Class 10

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 4.1 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 4 Financial Planning.

Practice Set 4.1 Algebra 10th Std Maths Part 1 Answers Chapter 4 Financial Planning

Financial Planning Class 10 Practice Set 4.1 Question 1.
‘Pawan Medical’ supplies medicines. On some medicines the rate of GST is 12%, then what is the rate of CGST and SGST?
Solution:

Question 2.
On certain article if rate of CGST is 9% then what is the rate of SGST? and what is the rate of GST?
Solution:
Rate of CGST = 9%
But, rate of SGST = rate of CGST
∴ Rate of SGST = 9%
Rate of GST = Rate of SGST + Rate of CGST = 9% + 9%
∴ Rate of GST = 18%

Financial Planning Class 10 Question 3.
‘M/s. Real Paint’ sold 2 tins of lustre paint and taxable value of each tin is ₹ 2800. If the rate of GST is 28%, then find the amount of CGST and SGST charged in the tax invoice.
Solution:
Taxable value of 1 tin = ₹ 2800
∴ Taxable value of 2 tins = 2 × 2800
= ₹ 5600
Rate of GST = 28 %
∴ Rate of CGST = Rate of SGST = 14 %
CGST = 14% of taxable value 14
= \(\frac { 14 }{ 100 } \) × 5600
∴ CGST = ₹ 784
∴ SGST = CGST = ₹ 784
∴ The amount of CGST and SGST charged in the tax invoice is ₹ 784 each.

Question 4.
The taxable value of a wrist watch belt is 7 586. Rate of GST is 18%. Then what is price of the belt for the customer?
Solution:
Taxable value of wrist watch belt = ₹ 586
Rate of GST = 18%
∴ GST = 18% of taxable value
= \(\frac { 18 }{ 100 } \) × 586
∴ GST = ₹ 105.48
∴ Amount paid by customer = Taxable value of wrist watch belt + GST
= 586+ 105.48
= ₹ 691.48
∴ The price of the belt for the customer is ₹ 691.48.

Question 5.
The total value (with GST) of a remote-controlled toy car is ₹ 1770. Rate of GST is 18% on toys. Find the taxable value, CGST and SGST for this toy-car.
Solution:
Let the amount of GST be ₹ x.
Total value of remote controlled toy car = ₹ 1770
∴ Taxable value of remote controlled toy car = ₹ (1770 – x)
Now, GST = 18% of taxable value

∴ Taxable value of toy car is ₹ 1500 and CGST and SGST for it is ₹ 135 each.

Question 6.
‘Tiptop Electronics’ supplied an AC of 1.5 ton to a company. Cost of the AC supplied is ₹ 51,200 (with GST). Rate of CGST on AC is 14%. Then find the following amounts as shown in the tax invoice of Tiptop Electronics.
i. Rate of SGST
ii. Rate of GST on AC
iii. Taxable value of AC
iv. Total amount of GST
v. Amount of CGST
vi. Amount of SGST
Solution:
i. Rate of CGST = 14%
But, Rate of SGST = Rate of CGST
∴ Rate of SGST = 14%

ii. Rate of GST on AC
= Rate of SGST + Rate of CGST
= 14% + 14% = 28%
∴ Rate of GST on AC is 28%.

iii. Let the cost (Taxable value) of AC be ₹ 100.
Given, GST = 28%
∴ The cost of AC with GST is ₹ 128.
For the total value of ₹ 128, the taxable value is ₹ 100.
For the total value of ₹ 51200, let the taxable value be ₹ x

∴ Taxable value of AC is ₹ 40,000.

iv. Total amount of GST = 28% of taxable value
= \(\frac { 28 }{ 100 } \) × 40000
= ₹ 11,200
∴ Total amount of GST is ₹ 11,200.

∴ Amount of CGST is ₹ 5600.

vi. Amount of SGST = Amount of CGST
= ₹ 5600
Amount of SGST is ₹ 5600.

Question 7.
Prasad purchased a washing-machine from ‘Maharashtra Electronic Goods’. The discount of 5% was given on the printed price of ₹ 40,000. Rate of GST charged was 28%. Find the purchase price of washing machine. Also find the amount of CGST and SGST shown in the tax invoice.
Solution:
Printed price of washing machine = ₹ 40,000
Rate of discount = 5%
Amount of discount = 5% of printed price
= \(\frac { 5 }{ 100 } \) × 40000 = ₹ 2000
∴ Taxable value = Printed price – Discount
= 40000 – 2000 = ₹ 38000
Rate of GST = 28%
∴ Rate of CGST = 14% and
Rate of SGST = 14%
CGST = 14% of taxable value
= \(\frac { 14 }{ 100 } \) × 38000
∴ CGST = ₹ 5320
∴ CGST = SGST = ₹ 5320
Purchase price of washing machine
= Taxable value + CGST + SGST
= 38000 + 5320 + 5320
= ₹ 48,640
∴ Purchase price of washing machine is ₹ 48640. Amount of CGST and SGST in tax invoice is ₹ 5320 each.

Question 1.
Observe the given bill and fill in the boxes with the appropriate number. (Textbook pg. no. 82 and 83)

Solution:
i. Price of 1 kg of Pedhe is ₹ 400, therefore cost of 500 gm. of Pedhe is ₹ 200.
CGST for pedhe at the rate of 2.5% is ₹ [5] and SGST at the rate of [2.5| % is ₹ 5.00. It means that the rate of GST on Pedhe is 2.5% + 2.5% = 5% and hence the total GST is ₹ 10.
ii. The rate of GST on chocolate is [28] % and hence the total GST is ₹ [22.40]
iii. Rate of GST on Ice-cream is [18] %, hence the total cost of ice-cream is ₹ 236
iv. On butter CGST rate is [6] % and SGST rate is also [6] %. So GST rate on butter is [12]%.

Question 2.
Fill in the blanks with the help of given information for the table given below. (Textbook pg. no. 83)
Solution:

Question 3.
Make a list of ten things you need in your daily life. Find the GST rates with the help of GST rate chart given in the textbook, news papers or books, internet, or the bills of purchases. Verify these rates with the list prepared by your friends. (Textbook pg. no. 85)
Solution:

Question 4.
Make a list of ten services and their GST rates as per activity 1. (e.g. Railway and ST bus booking services etc.) You can also collect service bills and complete the given information (Textbook pg. no. 85)
Solution:

Question 5.
Complete the given table by writing remaining SAC and HSN codes with rates and add some more items in the list. (Textbook pg, no. 85)
Solution:

[Note : The above Activities has many answers students may write answers other than the ones given]

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 3.3 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 3 Arithmetic Progression.

Practice Set 3.3 Algebra 10th Std Maths Part 1 Answers Chapter 3 Arithmetic Progression

Arithmetic Progression Practice Set 3.3 Question 1.
First term and common difference of an A.P. are 6 and 3 respectively; find S₂₇.
Solution:

Arithmetic Progression Class 10 Practice Set 3.3 Question 2.
Find the sum of first 123 even natural numbers.
Solution:
The even natural numbers are 2, 4, 6, 8,…
The above sequence is an A.P.
∴ a = 2, d = 4 – 2 = 2, n = 123

∴ The sum of first 123 even natural numbers is 15252.

Practice Set 3.3 Question 3.
Find the sum of all even numbers between 1 and 350.
Solution:
The even numbers between 1 and 350 are 2, 4, 6,…, 348.
The above sequence is an A.P.
∴ a = 2, d = 4 – 2 = 2, t_n = 348
Since, t_n = a + (n – 1)d
∴ 348 = 2 + (n – 1)2
∴ 348 – 2 = (n – 1)2
∴ 346 = (n – 1)2
∴ n – 1 = \(\frac { 346 }{ 2 } \)
∴ n – 1 = 173
∴ n = 173 + 1 = 174
Now, S_n = \(\frac { n }{ 2 } \) [2a + (n – 1)d]
∴ S₁₇₄ = \(\frac { 174 }{ 2 } \) [2 (2) + (174 – 1)2]
= 87(4 + 173 × 2)
= 87(4 + 346)
= 87 × 350
∴ S₁₇₄ = 30450
∴ The sum of all even numbers between 1 and 350 is 30450.

Arithmetic Progression 3.3 Question 4.
In an A.P. 19^th term is 52 and 38^th term is 128, find sum of first 56 terms.
Solution:
For an A.P., let a be the first term and d be the common difference.
t₁₉ = 52, t₃₈ = 128 …[Given]
Since, t_n = a + (n – 1)d
∴ t₁₉ = a + (19 – 1)d
∴ 52 = a + 18d
i.e. a + 18d = 52 …(i)
Also, t₃₈ = a + (38 – 1)d
∴ 128 = a + 37d
i.e. a + 37d = 128 …(ii)
Adding equations (i) and (ii), we get

3 Arithmetic Progression Question 5.
Complete the following activity to find the sum of natural numbers between 1 to 140 which are divisible by 4.
Solution:

10th Algebra Practice Set 3.3 Question 6.
Sum of first 55 terms in an A.P. is 3300, find its 28^th term.
Solution:
For an A.P., let a be the first term and d be the common difference.
S₅₅ =3300 …[Given]
Since, Sn = \(\frac { n }{ 2 } \) [2a + (n – 1)d]
∴ S₅₅ = \(\frac { 55 }{ 2 } \) [2a + (55 – 1)d]

Arithmetic Practice Set Question 7.
In an A.P. sum of three consecutive terms is 27 and their product is 504, find the terms. (Assume that three consecutive terms in A.P. are a – d, a, a + d.)
Solution:
Let the three consecutive terms in an A.P. be
a – d, a and a + d.
According to the first condition,
a – d + a + a + d = 27
∴ 3a = 27
∴ a = \(\frac { 27 }{ 3 } \)
∴ a = 9 ….(i)
According to the second condition,
(a – d) a (a + d) = 504
∴ a(a² – d²) = 504
∴ 9(a² – d²) = 504 …[From (i)]
∴ 9(81 – d²) = 504
∴ 81 – d² = \(\frac { 504 }{ 9 } \)
∴ 81 – d² = 56
∴ d² = 81 – 56
∴ d² = 25
Taking square root of both sides, we get
d = ± 5
When d = 5 and a =9,
a – d 9 – 5 = 4
a = 9
a + d 9 + 5 = 14
When d = -5 and a = 9,
a – d = 9 – (-5) = 9 + 5 = 14
a = 9
a + d = 9 – 5 = 4
∴ The three consecutive terms are 4, 9 and 14 or 14, 9 and 4.

10th Maths 1 Practice Set 3.3 Question 8.
Find four consecutive terms in an A.P. whose sum is 12 and sum of 3^rd and 4^th term is 14. (Assume the four consecutive terms in A.P. are a – d, a, a + d, a + 2d.)
Solution:
Let the four consecutive terms in an A.P. be
a – d, a, a + d and a + 2d.
According to the first condition,
a – d + a + a + d + a + 2d = 12
∴ 4a + 2d =12
∴ 2(2a + d) = 12
∴ 2a + d = \(\frac { 12 }{ 2 } \)
∴ 2a + d = 6 …(i)
According to the second condition,
a + d + a + 2d = 14
∴ 2a + 3d = 14 …(ii)
Subtracting equation (i) from (ii), we get

∴ The four consecutive terms are -3,1,5 and 9.

Math 1 Practice Set 3.3 Question 9.
If the 9^th term of an A.P. is zero, then show that the 29^th term is twice the 19th term.
To prove: t₂₉ = 2t₁₉
Proof:
For an A.P., let a be the first term and d be the common difference.
t₉ = 0 …[Given]
Since, tn = a + (n – 1)d
∴ t₉ = a + (9 – 1)d
∴ 0 = a + 8d
∴ a = -8d …(i)
Also, t₁₉ = a + (19 – 1)d
= a + 18d
= -8d + 18d … [From (i)]
∴ t₁₉ = 10d …(ii)
and t₂₉ = a + (29 – 1)d
= a + 28d
= -8d + 28d …[From (i)]
∴ t₂₉ = 20d = 2(10d)
∴ t₂₉ = 2(t₁₉) … [From (ii)]
∴ The 29^th term is twice the 19th term.

10 Class Math Part 1 Practice Set 3.3 Question 1.
Find the sum of all odd numbers from 1 to 150. (Textbook pg, no. 71)
Solution:
Odd numbers from 1 to 150 are 1,3, 5, 7,…, 149
Here, difference between any two consecutive terms is 2.
∴ It is an A.P.
∴ a = 1, d = 2
Let us find how many odd numbers are there from 1 to 150, i.e. find the value of n if
t_n = 149
t_n = a + (n – 1)d
∴ 149 = 1 + (n – 1)2
∴ 149 – 1 = (n – 1)2
∴ \(\frac { 148 }{ 2 } \) = n – 1
∴ 74 = n – 1
∴ n = 74 + 1 = 75

ii. Now, let’s find the sum of 75 numbers
i. e. 1 + 3 + 5 + 7 + … + 149

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

Problem Set 2 Algebra 10th Std Maths Part 1 Answers Chapter 2 Quadratic Equations

Question 1.
Choose the correct answers for the following questions.

i. Which one is the quadratic equation?

Answer:
(B)

ii. Out of the following equations which one is not a quadratic equation?
(A) x² + 4x = 11 + x²
(B) x = 4x
(C) 5x² = 90
(D) 2x – x² = x² + 5
Answer:
(A)

iii. The roots of x² + kx + k = 0 are real and equal, find k.
(A) 0
(B) 4
(C) 0 or 4
(D) 2
Answer:
(C)

iv. For √2 x² – 5x + √2 = 0, find the value of the discriminant.
(A) -5
(B) 17
(C) √2
(D) 2 √2 – 5
Answer:
(B)

v. Which of the following quadratic equations has roots 3,5?
(A) x² – 15x + 8 = 0
(B) x² – 8x + 15 = 0
(C) x² + 3x + 5 = 0
(D) x² + 8x – 15 = 0
Answer:
(B)

vi. Out of the following equations, find the equation having the sum of its roots -5.
(A) 3x² – 15x + 3 = 0
(B) x² – 5x + 3 = 0
(C) x² + 3x – 5 = 0
(D) 3x² + 15x + 3 = 0
Answer:
(D)

vii. √5m² – √5 m + √5 =0 which of the following statement is true for this given equation?
(A) Real and unequal roots
(B) Real and equal roots
(C) Roots are not real
(D) Three roots
Answer:
(C)

viii. One of the roots of equation x² + mx – 5 = 0 is 2; find m.
(A) -2
(B) – \(\frac { 1 }{ 2 } \)
(C) \(\frac { 1 }{ 2 } \)
(D) 2
Answer:
(C)

Question 2.
Which of the following equations is quadratic
i. x² + 2x + 11 = 0
ii. x² – 2x + 5 = x²
iii. (x + 2)² = 2x²
Solution:
i. The given equation is
x² + 2x + 11 = 0
Here, x is the only variable and maximum index of the variable is 2.
a = 1, b = 2, c = 11 are real numbers and
a ≠ 0.
The given equation is a quadratic equation.

ii. The given equation is
x² – 2x + 5 = x²
∴ x² – x² + 2x – 5 = 0
∴ 2x – 5 = 0
Here, x is the only variable and maximum index of the variable is not 2.
∴ The given equation is not a quadratic equation.

iii. The given equation is
(x + 2)² = 2x²
∴ x² + 4x + 4 = 2x²
∴ 2x² – x² – 4x – 4 = 0
∴ x² – 4x – 4 = 0
Here, x is the only variable and maximum index of the variable is 2.
a = 1, b = -4, c = —4 are real numbers and
a ≠ 0.
∴ The given equation is a quadratic equation.

Question 3.
Find the value of discriminant for each of the following equations.
i. 2y² – y + 2 = 0
ii. 5m² – m = 0
iii. √5 x² – x – √5 = 0
Solution:
i. 2y² – y + 2 = 0
Comparing the above equation with
ay² + by + c = 0, we get
a = 2, b = -1, c = 2
∴ b² – 4ac = (-1)² – 4 × 2 × 2
= 1 – 16
∴ b² – 4ac = -15

ii. 5m² – m = 0
∴ 5m² – m + 0 = 0
Comparing the above equation with
am² + bm + c = 0, we get
a = 5, b = -1, c = 0
∴ b² – 4ac = (-1)² – 4 × 5 × 0
= 1 – 0
∴ b² – 4ac = 1

iii. √5x² – x – √5 = 0
Comparing the above equation with
ax² + bx + c = 0, we get
a = √5, b = -1, c = -√5
∴ b² – 4ac = (-1)² – 4 × √5 × √5
= 1 + 20
∴ b² – 4ac = 21

Question 4.
One of the roots of quadratic equation 2x² + kx – 2 = 0 is – 2, find k.
Solution:
-2 is one of the roots of the equation
2x² + kx – 2 = 0.
∴ Putting x = – 2 in the given equation, we get
2(-2)² + k(-2) -2 = 0
∴ 8 – 2k – 2 = 0
∴ 6 – 2k = 0
∴ 2k = 6
∴ k = \(\frac { 6 }{ 2 } \)
∴ k = 3

Question 5.
Two roots of quadratic equations are given; frame the equation.
i. 10 and -10
ii. 1 – 3√5 and 1 + 3√5
iii. 0 and 7
Solution:
i. Let α = 10 and β = -10
∴ α + β = 10 – 10 = 0
and α × p = 10 × -10 = -100
∴ The required quadratic equation is
x² – (α + β)x + αβ = 0
∴ x² – 0x + (-100) = 0
∴ x² – 100 = 0

ii. Let α = 1 – 3 √5 and β = 1 + 3 √5
α + β = 1 – 3 √5 + 1 + 3 √5 = 2
and α × β = (1 – 3√5) (1 + 3 √5)
= (1)2 – (3√5)²
= 1 – 45
= -44
∴ The required quadratic equation is
x² – (α + β)x + αβ = 0
∴ x² – 2x – 44 = 0

iii. Let α = 0 and β = 7
∴ α + β = 0 + 7 = 7
and α × β = 0 × 7 = 0
∴ The required quadratic equation is
x² – (α + β)x + αβ = 0
∴ x² – 7x + 0 = 0
∴ x² – 7x = 0

Question 6.
Determine the nature of roots for each of the quadratic equation.
i. 3x² – 5x + 7 = 0
ii. √3 x² + √2 x – 2 √3 = 0
iii. m² – 2m + 1 = 0
Solution:
i. 3x² – 5x + 7 = 0
Comparing the above equation with
ax² + bx + c = 0, we get
a = 3, b = -5, c = 7
∴ ∆ = b² – 4ac
= (-5)² -4 × 3 × 7
= 25 – 84
∴ ∆ = -59
∴ ∆ < 0
∴ Roots of the given quadratic equation are not real.

ii. √3 x² + √2 x – 2 √3 = 0
Comparing the above equation with
ax² + bx + c = 0, we get
a = √3 , b = √2, c = -2√3
∴ ∆ = b² – 4ac
= (√2)² – 4 × √3 × (-2√3)
= 2 + 24
∴ ∆ = 26
∴ ∆ > 0
∴ Roots of the given quadratic equation are real and unequal.

iii. m² – 2m + 1 = 0
Comparing the above equation with
am² + bm + c = 0, we get
a = 1, b = -2, c = 1
∴ ∆ = b² – 4ac
= (-2)² – 4 × 1 × 1
= 4 – 4
∴ ∆ = 0
∴ Roots of the given quadratic equation are real and equal

Question 7.
Solve the following quadratic equations.

Solution:

ii. x² – \(\frac { 3x }{ 10 } \) – \(\frac { 1 }{ 10 } \) = 0
∴ 10x² – 3x – 1 = 0
…[Multiplying both sides by 10]
∴ 10x² – 5x + 2x – 1 = 0
∴ 5x(2x – 1) + 1(2x – 1) = 0
∴ (2x – 1)(5x + 1) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴2x – 1 = 0 or 5x + 1 = 0
∴2x = 1 or 5x = -1
∴ x = –\(\frac { 1 }{ 2 } \) or x = \(\frac { -1 }{ 5 } \)
∴ The roots of the given quadratic equation are \(\frac { 1 }{ 2 } \) and \(\frac { -1 }{ 5 } \)

iii. (2x + 3)² = 25
∴ (2x + 3)² – 25 = 0
∴ (2x + 3)² – (5)² = 0
∴ (2x + 3 – 5) (2x + 3 + 5) = 0 ….. [∵ a² – b² = (a – b) (a + b)]
∴ (2x – 2) (2x + 8) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ 2x – 2 = 0 or 2x + 8 = 0
∴ 2x = 2 or 2x = -8
∴ x = \(\frac { 2 }{ 2 } \) or x = \(\frac { -8 }{ 2 } \)
∴ x = 1 or x = -4
∴ The roots of the given quadratic equation are 1 and -4.

iv. m² + 5m + 5 = 0
Comparing the above equation with
am² + bm + c = 0, we get
a = 1, b = 5, c = 5
∴ b² – 4ac = (5)² – 4 × 1 × 5
= 25 – 20 = 5

v. 5m² + 2m+1 = 0
Comparing the above equation with
am² + bm + c = 0, we get
a = 5, b = 2, c = 1
∴ b² – 4ac = (2)² -4 × 5 × 1
= 4 – 20
= -16
∴ b² – 4ac < 0
∴ Roots of the given quadratic equation are not real.

vi. x² – 4x – 3 = 0
Comparing the above equation with
ax² + bx + c = 0, we get
a = 1, b = -4, c = -3
∴ b² – 4ac = (-4)² – 4 × 1 × -3
= 16 + 12
= 28

Question 8.
Find m, if (m – 12) x² + 2(m – 12) x + 2 = 0 has real and equal roots.
Solution:
(m – 12) x² + 2(m – 12)x + 2 = 0
Comparing the above equation with
ax² + bx + c = 0, we get
a = m – 12, b = 2(m – 12), c = 2
∴ ∆ = b² – 4ac
= [2(m -12)]² – 4 × (m – 12) × 2
= 4(m – 12)² – 8(m – 12)
= 4(m – 12) (m – 12 – 2)
∴ ∆ = 4(m – 12) (m – 14)
Since, the roots are real and equal.
∴ ∆ = 0
∴ 4(m – 12) (m – 14) = 0 (m – 12) (m – 14) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ m – 12 = 0 or m – 14 = 0
∴ m = 12 or m = 14
But ,if m = 12, then quadratic coefficient becomes zero.
∴ m ≠ 12
∴m = 14

Question 9.
The sum of two roots of a quadratic equation is 5 and sum of their cubes is 35, find the equation.
Solution:
Let α and β be the roots of the quadratic equation.
According to the given conditions,
α + β = 5 and α³ + β³ = 35
Now, (α + β)³ = α³ + 3α²β + 3αβ² + β3
∴ (α + β)³ = α³ + β³ + 3αβ (α + β)
∴ (5)³ = 35 + 3αβ(5)
∴ 125 = 35 + 15αβ
∴ 125 – 35 = 15αβ
∴ 15αβ = 90
∴ αβ = \(\frac { 90 }{ 15 } \)
∴ αβ = 6
∴ The required quadratic equation is
x² – (α + β)x + αβ = 0
∴ x² – 5x + 6 = 0

Question 10.
Find quadratic equation such that its roots are square of sum of the roots and square of difference of the roots of equation
2x² + 2(p + q)x + p² + q² = 0.
Solution:
The given quadratic equation is


According to the given condition,
Roots of the required quadratic equation are

Question 11.
Mukund possesses ₹ 50 more than what Sagar possesses. The product of the amount they have is 15,000. Find the amount each one has.
Solution:
Let the amount Sagar possesses be ₹ x.
∴ the amount Mukund possesses = ₹ (x + 50)
According to the given condition,
x(x +50)= 15000
∴ x² + 50x – 15000 = 0
∴ x² + 150x- 100x- 15000 = 0
∴ x(x + 150) – 100(x + 150) = 0
∴ (x + 150)(x – 100) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ x + 150 = 0 or x – 100 = 0
∴ x = -150 or x = 100
But, amount cannot be negative.
∴ x= 100 and x + 50 = 100 + 50 = 150
∴ The amount possessed by Sagar and Mukund are ₹ 100 and ₹150 respectively.

Question 12.
The difference between squares of two numbers is 120. The square of smaller number is twice the greater number. Find the numbers.
Solution:
Let the numbers be x and y (x > y).
According to the given condition,
x² – y² = 120 …(i)
y² = 2x …(ii)
Substituting y² = 2x in equation (i), we get
x² – 2x = 120
∴ x² – 2x – 120 = 0
∴ 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 x ≠ -10
as, y² = 2x = 2(-10) = -20 …[Since, the square of number cannot be negative]
∴ x = 12
Smaller number = y² =2x
∴ y² = 2 × 12
∴ y² = 24
∴ y = ± √24 …[Taking square root of both sides]
∴ The smaller number is √24 and greater number is 12 or the smaller number is – √24 and greater number is 12.

Question 13.
Ranjana wants to distribute 540 oranges among some students. If 30 students were more each would get 3 oranges less. Find the number of students.
Solution:
Let the number of students be x.
Total number of oranges = 540
∴ the number of oranges each student gets = \(\frac { 540 }{ x } \)
If there were 30 more students, the total number of students = (x + 30) and the total number of oranges each student gets
= (\(\frac { 540 }{ x+30 } \)
According to the given condition,

∴ 30 × 540 = 3x² + 90 x
∴ 3x² + 90x= 16200
∴ x² + 30x – 5400 = 0
…[Dividing both sides by 3]
∴ x² + 90x – 60x – 5400 = 0
∴ x(x + 90) – 60(x + 90) = 0
∴ (x + 90) (x – 60) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ x + 90 = 0 or x – 60 = 0
∴ x = – 90 or x = 60
But, number of students cannot be negative,
x = 60
∴ The total number of students is 60.

Question 14.
Mr. Dinesh owns an rectangular agricultural farm at village Talvel. The length of the farm is 10 metre more than twice the breadth. In order to harvest rain water, he dug a square shaped pond inside the farm. The side of pond is \(\frac { 1 }{ 3 } \) of the breadth of the farm. The
area of the farm is 20 times the area of the pond. Find the length and breadth of the farm and side of the pond.
Solution:
Let the breadth of the rectangular farm be x m.
∴ Length of rectangular farm = (2x + 10) m
Area of rectangular farm = Length × Breadth
= (2x + 10) × x
= (2x²+ 10x) sq. m
Now ,side of square shaped pond = \(\frac { x }{ 3 } \) m
∴ Area of square shaped pond = (side)²
= (\(\frac { x }{ 3 } \))2 m
= \(\frac { { x }^{ 2 } }{ 9 } \) m
According to the given condition,
Area of rectangular farm = 20 × Area of pond

By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ x = 0 or x – 45 = 0
x = 0 or x = 45
But, breadth of the rectangular farm cannot be zero,
∴ x = 45
Length of rectangular farm
= 2x + 10 = 2(45) + 10 = 100 m
Side of the pond = \(\frac { x }{ 3 } \) = \(\frac { 45 }{ 3 } \) = 15 m
∴ Length and breadth of the farm and the side of the pond are 100 m, 45 m and 15 m respectively.

Question 15.
A tank fills completely in 2 hours if both the taps are open. If only one of the taps is open at the given time, the smaller tap takes 3 hours more than the larger one to fill the tank. How much time does each tap take to fill the tank completely?
Solution:
Let the larger tap take x hours to fill the tank completely.
∴ Part of tank filled by the larger tap in 1 hour = \(\frac { 1 }{ x } \)
Also, the smaller tap takes (x + 3) hours to fill the tank completely.
∴ Part of tank filled by the smaller tap in 1 hour = \(\frac { 1 }{ x+3 } \)
∴Part of tank filled by both the taps in 1 hour
= (\(\frac { 1 }{ x } \) + \(\frac { 1 }{ x+3 } \))
But, the tank gets filled in 2 hours by both the taps.
∴ Part of tank filled by both the taps in 1 hour = \(\frac { 1 }{ 2 } \)
According to the given condition,

∴ 2(2x + 3) = x(x + 3)
∴ 4x + 6 = x² + 3x
∴ x² + 3x – 4x – 6 = 0
∴ x² – x – 6 = 0
∴ x² – 3x + 2x – 6 = 0
∴ x(x – 3) + 2(x – 3) = 0
∴ (x – 3)(x + 2) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ x – 3 = 0 or x + 2 = 0
∴ x = 3 or x = -2
But, time cannot be negative.
∴ x = 3 and x + 3 = 3 + 3 = 6
∴ The larger tap takes 3 hours and the smaller tap takes 6 hours to fill the tank completely.

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 3.1 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 3 Arithmetic Progression.

Practice Set 3.1 Algebra 10th Std Maths Part 1 Answers Chapter 3 Arithmetic Progression

Question 1.
Which of the following sequences are A.P.? If they are A.P. find the common difference.

Solution:
i. The given sequence is 2, 4, 6, 8,…
Here, t₁ = 2, t₂ = 4, t₃ = 6, t₄ = 8
∴ t₂ – t₁ = 4 – 2 = 2
t₃ – t₂ = 6 – 4 = 2
t₄ – t₃ = 8 – 6 = 2
∴ t₂ – t₁ =  t₃ – t₂ = … = 2 = d = constant
The difference between two consecutive terms is constant.
∴ The given sequence is an A.P. and common difference (d) = 2.

The difference between two consecutive terms is constant.
∴ The given sequence is an A.P. and common difference (d) = \(\frac { 1 }{ 2 } \).

iii. The given sequence is -10, -6, -2, 2,…
Here, t₁ = -10, t₂ = – 6, t₃ = -2, t₄ = 2
∴ t₂ – t₁ = -6 – (-10) = -6 + 10 = 4
t₃ – t₂ = -2 -(-6) = -2 + 6 = 4
t₄ – t₃ = 2 – (-2) = 2 + 2 = 4
∴ t₂ – t₁ = t₃ – t₂ = … = 4 = d = constant
The difference between two consecutive terms is constant.
∴ The given sequence is an A.P. and common difference (d) = 4.

iv. The given sequence is 0.3, 0.33, 0.333,…
Here, t₁ = 0.3, t₂ = 0.33, t₃ = 0.333
∴ t₂ -t₁ = 0.33 – 0.3 = 0.03
t₃ – t₂ = 0.333 – 0.33 = 0.003
∴ t₂ – t₁ ≠ t₃ – t₂
The difference between two consecutive terms is not constant.
∴ The given sequence is not an A.P.

v. The given sequence is 0, -4, -8, -12,…
Here, t₁ = 0, t₂ = -4, t₃ = -8, t₄ = -12
∴ t₂ – t₁ = -4 – 0 = -4
t₃ – t₂ = -8 – (-4) = -8 + 4 = -4
t₄ – t₃ = -12 – (-8) = -12 + 8 = -4
∴ t₂ – t₁ = t₃ – t₂ = … = —4 = d = constant
The difference between two consecutive terms is constant.
∴ The given sequence is an A.P. and common difference (d) = -4.

The difference between two consecutive terms is constant.
∴ The given sequence is an A.P. and common difference (d) = 0.

The difference between two consecutive terms is constant.
∴ The given sequence is an A.P. and common difference (d) = √2.

viii. The given sequence is 127, 132, 137,…
Here, t₁ = 127, t₂ = 132, t₃ = 137
∴ t₂ – t₁ = 132 – 127 = 5
t₃ – t₂ = 137 – 132 = 5
∴ t₂ – t₁ = t₃ – t₂ = … = 5 = d = constant
The difference between two consecutive terms is constant.
∴ The given sequence is an A.P. and common difference (d) = 5.

Question 2.
Write an A.P. whose first term is a and common difference is d in each of the following.
i. a = 10, d = 5
ii. a = -3, d = 0
iii. a = -7, d = \(\frac { 1 }{ 2 } \)
iv. a = -1.25, d = 3
v. a = 6, d = -3
vi. a = -19, d = -4
Solution:
i. a = 10, d = 5 …[Given]
∴ t₁ = a = 10
t₂ = t₁ + d = 10 + 5 = 15
t₃ = t₂ + d = 15 + 5 = 20
t₄ = t₃ + d = 20 + 5 = 25
∴ The required A.P. is 10,15, 20, 25,…

ii. a = -3, d = 0 …[Given]
∴ t₁ = a = -3
t₂ = t₁ + d = -3 + 0 = -3
t₃ = t₂ + d = -3 + 0 = -3
t₄ = t₃ + d = -3 + 0 = -3
∴ The required A.P. is -3, -3, -3, -3,…

∴ The required A.P. is -7, – 6.5, – 6, – 5.5,

iv. a = -1.25, d = 3 …[Given]
t₁ = a = -1.25
t₂ = t₁ + d = – 1.25 + 3 = 1.75
t₃ = t₂ + d = 1.75 + 3 = .4.75
t₄ = t₃ + d = 4.75 + 3 = 7.75
∴ The required A.P. is -1.25, 1.75, 4.75, 7.75,…

v. a = 6, d = -3 …[Given]
∴ t₁ = a = 6
t₂ = t₁ + d = 6 – 3 = 3
t₃ = t₂ + d = 3 – 3 = 0
t₄ = t₃ + d = 0- 3 = -3
∴ The required A.P. is 6, 3, 0, -3,…

vi. a = -19, d = -4 …[Given]
t₁ = a = -19
t₂ = t₁ + d = -19 – 4 = -23
t₃ = t₂ + d = -23 – 4 = -27
t₄ = t₃ + d = -27 – 4 = -31
∴ The required A.P. is -19, -23, -27, -31,…

Question 3.
Find the first term and common difference for each of the A.P.

Solution:
i. The given A.P. is 5, 1,-3,-7,…
Here, t₁ = 5, t₂ = 1
∴ a = t₁ = 5 and
d = t₂ – t₁ = 1 – 5 = -4
∴ first term (a) = 5,
common difference (d) = -4

ii. The given A.P. is 0.6, 0.9, 1.2, 1.5,…
Here, t₁ = 0.6, t₂ = 0.9
∴ a = t₁ = 0.6 and
d = t₂ – t₁ = 0.9 – 0.6 = 0.3
∴ first term (a) = 0.6,
common difference (d) = 0.3

iii. The given A.P. is 127, 135, 143, 151,…
Here, t₁ = 127, t₂ = 135
∴ a = t₁ = 127 and
d = t₂ – t₁ = 135 – 127 = 8
∴ first term (a) = 127,
common difference (d) = 8

Question 1.
Complete the given pattern. Look at the pattern of the numbers. Try to find a rule to obtain the next number from its preceding number. Write the next numbers. (Textbook pg, no. 55 and 56)
Answer:

Every pattern is formed by adding a circle in horizontal and vertical rows to the preceding pattern.
∴ The sequence for the above pattern is 1,3, 5, 7, 9,11,13,15,17,….

Every pattern is formed by adding 2 triangles horizontally and 1 triangle vertically to the preceding pattern.
∴ The sequence for the above pattern is 5,8,11,14,17,20,23,…

Question 2.
Some sequences are given below. Show the positions of the terms by t₁, t₂, t₃,…
Answer:

Question 3.
Some sequences are given below. Check whether there is any rule among the terms. Find the similarity between two sequences. To check the rule for the terms of the sequence look at the arrangements and fill the empty boxes suitably. (Textbook pg. no. 56 and 57)
i. .1,4,7,10,13,…
ii. 6,12,18,24,…
iii. 3,3,3,3,…
iv. 4, 16, 64,…
v. -1, -1.5, -2, -2.5,…
vi. 1³, 2³, 3³, 4³
Answer:


The similarity in the sequences i., ii., iii. and v. is that the next term is obtained by adding a particular fixed number to the previous term.

Note : A Geometric Progression is a sequence in which the ratio of any two consecutive terms is a constant,

Sequence iv. is a geometric progression.

Question 4.
Write one example of finite and infinite A.P. each. (Textbook pg. no. 59)
Answer:
Finite A.P.:
Even natural numbers from 4 to 50:
4, 6, 8, ………………. 50.
Infinite A. P.:
Positive multiples of 5:
5, 10, 15, ……………..

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 2.5 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 2 Quadratic Equations.

Practice Set 2.5 Algebra 10th Std Maths Part 1 Answers Chapter 2 Quadratic Equations

Question 1.
Fill in the gaps and complete.

Answer:

Question 2.
Find the value of discriminant.
i. x² + 7x – 1 = 0
ii. 2y² – 5y + 10 = 0
iii. √2 x² + 4x + 2√2 = 0
Solution:
i. x² +7 x – 1 = 0
Comparing the above equation with
ax² + bx + c = 0, we get
a = 1, b = 7, c = -1
∴ b²– 4ac = (7)² – 4 × 1 × (-1)
= 49 + 4
∴ b² – 4ac = 53

ii. 2y² – 5y + 10 = 0
Comparing the above equation with
ay² + by + c = 0, we get
a = 2, b = -5, c = 10
∴ b² – 4ac = (-5)2 -4 × 2 × 10
= 25 – 80
∴ b² – 4ac = -55

iii. √2 x² + 4x + 2√2 = 0
Comparing the above equation with
ax + bx + c = 0, we get
a = √2,b = 4, c = 2√2
∴ b² – 4ac = (4)2 – 4 × √2 × 2√2
= 16 – 16
∴ b² – 4ac =0

Question 3.
Determine the nature of roots of the following quadratic equations.
i. x² – 4x + 4 = 0
ii. 2y² – 7y + 2 = 0
iii. m² + 2m + 9 = 0
Solution:
i. x² – 4x + 4= 0
Comparing the above equation with
ax² + bx + c = 0, we get
a = 1,b = -4, c = 4
∴ ∆ = b² – 4ac
= (- 4)² – 4 × 1 × 4
= 16 – 16
∴ ∆ = 0
∴ Roots of the given quadratic equation are real and equal.

ii. 2y² – 7y + 2 = 0
Comparing the above equation with
ay² + by + c = 0, we get
a = 2, b = -7, c = 2
∴ ∆ = b² – 4ac
= (- 7)² – 4 × 2 × 2
= 49 – 16
∴ ∆ = 33
∴ ∆ > 0
∴ Roots of the given quadratic equation are real and unequal.

iii. m² + 2m + 9 = 0
Comparing the above equation with
am² + bm + c = 0, we get
a = 1,b = 2, c = 9
∴ ∆ = b² – 4ac
= (2)² – 4 × 1 × 9
= 4 – 36
∴ ∆ = -32
∴ ∆ < 0
∴ Roots of the given quadratic equation are not real.

Question 4.
Form the quadratic equation from the roots given below.
i. 0 and 4
ii. 3 and -10
iii. \(\frac { 1 }{ 2 } \) , \(\frac { 1 }{ 2 } \)
iv. 2 – √5, 2 + √5
Solution:
i. Let a = 0 and β = 4
∴ α + β = 0 + 4 = 4
and α × β = 0 × 4 = 0
∴ The required quadratic equation is
x² – (α + β) x + αβ = 0
∴ x² – 4x + 0 = 0
∴ x² – 4x = 0

ii. Let α = 3 and β = -10
∴ α + β = 3 – 10 = -7
and α × β = 3 × -10 = -30
∴ The required quadratic equation is
x² – (α + β)x + αβ = 0
∴ x² – (-7) x + (-30) = 0

∴ The required quadratic equation is
x² – (α + β)x + αβ = 0
∴ x² – 4x – 1 = 0

Question 5.
Sum of the roots of a quadratic equation is double their product. Find k if equation is x² – 4kx + k + 3 = 0.
Solution:
x² – 4kx + k + 3 = 0
Comparing the above equation with
ax² + bx + c = 0, we get
a = 1, b = – 4k, c = k + 3
Let α and β be the roots of the given quadratic equation.
Then, α + β  = \(\frac { -b }{ a } \) and αβ = \(\frac { c }{ a } \)
According to the given condition,

Question 6.
α, β are roots of y² – 2y – 7 = 0 find,
i. α² + β²
ii. α³ + β³
Solution:
y² – 2y – 7 = 0
Comparing the above equation with
ay² + by + c = 0, we get
a = 1, b = -2, c = -7

Question 7.
The roots of each of the following quadratic equations are real and equal, find k.
i. 3y² + ky + 12 = 0
ii. kx (x-2) + 6 = 0
Solution:
i. 3y² + kg + 12 = 0
Comparing the above equation with
ay² + by + c = 0, we get
a = 3, b = k, c = 12
∴ ∆ = b² – 4ac
= (k)² – 4 × 3 × 12
= k² – 144 = k² – (12)²
∴ ∆ = (k + 12) (k – 12) …[∵ a² – b² = (a + b) (a – b)]
Since, the roots are real and equal.
∴ ∆ = 0
∴ (k + 12) (k – 12) = 0
∴ k + 12 = 0 or k – 12 = 0
∴ k = -12 or k = 12

ii. kx (x – 2) + 6 = 0
∴ kx² – 2kx + 6 = 0
Comparing the above equation with
ax² + bx + c = 0, we get
a = k, b = -2k, c = 6
∴ ∆ = b² – 4ac
= (-2k)² – 4 × k × 6
= 4k² – 24k
∴ ∆ = 4k (k – 6)
Since, the roots are real and equal.
∴ ∆ = 0
∴ 4k (k – 6) = 0
∴ k(k – 6) = 0
∴ k = 0 or k – 6 = 0
But, if k = 0 then quadratic coefficient becomes zero.
∴ k ≠ 0
∴ k = 6

Question 1.
Fill in the blanks. (Textbook pg. no. 44)
Solution:

Question 2.
Determine nature of roots of the quadratic equation: x² + 2x – 9 = 0 (Textbook pg. no. 45)
Solution:

∴ The roots of the given equation are real and unequal.

Question 3.
Fill in the empty boxes properly. (Textbook pg. no. 46)
Solution:
10x² + 10x + 1 = 0
Comparing the above equation with
ax² + bx + c = 0, we get
a = 10, b = 10, c = 1

Question 4.
Write the quadratic equation if addition of the roots is 10 and product of the roots is 9. (Textbook pg, no. 48)
Answer:

Question 5.
What will be the quadratic equation if α = 2, β = 5. (Textbook pg. no, 48)
Solution:

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 2.2 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 2 Quadratic Equations.

Practice Set 2.2 Algebra 10th Std Maths Part 1 Answers Chapter 2 Quadratic Equations

Question 1.
Solve the following quadratic equations by factorisation.
i. x² – 15x + 54 = 0
ii. x² + x – 20 = 0
iii. 2y² + 27y + 13 = 0
iv. 5m² = 22m + 15
v. 2x² – 2x + \(\frac { 1 }{ 2 } \) = 0
vi. 6x – \(\frac { 2 }{ x } \) = 1
vii. √2x² + 7x + 5√2 = 0 to solve this quadratic equation by factorisation complete the following activity
viii. 3x² – 2√6x + 2 = 0
ix. 2m(m – 24) = 50
x. 252 = 9
xi. 7m² = 21 m
xii. m² – 11 = 0
Solution:

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 x – 6 = 0
∴ x = 9 or x = 6
∴ The roots of the given quadratic equation are 9 and 6.

By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ x + 5 = 0 or x – 4 = 0
∴ x = -5 or x = 4
∴ The roots of the given quadratic equation are -5 and 4.

By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ y + 13 = 0 or 2y + 1 = 0
∴ y = – 13 or 2y = -1
∴ y = -13 or y = –\(\frac { 1 }{ 2 } \)
∴ The roots of the given quadratic equation are -13 and – \(\frac { 1 }{ 2 } \)

By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ m – 5 = 0 or 5m + 3 = 0
∴ m = 5 or 5m = -3
∴ m = 5 or m = \(\frac { -3 }{ 5 } \)
∴ The roots of the given quadratic equation are 5 and – \(\frac { 3 }{ 5 } \)

By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ 3x – 2 = 0 or 2x + 1 = 0
∴ 3x = 2 or 2x = -1
∴ x = \(\frac { 2 }{ 3 } \) or 2x = -1
∴ The roots of the given quadratic equation are \(\frac { 2 }{ 3 } \) and \(\frac { -1 }{ 2 } \).

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

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

ix. 2m (m – 24) = 50
∴ 2m² – 48m = 50
∴ 2m² – 48m – 50 = 0
∴m² – 24m – 25 = 0 …[Dividing both sides by 2]

∴ m – 25 = 0 or m + 1 = 0
∴ m = 25 or m = -1
∴ The roots of thes given quadratic equation are 25 and -1.

x. 25m² = 9
∴ 25m² – 9 = 0
∴ (5m)² – (3)² = 0
∴ (5m + 3) (5m – 3) = 0
…. [∵a² – b² = (a + b) (a – b)]
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ 5m + 3 = 0 or 5m – 3 = 0
∴ 5m = -3 or 5m = 3
∴ m = \(\frac { -3 }{ 5 } \) or m = \(\frac { 3 }{ 5 } \)
∴ The roots of the given quadratic equation are \(\frac { -3 }{ 5 } \) and \(\frac { 3 }{ 5 } \).

xi. 7m² = 21m
∴ 7m² – 21m = 0
∴ m² – 3m = 0 …[Dividing both sides by 7]
∴ m(m – 3) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ m = 0 or m – 3 = 0
∴ m = 0 or m = 3
∴ The roots of the given quadratic equation are 0 and 3.

By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ m + √11 = 0 or m – √11 = 0
∴ m = -√11 or m = √11
∴ The roots of the given quadratic equation are – √11 and √11

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 2.4 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 2 Quadratic Equations.

Practice Set 2.4 Algebra 10th Std Maths Part 1 Answers Chapter 2 Quadratic Equations

Question 1.
Compare the given quadratic equations to the general form and write values of a, b, c.
i. x² – 7x + 5 = 0
ii. 2m² = 5m – 5
iii. y² = 7y
Solution:
i. x² – 7x + 5 = 0
Comparing the above equation with
ax² + bx + c = 0, we get
a = 1, b = -7, c = 5

ii. 2m² = 5m – 5
∴ 2m² – 5m + 5 = 0
Comparing the above equation with
am² + bm + c = 0, we get
a = 2, b = -5, c = 5

iii. y² = 7y
∴ y² – 7y + 0 = 0
Comparing the above equation with
ay² + by + c = 0, we get
a = 1, b = -7, c = 0

Question 2.
Solve using formula.
i. x² + 6x + 5 = 0
ii. x² – 3x – 2 = 0
iii. 3m² + 2m – 7 = 0
iv. 5m² – 4m – 2 = 0
v. y² + \(\frac { 1 }{ 3 } \) y = 2
vi. 5x² + 13x + 8 = 0
Solution:
i. x² + 6x + 5 = 0
Comparing the above equation with
ax² + bx + c = 0, we get
a = 1, b = 6, c = 5
∴ b² – 4ac = (6)2 – 4 × 1 × 5
= 36 – 20 = 16

∴ x = -3 + 2 or x = -3 -2
∴ x = -1 or x = -5
∴ The roots of the given quadratic equation are -1 and -5.

ii. x² – 3x – 2 = 0
Comparing the above equation with
ax² + bx + c = 0, we get
a = 1, b = -3, c = -2
∴ b² – 4ac = (-3)2 – 4 × 1 × (-2)
= 9 + 8 = 17

iii. 3m² + 2m – 7 = 0
Comparing the above equation with
am² + bm + c = 0, we get
a = 3, b = 2, c = -7
∴ b² – 4ac = (2)² – 4 × 3 × ( -7)
= 4 + 84 = 88

iv. 5m² – 4m – 2 = 0
Comparing the above equation with
am² + bm + c = 0, we get
a = 5, b = -4, c = -2
∴ b² – 4ac = (-4)² – 4 × 5 × (-2)
= 16 + 40 = 56

v. y² + \(\frac { 1 }{ 3 } \)y = 2
∴ 3y² + y = 6 …(Multiplying both sides by 3]
∴ 3y² + y – 6 = 0
Comparing the above equation with
ay² + by + c = 0, we get
a = 3, b = 1, c = -6
∴ b² – 4ac = (1)² – 4 × 3 × (-6)
= 1 + 72 = 73

vi. 5x² + 13x + 8 = 0
Comparing the above equation with
ax² + bx + c = 0, we get
a = 5, b = 13, c = 8
∴ b² – 4ac = (13)2 – 4 × 5 × 8
= 169 – 160 = 9


The roots of the given quadratic equation are -1 and \(\frac { -8 }{ 5 } \).

Question 3.
With the help of the flow chart given below solve the equation x² + 2√3 x + 3 = 0 using the formula.

Solution:
i. x² + 2√3 x + 3 = 0
Comparing the above equation with
ax² + bx + c = 0, we get
a = 1, b = 2√3 ,c = 3

ii. b² – 4ac = (2√3)2 -4 × 1 × 3
= 12 – 12
= 0

Question 1.
Solve the equation 2x² + 13x + 15 = 0 by factorisation method, by completing the square method and by using the formula. Verify that you will get the same roots every time. (Textbook pg. no. 43)
Solution:

By using the property, if the product of two numbers is zero, then at least zero, we get
∴ x + 5 = 0 or 2x + 3 = 0
∴ x + -5 = 0 or 2x = -3 = 0
∴ x + -5 = or x = \(\frac { -3 }{ 2 } \)
∴ The roots of the given quadratic equation are \(\frac { -3 }{ 2 } \) and -5.

ii. Completing the square method:
2x² + 13x + 15 = 0


∴ The roots of the given quadratic equation are \(\frac { -3 }{ 2 } \) and -5.

iii. Formula method:
2x² + 13x + 15 = 0
Comparing the above equation with
ax² + bx + c = 0, we get
a = 2, b = 13, c = 15
∴ b² – 4ac = (13)2 – 4 × 2 × 15
= 169 – 120 = 49

∴ The roots of the given quadratic equation are \(\frac { -3 }{ 2 } \) and -5.
∴ By all the above three methods, we get the same roots of the given quadratic equation.

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.

Practice Set 2.3 Algebra 10th Std Maths Part 1 Answers Chapter 2 Quadratic Equations

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

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.

Practice Set 2.6 Algebra 10th Std Maths Part 1 Answers Chapter 2 Quadratic Equations

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

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 3.2 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 3 Arithmetic Progression.

Practice Set 3.2 Algebra 10th Std Maths Part 1 Answers Chapter 3 Arithmetic Progression

Question 1.
Write the correct number in the given boxes from the following A.P.

Solution:

Question 2.
Decide whether following sequence is an A.P., if so find the 20^th term of the progression.
-12, -5, 2, 9,16, 23,30,…
Solution:
i. The given sequence is
-12, -5,2, 9, 16, 23,30,…
Here, t₁ = -1₂, t₂ = -5, t₃ = 2, t₄ = 9
∴ t₂ – t₁ – 5 – (-12) – 5 + 12 = 7
t₃ – t₂ = 2 – (-5) = 2 + 5 = 7
∴ t₄ – t₃ – 9 – 2 = 7
∴ t₂ – t₁ = t₃ – t₂ = … = 7 = d = constant
The difference between two consecutive terms is constant.
∴ The given sequence is an A.P.

ii. t_n = a + (n – 1)d
∴ t₂₀ = -12 + (20 – 1)7 …[∵a = -12, d = 7]
= -12 + 19 × 7
= -12 + 133
∴ t₂₀ = 121
∴ 20^th term of the given A.P. is 121.

Question 3.
Given Arithmetic Progression is 12, 16, 20, 24, … Find the 24^th term of this progression.
Solution:
The given A.P. is 12, 16, 20, 24,…
Here, a = 12, d = 16 – 12 = 4 Since,
t_n = a + (n – 1)d
∴ t₂₄ = 12 + (24 – 1)4
= 12 + 23 × 4
= 12 + 92
∴ t₂₄ = 104
∴ 24th term of the given A.P. is 104.

Question 4.
Find the 19th term of the following A.P. 7,13,19,25…..
Solution:
The given A.P. is 7, 13, 19, 25,…
Here, a = 7, d = 13 – 7 = 6
Since, t_n = a + (n – 1)d
∴ t₁₉ = 7 + (19 – 1)6
= 7 + 18 × 6
= 7 + 108
∴ t₁₉ = 115
∴ 19^th term of the given A.P. is 115.

Question 5.
Find the 27^th term of the following A.P. 9,4,-1,-6,-11,…
Solution:
The given A.P. is 9, 4, -1, -6, -11,…
Here, a = 9, d = 4- 9 = -5
Since, t_n = a + (n – 1)d
∴ t₂₇ = 9 + (27 – 1)(-5)
= 9 + 26 × (-5)
= 9 – 130
∴ t₂₇ = -121
∴ 27^th term of the given A.P. is -121.

Question 6.
Find how many three digit natural numbers are divisible by 5.
Solution:
The three digit natural numbers divisible by
5 are 100, 105, 110, …,995
The above sequence is an A.P.
∴ a = 100, d = 105 – 100 = 5
Let the number of terms in the A.P. be n.
Then, t_n = 995
Since, t_n = a + (n – 1)d
∴ 995 = 100 +(n – 1)5
∴ 995 – 100 = (n – 1)5
∴ 895 = (n – 1)5
∴ n – 1 = \(\frac { 895 }{ 5 } \)
∴ n – 1 = 179
∴ n = 179 + 1 = 180
∴ There are 180 three digit natural numbers which are divisible by 5.

Question 7.
The 11th term and the 21st term of an A.P. are 16 and 29 respectively, then find the 41st term of that A.P.
Solution:
Bor an A.P., let a be the first term and d be the common difference,
t₁₁ = 16, t₂₁ = 29 …[Given]
t_n = a + (n – 1)d
∴ t₁₁, = a + (11 – 1)d
∴ 16 = a + 10d
i.e. a + 10d = 16 …(i)
Also, t₂₁ = a + (21 – 1)d
∴ 29 = a + 20d
i.e. a + 20d = 29 …(ii)
Subtracting equation (i) from (ii), we get a

Question 8.
8. 11, 8, 5, 2, … In this A.P. which term is number-151?
Solution:
The given A.P. is 11, 8, 5, 2,…
Here, a = 11, d = 8 – 11 = -3
Let the n^th term of the given A.P. be -151.
Then, t_n = – 151
Since, t_n = a + (n – 1)d
∴ -151= 11 + (n – 1)(-3)
∴ -151 – 11 =(n – 1)(-3)
∴ -162 = (n – 1)(-3)
∴ n – 1 = \(\frac { -162 }{ -3 } \)
∴ n – 1 = 54
∴ n = 54 + 1 = 55
∴ 55th term of the given A.P. is -151.

Question 9.
In the natural numbers from 10 to 250, how many are divisible by 4?
Solution:
The natural numbers from 10 to 250 divisible
by 4 are 12, 16, 20, …,248
The above sequence is an A.P.
∴ a = 12, d = 16 – 12 = 4
Let the number of terms in the A.P. be n.
Then, t_n = 248
Since, t_n = a + (n – 1)d
∴ 248 = 12 + (n – 1)4
∴ 248 – 12 = (n – 1)4
∴ 236 = (n – 1)4
∴ n – 1 = \(\frac { 236 }{ 4 } \)
∴ n – 1 = 59
∴ n = 59 + 1 = 60
∴ There are 60 natural numbers from 10 to 250 which are divisible by 4.

Question 10.
In an A.P. 17^th term is 7 more than its 10th term. Find the common difference.
Solution:
For an A.P., let a be the first term and d be the common difference.
According to the given condition,
t₁₇ = t₁₀ + 7
∴ a + (17 – 1)d = a + (10 – 1)d + 7 …[∵ t_n = a + (n – 1)d]
∴ a + 16d = a + 9d + 7
∴ a + 16d – a – 9d = 7
∴ 7d = 7
∴ d = \(\frac { 7 }{ 7 } \) = 1
∴ The common difference is 1.

Question 1.
Kabir’s mother keeps a record of his height on each birthday. When he was one year old, his height was 70 cm, at 2 years he was 80 cm tall and 3 years he was 90 cm tall. His aunt Meera was studying in the 10th class. She said, “it seems like Kabir’s height grows in Arithmetic Progression”. Assuming this, she calculated how tall Kabir will be at the age of 15 years when he is in 10th! She was shocked to find it. You too assume that Kabir grows in A.P. and find out his height at the age of 15 years. (Textbook pg. no. 63)
Solution:
Height of Kabir when he was 1 year old = 70 cm Height of Kabir when he was 2 years old = 80 cm
Height of Kabir when he was 3 years old = 90 cm The heights of Kabir form an A.P.
Here, a = 70, d = 80 – 70 = 10
We have to find height of Kabir at the age of 15years i.e. t₁₅.
Now, tn = a + (n – 1)d
∴ t₁₅ = 70 + (15 – 1)10
= 70 + 14 × 10 = 70 + 140
∴ t₁₅ = 210
∴ The height of Kabir at the age of 15 years will be 210 cm.

Question 2.
Is 5, 8, 11, 14, …. an A.P.? If so then what will be the 100^th term? Check whether 92 is in this A.P.? Is number 61 in this A.P.? (Textbook pg. no, 62)
Solution:
i. The given sequence is
5, 8,11,14,…
Here, t₁ = 5, t₂ = 8, t₃ = 11, t₄ = 14
∴ t₂ – t₁ = 8 – 5 = 3
t₃ – t₂ = 11 – 8 = 3
t₄ – t₃ = 14 – 11 = 3
∴ t₂ – t₁ = t₃ – t₂ = t₄ – t₃ = 3 = d = constant
The difference between two consecutive terms is constant
∴ The given sequence is an A.P.

ii. t_n = a + (n – 1)d
∴ t₁₀₀ = 5 + (100 – 1)3 …[∵ a = 5, d = 3]
= 5 + 99 × 3
= 5 + 297
∴ t₁₀₀ = 302
∴ 100^th term of the given A.P. is 302.

iii. To check whether 92 is in given A.P., let t_n = 92
∴ t_n = a + (n – 1)d
∴ 92 = 5 + (n – 1)3
∴ 92 = 5 + 3n – 3
∴ 92 = 2 + 3n
∴ 90 = 3n
∴ n = \(\frac { 90 }{ 3 } \) = 30
∴ 92 is the 30th term of given A.P.

iv. To check whether 61 is in given A.P., let t_n = 61
61 = 5 + (n – 1)3
∴ 61 = 5 + 3n – 3
∴ 61 = 2 + 3n
∴ 61 – 2 = 3n
∴ 59 = 3n
∴ n = \(\frac { 59 }{ 3 } \)
But, n is natural number 59
∴ n ≠ \(\frac { 59 }{ 3 } \)
∴ 61 is not in given A.P.