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

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

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4B. You must go through NCERT Solutions for Class 10 Maths to get better score in CBSE Board exams along with RS Aggarwal Class 10 Solutions.

Question 1.
Solution:
(i) 36, 84
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 1
36 = 2 x 2 x 3 x 3 = 2² x 3²
84 = 2 x 2 x 3 x 7 = 2² x 3 x 7
HCF = 2² x 3 = 2 x 2 x 3 = 12
LCM = 2² x 3² x 7 = 2 x 2 x 3 x 3 x 7 = 252
Now HCF x LCM = 12 x 252 = 3024
and product of number = 36 x 84 = 3024
HCF x LCM = Product of given two numbers.
(ii) 23, 31
23 = 1 x 23
31 = 1 x 31
HCF= 1
and LCM = 23 x 31 = 713
Now HCF x LCM = 1 x 713 = 713
and product of numbers = 23 x 31 = 713
HCF x LCM = Product of given two numbers
(iii) 96, 404
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 2
96 = 2 x 2 x 2 x 2 x 2 x 3 = 25 x 3
404 = 2 x 2 x 101 = 2² x 101
HCF = 2² = 2 x 2 = 4
LCM = 25 x 3 x 101 = 32 x 3 x 101 = 9696
Now HCF x LCM = 4 x 9696 = 38784
and product of two numbers = 96 x 404 = 38784
HCF x LCM = Product of given two numbers
(iv) 144, 198
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 3
144 = 2 x 2 x 2 x 2 x 3 x 3 = 24 x 32
198 = 2 x 3 x 3 x 11 = 2 x 3² x 11
HCF = 2 x 32 = 2 x 3 x 3 = 18
LCM = 24 x 3² x 11 = 16 x 9 x 11 = 1584
and product of given two numbers = 144 x 198 = 28512
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 4
and HCF x LCM = 18 x 1584 = 28512
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 5
HCF x LCM = Product of given two numbers
(v) 396, 1080
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 6
396 = 2 x 2 x 3 x 3 x 11 = 2² x 3² x 11
1080 = 2 x 2 x 2 x 3 x 3 x 3 x 5 = 23 x 33 x 5
HCF = 2² x 3² = 2 x 2 x 3 x 3 = 36
LCM = 23 x 33 x 11 x 5 = 2 x 2 x 2 x 3 x 3 x 3 x 5 x 11 = 11880
Now HCF x LCM = 36 x 11880 = 427680
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 7
Product of two numbers = 396 x 1080 = 427680
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 8
HCF x LCM = Product of two given numbers.
(vi) 1152, 1664
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 9
1152 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 = 27 x 3²
1664 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 13 = 27 x 13
HCF = 27 = 2 x 2 x 2 x 2 x 2 x 2 x 2 = 128
LCM = 27 x 3² x 13 = 128 x 9 x 13 = 128 x 117= 14976
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 10
Now HCF x LCM = 128 x 14976= 1916928
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 11
and product of given two numbers = 1152 x 1664 = 1916928
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 12
HCF x LCM = Product of given two numbers.

Question 2.
Solution:
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 13
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 14
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 15
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 16

Question 3.
Solution:
HCF of two numbers = 23
LCM =1449
One number = 161
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 17
Second number = 207

Question 4.
Solution:
HCF of two numbers = 145
LCM = 2175
One number = 725
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 18
Second number = 435

Question 5.
Solution:
HCF of two numbers = 18
and product of two numbers = 12960
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 19
LCM of two numbers = 720

Question 6.
Solution:
HCF= 18
LCM = 760
HCF always divides the LCM completely
760 – 18 = 42 and remainder 4
Hence, it is not possible.

Question 7.
Solution:
(a) \(\frac { 69 }{ 92 }\)
HCF of 69 and 92 = 23
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 20
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 21
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 22
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 23

Question 8.
Solution:
Numbers are 428 and 606 and remainder in each case = 6
Now subtracting 6 from each number, we get 438 – 6 = 432
and 606 – 6 = 600
Required number = HCF of 432 and 600 = 24
The largest required number is 24
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 24

Question 9.
Solution:
The numbers are 320 and 457
and remainders are 5 and 7 respectively
320 – 5 = 315 and 457 – 7 = 450
Now the required greatest number of 315 and 450 is their HCF
Now HCF of 315 and 450 = 45
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 25

Question 10.
Solution:
The numbers are given = 35, 56, 91 and the remainder = 7 in each case,
Now the least number = LCM of 35, 56, 91 = 3640
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 26
LCM = 7 x 5 x 8 x 13 = 3640
Required least number = 3640 + 7 = 3647

Question 11.
Solution:
Given numbers are 28 and 32
Remainders are 8 and 12 respectively
28 – 8 = 20
32 – 12 = 20
Now, LCM of 28 and 32 = 224
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 27
LCM = 2 x 2 x 7 x 8 = 224
Least required number = 224 – 20 = 204

Question 12.
Solution:
The given numbers are 468 and 520
Now LCM of 468 and 520 = 4680
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 28
LCM = 2 x 2 x 13 x 9 x 10 = 4680
When number 17 is increase then required number = 4680 – 17 = 4663

Question 13.
Solution:
LCM of 15, 24, 36 = 360
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 29
Required number = 9999 – 279 = 9720

Question 14.
Solution:
Greatest number of 4 digits is 9999
LCM of 4, 7 and 13 = 364
On dividing 9999 by 364, remainder is 171
Greatest number of 4 digits divisible by 4, 7 and 13 = (9999 – 171) = 9828
Hence, required number = (9828 + 3) = 9831

Question 15.
Solution:
LCM of 5, 6, 4 and 3 = 60
On dividing 2497 by 60, the remainder is 37
Number to be added = (60 – 37) = 23

Question 16.
Solution:
We can represent any integer number in the form of: pq + r, where ‘p is divisor, ‘q is quotient’, r is remainder*.
So, we can write given numbers from given in formation As :
43 = pq1 + r …(i)
91 = pq2 + r …(ii)
And 183 = pq3 + r …(iii)
Here, we want to find greatest value of ‘p’ were r is same.
So, we subtract eq. (i) from eq. (ii), we get
Pq2 – Pq1 = 48
Also, subtract eq. (ii) from eq. (iii), we get
pq3 – pq2 = 92
Also, subtract eq. (i) from eq. (iii), we get
Pq3 – Pq1 = 140
Now, to find greatest value of ‘p’ we find HCF of 48, 92 and 140 as,
48 = 2 x 2 x 2 x 2 x 3
92 = 2 x 2 x 2 x 2 x 2 x 3
and 140 = 2 x 2 x 5 x 7
So, HCF (48, 92 and 140) = 2 x 2 = 4
Greatest number that will divide 43, 91 and 183 as to leave the same remainder in each case = 4.

Question 17.
Solution:
Remainder in all the cases is 6, i.e.,
20 – 14 = 6
25 – 19 = 6
35 – 29 = 6
40 – 34 = 6
The difference between divisor and the corresponding remainder is 6.
Required number = (LCM of 20, 25, 35, 46) – 6 = 1400 – 6 = 1394

Question 18.
Solution:
Number of participants in Hindi = 60
Number of participants in English = 84
Number of participants in Mathematics =108
Minimum number of participants in one room = HCF of 60, 84 and 108 = 12
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 30
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 31

Question 19.
Solution:
Number of books in English = 336
Number of books in Mathematics = 240
Number of books in Science = 96
Minimum number of books of each topic in a stack = HCF of 336, 240 and 96 = 48
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 32

Question 20.
Solution:
Length of first piece of timber = 42 m
Length of second piece of timber = 49 m
and length of third piece of timber = 63 m
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 33
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 34

Question 21.
Solution:
Lengths are given as 7 m, 3 m 85 cm and 12 m 95 cm = 700 cm, 385 cm and 1295 cm
Greatest possible length that can be used to measure exactly = HCF of 700, 385, 1295 = 35 cm
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 35

Question 22.
Solution:
Number of pens =1001
and number of pencils = 910
Maximum number of pens and pencils equally distributed to the students = HCF of 1001 and 910 = 91
Number of students = 91
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 36

Question 23.
Solution:
Length of the room = 15 m 17 cm = 1517 cm
and breadth = 9 m 2 cm = 902 cm
Maximum side of square tile used = HCF of 1517 and 902 = 41 cm
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 37

Question 24.
Solution:
Measures of three rods = 64 cm, 80 cm and 96 cm
Least length of cloth that can be measured an exact number of times
= LCM of 64, 80, 96
= 960 cm
= 9 m 60 cm
= 9.6 m
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 38
LCM = 2 x 2 x 2 x 2 x 2 x 2 x 5 x 3 = 960

Question 25.
Solution:
Beep made by first devices after every = 60 seconds
Second device after = 62 seconds
Period after next beep together = LCM of 60, 62
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 39
LCM = 2 x 30 x 31 = 1860 = 1860 seconds = 31 minutes
Time started beep together, first time together = 10 a.m.
Time beep together next time = 10 a.m. + 31 minutes = 10 : 31 a.m.

Question 26.
Solution:
The traffic lights of three roads change after
48 sec., 72 sec. and 108 sec. simultaneously
They will change together after a period of = LCM of 48 sec., 72 sec. and 108 sec.
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 40
= 7 minutes, 12 seconds
First time they light together at 8 a.m. i.e., after 8 hr.
Next time they will light together = 8 a.m. + 7 min. 12 sec. = 8 : 07 : 12 hrs.

Question 27.
Solution:
Tolling of 6 bells = 2, 4, 6, 8, 10, 12 minutes
They take time tolling together = LCM of 2, 4, 6, 8, 10, 12 = 120 minutes = 2 hours
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 41
LCM of 2 x 2 x 2 x 3 x 5 = 120 min. (2 hr)
They will toll together after every 2 hours Total time given = 30 hours
Number of times, there will toll together in 30 hours = \(\frac { 30 }{ 2 }\) = 15 times
Total numbers of times = 15 + 1 (of starting time) = 16 times

Hope given RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Ex 1B 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 1 Real Numbers Ex 1A

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

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
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1A 1
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1A 2

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:
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1A 3
(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

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

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
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11A 1
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11A 2

Question 2.
Solution:
(i) AP is 9, 13, 17, 21, ……
Here, a = 9, d = 13 – 9 = 4
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11A 3
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11A 4
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11A 5

Question 3.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11A 6

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
Tn = 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
Tn = 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
Tn = 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 = Tn = 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
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11A 7
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
Tn = 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
Tn = 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:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11A 8
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
Tn = 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
Tn = 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
Tn = 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
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11A 9

Question 17.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11A 10
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 = Tn = 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:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11A 11
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11A 12

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
Tn = 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,…
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11A 13

Question 24.
Solution:
Let nth of the AP 121, 117, 113,… is negative
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11A 14

Question 25.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11A 15
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11A 16

Question 26.
Solution:
Let a be the first term and d be the common difference of an AP
Tn = 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.
Tn = 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.
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11A 17

Question 31.
Solution:
Let a be the first term and d be the common difference, then
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11A 18

Question 32.
Solution:
In an AP,
Let a be the first term and d be the common difference, then
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11A 19

Question 33.
Solution:
Let a be the first term and d be the common difference in an AP, then
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11A 20

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,
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11A 21

Question 35.
Solution:
Let a be the first term and d be the common difference in an AP, then
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11A 22
a = 1, d = 4
AP = 1, 5, 9, 13, 17, …

Question 36.
Solution:
In AP 63, 65, 67, …..
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11A 23

Question 37.
Solution:
Let first term of AP = a
and common difference = d
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11A 24
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11A 25

Question 38.
Solution:
Let a be the first term and d be the common difference, then
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11A 26

Question 39.
Solution:
Let a be the first term and d be the common difference, then
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11A 27
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11A 28

Question 40.
Solution:
Let a be the first term and d be the common difference, then
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11A 29

Question 41.
Solution:
Let a be the first term, d be the common difference, then
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11A 30

Question 42.
Solution:
Two-digit numbers are 10 to 99 and two digit numbers divisible by 6 will be
12, 18, 24, 30, …, 96
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11A 31

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
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11A 32

Question 44.
Solution:
Three digit numbers are 100 to 999 and numbers divisible by 9 will be
108, 117, 126, 999
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11A 33

Question 45.
Solution:
Numbers between 101 and 999 which are divisible both by 2 and 5 will be
110, 120, 130,…, 990
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11A 34

Question 46.
Solution:
Let number of from a rows are in the flower bed, then
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11A 35

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

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

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
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 1

Question 9.
Solution:
In equation 3x² – 10x + 3 = 0
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 2

Question 10.
Solution:
In equation 5x² + 13x + k = 0
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 3

Question 11.
Solution:
In equation kx² + 2x + 3k = 0
Sum of roots = \(\frac { -b }{ a }\) = \(\frac { -2 }{ k }\)
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 4

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
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 5
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 6

Question 15.
Solution:
In equation ax² + bx + c = 0
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 7

Question 16.
Solution:
In equation ax² + bx + c = 0
Let α and β are the roots, then
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 8

Question 17.
Solution:
In equation 9x² + 6kx + 4 = 0, roots are equal
Let roots be α, α then
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 9

Question 18.
Solution:
In equation x² + 2 (k + 2) x + 9k = 0
Roots are equal
Let α, α be the roots, then
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 10

Question 19.
Solution:
In the equation
4x² – 3kx + 1 = 0 roots are equal
Let α, α be the roots
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 11

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
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 12

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
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 13

Question 28.
Solution:
Let the number be = x
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 14

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
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 15

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
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 16
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 17

Question 36.
Solution:
2x² – x – 6 = 0
⇒ 2x² – 4x + 3x – 6 = 0
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 18

Question 37.
Solution:
3√3 x² + 10x + √3 = 0
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 19
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 20

Question 38.
Solution:
Roots of the quadratic equation 2x² + 8x + k = 0 are equal
Let α, α be its roots, then
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 21

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
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 22

Question 42.
Solution:
In the quadratic equation 3x² – 10x + k = 0
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 23

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
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 24

Question 47.
Solution:
2x² + ax – a² = 0
D = B² – 4AC = a² – 4 x 2(-a)² = a² + 8a² = 9a²
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 25

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
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 26
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 27

Question 49.
Solution:
√3 x² + 10x – 8√3 = 0
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 28

Question 50.
Solution:
√3 x² – 2√2 x – 2√3 = 0
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 29
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 30

Question 51.
Solution:
4√3 x² + 5x – 2√3 = 0
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 31
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 32

Question 52.
Solution:
4x² + 4bx – (a² – b²) = 0
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 33

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
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 34
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Test Yourself 35

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.

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

RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3D

NCERT Maths Solutions for Chapter 3 Ex 3.4 Class 10 acts as the best resource during your learning and helps you score well in your board exams.

RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3D

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3D. You must go through NCERT Solutions for Class 10 Maths to get better score in CBSE Board exams along with RS Aggarwal Class 10 Solutions.

RS Aggarwal Solutions Class 10 Chapter 3

Show that each of the following systems of equations has a unique solution and solve it:
Question 1.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3D 1

Question 2.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3D 2
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3D 3

Question 3.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3D 4
This system has a unique solution.
From (ii), x = 2 + 2y
Substituting the value of x in (i),
2(2 + 2y) + 3y = 18
=> 4 + 4y + 3y = 18
=> 7y = 18 – 4 = 14
=> y = 2
and x = 2 + 2 x 2 = 2 + 4 = 6
x = 6, y = 2

Find the value of k for which each of the following systems of equations has a unique solution:
Question 4.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3D 5

Question 5.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3D 6

Question 6.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3D 7

Question 7.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3D 8

Question 8.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3D 9

Question 9.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3D 10
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3D 11

Question 10.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3D 12

Question 11.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3D 13

Question 12.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3D 14

Question 13.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3D 15
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3D 16

Question 14.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3D 17

Find the value of k for which each of the following systems of linear equations has an infinite number of solutions:
Question 15.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3D 18

Question 16.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3D 19
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3D 20

Question 17.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3D 21

Question 18.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3D 22
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3D 23

Question 19.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3D 24
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3D 25

Question 20.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3D 26
=> k (k – 6) = 0
Either k = 0, which is not true, or k – 6 = 0, then k = 6
k = 6

Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:
Question 21.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3D 27

Question 22.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3D 28
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3D 29

Question 23.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3D 30
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3D 31

Question 24.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3D 32
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3D 33

Question 25.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3D 34

Question 26.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3D 35

Find the value of k for which each of the following systems of equations has no solution:
Question 27.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3D 36

Question 28.
Solution:
kx + 3y = 3
12x + ky = 6
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3D 37

Question 29.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3D 38

Question 30.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3D 39
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3D 40

Question 31.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3D 41

Hope given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3D 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 10 Quadratic Equations Ex 10A

NCERT Maths Solutions for Ex 4.1 class 10 Quadratic Equations is the perfect guide to boost up your preparation during CBSE 10th Class Maths Examination.

RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 10 Quadratic Equations Ex 10A.

Other Exercises

Question 1.
Solution:
We know that a second degree of equation is called a quadratic equation. Therefore,
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 1
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 2
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 3
It is not a quadratic equation. It is of degree 5.
So, (i), (ii), (iii), (iv), (vi) and (ix) are quadratic equations.

Question 2.
Solution:
3x² + 2x – 1
= 3x² + 3x – x – 1
= 3x (x + 1) – 1 (x + 1)
= (x + 1) (3x – 1)
Either, x + 1 = 0 ⇒ x = -1
or 3x – 1 =0
⇒ 3x = 1
⇒ x = \(\frac { 1 }{ 3 }\)
Hence, (-1) and \(\frac { 1 }{ 3 }\) are its roots.

Question 3.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 4
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 5

Question 4.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 6
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 7

Solve each of the following quadratic equations.

Question 5.
Solution:
Given : (2x – 3)(3x + 1) = 0
Either 2x – 3 = 0, then 2x = 3 ⇒ x = \(\frac { 3 }{ 2 }\)
or 3x + 1 = 0, then 3x = -1 ⇒ x = \(\frac { -1 }{ 3 }\)
x = \(\frac { 3 }{ 2 }\) , \(\frac { -1 }{ 3 }\)

Question 6.
Solution:
4×2 + 5x = 0 ⇒ x (4x + 5) = 0
Either x = 0
or 4x + 5 = 0, then 4x = -5 ⇒ x = \(\frac { -5 }{ 4 }\)
x = \(\frac { -5 }{ 4 }\) or 0

Question 7.
Solution:
3x² – 243 = 0
x² – 81 =0 (Dividing by 3)
⇒ (x)² – (9)² = 0
⇒ (x + 9) (x – 9) = 0
Either, x + 9 = 0, then x = -9
or x – 9 = 0, then x = 9
Hence, x = 9 or -9

Question 8.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 8

Question 9.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 9

Question 10.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 10
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 11

Question 11.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 12

Question 12.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 13

Question 13.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 14
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 15

Question 14.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 16

Question 15.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 17

Question 16.
Solution:
4x² – 9x = 100
4x² – 9x – 100 = 0
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 18

Question 17.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 19

Question 18.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 20
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 21

Question 19.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 22

Question 20.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 23

Question 21.
Solution:
√3 x² + 10x + 7√3 = 0
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 24

Question 22.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 25

Question 23.
Solution:
3√7 x² + 4x + √7 = 0
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 26

Question 24.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 27
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 28

Question 25.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 29
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 30

Question 26.
Solution:
3x² – 2√6x + 2 = 0
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 31

Question 27.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 32

Question 28.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 33

Question 29.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 34

Question 30.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 35
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 36

Question 31.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 37

Question 32.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 38

Question 33.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 39

Question 34.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 40

Question 35.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 41

Question 36.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 42

Question 37.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 43

Question 38.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 44

Question 39.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 45

Question 40.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 46
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 47

Question 41.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 48

Question 42.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 49
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 50

Question 43.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 51

Question 44.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 52

Question 45.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 53

Question 46.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 54
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 55

Question 47.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 56

Question 48.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 57

Question 49.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 58
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 59

Question 50.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 60

Question 51.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 61

Question 52.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 62

Question 53.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 63

Question 54.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 64
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 65

Question 55.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 66
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 67
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 68

Question 56.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 69

Question 57.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 70
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 71

Question 58.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 72
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 73
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 74

Question 59.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 75
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 76
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 77
⇒ x = -2
Roots, x = -2

Question 60.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 78
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 79

Question 61.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 80

Question 62.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 81
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 82

Question 63.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 83
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 84

Question 64.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 85
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 86
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 87

Question 65.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 88
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 89

Question 66.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 90
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 91

Question 67.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 92
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 93

Question 68.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 94

Question 69.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 95
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 96

Question 70.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 97
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 98

Question 71.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 99

Question 72.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 100
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 101

Question 73.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10A 102

Hope given RS Aggarwal Solutions Class 10 Chapter 10 Quadratic Equations Ex 10A 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 2 Polynomials Ex 2A

NCERT Maths Solutions for Ex 2.1 class 10 Polynomials is the perfect guide to boost up your preparation during CBSE 10th Class Maths Examination.

RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Ex 2A

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Ex 2A

Other Exercises

Find the zeros of the following quadratic polynomials and verify the relationship between the zeros and the coefficients:
Question 1.
Solution:
RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Ex 2A 1

Question 2.
Solution:
x² – 2x – 8
Let f(x) = x² – 2x – 8
RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Ex 2A 2

Question 3.
Solution:
RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Ex 2A 3

Question 4.
Solution:
4x² – 4x – 3
RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Ex 2A 4
RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Ex 2A 5

Question 5.
Solution:
RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Ex 2A 6
RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Ex 2A 7

Question 6.
Solution:
RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Ex 2A 8
RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Ex 2A 9
RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Ex 2A 10

Question 7.
Solution:
RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Ex 2A 11

Question 8.
Solution:
RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Ex 2A 12
RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Ex 2A 13

Question 9.
Solution:
RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Ex 2A 14

Question 10.
Solution:
RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Ex 2A 15
RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Ex 2A 16

Question 11.
Solution:
RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Ex 2A 17

Question 12.
Solution:
RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Ex 2A 18

Question 13.
Solution:
Zeros of a quadratic polynomial are 2, -6
RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Ex 2A 19

Question 14.
Solution:
RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Ex 2A 20

Question 15.
Solution:
Sum of zeros = 8
Product of zeros = 12
Quadratic equation will be x² – (Sum of zeros) x + Product of zeros = 0
=> x² – 8x + 12 = 0
=> x² – 6x – 2x + 12 = 0
=> x (x – 6) – 2 (x – 6) = 0
=> (x – 6) (x – 2) = 0
Either x – 6 = 0, then x = 6
or x – 2 = 0, then x = 2
Zeros are 6, 2
and quadratic polynomial is x² – 8x + 12

Question 16.
Solution:
Sum of zeros = 0
and product of zeros = -1
Quadratic equation will be
x² – (Sum of zeros) x + Product of zeros = 0
=> x² – 0x – 1 = 0
=> x² – 1= 0
(x + 1)(x – 1) = 0
Either x + 1 = 0, then x = -1 or x – 1 =0, then x = 1
Zeros are 1, -1
and quadratic polynomial is x² – 1

Question 17.
Solution:
Sum of zeros = \(\frac { 5 }{ 2 }\)
Product of zeros = 1
Quadratic equation will be
x² – (Sum of zeros) x + Product of zeros = 0
RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Ex 2A 21
and quadratic polynomial is 2x² – 5x + 2

Question 18.
Solution:
RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Ex 2A 22

Question 19.
Solution:
RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Ex 2A 23

Question 20.
Solution:
RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Ex 2A 24

Question 21.
Solution:
One zero of the given polynomial is \(\frac { 2 }{ 3 }\)
RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Ex 2A 25
=> (x + 3) (x + 3) = 0
x = -3, -3
Hence, other zeros are -3, -3

Hope given RS Aggarwal Solutions Class 10 Chapter 2 Polynomials Ex 2A 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 3 Linear equations in two variables Ex 3B

Are you looking for the best Maths NCERT Solutions Chapter 3 Ex 3.2 Class 10? Then, grab them from our page and ace up your preparation for CBSE Class 10 Exams.

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3B. You must go through NCERT Solutions for Class 10 Maths to get better score in CBSE Board exams along with RS Aggarwal Class 10 Solutions.

RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B

Solve for x and y:
Question 1.
Solution:
x + y = 3 …..(i)
4x – 3y = 26 …(ii)
From (i), x = 3 – y
Substituting the value of x in (ii),
4(3 – y) – 3y = 26
=> 12 – 4y – 3y = 26
-7y = 26 – 12 = 14
y = -2
x = 3 – y = 3 – (-2) = 3 + 2 = 5
Hence, x = 5, y = -2

Question 2.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 1

Question 3.
Solution:
2x + 3y= 0 ……..(i)
3x + 4y = 5 …….(ii)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 2

Question 4.
Solution:
2x – 3y = 13 ……(i)
7x – 2y = 20 ……….(ii)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 3

Question 5.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 4
=> x = -2, y = -5

Question 6.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 5

Question 7.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 6

Question 8.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 7

Question 9.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 8
Hence, x = \(\frac { 3 }{ 2 }\) , y = \(\frac { -2 }{ 3 }\)

Question 10.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 9

Question 11.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 10
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 11

Question 12.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 12

Question 13.
Solution:
0.4x + 0.3y = 1.7
0.7x – 0.2y = 0.8
Multiplying each term by 10
4x + 3y = 17
0.7x – 2y = 8
Multiply (i) by 2 and (ii) by 3,
8x + 6y = 34
21x – 6y = 24
Adding, we get
29x = 58
x = 2
From (i) 4 x 2 + 3y = 17
=> 8 + 3y = 17
=> 3y = 17 – 8 = 9
y = 3
x = 2, y = 3

Question 14.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 13

Question 15.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 14
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 15

Question 16.
Solution:
6x + 5y = 7x + 3y + 1 = 2(x + 6y – 1)
6x + 5y = 7x + 3y + 1
=> 6x + 5y – 7x – 3y = 1
=> -x + 2y = 1
=> 2y – x = 1 …(i)
7x + 3y + 1 = 2(x + 6y – 1)
7x + 3y + 1 = 2x + 12y – 2
=> 7x + 3y – 2x – 12y = -2 – 1
=> 5x – 9y = -3 …..(ii)
From (i), x = 2y – 1
Substituting the value of x in (ii),
5(2y – 1) – 9y = -3
=> 10y – 5 – 9y = -3
=> y = -3 + 5
=> y = +2
x = 2y – 1 = 2 x 2 – 1 = 4 – 1 = 3
x = 3, y = 2

Question 17.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 16
and x = y – 4 = 6 – 4 = 2
x = 2, y = 6

Question 18.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 17

Question 19.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 18
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 19

Question 20.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 20
x = 3, y = -1

Question 21.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 21
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 22

Question 22.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 23
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 24

Question 23.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 25
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 26

Question 24.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 27
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 28

Question 25.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 29
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 30

Question 26.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 31
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 32

Question 27.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 33
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 34
x = 3, y = 2

Question 28.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 35
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 36

Question 29.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 37
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 38

Question 30.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 39
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 40

Question 31.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 41
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 42
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 43

Question 32.
Solution:
71x + 37y = 253
37x + 71y = 287
Adding, we get
108x + 108y = 540
x + y = 5 ………(i) (Dividing by 108)
and subtracting,
34x – 34y = -34
x – y = -1 ……..(ii) (Dividing by 34)
Adding, (i) and (ii)
2x = 4 => x = 2
and subtracting,
2y = 6 => y = 3
Hence, x = 2, y = 3

Question 33.
Solution:
217x + 131y = 913 …(i)
131x + 217y = 827 …..(ii)
Adding, we get
348x + 348y = 1740
x + 7 = 5 …..(iii) (Dividing by 348)
and subtracting,
86x – 86y = 86
x – y = 1 …(iv) (Dividing by 86)
Now, adding (iii) and (iv)
2x = 6 => x = 3
and subtracting,
2y = 4 => y = 2
x = 3, y = 2

Question 34.
Solution:
23x – 29y = 98 ……(i)
29x – 23y = 110 ……(ii)
Adding, we get
52x – 52y = 208
x – y = 4 ……(iii) (Dividing by 52)
and subtracting
-6x – 6y = -12
x + y = 2 …..(iv)
Adding (iii), (iv)
2x = 6 => x = 3
Subtracting (iii) from (iv)
2y = -2 => y = -1
x = 3, y = -1

Question 35.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 44
x = 1 and y = 2

Question 36.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 45
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 46

Question 37.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 47
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 48
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 49

Question 38.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 50
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 51

Question 39.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 52
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 53

Question 40.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 54

Question 41.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 55
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 56

Question 42.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 57

Question 43.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 58
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 59

Question 44.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 60
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 61
a = \(\frac { 1 }{ 2 }\) , b = \(\frac { 1 }{ 3 }\)

Question 45.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 62

Question 46.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 63

Question 47.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 64

Question 48.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 65

Question 49.
Solution:
a²x + b²y = c² ……(i)
b²x + a²y = d² …….(ii)
Multiply (i) by a² and (ii) by b²,
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 66
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 67

Question 50.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 68
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 69
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3B 70

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RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3E

NCERT Maths Solutions for Ex 3.5 class 10 Linear equations is the perfect guide to boost up your preparation during CBSE 10th Class Maths Examination.

RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3E

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3E. You must go through NCERT Solutions for Class 10 Maths to get better score in CBSE Board exams along with RS Aggarwal Class 10 Solutions.

Question 1.
Solution:
Let cost of one chair = ₹ x
and cost of one table = ₹ y
According to the conditions,
5x + 4y = ₹ 5600 …(i)
4x + 3y = ₹ 4340 …(ii)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3E 1

x = -560
and from (i)
5 x 560 + 4y = 5600
2800 + 4y = 5600
⇒ 4y = 5600 – 2800
⇒ 4y = 2800
⇒ y = 700
Cost of one chair = ₹ 560
and cost of one table = ₹ 700

Question 2.
Solution:
Let the cost of one spoon = ₹ x and cost of one fork = ₹ y
According to the conditions,
23x + 17y = 1770 …(i)
17x + 23y = 1830 …(ii)
Adding, we get
40x + 40y = 3600
Dividing by 40,
x + y = 90 …(iii)
and subtracting,
6x – 6y = -60
Dividing by 6,
x – y = -10 …(iv)
Adding (iii) and (iv)
2x = 80 ⇒ x = 40
and subtracting,
2y = 100 ⇒ y = 50
Cost of one spoon = ₹ 40
and cost of one fork = ₹ 50

Question 3.
Solution:
Let number of 25-paisa coins = x
and number 50-paisa coins = y
Total number of coins = 50
and total amount = ₹ 19.50 = 1950 paisa
x + y = 50 …(i)
25x + 50y = 1950
⇒ x + 2y = 78 …(ii)
Subtracting (i) from (ii), y = 28
x = 50 – y = 50 – 28 = 22
Number of 25-paisa coins = 22
and 50-paisa coins = 28

Question 4.
Solution:
Sum of two numbers = 137
and difference = 43
Let first number = x
and second number = y
x + y = 137 …..(i)
x – y = 43 ……(ii)
Adding, we get
2x = 180 ⇒ x = 90
and subtracting,
2y = 94
y = 47
First number = 90
and second number = 47

Question 5.
Solution:
Let first number = x
and second number = y
According to the conditions,
2x + 3y = 92 …(i)
4x – 7y = 2 …(ii)
Multiply (i) by 2 and (ii) by 1
4x + 6y = 184 …..(iii)
4x – 7y = 2 …….(iv)
Subtracting (iii) from (iv),
13y = 182
y = 14
From (i), 2x + 3y = 92
2x + 3 x 14 = 92
⇒ 2x + 42 = 92
⇒ 2x = 92 – 42 = 50
⇒ x = 25
First number = 25
Second number = 14

Question 6.
Solution:
Let first number = x
and second number = y
According to the conditions,
3x + y=142 …(i)
4x – y = 138 …(ii)
Adding, we get
7x = 280
⇒ x = 40
and from (i)
3 x 40 + y = 142
⇒ 120 + y = 142
⇒ y = 142 – 120 = 22
First number = 40,
second number = 22

Question 7.
Solution:
Let first greater number = x
and second smaller number = y
According to the conditions,
2x – 45 = y …(i)
2y – 21 = x …(ii)
Substituting the value of y in (ii),
2 (2x – 45) – 21 = x
⇒ 4x – 90 – 21 = x
⇒ 4x – x = 111
⇒ 3x = 111
⇒ x = 37
From (i),
y = 2 x 37 – 45 = 74 – 45 = 29
The numbers are 37, 29

Question 8.
Solution:
Let larger number = x
and smaller number = y
According to the conditions,
3x = 4 x y + 8 ⇒ 3x = 4y + 8 …….(i)
5y = x x 3 + 5 ⇒ 5y = 3x + 5 …(ii)
Substitute the value of 3x in (ii),
5y = 4y + 8 + 5
⇒ 5y – 4y = 13
⇒ y = 13
and 3x = 4 x 13 + 8 = 60
⇒ x = 20
Larger number = 20
and smaller number = 13

Question 9.
Solution:
Let first number = x and
second number = y
According to the conditions,
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3E 2
⇒ 11x – 44 = 5(2x + 2) – 20
⇒ 11x – 44 = 10x + 10 – 20
⇒ 11x – 10x = 10 – 20 + 44
⇒ x = 34
and y = 2 x 34 + 2 = 68 + 2 = 70
Numbers are 34 and 70

Question 10.
Solution:
Let first number = x
and second number (smaller) = y
According to the conditions,
x – y = 14
and x² – y² = 448
⇒ (x + y) (x – y) = 448
⇒ (x + y) x 14 = 448
⇒ x + y = 32 ……(i)
and x – y = 14 ……(ii)
Adding (i) and (ii),
2x = 46 ⇒ x = 23
and subtracting (i) and (ii),
2y = 18 ⇒ y = 9
Numbers are 23, 9

Question 11.
Solution:
Let ones digit of a two digit number = x
and tens digit = y
Number = x + 10y
By interchanging the digits,
Ones digit = y
and tens digit = x
Number = y + 10x
According to the conditions,
x + y = 12 ………. (i)
y + 10x = x + 10y + 18
⇒ y + 10x – x – 10y = 18
⇒ 9x – 9y = 18
⇒ x – y = 2 …(ii) (Dividing by 9)
Adding (i) and (ii),
2x = 14 ⇒ x = 7
and subtracting,
2y = 10 ⇒ y = 5
Number = 7 + 10 x 5 = 7 + 50 = 57

Question 12.
Solution:
Let one’s digit of a two digit number = x
and ten’s digit = y
Then number = x + 10y
After reversing the digits,
Ones digit = y
and ten’s digit = x
and number = y + 10x
According to the conditions,
x + 10y – 27 = y + 10x
⇒ y + 10x – x – 10y = -27
⇒ 9x – 9y = -27
⇒ x – y = -3 …(i)
and 7 (x + y) = x + 10y
7x + 7y = x+ 10y
⇒ 7x – x = 10y – 7y
⇒ 6x = 3y
⇒ 2x = y …(ii)
Substituting the value of y in (i)
x – 2x = -3
⇒ -x = -3
⇒ x = 3
y = 2x = 2 x 3 = 6
Number = x + 10y = 3 + 10 x 6 = 3 + 60 = 63

Question 13.
Solution:
Let one’s digit of a two digit number = x
and ten’s digit = y
Then number = x + 10y
After interchanging the digits,
One’s digit = y
and ten’s digit = x
Then number = y + 10x
According to the conditions,
y + 10x = x + 10y + 9
⇒ y + 10x – x – 10y = 9
⇒ 9x – 9y = 9
⇒ x – y = 1 …(i)
and x + y= 15 …(ii)
Adding, we get
2x = 16
x = 8
and subtracting,
2y = 14
⇒ y = 7
Number = x + 10y = 8 + 10 x 7 = 8 + 70 = 78

Question 14.
Solution:
Let one’s digit of the two digit number = x
and ten’s digit = y
Then number = x + 10y
By reversing the digits,
One’s digit = y
and ten’s digit = x
Then number = y + 10x
Now, according to the conditions,
x + 10y + 18 = y + 10x
⇒ 18 = y + 10x – x – 10y
⇒ 9x – 9y = 18
⇒ x – y = 2 …(i)
and 4(x + y) + 3 = x + 10y
4x + 4y + 3 = x + 10y
⇒ 4x + 4y – x – 10y = -3
3x – 6y = -3
⇒ x – 2y = -1 ……..(ii)
Subtracting,
y = 3
and x = 2y – 1 = 2 x 3 – 1 = 6 – 1 = 5
Number = x + 10y = 5 + 10 x 3 = 5 + 30 = 35

Question 15.
Solution:
Let ones digit of a two digit number = x
and tens digit = y
Then number = x + 10y
By reversing the digits,
One’s digit = y
and ten’s digit = x
and number = y + 10x
According to the conditions,
x + 10y – 9 = y + 10x
⇒ x + 10y – y – 10x = 9
⇒ -9x + 9y = 9
⇒x – y = -1 …(i) (Dividing by -9)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3E 3

Question 16.
Solution:
Let the one’s digit of a two digit number = x
and ten’s digit = y
Then number = x + 10y
By interchanging the digits,
One’s digit = y
and ten’s digit = x
Then number = y + 10x
According to the conditions,
x + 10y + 18 = y + 10x
⇒ 18 = y + 10x – x – 10y
⇒ 9x – 9y = 18
⇒ x – y = 2 …(i)
and xy = 35 …(ii)
Now, (x + y)² = (x – y)² + 4xy = (2)² + 4 x 35 = 4 + 140 = 144 = (12)²
⇒ (x + y) = 12 …(iii)
Subtracting (i) from (iii), we get
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3E 4

Question 17.
Solution:
Let one’s digit of a two digit number = x
and ten’s digit = y
Then number = x + 10y
After interchanging the digits One’s digit = y
Ten’s digit = x
Then number = y + 10x
According to the conditions,
x + 10y – 63 = y + 10x
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3E 5

Question 18.
Solution:
Let one’s digit of a two digit number = x
and ten’s digit = y
Number = x + 10y
By reversing the digits,
One’s digit = y
and ten’s digit = x
Number = y + 10x
According to the conditions,
x + 10y + y + 10x = 121
⇒ 11x + 11y = 121
⇒ x + y = 11 …(i)
x – y = 3 …(ii)
Adding, we get
2x = 14 ⇒ x = 7
Subtracting,
2y = 8 ⇒ y = 4
Number = 7 + 10 x 4 = 7 + 40 = 47
or 4 + 10 x 7 = 4 + 70 = 74

Question 19.
Solution:
Let numerator of a fraction = x
and denominator = y
Then fraction = \(\frac { x }{ y }\)
According to the conditions,
x + y = 8 …(i)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3E 6

Question 20.
Solution:
Let numerator of a fraction = x
and denominator = y
Then fraction = \(\frac { x }{ y }\)
According to the conditions,
\(\frac { x + 2 }{ y }\) = \(\frac { 1 }{ 2 }\)
\(\frac { x }{ y – 1 }\) = \(\frac { 1 }{ 3 }\)
⇒ 2x + 4 = y …(i)
3x = y – 1 …(ii)
⇒ 3x = 2x + 4 – 1
⇒ 3x = 2x + 3
⇒ 3x – 2x = 3
⇒ x = 3
and y = 2x + 4 = 2 x 3 + 4 = 6 + 4 = 10
Fraction = \(\frac { 3 }{ 10 }\)

Question 21.
Solution:
Let numerator of a fraction = x
and denominator = y
Then fraction = \(\frac { x }{ y }\)
According to the conditions,
y – x = 11
y = 11 + x …(i)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3E 7

Question 22.
Solution:
Let numerator of a fraction = x
and denominator = y
Then fraction = \(\frac { x }{ y }\)
According to the conditions,
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3E 8

Question 23.
Solution:
Let numerator of a fraction = x
and denominator = y
Then fraction =
According to the conditions,
x + y = 4 + 2x
⇒ y = 4 + x …(i)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3E 9
Fraction = \(\frac { x }{ y }\) = \(\frac { 5 }{ 9 }\)

Question 24.
Solution:
Let first number = x
and second number = y
According to the conditions,
x + y = 16
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3E 10

Question 25.
Solution:
Let in classroom A, the number of students = x
and in classroom B = y
According to the conditions,
x – 10 = y + 10
⇒ x – y = 10 + 10 = 20
⇒ x – y = 20 …(i)
and x + 20 = 2 (y – 20)
⇒ x + 20 = 2y – 40
⇒ x – 2y = -(40 + 20) = -60
x – 2y = -60 …(ii)
Subtracting, y = 80
and x – y = 20
⇒ x – 80 = 20
⇒ x = 20 + 80 = 100
Number of students in classroom A = 100 and in B = 80

Question 26.
Solution:
Let fixed charges = ₹ x
and other charges = ₹ y per km
According to the conditions,
For 80 km,
x + 80y = ₹ 1330 …(i)
and x + 90y = ₹ 1490 …(ii)
Subtracting (i) from (ii),
10y = 160 ⇒ y = 16
and from (i)
x + 80 x 16 = 1330
⇒ x + 1280 = 1330
⇒ x = 1330 – 1280 = 50
Fixed charges = ₹ 50
and rate per km = ₹ 16

Question 27.
Solution:
Let fixed charges of the hostel = ₹ x
and other charges per day = ₹ y
According to the conditions,
x + 25y = 4500 ……..(i)
x + 30y = 5200 ……(ii)
Subtracting (i) from (ii),
5y = 700
y = 140
and from (i),
x + 25 x 140 = 4500
⇒ x + 3500 = 4500
⇒ x = 4500 – 3500 = 1000
Fixed charges = ₹ 1000
and per day charges = ₹ 140

Question 28.
Solution:
Let first investment = ₹ x
and second investment = ₹ y
Rate of interest = 10% p.a. for first kind and 8% per second
Interest is for the first investment = ₹ 1350
and for the second = ₹ 1350 – ₹45 = ₹ 1305
According to the conditions,
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3E 11
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3E 12

Question 29.
Solution:
Ratio in the income of A and B = 5 : 4
Let A’s income = ₹ 5x and
B’s income = ₹ 4x
and ratio in their expenditures = 7 : 5
Let A’s expenditure = 7y
and B’s expenditure = 5y
According to the conditions,
5x – 7y = 9000 …(i)
and 4x – 5y = 9000 …(ii)
Multiply (i) by 5 and (ii) by 7,
25x – 35y = 45000
28x – 37y = 63000
Subtracting, we get
3x = 18000
⇒ x = 6000
A’s income = 5x = 5 x 6000 = ₹ 30000
and B’s income = 4x = 4 x 6000 = ₹ 24000

Question 30.
Solution:
Let cost of one chair = ₹ x
and cost of one table = ₹ y
In first case,
Profit on chair = 25%
and on table = 10%
and selling price = ₹ 1520
According to the conditions,
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3E 13

Question 31.
Solution:
Distance between two stations A and B = 70 km
Let speed of first car (starting from A) = x km/hr
and speed of second car = y km/hr
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3E 14
According to the conditions,
7x – 7y = 70
⇒ x – y = 10 …(i)
and x + y = 70 …(ii)
Adding (i) and (ii),
2x = 80 ⇒ x = 40
Subtracting (i) and (ii),
2y = 60 ⇒ y = 30
Speed of car A = 40 km/hr
and speed of car B = 30 km/hr

Question 32.
Solution:
Let uniform speed of the train = x km/hr
and time taken = y hours
Distance = x x y = xy km
Case I:
Speed = (x + 5) km/hr
and Time = (y – 3) hours
Distance = (x + 5) (y – 3)
(x + 5) (y – 3) = xy
⇒ xy – 3x + 5y – 15 = xy
-3x + 5y = 15 …(i)
Case II:
Speed = (x – 4) km/hr
and Time = (y + 3) hours
Distance = (x – 4) (y + 3)
(x – 4) (y + 3) = xy
⇒ xy + 3x – 4y – 12 = xy
3x – 4y = 12 …(ii)
Adding (i) and (ii),
y = 27
and from (i),
-3x + 5 x 27 = 15
⇒ -3x + 135 = 15
⇒ -3x = 15 – 135 = -120
⇒ x = 40
Speed of the train = 40 km/hr
and distance = 27 x 40 = 1080 km

Question 33.
Solution:
Let the speed of the train = x km/hr
and speed of taxi = y km/hr
According to the conditions,
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3E 15
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3E 16

Question 34.
Solution:
Distance between stations A and B = 160 km
Let the speed of the car starts from A = x km/hr
and speed of car starts from B = y km/hr
8x – 8y = 160
⇒ x – y = 20 …(i)
and 2x + 2y = 160
⇒ x + y = 80 …(ii)
Adding (i) and (ii)
2x = 100 ⇒ x = 50
and subtracting,
2y = 60 ⇒ y = 30
Speed of car starting from A = 50 km/hr
and from B = 30 km/hr

Question 35.
Solution:
Distance = 8 km
Let speed of sailor in still water = x km/hr
and speed of water = y km/hr
According to the conditions,
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3E 17
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3E 18

Question 36.
Solution:
Let speed of a boat = x km/hr
and speed of stream = y km/hr
According to the conditions,
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3E 19
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3E 20

Question 37.
Solution:
Let a man can do a work in x days
His 1 day’s work = \(\frac { 1 }{ x }\)
and a boy can do a work in y days
His 1 day’s work = \(\frac { 1 }{ y }\)
According to the conditions,
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3E 21
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3E 22

Question 38.
Solution:
Let length of a room = x m
and breadth = y m
and area = xy m²
According to the conditions,
x = y + 3 …(i)
(x + 3) (y – 2) = xy
xy – 2x + 3y – 6 = xy
-2x + 3y = 6 …(ii)
-2 (y + 3) + 3y = 6 [From (i)]
-2y – 6 + 3y = 6
⇒ y = 6 + 6 = 12
x = y + 3 = 12 + 3 = 15 …(ii)
Length of room = 15 m
and breadth = 12 m

Question 39.
Solution:
Let length of a rectangle = x m
and breadth = y m
Then area = x x y = xy m²
According to the conditions,
(x – 5) (y + 3) = xy – 8
⇒ xy + 3x – 5y – 15 = xy – 8
⇒ 3x – 5y = -8 + 15 = 7 …..(i)
and (x + 3) (y + 2) = xy + 74
⇒ xy + 2x + 3y + 6 = xy + 74
⇒ 2x + 3y = 74 – 6 = 68 …(ii)
Multiply (i) by 3 and (ii) by 5
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3E 23

Question 40.
Solution:
Let length of a rectangle = x m
and breadth = y m
Then area = xy m²
According to the conditions,
(x + 3) (y – 4) = xy – 67
⇒ xy – 4x + 3y – 12 = xy – 67
⇒ -4x + 3y = -67 + 12 = -55
⇒ 4x – 3y = 55 …(i)
and (x – 1) (y + 4) = xy + 89
⇒ xy + 4x – y – 4 = xy + 89
⇒ 4x – y = 89 + 4 = 93 ….(ii)
⇒ y = 4x – 93
Substituting the value of y in (i),
4x – 3(4x – 93) = 55
⇒ 4x – 12x + 279 = 55
⇒ -8x = 55 – 279 = -224
⇒ x = 28
and y = 4x – 93 = 4 x 28 – 93 = 112 – 93 = 19
Length of rectangle = 28 m
and breadth = 19 m

Question 41.
Solution:
Let reservation charges = ₹ x
and cost of full ticket from Mumbai to Delhi
According to the conditions,
x + y = 4150 …(i)
2x + \(\frac { 3 }{ 2 }\) y = 6255
⇒ 4x + 3y = 12510 …(ii)
From (i), x = 4150 – y
Substituting the value of x in (ii),
4 (4150 – y) + 3y = 12510
⇒ 16600 – 4y + 3y = 12510
-y = 12510 – 16600
-y = -4090
⇒ y = 4090
and x = 4150 – y = 4150 – 4090 = 60
Reservation charges = ₹ 60
and cost of 1 ticket = ₹ 4090

Question 42.
Solution:
Let present age of a man = x years
and age of a son = y years
5 year’s hence,
Man’s age = x + 5 years
and son’s age = y + 5 years
x + 5 = 3 (y + 5) = 3y + 15
⇒ x – 3y = 15 – 5 = 10
x = 10 + 3y …(i)
and 5 years ago,
Man’s age = x – 5 years
and son’s age = y – 5 years
x – 5 = 7 (y – 5) = 7y – 35
x = 7y – 35 + 5 = 7y – 30 …(ii)
From (i) and (ii),
10 + 3y = 7y – 30
⇒ 7y – 3y = 10 + 30
⇒ 4y = 40
⇒ y = 10
and x = 10 + 3y = 10 + 3 x 10 = 10 + 30 = 40
Present age of a man = 40 years
and of son’s age = 10 years

Question 43.
Solution:
Let present age of a man = x years
and age of his son = y years
2 years ago,
Man’s age = x – 2 years
Son’s age = y – 2 years
x – 2 = 5 (y – 2)
⇒ x – 2 = 5y – 10
x = 5y – 10 + 2 = 5y – 8 …(i)
2 years later,
Man’s age = x + 2 years
and son’s age = y + 2 years
x + 2 = 3(y + 2) + 8
x + 2 = 3y + 6 + 8
⇒ x = 3y + 6 + 8 – 2 = 3y + 12 …(ii)
From (i) and (ii),
5y – 8 = 3y + 12
⇒ 5y – 3y = 12 + 8
⇒ 2y = 20
⇒ y = 10
and x = 5y – 8 = 5 x 10 – 8 = 50 – 8 = 42
Present age of man = 42 years
and age of son = 10 years

Question 44.
Solution:
Let age of father = x years
and age of his son = y years
According to the conditions,
2y + x = 10 …(i)
2x + y = 95 …(ii)
From (i),
x = 70 – 2y
Substituting the value of x in (ii),
2 (70 – 2y) + y = 95
⇒ 140 – 4y + y = 95
⇒ -3y = 95 – 140 = -45
⇒ -3y = -45
⇒ y = 15
and x = 70 – 2y = 70 – 2 x 15 = 70 – 30 = 40
Age of father = 40 years
and age of his son = 15 years

Question 45.
Solution:
Let present age of a woman = x years
and age of her daughter = y years
According to the conditions,
x = 3y + 3 …(i)
3 years hence,
Age of woman = x + 3 years
and age of her daughter = y + 3 years
x + 3 = 2 (y + 3) + 10
⇒ x + 3 = 2y + 6 + 10
⇒x = 2y + 16 – 3 = 2y + 13 …(ii)
From (i),
3y + 3 = 2y + 13
⇒ 3y – 2y = 13 – 3
⇒ y = 10
and x = 3y + 3 = 3 x 10 + 3 = 30 + 3 = 33
Present age of woman = 33 years
and age of her daughter = 10 years

Question 46.
Solution:
Let cost price of tea set = ₹ x
and of lemon set = ₹ y
According to the conditions,
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3E 24

Question 47.
Solution:
Let fixed charges = ₹ x (for first three days)
and then additional charges for each day = ₹ y
According to the conditions,
Mona paid ₹ 27 for 7 dyas
x + (7 – 3) x y = 27
⇒ x + 4y = 27
and Tanvy paid ₹ 21 for 5 days
x + (5 – 3) y = 21
⇒ x + 2y = 21 …(ii)
Subtracting,
2y = 6 ⇒ y = 3
But x + 2y = 21
⇒ x + 2 x 3 = 21
⇒ x + 6 = 21
⇒ x = 21 – 6 = 15
Fixed charges = ₹ 15
and additional charges per day = ₹ 3

Question 48.
Solution:
Let x litres of 50% solution be mixed with y litres of 25% solution, then
x + y = 10 …(i)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3E 25
Subtracting (i) from (ii),
x = 6
and x + y = 10
⇒ 6 + y = 10
⇒ y = 10 – 6 = 4
50% solution = 6 litres
and 25% solution = 4 litres

Question 49.
Solution:
Let x g of 18 carat be mixed with y g of 12 carat gold to get 120 g of 16 carat gold, then
x + y = 120 …(i)
Now, gold % in 18-carat gold = \(\frac { 18 }{ 24 }\) x 100 = 75%
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3E 26
⇒ 3x + 2y = 320 …(ii)
From (i),
x = 120 – y
Substituting the value of x in (ii),
3 (120 – y) + 2y = 320
⇒ 360 – 3y + 2y = 320
⇒ -y = 320 – 360
⇒ -y = -40
⇒ y = 40
and 40 + x = 120
⇒ x = 120 – 40 = 80
Hence, 18 carat gold = 80 g
and 12-carat gold = 40 g

Question 50.
Solution:
Let x litres of 90% pure solution be mixed withy litres of 97% pure solution to get 21 litres of 95% pure solution. Then,
x + y = 21 …(i)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3E 27
⇒ 90x + 97y = 1995
From (i), x = 21 – y
Substituting the value of x in (ii),
90 (21 – y) + 97y = 1995
⇒ 1890 – 90y + 97y = 1995
⇒ 7y = 1995 – 1890 = 105
⇒ y =15
and x = 21 – y = 21 – 15 = 6
90% pure solution = 6 litres
and 97% pure solution = 15 litres

Question 51.
Solution:
Let larger supplementary angle = x°
and smaller angle = y°
According to the conditions,
x + y = 180° …(i)
x = y + 18° …(ii)
From (i),
y + 18° + y = 180°
⇒ 2y = 180° – 18° = 162°
⇒ 2y = 162°
⇒ y = 81°
and x= 180°- 81° = 99°
Hence, angles are 99° and 81°

Question 52.
Solution:
In ∆ABC,
∠A = x, ∠B = (3x – 2)°, ∠C = y°, ∠C – ∠B = 9°
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3E 28

Question 53.
Solution:
In a cyclic quadrilateral ABCD,
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3E 29

Hope given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3E are helpful to complete your math homework.

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RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Test Yourself

RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Test Yourself

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Test Yourself.

RS Aggarwal Solutions Class 10 Chapter 3

MCQ
Question 1.
Solution:
(a)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Test Yourself 1

Question 2.
Solution:
(d)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Test Yourself 2
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Test Yourself 3

Question 3.
Solution:
(a)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Test Yourself 4

Question 4.
Solution:
(d)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Test Yourself 5

Short-Answer Questions
Question 5.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Test Yourself 6

Question 6.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Test Yourself 7

Question 7.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Test Yourself 8
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Test Yourself 9

Question 8.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Test Yourself 10

Question 9.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Test Yourself 11

Question 10.
Solution:
Let first number = x
and second number = y
According to the conditions, x – y = 26 …(i)
and x = 3y …..(ii)
From (i),
3y – y = 26
⇒ 2y = 26
⇒ y = 13
and x = 3 x 13 = 39
Numbers are 39 and 13

Short-Answer Questions (3 marks)
Question 11.
Solution:
23x + 29y = 98 …..(i)
29x + 23y = 110 …..(ii)
Adding, we get 52x + 52y = 208
x + y = 4 …..(iii) (Dividing by 52)
and subtracting,
-6x + 6y = -12
x – y = 2. …..(iv) (Dividing by -6)
Adding (iii) and (iv),
2x = 6 ⇒ x = 3
Subtracting,
2x = 2 ⇒ y = 1
Hence, x = 3, y = 1

Question 12.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Test Yourself 12
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Test Yourself 13
x = 1, y = \(\frac { 3 }{ 2 }\)

Question 13.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Test Yourself 14
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Test Yourself 15

Question 14.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Test Yourself 16

Question 15.
Solution:
Let cost of one pencil = ₹ x
and cost of one pen = ₹ y
According to the condition,
5x + 7y = 195 …(i)
7x + 5y= 153 …(ii)
Adding, (i) and (ii)
12x + 12y = 348
x + y = 29 ….(iii) (Dividing by 12)
and subtracting,
-2x + 2y = 42
-x + y = 21 …..(iv) (Dividing by -2)
Now, Adding (iii) and (iv),
2y = 50 ⇒ y = 25
and from (iv),
-x + 25 = 21 ⇒ -x = 21 – 25 = -4
x = 4
Cost of one pencil = ₹ 4
and cost of one pen = ₹ 25

Question 16.
Solution:
2x – 3y = 1, 4x – 3y + 1 = 0
2x – 3y = 1
2x = 1 + 3y
x = \(\frac { 1 + 3y }{ 2 }\)
Giving some different values to y, we get corresponding values of x as given below
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Test Yourself 17
Now plot the points (2, 1), (5, 3) and (-1, -1) on the graph and join them to get a line.
Similarly,
4x – 3y + 1 = 0
⇒ 4x = 3y – 1
⇒ x = \(\frac { 3y – 1 }{ 4 }\)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Test Yourself 18
Now plot the points (-1, -1), (-4, -5) and (2, 3) on the graph and join them to get another line which intersects the first line at the point (-1, -1).
Hence, x = -1, y = -1
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Test Yourself 19

Long-Answer Questions
Question 17.
Solution:
We know that opposite angles of a cyclic quadrilateral are supplementary.
∠A + ∠C = 180° and ∠B + ∠D = 180°
Now, ∠A = 4x° + 20°, ∠B = 3x° – 5°, ∠C = 4y° and ∠D = 7y° + 5°
But ∠A + ∠C = 180°
4x + 20° + 4y° = 180°
⇒ 4x + 4y = 180° – 20 = 160°
x + y = 40° …(i) (Dividing by 4)
and ∠B + ∠D = 180°
⇒ 3x – 5 + 7y + 5 = 180°
⇒ 3x + 7y = 180° …(ii)
From (i), x = 40° – y
Substituting the value of x in (ii),
3(40° – y) + 7y = 180°
⇒ 120° – 3y + 7y = 180°
⇒ 4y = 180°- 120° = 60°
y = 15°
and x = 40° – y = 40° – 15° = 25°
∠A = 4x + 20 = 4 x 25 + 20 = 100 + 20= 120°
∠B = 3x – 5 = 3 x 25 – 5 = 75 – 5 = 70°
∠C = 4y = 4 x 15 = 60°
∠D = 7y + 5 = 7 x 15 + 5 = 105 + 5 = 110°

Question 18.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Test Yourself 20
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Test Yourself 21

Question 19.
Solution:
Let numerator of a fraction = x
and denominator = y
Fraction = \(\frac { x }{ y }\)
According to the conditions,
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Test Yourself 22
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Test Yourself 23

Question 20.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Test Yourself 24

Hope given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Test Yourself 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 3 Linear equations in two variables MCQS

RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables MCQS

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables MCQS.

RS Aggarwal Solutions Class 10 Chapter 3

Choose the correct answer in each of the following questions.
Question 1.
Solution:
(c)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables MCQS 1

Question 2.
Solution:
(c)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables MCQS 2

Question 3.
Solution:
(a)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables MCQS 3

Question 4.
Solution:
(d)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables MCQS 4

Question 5.
Solution:
(a)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables MCQS 5
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables MCQS 6

Question 6.
Solution:
(b)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables MCQS 7
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables MCQS 8

Question 7.
Solution:
(c) 4x + 6y = 3xy, 8x + 9y = 5xy
Dividing each term by xy,
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables MCQS 9
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables MCQS 10

Question 8.
Solution:
(a)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables MCQS 11

Question 9.
Solution:
(c)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables MCQS 12

Question 10.
Solution:
(b)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables MCQS 13
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables MCQS 14

Question 11.
Solution:
(d)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables MCQS 15

Question 12.
Solution:
(b)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables MCQS 16
RS AggarRS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables MCQS 17wal Class 10 Solutions Chapter 3 Linear equations in two variables MCQS 16

Question 13.
Solution:
(a)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables MCQS 18

Question 14.
Solution:
(d)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables MCQS 19
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables MCQS 20

Question 15.
Solution:
(d)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables MCQS 21

Question 16.
Solution:
(d)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables MCQS 22
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables MCQS 23

Question 17.
Solution:
(d)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables MCQS 24

Question 18.
Solution:
(d) The system of equations is consistent then their graph lines will be either intersecting or coincident.

Question 19.
Solution:
(a) The pair of lines of equation is inconsistent then the system will not have no solution i.e., their lines will be parallel.

Question 20.
Solution:
(b)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables MCQS 25

Question 21.
Solution:
(b) ABCD is a cyclic quadrilateral
∠A = (x + y + 10)°, ∠B = (y + 20)°, ∠C = (x + y – 30)° and ∠D = (x + y)°
∠A + ∠C = 180°
Now, x + y + 10°+ x + y – 30° = 180°
⇒ 2x + 2y – 20 = 180°
⇒ 2x + 2y = 180° + 20° = 200°
⇒ x + y = 100° …(i)
and ∠B + ∠D = 180°
⇒ y + 20° + x + y = 180°
⇒ x + 2y = 180° – 20° = 160° …(ii)
Subtracting,
-y = -60° ⇒ y = 60°
and x + 60° = 100°
⇒ x = 100° – 60° = 40°
Now, ∠B = y + 20° = 60° + 20° = 80°

Question 22.
Solution:
(d) Let one’s digit of a two digit number = x
and ten’s digit = y
Number = x + 10y
By interchanging the digits,
One’s digit = y
and ten’s digit = x
Number = y + 10x
According to the conditions,
x + y = 15 …(i)
y + 10x = x + 10y + 9
⇒ y + 10x – x – 10y = 9
⇒ 9x – 9y = 9
⇒ x – y = 1 …(ii)
Adding (i) and (ii),
2x = 16 ⇒ x = 8
and x + y = 15
⇒ 8 + y = 15
⇒ y = 15 – 8 = 7
Number = x + 10y = 8 + 10 x 7 = 8 + 70 = 78

Question 23.
Solution:
(b) Let the numerator of a fractions = x
and denominator = y
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables MCQS 26

Question 24.
Solution:
(d) Let present age of man = x years
and age of his son = y years
5 years hence,
Age of man = (x + 5) years
and age of son = y + 5 years
(x + 5) = 3 (y + 5)
⇒ x + 5 = 3y + 15
x = 3y + 15 – 5
x = 3y + 10 ……(i)
and 5 years earlier
Age of man = x – 5 years
and age of son = y – 5 years
x – 5 = 7 (y – 5)
x – 5 = 7y – 35
⇒ x = 7y – 35 + 5
x = 7y – 30 ……….(ii)
From (i) and (ii),
7y – 30 = 3y + 10
⇒ 7y – 3y = 10 + 30
⇒ 4y = 40
y = 10
x = 3y + 10 = 3 x 10 + 10 = 30 + 10 = 40
Present age of father = 40 years

Question 25.
Solution:
(b)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables MCQS 27

Question 26.
Solution:
(c)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables MCQS 28

Question 27.
Solution:
(a)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables MCQS 29
The system has infinitely many solutions.
The lines are coincident.

Hope given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables MCQS 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.