Category: CBSE Class 8

RS Aggarwal Class 8 Solutions Chapter 5 Playing with Numbers Ex 5C

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 5 Playing with Numbers Ex 5C.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 5 Playing with Numbers Ex 5A
• RS Aggarwal Solutions Class 8 Chapter 5 Playing with Numbers Ex 5B
• RS Aggarwal Solutions Class 8 Chapter 5 Playing with Numbers Ex 5C
• RS Aggarwal Solutions Class 8 Chapter 5 Playing with Numbers Ex 5D

Replace A, B, C by suitable numerals.

Question 1.
Solution:
Here A can be as 6 + 7 = 13
Now 1 + 5 + 8 = 14
∴C = 1, B = 4, A = 6

Question 2.
Solution:
Here A can be 7, as 6+7 = 13
1 + B + 9 = 10 + B
∴B can be 7
∴10 + 7 = 17
1 + C + 6 = 7 + C
∴C can be 4
∴1 + 4 + 6 = 11
and 1 + 4 + 3 = 8
∴A = 7, B = 7, C = 4

Question 3.
Solution:
Here A + A + A = A
∴A can = 5
∴5 + 5 + 5 = 15
∴B = 1
Hence A = 5, B = 1

Question 4.
Solution:
6 – A = 3
1 + 5 – A = 3
5 – A = 3
∴A = 5 – 3 = 2
Now 2 – B = 7
=>12 – B = 7
∴B = 5
Hence A = 2, B = 5

Question 5.
Solution:
– 5 – A = 9 =>A = 5 – 9 or 15 – 9
= 6
∴A = 6
Now B – 1 – 8 = 5 =>B – 9 = 5
=>B = 5 + 9 = 14
∴B = 4
Now C – 1 – 2 = 2 =>C – 3 = 2
C = 2 + 3 = 5
∴A = 6, B = 4, C = 5

Question 6.
Solution:
B x 3 = B
∴B can be 5 or 0
∴5 x 3 = 15 => B = 5 or 3 x 0 =0
If B = 0, then A can be 5
∴3 x 5 = 15
∴A = 5 and C = 1
Hence A = 5, B = 0, C = 1

Question 7.
Solution:

∴AB = B
=>A = 1
and A² + B² – 1 + B² + C
∴B² +1 = C
∴B² in one digit
If B = 3
∴3² + 1 = 9 + 1 = 10 = C
∴C = 0
B x 1 + 1 = B + 1 = 3 + 1
Hence A = 1, B = 3, C = 0

Question 8.
Solution:
Here we see that 6 x 9 = 54
∴A – 4 = 3 => A = 3 + 4 = 7
and 6 x 6 = 36
3B = 36 => B = 6
and C = 6
Hence A = 7, B = 6, C = 6

Question 9.
Solution:
Product of two numbers = 1 -digit number
and sum = 2-digit numbers
Let first number = x
and second number = y
∴x X y = 1-digit number
x + y = 2-digit number
By hit and hail, we sec that
1 x 9 = 9 which is I-digit number
and 1 + 9 = 10 which is 2-digit number

Question 10.
Solution:
By hit and trail method, we see that
1 + 2 + 3 = 6 and 1 x 2 x 3 = 6
1, 2 and 3 are the required whole numbers
whose sum and product is same

Question 11.
Solution:
In the given square, we have to interest the numbers from 1 to 9, such that the sum in each raw, column on diagonal to be 15
So, we complete it as given here

Question 12.
Solution:
We shall complete the triangle by intersecting the numbers from 1 to 6 without repetition so that the sum in each side be 12

Question 13.
Solution:
The given numbers are
a, b (a + b), (a + 2b), (2a + 3b), (3a + 5b), (5a + 8b), (8a + 13b), (13a + 21b), and (21a + 34b)
Sum of there numbers = 11 (5a + 8b)
= 11 x 7th number
Now taking a = 8, b = 13, then the 10 number be 8, 13, 21, 34, 55, 89, 144, 233, 377, 610
Whose 7th number = 144
By adding these 10 numbers, we get the
sum
= 8+ 13 + 21 + 34 + 55 + 89 + 144 + 233 + 377 + 610 = 1584 and 11 x 7th number =11 x 144 , = 1584
Which is same in each case

Question 14.
Solution:
We see that in the magic box sum of 0 + 11 + 7 + 12 = 30
Now we shall complete this magic square, to get 30 as the sum in each row and column and also diagonal wise

Hope given RS Aggarwal Solutions Class 8 Chapter 5 Playing with Numbers Ex 5C are helpful to complete your math homework.

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RS Aggarwal Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3H

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3H.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3A
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3B
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3C
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3D
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3E
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3F
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3G
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3H

Question 1.
Solution:
5478 as it has 8 at in the end (c)

Question 2.
Solution:
2222 as it has 2 in the end (d)

Question 3.
Solution:
1843 as it has 3 in the end (a)

Question 4.
Solution:
4787 as it has 7 in the end (b)

Question 5.
Solution:
81000 as it has an odd number of zeros at its end (c)

Question 6.
Solution:
8, as the number with 8, in the end, cannot be a perfect square. (d)

Question 7.
Solution:
The square of a proper fraction is smaller than the given fraction. (b)

Question 8.
Solution:
1 + 3 + 5 + 7 + … to n terms when n is an odd is equal to n² (c)
Sum of first n odd natural numbers is n²

Question 9.
Solution:
Answer = (d)
(8)² + (15)²
= 64 + 225
= 289 = (17)²

Question 10.
Solution:
Answer = (c)

7 must be subtracted

Question 11.
Solution:
Finding the square root of 526 by division method

We get remainder = 41
Now (22)² = 484 and (23)² = 529
The least number to be added = 529 – 526 = 3 (a)

Question 12.
Solution:
Finding the square root of 15370 by division method
We get remainder = 261
Now (123)² = 15129
and (124)² = 15376
The least number to be added
= 15376 – 15370 = 6 (b)

Question 13.
Solution:
Answer = (d)

Question 14.
Solution:
√0.1 = 0.316 (c)

Question 15.
Solution:
Answer = (b)
\(\sqrt { 0.9\times 1.6 } \)
= \(\sqrt { 1.44 } \)
= \(\sqrt { 1.2\times 1.2 } \)
= 1.2

Question 16.
Solution:
Answer = (c)

Question 17.
Solution:
\(\sqrt { 2\frac { 1 }{ 4 } } \)
= \(\sqrt { \frac { 9 }{ 4 } } =\frac { 3 }{ 2 } \)
= \(1\frac { 1 }{ 2 } \) (b)

Question 18.
Solution:
We know that the square on an even number is also an even number. 196 is the square of an even number (a)

Question 19.
Solution:
We know that the square of an odd number is also an odd number

1369 is an odd number

Hope given RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3H are helpful to complete your math homework.

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RS Aggarwal Class 8 Solutions Chapter 6 Operations on Algebraic Expressions Ex 6E

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6E.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6A
• RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6B
• RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6C
• RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6D
• RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6E

Objective questions :
Tick the correct answer in each of the following :

Question 1.
Solution:

Question 2.
Solution:
(3q + 7p² – 2r³ + 4) – (4p² – 2q + 7r³ – 3)
= 3q + 7p² – 2r³ + 4 – 4p² + 2q – 7r³ + 3
= 5q + 3p² – 9r³ + 7 = 3p² + 5q – 9r³ + 7 (d)

Question 3.
Solution:
(x + 5) (x – 3) = x² + (5 – 3) x + 5 X ( – 3)
= x² + 2x – 15 (d)

Question 4.
Solution:
(2x + 3) (3x – 1)
= 6x² – 2x + 9x – 3
= 6x² + 7x – 3 (b)

Question 5.
Solution:
(x + 4) (x + 4)
= x² + (4 + 4) x + 4 X 4
= x² + 8x + 16 (c)

Question 6.
Solution:
(x – 6) (x – 6)
= x² + ( – 6 – 6) x + ( – 6) ( – 6)
= x² – 12x + 36 (d)

Question 7.
Solution:
(2x + 5) (2x – 5)
= (2x)² – (5)²
= 4x² – 25

Question 8.
Solution:

Question 9.
Solution:
(2x² + 3x + 1) ÷ (x + 1)

= 2x +1 (b)

Question 10.
Solution:
(x² – 4x + 4) ÷ (x – 2)

= x – 2 (a)

Question 11.
Solution:
(a + 1) (a – 1) (a² + 1)
= (a² – 1) (a² + 1)
= a⁴ (c)

Question 12.
Solution:

Question 13.
Solution:
\(\left( { x }+\frac { 1 }{ { x } } \right) =5\)
Squaring on both sides

Question 14.
Solution:
\(\left( { x }-\frac { 1 }{ { x } } \right) =6\)
Squaring on both sides

Question 15.
Solution:
(82)² – (18)²
= (82 + 18) (82 – 18)
= 100 x 64
= 6400 (c)

Question 16.
Solution:
(197 x 203)
= (200 – 3) (200 + 3)
= (200)² – (3)²
= 40000 – 9
= 39991 (a)

Question 17.
Solution:
a + b = 12, ab = 14
a² + b²
= (a + b)² – 2ab
= (12)² – 2 x 14
= 144 – 28
= 16 (b)

Question 18.
Solution:
a – b = 7, ab = 9
a² + b²
= (a – b)² + 2ab
= (7)² + 2 x 9
= 49 + 18
= 67 (a)

Question 19.
Solution:
x = 10
4x² + 20x + 25
= (2x)² + 2 x 2x x 5 + (5)²
= (2x + 5)²
= (2 x 10 + 5)
= (20 + 5)²
= (25)²
= 625 (c)

Hope given RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6E are helpful to complete your math homework.

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RS Aggarwal Class 8 Solutions Chapter 5 Playing with Numbers Ex 5A

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 5 Playing with Numbers Ex 5A.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 5 Playing with Numbers Ex 5A
• RS Aggarwal Solutions Class 8 Chapter 5 Playing with Numbers Ex 5B
• RS Aggarwal Solutions Class 8 Chapter 5 Playing with Numbers Ex 5C
• RS Aggarwal Solutions Class 8 Chapter 5 Playing with Numbers Ex 5D

Question 1.
Solution:
Let tens digit = x
Units digit = 3
∴Number = 3 + 10x
According to the condition,
7 (x + 3) = 3 + 10x
7x + 21 = 3 + 10x
21 – 3 = 10x – 7x
=> 3x = 18
x = \(\\ \frac { 18 }{ 3 } \)
∴Number = 3 + 10x
= 3 + 10 x 6 = 3 + 60 = 63

Question 2.
Solution:
Let ten’s digit = x
Then units digit = 2x
and number = 10x + 2x = 12x
According to the condition,
12x = x + 2x + 18
12x – x – 2x = 18
=> 9x = 18
x = \(\\ \frac { 18 }{ 9 } \) = 2
∴Number = 12x = 2 x 12 = 24

Question 3.
Solution:
Let units digit = x
and tens digit = y
Number = x + 10y
Now x + 10y = 4 (x + y) + 3
=> x + 10y = 4x + 4y + 3
10y – 4y – 4x + x = 3
=> 6y – 3x = 3
2y – x = 1 ….(i)
∴Number by reversing the order of digits = y + 10x
=>x + 10y + 18 = y + 10x
=>10x – x + y – 10y = 18
=> 9x – 9y = 18
x – y = 2 ….(ii)
∴Adding (i) and (ii)
=> 2y – y = 3
y = 3
x = 2y – 1 = 2 x 3 – 1 = 6 – 1 = 5
∴Number = x + 10y = 5 + 3 x 10
= 5 + 30 = 35

Question 4.
Solution:
Sum of two digits of a number =15
Let units digit = x
Then tens digit = 15 – x
∴Number = 10 (15 – x) + x
= 150 – 10x + x = 150 – 9x
By interchanging the digits, the new number will be
= 10x + 15 – x = 9x + 15
According to the condition,
9x + 15 = 9 + 150 – 9x
9x + 9x = 159-15 = 144
18x = 144
=>x = \(\\ \frac { 144 }{ 18 } \) = 8
∴Number = 150 – 9x = 150 – 9 x 8
= 150 – 72 = 78

Question 5.
Solution:
Let units place digit = x
and tens place digit = y
Then number = x + 10y
By interchanging the positions of the digits then
Units digits = y
and tens digit = x
∴Number = y + 10x
(x + 10y) – (y + 10x) = 63
=> x + 10y – y – 10x = 63
9y – 9x = 63
=> 9(y – x) = 63
y – x = \(\\ \frac { 63 }{ 9 } \) = 7
∴Hence, difference of its digits = 7 Ans.

Question 6.
Solution:
Sum of three digits of a number = 16
Let units digit of a three-digit number = x
Then tens digit = 3x
and hundreds digit = 4x
∴Number = x + 10 x 3x + 100 x 4x
= x + 30x + 400x = 431x
But x + 3x + 4x = 16 => 8x = 16
∴x = \(\\ \frac { 16 }{ 8 } \) = 2
∴Number = 431 x 862

 

Hope given RS Aggarwal Solutions Class 8 Chapter 5 Playing with Numbers Ex 5A 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 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4B

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 4 Cubes and Cube Roots Ex 4B.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 4 Cubes and Cube Roots Ex 4A
• RS Aggarwal Solutions Class 8 Chapter 4 Cubes and Cube Roots Ex 4B
• RS Aggarwal Solutions Class 8 Chapter 4 Cubes and Cube Roots Ex 4C
• RS Aggarwal Solutions Class 8 Chapter 4 Cubes and Cube Roots Ex 4D

Find the value of each of the following using the short-cut method :

Question 1.
Solution:
(25)³ = We know that short-cut method for finding the cube of any two digit number is as given.

Question 2.
Solution:
(47)³ = Here a = 4, b = 7

Question 3.
Solution:
(68)³ = Here a = 6, b = 8

Question 4.
Solution:
(84)³ Here a = 8, b = 4

Hope given RS Aggarwal Solutions Class 8 Chapter 4 Cubes and Cube Roots Ex 4B are helpful to complete your math homework.

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RS Aggarwal Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4D

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 4 Cubes and Cube Roots Ex 4D.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 4 Cubes and Cube Roots Ex 4A
• RS Aggarwal Solutions Class 8 Chapter 4 Cubes and Cube Roots Ex 4B
• RS Aggarwal Solutions Class 8 Chapter 4 Cubes and Cube Roots Ex 4C
• RS Aggarwal Solutions Class 8 Chapter 4 Cubes and Cube Roots Ex 4D

Objective Questions
Tick the correct answer in each of the following:

Question 1.
Solution:
(a) 141
= 3 x 47

Question 2.
Solution:
1152
= 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3

Question 3.
Solution:
\(\sqrt [ 3 ]{ 512 }\)

Question 4.
Solution:
\(\sqrt [ 3 ]{ 125\times 64 }\)

Question 5.
Solution:
\(\sqrt [ 3 ]{ \frac { 64 }{ 343 } } \)

Question 6.
Solution:
\(\sqrt [ 3 ]{ \frac { -512 }{ 729 } } \)

Question 7.
Solution:
Factorising 648, we get

Question 8.
Solution:
Factorising 1536, we get

Question 9.
Solution:
\({ \left( 1\frac { 3 }{ 10 } \right) }^{ 3 } \)

Question 10.
Solution:
(0.8)³
= 0.8 x 0.8 x 0.8 = 0.512 (c)

Hope given RS Aggarwal Solutions Class 8 Chapter 4 Cubes and Cube Roots Ex 4D are helpful to complete your math homework.

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RS Aggarwal Class 8 Solutions Chapter 5 Playing with Numbers Ex 5B

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 5 Playing with Numbers Ex 5B.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 5 Playing with Numbers Ex 5A
• RS Aggarwal Solutions Class 8 Chapter 5 Playing with Numbers Ex 5B
• RS Aggarwal Solutions Class 8 Chapter 5 Playing with Numbers Ex 5C
• RS Aggarwal Solutions Class 8 Chapter 5 Playing with Numbers Ex 5D

Question 1.
Solution:
We know that a number is divisible by 2 if its unit digit is 0, 2, 4, 6 or 8
Therefore, (i) 94, (ii) 570, (iv) 2398,(v) 79532 and (vi) 13576 are divisible by 2.

Question 2.
Solution:
We know that a number is divisible by 5 if its unit digit is 0 or 5.
Therefore, (i) 95, (ii) 470, (iv) 2735, (vi) 35790, (vii) 98765 and (ix) 77990 are divisible by 5.

Question 3.
Solution:
We know that a number is divisible by 10 if its unit digit is zero.
Therefore, (ii) 90 and (iv) 57930 are divisible by 10.

Question 4.
Solution:
We know that a number is divisible by 3 if the sum of its digits is divisible by 3. Therefore
(i) 83 – 8 + 3 = 11,not divisible by 3
(ii) 378 – 3 + 7 + 8 = 18, is divisible by 3
(iii) 474 – 4 + 7 + 4 = 15, is divisible by 3
(iv) 1693 – 1 + 6 + 9 + 3 = 19, is divisible by 3
(v) 20345 – 2 + 0 + 3 + 4 + 5 = 14 is not divisible by 3
(vi) 67035 – 6 + 7 + 0 + 3 + 5 = 21 is divisible by 3
(vii)591282 – 5 + 9 + 1 + 2 + 8 = 27 is divisible by 3
(viii)903164 – 9 + 0 + 3 + 1 + 6 + 4 = 23,is not divisible by 3
(ix) 100002 – 1 + 0 + 0 + 0 + 0 + 2 = 3,is divisible by 3

Question 5.
Solution:
We know that a number is divisible by 9, if the sum of its digits is divisible by 9. Therefore,
(i) 327 = 3 + 2 + 7 = 12,is not divisible by 9
(ii) 7524 = 7 + 5 + 2 + 4 = 18, is divisible by 9
(iii) 32022 = 3 + 2 + 0 + 2 + 2 = 9,is divisible by 9
(iv) 64302 = 6 + 4 + 3 + 0 + 2 = 15, is not divisible by 9
(v) 89361= 8 + 9 + 3 + 6 + 1 = 27 is divisible by 9
(vi)14799 = 1 + 4 + 7 + 9 + 9 = 30,is not divisible by 9
(vii) 66888 = 6 + 6 + 8 + 8 + 8 = 36, is divisible by 9
(viii) 30006 = 3 + 0 + 0 + 0 + 6 = 9, is divisible by 9
(ix) 33333 = 3 + 3 + 3 + 3 + 3 = 15 is not divisible by 9

Question 6.
Solution:
We know that a number is divisible by 4, only when the number formed by its last two digits is divisible by 4.
Therefore,
(i) 134, is not divisible by 4 as last two digits 34 is not divisible by 4.
(ii) 618, is not divisible by 4 as last two digits 18 is not divisible by 4.
(iii) 3928, is divisible by 4 as last two digits 28 is divisible by 4.
(iv) 50176, is not divisible by 4 as last two digits 76 is not divisible by 4.
(y) 39392, is not divisible by 4 as last two digits 92 is not divisible by 4.
(vi) 56794, is not divisible by 4 as last two digits 94 is not divisible by 4.
(vii) 86102, is not divisible by 4 as last two digits 02 is not divisible by 4.
(viii) 66666, is not divisible by 4 as last two digits 66 is not divisible by 4.
(ix) 99918, is not divisible by 4 as last two digits 18 is not divisible by 4.
(x) 77736, is divisible by 4 as last two digits 36 is divisible by 4.

Question 7.
Solution:
A given number is divisible by 8 only when the number formed by its last three digits is divisible by 8.
(i) 6132, is not divisible by 8 as last three digits 132 is not divisible by 8.
(ii) 7304, is divisible by 8 as last three digits 304 is not divisible by 8.
(iii) 59312, is divisible by 8 as last three digits 312 is divisible by 8.
(iv) 66664, is divisible by 8 as last three digits 664 is divisible by 8.
(v) 44444, is not divisible by 8 as last three digits 444 is not divisible by 8.
(vi) 154360, is divisible by 8 as last three digits 360 is not divisible by 8.
(vii) 998818, is not divisible by 8 as last three digits 818 is not divisible by 8.
(viii) 265472, is divisible by 8 as last three digits 472 is divisible by 8.
(ix) 7350162, is not divisible by 8 as last three digits 162 is not divisible by 8.

Question 8.
Solution:
A given number is divisible by 11, if the difference between the sum of its digits at odd places and the sum of its digits at even places, is either O or a number divisible by 11.
(i) 22222
Sum of digit at odd places = 2 + 2 + 2 = 6
Sum of digit at even places = 2 + 2 = 4
Difference of the above sum = 6 – 4 =2,
which is not divisible by 11
22222 is not divisible by 11

(ii) 444444
Sum of digit at odd places = 4 + 4 + 4 = 12
Sum of digit at even places = 4 + 4 + 4 = 12
Difference of the above sum =(12 – 12) = O
444444 is divisible by 11

(iii) 379654
Sum of digit at odd places = 7 + 6 + 4 = 17
Sum of digit at even places = 3 + 9 + 5 = 17
Difference of the above sum = (17 – 17) = 0
379654 is divisible by 11

(iv) 1057982
Sum of digit at odd places = 1 + 5 + 9 + 2 = 17
Sum of digit at even places = 0 + 7 + 8 = 15
Difference of the above sum = (17 – 15) = 2, which is not divisible by 11
1057982 is not divisible by 11

(v) 6543207
Sum of digit at odd places = 6 + 4 + 2 + 7 = 19
Sum of digit at even places = 5 + 3 + 0 = 8
Difference of the above sum = (19 – 8) = 11, Which is divisible by 11
6543207 is divisible by 11

(vi) 818532
Sum of digital to odd places = 1 + 5 + 2 = 8
Sum of digit at even places = 8 + 8 + 3 = 19
Difference of the above sum = 19 – 8 = 11, which is divisible by 11
818532 is divisible by 11

(vii) 900163
Sum of digit at odd places = 0 + 1 + 3 = 4
Sum of digit at even places = 9 + 0 + 6 = 15
Difference of the above sum = (15 – 4) = 11, which is divisible by 11
900163 is divisible by 11

(viii) 7531622
Sum of digit at odd places = 7 + 3 + 6 + 2 = 18
Sum of digit at even places = 5 + 1 + 2 = 8
Difference of the above sum = (18 – 8) = 10, which is not divisible by 11
7531622 is not divisible by 11

Question 9.
Solution:
For testing the divisibility of a number by 7, we proceed according to the
following steps:
Step 1: Double the unit digit of the given number.
Step 2 : Subtract the above number from the number formed by excluding the unit digit of the given number.
Step 3 : 1f the number so obtained is divisible by 7 then the given number is divisible by 7.
(i) 693
Now, 69 – (2 x 3) = 63, which is divisible by 7
693 is divisible by 7

(ii) 7896
Now 789 – (6 x 2) = 777, which is divisible by 7
7896 is divisible by 7

(iii) 3467
Now, 346 – (7 x 2) = 332, which is not divisible by 7
3467 is not divisible by 7

(iv) 12873
Now,1287 – (3 x 2) = 1281, which is divisible by 7
12873 is divisible by 7

(v) 65436
Now, 6543 – (6 x 2) = 6531, which is divisible by 7
65436 is divisible by 7

(vi) 54636
Now, 5463 – (6 x 2) 5451, which is not divisible by 7
54636 is not divisible by 7

(vii) 98175
Now, 9817 – (5 x 2) 9807, which is divisible by 7
98175 is divisible by7

(viii) 88777
Now, 8877 – (7 x 2) = 8863, which is not divisible by 7
88777 is not divisible by 7

Question 10.
Solution:
The given number 7×3 is divisible by 3
The sum of its digits is divisible by 3
7 + x + 3 =>10 + x is divisible by 3
Value of x can be 2, 5, 8
The numbers can be 723, 753, 783

Question 11.
Solution:
The given number 53yl is divisible by 3
Sum of its digits is divisible by 3
i.e., 5 + 3 + y + 1 or 9 + y is divisible by 3
Values of y can be 0, 3, 6, 9
Then the numbers can be 5301, 5331, 5361, 5391

Question 12.
Solution:
Number x806 is divisible by 9
The sum of its digits is also divisible by 9
or x + 8 + 0 + 6 or 14 + x is divisible by 9
x can be 4
Number will be 4806

Question 13.
Solution:
The number 471z8 is divisible by 9
The sum of its digits is also divisible by 9
471z8 = 4 + 7 + 1 + z + 8
=> 20 + z is divisible by 9
Value of z can be 7
Number will be 47178

Question 14.
Solution:
Let the number 21, sum of digits 2 + 1 = 3
which is divisible by 3 not by 9
Let the number 24, sum of digits 2 + 4 = 6
which is divisible by 3 not by 9
Let the number 30, sum of digits 3+0 = 3
which is divisible by 3 not by 9
Let the number 33, sum of digits 3 + 3 = 6
which is divisible by 3 not by 9
Let the number by 39 sum of digits 3 + 9 = 12
which is divisible by 3 not by 9

Question 15.
Solution:
Consider numbers as 28, 36,44, 52,60 as these numbers are divisible by 4 not by 8.
Let the number 39, sum of digits 3 + 9 = 12
which is divisible by 3 not by 9

Hope given RS Aggarwal Solutions Class 8 Chapter 5 Playing with Numbers Ex 5B are helpful to complete your math homework.

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RS Aggarwal Class 8 Solutions Chapter 7 Factorisation Ex 7A

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7A.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7A
• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7B
• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7C
• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7D
• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7E

Factorize :

Question 1.
Solution:
(i) 12x + 15 = 3(4x + 5) {HCF of 12 and 15 = 3)
(ii) 14m – 21 = 7(2m – 3) {HCF of 14 and 21 = 7)
(iii) 9n – 12n² = 3n(3 – 4n) Ans.{HCF of 9 and 12 = 3}

Question 2.
Solution:
(i) 16a² – 24ab = 8a(2a – 3b) { HCF of 16 and 24 = 8)
(ii) 15ab² – 20a²b = 5ab(3b – 4a) {HCF of 15 and 20 5}
(iii) 12x²y³ – 21x³y²= 3x²y²(4y – 7x) Ans. {HCF of 12 and 21 = 3)

Question 3.
Solution:
(i) 24x³ – 36x²y = 12x²(2x – 3y) {HCF of 24 and 36 = 12}
(ii) 10x³ – 15x² = 5x²(2x – 3) {HCF of 10, 15 = 5}
(iii) 36x³y – 60x²y³z = 12x²y (3x – 5y²z) {HCF of 36 and 60 = 12}

Question 4.
Solution:
(i)9x³ – 6x² + 12x
= 3x(3x² – 2x + 4) {HCF of 9, 6, 12 = 3)
(ii)8x² – 72xy + 12x
= 4x(2x – 18y + 3) (HCF of 8, 72, 124)
(iii)18a³b³ – 27a²b³ + 36a³b²
= 9a²b²(2ab – 3b + 4a) Ans.{HCF of 10, 27, 36 = 9)

Question 5.
Solution:
(i) 14x³ + 21x⁴y – 28x²y²
= 7x² (2x + 3x²y – 4y²) {HCF of 14, 21 28 = 7)
(ii) – 5 – 10t + 20t²
= – 5(1 + 2t – 4t²) Ans. {HCF of 5, 10, 20=5)

Question 6.
Solution:
(i) x(x + 3) + 5(x + 3)
=(x + 3)(x + 5)
(ii) 5x(x – 4) – 7(x – 4)
= (x – 4) (5x – 7)
(iii) 2m (1 – n) + 3(1 – n)
= (1 – n) (2m + 3) Ans.

Question 7.
Solution:
6a(a – 2b) + 5b(a – 2b)
= (a – 2b) (6a + 5b) Ans.

Question 8.
Solution:
x³(2a – b) + x²(2a – b)
=x²(2a – b)(x + 1)Ans.

Question 9.
Solution:
9a(3a – 5b) – 12a²(3a – 5b)
= 3a (3a – 5b) (3 – 4a) Ans.

Question 10.
Solution:
(x + 5)² – 4(x + 5)
= (x + 5)(x + 5 – 4)
=(x + 5)(x + 1)Ans.

Question 11.
Solution:
3(a – 2b)² – 5 (a – 2b)
= (a – 2b) {3(a – 2b) – 5}
= (a – 2b) (3a – 6b – 5) Ans.

Question 12.
Solution:
2a + 6b – 3 (a + 3b)²
= 2(a + 3b) – 3(a + 3b)²
= (a + 3b){ 2 – 3 (a + 3b)}
= (a + 3b) (2 – 3a – 9b) Ans.

Question 13.
Solution:
16(2p – 3q)² – 4 (2p – 3q)
= 4(2p – 3q) {4(2p – 3q) – 1}
= 4(2p – 3q) (8p – 12q – 1) Ans.

Question 14.
Solution:
x(a – 3) + y (3 – a)
= x(a – 3) – y(a – 3)
= (a – 3) (x – y) Ans.

Question 15.
Solution:
12(2x – 3y)² – 16(3y – 2x)
= 12(2x – 3y)² + 16(2x – 3y)
= 4(2x – 3y) {3(2x – 3y) + 4}
= 4(2x – 3y) (6x – 9y + 4) Ans.

Question 16.
Solution:
(x + y)(2x + 5) – (x + y)(x + 3)
= (x + y)(2 + 5 – x – 3)
= (x + y)(x + 2) Ans.

Question 17.
Solution:
ar + br + at + bt
= r(a + b) + t(a + b)
= (a + b) (r + t) Ans.

Question 18.
Solution:
x² – ax – bx + ab
= x(x – a) – b(x – a)
= (x – a)(x – b)Ans.

Question 19.
Solution:
ab² – bc² – ab + c²
= ab² – ab – bc² + c²
= ab(b – 1) – c²(b – 1)
= (b – 1) (ab – c²) Ans.

Question 20.
Solution:
x² – xz + xy – yz
= x(x – z) + y(x – z)
= (x – z)(x + y)Ans.

Question 21.
Solution:
6ab – b² + 12ac – 2bc
6ab + 12ac – b² – 2bc
= 6a(b + 2c) – b(b + 2c)
= (b + 2c) (6a – b) Ans.

Question 22.
Solution:
(x – 2y)² + 4x – 8y
= (x – 2y)² + 4(x – 2y)
= (x – 2y)(x – 2y + 4)Ans.

Question 23.
Solution:
y² – xy(1 – x) – x³
y² – xy + x²y – x²
= y(y – x) + x² (y – x)
= (y – x)(y + x²)Ans.

Question 24.
Solution:
(ax + by)² + (bx – ay)²
= a²x² + b²y² + 2abxy + b²x² + a²y² – 2abxy
= a²x² + b²y² + b²x² + a²y²
= a²x² + b²x² + a²y² + b²y²
= x²(a² + b²) + y²(a² + b²)
= (a² + b²) (x² + y²) Ans.

Question 25.
Solution:
ab² + (a – 1)b – 1
= ab² + ab – b – 1
= ab(b + 1) – 1(b + 1)
= (b + 1) (ab – 1) Ans.

Question 26.
Solution:
x³ – 3x² + x – 3
= x²(x – 3) + 1(x – 3)
(x – 3)(x² + 1)Ans.

Question 27.
Solution:
ab (x² + y²) – xy (a² + b²)
= abx² + aby² + xya² – xyb²
= abx² – xya² – xyb² + aby2
=ax(bx – ay) – by(bx – ay)
= (bx – ay) (ax – by) Ans.

Question 28.
Solution:
x² – x (a + 2b) + 2ab
= x² – xa – 2bx + 2ab
= x(x – a) – 2b(x – a)
= (x – a)(x – 2b)

 

Hope given RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7A 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 8 Solutions Chapter 3 Squares and Square Roots Ex 3G

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3G.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3A
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3B
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3C
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3D
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3E
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3F
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3G
• RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3H

Evaluate:

Question 1.
Solution:
\(\sqrt { \frac { 16 }{ 81 } } \)
= \(\frac { \sqrt { 16 } }{ \sqrt { 81 } } \)
= \(\sqrt { \frac { 4X4 }{ 9X9 } } \)
= \(\\ \frac { 4 }{ 9 } \)

Worksheets for Class 8 Maths

Question 2.
Solution:
\(\sqrt { \frac { 64 }{ 225 } } \)
= \(\frac { \sqrt { 64 } }{ \sqrt { 225 } } \)
= \(\sqrt { \frac { 8X8 }{ 15X15 } } \)
= \(\\ \frac { 8 }{ 15 } \)

Question 3.
Solution:
\(\sqrt { \frac { 121 }{ 256 } } \)
= \(\frac { \sqrt { 121 } }{ \sqrt { 256 } } \)
= \(\sqrt { \frac { 11X11 }{ 16X16 } } \)
= \(\\ \frac { 11 }{ 16 } \)

Question 4.
Solution:
\(\sqrt { \frac { 625 }{ 729 } } \)

Question 5.
Solution:
\(\sqrt { 3\frac { 13 }{ 36 } } \)
= \(\sqrt { \frac { 3X36+13 }{ 36 } } \)
= \(\sqrt { \frac { 108+13 }{ 36 } } \)
= \(\sqrt { \frac { 121 }{ 36 } } \)
= \(\sqrt { \frac { 11X11 }{ 6X6 } } \)
= \(\\ \frac { 11 }{ 6 } \)

Question 6.
Solution:
\(\sqrt { 4\frac { 73 }{ 324 } } \)
= \(\sqrt { \frac { 4X324+73 }{ 324 } } \)
= \(\sqrt { \frac { 1296+73 }{ 324 } } \)
= \(\sqrt { \frac { 1369 }{ 324 } } \)

Question 7.
Solution:
\(\sqrt { 3\frac { 33 }{ 289 } } \)
= \(\sqrt { \frac { 3X289+33 }{ 289 } } \)
= \(\sqrt { \frac { 867+33 }{ 289 } } \)
= \(\sqrt { \frac { 900 }{ 289 } } \)

Question 8.
Solution:
\(\frac { \sqrt { 80 } }{ \sqrt { 405 } } \)
= \(\sqrt { \frac { 80 }{ 405 } } \)
= \(\sqrt { \frac { 16 }{ 81 } } \)
= \(\frac { \sqrt { 16 } }{ \sqrt { 81 } } \)
= \(\\ \frac { 4 }{ 9 } \)

Question 9.
Solution:
\(\frac { \sqrt { 1183 } }{ \sqrt { 2023 } } \)
= \(\sqrt { \frac { 1183 }{ 2023 } } \)
= \(\sqrt { \frac { 1183\div 7 }{ 2023\div 7 } } \)
= \(\frac { \sqrt { 169 } }{ \sqrt { 289 } } \)
= \(\frac { \sqrt { 13X13 } }{ \sqrt { 17X17 } } \)
= \(\\ \frac { 13 }{ 17 } \)

Question 10.
Solution:
\(\sqrt { 95 } \times \sqrt { 162 } \)
= \(\sqrt { 98\times 162 }\)
= \(\sqrt { 2\times 7\times 7\times 2\times 3\times 3\times 3\times 3 } \)

Hope given RS Aggarwal Solutions Class 8 Chapter 3 Squares and Square Roots Ex 3G 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 8 Solutions Chapter 5 Playing with Numbers Ex 5D

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 5 Playing with Numbers Ex 5D.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 5 Playing with Numbers Ex 5A
• RS Aggarwal Solutions Class 8 Chapter 5 Playing with Numbers Ex 5B
• RS Aggarwal Solutions Class 8 Chapter 5 Playing with Numbers Ex 5C
• RS Aggarwal Solutions Class 8 Chapter 5 Playing with Numbers Ex 5D

Question 1.
Solution:
5×6 is exactly divisible by 3
Sum of its digits must be divisible by 3
5 + x + 6 = 11 + x is divisible by 3
Least value of x = 1 as 12 is divisible by 3 (b)

Question 2.
Solution:
64y8 is exactly divisible by 3 then the sum of its digits must be divisible by 3
6 + 4 + y + 8 or 18 + y is divisible by 3
Least value of y = 0
18 is divisible by 3 (a)

Question 3.
Solution:
7 x 8 is exactly divisible by 9
Sum of its digits must be divisible by 9
7 + x + 8 = 15 + x must be divisible by 9
Least value of x = 3 as 15 + 3 = 18 is divisible by 9 (c)

Question 4.
Solution:
37y4 is exactly divisible by 9
The sum of its digits must be divisible by
3 + 7 + y + 4 or 14 + y is divisible by 9
Least value of y = 4
As 14 + 4 = 18 is divisible by 9 (d)

Question 5.
Solution:
4xy7 is exactly divisible by 3
The sum of its digits must be divisible by 9
or 4 + x + y + 7 or 11 + (x + y) is divisible by 9
Least value of x + y = 7
as 11 + 7 = 18 is divisible by 9 (d)

Question 6.
Solution:
x7y5 is exactly divisible by 3
Sum of its digits must be divisible by 3
x + 7 + y + 512 + (x + y) is divisible by 3
Least value of x + y = 0 as
12 + 0 = 12 is divisible by 3 (b)

Question 7.
Solution:
x4y5z is exactly divisible by 9
The sum of its digits must be divisible by 9
x + 4 + y + 5 + z or 9 + (x + y + z) must be divisible by 9
Least value of x + y + z = 0 as 9 + 0 = 9 is divisible by 9 (d)

Question 8.
Solution:
A2B5 is exactly divisible by 9
Sum of its digits must be divisible by 9
A + 2 + B + 5 = 7 + A + B is divisible by 9
Least value of A + B = 2 as 7 + 2 = 9 is divisible by 9

Question 9.
Solution:
x27y is exactly divisible by 9
The sum of its digits must be divisible by 9
x + 2 + 7 + y = x + y + 9 is divisible by 9
Least value of x + y = 0 as 0 + 99 is exactly divisible by 9 (a)

Hope given RS Aggarwal Solutions Class 8 Chapter 5 Playing with Numbers Ex 5D 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 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4A

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 4 Cubes and Cube Roots Ex 4A.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 4 Cubes and Cube Roots Ex 4A
• RS Aggarwal Solutions Class 8 Chapter 4 Cubes and Cube Roots Ex 4B
• RS Aggarwal Solutions Class 8 Chapter 4 Cubes and Cube Roots Ex 4C
• RS Aggarwal Solutions Class 8 Chapter 4 Cubes and Cube Roots Ex 4D

Question 1.
Solution:
(i) (8)³ = 8 x 8 x 8 = 512
(ii) (15)³ = 15 x 15 x 15 = 3375
(iii) (21)³ = 21 x 21 x 21 = 9261
(iv) (60)³ = 60 x 60 x 60 = 216000 Ans.

Question 2.
Solution:
(i)(1.2)³= 1.2 x 1.2 x 1.2 = 1.728
(ii) (3.5)³ = 3.5 x 3.5 x 3.5 = 42.875
(iii) (0.8)³ = 0.8 x 0.8 x 0.8 = 0.512
(iv) (0.05)³ = 0.05 x 0.05 x 0.05 = 0.000125

Question 3.
Solution:

Question 4.
Solution:
(i) 125
= \(\overline { 5\times 5\times 5 } ={ \left( 5 \right) }^{ 3 }\)

Question 5.
Solution:
We know that cube of an even number is also even.
216, 1000 and 512 are the cubes of even numbers as these are all even numbers. Ans.

Question 6.
Solution:
We know that cube of an odd number is also odd.
125, 343 and 9261 are the cubes of odd natural numbers as these are also odd numbers. Ans.

Question 7.
Solution:
Factorising 1323 into prime factors,

Question 8.
Solution:
Factorising 2560 into prime factors.

Making them in groups of 3 equal factors, we are left 5
To make it into a group of 3, we have to multiply it by 5 x 5 i.e. by 25.
Hence, the smallest number by which it is multiplied = 25 Ans.

Question 9.
Solution:
Factorising 1600 into prime factors

Question 10.
Solution:
Factorising 8788 into prime factors

Making them in groups of 3 equal factors, we are left with 2 x 2
2 x 2 i.e. 4 is to be divided.
Hence least number to be divided for getting perfect cube = 4 Ans.

 

Hope given RS Aggarwal Solutions Class 8 Chapter 4 Cubes and Cube Roots Ex 4A 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.