Math Ncert Exemplar 2019 Solutions for Class 7 Maths Chapter 10 - Algebraic Expressions

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LHS = 3x + 23x² + 6y² + 2x + y² + ____________ 
        = (3x + 2x) + 23x² + (y² + 6y² )+____________
        = 5x + 23x² + 7y² +____________

RHS = 5x + 7y²

Now on equating LHS and RHS, we get to know that the term 23x² needs to be vanished.
i.e., we need to add  -23x²in the LHS.
So, the missing term is   -23x²
     

Answer:

If Rohit has 5xy toffees and Shantanu has 20yx toffees, then Shantanu has (20yx -  5xy) more toffees.
Now, 20yx = 20xy
Therefore,  20yx -  5xy = 20xy - 5xy = 15xy 
Hence, Shantanu has 15xy more toffees.
 

Answer:

A polynomial is an algebraic expression in which the power of the variable is a whole number.
In the given expression the power of all the variable is a whole number.

Therefore, $1+\frac{x}{2}+x^3$ is a polynomial.
Hence the given statement is true.

Answer:

(3a – b + 3) – (a + b)
= (3a – a) + (–b – b) + 3
= 2a – 2b + 3 has three terms thus it is a trinomial

Hence, the given statement is false.

Answer:

A polynomial can have any (finite) number of terms.
A polynomial having three terms is called a trinomial.

Hence, the statement is true.

Answer:

A polynomial with three terms is called a trinomial.

Hence, the given statement is false.

Answer:

Sum of x and y is x + y.

Hence, the given statement is true.

Answer:

Sum of 2 and p is 2 + p.

Hence, the given statement is False.

Answer:

A binomial has two terms.

Hence, the given statement is false.

Answer:

A trinomial has exactly three terms.

Hence, the given statement is true.

Answer:

In like terms, variables and their powers are the same.

Hence, the given statement is true.

Answer:

The expression x + y + 5x can be simplified as 6x + y, which is a binomial.

Hence, the given statement is false.

Answer:

–4 is the numerical coefficient of q² in –4pq²
Hence, the given statement is false.
 

Answer:

Unlike terms are two or more terms that are not like terms, i.e. they do not have the same variables or powers

5a and 5b have different variables in them, thus, they are unlike terms.

Hence, the given statement is true.

Answer:

Sum of x² + x and y + y² is x² + x + y + y² =  x²+ y² + x + y  which is not equal to 2x² + 2y² 
Hence, the given statement is false.
 

Answer:

We know that, additive inverse is the number that is added to a given number to make the sum zero.

It is the negotiation of a number or expression.

For example 3 + (-3)
Additive Inverse of 3 is -3.
By subtraction the answer will be zero.

Hence, the given statement is true.

Answer:

The total number of planets of Sun are fixed and therefore can not be denoted by the variable n.

Hence, the given statement is false.

Answer:

In like terms, the numerical coefficients need not to be the same.

Hence, the given statement is false.

Answer:

If we add a monomial and binomial, then answer will either be a binomial or a trinomial.

Hence, the given statement is true

Answer:

If we subtract a monomial from a binomial, then answer can either be a monomial or a trinomial.

Hence, the given statement is false

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Answer:

When we add a monomial and a trinomial, then answer can either be a binomial or a quadrinomial.
It can never be a monomial.

Hence, the given statement is false.
 

Answer:

(a) x is multiplied by itself and then added to the product of x and y.
The above statement can be written in the form of algebraic expressions as
x × x + x × y  = x² + xy, which is a binomial.

(b) Three times of p and two times of q are multiplied and then subtracted from r.
The above statement can be written in the form of algebraic expressions as
r - (3p × 2q) = r - (6pq) , which is a binomial.

(c) Product of p, twice of q and thrice of r .
The above statement can be written in the form of algebraic expressions as
(p × 2q × 3r) = 6pqr , which is a monomial.

(d) Sum of the products of a and b, b and c and c and a.
The above statement can be written in the form of algebraic expressions as
ab + bc + ca , which is a trinomial.


(e) Perimeter of an equilateral triangle of side x.
The above statement can be written in the form of algebraic expressions as
x + x + x = 3x , which is a monomial.


(f) Perimeter of a rectangle with length p and breadth q.
The above statement can be written in the form of algebraic expressions as
2(p + q) = 2p + 2q, which is a binomial.


(g) Area of a triangle with base m and height n.
The above statement can be written in the form of algebraic expressions as
 $\frac{1}{2}\times m\times n=\frac{mn}{2}$, which is a monomial.

(h) Area of a square with side x.
The above statement can be written in the form of algebraic expressions as
 $\frac{1}{2}\times m\times n=\frac{mn}{2}$, which is a monomial.


(i) Cube of s subtracted from cube of t.
The above statement can be written in the form of algebraic expressions as
 $t^3-s^3$, which is a binomial.






 

Answer:

(i) x² – x + 4  

The coefficient of x² is 1.

(ii) x³ – 2x² + 3x + 1

The coefficient of x² is – 2.

(iii) 1 + 2x + 3x² + 4x³  

The coefficient of x² is 3.


(iv) y + y²x + y³x² + y⁴x³

The coefficient of x² is y³.



 

Answer:

(i) x³y²z, xy²z³, –3xy²z³, 5x³y²z, –7x²y²z²
The numerical coefficient of x³y²z is 1, 
The numerical coefficient of xy²z³ is 1, 
The numerical coefficient of –3xy²z³ is –3,
The numerical coefficient of –3xy²z³ is –3,
The numerical coefficient of 5x³y²z is 5


(ii) 10xyz, –7xy²z, –9xyz, 2xy²z, 2x²y²z²

The numerical coefficient of 10xyz is 10, 
The numerical coefficient of –7xy²z is –7, 
The numerical coefficient of –9xyz is –9,
The numerical coefficient of 2xy²z is 2,
The numerical coefficient of  2x²y²z² is 2

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Answer:

(a) Sum of  p² – 7pq – q² and – 3p² – 2pq + 7q²
(p² – 7pq – q²)+ (–3p² – 2pq + 7q²)
= (p²– 3p²) + (– 7pq – 2pq) + (–q² + 7q²)
= (–2p²) + (– 9pq) + (6q²)

(b)  Sum of x³ – x²y – xy² – y³ and x³ – 2x²y + 3xy² + 4y
(x³ – x²y – xy² – y³ ) + (x³ – 2x²y + 3xy² + 4y)

(c) Sum of ab + bc + ca and – bc – ca – ab
(ab + bc + ca)+ (– bc – ca – ab)
= 0

(d) Sum of p² – q + r, q² – r + p and r² – p + q
= p² – q + r + q² – r + p + r² – p + q
= p² + q²  + r² 

(e) Sum of  x³y² + x²y³ + 3y⁴ and x⁴ + 3x²y³ + 4y⁴
= x³y² + x²y³ + 3y⁴ + x⁴ + 3x²y³ + 4y⁴
=  x³y² + 4x²y³ +7y⁴ + x⁴ 

(f) Sum of p²qr + pq²r + pqr² and – 3pq²r –2pqr²
(p²qr + pq²r + pqr²) + (– 3pq²r –2pqr²)
= (p²qr + pq²r – 3pq²r+ pqr²–2pqr²)
= p²qr – 2pq²r – pq²r

(g) Sum of uv – vw, vw – wu and wu – uv
uv – vw + vw – wu + wu – uv
= 0

(h) Sum of a² + 3ab – bc, b² + 3bc – ca and c² + 3ca – ab
= a² + 3ab – bc + b² + 3bc – ca + c² + 3ca – ab
= a² + 2ab + b² + 2bc + c² + 2ca

$\left(\mathrm{i}\right) \mathrm{Sum} \mathrm{of} \frac{5}{8}{p}^4+2p^2+\frac{5}{8};\frac{1}{8}-17p+\frac{9}{8}{p}^2 \mathrm{and} p^5-p^3+7$ 
 $$\begin{gathered} \frac{5}{8}{p}^4+2p^2+\frac{5}{8}+\frac{1}{8}-17p+\frac{9}{8}{p}^2+ p^5-p^3+7 \\ =p^5+\frac{5}{8}{p}^4-p^3+2p^2+\frac{9}{8}{p}^2-17p+\frac{6}{8}+7 \\ =p^5+\frac{5}{8}{p}^4-p^3+\frac{25}{8}{p}^2-17p+\frac{62}{8} \end{gathered}$$

(j) Sum of t – t² – t³ – 14; 15t³ + 13 + 9t – 8t²; 12t² – 19 – 24t and 4t – 9t² + 19t³
 = t – t² – t³ – 14 + 15t³ + 13 + 9t – 8t²  + 12t² – 19 – 24t + 4t – 9t² + 19t³
=  33t³ – 6t² – 10t – 20

 




 



 

Answer:

(a) – 7p²qr from – 3p²qr
 – 3p²qr – (–7p²qr) = – 3p²qr +7p²qr = 4p²qr

(b) –a² – ab from b² + ab 
 (b² + ab) – (–a² – ab) =  (b² + a² + 2ab)

(c) –4x²y – y³ from x³ + 3xy² – x²y.
(x³ + 3xy² – x²y) – (–4x²y – y³) = (x³ + 3xy² – x²y + 4x²y + y³) =  (x³ + 3xy² + 3x²y + y³)

(d) x⁴ + 3x³y³ + 5y⁴ from 2x⁴ – x³y³ + 7y⁴
(2x⁴ – x³y³ + 7y⁴) – (x⁴ + 3x³y³ + 5y⁴) = (x⁴ – 4x³y³ + 2y⁴)

(e) ab – bc – ca from – ab + bc + ca.
 (– ab + bc + ca) – (ab –  bc –  ca) = –2ab + 2bc + 2ca

(f)  –2a² – 2b² from – a² – b² + 2ab
– a² – b² + 2ab – (–2a² – 2b²)
= –a² – b² + 2ab + 2a² + 2b²
= a² + b² + 2ab

(g) x³y² + 3x²y² – 7xy³ from x⁴ + y⁴ + 3x²y² – xy³
(x⁴ + y⁴ + 3x²y² – xy³) – (x³y² + 3x²y² – 7xy³)
= (x⁴ + y⁴ + 3x²y² – xy³ – x³y²  –  3x²y² + 7xy³)
= (x⁴ + y⁴  – 6xy³ + 6x³y²)

(h) 2(ab + bc + ca) from –ab – bc – ca.
(–ab – bc – ca)  – 2(ab + bc + ca)
= (–ab – bc – ca – 2ab – 2bc – 2ca)
= (–3ab – 3bc – 3ca)

(i) 4.5x⁵ – 3.4x² + 5.7 from 5x⁴ – 3.2x² – 7.3x
 5x⁴ – 3.2x² – 7.3x – 4.5x⁵ + 3.4x² – 5.7
= –4.5x⁵  + 5x⁴  + 0.2x² – 7.3x – 5.7

(j) 11 – 15y² from y³ – 15y² – y – 11
y³ – 15y² – y – 11 – 11 + 15y² 
= y³ – y – 22
   


 

     

 
 

Answer:

(a) x³ + y³ − (x³ + 3x²y + 3xy² + y³)
=  x³ + y³ − x³ − 3x²y − 3xy² −  y^3 
= − 3x²y − 3xy²
So,  − 3x²y − 3xy²should be added tox³ + 3x²y + 3xy² + y³ to get x³ + y³ 

(b) p³ + 2p²q² + 4pq − (3pq + 5p²q² + p³) 
=  p³ + 2p²q² + 4pq − 3pq − 5p²q² − p³ 
=  −3p²q² − pq
So, −3p²q² − pq should be added to 3pq + 5p²q² + p³ to get  p³ + 2p²q² + 4pq.

Answer:

a) 2x³ – 3x²y + 2xy² + 3y³  – (x³ – 2x²y + 3xy² + 4y³)
= 2x³ – 3x²y + 2xy² + 3y³  – x³ + 2x²y  – 3xy²  – 4y³
= x³ – x²y – xy²  – y³


b) –7mn + 2m² + 3n² – (m² + 2mn + n²)
= –7mn + 2m² + 3n² – m² – 2mn – n²
= –9mn + m² + 2n² 
 

Answer:

89a³ – 64a² + 6a + 16 – (21a³ – 17a² )
= 89a³ – 64a² + 6a + 16 – 21a³ + 17a² 
= 89a³  – 21a³  + 17a² – 64a² + 6a + 16
= 68a³ – 47a² + 6a + 16

Hence, 21a³ – 17a² is 68a³ – 47a² + 6a + 16 less than 89a³ – 64a² + 6a + 16.

Answer:

y⁴ – 12y² + y + 14 – (17y³ + 34y² – 51y + 68)
= y⁴ – 12y² + y + 14 – 17y³ – 34y² + 51y – 68
= y⁴ – 17y³ – 12y² – 34y² + y + 51y – 68 + 14
= y⁴ – 17y³ – 46y² + 52y – 54

Hence, y⁴ – 12y² + y + 14 is y⁴ – 17y³ – 46y² + 52y – 54 greater than 17y³ + 34y² – 51y + 68.

Answer:

 93p² – 55p + 41 – (3p³ – 5p² + 17p – 90)
= 93p² – 55p + 41 – 3p³ + 5p² –  17p + 90
= – 3p³  + 98p² – 72p + 131

Hence, 93p² – 55p + 4 exceed 13p³ – 5p² + 17p – 90 by  – 3p³  + 98p² – 72p + 131

Answer:

( –99x³ + 33x² + 13x + 41) can be added to 99x³ – 33x² – 13x – 41 to make the sum zero.
As 99x³ – 33x² – 13x – 41 –99x³ + 33x² + 13x + 41 = 0
 

Answer:

 (9a² – 15a + 3) – (1) = 9a² – 15a + 2 

Answer:

(a)  (–2)² + 2(–2)(3) + 3² = 4 – 12 + 9 = 1
(b) (–2)² –2(–2)(3) + 3² = 4 + 12 + 9 = 25
(c) (– 2)³ + 3(– 2)² × 3 + 3(3)² × (–2) + 3³ = –8 + 36 –54 + 27 = 9
(d) (– 2)³ – 3(– 2)² × 3 + 3(3)² × (–2) – 3³ = –8 – 36 –54 + 27 = -71
(e) $\frac{a^2+b^2}{3}=\frac{{(-2)}^2+{\left(3\right)}^2}{3}=\frac{4+9}{3}=\frac{13}{3}$
(f) $\frac{a^2-b^2}{3}=\frac{{(-2)}^2-{\left(3\right)}^2}{3}=\frac{4-9}{3}=\frac{-5}{3}$
(g) $\frac{a}{b}+\frac{b}{a}=\frac{-2}{3}+\frac{3}{-2}=\frac{-4+9}{-6}=\frac{-5}{6}$

(h) (-2)² + (3)² – (-2)(3) – (3)² – (-2)²  = 4 + 9 +6 -9-4 = 6 

 

Answer:

(a) (1) + (–1) + (2) =  2
(b) (1)² + (–1)² + (2)² = 1 + 1 + 4 = 6
(c) (1)³ + (– 1)³ + (2)³ = 1 – 1 + 8 = 8
(d) (1) × (–1) + (–1) × (2) + (2) × (1)  = –1 – 2 + 2 = –1
(e) (1)³ + (–1)³ + (2)³ – 3 × (1) × (–1) × ( 2) = 1 – 1 + 8 + 6 = 14
(f)  (1)² (–1)² + (–1)²(2)² + (2)²(1)² = 1 + 4 + 4 = 9
 

Answer:

(i) (A + B) – C
= (3x² – 4x + 1 + 5x² + 3x – 8) – (4x² – 7x + 3)
= 3x² – 4x + 1 + 5x² + 3x – 8 – 4x² + 7x – 3
= 8x² + 6x – 10

(ii) B + C – A
= 5x² + 3x – 8 + 4x² – 7x + 3 – (3x² – 4x + 1)
= 5x² + 3x – 8 + 4x² – 7x + 3 – 3x² + 4x – 1
= 6x² – 6


(iii) A + B + C
= 3x² – 4x + 1 + 5x² + 3x – 8 + 4x² – 7x + 3
= 12x² – 8x – 4 

Answer:

P + Q + R = ax
​⇒ –(x – 2) –2(y +1) – x + 2y = ax
⇒ –x + 2 –2y –2 – x + 2y = ax
⇒ –2x = ax
⇒a = –2



 

Answer:

(x² – y² – 1) + (y² – x² – 1) + (1 – x² – y² ) – [–(1 + y²)]
= x² – y² – 1 + y² – x² – 1 + 1 – x² – y² + (1 + y²)
= x² – y² – 1 + y² – x² – 1 + 1 – x² – y² + 1 + y²
= –x² 

 

Answer:

(ab + 2b² + 3b² – a²)  – (12ab –10b² –18a²  + 9ab + 12b² + 14a²)
= ab + 2b² + 3b² – a² – 12ab +10b² + 18a²  – 9ab – 12b² – 14a²
= 3a² + 3b² – 20ab

Answer:

$$\begin{gathered} △=2x^2 + 3y, ○=5x^2+3x, \square =8y^2 +3x^2+2x+3y \\ \Rightarrow △+○-\square =\left(2x^2 + 3y\right)+\left(5x^2+3x\right)-\left(8y^2 +3x^2+2x+3y\right) \\ = 7x^2 + 3y+3x-8y^2 -3x^2-2x-3y \\ =4x^2-8y^2+x \end{gathered}$$

Answer:

(a) y units of potatoes and 2 units of rajma
22×y +2×60 = 22y+120

(b) 2x units tomatoes and y units apples
2x×4+14×y=8x+14y
 

Answer:

​Arjun bought a rectangular plot with length x and breadth y = xy
Area of  triangular part of it whose base is y and height is z = $\frac{1}{2}yz$
The area of the remaining part of the plot = $xy-\frac{1}{2}yz$

Answer:

​Amisha has a square plot of side m and another triangular plot with base and height each equal to m.
Total area of both plots = m² + $\frac{1}{2}{m}^2$ = $\frac{3}{2}{m}^2$

Answer:

Algebraic expression for the given situation, if the taxi is hired for x km
= 8 + 50x

Answer:

 Algebraic expression for the money paid to him for both the weeks
= (50×7) + 50x
= 350 + 50x
 

Answer:

Algebraic expression to find the total number of leaves collected by both of them
= 12 + x - 3 + 2x
= 9 + 3x
 

Answer:

Total perimeter of both of them combined together
= x × y + x²

Answer:

Cost of planting the grass on a triangular lawn
= $\frac{1}{2}\times xyz$

Answer:

 The perimeter of the figure
= Length × Breadth
= (5x - y) × 2(x + y)
= [5x × 2(x + y)] - [y × 2(x + y)]
= [10x × (x + y)] - [2y × (x + y)]
= â€‹(10x² + 10xy) - (2yx + 2y²)
= â€‹(10x² - 8xy - 2y²)

Answer:

The total cost of fencing the flower beds
= 5 × 4(x + 2) × $\frac{1}{2}$
= ₹ 10(x + 2)
 

Answer:

(a) Wire is left = (7x – 3) – (3x – 4) = 4x + 1 metres
(b) Length of the triangle formed = $\frac{4x + 1}{3}$ metres


 

Answer:

Money is left with him = â‚¹(10 – 3xy²)  –  [₹3xy² + â‚¹5(xy² + 2)]
                                     =  â‚¹(10 – 3xy²)  –  [₹3xy² + â‚¹5(xy² + 2)]
                                     = â‚¹(10 – 3xy²)  –  ₹3xy² –   â‚¹5(xy² + 2)
                                     = â‚¹10 – ₹3xy²  –  ₹3xy² –   â‚¹5xy² + ₹10
                                     =  â‚¹20 – â‚¹11xy²

Answer:

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Answer:

The sum of first n natural numbers is given by $\frac{1}{2}{n}^2+\frac{1}{2}n.$
​(i) The sum of first 5 natural numbers.
$$\begin{gathered} \frac{1}{2}{\left(5\right)}^2+\frac{1}{2}\left(5\right) \\ =\frac{25}{2}+\frac{5}{2} \\ =\frac{30}{2} \end{gathered}$$

(ii) The sum of first 11 natural numbers.
$$\begin{gathered} \frac{1}{2}{\left(11\right)}^2+\frac{1}{2}\left(11\right) \\ =\frac{121}{2}+\frac{11}{2} \\ =\frac{132}{2} \end{gathered}$$


(iii) The sum of natural numbers from 11 to 30.
$$\begin{gathered} \frac{1}{2}{\left(30\right)}^2+\frac{1}{2}\left(30\right)-\left[\frac{1}{2}{\left(11\right)}^2+\frac{1}{2}\left(11\right)\right] \\ =\frac{900}{2}+\frac{30}{2}-\frac{121}{2}+\frac{11}{2} \\ =\frac{930}{2}-\frac{132}{2} \\ =\frac{798}{2} \\ =399 \end{gathered}$$

 

Answer:

The sum of squares of the first 10 natural numbers = $\frac{1}{2}10(10+1)\left[(2\times 10)+1\right] = 5\times 11\times 21=1155$

Answer:

The sum of the multiplication table of 7 till natural number ‘n’ is given by 55 × 7 = 385 
The sum of the multiplication table of 7 till natural number ‘n’ is given by 55 × 10 = 550
The sum of the multiplication table of 7 till natural number ‘n’ is given by 55 × 19 = 1045
 

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4b – 3 into words can be written as 3 is subtracted from 4 times b.

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​8(m + 5) can be written in word form as 5 more than m is multiplied by 8.

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$\frac{7}{8-x}$ can be written in word form as 7 is divided by the difference of 8 and x.

 

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$17\left(\frac{16}{w}\right)$ can be written as 17 multiplied by the quotient of 16 and w.

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(i) Critical Thinking Write two different algebraic expressions for the word phrase “$\left(\frac{1}{4}\right)$ of the sum of x and 7.”
$\frac{1}{4}\times (x+7)$ and $\left(\frac{x}{4}+\frac{7}{4}\right)$

(ii) What’s the Error? A student wrote an algebraic expression for “5 less than a number n divided by 3” as $\frac{n}{3}-5$. What error did the student make?

The correct expression is $\frac{n -5}{3}$


(iii) Write About it Shashi used addition to solve a word problem about the weekly cost of commuting by toll tax for â‚¹ 15 each day. Ravi solved the same problem by multiplying. They both got the correct answer. How is this possible?

Consider Shashi, he used addition
Total weekly cost = 15 + 15 + 15 + 15 + 15 + 15 + 15 = 105

Consider Ravi, he used addition
Total weekly cost = 15 × 7 = 105

  

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Expression to sum of 1 and twice a number n is 1 + 2n.
If n is odd number then twice of n will be an even number, adding 1 to an even number gives an odd number.
So, if n is any odd number, then the result will always be an odd number.
 

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11x for x = –5 
⇒ 11 × –5 = –55 < 11

Note: 11 multiplied by any negative number is a negative number which will always be less than 11.
 

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At age of 2 years, a cat or a dog is considered 24 “human” years old. Each year, after age 2 is equivalent to 4 “human” years.
 

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(i) Commutative property of addition
x + y = y + x

(ii) Commutative property of multiplication
x × y = y × x

(iii) Associative property of addition
x + (y + z) = (x + y) + z

(iv) Associative property of multiplication
x × (y × z) = (x × y) × z

(v) Distributive property of multiplication over addition
x ÷ (y ÷ z) = (x ÷ y) ÷ z