Class 7 Maths

Algebra Chapter 5 Algebraic Formula Exercise

Question 1. Choose the correct answer.

1. If (25x² + pxy+ 36y²) is a perfect square then the Value of P is
Solution:

25x² + Pxy + 36y²

⇒ (5x)² + pxy + (6y)²

Now this expression is in the form of (a+b)²= a²+2ab+b²

Here, a = 5x, b = 6y

∴ 2ab = рху

2(5x)(6x) = pxy

2(30xy) = pxy

60xy = pxy

P = 60

∴ The value of the ‘p’ is 60′

Read and Learn More Class 7 Maths Solutions

2. If x² – 12x + 36 = (x+m)² then the value of ‘m’?
Solution:

⇒ x² – 12x+36 = (x+m)²

⇒ x² – 2.6x + (6)² = (x+m)²

⇒ (x-6)² = (x+m)²

Cancel the equal terms on both sides

⇒ (x-6)² = (x+m)²

⇒ x – 6 – x – m = 0

⇒ -6 – m =0

⇒ -(6+m)=0

m + 6 =0

m = -6

∴ x² – 12x+36= (X+m)² The value of m is -6′

Algebra Formulas For Class 7 WBBSE

3. If a+b+c= 6 and a² + b² + c² = 8 then the value of (ab+ bc+ca) is

1. 28
2. 24
3. 14
4. 12

Solution:

a+b+c= 6 and a²+ b²+c²=8.

Squaring the equation a+b+c=6;

⇒ (a+b+c)² = 62

⇒ a² + b²+c² + 2(ab+bc+ca) =36

Given that a² + b² + c² = 8 substitute in above expression

⇒ 8 + 2(ab+bc+ca) = 36

⇒ 2(ab+bc+ca)=36-8

⇒ (ab+bc+ca) = \(\frac{28}{2}\)

∴ ab+bc + ca = 14

The value of (ab+bc+ca) is 14

4. If a² – b² = 55 [a>b] then the value of a and b are respectively.

1. 8,3
2. 11,5
3. 3,8
4. 5,11

Solution:

a² – b² =55

⇒ (a+b)(a-b) = 55

By checking the options in a given condition.

⇒ 8, 3

a² – b² = 55

(8)² – (3)² = 55

64 – 9 =55

55 = 55

∴ The values of a and b are 8, 3

Class 7 Maths Chapter 5 Solved Exercises

5. If \(x^4+\frac{1}{x^4}=119\) then the value of \(\left(x-\frac{1}{x}\right)\)

1. 11
2. 121
3. 9
4. 3

Solution:

⇒ \(x^4+\frac{1}{x^4}=119\)

⇒ \(\left(x^2\right)^2+2 \cdot x^2 \frac{1}{-x^2}+\frac{1}{\left(-x^2\right)^2}=119\)

⇒ \(\left(x^2\right)^2+\frac{1}{\left(-x^2\right)^2}=119+2\)

⇒ \(\left(x^2\right)^2+\frac{1}{\left(x^2\right)^2}=121\)

Applying square root on both sides.

⇒ \(\sqrt{\left(x^2\right)^{x^1}+\left(\frac{1}{x^2}\right)^2}=\sqrt{121}\)

⇒ \(x^2+\frac{1}{x^2}=11\)

⇒ \(x^2+2 x \cdot \frac{1}{-x}+\left(\frac{1}{-x^2}\right)^2=11\)

⇒ \(x^2-\frac{1}{x^2}=11-2\)

⇒ \(x^2-\left(\frac{1}{x}\right)^2=9\)

Applying square root on both sides.

⇒ \(\sqrt{(x)^2-\left(\frac{1}{x}\right)^x}=\sqrt{9}\)

⇒ \(x-\frac{1}{x}=3\)

∴ Option (4) 3 is the correct Answer.

6. If 3a-\(\frac{1}{5a}\) = 12 then the value of (25a² + \(\frac{1}{9a^2}\)) is

1. 147\(\frac{1}{3}\)
2. 403\(\frac{1}{3}\)
3. 7\(\frac{1}{5}\)
4. 5\(\frac{1}{7}\)

Solution:

3a – \(\frac{1}{5a}\) Multiplying \(\frac{5}{3}\) on both sides

⇒ \(\frac{5}{3}\left(3 a-\frac{1}{5 a}\right)=14 \times \frac{5}{3}\)

⇒ \(\frac{5}{7} \times 3 a-\frac{8}{3} \times \frac{1}{5 a}=20\)

5a – \(\frac{1}{3 a}=20\)

Squaring on Both sides

⇒ \(\left(5 a-\frac{1}{3 a}\right)^2=(20)^2\)

(5a)\(^2+\left(\frac{1}{3 a}\right)^2-2(5 a) \times\left(\frac{1}{3 a}\right)=400\)

25 \(a^2+\frac{1}{9 a^2}=400+2\left(\frac{5}{3}\right)\)

= \(\frac{3(400)+10}{3}\)

= \(\frac{1200+10}{3}\)

= \(\frac{1210}{3}\)

= \(403 \cdot 33\)

∴ 25 \(a^2+\frac{1}{9 a^2}=403 \cdot \frac{1}{3}\)

Question 2. Write true or false

1. If x² + \(\frac{1}{x^2}\) = 2 then the value of (x-\(\frac{1}{x}\)) is 0
Solution:

x² + \(\frac{1}{x^2}\) -2(x)(\(\frac{1}{x}\))=0

a² + b² -2ab = (a – b)²

(x-\(\frac{1}{x}\))² = 0

(x-\(\frac{1}{x}\)) = 0

∴ True

2. If 3a – 5b then the value of (9a² – 30ab+25b²) is 64
Solution:

9a² – 30ab+ 25b² = 64

(3a)² – 2(30) (5b)+(5b)² = 64

(3a-5b)² = 64

∴ 3a=5b

(5b-5b)² = 64

(0)²=64

∴ False

Class 7 Algebra Problems With Solutions 

3. If x² – 6x + 9=0 then the value of (x² + 6x + 9) is 16
Solution:

x² – 6x + 9 = 0

x² – 2·x·3 + (3)² = 0

(x – 3)² = 0

x – 3 =0

∴ x = 3

x² + 6x + 9 = 16

⇒ (3)² + 6(3) + 9 = 16

⇒ 9+18+9 = 16

⇒ 36 = 16

∴ False

Question 3. Fill in the blanks

1. (a+2b)² =(a-2b)² + 8ab
Solution:

(a+b)² = (a−b)² + 4ab

(a+2b)² = (a-2b)² + 4a(2b)

= (a-2b)²+ 8ab

(a+b)² = (a-2b)² + 8ab

2. The value of (7.3 x 7.3 + 2·3 × 2·3 – 7.3 × 4.6) is
Solution:

⇒ 7.3 x 7.3 + 2·3 × 2·3 – 7.3 × 4.6

⇒ 53.29 + 5.29 – 33.58

⇒ 58.58 – 33.58

⇒ 25

∴ (7·3 × 7·3 + 2.3 × 2·3 – 7·3 × 4.6) = 25

3. If (x-y)² = x² – 10x +25 then the value of y is
Solution:

= (x-y)² = x² – 10x + 25

(x-y)² = x² – 2·x·5 + (5)²

∴ The value of ‘y’ is ‘5’

(x-y)² = x² – 2xy + y²

2xy = 2x.5

∴ y =5

Question 4. Find the Square

1. 995
2. 805
3. 2a-3b+4c

Solution:

1. 995 x 995 = 990025

2. 805 x 805 = 648025

3. (2a-3b+4c)²

⇒ (2a+(-3b)+4c)²

⇒ (a+b+c)²= a² + b²+c² + 2ab+ 2bc + 2ca

Here a= 2a, b= -3b, c = 4c

⇒ (2a+(-3b)+4c)² = (2a)² + (-3b)² +(4c)² + 2(2a)(−3b) + 2(3b)(4c) + 2(4c)(2a)

= 4a² + 9b² + 16c² – 12ab – 24bc + 16ac

Class 7 Maths Exam Preparation WBBSE

Question 5. Find the value of 998 × 1002
Solution:

⇒ \(\begin{array}{r}
1002 \\
\times 998 \\
\hline 8016 \\
9018 \times \\
9018 \times \\
\hline 999996 \\
\hline
\end{array}\)

∴ 998 × 1002 = 999996

Question 6. Express (a²+b²) (c²+ d²) as a sum of two square
Solution:

(a²+b²)(c²+d²)

Let us expand it a²c² + a²d² + c²b² + b²d² ⇒ (a²c² + b²d²) + (a²d² + b²c²)

Notice that both pairs have the same structure but with different terms. Each pair is a perfect square of two terms.

⇒ a²d²+ b²c²

⇒ (ad)² + (bc)²

Or,

⇒ (a²c²+b²d²)

⇒ (ac)² + (bd)²

∴ (a² + b²) (c² + d²) can be expressed as the

Sum of two squares: (ac)² +(bd)² + (ad)² + (bc)²

Question 7. If (a-3)² + (b-2)² = 0 then find the value of (a+3)² + (b+2)²
Solution:

(a-3)² + (b-2)² = 0

(a – 3)² =0,

a – 3 = 0

a = 3

and

(b – 2)² =0

b – 2 = 0

b = 2

(a+3)² + (b+2)²

Now Substitute the a and b values in the above equation.

(3+3)²+(2+2)²

(6)² +(4)²

⇒ 36+16

⇒ 52

∴ (a+3)² + (b+2)² = 52

Class 7 Maths Algebra Solutions WBBSE

Question 8. If x² – ax – 1 = 0 then find the value of \(\left(x^4+\frac{1}{x^4}\right)\)
Solution:

x² – ax – 1 = 0

⇒ x² – 1 = ax

⇒ \(\frac{x^2-1}{x}=a, \Rightarrow x-\frac{1}{x}=a\)

squaring on both sides

⇒ \(\left(x-\frac{1}{x}\right)^2=(a)^2\)

⇒ \(x^2-2 \cdot x\left(\frac{1}{x x}\right)+\frac{1}{x^2}=a^2\)

⇒ \(x^2+\frac{1}{x^2}=a^2+2\)

⇒ \(\left(x^2+\frac{1}{x^2}\right)^2=\left(a^2+2\right)^2\) (squaring on both sides)

⇒ \(\left(x^2\right)^2+2\left(x^2\right)\left(\frac{1}{x^2}\right)+\left(\frac{1}{x^2}\right)^2=\left(a^2\right)^2+(2)^2+2\left(a^2\right)(2)\)

⇒ \(\left(x^4+\frac{1}{x^4}\right)=a^4+4+4 a^2-2\)

⇒ \(x^4+\frac{1}{x^4}=a^4+4 a^2+2\)

∴ \(\left(x^4+\frac{1}{x^4}\right)=a^4+4 a^2+2\)

Question 9. If a = 2017, b = 2018, and c = 2019 then Find the value of (a²+b+c²-ab-bc-ca)
Solution:

a = 2017, b = 2018, C = 2019

(a² + b² + c² – ab – bc – ca)

⇒ a² + b² + c² – (ab + bc + ca)

⇒ (2017) ‍+ (2018) + (2019)² – {(2017x 2018) + (2018x 2019) + (2019×2019)}

⇒ {(4068289 + 4072324 + 407636)} – {(4070306) + (4074342) + (4072323)}

⇒ 12216974 – 12216971

⇒ 3

∴ (a² + b + c² – ab – bc – ca) = 3

Question 10. If 4x – \(\frac{1}{4x}\) = 16 then find the value of x² + \(\frac{1}{256 x^2}\)
Solution:

4x – \(\frac{1}{4x}\) = 16

Multiply with \(\frac{1}{4}\) on Both sides

⇒ \(\left(4 x-\frac{1}{4 x}\right) \times \frac{1}{4}=46 \times \frac{1}{4}\)

⇒ \(\frac{4 x}{4}-\frac{1}{16 x}=4\)

squaring on Both sides

⇒ \(\left(x-\frac{1}{16 x}\right)^2=(4)^2 \quad(a-b)^2=a^2-2 a b+b^2\)

⇒ \(x^2-2 \cdot x \cdot\left(\frac{1}{16 x}\right)+\left(\frac{1}{16 x}\right)^2=16\)

⇒ \(x^2+\frac{1}{256 x^2}=\frac{16+\frac{1}{8}}{8}\)

⇒ \(x^2+\frac{1}{256 x^2}=\frac{128+1}{8}\)

⇒ \(x^2+\frac{1}{256 x^2}=\frac{129}{8}\)

⇒ \(x^2+\frac{1}{256 x^2}=16.125\)

⇒ \(16 \frac{1}{8}\)

∴ \(x^2+\frac{1}{256 x^2}=16 \cdot \frac{1}{8}\)

Class 7 Maths Algebra Solutions WBBSE

Question 11. If 3(a² + b² + c²) = (a + b + c)² then Find a:b:c
Solution:

3(a² + b² + c²) = (a + b + c)²

3a² + 3b² + 3c² = a² + b² + c² + 2(ab + bc + ca)

3a² + 3b² + 3c² – (a² + b² + c) = 2(ab + bc + ca)

3a² + 3b² + 3c² – a² – b² – c² = 2(ab + bc + ca)

2a² + 2b² + 2c² = 2(ab+bc+ca)

Dividing the equation by ‘2’ on Both sides

a² + b² + c² = ab + bc + ca

a² + b² + c² = ab – bc + ca = 0

Now Let’s Factorize and rearrange.

a² – 2ab + b² + a² – 2ac + c² + b² + 2bc + c² = 0

(a – b)² + (a-c)² + (b – c)² = 0

For the expression to be equal to zero

(a-b)² = 0

a – b = 0

and

(a-c)² = 0

a – c = 0

and

(b-c)² = 0

b – c = 0

This implies. a=b = c

∴ The ratio of a:b:c is 1:1:1

Question 13. Express \(\left(x^2-\frac{1}{y^2}\right)\left(y^2-\frac{1}{x^2}\right)\) as a perfect square
Solution:

⇒ \(\left(x^2-\frac{1}{y^2}\right)\left(y^2-\frac{1}{x^2}\right)\)

⇒ \(x^2\left(y^2-\frac{1}{x^2}\right)-\frac{1}{y^2}\left(y^2-\frac{1}{x^2}\right)\)

⇒ \(x^2 y^2-\frac{x^2}{x^2}-\frac{y^2}{x^2}+\frac{1}{x^2 y^2}\)

⇒ \(x^2 y^2-1-1+\frac{1}{x^2 y^2}\)

⇒ \(x^2 y^2-2+\frac{1}{x^2 y^2}\)

⇒ \(x^2 y^2-2(x y)\left(\frac{1}{x^2 y}\right)+\frac{1}{x^2 y^2}\)

(\(a^2-2 a b+b^2=(a-b)^2\); \(a^2=x^2 x^2\); \(b^2=\frac{1}{x^2 y^2}\))

⇒ \(\left(x y-\frac{1}{x y}\right)^2\)

∴ \(\left(x^2-\frac{1}{y^2}\right)\left(y^2-\frac{1}{x^2}\right)=\left(x y-\frac{1}{x y}\right)^2\)

Question 15. Express the following as a perfect square and hence Find the values

1. \(\frac{169}{a^2}-\frac{104}{a b}+\frac{16}{b^2}\) when a = 13, and b=-4
Solution:

⇒ \(\frac{169}{(13)^2}-\frac{104}{(13)(-4)}+\frac{16}{(-4)^2}\)

⇒ \(\frac{169}{169}-\frac{104}{-52}+\frac{16}{16}\)

⇒ \(1+\frac{104}{52}+1\)

⇒ 1 + 2 + 1

⇒ 4

∴ \(\frac{169}{a^2}-\frac{104}{a b}+\frac{16}{b^2}=4\)

2. \(121 a^2 b^2-66 a b+9\) when a = 1 b = -1
Solution:

⇒ \(121(1)^2(-1)^2-66(1)(-1)+9\)

⇒ 121+66+9

⇒ 196

∴ \(121 a^2 b^2-66 a b+9=196\)

WBBSE Class 7 Algebra Exercise Solutions

Question 16. Multiply: \(\left(a^2+a b+b^2\right)\left(a^2-a b+b^2\right)\left(a^4+a^2 b^2+b^4\right)\)
Solution:

⇒ \(\left(a^2+a b+b^2\right)\left(a^2-a b+b^2\right)\left(a^4+a^2 b^2+b^4\right)\)

⇒ \(\left\{\left(a^2+b^2\right)+a b\right\}\left\{\left(a^2+b^2\right)-a b\right\}\left(a^4+a^2 b^2+b^4\right)\)

⇒ \(\left\{\left(a^2+b^2\right)^2-(a b)^2\right\}\left\{a^2+a^2 b^2+b^4\right\}\)

((a+b)(a-b)=\(a^2-b^2\)

(a+b)\(^2=a^2+b^2+2 a b\))

⇒ \(\left.\Rightarrow\left\{\left(a^4+b^4+2 a^2 b^2\right)\right\}-a^2 b^2\right\}\left\{a^4+a^2 b^2+b^4\right\}\)

⇒ \(\left\{\left(a^4+a^2 b^2+b^4\right)\right\}\left\{\left(a^4+a^2 b^2+b^4\right)\right\}\)

⇒ \(\left(a^4+a^2 b^2+b^4\right)^2\)

⇒ \(\left(a^4\right)^2+\left(a^2 b^2\right)^2+\left(b^4\right)^2 \Rightarrow\left(a^4 \times a^4\right)+a^2 b^2 \times a^2 b^2+b^4 \times b^4\)

⇒ \(a^8+a^4 \cdot b^4+b^8\) (\(a^m \cdot a^n=a^{m+n}\))

∴ \(\left(a^2+a b+b^2\right)\left(a^2-a b+b^2\right)\left(a^4+a^2 b^2+b^4\right)=a^8+a^2 b^4+b^8\)

Question 17. If a + b = 10, and a – b = 4, then find the value of 7ab(a² + b²)
Solution:

a + b = 10

Squaring on Both sides

(a + b)² = (10)²

⇒ a² + b² + 2ab = 100

⇒ a² + b² = (100 – 2ab)

⇒ (a – b)2 2ab = (100 – 2ab)

⇒ (4)² + 2ab = 100 – 2b

⇒ 2ab + 2ab = 100 – 16

⇒ 4ab = 84

ab = \(\frac{84}{4}\)

∴ ab = 21

a² + b² = 100 – 2ab

= 100 – 2(21)

=100 – 42

∴ a² + b² = 58

7ab(a² + b²) = 7(21)(58)

7ab(a²+b²)= 8526

WBBSE Class 7 Algebra Exercise Solutions

Question 18. If \(\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}=1\) then find the value of \(\left(\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}\right)\)
Solution:

⇒ \(\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}=1\)

Adding 3 on Both sides

1 + \(\frac{a}{1-a}+1+\frac{b}{1-b}+1+\frac{c}{1-c}=1+3\)

⇒ \(\frac{1-a+a}{1-a}+\frac{1-b+b}{1-b}+\frac{1-c+c}{1-c}=4\)

⇒ \(\frac{1}{(1-a)}+\frac{1}{(1-b)}+\frac{1}{(1-c)}=4\)

Question 19. (x+y+z) (x-y+z) (x+y-z) (y+z-x)
Solution:

{(x+z)+y}{(x+z)-y}{y-(z-x)}{y+(z-x)} [(a+b)(a−b) = a² – b²]

{(x+z)² + y2²} {(y)² -(z-x)²}

{x²+ z²+2xz-y²} {y²-(z²+x²- 2xz)}

{x²+z²+2xz-y²} {y²-z²-x²+ 2xz}

{(2xz)² + (x² – y+ z²)} {(2xz)-(x²- y²+z²)}

(2xz)² – (x²-y²+z²)²

4x²z² – ((x²- y²)² + 2(x² – y²)z² + (z²)²}

4x²z² – {(x⁴ +y⁴ – 2x²y² + 2x²z² – 2y²z² + z⁴

4x²z² – x⁴ – y⁴ + 2x²y² – 2x²z² + 2y²z² – z⁴

2x²z² + 2 y²z² + 2x²y² =x⁴ – y⁴ – z⁴

∴ (x + y +z) (z – y + z) (x + y – z) (y + z – x) = 2x²z² + 2y²z² + 2x²y² – x⁴ – y⁴ – z⁴

Algebra Chapter 4 Algebraic Operations

Question 1. Choose the correct answer.

1. Sum of x²y – 3xy and 7xy² is

1. 7xy² + 2x²y
2. 7xy² – 2x²y
3. 5x²y
4. 8xy² – 3x²y

Solution:

x²y – 3x²y and 7xy²

⇒ 7xy² – 2x²y

∴ (x²y-3x²y) and 7xy² = 7xy² -2x²y

Option (2) 7xy² – 2x²y is the correct answer.

Read and Learn More Class 7 Maths Solutions

WBBSE Class 7 Algebra Chapter 4

2. The numerical co-efficient of the term -3x²+5x-7 other than the constant term is

1. (-3,5)
2. (-3,5,-7)
3. (5, -7)
4. (-3,-7)

Solution: -3x² + 5x -7

The numerical coefficient of the term (-3,5)

∴ Option (1) (3,5) is the correct answer,

3. The product of (5abc) and (-3abc) is

1. -15a⁸b¹²c¹⁰
2. 15 a⁶b⁷c⁷
3. -15 a⁶b⁷c⁷
4. 15 a⁸b¹²c¹⁰

Solution:

⇒ \(\left(5 a^4 b^3 c^2\right) \text { and }\left(-3 a^2 b^4 c^5\right)\)

⇒ \(\left(5 a^4 b^3 c^2\right) \times\left(-3 a^2 b^4 c^5\right)\)

⇒ \(5 a^4 \times\left(-3 a^2\right) \times b^3 \times b^4 \times c^2 \times c^5\)

⇒ \(-15 a^6 b^7 c^7\)

⇒ \(\left(5 a^4 b^3 c^2\right) \text { and }\left(-3 a^2 b^4 c^5\right)=-15 a^6 b^7 c^7\)

∴ Option (3) \(-15 a^6 b^7 c^7\) is the correct Answer.

Class 7 Maths Algebra Solutions WBBSE

4. The value of \(\frac{(-8 y) \div(4 z)}{4}\) is

1. -8yz
2. -16yz
3. \(\frac{2z}{y}\)
4. \(\frac{-y}{2z}\)

Solution:

⇒ \(\frac{(-8 y) \div(4 z)}{4}\)

⇒ \(\frac{\frac{-8 y}{4 z}}{4}\)

⇒ \(\frac{-8 y}{4 z} \times \frac{1}{4}\)

⇒ \(\frac{-8 y}{\frac{16 z}{2}}\)

⇒ \(\frac{-y}{2 z}\)

∴ \(\frac{(-8 y) \div(4 z)}{4}=\frac{-y}{2 z}\)

∴ Option (4) \(\frac{-y}{2 z}\) is correct answer.

Question 2. Write true or False:

1. a ÷ b of c = a ÷ b x c
2. a (b x c) = ab x ac
3. If the HFC of a and b is c then their LCM is \(\frac{ab}{c}\)

Solution:

1. a ÷ b of c = a ÷ b x c → False
2. a (b x c) = ab x ac → False
3. If the HFC of a and b is then their LCM is \(\frac{ab}{c}\) → True

Question 3. Fill in the blanks.

1. The value of a^x-y x a^y-z x a^z-x is _______
2. The sum of numerical coefficients of the term \(-\frac{1}{2} x^2, \frac{3}{4} y^2, \frac{1}{4} z^2 \text { and }-\frac{5}{2} x y\) is _______
3. The Subtraction of (5x² -4xy-y²) from O is ______

Solution:

1. \(a^{x-y} \times a^{y-z} \times a^{z-x}\)

⇒ \(a^{x-y+y-z+z+x} \quad a^m \times a^n \times a^0=a^{m+n+0}\)

⇒ \(a^{\prime}=0\)

⇒ \(a^{x-y} \times a^{y-z} \times a^{z-x}=0\)

2. The Sum of numerical coefficients of the term \(-\frac{1}{2} x^2, \frac{3}{4} x^2, \frac{1}{4} z^2\) and \(-\frac{5}{2} x y\) is

⇒ \(-\frac{1}{2}+\frac{3}{4}+\frac{1}{4}+\left(-\frac{5}{2}\right)\)

⇒ \(\frac{-2+3+1-10}{4}\)

⇒ \(\frac{-12+4}{4}\)

⇒ \(2 \times 2=4\)

∴ \(\frac{-8}{4}\)

∴ -2

∴ \(-\frac{1}{2} x^2, \frac{3}{4} y^2, \frac{1}{4} z^2 \text { and }-\frac{5}{2} x y=-2\)

3. The Subtraction of (5x²-4xy-y²) from 0 is

⇒ 0-(5x²-4xy-y²)

⇒ -5x² + 4ху + y²

∴ -5x² + 4xy + y²

WBBSE Class 7 Maths Chapter 4 Answers

Question 4. Identify the like terms. -a²b, 7ab², 1/2a²b, 5/2ab, 7a²b
Solution:

Like terms = -a²b, 7a²b, 1/2 a²b.

Question 5. Find the numerical coefficient of the terms other than the constant term.

1. 3x² – 5xy + 7
2. x – 3xy + 7y² – 6
3. \(\frac{5x^3}{7}\) + 4x – 8

Solution:

1. 3x² – 5xy + 7

The numerical coefficients are (3,-5).

2. x – 3xy + 7y² – 6

The numerical coefficients are (1,-3, +7)

3. \(\frac{5x^3}{7}\) + 4x – 8

The numerical coefficient is (5/7,4)

Question 6. Represent the following algebraic expressions into a factor tree type of figure mentioning the prime factors of each term.

1. 7x
2. 3y²+27+5

Solution:

WBBSE Class 7 Algebra Exercise Solutions

Question 7. How much must be added to (x² – 7x + 8) to get (3x² + 8x – 9)
Solution:

(x² +7x +8) to get (3x² + 8x -9)

(3x² + 8x-9) – (x² – 7x + 8)

3x² + 8x – 9 – x² + 7x – 8

2x² + 15x – 17

∴ 2x² + 15x – 17 must be added to (x² – 7x + 8) to get (3x² + 8x – 9).

Question 8. Multiply

1. (5x – 7y) (5x + 7y)
2. \(\left(-\frac{2}{3} a^2 b c^2\right) \times\left(\frac{7}{2} a b^5 c^5\right) \times\left(\frac{3}{7} a^3 b c^4\right)\)

Solution:

1. (5x – 7y)(5x + 7y)

Here a = 5, b = 7y

⇒ (5x)² – (7y)²

⇒ 25x² – 49y²

∴ (5x-7y)(5x+7y) = 25x² – 49y²

2. \(\left(-\frac{2}{3} a^2 b c^2\right) \times\left(\frac{7}{2} a b^5 c^5\right) \times\left(\frac{3}{7} a^3 b c^4\right)\)

⇒ \(-\left(\frac{2}{3} \times \frac{7}{2} \times \frac{3}{7}\right) \times\left(a^2 b c^2\right) \times\left(a b^5 c^5\right) \times\left(a^3 b c^4\right)\)

⇒ \(-\left(a^{2+1+3} \cdot b^{1+5+1} \cdot c^{2+5+4}\right)\)

⇒ \(-\left(a^6 \cdot b^7 \cdot c^{11}\right)\)

∴ \(\left(-\frac{2}{3} a^2 b c^2\right) \times\left(\frac{7}{2} a b^5 c^5\right) \times\left(\frac{3}{7} a^3 b c^4\right)=-\left(a^6 \cdot b^7 \cdot c^{11}\right)\)

Question 9. Divide:

1. \(\left(22 x^5-11 x^3+33 x^2 y\right)\) by \(11 x^4\)
2. \(169 a^2 b c^4-52 a b^4 c^5-78 a^3 b^3 c^3 b y-13 a b c^4\)

Solution:

1. \(\left(22 x^5-11 x^3+33 x^2 y\right)\) by \(11 x^4\)

⇒ \(\frac{22 x^5}{11x^3}-\frac{11 x^4}{11 x^4}+\frac{33 x^2 y}{11 x^4}\)

⇒ \(2 x-\frac{1}{x}+\frac{3 y}{x^2}\)

∴ \(\left(22 x^5-11 x^3+33 x^2 y\right) \text { by } 11 x^4\) = \(2 x-\frac{1}{x}+\frac{3 y}{x^2}\)

2. \(169 a^2 b c^4-52 a b^4 c^5-78 a^3 b^3 c^3 b y-13 a b c^4\)

⇒ \(\frac{169 a^2 bc^4}{-13 a b c^4}-\frac{52 a b^4 c^5}{-13 a b c^4}-\frac{78 a^3 b^3 c^3}{-13 a b c^4}\)

⇒ \(-13 a+4 b^3 c+6 a^2 b^2 c^{-1}\)

⇒ \(-13 a+4 b^3 c+\frac{6 a^2 b^2}{c^-1}\)

∴ \(169 a^2 b c^4-52 a b^4 c^5-78 a^3 b^3 c^3 by -13 a b c^4\) is \(-13 a+4 b^3 c+\frac{6 a^2 b^2}{c}\)

Class 7 Algebra Problems With Solutions

Question 10. If the perimeter of an isosceles triangle is (3a-4b+5c) cm and the length of its base is (a+2b-c)cm then Find the length of the other sides.
Solution:

Given Data, Isosceles Triangle

Perimeter = (3a-4b+5c)cm

Base length = (a+2b-c)cm

length of other sides =?

In the Isoceles Triangle two sides are equal in length

perimeter = 2a+b; a = side , b = base

(3a-4b+5c) = 2a +(a+2b-c)

(3a-4b+5c) – (a+2b-c) = 2a

3a – 4b + 5c – a – 2b + c = 2a

3a – a – 4b -2b + 5c + c = 2a

2a – 6b + 6c = 2a

2(a-3b+3c) = 2a

∴ Length of other sides are a = (a-3b+3c) cm

Question 11. Simplify

1. \(\frac{a-b}{a b}+\frac{b-c}{b c}+\frac{c-a}{c a}\)
Solution:

⇒ \(\frac{a-b}{a b}+\frac{b-c}{b c}+\frac{c-a}{c a}\)

⇒ \(\frac{c(a-b)+a(b-c)+b(c-a)}{a b c}\)

⇒ \(\frac{a c-b c+a b-ac c+b c-a b}{a b c}\)

⇒ \(\frac{1}{a b c}\)

∴ \(\frac{a-b}{a b}+\frac{b-c}{b c}+\frac{c-a}{c a}=\frac{1}{a b c}\)

2. \((x-5)\left(x^2+5 x+25\right)+(2 x-1)\left(4 x^2+2 x+1\right)-(3 x-2)\left(9 x^2+6 x+4\right)\)
Solution:

⇒ \(x^3+5 x^2+25 x-5 x^2-25 x-125+8 x^3+4 x^2+2 x-4 x^2-2 x-1\)–\((27 x^3+18 x^2+12 x-18 x^2-12 x-8)\)

⇒ \(x^3+8 x^3-27 x^3+5 x^2-5 x^2+4 x^2-4 x^2\)–\(18 x^2+18 x^2+25 x-25 x+2 x-2 x+12 x-12 x-125-1+18\)

∴ \(-18 x^3-118\)

3. (a – b)(b – 2c + a)+ (b – c)(c – 2a + b) + (c – a) (a – 2b + c)
Solution:

⇒ (a – b)(b – 2c + a)+ (b – c)(c – 2a + b) + (c – a) (a – 2b + c)

⇒ ab-2ac+a²-b²+2bc-ab+bc-2ab+b²-c²+2ac-bc+ac-2bc+c²-a²+2ab-ac

∴ (a-b) (b-2c+a)+(b-c)(c-2a+b) + (c-a)(a-2b+c) = 0

Class 7 Maths Chapter 4 Solved Exercises

Question 12. If A = 2x+3y-4z; B = 2y+3x-4z and C = 2z + 3x – 4y then find the value of (A-2B-3C)
Solution:

Given that.

A = 2x+3y-4z

B = 2y+3x-4z

C = 2z+3x-4y

(A-2B-3c) = {(2x+3y-4z) – 2(2x+3x-4z)-3(2z +3x-4y)}

= {(2x+3y-4z)-4y-6x+8z-6z-9x+12y}

= (2x+3x-4z-4y-6x+8z-6z-9x+12y)

= (2x-6y-9x + 3y – 4y + 12y -4z + 8z – 62)

(A-2B-3C) = (-13x + 11y – 2z)

Algebra Chapter 3 Concept of Index

Question 1. Choose the correct Answer

1. If we express the number 1234500000 in an index. form as power of 10 then we get

1. 1234.5 x 10⁶
2. 12345 x 10⁶
3. 123.45 x 10⁶
4. 12.345 x 10⁶

Solution:

Given number 1234500000

Now Count the Zero one in the number is ‘5’

we must be present in a form as the power of 10.

∴ 12345 x 10⁵ = 1234.5x 10⁶ (Power should not be Odd)

Option (1) 1234.5 x 10⁶ is the correct answer.

Read and Learn More Class 7 Maths Solutions

2. The value of \(\frac{3^4 \times 3^{-5}}{3^{-6}}\) is

1. 243
2. 343
3. 81
4. 1/9

⇒ \(\frac{3^4 \times 3^{-5}}{3^{-6}}\) is

Solution:

∵ a^m x aⁿ = a^m+n

Here a = 3, m = 4, n =-5

⇒ \(\frac{3^{4+(-5)}}{3^{-6}}\)

⇒ \(\frac{3^{-1}}{3^{-6}}\)

⇒ \(3^{-1)-(-6)}\) ∵ \(\frac{a^m}{a^n}=a^{m-n}\)

Here, a = 3, m = -1, n = -6

⇒  \(3^{-1+6}\)

⇒  \(3^5\)

⇒ 3 x 3 x 3 x 3 x 3

⇒ 243

∴ \(\frac{3^4 \times 3^{-5}}{3^{-6}}\) = 243

Option 1 is 243 is the correct answer

Class 7 Algebra Problems With Solutions

3. The value of (a³ x b)² x (a² x b^-3)³ is

1. \(\frac{a^6}{b^7}\)
2. \(a^6 \cdot b^7\)
3. \(a^7 \cdot b^6\)
4. \(\frac{a^7}{b^6}\)

Solution:

(a⁰ x b)² x (a² x b^-3)³  ∵ a = 1

⇒ (1x b)² x (a² x b^-3)³  ∵  (a x b)^m = a^m x b^m

⇒ (1)² x (b)² x (a²)³ x (b^-3)³

In first bracket (1 x b)²

a = 1, b = b, m = 2

In the second bracket (a² x b^-3)³

∵ a = a², b = b^-3, m = 3

⇒ 1 x b² x a^2×3 x b^-3×3

⇒ b² x a⁶ x b^-9

⇒ a⁶ x b² x b^-9

⇒ a⁶ × b^2-9

⇒ a⁶ x b^-7

Here, a = b,

m = 2

n = -9

⇒ a⁶ x \(\frac{1}{b^7}\)

⇒ \(\frac{a^6}{b^7}\)

Option 1 is correct

Question 2. Write true or false.

1. 2 x 10⁵ + 3 × 10⁴ +4 × 10³ + 5 × 10² + 6 × 10 + 7 = 765433
2. The expression of 6x6x6x6x6 in terms of the index of a prime number is (2⁵ x 3⁵)
3. The value of \(\frac{2^6 \times 8 \times 16}{2^{11} \times 2^2}\) is 1

Solution:

1. 2 x 10⁵ + 3 × 10⁴ + 4 x 10³ + 5 × 10² +6 × 10 + 7= 765433

⇒ 200000+30000+4000 + 500+60 +7 = 765433

⇒ 234,567 = 765433

∴ 2 x 10⁵ + 3 × 10⁴ + 4 × 10³ + 5 x 10² + 6 x 10 + 7 = 76.5433; False

2. 6 × 6 x 6 x 6 x 6 = (2⁵ х 3⁵)

⇒ 2×3×2×3×2×3×2×3×2×3 = 2⁵ × 3⁵

⇒ 2×2×2×2×2×3×3×3×3×3 = 2⁵ x 3⁵

⇒ 2⁵ x 3⁵ = 2⁵ x 3⁵

∴ 6x6x6x6x6 =(2⁵ x 3⁵) → True

WBBSE Class 7 Algebra Chapter 3 

3. The value of \(\frac{2^6 \times 8 \times 16}{2^{11} \times 2^2}\) is 1

⇒ \(\frac{2^8 \times 8 \times 16}{\frac{2^{14}}{2^5} \times 2^2}\)

⇒ \(\frac{128}{32 \times 4}\)

⇒ \(\frac{128}{128}=1\)

∴ \(\frac{2^6 \times 8 \times 16}{2^{11} \times 2^2}=1\); True

Question 3. Fill in the blanks:

1. (-5)² x (6)² = ( )⁵
2. 0.53= 0·053 x 10
3. The value of \(\frac{3 \times 7^2 \times 2^5}{14 \times 42}\) is

Solution:

1. (-5)² x(6)² = ( )²

⇒ 2.5×36

⇒ 900

⇒ (30)²

∴ (-5)² x (6)² = (30)²

2. 0.53 = 0.053 × 10¹

⇒ 0.053×10¹

⇒ 0.53

∴ 0.53= 0.053×10¹

3. The value of \(\frac{3 \times 7^2 \times 2^5}{14 \times 42}\)

⇒ \(\frac{3 \times 7^2 \times 2^5}{14 \times 42}\)

⇒ \(\frac{3 \times 49 \times 32}{588}\)

⇒ \(\frac{4.704}{588}\)

= 8

∴ \(\frac{3 \times 7^2 \times 2^5}{14 \times 42}=8\)

Class 7 Maths Algebra Solutions WBBSE

Question 4. Express the following number in terms of the index

1. 4672
2. 12045
3. 400608

Solution:

1. 4672

Number in terms of index 4 ×10³+6×10²+7×10¹ +2

⇒ 4000+600+ 70+2

⇒ 4672

2. 12045

Number in terms of index 19 x 10⁴ +2 ×10³+ 4 ×10 + 5

⇒ 10000+2.000+40+5

⇒ 12,045

3. 4000608

Number in terms of index

⇒ 4×10⁶+6×10² +8

⇒ 4000000 +600+8

⇒ 4000608.

WBBSE Class 7 Maths Chapter 3 Answers

Question 5. Express in terms of the index of a prime number.

1. 1800
2. 882
3. 80

Solution:

1. 1800

Taking LCM of 1800

∴ 2 x 2 x 2 x -3 x 3 x 5 x 5

∴ 2³×3² x 5²

∴ 1800 = 2³ x 3² x 5²

Index of prime number.

2. 882

Taking LCM of 882

∴ 2 x3 × 3 ×7 x 7

∴2 x 3² x 7²

∴ 882=2 x 3² x 7² index of prime number.

3. 80

Taking LCM of 80.

∴ 2 x 2 x 2 x 2 x 5

∴ 2⁴ x 5

∴ 80 = 2⁴ x 5 index of prime number.

Class 7 Maths Chapter 3 Solved Exercises

Question 6. Simplify

1. \(\frac{x^{-2} \times x^7}{x^8 \times x^{-4}}\)
2. \(\frac{(18)^3 \times(2.5)^2}{(30)^4}\)

Solution:

1. \frac{x^{-2} \times x^7}{x^8 \times x^{-4}} \quad a^m \times d^n=a^{m-n}[/latex]

Here a = x, m = -2, n = 7 → Numerator

a = x, m = 8, n = -4 → Denominatore

⇒ \(\frac{x^{-2+7}}{x^{8+(-4)}}\)

⇒ \(\frac{x^5 x}{x^4}\)

∴ x

∴ \(\frac{x^{-2} \times x^7}{x^8 \times x^{-4}}=x\)

2. \(\frac{(18)^3 \times(2.5)^2}{(30)^4}\)

⇒ \(\frac{18 \times 18 \times 18 \times 25 \times 25}{30 \times 30 \times 30 \times 30}\)

⇒ \(\frac{58,32 \times 625}{810,000}\)

⇒ \(\frac{3.64,5,000}{810,000} \Rightarrow \frac{364.5}{810}=4.5\)

∴ \(\frac{(18)^3 \times(2.5)^2}{(30)^4}=4.5\)

Question 7. Find the value of

1. \(\frac{7^3 \times 2^6 \times 10}{3.5 \times 224}\)
2. \(\frac{3^4 \times 6^5}{72}\)

Solution:

1. \(\frac{7^3 \times 2^6 \times 10}{35 \times 224}\)

⇒ \(\frac{343 \times 64 \times 10}{35 \times 224}\)

⇒ \(\frac{219,520}{7840}\)

= 28

∴ \(\frac{7^3 \times 2^6 \times 10}{35 \times 224}=28\)

2. \(\frac{3^4 \times 6^5}{72}\)

⇒ \(\frac{81 \times 776}{72}\)

⇒ \(\frac{629,856}{72}=8748\)

⇒ \(\frac{3^4 \times 6^5}{72}=8748\)

WBBSE Maths Study Material Class 7

Question 8. Express in terms of index.

1. \(\frac{(a b)^6 \times\left(a^2 c\right)^2}{\left(a^4\right)^2 \times(b c)^5}\)
2. \(9 \times 27 \times 81 \times 243\) (index of 3)

Solution:

1. \(\frac{a^6 \times b^6 \times a^2 \times a^2 \times c^2}{a^4 \times a^4 \times b^5 \times c^5 c^3}\)

⇒ \(\frac{a^2 \times b}{c^3}\)

⇒ \(a^2 \times b \times c^{-3}\)

∴ \(\frac{(a b)^6 \times\left(a^2 c\right)^2}{\left(a^4\right)^2 \times(b c)^2}=a^2 \times b \times c^{-3}\)

2. \(9 \times 27 \times 81 \times 243\) (index of 3)

⇒ 3 x 3 x 9 x 3 x 9 x 9 x 3 x 81

⇒ 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 9 x 9

⇒ 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3

⇒ 3¹⁴

∴ 9 x 27 x 81 x 243 = 3¹⁴

Algebra Chapter 2 Exercise

Question 1. Choose the correct answers:

1. The value of (-6) x (-12) -(-4) x (+3) is

1. 60
2. -84
3. 84
4. -60

Solution:

⇒ (-6) x (-12)  – (-4) x (+3)

⇒+72 +12

⇒ +84

∴ The value of (-6) x (-12) – (-4) x (+3) is 84

Option (3) → 84 is the correct Answer

Read and Learn More Class 7 Maths Solutions

2. If x = -2, y = 3 and z = -5 then the value of (x-y+z) is

1. 0
2. -4
3. -10
4. -4

Solution:

Given that

x = -2,

y = 3

z = -5

(x-y+z)=?

∴ (x-y+z) = {(-2) – (3) + (-5)}

Here Substituted the values of x,y,z

= {(-2-3-5)}

∴ (x-y+z) = -10

Option (3)=-10 is correct.

Class 7 Algebra Problems With Solutions 

3. The value of 4(3+1)÷(-4) of (2) is

1. 2
2. -2
3. 8
4. -8

Solution:

⇒ 4(3+1) ÷ (-4) Of (-2)

⇒  4(3+1) ÷ (-4)

⇒ 4(4) ÷ (-4)

⇒ 16÷(-4)

⇒ -4

∴ -4 of (-2)

Question 2. Write true or false:

1. a ÷ (b + c) = \(\frac{a}{b} + {a}{c}\)
2. If a and b are two different integers then a + b = b + a but a-b ≠ b-a
3. For any three integers, the law of association of integers is not valid in the case of subtraction.

Solution:

1. False
2. a+b = b+a but a-b ≠ b-a; True
3. True

Question 3. Fill in the blanks

1. (-15) ÷ (-5) ÷ (-3)=
2. (-1) x (-2) x (-3) × ____ = (-5) × (-6)
3. (-20) ÷ (-4) of 5 x 2 = _____

Solution:

1. (-15)÷(+5)=(-3)

⇒ \(\frac{-15}{-5}{\frac{-3}{-1}}\)

⇒ \(\frac{-15}{-5} \times \frac{1}{-3}\)

⇒ -1

∴ (-15) ± (-5) ÷ (-3) = -1

2. (-1) x (-2 )x (-3) × _____ = (-5) x (-6)

(-1) x (-2)× (-3) × x = (-5)× (-6)

Let the number be x

⇒ 2 × (-3) x x = 30.

⇒ -3x = \(\frac{30}{2}\)

x = \(\frac{15}{-3}\)

∴ x = -5

∴ (-1) × (-2) x (-3) × (-5) = (-5) x (-6)

3. (-20) ÷ (-4) of (5 x 2 )= _____

⇒ (-20) ÷ (-4) of (5×2)

⇒ (-20) ÷ (-20) x (2)

⇒ (+1) x (2)

⇒ 2

∴ (-20) ÷ (-4) of (5 x 2) = 2

Class 7 Maths Algebra Solutions WBBSE

Question 4. Find the Sum with the help of a number line.

1. (-8)+(+5)+(-4)
2. (-7)+(-3)+(+10).

Solution:

1. (-8)+(+5) + (-4)

⇒ -8 + 5 -4

⇒ -12 + 5

⇒ -7

2. (-7) + (-3) + (+10)

⇒ (-7-3+10)

⇒ -10 + 10

⇒ 0

Question 5. Find the values of the following on the number line:

1. (-4)-(-3)
2. (-5)x(+4)
3. (-3)+{(+4)+(-2)}
4. (-8)÷(+2)

Solution:

1. (-4) – (-3)

⇒ -4 + 3

⇒ -1

2. (-5) x (+4)

⇒ -20

3. (-3)+{(+4)+(-2)}

⇒ (-3) + {4-2}

⇒ -3 +2

⇒ -1

4. (-8) ÷ (+2)

⇒ -4

WBBSE Class 7 Maths Chapter 2 Answers

Question 6. Verify if the distributive law of multiplication holds for integers:

1. (-5) x {(-2)+(-4)}
2. (+8) ÷ {(+4)+(-2)}

Solution:

(-5) x {(-2)+(-4)}

⇒ (-5) x {-6}

⇒ +30

Or,

(-5) x (-2) + (-5) x (-4)

⇒ (-10) + (-20)

⇒ +30

∴ (-5) × {(-2) + (-4)}

= (-5) X (2) + (-5) x (-4)

∴ The distributive face of multiplication is verified

Question 7. The temperature of Kashmir is 24°C If the temperature reduces uniformly every hour and reaches to -4°C after 7 hours. Find the rate of reduction of temperature. per hour.
Solution:

Given Data,

The temperature of Kashmir is (T_i) = 24°C

Final Temperature after 7hrs (T_c) = -4°C

Change in temperature is T_f-T_i = -4-24 = -28°C

∴ Rate of Reduction = \(\frac{\text { change in temperature }}{\text { Time }}\)

= \(\frac{-28}{7}\)

= -4°c/hour.

∴ The rate of reduction of temperature per hour is 4°C

WBBSE Class 7 Algebra Exercise Solutions

Question 8. Draw a number line and verify with examples that the commutative law of Subtraction does not hold.
Solution:

Let us take a simple example

1. 5-3=2 and also
2. 3-5=-2

These results are not the same.

∴ The commutative law of Subtraction does not hold true for these numbers.

Question 9. By what number should be added to ((-5)x(4)-(+3) to get the number [{(-12) (-3)}x (-2)]
Solution:

⇒ [{(-12) ÷ (-3)} x (-2)] – {(5)x(4)-(+3)}

⇒ [{\(\frac{-12}{-3}\) x(-2)] – {(20-3)}

⇒ [4 x (2)] – (17)

⇒ -8-17

⇒ -25

∴ {(-5)× (4) −(+3)} to get the number [{(-12) ÷ (-3)} × (-2)] is -25

Algebra Formulas For Class 7 WBBSE

Question 10. Find the value of [720÷ (21+3) ÷ (-6) x 5] ÷ (-25)  
Solution:

⇒ \([720 \div(21+3) \div(-6) \times 5] \div(-25)\)

⇒ \([720 \div(24) \div(-30)] \div(-25)\)

⇒ \(\left[\frac{\frac{720}{24}}{(-30)}\right] \div(-2.5)\)

⇒ \(\left[\frac{720}{24} \times \frac{1}{(-30)}\right] \div(-25)\)

⇒ \(\left(\frac{720}{-720}\right) \div(-25)\)

⇒ \((-1) \div(-25)\)

⇒ \(\frac{+1}{+25}\)

∴ 0.04

∴ \([720 \div(21+3) \div(-6) \times 5] \div(-25)=0.04\)

Algebra Chapter 1 Revision

Question 1. The sum of 9 and (-y) is
Solution:

Sum of ‘a’ and ‘b’ is (a + b)

⇒ 9 + (-y)

⇒ a + (b)

⇒ Here a = 9;

b = -y

⇒ a – b

∴ The sum of 9 and (-y) is a-b

Question 2. What must be done added to (-17) to get 12?
Solution:

The number which is added to be (-17) to get 2 is {a+b}

Let us assume a = -17

b = 12

⇒ {12 -(-17)}

⇒ {12 + 14}

⇒ 29

∴ Added to (-17) to get (12) is 29

Read and Learn More Class 7 Maths Solutions

Class 7 Algebra Problems With Solutions

Question 3. The value of (-2) x (-3) x (-5) is
Solution:

⇒ (-2)² x (-3)² x (-5) ∵ a² = a x a

⇒ (-2) x (-2) x (-3)² x (-3) x (-5)

⇒ 4 x 9 x (-5)

⇒ 36 x (-5)

⇒ -180

∴ The value of (-2)² x (-3)² x (-5)² is -180

Question 4. Write ‘true or false’

1. A profit of -10 rupees means that loss of 10 rupees.
Solution: The statement is true

2. If the length and breadth of a rectangle are ‘ and ‘y’ respectively, its semi-perimeter is 2(x+y).
Solution:

Semi-Perimeter is (x + y),

So the statement is false

3. The difference between the two numbers is x. If the greater number is y then the least number is (x+y).
Solution:

The least number is (y-x)

So the statement is false

Algebra Questions For Class 7 WBBSE

Question 3. Fill in the blanks.

1. The value of (-5)² x (-7) x (-6) is
Solution:

⇒ (-5)² x (-7) x (-6)

⇒ (-5)x (-5) × (7) ×(-6)  ∵ a² = a x a

⇒ 25×42

⇒ 1050

∴ The value of (-5)² x (-7) x (-6) is 1050

2. If the perimeter of a square is x cm, then its area is ________ Sq. cm.
Solution:

The perimeter of the square is x cm.

Length of each side is \(\frac{x}{4}\) cm

Area is \(\left(\frac{x}{4}\right)^2\) sq cm

Area = \(\frac{x^2}{16}\) Sq.cm

3. The absolute value of (-3) is
Solution: 3

WBBSE Class 7 Algebra Chapter 1

Question 4. Write in language the following expressions.

1. x/4 – 3
2. a </ 4
3. 3P-2

Solution:

1. x/4 -3 ⇒  Three less than one fourth of ‘x’
2. a </ 4 ⇒ ‘a’ is not less than four.
3. 3P-2 ⇒ 2 is less trim three times of p

Question 5. Form the algebraic expansion with signs and Symbols:

1. 5 is subtracted from 4 times y
2. 4 is not less than x
3. x is not equal to ‘y’.
4. The sum of Five times y and 6.

Solution:

1. 5 is subtracted from 4 times y ⇒ 4y-5
2. 4 is not less than ‘x’ ⇒ 4 </ x
3. ‘x’ is not equal to ‘y’ ⇒ x ≠ y
4. The sum of five times y and ‘6’ ⇒ 5y+6

Question 6. Subtract using the Concept of opposite number:-

1. (-13)-(-16)
2. (+12)-(-15)
3. (-17)-(+18)
4. (+10)-(+15)

Solution:

1. (-13)-(-16) ⇒ Subtract with opposite number.

⇒ (-13)+(+16)

⇒ +3

⇒ (-13)-(-16)=+3

2. (+12) – (-15)

⇒ (+12) + (opposite number (-15))

⇒ (+12)+(+15)

⇒ +27

∴ (+12)-(-15) = +27

3. (-17)-(+18)

⇒ (-17)+(opposite number +18)

⇒ (-17)+(-18)

⇒ -35

∴ (-17)-(+18)= -35

4. (10) – (+15).

⇒ (+10)+(opposite number of +15)

⇒ (+10) + (-15)

⇒ -5

∴ (10)-(+15)=-5

Class 7 Maths Algebra Solutions WBBSE

Example: 7 Simplify: 10(opposite number of -25)-(opposite number of +12)- (opposite number of -18) – (-6).
Solution:

⇒ 10 -(+25)-(-12) – (+18)-(-6).

⇒ 10-25+12-18+6

⇒ (10+12+6) – (25+18)

⇒ 28-43

⇒ -15

⇒ 10-(+25)-(-12)-(+18) -(-6)=-15

Question 8. Add the following on the number line:

1. (-7), (+2);
2. (+ 4), (-8).

Solution:

1. (-7), (+2)

⇒ -7 +2

= -5

2. (+4), (-8)

⇒ (+4)+(-8)

⇒ 4-8

⇒ -4

WBBSE Class 7 Algebra Exercise Solutions

Question 9. Verify associative property of addition. (-5), (-3), (+2)
Solution:

⇒ {(-5) + (-3)} + (+2)

⇒ {(-8) + (+2)}

∴ -6

Or,

⇒ (-5) + {(-3) + (+2)}

⇒ (-5) + {(-1)}

∴ -6

So, {(-5) + (-3)} + (+12) = (-5) + {(-3) + (+2)}

∴ The associative property of addition is verified.

Question 10. Find what must be added to the first to get the Second.

1. (-15), (-10)
2. (+6), (-18).

Solution:

1. The number which added to the (-15) to get (-10) is

⇒ (-10) -(-15)

⇒ (-10) + 15

⇒ +5

∴ (-15), (-10) = +5

2. (+6), (-18)

The number added to the (+6) to get (18) is.

⇒ (-18) – (+6)

⇒ -18-6

⇒ -24

The required number is (-18) – (+6) = -24

Algebra Chapter 1 Exercise 2

Question 1. Choose the correct answer:

1. The sum of two numbers is a; if the least number is b. then the greater number is

1. a-b
2. a+b
3. b-a
4. None of these.

Solution:

Let us assume the two numbers are x, y

Given that,

⇒ The sum of two numbers is ‘a’

⇒ x+y=a ……(1)

∵ Given condition.

If the least number is ‘b’

Now, y = b

∴ x + b = a

x = a-b

∴ The greater number is x = a -b so, option (1) is correct.

2. If the length and breadth of a rectangle are X cm and Y cm then its Area is

1. 2(x+y) Sq cm.
2. хуcm
3. xy sq. cm
4. x/y sq. cm

Solution:

Given Rectangle Dimensions.

Length (l) = x cm

breadth (b) = y cm

Area =?

we know that,

⇒ The area of the rectangle is length x breadth.

⇒ l x b

⇒ х х у

∴ xy sq. cm

∴ The area of the rectangle is xy sq cm.

∴ The correct answer is option 3

Class 7 Maths Chapter 1 Solved Exercises

3. If the digits in the unit’s place and ten’s place are x and y respectively, then the number is.

1. x+y
2. 10x+y
3. 10y+2
4. 10(x+y)

Solution:

This problem is solved by verifying the options.

Given condition:

⇒ The digits in the unit’s place and ten’s Place are x and y.

∴ The number is y+x

The unit’s place digit is ‘x’

The ten’s place digit is y.

Here the equation is 10y+x.

The correct answer is ‘3’

Question 2. Write true or false:

1. The value of (+3)+(-6)+(+9) Is 0.
2. The value of (-2)2 x 52x (-3)2 is 900
3. \(\frac{1}{x}+\frac{1}{y}=\frac{x+y}{x y}\)

Solution:

1. The value of (+3) + (-6) + (+9) = 0

⇒ +3+(-6)+(+9)

⇒ 3-6+9

⇒ 12-6

⇒ 6 ≠ 0

∴ The value of (13) + (-6) + (+9) is ‘0’ → False

2. The value of (-2)² x (5)² x (-3)²

⇒  (-2)² x (5)² × (-3)²

⇒ 4 × 25 × 9

⇒  100×9

⇒  900

∴ The value of (-2)x(5) x (-3)² = 900 True

3. \(\frac{1}{x}+\frac{1}{y}=\frac{x+y}{x y}\)

⇒ \(\frac{1}{x}+\frac{1}{y}\) {Lcm of x, y}

⇒ \(\frac{y+x}{x y} \Rightarrow \frac{x+y}{x y}\)

∴ The value of \(\frac{1}{x}+\frac{1}{y}=\frac{x+y}{x y}\) is True.

Question 3 Fill in the blanks:

1. The absolute value of (-15) is
2. The Value of (-18) ÷ (-3) is
3. The difference between two numbers is ‘x’; if the greater number is ‘p’ then the last one is ________

Solution:

1. The absolute value of (-15) is 15

2. The Value of (-18)= ÷ (-3) is 6

3. The difference between two numbers is ‘x’; if the greater number is ‘p’ then the last one is p-x

a-b = x  ∵ Let the two numbers be ab

p-b = x ∵Greater number is p’ then a = P

∴ Least number = b

b= p – x

Class 7 Algebra Problems With Solutions 

Question 4. From two Algebraic expressions.

1. 5 is not greater than ‘y’
2. x is Subtracted from 6 times p.
3. The difference of 7 times between ‘a’ and ‘b’

Solution:

1. 5 >/ y
2. 6p – x
3. 7(a-b)

Question 5. Write in language the following expression:

1. \(\frac{p}{6}\) – 8
2. x + y + 7
3. 3a-b
4. x >/ Y

Solution:

1. \(\frac{p}{6}\) ⇒ 8 ‘g’ is subtracted from one-sixth of “p’.
2. (x+y+z) ⇒ The sum of x,y,z
3. (3a-b) ⇒ ‘b’ is subtracted from three times ‘a’
4. (x >/ y) ) ⇒ ‘x’ is not greater than ‘y’

Question 6. Subtract using the concept of opposite numbers.

1. (+16)-(-25)
2. (-15)-(+36)
3. (-18)-(-6)

Solution:

1. (+16)-(-25)

⇒ (+16)+(opposite number of (-25)).

⇒ (+16)+(25)

⇒ +41

∴ (+16)-(-25)= +41.

2. (-15)-(+36)

⇒ (-15)+(opposite number of (+36))

⇒ (-15)+(-36)

⇒ -51

∴ (-15)-(+36) = -51

3. (-18)-(-6)

⇒ (-18)+(opposite number of (-6))

⇒ (-18) + (+6)

⇒ -12

∴ (-18)-(-6) =-12

Algebra Questions For Class 7 WBBSE

Question 7. Adding on a number line

1. (-7) + (+15)
2. (+25)+(-20)
3. (-3)+(-2)

Solution:

1. (-7)+(+15)

⇒ -7+15

⇒ +8

2. (+25)+(-20)

⇒ +25-20

⇒ +5

3. (-3)+(-2)

WBBSE Class 7 Algebra Chapter 1

Question 8. Add the following

1. (-14), (+12), (-16)
2. (+13), (-4), (-9)
3. (-18), (-12) (+19)

Solution:

1. (-14) + (+12) + (-16)

⇒ -14 +12-16

⇒ +12-30

⇒ -18

2. (+13)+(-4)+(-9)

⇒ 13-4-9

⇒ 13-13

⇒ 0

3. (-18) + (-12)+(+19)

⇒ -18-12+19

⇒ -30+19

⇒ -11

Question 9. Find what must be added to the first to get the second:

1. (-13), (+15)
2. (+18), (-19)
3. (+18), (-17)

Solution:

1. (-13), (+15)

The number added to the (-13) to get (+15) is

⇒ (15)-(-13)

⇒ 15+13

⇒ +28

∴  (-13), (+15)= +28

The number that must be added to the First to get the second is +28

2. (+18), (-19)

The required number is (-19)-(+18)

⇒ (-19)-(+18)

⇒-19-18

⇒ -37

3. (+18), (+7).

The required number is (+7) – (+18)

⇒ +7-18

⇒ -11

The number that must be added to the first to get the second is -11

Class 7 Maths Algebra Solutions WBBSE

Question 10. Verify the Associative property of addition for the following.

1. (-3)(-2), (-5)
2. (-7) (+9), (-8)
3. (+4) (-6) (-10)

Solution:

1. (-3)(-2), (-5)

Associative Property {(a) + (b)} + c = (a)+{(b) + (c)}

⇒ {(-3)+(-2)}+(-5)

⇒ {-3-2} – 5

⇒ {-5-5}

⇒ -10

or,

⇒ (-3) + {(-2)+(-5)}

⇒ (3)+ {-2-5}

⇒ (-3)+{-7}

⇒ -3-7

⇒ -10

So, {(-3) + (-2)} +(-5) = (-3)+ {(-2)+(-5)}

2. (-7) (+9), (-8)

⇒ {(-7)+(+9)}+(-8)

⇒ {(-7 +9)} +(-8)

⇒ (+2)+(-8)

⇒ +2-8

⇒ -6

Or,

(-7) + {(+9) + (-8)}

⇒ (-7) + {9-8}

⇒ (-7) + (+1)

⇒ -7 + 1

⇒ -6

So, {(7) + (+9)} + (-8) = (-7) + {(+9) + (-8)}

3. (-14), (-6), (-10)

⇒ {(-14)+(-6)}+(-10)

⇒ {(-14-6)} +(-10)

⇒ (-20)-10

⇒ -30

Or,

⇒ (-14)+{(-6)+(-10)}

⇒ (-14)+{-6-10}

⇒ (-14)+(-16)

⇒ -30

So, {(-14)+(-6)} +(-10) = (-14) + {(-6) + (-10)}

WBBSE Class 7 Algebra Exercise Solutions

Question 11. Simplify 18-(-7)+(opposite number of -15)-(opposite number -6) -(opposite number of +14).
Solution:

⇒ 18-(-7) + (+15) −(+6) −(-14)

⇒ 18+7+15-6+14

⇒ 25+ 15-6 +14

⇒ 54-6

⇒ 48.

∴ 18-(-7)+(+15)-(+6)-(-14)=48

Question 12.If a= -2, b=-3, c = +6, then Find the values of

1. (a – b + c)
2. (a x b) ÷ c
3. a ÷ b x c
4. a + b ÷ c

Solution:

1. (a – b + c)

⇒ (-2) – (-3) + 6

⇒ (-2)+3+6

⇒ -2+9

⇒ 7

2. (a x b) ÷ c

⇒ (-2)+(-3)÷6

⇒ (2)x(-0.5)

⇒ 1

3. a ÷ b x c

⇒ (-2) ÷ (-3) x (+6)

⇒ 0.6 x 6

4

4. a + b ÷ c

⇒ (-2) + (-3) ÷ 6

⇒ (-2) – 0.5

= -2.5

⇒ -5/2