WBBSE Class 7 Maths Solutions For Algebra Chapter 1

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