Tag: Maths Class 10 RS Aggarwal Solutions

RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1A

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Ex 1A.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Ex 1A
• RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Ex 1B
• RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Ex 1C
• RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Ex 1D
• RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Ex 1E
• RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers MCQs
• RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Test Yourself

Question 1.
Solution:
For any two given positive integers a and b, there exist unique whole numbers q and r such that
a = bq + r, when 0 ≤ r < b.
Here, a is called dividend, b as divisor, q as quotient and r as remainder.
Dividend = (Divisor x Quotient) + Remainder.

Question 2.
Solution:
Using Euclid’s divison Lemma
Dividend = (Divisor x Quotient) + Remainder
= (61 x 27) + 32
= 1647 + 32
= 1679
Required number = 1679

Question 3.
Solution:
Let the required divisor = x
Then by Euclid’s division Lemma,
Dividend = (Divisor x Quotient) + remainder
1365 = x x 31 + 32
=> 1365 = 31x + 32
=> 31x= 1365 – 32 = 1333
x = \(\frac { 1331 }{ 31 }\) = 43
Divisor = 43

Question 4.
Solution:
(i) 405 and 2520
HCF of 405 and 2520 = 45

Question 5.
Solution:
Let n be an arbitrary positive integer.
On dividing n by 2, let m be the quotient and r be the remainder, then by Euclid’s division lemma
n = 2 x m + r = 2m + r, 0 ≤ r < 2
n = 2m or 2m + 1 for some integer m.
Case 1 : When n = 2m, then n is even
Case 2 : When n = 2m + 1, then n is odd.
Hence, every positive integer is either even or odd.

Question 6.
Solution:
Let n be a given positive odd integer.
On dividing n by 6, let m be the quotient and r be the remainder, then by Euclid’s division Lemma.
n = 6m + r, where 0 ≤ r < 6 => n = 6m + r, where r = 0, 1, 2, 3, 4, 5
=> n = 6m or (6m + 1) or (6m + 2) or (6m + 3) or (6m + 4) or (6m + 5)
But n = 6m, (6m + 2) and (6m + 4) are even.
Thus when n is odd, it will be in the form of (6m + 1) or (6m + 3) or (6m + 5) for some integer m.

Question 7.
Solution:
Let n be an arbitrary odd positive integer.
On dividing by 4, let m be the quotient and r be the remainder.
So by Euclid’s division lemma,
n = 4m + r, where 0 ≤ r < 4
n = 4m or (4m + 1) or (4m + 2) or (4m + 3)
But 4m and (4m + 2) are even integers.
Since n is odd, so n ≠ 4m or n ≠ (4m + 2)
n = (4m + 1) or (4m + 3) for some integer m.
Hence any positive odd integer is of the form (4m + 1) or (4m + 3) for some integer m.

Question 8.
Solution:
Let a = n³ – n
=> a = n (n² – 1)
=> a = n (n – 1) (n + 1) [(a² – b²) = (a – b) (a + b)]
=> a = (n – 1 ) n (n + 1)
We know that,
(i) If a number is completely divisible by 2 and 3, then it is also divisible by 6.
(ii) If the sum of digits of any number is divisible by 3, then it is also divisible by 3.
(iii) If one of the factor of any number is an even number, then it is also divisible by 2.
a = (n – 1) n (n + 1) [From Eq. (i)]
Now, sum of the digits
= n – 1 + n + n + 1 = 3n
= Multiple of 3, where n is any positive integer.
and (n – 1) n (n +1) will always be even, as one out of (n – 1) or n or (n + 1) must be even.
Since, conditions (ii) and (iii) is completely satisfy the Eq. (i).
Hence, by condition (i) the number n3 – n is always divisible by 6, where n is any positive integer.
Hence proved.

Question 9.
Solution:
Let x = 2m + 1 and y = 2m + 3 are odd positive integers, for every positive integer m.
Then, x² + y² = (2m + 1)² + (2m + 3)²
= 4m² + 1 + 4 m + 4m² + 9 + 12m [(a + b)² = a² + 2ab + b²]
= 8m² + 16m + 10 = even
= 2(4m² + 8m + 5) or 4(2m² + 4m + 2) + 1
Hence, x² + y² is even for every positive integer m but not divisible by 4.

Question 10.
Solution:
We find HCF (1190, 1145) using the following steps:

(i) Since 1445 > 1190, we divide 1445 by 1190 to get 1 as quotient and 255 as remainder.
By Euclid’s division lemma, we get
1445 = 1190 x 1 + 255 …(i)
(ii) Since the remainder 255 ≠ 0, we divide 1190 by 255 to get 4 as a quotient and 170 as a remainder.
By Euclid’s division lemma, we get
1190 = 255 x 4 + 170 …(ii)
(iii) Since the remainder 170 ≠ 0, we divide 255 by 170 to get 1 as quotient and 85 as remainder.
By Euclid’s division lemma, we get
255 = 170 x 1 +85 …(iii)
(iv) Since the remainder 85 ≠ 0, we divide 170 by 85 to get 2 as quotient and 0 as remainder.
By Euclid’s division lemma, we get
170 = 85 x 2 + 0 …(iv)
The remainder is now 0, so our procedure steps
HCF (1190, 1445) = 85
Now, from (iii), we get
255 = 170 x 1 + 85
=> 85 = 255 – 170 x 1
= (1445 – 1190) – (1190 – 255) x 4
= (1445 – 1190) – (1190 – 255) x 4
= (1445 – 1190) x 2 + (1445 – 1190) x 4
= 1445 – 1190 x 2 + 1445 x 4 – 1190 x 4
= 1445 x 5 – 1190 x 6
= 1190 x (-6) + 1445 x 5
Hence, m = -6, n = 5

Hope given RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Ex 1A are helpful to complete your math homework.

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

RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11A

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11A

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11A
• RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11B
• RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C
• RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11D
• RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions MCQS

Question 1.
Solution:
(i) 9, 15, 21, 27, …
Here, 15 – 9 = 6,
21 – 15 = 6,
27 – 21 = 6
d = 6 and a = 9
Next term = 27 + 6 = 33
(ii) 11, 6, 1, -4, …
Here, 6 – 11 = -5,
1 – 6 = -5,
-4 – 1 = -5
d = -5 and a = 11
Next term = -4 – 5 = -9

Question 2.
Solution:
(i) AP is 9, 13, 17, 21, ……
Here, a = 9, d = 13 – 9 = 4

Question 3.
Solution:

Question 4.
Solution:
If the terms are in AP, then
a₂ – a₁ = a₃ – a₂ = …….
a₂ = 3p + 1
a₁ = 2p – 1
a₃ = 11
⇒ (3p + 1) – (2p – 1) = 11 – (3p + 1)
⇒ 3p + 1 – 2p + 1 = 11 – 3p – 1
⇒ p + 2 = 10 – 3p
⇒ 4p = 8
⇒ p = 2
Then for p = 2, these terms are in AP.

Question 5.
Solution:
(i) AP is 5, 11, 17, 23, ……
Here, a = 5, d = 11 – 5 = 6
T_n = a (n – 1)d = 5 + (n – 1) x 6 = 5 + 6n – 6 = (6n – 1)
(ii) AP is 16, 9, 2, -5, ……
Here, a = 16 d = 9 – 16 = -7
T_n = a + (n – 1)d = 16 + (n – 1) (-7)
= 16 – 7n + 7 = (23 – 7n)

Question 6.
Solution:
nth term = 4n – 10
Substituting the value of 1, 2, 3, 4, …, we get
4n – 10
= 4 x 1 – 10 = 4 – 10 = -6
= 4 x 2 – 10 = 8 – 10 = -2
= 4 x 3 – 10 = 12 – 10 = 2
= 4 x 4 – 10 = 16 – 10 = 6
We see that -6, -2, 2, 6,… are in AP
(i) Whose first term = -6
(ii) Common difference = -2 – (-6) = -2 + 6 = 4
(iii) 16th term = 4 x 16 – 10 = 64 – 10 = 54

The common difference calculator takes the input values of sequence and difference and shows you the actual results.

Question 7.
Solution:
In AP 6, 10, 14, 18,…, 174
Here, a = 6, d= 10 – 6 = 4
nth or l = 174
T_n = a + (n – 1)d
⇒ 174 = 6 + (n – 1) x 4
⇒ 174 – 6 = (n – 1) x 4
⇒ n – 1 = \(\frac { 168 }{ 4 }\) = 42
n = 42 + 1 = 43
Hence, there are 43 terms in the given AP.

Question 8.
Solution:
In AP 41, 38, 35,…, 8
a = 41, d = 38 – 41 = -3, l = 8
Let l be the nth term
l = T_n = a + (n – 1) d
⇒ 8 = 41 + (n – 1)(-3)
⇒ 8 – 41 = (n – 1)(-3)
⇒ n – 1 = 11
⇒ n = 11 + 1 = 12
There are 12 terms in the given AP.

Question 9.
Solution:
the AP is 8, 15\(\frac { 1 }{ 2 }\) , 13, …, -47

There are 27 terms in the given AP.

Question 10.
Solution:
Let 88 be the nth term
Now, in AP 3, 8, 13, 18, …
a = 3, d = 8 – 3 = 5
T_n = a + (n – 1) d
88 = 3 + (n – 1)(5)
⇒ 88 – 3 = (n – 1) x 5
⇒ \(\frac { 88 }{ 5 }\) = n – 1
⇒ 17 = n – 1
n= 17 + 1 = 18
88 is the 18th term.

Question 11.
Solution:
In the AP 72, 68, 64, 60, …..
Let 0 be the nth term
Here, a = 72, d = 68 – 72 = -4
T_n = a + (n – 1)d
0 = 72 + (n – 1)(-4)
⇒ -72 = -4(n – 1)
⇒ n – 1 = 18
⇒ n = 18 + 1 = 19
0 is the 19th term.

Question 12.
Solution:

n = 13 + 1 = 14
3 is the 14th term.

Question 13.
Solution:
In the AP 21, 18, 15, ……
Let -81 is the nth term
a = 21, d = 18 – 21 = -3
T_n = a + (n – 1)d
⇒ -81 = 21 + (n – 1)(-3)
⇒ -81 – 21 = (n – 1)(-3)
⇒ -102 = (n – 1)(-3)
⇒ n – 1 = 34
n = 34 + 1 = 35
-81 is the 35th term

Question 14.
Solution:
In the given AP 3, 8, 13, 18,…
a = 3, d = 8 – 3 = 5
T₂₀ = a + (n – 1)d = 3 + (20 – 1) x 5 = 3 + 19 x 5 = 3 + 95 = 98
The required term = 98 + 55 = 153
Let 153 be the nth term, then
T_n = a + (n – 1)d
⇒ 153 = 3 + (n – 1) x 5
⇒ 153 – 3 = 5(n – 1)
⇒ 150 = 5(n – 1)
⇒ n – 1 = 30
⇒ n = 30 + 1 = 31
Required term will be 31st term.

Question 15.
Solution:
AP is 5, 15, 25,…
a = 5, d = 15 – 5 = 10
T₃₁ = a + (n – 1)d = 5 + (31 – 1) x 10 = 5 + 30 x 10 = 5 + 300 = 305
Now the required term = 305 + 130 = 435
Let 435 be the nth term, then
T_n = a + (n – 1)d
⇒ 435 = 5 + (n – 1)10
⇒ 435 – 5 = (n – 1)10
⇒ n – 1 = 43
⇒ n = 43 + 1 = 44
The required term will be 44th term.

Question 16.
Solution:
Let a be the first term and d be the common difference, then

Question 17.
Solution:

T₁₆ = 6 + (16 – 1)7 = 6+ 15 x 7 = 6 + 105 = 111

Question 18.
Solution:
AP is 10, 7, 4, …, (-62)
a = 10, d = 7 – 10 = -3, l = -62
l = T_n = a + (n – 1)d
⇒ -62 = 10 + (n – 1) x (-3)
⇒ -62 – 10 = -3(n- 1)
-72 = -3(n – 1)
n = 24 + 1 = 25
Middle term = \(\frac { 25 + 1 }{ 2 }\) th = 13th term
T₁₃ = 10 + (13 – 1)(-3) = 10+ 12 x (-3)= 10 – 36 = -26

Question 19.
Solution:

Question 20.
Solution:
AP is 7, 10, 13,…, 184
a = 7, d = 10 – 7 = 3, l = 184
nth term from the end = l – (n – 1)d
8th term from the end = 184 – (8 – 1) x 3 = 184 – 7 x 3 = 184 – 21 = 163

Question 21.
Solution:
AP is 17, 14, 11, …,(-40)
Here, a = 17, d = 14 – 17 = -3, l = -40
6th term from the end = l – (n – 1)d
= -40 – (6 – 1) x (-3)
= -40 – [5 x (-3)]
= -40 + 15
= -25

Question 22.
Solution:
Let 184 be the nth term of the AP
3, 7, 11, 15, …
Here, a = 3, d = 7 – 3 = 4
T_n = a + (n – 1)d
⇒ 184 = 3 + (n – 1) x 4
⇒ 184 – 3 = (n – 1) x 4
⇒ \(\frac { 181 }{ 4 }\) = n – 1
⇒ n = \(\frac { 181 }{ 4 }\) + 1 = \(\frac { 185 }{ 4 }\) = 46\(\frac { 1 }{ 4 }\)
Which is in fraction.
184 is not a term of the given AP.

Question 23.
Solution:
Let -150 be the nth term of the AP
11, 8, 5, 2,…

Question 24.
Solution:
Let nth of the AP 121, 117, 113,… is negative

Question 25.
Solution:

Question 26.
Solution:
Let a be the first term and d be the common difference of an AP
T_n = a + (n – 1)d
T₇ = a + (7 – 1)d = a + 6d = -4 …(i)
T₁₃ = a + 12d = -16 …..(ii)
Subtracting (i) from (ii),
6d = -16 – (-4) = -16 + 4 = -12
From (i), a + 6d = -4
a + (-12) = -4
⇒ a = -4 + 12 = 8
a = 8, d = -2
AP will be 8, 6, 4, 2, 0, ……

Question 27.
Solution:
Let a be the first term and d be the common difference of an AP.
T₄ = a + (n- 1)d = a + (4 – 1)d = a + 3d
a + 3d = 0 ⇒ a = -3d
Similarly,
T₂₅ = a + 24d and T₁₁ = a + 10d = -3d + 24d = 21d
It is clear that T₂₅ = 3 x T₁₁

Question 28.
Solution:
Given, a₆ = 0
⇒ a + 5d = 0
⇒ a = -5 d
Now, a₁₅ = a + (n – 1 )d
= a + (15 – 1)d = -5d + 14d = 9d
and a₃₃ = a + (n – 1 )d = a + (33 – 1)d = -5d + 32d = 27d
Now, a₃₃ : a₁₂
⇒ 27d : 9d
⇒ 3 : 1
a₃₃ = 3 x a₁₅

Question 29.
Solution:
Let a be the first term and d be the common difference of an AP.
T_n = a + (n – 1)d
T₄ = a + (4 – 1)d = a + 3d
a + 3d = 11 …(i)
Now, T₅ = a + 4d
and T₇ = a + 6d
Adding, we get T₅ + T₇ = a + 4d + a + 6d = 2a + 10d
2a + 10d = 34
⇒ a + 5d = 17 …(ii)
Subtracting (i) from (ii),
2d = 17 – 11 = 6
⇒ d = 3
Hence, common difference = 3

Question 30.
Solution:
Let a be the first term and d be the common difference of an AP.

Question 31.
Solution:
Let a be the first term and d be the common difference, then

Question 32.
Solution:
In an AP,
Let a be the first term and d be the common difference, then

Question 33.
Solution:
Let a be the first term and d be the common difference in an AP, then

Question 34.
Solution:
In an AP,
Let d be the common difference,
First term (a) = 5
Sum of first 4 terms
= a + a + d + a + 2d + a + 3d = 4a + 6d
Sum of next 4 terms
= a + 4d + a + 5d + a + 6d + a + 7d = 4a + 22d
According to the condition,

Question 35.
Solution:
Let a be the first term and d be the common difference in an AP, then

a = 1, d = 4
AP = 1, 5, 9, 13, 17, …

Question 36.
Solution:
In AP 63, 65, 67, …..

Question 37.
Solution:
Let first term of AP = a
and common difference = d

Question 38.
Solution:
Let a be the first term and d be the common difference, then

Question 39.
Solution:
Let a be the first term and d be the common difference, then

Question 40.
Solution:
Let a be the first term and d be the common difference, then

Question 41.
Solution:
Let a be the first term, d be the common difference, then

Question 42.
Solution:
Two-digit numbers are 10 to 99 and two digit numbers divisible by 6 will be
12, 18, 24, 30, …, 96

Question 43.
Solution:
Two digit numbers are 10 to 99 and
Two digit numbers which are divisible by 3 are
12, 15, 18, 21, 24, … 99

Question 44.
Solution:
Three digit numbers are 100 to 999 and numbers divisible by 9 will be
108, 117, 126, 999

Question 45.
Solution:
Numbers between 101 and 999 which are divisible both by 2 and 5 will be
110, 120, 130,…, 990

Question 46.
Solution:
Let number of from a rows are in the flower bed, then

Question 47.
Solution:
Total amount = ₹ 2800
and number of prizes = 4
Let first prize = ₹ a
Then second prize = ₹ a – 200
Third prize = a – 200 – 200 = a – 400
and fourth prize = a – 400 – 200 = a – 600
But sum of there 4 prizes are ₹ 2800
a + a – 200 + a – 400 + a – 600 = ₹ 2800
⇒ 4a – 1200 = 2800
⇒ 4a = 2800 + 1200 = 4000
⇒ a = 1000
First prize = ₹ 1000
Second prize = ₹ 1000 – 200 = ₹ 800
Third prize = ₹ 800 – 200 = ₹ 600
and fourth prize = ₹ 600 – 200 = ₹ 400

Question 48.
Solution:
The first term between 200 and 500 divisible by 8 is 208, and last term is 496.
So, first term (a) = 208
Common difference (d) = 8
Now, an = a + (n – 1 )d
⇒ 496 = 208 + (n – 1) x 8
⇒ (n – 1) = \(\frac { 288 }{ 8 }\)
⇒ n – 1 = 36
⇒ n = 36 + 1 = 37
Hence, there are 37 integers between 200 and 500 which are divisible by 8.

Hope given RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11A are helpful to complete your math homework.

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

Online Education for RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself

These Solutions are part of Online Education RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 10 Quadratic Equations Test Yourself.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 10 Quadratic Equations Ex 10A
• RS Aggarwal Solutions Class 10 Chapter 10 Quadratic Equations Ex 10B
• RS Aggarwal Solutions Class 10 Chapter 10 Quadratic Equations Ex 10C
• RS Aggarwal Solutions Class 10 Chapter 10 Quadratic Equations Ex 10D
• RS Aggarwal Solutions Class 10 Chapter 10 Quadratic Equations Ex 10E
• RS Aggarwal Solutions Class 10 Chapter 10 Quadratic Equations Test Yourself

Objective Questions (MCQ)
Question 1.
Solution:
(a) x² – 3√x + 2 = 0
It is not a quadratic equation, it has a fractional power of √x
(b) x + \(\frac { 1 }{ x }\) = x²
⇒ x² + 1 = x³
It is not a quadratic equation.
(c) x² + \(\frac { 1 }{ { x }^{ 2 } }\) = 5
⇒ x⁴ + 1 + 5x²
It is not a quadratic equation.
(d) 2x² – 5x = (x – 1)²
⇒ 2x² – 5x = x² – 2x + 1
⇒ x² – 3x – 1 = 0
It is a quadratic equation. (d)

Question 2.
Solution:
(a) (x² + 1) = (2 – x)² + 3
⇒ x² + 1 = 4 + x² – 4x + 3 is not a quadratic equation.
(b) x³ – x² = (x – 1)³
⇒ x³ – x² = x³ – 3x² + 3x – 1
⇒ 3x² – x² – 3x + 1 = 0
⇒ 2x² – 3x + 1 = 0
It is a quadratic equation.
(c) 2x² + 3 = 10x – 15 + 2x² – 3x
⇒ 3x – 15 – 3 = 0
It is not a quadratic equation. (b)

Question 3.
Solution:
(a) It is a quadratic equation.
(b) (x + 2)² = 2(x² – 5)
⇒ x² + 4x + 4 = 2x² – 10
⇒ x² – 4x – 14 = 0
It is a quadratic equation.
(c) (√2 x + 3)² = 2x² + 6
⇒ 2x² + 3√2 x + 9 = 2x² + 6
⇒ 3√2 + 3 = 0
It is not a quadratic equation.
(d) (x – 1)² = 3x² + x – 2
⇒ x² – 2x +1 = 3x² + x – 2
⇒ 2x² + 3x – 3 = 0
It is a quadratic equation. (c)

Question 4.
Solution:
x = 3 is solution of 3x² + (k – 1)x + 9 = 0
It will satisfy it
3(3)² + (k – 1)(3) + 9 = 0
⇒ 27 + 3k – 3 + 9 = 0
⇒ 3k + 33 = 0
⇒ k = -11 (b)

Question 5.
Solution:
2 is one root of equation 2x² + ax + 6 = 0
It will satisfy it
2(2)² + a(2) + 6 = 0
⇒ 8 + 2a + 6 = 0
⇒ 2a = -14
⇒ a = -7
a = -7 (b)

Question 6.
Solution:
In equation x² – 6x + 2 = 0
Sum of roots = \(\frac { -b }{ a }\) = \(\frac { -(-6) }{ 1 }\) = 6 (c)

Question 7.
Solution:
In equation x² – 3x + k = 10
x² – 3x + (k – 10) = 0
Product of roots = \(\frac { c }{ a }\) = \(\frac { k – 10 }{ 1 }\) = k – 10
k – 10 = -2 then k = 10 – 2 = 8 (c)

Question 8.
Solution:
In equation 7x² – 12x + 18 = 0

Question 9.
Solution:
In equation 3x² – 10x + 3 = 0

Question 10.
Solution:
In equation 5x² + 13x + k = 0

Question 11.
Solution:
In equation kx² + 2x + 3k = 0
Sum of roots = \(\frac { -b }{ a }\) = \(\frac { -2 }{ k }\)

Question 12.
Solution:
Roots of an equation are 5, -2
Sum of roots (S) = 5 – 2 = 3
and product (P) = 5 x (-2) = -10
Equation will be
x² – (S)x + (P) = 0
⇒ x² – 3x – 10 = 0 (b)

Question 13.
Solution:
Sum of roots (S) = 6
Product of roots (P) = 6
Equation will be x² – (S)x + (P) = 0
x² – 6x + 6 = 0 (a)

Question 14.
Solution:
α and β are the roots of the equation 3x² + 8x + 2 = 0

Question 15.
Solution:
In equation ax² + bx + c = 0

Question 16.
Solution:
In equation ax² + bx + c = 0
Let α and β are the roots, then

Question 17.
Solution:
In equation 9x² + 6kx + 4 = 0, roots are equal
Let roots be α, α then

Question 18.
Solution:
In equation x² + 2 (k + 2) x + 9k = 0
Roots are equal
Let α, α be the roots, then

Question 19.
Solution:
In the equation
4x² – 3kx + 1 = 0 roots are equal
Let α, α be the roots

Question 20.
Solution:
Roots of ax² + bx + c = 0, a ≠ 0 are real and unequal if D > 0
⇒ b² – 4ac > 0 (a)

Question 21.
Solution:
In the equation ax² + bx + c = 0
D = b² – 4ac > 0, then roots are real and unequal. (b)

Question 22.
Solution:
In the equation 2x² – 6x + 7 = 0
D = b² – 4ac = (-6)² – 4 x 2 x 7 = 36 – 56 = -20 < 0
Roots are imaginary (not real) (d)

Question 23.
Solution:
In equation 2x² – 6x + 3 = 0
D = b² – 4ac = (-6)² – 4 x 2 x 3 = 36 – 24 = 12 > 0
Roots are real, unequal and irrational, (b)

Question 24.
Solution:
In equation 5x² – kx + 1 = 0
D = b² – 4ac = (-k)² – 4 x 5 x 1 = k² – 20
Roots are real and distinct
D > 0
⇒ k² – 20 > 0
⇒ k² > 20
⇒ k > √±20
⇒ k > ±2√5
⇒ k > 2√5 or k < -2√5 (d)

Question 25.
Solution:
In equation x² + 5kx + 16 = 0
D = b² – 4ac = (5k)² – 4 x 1 x 16

Question 26.
Solution:
The equation x² – kx + 1 = 0
D = b2 – 4ac = (-k)² – 4 x 1 x 1 ⇒ k² – 4
Roots are not real
D < 0
⇒ k² – 4 < 0
⇒ k² < 4
⇒k < (±2)²
⇒ k < ±2
-2 < k < 2 (c)

Question 27.
Solution:
In the equation kx² – 6x – 2 = 0

Question 28.
Solution:
Let the number be = x

Question 29.
Solution:
Perimeter of a rectangle = 82 m
and Area = 400
Let breadth (b) = x, then
Length = \(\frac { P }{ 2 }\) – x = \(\frac { 82 }{ 2 }\) – x = 41 – x
Area = lb
400 = x (41 – x) = 41x – x²
⇒ x² – 41x + 400 = 0
⇒ x² – 25x – 16x + 400 = 0
⇒ x (x – 25) – 16(x – 25) = 0
⇒ (x – 25) (x – 16) = 0
Either, x – 16 = 0, then x = 16
or x – 25 = 0, then x = 25
Breadth = 16 m (c)

Question 30.
Solution:
Let breadth of a rectangular field = x m
Then length = (x + 8) m
and area = 240 m²
x (x + 8) = 240
⇒ x² + 8x – 240 = 0
⇒ x² + 20x – 12x – 240 = 0
⇒ x (x + 20) – 12 (x + 20) = 0
⇒ (x + 20) (x – 12) = 0
Either, x + 20 = 0, then x = -20 which is not possible being negative,
or x – 12 = 0, then x = 12
Breadth = 12 m (c)

Question 31.
Solution:
2x² – x – 6 = 0

Very-Short-Answer Questions
Question 32.
Solution:
Sum of two natural numbers = 8
Let first number – x
Then second number = 8 – x
According to the condition,
x (8 – x) = 15
⇒ 8x – x² = 15
⇒ x² – 8x + 15 = 0
⇒ x² – 3x – 5x + 15 = 0
⇒ x(x – 3) – 5(x – 3) = 0
⇒ (x – 3)(x – 5) = 0
Either, x – 3 = 0, then x = 3
or x – 5 = 0, then x = 5
Natural numbers are 3, 5

Question 33.
Solution:
x = -3 is a solution of equation x² + 6x + 9 = 0 Then it will satisfy it
LHS = x² + 6x + 9 = (-3)² + 6(-3) + 9 = 9 – 18 + 9 = 0 = RHS

Question 34.
Solution:
3x² + 13x + 14 = 0
If x = -2 is its root then it will satisfy it
LHS = 3(-2)² + 13(-2) + 14 = 3 x 4 – 26 + 14 = 12 – 26 + 14 = 26 – 26 = 0 = RHS

Question 35.
Solution:
x = y is a solution of equation 3x² + 2kx – 3 = 0, then it will satisfy it

Question 36.
Solution:
2x² – x – 6 = 0
⇒ 2x² – 4x + 3x – 6 = 0

Question 37.
Solution:
3√3 x² + 10x + √3 = 0

Question 38.
Solution:
Roots of the quadratic equation 2x² + 8x + k = 0 are equal
Let α, α be its roots, then

Question 39.
Solution:
px² – 2√5 px + 15 = 0
Here, a = p, b = 2√5 p, c = 15
D = b² – 4ac = (-2√5 p)² – 4 x p x 15 = 20p² – 60p
Roots are equal.
D = 0
⇒ 20p² – 60p = 0
⇒ p² – 3p = 0
⇒ p (p – 3) = 0
p – 3 = 0, then p = 3

Question 40.
Solution:
1 is a root of equation
ay² + ay + 3 = 0 and y² + y + b = 0
Then a(1)² + a(1) + 3 = 0
⇒ a + a + 3 = 0
⇒ 2a + 3 = 0
⇒ a = \(\frac { -3 }{ 2 }\)
and 1 + 1 + b = 0
⇒ 2 + b = 0
⇒ b = -2
ab = \(\frac { -3 }{ 2 }\) x (-2) = 3
Hence, ab = 3

Question 41.
Solution:
The polynomial is x² – 4x + 1
Here, a = 1, b = -4, c = 1

Question 42.
Solution:
In the quadratic equation 3x² – 10x + k = 0

Question 43.
Solution:
The quadratic equation is
px (x – 2) + 6 = 0
⇒ px² – 2px + 6 = 0
D = b² – 4ac = (-2p)² – 4 x p x 6 = 4p² – 24p
Roots are equal
D = 0
Then 4p² – 24p = 0
⇒ 4p (p – 6) = 0
⇒ p – 6 = 0
⇒ p = 6

Question 44.
Solution:
x² – 4kx + k = 0
D = b² – 4ac = (-4k)² – 4 x 1 x k = 16k² – 4k
Roots are equal
D = 0
16k² – 4k = 0
⇒ 4k (4k – 1) = 0
⇒ 4k – 1 = 0
⇒ k = \(\frac { 1 }{ 4 }\)

Question 45.
Solution:
9x² – 3kx + k = 0
D = b² – 4ac = (-3k)² – 4 x 9 x k = 9k² – 36k
Roots are equal
D = 0
9k² – 36k = 0
9k (k – 4) = 0
Either, 9k = 0, then k = 0
or (k – 4) = 0 ⇒ k = 4
k = 0, 4

Short-Answer Questions
Question 46.
Solution:
x² – (√3 + 1) x + √3 = 0
D = b² – 4ac
= [-(√3 + 1)]² – 4 x 1 x √3
= 3 + 1 + 2√3 – 4√3
= 4 + 2√3 – 4√3
= 4 – 2√3
= 3 + 1 – 2√3
= (√3 – 1)²

Question 47.
Solution:
2x² + ax – a² = 0
D = B² – 4AC = a² – 4 x 2(-a)² = a² + 8a² = 9a²

Question 48.
Solution:
3x² + 5√5 x – 10 = 0
D = b² – 4ac = (5√5)² – 4 x 3 x (-10)
= 125 + 120 = 245 = 49 x 5 = (7√5)²

Question 49.
Solution:
√3 x² + 10x – 8√3 = 0

Question 50.
Solution:
√3 x² – 2√2 x – 2√3 = 0

Question 51.
Solution:
4√3 x² + 5x – 2√3 = 0

Question 52.
Solution:
4x² + 4bx – (a² – b²) = 0

Question 53.
Solution:
x² + 5x – (a² + a – 6) = 0
a² + a – 6 = a² + 3a – 2a – 6 = a(a + 3) – 2(a + 3) = (a + 3)(a – 2)
and 6 = (a + 3) – (a – 2)
x² + (a + 3)x – (a – 2)x – (a + 3) (a – 2) = 0
x (x + a + 3) – (a – 2) (x + a + 3) = 0
(x + a + 3)(x – a + 2) = 0
Either, x + a + 3 = 0, then x = -(a + 3)
or x – a + 2 = -0 then x = (a – 2)
x = -(a + 3) or (a – 2)

Question 54.
Solution:
x² + 6x – (a² + 2a – 8) = 0

Question 55.
Solution:
x² – 4ax + 4a² – b² = 0
4a² – b² = (2a)² – (b)² = (2a + b)(2a – b) – 4ax = (2a + b)x + (2a – b)x
x² – 4ax + 4a² – b² = x² – (2a + b)x – (2a – b)x + (2a + b)(2a – b) = 0
⇒ x (x – 2a – b) – (2a – b)(x – 2a – b) = 0
⇒ (x – 2a – b)(x – 2a + b)
Either, x – 2a – b = 0, then x = 2a + b
or x – 2a + b = 0, then x = 2a – b
Hence, x = (2a + b) or (2a – b)

Hope given RS Aggarwal Solutions Class 10 Chapter 10 Quadratic Equations Test Yourself are helpful to complete your math homework.

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