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.
- x/4 – 3
- a </ 4
- 3P-2
Solution:
- x/4 -3 ⇒ Three less than one fourth of ‘x’
- a </ 4 ⇒ ‘a’ is not less than four.
- 3P-2 ⇒ 2 is less trim three times of p
Question 5. Form the algebraic expansion with signs and Symbols:
- 5 is subtracted from 4 times y
- 4 is not less than x
- x is not equal to ‘y’.
- The sum of Five times y and 6.
Solution:
- 5 is subtracted from 4 times y ⇒ 4y-5
- 4 is not less than ‘x’ ⇒ 4 </ x
- ‘x’ is not equal to ‘y’ ⇒ x ≠ y
- The sum of five times y and ‘6’ ⇒ 5y+6
Question 6. Subtract using the Concept of opposite number:-
- (-13)-(-16)
- (+12)-(-15)
- (-17)-(+18)
- (+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:
- (-7), (+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.
- (-15), (-10)
- (+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
- a-b
- a+b
- b-a
- 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
- 2(x+y) Sq cm.
- хуcm
- xy sq. cm
- 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.
- x+y
- 10x+y
- 10y+2
- 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:
- The value of (+3)+(-6)+(+9) Is 0.
- The value of (-2)2 x 52x (-3)2 is 900
- \(\frac{1}{x}+\frac{1}{y}=\frac{x+y}{x y}\)
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:
- The absolute value of (-15) is
- The Value of (-18) ÷ (-3) is
- 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.
- 5 is not greater than ‘y’
- x is Subtracted from 6 times p.
- The difference of 7 times between ‘a’ and ‘b’
Solution:
- 5 >/ y
- 6p – x
- 7(a-b)
Question 5. Write in language the following expression:
- \(\frac{p}{6}\) – 8
- x + y + 7
- 3a-b
- x >/ Y
Solution:
- \(\frac{p}{6}\) ⇒ 8 ‘g’ is subtracted from one-sixth of “p’.
- (x+y+z) ⇒ The sum of x,y,z
- (3a-b) ⇒ ‘b’ is subtracted from three times ‘a’
- (x >/ y) ) ⇒ ‘x’ is not greater than ‘y’
Question 6. Subtract using the concept of opposite numbers.
- (+16)-(-25)
- (-15)-(+36)
- (-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
- (-7) + (+15)
- (+25)+(-20)
- (-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
- (-14), (+12), (-16)
- (+13), (-4), (-9)
- (-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:
- (-13), (+15)
- (+18), (-19)
- (+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.
- (-3)(-2), (-5)
- (-7) (+9), (-8)
- (+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
- (a – b + c)
- (a x b) ÷ c
- a ÷ b x c
- 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