5 dozen = 5 x 12 units = 60 units

iv. 4 sq.m, 800 sq.cm
4 sq.m = 4 x 10000 sq.cm = 40000 sq.cm

v. 1.5 kg, 2500 gm
1.5 kg = 1.5 x 1000 gm = 1500gm

Question 8.
If \(\frac { a }{ b }\) = \(\frac { 2 }{ 3 }\), then find the values of the following expressions.

Solution:

Question 9.
If a, b, c, d are in proportion, then prove that

Solution:
a, b, c are in continued proportion. …[Given]

Question 10.
If a, b, c are ¡n continued proportion, then prove that

Solution:
a, b, c are in continued proportion. … [Given]
∴ \(\frac { a }{ b }\) = \(\frac { b }{ c }\)
Let \(\frac { a }{ b }\) = \(\frac { b }{ c }\) = k
∴ b = ck
∴ a = bk
= (ck)k . .. [From (j)]
a = ck² …(ii)

Question 11.
Solve:
\( \frac{12 x^{2}+18 x+42}{18 x^{2}+12 x+58}=\frac{2 x+3}{3 x+2}\)
Solution:


∴ 29(2x + 3) = 21 (3x + 2)
∴ 5 + 87= 63x + 42
∴ 87 – 42 = 63x – 58x
∴ 45 = 5x
∴ x = 9
∴ x = 9 is the solution of the given equation.

Question 12.
If \( \frac{2 x-3 y}{3 z+y}=\frac{z-y}{z-x}=\frac{x+3 z}{2 y-3 x}\) ,then prove that every ratio = \(\frac { x }{ y }\).
Solution:

Question 13.
If \(\frac{b y+c z}{b^{2}+c^{2}}=\frac{c z+a x}{c^{2}+a^{2}}=\frac{a x+b y}{a^{2}+b^{2}}\), then prove \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)
Solution:

Question 1.
Take 5 pieces of card paper. Write the following statements, one on each paper.

a, b, c, d are positive numbers and \(\frac{a}{b}=\frac{c}{d}\) is given. Which of the above statements are true or false, write at the back of each card, if false explain why. (Textbook pg. no. 70)
Answer:
i. True
ii. True
iii. False
Here, numerator and denominator are multiplied by two different numbers a and b.
iv. False
Here, different numbers a and b are subtracted from numerator and denominator.
v. True

Question 2.
In the following activity, the values of a and b can be changed. That is by changing a : b we can create many examples. Teachers should give lot of practice to the students and encourage them to construct their own examples. (Textbook pg. no. 70)

Question 3.
Observe the political map of India from a Geography text book. Study the scale of this map.
From the given scale find the straight line distances between various cities like
i. New Delhi to Bengaluru
ii. Mumbai to Kolkata
iii. Jaipur to Bhuvaneshvar. (Textbook pg. no. 77)
[Students should attempt the above activity on their own.]

Class 9 Maths Solution Maharashtra Board 

• Ratio and Proportion Practice Set 4.1 Class 9 Maths Solutions
• Ratio and Proportion Practice Set 4.2 Class 9 Maths Solutions
• Ratio and Proportion Practice Set 4.3 Class 9 Maths Solutions
• Ratio and Proportion Practice Set 4.4 Class 9 Maths Solutions
• Ratio and Proportion Practice Set 4.5 Class 9 Maths Solutions
• Ratio and Proportion Problem Set 4 Class 9 Maths Solutions

10th Standard Maths 1 Practice Set 2.3 Chapter 2 Quadratic Equations Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 2.3 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 2 Quadratic Equations.

Class 10 Maths Part 1 Practice Set 2.3 Chapter 2 Quadratic Equations Questions With Answers Maharashtra Board

Question 1.
Solve the following quadratic equations by completing the square method.
1. x² + x – 20 = 0
2. x² + 2x – 5 = 0
3. m² – 5m = -3
4. 9y² – 12y + 2 = 0
5. 2y² + 9y + 10 = 0
6. 5x² = 4x + 7
Solution:
1. x² + x – 20 = 0
If x² + x + k = (x + a)², then
x² + x + k = x² + 2ax + a²
Comparing the coefficients, we get
1 = 2a and k = a²

∴ The roots of the given quadratic equation are 4 and -5.

2. x² + 2x – 5 = 0
If x² + 2x + k = (x + a)², then
x² + 2x + k = x² + 2ax + a²
Comparing the coefficients, we get
2 = 2a and k = a²
∴ a = 1 and k = (1)² = 1
Now, x² + 2x – 5 = 0
∴ x² + 2x + 1 – 1 – 5 = 0
∴ (x + 1)² – 6 = 0
∴ (x + 1)² = 6
Taking square root of both sides, we get
x + 1 = ± √6
∴ x + 1 √6 or x + 1 = √6
∴ x = √6 – 1 or x = -√6 – 1
∴ The roots of the given quadratic equation are √6 -1 and – √6 -1.

3. m² – 5m = -3
∴ m² – 5m + 3 = 0
If m² – 5m + k = (m + a)², then
m² – 5m + k = m² + 2am + a²
Comparing the coefficients, we get
-5 = 2a and k = a²

4. 9y² – 12y + 2 = 0

5. 2y² + 9y + 10 = 0

Taking square root of both sides, we get

∴ The roots of the given quadratic equation are -2 and \(\frac { -5 }{ 2 } \).

6. 5x² = 4x + 7
∴ 5x² – 4x – 7 = 0

Comparing the coefficients, we get

Maharashtra State Board Class 10 Maths Solutions Part 1

• Linear Equations in Two Variables Practice Set 1.1 Class 10 Maths Solutions
• Linear Equations in Two Variables Practice Set 1.2 Class 10 Maths Solutions
• Linear Equations in Two Variables Practice Set 1.3 Class 10 Maths Solutions
• Linear Equations in Two Variables Practice Set 1.4 Class 10 Maths Solutions
• Linear Equations in Two Variables Practice Set 1.5 Class 10 Maths Solutions
• Linear Equations in Two Variables Problem Set 1 Class 10 Maths Solutions
• Quadratic Equations Practice Set 2.1 Class 10 Maths Solutions
• Quadratic Equations Practice Set 2.2 Class 10 Maths Solutions
• Quadratic Equations Practice Set 2.3 Class 10 Maths Solutions
• Quadratic Equations Practice Set 2.4 Class 10 Maths Solutions
• Quadratic Equations Practice Set 2.5 Class 10 Maths Solutions
• Quadratic Equations Practice Set 2.6 Class 10 Maths Solutions
• Quadratic Equations Problem Set 2 Class 10 Maths Solutions

10th Standard Maths 1 Practice Set 2.6 Chapter 2 Quadratic Equations Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 2.6 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 2 Quadratic Equations.

Class 10 Maths Part 1 Practice Set 2.6 Chapter 2 Quadratic Equations Questions With Answers Maharashtra Board

Question 1.
Product of Pragati’s age 2 years ago and years hence is 84. Find her present age.
Solution:
Let the present age of Pragati be x years.
∴ 2 years ago,
Age of Pragati = (x – 2) years
After 3 years,
Age of Pragati = (x + 3) years
According to the given condition,
(x – 2) (x + 3) = 84
∴ x(x + 3) – 2(x + 3) = 84
∴ x² + 3x – 2x – 6 = 84
∴ x² + x – 6 – 84 = 0
∴ x² + x – 90 = 0
x² + 10x – 9x – 90 = 0
∴ x(x + 10) – 9(x + 10) = 0
∴ (x + 10)(x – 9) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ x + 10 = 0 or x – 9 = 0
∴ x = -10 or x = 9
But, age cannot be negative.
∴ x = 9
∴ Present age of Pragati is 9 years.

Question 2.
The sum of squares of two consecutive even natural numbers is 244; find the numbers.
Solution:
Let the first even natural number be x.
∴ the next consecutive even natural number will be (x + 2).
According to the given condition,
x² + (x + 2)2 = 244
∴ x² + x² + 4x + 4 = 244
∴ 2x² + 4x + 4 – 244 = 0
∴ 2x² + 4x – 240 = 0
∴ x² + 2x – 120 = 0 …[Dividing both sides by 2]
∴ x² + 12x – 10x – 120 = 0
∴ x(x + 12) – 10 (x + 12) = 0
∴ (x + 12) (x – 10) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ x + 12 = 0 or x – 10 = 0
∴ x = -12 or x = 10
But, natural number cannot be negative.
∴ x = 10 and x + 2 = 10 + 2 = 12
∴ The two consecutive even natural numbers are 10 and 12.

Question 3.
In the orange garden of Mr. Madhusudan there are 150 orange trees. The number of trees in each row is 5 more than that in each column. Find the number of trees in each row and each column with the help of following flow chart.

Solution:
i. Number of trees in a column is x.
ii. Number of trees in a row = x + 5
iii. Total number of trees = x x (x + 5)
iv. According to the given condition,
x(x + 5) = 150
∴ x² + 5x = 150
∴ x² + 5x – 150 = 0
v. x² + 15x – 10x – 150 = 0
∴ x(x+ 15) – 10(x + 15) = 0
∴ (x + 15)(x – 10) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ x + 15 = 0 or x – 10 = 0
∴ x = -15 or x = 10
But, number of trees cannot be negative.
∴ x = 10
vi. Number of trees in a column is 10.
vii. Number of trees in a row = x + 5 = 10 + 5 = 15
∴ Number of trees in a row is 15.

Question 4.
Vivek is older than Kishor by 5 years. The Find their present ages is \(\frac { 1 }{ 6 } \) Find their Present ages
Solution:
Let the present age of Kishor be x.
∴ Present age of Vivek = (x + 5) years
According to the given condition,

∴ 6(2x + 5) = x(x + 5)
∴ 12x + 30 = x2 + 5x
∴ x² + 5x – 12x – 30 = 0
∴ x² – 7x – 30 = 0
∴ x² – 10x + 3x – 30 = 0
∴ x(x – 10) + 3(x – 10) = 0
∴ (x – 10)(x + 3) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ x – 10 = 0 or x + 3 = 0
∴ x = 10 or x = – 3
But, age cannot be negative.
∴ x = 10 andx + 5 = 10 + 5 = 15
∴ Present ages of Kishor and Vivek are 10 years and 15 years respectively.

Question 5.
Suyash scored 10 marks more in second test than that in the first. 5 times the score of the second test is the same as square of the score in the first test. Find his score in the first test.
Solution:
Let the score of Suyash in the first test be x.
∴ Score in the second test = x + 10 According to the given condition,
5(x + 10) = x²
∴ 5x + 50 = x²
∴ x² – 5x – 50 = 0
∴ x² – 10x + 5x – 50 = 0
∴ x(x – 10) + 5(x – 10) = 0
∴ (x – 10) (x + 5) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ x – 10 = 0 or x + 5 = 0
∴ x = 10 or x = – 5
But, score cannot be negative.
∴ x = 10
∴ The score of Suyash in the first test is 10.

Question 6.
‘Mr. Kasam runs a small business of making earthen pots. He makes certain number of pots on daily basis. Production cost of each pot is ₹ 40 more than 10 times total number of pots, he makes in one day. If production cost of all pots per day is ₹ 600, find production cost of one pot and number of pots he makes per day.
Solution:
Let Mr. Kasam make x number of pots on daily basis.
Production cost of each pot = ₹ (10x + 40)
According to the given condition,
x(10x + 40) = 600
∴ 10x² + 40x = 600
∴ 10x² + 40x- 600 = 0
∴ x² + 4x – 60 = 0 …[Dividing both sides by 10]
∴ x² + 10x – 6x – 60 = 0
∴ x(x + 10) – 6(x + 10) = 0
∴ (x + 10) (x – 6) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ x + 10 = 0 or x – 6 = 0
∴ x = – 10 or x = 6
But, number of pots cannot be negative.
∴ x = 6
∴ Production cost of each pot = 7(10 x + 40)
= ₹ [(10×6)+ 40]
= ₹(60 + 40) = ₹ 100
Production cost of one pot is ₹ 100 and the number of pots Mr. Kasam makes per day is 6.

Question 7.
Pratik takes 8 hours to travel 36 km downstream and return to the same spot. The speed of boat in still water is 12 km. per hour. Find the speed of water current.
Solution:
Let the speed of water current be x km/hr. Speed of boat is 12 km/hr. (x < 12)
In upstream, speed of the water current decreases the speed of the boat and it is the opposite in downstream.
∴ speed of the boat in upstream = (12 – x) km/hr and speed of the boat in downstream = (12 + x) km/hr.

∴ The speed of water current is 6 km/hr.

Question 8.
Pintu takes 6 days more than those of Nishu to complete certain work. If they work together they finish it in 4 days. How many days would it take to complete the work if they work alone.
Solution:
Let Nishu take x days to complete the work alone.
∴ Total work done by Nishu in 1 day = \(\frac { 1 }{ x } \)
Also, Pintu takes (x + 6) days to complete the work alone.
∴ Total work done by Pintu in 1 day = \(\frac { 1 }{ x+6 } \)
∴ Total work done by both in 1 day = (\(\frac { 1 }{ x } \) + \(\frac { 1 }{ x+6 } \))
But, both take 4 days to complete the work together.
∴ Total work done by both in 1 day = \(\frac { 1 }{ 4 } \)
According to the given condition,

∴ 4(2x + 6) = x(x + 6)
∴ 8x + 24 = x² + 6x
∴ x² + 6x – 8x – 24 = 0
∴ x² – 2x – 24 = 0
∴ x² – 6x + 4x – 24 = 0
∴ x(x – 6)+ 4(x – 6) = 0
∴ (x – 6) (x + 4) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ x – 6 = 0 or x + 4 = 0
∴ x = 6 or x = -4
But, number of days cannot be negative,
∴ x = 6 and x + 6 = 6 + 6 = 12
∴ Number of days taken by Nishu and Pintu to complete the work alone is 6 days and 12 days respectively.

Question 9.
If 460 is divided by a natural number, quotient is 6 more than five times the divisor and remainder is 1. Find quotient and divisor.
Solution:
Let the natural number be x.
∴ Divisor = x, Quotient = 5x + 6
Also, Dividend = 460 and Remainder = 1
Dividend = Divisor × Quotient + Remainder

By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ x – 9 = 0 or 5x + 51 = 0
∴ x = 9 or x = \(\frac { -51 }{ 5 } \)
But, natural number cannot be negative,
∴ x = 9
∴ Quotient = 5x + 6 = 5(9) + 6 = 45 + 6 = 51
∴ Quotient is 51 and Divisor is 9.

Question 10.
In the given fig. []ABCD is a trapezium, AB || CD and its area is 33 cm2. From the information given in the figure find the lengths of all sides of the []ABCD. Fill in the empty boxes to get the solution.

Solution:

By using the property, if the product of two numbers is zero, then at least one of them is zero, we get

Maharashtra State Board Class 10 Maths Solutions Part 1

• Linear Equations in Two Variables Practice Set 1.1 Class 10 Maths Solutions
• Linear Equations in Two Variables Practice Set 1.2 Class 10 Maths Solutions
• Linear Equations in Two Variables Practice Set 1.3 Class 10 Maths Solutions
• Linear Equations in Two Variables Practice Set 1.4 Class 10 Maths Solutions
• Linear Equations in Two Variables Practice Set 1.5 Class 10 Maths Solutions
• Linear Equations in Two Variables Problem Set 1 Class 10 Maths Solutions
• Quadratic Equations Practice Set 2.1 Class 10 Maths Solutions
• Quadratic Equations Practice Set 2.2 Class 10 Maths Solutions
• Quadratic Equations Practice Set 2.3 Class 10 Maths Solutions
• Quadratic Equations Practice Set 2.4 Class 10 Maths Solutions
• Quadratic Equations Practice Set 2.5 Class 10 Maths Solutions
• Quadratic Equations Practice Set 2.6 Class 10 Maths Solutions
• Quadratic Equations Problem Set 2 Class 10 Maths Solutions

9th Standard Maths 1 Practice Set 4.4 Chapter 4 Ratio and Proportion Textbook Answers Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 4.4 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 4 Ratio and Proportion.

Class 9 Maths Part 1 Practice Set 4.4 Chapter 4 Ratio and Proportion Questions With Answers Maharashtra Board

Question 1.
Fill in the blanks of the following.

Solution:

Question 2.
5m – n = 3m + 4n, then find the values of the following expressions.

Solution:
5m – n = 3m + 4n … [Given]
∴ 5m – 3m = 4n + n
∴ 2m = 5n

Question 3.
Solve:
i. If a(y + z) = b(z + x) = c(x + y) and out of a, b, c no two of them are equal, then show that, \( \frac{y-z}{a(b-c)}=\frac{z-x}{b(c-a)}=\frac{x-y}{c(a-b)}\).
Solution:
Here, no two of a, b and c are equal.
∴ values of (b – c), (c – a) and (a – b) are not zero.
a(y + z) = b(z + x) = c(x + y) … [Given]

ii. If \(\frac{x}{3 x-y-z}=\frac{y}{3 y-z-x}=\frac{z}{3 z-x-y}\) and x + y + z ≠ 0, then show that the value of each ratio is equal to 1.
Solution:

iii. \(\frac{x}{3 x-y-z}=\frac{y}{3 y-z-x}=\frac{z}{3 z-x-y}\) and x + y + z ≠ 0,then show that \(\frac { a+b }{ 2 }\).
Solution:

iv. If \(\frac{y+z}{a}=\frac{z+x}{b}=\frac{x+y}{c}\) , then show that \(\frac{x}{b+c-a}=\frac{y}{c+a-b}=\frac{z}{a+b-c}\).
Solution:

v. If \(\frac{3 x-5 y}{5 z+3 y}=\frac{x+5 z}{y-5 x}=\frac{y-z}{x-z}\) , then show that every ratio = \(\frac { x }{ y }\).
Solution:

Ratio And Proportion Class 9 Practice Set 4.4 Question 4.
Solve:

Solution:


∴ 7(4x – 5)3(2x + 3)
∴ 28x – 35 = 6x + 9
∴ 28x – 6x = 9 + 35
∴ 22x = 44
∴ x = 2
∴ x = 2 is the solution of the given equation.

ii. \(\frac{5 y^{2}+40 y-12}{5 y+10 y^{2}-4}=\frac{y+8}{1+2 y}\)


∴ y + 8 = 3(1 + 2y)
∴ y + 8 = 3 + 6y
∴ 8 – 3 = 6y – y
∴ 5 = 5y
∴ y = 1
∴ y = 1 is the solution of the given equation.

Class 9 Maths Solution Maharashtra Board 

• Ratio and Proportion Practice Set 4.1 Class 9 Maths Solutions
• Ratio and Proportion Practice Set 4.2 Class 9 Maths Solutions
• Ratio and Proportion Practice Set 4.3 Class 9 Maths Solutions
• Ratio and Proportion Practice Set 4.4 Class 9 Maths Solutions
• Ratio and Proportion Practice Set 4.5 Class 9 Maths Solutions
• Ratio and Proportion Problem Set 4 Class 9 Maths Solutions