CBSE Class 7

These NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Exercise 12.3

Question 1.
If m – 2, find the value of:
(i) m – 2
(ii) 3m – 5
(iii) 9 – 5m
(iv) 3m² – 2m – 7
(v) \(\frac { 5m }{ 2 }\) – 4
Answer:
(i) m- 2 = 2 – 2 (m = = 2)
= 0

(ii) 3m – 5 = 3 (2) – 5 (m = – 2)
= 6 – 5 = 1

(iii) 9 – 5m = 9 – 5(2) (m = = 2)
= 9- 10
= -1

(iv) 3m² – 2m – 7
= 3 (2)² – 2(2) – 7
(m = 2)
= 3 (4) – 4 – 7
= 12 – 4 – 7
= 12 – 11
= 1

(v) \(\frac { 5m }{ 2 }\) – 4 = \(\frac{5(2)}{2}\) – 4 (m = 2)
= 5 – 4
= 1

Question 2.
If p = -2, find the value of:
(i) 4p + 7
(ii) – 3p² + 4p + 7
(iii) – 2p² – 3p² + 4p + 7
Answer:
(i) 4p + 7 = 4(-2) + 7
= – 8 + 7 = -1

(ii) -3p² + 4p + 7
= – 3 (-2)² + 4(-2) + 7
(when p = -2)
= – 3 (4) + (- 8) + 7
= -12 – 8 + 7
= -20 +7
= -13

(iii) -2p³ – 3p² + 4p + 7 =
-2 (-2)³ – 3 (-2)² + 4 (-2) + 7
(when p = -2)
= – 2(- 8) — 3 (4) + (— 8) + 7 = + 16 – 12 – 8 + 7 = 23 – 20 = 3

Question 3.
Find the value of the following expressions, when x = -1:
(i) 2x – 7
(ii) -x + 2
(iii) x² + 2x + 1
(iv) 2x² – x -2
Answer:
(i) 2x – 7 = 2(-1) – 7 (whenx = -1)
= -2 – 7 = -9

(ii) -x + 2 = -(-1) + 2 (When x = -1)
= 1 + 2 = 3

(iii) x² + 2x + 1 = (-1)² + 2 (-1) + 1
(When x = – 1)
= 1 – 2 + 1
= 2 – 2 = 0

(iv) 2x² – x – 2 =2 (-1)² – (-1) – 2
(When x = -1)
= 2 + 1 – 2
= 3 – 2 = 1

Question 4.
If a = 2, b = – 2, find the value of:
(i) a² + b²
(ii) a² + ab + b²
(iii) a² – b²
Answer:
(i) a² + b² = (2)² + (-2)²
(When a = 2, b = -2)
= 4 + 4 = 8

(ii) a² + ab + b²
= (2)² + 2(-2) + (-2)²
(When a = 2, b = – 2)
= 4 – 4 + 4
= 8 – 4 = 4

(iii) a2 – b2 = 22 – (-2)2
(When a = 2, b = -2) = 4-(4) = 4- 4 = 0

Question 5.
When a = 0, b = – 1, find the value of the given expressions:
(i) 2a + 2b
(ii) 2a² + b² + 1
(iii) 2a²b + 2ab² + ab
(iv) a² + ab + 2
Answer:
(i) 2a + 2b = 2 (0) + 2(—1)
(When a = 0, b = -1)
= 0 – 2 = -2

(ii) 2a² + b² +1
= 2 (0)² + (-1)² + 1
(When a = 0, b = -1)
= 0 + (1) + (1)
= 1 + 1 = 2

(iii) 2a²b + 2ab² + ab
= 2(0)² (-1) + 2(0) (-1)² + (0) (-1)
(When a = 0, b = -1)
= 2(0) (-1) + 2(0) (1) + 0
= 0 + 0 + 0
= 0

(iv) a² + ab + 2 = (0)² + 0 (-1) + 2
(When a = 0, b = -1)
= 0 – 0 + 2 = 2

Question 6.
Simplify the expressions and find the value if x is equal to 2.
(i) x + 7 + 4 (x – 5)
(ii) 3 (x + 2) + 5x – 7
(iii) 6x + 5 (x – 2)
(iv) 4(2x – l) + 3x+ 11
Answer:
(i) x + 7 + 4 (x -5)
= x + 7 + 4x – 20
= x + 4x + 7 – 20
= 5x – 13 (When x = 2)
= 5 (2) – 13
= 10- 13
= -3

(ii) 3 (x + 2) + 5x – 7 = 3x + 6 + 5x – 7
= 3x + 5x + 6 – 7
= 8x – 1 (When x = 2)
= 16 – 1 = 15

(iii) 6x + 5 (x – 2) = 6x + 5x – 10
= 1 lx – 10 (When x = 2)
= 11 (2) – 10 = 22-
= 12

(iv) 4(2x – 1)+ 3x + 11
= 8x + 3x – 4 + 11
= 8x + 3x + 11 – 4
= 11x + 7 (When x = 2)
h = 11(2) + 7
= 22 + 7 = 29

Question 7.
Simplify these expressions and find their values if x = 3, a = -1, b = -2.
(i) 3x – 5 – x + 9
(ii) 2 – 8x + 4x + 4
(iii) 3a + 5 – 8a + 1
(iv) 10 – 3b – 4 – 5b
(v) 2a – 2b – 4 – 5 + a
Answer:
(i) 3x-5-x + 9
= 3x-x-5 + 9
= 2x + 4 (When x = 3)
= 2(3)+ 4
= 6 + 4
= 10

(ii) 2-8x + 4x + 4
= -8x + 4x + 4 + 2
= (- 8 + 4) x + 6
= – 4x + 6
(When x = 3)
= – 4(3) + 6
= -12+ 6 =-6

(iii) 3a + 5 – 8a + 1
= 3a – 8a + 5 + 1
= (3 – 8) a + 6
= -5a + 6
(When a = – 1)
= -5 (-1) + 6
= 5 + 6= 11

(iv) 10 – 3b – 4 – 5b = -3b -5b +10-4
= (-3 -5) b + 6 = – 8 b + 6
(When b = – 2) = – 8 (-2) + 6
= 16 + 6 = 22

(v) 2a – 2b – 4 – 5 + a =
2a + a – 2b – 4 -5 = 3a – 2b – 9
(When a = – 1 and b = -2) = 3 (-1) -2 (-2) -9 = -3 + 4 -9 = -12+ 4 =-8

Question 8.
(i) If z = 10, find the value of z³ – 3 (z – 10).
(ii) If p = – 10, find the value of p² – 2p – 100
Answer:
(i) z³ – 3 [z – 10] = 10² – 3 [10- 10]
(When z = 10)
= 1000 – 3 (0)= 1000

(ii) p² – 2p – 100
= (-10)² – 2 (-10) – 100
(When p = -10)
= 100 + 20 – 100
= 120 – 100 = 20

Question 9.
What should be the value of a if the value of 2x² + x – a equals to 5, when x = 0? Ans: 2x²+ x – a = 5 (Given: x = 0)
2(0)² + 0 – a = 5
0 + 0 – a = 5
-a =5
a = – 5
The value of a = – 5

Question 10.
Simplify the expression and find its value when a = 5 and b = – 3.
2(a² + ab) + 3 – ab
Answer:
2(a² + ab) + 3 – ab
= 2 [5² + 5(-3)] + 3 -(5) (-3)
= 2 [25- 15] +3 + 15
= 2(10) + 18
= 20 + 18 = 38

These NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Exercise 4.1

Question 1.
Complete the last column of the table.
Answer:

Question 2.
Check whether the value given in the brackets is a solution to the given equation or not:
(a) n + 5 = 19 (n = 1)
(b) 7n + 5 = 19 (n = – 2)
(c) 7n + 5 = 19 (n = 2)
(d) 4p – 3 = 13 (p = 1)
(e) 4p – 3 = 13 (p = – 4)
(f) 4p – 3 = 13 (p = 0)
Answer:
(a) n + 5 = 19
1 + 5 =19 (Putting n = 1)
6 ≠ 19
∴ n = 1 is not a solution.

(b) 7n + 5 =19
7(- 2) + 5 =19 (Put n = – 2)
-14 + 5 = 19
-9 ≠ 19
n = -2 is not a solution.

(c) 7n + 5 =19
7(2) + 5=19 (Put n = 2)
14 + 5 = 19
19 = 19
∴ n = 2 is a solution.

(d) 4p – 3 = 13
4(1) – 3 =13 (Put p = 1)
4 – 3 = 13
1 ≠ 13
∴ p = 1 is not a solution.

(e) 4p – 3 = 13
4(- 4) – 3 = 13 (Put p = – 4)
-16 -3 = 13
-19 ≠ 13
∴ p = – 4 is not a solution.

(f) 4p – 3 = 13
4(0) – 3 = 13 (Put p = 0)
0 – 3 = 13
-3 ≠ 13
p = 0 is not a solution.

Question 3.
Solve the following equations by trial and error method:
(i) 5p + 2 = 17
(ii) 3m – 14 = 4
Answer:
(i) 5p + 2 = 17
Put p = 0
L.H.S = 5 (0) + 2
= 0 + 2
= 2 ≠ RHS

Put p = 1
L.H.S = 5(1)+ 2
= 5 + 2
= 7 ≠ RHS

Put p = -1
L.H.S = 5 (-1) + 2
= -5 + 2
= -3 ≠ RHS

Put p = 2
L.H.S = 5 (2) + 2
= 10 + 2
= 12 ≠ RHS

Put p = -2
L.H.S = 5 (-2) + 2
= -10 + 2
= -8 ≠ 17 RHS

Put p = 3
L.H.S = 5 (3) + 2
= 15 + 2
= 17 = RHS
p = 3 is the solution of 5P + 2 = 17

(ii) 3m – 14 =4
Put m = 0
L.H.S = 3m – 14
= 3 (0) – 14
= – 14
– 14 ≠ 4
L.H.S ≠ R.H.S

Put m = 1
L.H.S = 3m -14
= 3(1) – 14
= 3 – 14
= – 11 ≠ R.H.S
L.H.S ≠ R.H.S

Put m = 2
L.H.S = 3m – 14
= 3 (2) – 14
= 6 – 14
= – 8 ≠ R.H.S
L.H.S ≠ R.H.S

Put m = 3
L.H.S = 3(3) – 14
= 9 – 14
= – 5 ≠ R.H.S
L.H.S ≠ R.H.S

Put m = 4
L.H.S = 3(4) – 14
= 12 – 14
= – 2 ≠ R.H.S
L.H.S ≠ R.H.S

Put m = 5
L.H.S = 3(5) – 14
= 15 – 14
= 1 ≠ R.H.S
L.H.S ≠ R.H.S

Put m = 6
L.H.S = 3(6) – 14
= 18 – 14
= 4 = R.H.S
L.H.S = R.H.S
m = 6 is the solution to 3m – 14 = 4

Question 4.
Write equations for the following statements:
(i) The sum of numbers x and 4 is 9.
(ii) 2 subtracted from y is 8.
(iii) Ten times a is 70.
(iv) The number b divided by 5 gives 6.
(v) Three-fourth of t is 15.
(vi) Seven times m plus 7 gets you 77.
(vii) One-fourth of a number x minus 4 gives 4.
(viii) If you take away 6 from 6 times y, you get 60.
(ix) If you add 3 to one-third of z, you get 30.
Answer:
(i) x + 4 = 9
(ii) y – 2 = 8
(iii) 10 a = 70
(iv) \(\frac { b }{ 5 }\) = 6
(v) \(\frac { 3t }{ 4 }\) = 15
(vi) 7m + 7 = 77
(vii) \(\frac { x }{ 4 }\) x – 4 = 4
(viii) 6 y – 6 = 60
(ix) \(\frac { z }{ 3 }\) + 3 = 30

Question 5.
Write the following equations in statement forms:
(i) p + 4 = 15
(ii) m – 7 = 3
(iii) 2m = 7
(iv) \(\frac { m }{ 5 }\) = 3
(v) \(\frac { 3m }{ 5 }\) = 6
(vi) 3p + 4 = 25
(vii) 4p – 2 = 18
(viii) \(\frac { p }{ 2 }\) + 2 = 8
Answer:
(i) The sum of p and 4 is 15.
(ii) 7 subtracted from m is 3.
(hi) Twice a number m is 7.
(iv) One-fifth of a number m is 3.
(v) Three-fifth of a number m is 6.
(vi) Three times a number p when added to 4 gives 25.
(vii) 2 subtracted from four times a number p is 18.
(viii) 2 added to half of a number p is 8.

Question 6.
Set up an equation in the following cases:
(i) Irfan says that he has 7 marbles more than live times the marbles Parmit has. Irfan has 37 marbles. (Take m to be the number of Parmit’s marbles.)

(ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. (Take Laxmi’s age to be y years.)

(iii) The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest score is 87. (Take the lowest score to be 1.)

(iv) In an isosceles triangle, the verte x angle is twice either base angle. (Let the base angle be b in degrees. Remember that the sum of angles of a triangle is 180 degrees).
Answer:
(i) Let Parmit has m marbles.
Then, five time the marbles Parmit has = 5 m.
Irfan has 7 marbles more than five times the marbles parmit has
So, Irfan has (5m + 7) marbles
But it is given that Irfan has 37 marbles.
5m + 7 = 37

(ii) Let Laxmi’s age = y years
3 times Laxmi’s age = 3y years.
Age of Laxmi’s father = 3 times
Laxmi’s age + 4 years
= 3y + 4 years
But Laxmi’s father is 49 years old.
3y + 4 = 49

(iii) Let the lowest score (marks) = 1
Twice the lowest marks = 2 l
Since highest marks = (twice the lowest marks) + 7 = 2l + 7
But the highest marks = 87
2l + 7 = 87

(iv) Let the base angle be b degrees
The base angle of an isosceles triangle are equal.
The other base angle = b degrees
Since the vertex angle = Twice either base angle = 2b degrees.
Also, the sum of three angles of triangle = 180°
b + b + 2b = 180°
(OR)
4b = 180°

These NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Exercise 12.2

Question 1.
Simplify combining like terms:
(i) 21b – 32 + 7b – 20b
(ii) – z² + 13z² – 5z + 7z³ – 15z
(iii) p – (p – q) – q – (q – p)
(iv) 3a – 2b – ab – (a – b + ab) + 3ab + b – a
(v) 5x²y – 5x² + 3yx² – 3y² + x² – y² + 8xy² – 3y²
(vi) (3y² + 5y – 4) – (8y – y² – 4)
Answer:
(i) 21b – 32 + 76 – 20b
Combining the like terms, we have
(21b + 7b – 20b) + (- 32) = (21 + 7 – 20) b + (- 32)
= 8b – 32

(ii) -z² + 13z² – 5z + 7z³ – 15z
Combining like terms, we get
= 7z³ + (-z² + 13z²) + (-5z – 15z)
= 7z³ + (-1 + 13)z² + (- 5 – 15)z
= 7z³ + 12z² – 20z

(iii) p-(p-q)-q-(q-p) = p- p + q – q – q + p
Combining like terms, we get
= p – p + p + q – q – q
= (1 – 1 + 1) p + (1 – 1 – 1) q
= p + (-1)q
= p – q

(iv) 3a – 2b – ab – (a – b + ab) + 3ab + b – a
= 3a – 2b – ab – a + b – ab + 3ab + b -a
Combining like terms, we get
= 3a – a – a -2b + b + b – ab – ab + 3ab
= (3 – 1 – 1) a + (-2 + 1 + 1) b + (-1 -1 + 3)ab
= (3 – 2) a + (-2 + 2) b + (-2 + 3) ab
= (1) a + (0) b + (1) ab
= a + ab

(v) 5x²y – 5x² + 3yx² – 3y² + x² – y² + 8xy² – 3y²
Combining the like terms, we get
(5x²y + 3yx²) + 8xy² + (-5x² + x² ) + (-3y² – y² – 3y² )
(5 + 3) x²y + 8xy² + (-5 + 1) x² + (-3 – 1 – 3)y²
8x²y + 8xy² + (- 4) x² + (-7) y²
8x²y + 8xy² – 4x² – 7y²

(vi) (3y² + 5y – 4)-(8y – y² – 4)
= 3y² + 5y – 4 – 8y + y² + 4
Combining the like terms, we get
= 3y² + y² + 5y – 8y – 4 + 4
= (3 + 1)y² + (5-8)y + (-4 + 4)
= 4y² – 3y

Question 2.
Add:
(i) 3mn, – 5mn, 8mn, – 4mn
(ii) t – 8tz, 3tz – z, z – t
(iii) – 7mn + 5, 12mn + 2, 9mn – 8, – 2mn – 3
(iv) a + b-3, b-a + 3, a-b + 3
(v) 14x + lOy – 12xy – 13, 18 – 7x – 10y + 8xy, 4xy
(vi) 5m – 7n, 3n – 4m + 2, 2m – 3mn – 5
(vii) 4x²y, – 3xy², – 5xy², 5x²y
(viii) 3p²q² – 4pq + 5, – 10p²q², 15 + 9pq + 7p²q²
(ix) ab – 4a, 4b – ab, 4a – 4b
(x) x² – y² – 1, y² -1 – x², 1 – x² – y²
Answer:
(i) 3mn, – 5mn, 8mn, – 4mn
3mn + (-5mn) + 8mn + (- 4mn)
= 3mn – 5mn + 8mn – 4mn
= (3 – 5 + 8 – 4) mn = 2mn

(ii) t – 8tz, 3tz – z; z -t
t – 8tz + 3tz + (-z) + z + (-t)
= t – 8tz + 3tz – z + z – t
Combining like terms, we get
= t – t-z + z – 8tz + 3tz
= (1 – 1) t + (- 1 + 1) z + (- 8 + 3) tz
= (0) t + (0) z + (-5) tz
= 0 + 0 – 5tz
= -5tz

(iii) -7mn + 5; 12mn + 2; 9mn – 8; – 2mn -3
-7mn + 5 + 12mn + 2 + 9mn – 8 + (-2mn – 3)
– 7mn + 5 + 12mn + 2 + 9mn – 8 -2mn – 3
Combining like terms, we get
– 7mn + 12mn + 9mn – 2mn + 5 + 2 – 8 – 3
= (-7 + 12 + 9-2) mn + (5 + 2 – 8 -3)
= (21-9) mn + (7 – 11)
= (12) mn + (- 4)
= 12mn – 4

(iv) a + b-3;b-a + 3;a-b + 3
a + b – 3 + b – a + 3 + a – b + 3
Combining like terms, we get
= (a – a + a) + (b + b – b) + (-3 + 3 + 3)
= (1 – 1 + 1) a + (1 + 1 – 1)b + (6 – 3)
= (2 – 1) a + (2 -1) b + 3
= a + b + 3

(v) 14x + 10y – 12xy – 13; 18 – 7x – lOy + 8xy, 4xy
14x + 10y – 12xy – 13 + 18 – 7x – 10y + 8xy + 4xy
Combining like terms, we get
= (14x – 7x) + (10y – 10y) + (-12xy + 8xy + 4xy)
= (14 – 7)x + (10 – 10) y + (-12 + 8 + 4) xy+ (-13 + 18)
= 7x + (0)y + (0)xy + (+5)
= 7x + 5

(vi) 5m – 7n, 3n – 4m + 2; 2m – 3mn – 5
5m – 7n + 3n – 4m + 2 + 2m – 3mn – 5
Combining like terms, we get
= (5m – 4m + 2m) + (-7n + 3n) + (-3mn) + (2 – 5)
= (5 – 4 + 2) m + (-7 + 3) n – 3mn + (-3)
= (7 – 4) m + (- 4) n – 3mn -3
= 3m – 4n – 3mn – 3

(vii) 4x²y; – 3xy², – 5xy²; 5x²y
4x²y + (-3xy²) + (-5xy²) + 5x²y
= 4x²y – 3xy² – 5xy² + 5x²y
Combining like terms, we get
= 4x²y + 5x²y – 5xy² – 5xy²
= (4 + 5) x²y + (- 3 – 5) xy²
= 9x²y + (- 8) xy²
= 9x²y – 8xy²

(viii) 3p²q² – 4pq + 5; – 10p²q²; 15 + 9pq + 7p²q²
3p²q² + (- 4pq) + 5 + (-10p²q²) + 15 + 9pq + 7p²q²
= 3p²q² – 4pq + 5 – 10p²q² + 15 + 9pq + 7p²q²
Combining like terms, we get
= 3p²q² – 10p²q² + 7p²q² – 4pq + 9pq + 5 + 15
= (3 – 10 + 7) p²q² + (- 4 + 9) pq + (5 + 15)
= (10 – 10) p²q² + (5) pq + 20
= (0) p²q² + 5pq + 20
= 5pq + 20

(ix) ab – 4a; 4b – ab; 4a – 4b
ab – 4a + 4b – ab + 4a – 4b = (ab – ab) + (- 4a + 4a) + (4b – 4b)
= (1 – 1) ab + (- 4 + 4) a + (4 – 4) b
= (0) ab + (0) a + (0) b
= 0 + 0 + 0
= 0

(x) x² – y² – 1; y² – 1 – x²; 1 – x² – y²
x²– y² – 1 + y² – 1 – x² + 1 – x² – y²
= (x² – x² – x²) + (- y² + y² – y²) + (-1 – 1 + 1)
= (1 – 1 – 1)x² + (-2 + 1)y² + (-1)
= (1 – 2) x² + (-2 + 1)y² + (-1)
= (- 1) x² + (-1) y² +(-1)
= – x² – y² – 1

Question 3.
Subtract:
(i) – 5y² from y²
(ii) 6xy from -12xy
(iii) (a – b) from (a + b)
(iv) a (b – 5) from b (5 – a)
(v) -m² + 5 mn from 4m² – 3mn + 8
(vi) – x² + 10x – 5 from 5x – 10
(vii) 5a² – 7ab + 5b² from 3ab – 2a² – 2b²
(viii) 4pq – 5q² – 3p² from 5p² + 3q² – pq
Answer:
(i) Subtract – 5y² from y²
y² – (-5y² ) = y² + 5y² – 12xy – (6xy) = 6y²

(ii) Subtract 6xy from – 12xy = – 12xy – 6xy
= (-12 – 6)xy = -18xy

(iii) Subtract (a – b) from (a + b)
(a + b)-(a-b) = a + b- a + b =a-a+b+b
= (1 – 1)a + (1 + 1)b
= (0)a + 2b
= 0 + 2b
= 2b

(iv) Subtract a (b – 5) from b (5 – a) b (5 – a) – a (b – 5)
= 5b – ab – ab + 5a
= 5b + 5a – ab – ab
= 5a + 5b + (-1 -1) ab
= 5a + 5b – 2ab

(v) Subtract -m² + 5mn from 4m² – 3mn + 8
4m² – 3mn + 8 – (-m² + 5mn)
= 4m² – 3mn + 8 + m² – 5mn
= 4m² + m² – 3mn – 5mn + 8
= (4 + 1) m² + (-3 – 5) mn + 8 = 5m² + (- 8) mn + 8
= 5m² – 8mn + 8

(vi) Subtract -x² + 10x – 5 from 5x – 10
5x – 10 – (-x² + 10x – 5)
= 5x – 10 + x² – 10x + 5
= x² + 5x – 10x – 10 + 5
= x² + (5 – 10) x + (-10 + 5)
= x² + (-5) x + (-5)
= x² – 5x – 5

(vii) Subtract 5a² – 7ab + 5b² from 3ab – 2a² – 2b²
3ab – 2a² – 2b² – (5a² – 7ab + 5b²)
= 3ab – 2a² – 2b² – 5a² + 7ab – 5b²
= -2a² – 5a² – 2b² – 5b² + 3ab + 7ab
= (-2 -5) a² + (-2 – 5) b² + (3 + 7) ab
= -7a² – 7b² + 10ab (or) 10ab – 7a² – 7b²

(viii) Subtract 4pq – 5q² – 3p² from 5p² + 3q² – pq
5p² + 3q² – pq – (4pq – 5q² – 3p²)
= 5p² + 3q² – pq – 4pq + 5p² + 3p²
= (5 + 3) p² + (3 + 5)q² – pq – 4pq
= 8p² + 8q² -5pq

Question 4.
(a) What should be added to x² + xy + y² to obtain 2x² + 3xy?
(b) What should be subtracted from 2a + 8b + 10 to get – 3a + 7b + 16?
Answer:
(a) The required expression is
2x² + 3xy – (x² + xy + y²)
= 2x² + 3xy – x² – xy – y²
= (2x² – x²) – y² + 3xy – xy
= (2 – 1) x² – y² + (3 – 1) xy
= (1) x² – y² + (2) xy
= x² + 2xy – y²

(b) The required expression is
(2a + 8b + 10) – (- 3a + 7b + 16)
= 2a + 8b + 10 + 3a – 7b – 16
= (2a + 3a) + (8b – 7b) + 10- 16
= (2 + 3) a + (8 – 7) b + (10 – 16)
= (5)a + (l)b + (-6)
= 5a + b – 6

Question 5.
What should be taken away from 3x² – 4y² + 5xy + 20 to obtain -x² – y² + 6xy + 20?
Answer:
The required expression is
3x² – 4y² + 5xy + 20 –
(-x² – y² + 6xy + 20)
= 3x² – 4y² + 5xy + 20 – (-x2 -y2 + 6xy + 20)
= 3x² – 4y² + 5xy + 20 + x² + y² – 6xy – 20)
= (3x² + x²) + (- 4y² + y²) +
(+ 5xy – 6xy) + (20 – 20) = (3 + 1) x2 + (- 4 + 1) y2 +
(+ 5 – 6) xy + (0)
= (4) x² + (-3) y² + (-1xy)
= 4x² – 3y² – xy

Question 6.
(a) From the sum of 3x – y + 11 and – y -11, subtract 3x – y – 11.
(b) From the sum of 4 + 3x and 5 – 4x + 2x², subtract the sum of 3x² – 5x and -x² + 2x + 5.
Answer:
(a) The required expression is (3x-y+ 11) + (- y-11) –
(3x – y -11)
= 3x – y + 11 – y – 11 – 3x + y + 11
= (3x – 3x) + (-y – y + y) + (11 – 11 + 11)
= (3 – 3)x + (-1 -1 + 1) y + (11 – 11 + 11)
= (3 – 3)x + (-2 + 1)y + (22 – 11)
– (0) x + (-1) y + (11)
= 0 – y + 11 = -y + 11

(b) The required expression is
(4 + 3x) + (5 – 4x + 2x²) –
[(3x² – 5x + (-x² + 2x + 5)]
= 4 + 3x + 5 – 4x + 2x² – [3x² – 5x – x2 + 2x + 5]
= 4 + 3x + 5 – 4x + 2x² – 3x² +5x + x2 – 2x – 5
= (2x² – 3x² + x²) + (3x -4x + 5x- 2x) + (4 + 5 – 5)
= (2 – 3 + 1) x² + (3 – 4 + 5 – 2) x + (9-5)
= (3 – 3) x² + (8 – 6) x + 4
= (0)x² + (2) x + 4
= 0 + 2x + 4
= 2x + 4

These NCERT Solutions for Class 7 Maths Chapter 3 Data Handling InText Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 3 Data Handling InText Questions

NCERT In-text Question Page No. 59
Question 1.
Weigh (in kg) at least 20 children (girls and boys) of your class. Organise the date, and answer the following questions using this data.
(i) Who is the heaviest of all?
(ii) What is the most common weight?
(iii) What is the difference between your weight and that of your best friend?
Answer:
We assume that weights (in kg) of 20 children of the class are as follows:
27, 30, 28, 24, 25, 26, 21, 20, 18, 15,
28, 16, 27, 23, 24, 26, 22, 31, 18, 26
Organising the data in ascending order, we have.
15, 16, 18, 18, 20, 21, 22, 23, 24, 24,
25, 26, 26, 27, 27, 28,28, 30,31

(i) The child having weight 31 kg is the heaviest of all.
(ii) The most common weight is 26 Kg.
(iii) If my weight is 27, then required difference = 27 kg – 25 Kg = 2 Kg.

NCERT In-text Question Page No. 61
Question 1.
How would you find the average of your study hours for the whole week?
Answer:
In a week there are 7 days. The number of study hours everyday are different. Let the study hours for different days of the week are as follows:
Monday : 6 hours
Tuesday : 7 hours
Wednesday : 6 hours
Thursday : 8 hours
Friday : 8 hours
Saturday : 7 hours
Sunday : 7 hours
Total number of study hours.
= 6 + 7 + 6 + 8 + 8 + 7 + 7 = 40 hours
Number of total days = 7
∴ Arithmetic mean = \(\frac{\text { Total study hours }}{\text { Total Number of days }}\)
Thus, average study hours = 7 hours per day.

NCERT In-text Question Page No. 61
Question 1.
Find the mean of your sleeping hours daring one week?
Answer:
Let the sleeping hours be as follows:
Monday : 8 hours
Tuesday : 9 hours
Wednesday : 11 hours
Thursday : 8 hours
Friday : 10 hours
Saturday : 9 hours
Sunday : 8 hours
Sum of sleeping hours
= 8 + 9+11 + 8+10 + 9 + 8 = 63
Number of days = 7
∴ Arithmetic mean = \(\frac{\text { Sum of sleeping hours }}{\text { Number of days }}\)
= \(\frac{63}{7}\) hours = 9hours

Question 2.
Find at least 5 numbers between \(\frac{1}{2}\) and \(\frac{1}{3}\).
Answer:
The arithmetic mean of two given num¬bers is average between them

NCERT In-text Question Page No. 65
Question 1.
Find the mode of?
(i) 2, 6, 5, 3, 0, 3,4, 3, 2, 4, 5, 2, 4
(ii) 2, 14,16,12,14,14, 16, 14,10, 14,18,14
Answer:
(i) Given set of data is :
2, 6, 5, 3, 0, 3, 4, 3, 2, 4, 5, 2, 4
Writing the numbers with same value together, we have:
0, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 6
∵ 2,3, and 4 occur in the given data highest times (3 times).
∴ Modes of the set of data are 2, 3 and 4.

(ii) Given set of data is :
2, 14, 16, 12, 14, 14, 16, 14, 10, 14, 18, 14
Writing the numbers with same value together, we have:
2, 10, 12, 14, 14, 14, 14, 14, 14, 16, 16, 18
∵ 14 occurs highest times (i.e. six times).
∴ Required mode is 14.

NCERT In-text Question Page No. 65-66
Question 1.
Find the mode of the following data.
12, 14, 12, 16, 15, 13, 14, 18, 19, 12, 14, 15, 16, 15, 16, 16, 15, 17, 13, 16, 16, 15, 15, 13, 15, 17, 15, 14, 15, 13, 15, 14
Answer:
Here the given set of data has large number of observations so, we put them in a table
Number Tally Marks Frequency

The highest frequency is 10, corresponding value is 15.
The required mode is 15.

Question 2.
Heights (in cm) of 25 children are given below:
168, 165, 163, 160, 163, 161, 162, 164, 163, 162, 164, 163, 160, 164, 163, 160, 165, 163, 162
What is the mode of their heights? What do we understand by mode here?
Answer:
Putting the data ina tabular form, we have

Since, 163 is occuring the maximum number of times.
∴ Mode is 163 cm.
By the mode, we mean that most of stu¬dents have height 163 cm.

Note :
Mode is modal value of a group which occurs most frequently in the data whereas mean gives us the average of all observations of the data.

NCERT In-text Question Page No. 66
Question 1.
Discuss with your friends and give:
(a) Two situations where mean would be an appropriate representative value to use, and
(b) Two situations where mode would be an appropriate representative value to use.
Answer:
(a) (i) For measures of heights of all students of a class, and
(ii) For weights of wheat stored in similar containers, the mean would give us a better picture of the data.

(b) (i) If we are collecting the shoe size suitable for a particular age, and
(ii) A shopkeeper selling shirts needs to know suitable size to enrich the stock, then mode is more useful.

NCERT In-text Question Page No. 67
Question 1.
Your friend found the median and the mode of a given data. Describe and correct your friend’s error if any:
35, 32, 35, 42, 38, 32, 34
Median = 42, Mode = 32
Answer:
Set of data is: 35, 32, 35, 42, 38, 32, 34
Putting the given data in ascending order, we have:
32, 32, 34, 35, 35, 38, 42

(i)∵ The middle value of the data is 35.
∴ Median = 35

(ii)∵ The observations occuring maximum number of times are 32 and 35.
∴ The correct modes are 32 and 35.

NCERT In-text Question Page No. 71 & 72
Question 1.
The bar graph given below shows the result of a survey to test water resistant watches made by different companies.
Each of these companies claimed that their watches were water resistant. After a test the above results were revealed.
(a) Can you work out a fraction of the number of watches that leaked to the number tested for each company?
(b) Could you tell on this basis which company has better watches?

Answer:
(a) From the graph, we have :
Total number of watches tested for each company = 40
Fraction of ‘the numbers of watches that leaked’ to ‘the number of watches tested’ 20 1
For company A = \(\frac{20}{40}=\frac{1}{2}\)
For company B = \(\frac{10}{40}=\frac{1}{4}\)
For company C = \(\frac{15}{40}=\frac{3}{8}\)
For company D = \(\frac{25}{40}=\frac{5}{8}\)
(b) Obviously company B has better watches
∵ \(\frac { 1 }{ 4 }\) < \(\frac { 3 }{ 8 }\) > \(\frac { 1 }{ 2 }\) < \(\frac { 5 }{ 8 }\)

Question 2.
Sale of English and Hindi books in the years 1995, 1996, 1997 and 1998 are given below:

Draw a double bar graph and answer the following questions:
(a) In which year was the difference in the sale of the two language books least?
(b) Can you say that the demand for English books rose faster? Justify.
Answer:
(a) The difference in the sale of the two language books:
In 1995 is 500 – 350 = 150
In 1996 is 525 – 400 = 125
In 1997 is 600 – 450 = 150
In 1998 is 650 – 620 = 30
The difference in the scale of the two language books is least in the year 1998.

(b) Yes we can say that the demand of English books rose faster, during 1995 to 1998
(i) Sale of Hindi books rose from 500 to 650 i.e. it rose by 650 – 500 = 150
(ii) Sale of English books rose from 350 to 620 i.e. it rose by 620 – 350 = 270
270 is much greater than 150

NCERT In-text Question Page No. 74
Question 1.
Think of some situations, atleast 3 examples of each, that are certain to happen, some that are impossible and some that may or may not happen i.e., situations that have some chance of happening.
Answer:
(i) Situations that are certain to happen
(a) The sun rises in the east.
(b) Getting a number from 1 to 6 by throwing a dice.
(c) When a coin is tossed, then either a head or tail comes.

(ii) Situations that are impossible:
(a) Getting the number 8 by throwing a dice.
(b) To draw a five rupee coin from a bag containing one rupee coins.
(c) Drawing a red ball from a bag containing yellow and blue balls only.

(iii) Situations that have some chance of happening
(a) An ant rising to 5m height.
(b) To throw a dice and get an odd number.
(c) To toss a coin and get tail.

Note:
(i) A coin has two faces. The face with Ashoka Chakra is called ‘Head’ and other is called ‘Tail’.
(ii) A cube with six faces is called a dice. Its each face has making of dots numbering from 1 to 6.

NCERT In-text Question Page No. 75
Question 1.
Toss a coin 100 times and record the data. Find the number of times heads and tails occur in it.
Answer:
This is a group activity. Please do it yourself.

Question 2.
Aftaab threw a die 250 times and got the following table. Draw a bar graph for this data.
Answer:

NCERT In-text Question Page No. 76
Question 1.
Construct or think of five situations where outcomes do not have equal chances.
Answer:
In the following situations, the outcomes do not have equal chances:
(i) To draw red ball and to draw a yellow ball from a bag having 3 red balls and 5 yellow balls.
(ii) To throw a dice are getting 2 and getting an odd number.
(iii) To draw a flash card bearing ‘B’ and to draw a flash card bearing ‘C’ from a collection of 5 flash cards bearing A, B, C, D and E.
(iv) Choosing a boy and choosing a girl as class representative from a group of 10 boys and 15 girls.
(v) Choosing an even number and choosing an odd number from first five natural numbers.

These NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Exercise 12.1

Question 1.
Get the algebraic expressions in the following cases using variables, constants and arithmetic operations.
(i) Subtraction of z from y.
(ii) One-half of the sum of numbers x and y.
(iii) The number z multiplied by itself.
(iv) One-fourth of the product of numbers p and q.
(v) Numbers x and y both squared and added.
(vi) Number 5 added to three times the product of number m and n.
(vii) Product of numbers y and z subtracted from 10.
(viii) Sum of numbers a and b subtracted from their product.
Answer:
(i) y – z
(ii) \(\frac { 1 }{ 2 }\)(x + y)
(iii) z²
(iv) \(\frac { 1 }{ 4 }\)pq
(v) x² + y²
(vi) 5 + 3mn
(vii) 10 – yz
(viii) ab – (a + b)

Question 2.
(i) Identify the terms and their factors in the following expressions. Show the terms and factors by tree diagrams.
(a) x – 3
(b) 1 + x + x²
(c) y – y³
(d) 5xy² + 7x²y
(e) – ab + 2b² – 3a²

(ii) Identify terms and factors in the expressions given below:
(a) – 4x + 5
(b) – 4x + 5y
(c) 5y + 3y²
(d) xy + 2x²y²
(e) pq + q
(f) 1.2ab- 2.4 b + 3.6a
(g) \(\frac { 3 }{ 4 }\)x + \(\frac { 1 }{ 4 }\)
(h) 0.1p² + 0.2p²
Answer:
(i) (a) x – 3
Terms : x; – 3 Factors x; -3
1 + x + x²

(b) 1 + x + x²
Terms = 1; x; x²
Factors = 1; x; x, x (x² )

(c) y-y³
Terms: y; -y³
Factors y;-1,y,y,y

(d) 5xy² + 7x² y
Terms: 5xy²; 7x² y
Factors: 5, x, y, y; 7, x, x, y

(e) -ab + 2b² – 3a²
Terms: -ab; 2b², -3a²
Factors: -1, a, b; 2; b; b; 3, a, a

(ii)

Question 3.
Identify the numerical coefficients of terms (other than constants) in the following expressions:
(i) 5 – 3t²
(ii) 1 + t + t² + t³
(iii) x + 2xy + 3y
(iv) 100m + 1000n
(v) -p²q² + 7pq
(vi) 1.2a + 0.8b
(vii) 3.14r²
(viii) 2(l + b)
(ix) 0.1y + 0.01y²
Answer:

Question 4.
(a) Identify terms which contain x and give the coefficient of x.
(i) y²x + y
(ii) 13y² – 8yx
(iii) x + y + 2
(iv) 5 + z + zx
(iv) 1+x + xy
(vi) 12xy² + 25
(vii) 7x + xy²

(b) Identify terms which contain y² and give the coefficient of y².
(i) 8-xy²
(ii) 5y² + 7x
(iii) 2x²y – 15xy² + 7y²
Answer:

(b)

Question 5.
Classify into monomials, binomials and trinomials.
(i) 4y – 7z
(ii) y²
(iii) x + y – xy
(iv) 100
(v) ab – a – b
(vi) 5 – 3t
(vii) 4p²q – 4pq²
(viii) 7mn
(ix) z² – 3z + 8
(x) a² + b²
(xi) z² + z
(xii) 1 + x + x²
Answer:
(i) 4y – 7z
The expression 4y – 7z is having two unlike terms (4y and – 7z)
∴ It is a binomial.

(ii) y²
The expression y² is having only one term (y²)
∴ It is a monomial.

(iii) x + y – xy
The expression x + y – xy is having three terms (x, y and – xy)
∴ It is a trinomial.

(iv) 100
The expression 100 is having only one term (100)
∴ It is a monomial.

(v) ab – a – b.
The expression ab – a – b is having three terms (ab, -a, and -b)
∴ The expression is a trinomial.

(vi) 5 – 3t
The expression 5 – 3t is having two terms (5 and -3t)
∴ It is a binomial expression.

(vii) 4p²q – 4pq²
The expression 4p²q – 4pq² is having
two unlike terms (4p²q and – 4pq²)
∴ The expression is a binomial.

(viii) 7mn
The expression 7mn is having only one term (ie 7mn)
∴ The expression is a monomial.

(ix) z²– 3z + 8
The expression z² – 3z + 8 is having three terms (ie z², – 3z and 8)
∴ The expression is a trinomial.

(x) a² + b²
The expression (a² + b²) is having two unlike terms (a² and b²)
∴ It is a binomial expression.

(xi) z² + z
The expression z² + z is having two unlike terms (z2 and z)
∴ The expression in binomial.

(xii) 1 + x + x²
The expression 1 + x + x² is having three terms (1, x and x²)
∴ The expression is a trinomial.

Question 6.
State whether a given pair of terms is of like or unlike terms.
(i) 1, 100
(ii) -7x, \(\frac { 5 }{ 2 }\)x
(iii) -29x, -29y
(iv) 14xy, 42yx
(v) 4m²p, 4mp²
(vi) 12xz, 12x²z²
Answer:
(i) 1, 100 is a pair of like terms.
(ii) \(\frac { 5 }{ 2 }\)x is a pair of like terms.
(iii) – 29x, – 29y is a pair of unlike terms.
(iv) 14xy, 4²yx is a pair of like terms.
(v) 4m²p, 4mp² is a pair of unlike terms.
(vi) 12xz; 12x²z² is a pair of unlike terms.

Question 7.
Identify like terms in the following:
(a) -xy², – 4yx², 8x², 2xy², 7y, -11x², 100x, -11yx, 20x²y, – 6x², y, 2xy, 3x
(b) 10pq, 7p, 8q, – p²q², -7pq, -100q, -23, 12q²p², -5p², 41, 2405p, 78qp,
13p²q, qp²,701p²
Answer:
(a) -xy² and 2xy², – 4yx² and 20x² y, 8x², -11x² and – 6x²; 7y and y; lOOx and 3x; -1 lyx and 2xy are like terms.
(b) 10pq, – 7pq and 78qp.
7p and 2405p; 8q and 100q; – p²q² and 12q²p²; – 23 and 41; – 5p² and 701p²; 13p²q and qp² are like terms.

These NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Exercise 3.4

Question 1.
Tell whether the following is certain to happen, impossible, can happen but not certain.
(i) You are older today than yesterday.
(ii) A tossed coin will land heads up.
(iii) A die when tossed shall land up with 8 on top.
(iv) The next traffic light seen will be green.
(v) Tomorrow will be a cloudy day.
Answer:
The chances are
(i) It is certain to happen.
(ii) It can happen but not certain.
(iii) It is impossible.
(iv) It can happen but not certain.
(v) It can happen but not certain.

Question 2.
There are 6 marbles in a box with numbers from 1 to 6 marked on each of them.
(i) What is the probability of drawing a marble with number 2?
(ii) What is the probability of drawing a marble with number 5?
Answer:
Total number of marbles = 6
Total number of possible outcomes = 6
(i) Probability of drawing marble with number 2
p (2) = \(\frac{1}{6}\)
(ii) Probability of drawing marble with number 5
p (5) = \(\frac{1}{6}\)

Question 3.
A coin is flipped to decide which team starts the game. What is the probability that your team will start?
Answer:
Since there are two faces of a coin head (H) and tail (T), when a coin is tossed, then either a ‘H’ or a ‘T’ comes.
successfully achieved outcome = 1
Total number of possible outcomes= 2
[ Head + Tail = 1 + 1 = 2]
Required probability = \(\frac{1}{2}\)