SCERT AP 7th Class Maths Solutions Pdf Chapter 9 Algebraic Expressions InText Questions and Answers.

## AP State Syllabus 7th Class Maths Solutions 9th Lesson Algebraic Expressions InText Questions

Check your progress [Page No. 49]

Question 1.

How many number of terms are there in each of the following expressions ?

(i) 5x^{2} + 3y + 7

Answer:

Given expression is 5x^{2} + 3y + 7 Number of terms = 3

(ii) 5x^{2}y + 3

Answer:

Given expression is 5x^{2}y + 3

Number of terms = 2

(iii) 3x^{2}y

Answer:

Given expression is 3x^{2}y

Number of terms = 1

(iv) 5x – 7

Answer:

Given expression is 5x – 7

Number of terms = 2

(v) 7x^{3} – 2x

Answer:

Given expression is 7x^{3} – 2x

Number of terms = 2

Question 2.

Write numeric and algebraic terms in ‘ the above expressions.

(i) 5x^{2} + 3y + 7

Answer:

Given expression is 5x^{2} + 3y + 7

Numerical terms = 7

Algebraic terms = 5x^{2}, 3y

(ii) 5x^{2}y + 3

Answer:

Given expression is 5x^{2}y + 3

Numerical terms = 3

Algebraic terms = 5x^{2}y

(iii) 3x^{2}y

Answer:

Given expression is 3x^{2}y

Numerical terms = No

Algebraic terms = 3x^{2}y

(iv) 5x – 7

Answer:

Given expression is 5x – 7

Numerical terms = – 7

Algebraic terms = 5x

(v) 7x^{3} – 2x

Answer:

Given expression is 7x^{3} – 2x

Numerical terms = No

Algebraic terms = 7x^{3}, – 2x

Question 3.

Write the terms in the following expressions.

– 3x + 4, 2x – 3y, \(\frac{4}{3}\)a^{2} + \(\frac{5}{2}\)b,

1.2ab + 5.1b – 3.2a

Answer:

Expressions | Terms |

– 3x+4 | – 3x, 4 |

2x – 3y | 2x, – 3y |

\( \frac{4}{3} \)a^{2} + \( \frac{5}{2} \)b |
\( \frac{4}{3} \)a^{2} + \( \frac{5}{2} \)b |

1.2 ab + 5.1 b – 3.2 a | 1.2 ab, 5.1 b, – 3.2a |

Let’s Explore (Page No: 50)

Question 1.

Identify the terms which contain m^{2} and write the coefficients of m^{2}.

(i) mn^{2} + m^{2}n

Answer:

Given expression is mn^{2} + m^{2}n

m^{2} term |
Coefficeint |

m^{2}n |
n |

(ii) 7m^{2} – 5m – 3

Answer:

Given expression is 7m^{2} – 5m – 3

m^{2} term |
Coefficeint |

7m^{2} |
7 |

(iii) 11 – 5m^{2} + n + 8 mn

Answer:

Given expression is 11 – 5m^{2} + n + 8 mn

m^{2} term |
Coefficeint |

– 5m^{2} |
-5 |

Check Your Progress [Page No. 52]

Question 1.

Write like terms from the following :

– xy^{2}, – 4yx, 8x, 2xy^{2}, 7y, – 11x^{2}, – 100x, – 11yx, 20x^{2}y, – 6x^{2}, y, 2xy, 3x

Answer:

Given – xy^{2}, – 4yx, 8x, 2xy^{2}, 7y, – 11x^{2}, – 100x, – 11yx, 20x^{2}y, – 6x^{2}, y, 2xy, 3x

- – xy
^{2}, 2xy^{2}are like terms because they contain same algebraic factor xy^{2}. - – 4yx, – 11yx, 2xy are like terms because they contain same algebraic factor xy.
- 8x, – 100x, 3x are like terms because they contain same algebraic factor x.
- 7y, y are like terms because they contain same algebraic factor y.
- – 11x
^{2}, – 6x^{2}. are like terms because they contain same algebraic factor x^{2}.

Question 2.

Write 3 like terms for

(i) 3x^{2}y

Answer:

Like terms of 3x^{2}y are – 8x^{2}y, \(\frac{5}{3}\) x^{2}y, 2.9 x^{2}y

(ii) – ab^{2}c

Answer:

Like terms of – ab^{2}c are 3ab^{2}c, 5.8 ab^{2}c, \(\frac{-1}{4}\) ab^{2}c.

Let’s Explore [page No: 52]

Question 1.

Jasmin says that 3xyz is a trinomial. Is she right ? Give reason.

Answer:

Given expression 3xyz.

In this expression only one term is there. So, it is a monomial only. But, not trinomial.

So, Jasmin is wrong.

Question 2.

Give two examples each for Monomial and Binomial algebraic expression.

Answer:

Type of Expressions | Expressions |

Monomial | x, b^{2}c, xy^{2}z, ……………….. |

Binomial | x + 2y, 4b – 3c, x^{2}y – yz, …………… |

Check Your progress [Page No. 54]

Question 1.

Add the following like terms.

(i) 12ab, 9ab, ab

Answer:

Sum of 12ab, 9ab, ab

= 12ab + 9ab + ab

= (12 + 9 + 1) ab

= 22 ab.

(ii) 10x^{2}, – 3x^{2}, 5x^{2}

Answer:

Sum of 10x^{2}, – 3x^{2}, 5x^{2}

= 10x^{2} + (- 3x^{2}) + 5x^{2}

= 10x^{2} – 3x^{2} + 5x^{2}

= (10 – 3 + 5) x^{2}

= 12x^{2}

(iii) – y^{2}, 5y^{2}, 8y^{2}, – 14y^{2}

Answer:

Sum of – y^{2}, 5y^{2}, 8y^{2}, – 14y^{2}

= – y^{2} + 5y^{2} + 8y^{2} + (- 14y^{2})

= – 1y^{2} + 5y^{2} + 8y^{2} – 14y^{2}

= (- 1 + 5 + 8 – 14) y^{2}

= – 2y^{2}

(iv) 10mn, 6mn, – 2mn, – 7mn

Answer:

Sum of 10mn, 6mn, – 2mn, – 7mn

= 10mn + 6mn + (- 2mn) + (- 7mn)

= 10mn + 6mn – 2mn – 7mn

= (10 + 6 – 2 – 7) mn

= 7mn

Let’s Think [Page No: 54]

Reshma simplified the expression

4p + 6p + p like this.

4p + 6p + p = 10p. Is she correct ?

Answer:

4p + 6p + p = (4 + 6 + 1)p

= 11p ≠ 10p

So, Reshma simplified is wrong.

Check Your Progress [Page No: 54]

Question 1.

Write the standard form of the following expressions :

(i) – 5l + 2l^{2} + 4

Answer:

2l^{2} – 5l + A

(ii) 4b^{2} + 5 – 3b

Answer:

4b^{2} – 3b + 5

(iii) z – y – x

Answer:

– x – y + z.

Check Your Progress [Page No. 56]

Question 1.

Add the following expressions in both Row and Column methods:

(i) x – 2y, 3x + 4y

Answer:

(ii) 4m^{2} – 7n^{2} + 5mn, 3n^{2} + 5m^{2} – 2mn

Answer:

(iii) 3a – 4b, 5c – 7a + 2b

Answer:

Let’s Explore [Page No. 56]

Think of at least two situations in each of which you need to form two algebraic expressions and add them.

(i) Aditya and Madhu went to a store. Aditya bought 6 books and Madhu bought 2 books. All the books are same cost. How much money did both spend ₹

Answer:

Let the cost of each book ₹a.

Number of books Aditya bought = 6

Cost of 6 books = 6 × a = ₹ 6a

Number of books Madhu bought = 2

Cost of 2 books = 2 × a = ₹ 2a

Money spend on books bought by Aditya and Madhu

= 6a + 2a

= (6 + 2)a

= ₹ 8a

(ii) Sreeja and Sreekari went to icecream parlour. Sreeja bought Vineela icecreams 2 and Sreekari bought Butter Scotch 3. Cost of icecreams are different. How much money they gave to the shop keeper ₹

Answer:

Let the cost of Vineela is ₹x and cost of Butter Scotch is ₹y.

Number of Vineela icecreams Sreeja bought = 2

Cost of 2 iceqreams = 2 × x = ₹ 2x

Number of Butter Scotch icecreams Sreekari bought = 3

Cost of 3 icecreams = 3 × y = ₹ 3y

Money given to shopkeeper = 2x + 3y

= ₹(2x + 3y)

Check Your Progress [Page No. 57]

Question 1.

Subtract the first term from second term :

(i) 2xy, 7xy

Answer:

7xy – 2xy

= (7 – 2) xy

= 5xy

(ii) 4a^{2}, 10a^{2}

Answer:

10a^{2} – 4a^{2} = (10 – 4)a^{2} = 6a^{2}

(iii) 15p, 3p

Answer:

3p – 15p = (3 – 15)p = – 12p

(iv) 6m^{2}n, – 20m^{2}n

Answer:

– 20 m^{2}n – 6m^{2}n = (- 20 – 6) m^{2}n

= – 26m^{2}n

(v) a^{2}b^{2}, – a^{2}b^{2}

Answer:

– a^{2}b^{2} – a^{2}b^{2} = (- 1 – 1)a^{2}b^{2} = – 2a^{2}b^{2}

Let’s Explore [Page No. 58]

Add and subtract the following expressions in both Horizontal and Vertical method x – 4y + z, 6z – 2x + 2y

Answer:

Addition

Subtraction:

Let’s Explore [Page No. 60]

Question 1.

Write an expression whose value is – 15 when x = – 5.

Answer:

Given x = – 5 and value = – 15

Value = – 15 ,

= 3 × – 5 ,

= – 3 × x (∵ x = – 5)

∴ Expression = 3x

Question 2.

Write an expression whose value is 15 when x = 2.

Answer:

Given x = 2 and value = 15

Value = 15

= \(\frac{30}{2}\)

= \(\frac{1}{2}\) × 15 × 2

= \(\frac{1}{2}\) × 15 × x (∵ x = 2)

∴ Expression = \(\frac{15 x}{2}\)

Lets Think [Page No. 60]

While finding the value of an algebraic expression 5x when x = – 2, two stu¬dents solved as follows:

Can you guess who has done it correctly? Justify!

Answer:

Given expression is 5x when x = – 2

5x = 5(- 2) = – 10

Chaitanya is correct.

Here we have to multiply 5 and,- 2.

But Reeta subtracted.

So, Reeta is wrong.

Examples

Question 1.

How many number of terms are there in each of the following expressions ?

(i) a + b

Answer:

a + b ……….. 2 terms

(ii) 3t^{2}

Answer:

3t^{2} ……….. 1 term

(iii) 9p^{3} + 10q – 15

Answer:

9p^{3} + 10q – 15 ……….. 3 terms

(iv) \(\frac{5 \mathrm{~m}}{3 \mathrm{n}}\)

Answer:

\(\frac{5 \mathrm{~m}}{3 \mathrm{n}}\) ……….. 1 term

(v) 4x + 5y – 3z – 1

Answer:

4x + 5y – 3z – 1 ……….. 4 terms

Question 2.

In the following expressions, write the number of terms and identify numeri¬cal and algebraic expressions in them.

(i) 8p

Answer:

8p = 1 term – Algebraic expression

(ii) 5c + s – 7

Answer:

5c + s – 7 = 3 terms – Algebraic expression

(iii) – 6

Answer:

– 6 = 1 term – Numerical expression

(iv) (2 + 1) – 6

Answer:

(2 + 1) – 6 = 2 terms – Numerical expression

(v) 9t + 15

Answer:

9t + 15 = 2 terms – Algebraic expression

Question 3.

Write the coefficients of

(i) p in 8pq

Answer:

8pq = p(8q)

so, coefficient of p in 8pq is 8q.

(ii) x in \(\frac{x y}{3}\)

Answer:

\(\frac{x y}{3}\) = x\(\left(\frac{\mathrm{y}}{3}\right)\)

so, coefficient of x in \(\frac{x y}{3}\) is \(\left(\frac{\mathrm{y}}{3}\right)\)

(iii) abc in (- abc)

Answer:

(- abc) = – (abc)

so, coefficient of abc is – 1.

Question 4.

Identify like terms among the following and group them:

10ab, 7a, 8b, – a^{2}b^{2}, – 7ba, – 105b, 9b^{2}a^{2}, – 5a^{2}, 90a.

Answer:

(7a, 90a) are like terms because they

contain same algebraic factor ‘a’.

(10ab, – 7ba) are like terms as they have same algebraic factor ’ab’.

(8b, -105b) are like terms because they contain same algebraic factor ‘b’.

(- a^{2}b^{2}, 9b^{2}a^{2}) are like terms because they contain same algebraic factor ‘a^{2}b^{2}‘.

Question 5.

State with reasons, classify the following expressions into monomials, binomials, trinomials.

a + 4b, 3x^{2}y, px^{2} + qx + 2, qz^{2}, x^{2} + 2y, 7xyz, 7x^{2} + 9y^{3} – 10z^{4}, 3l^{2} – m^{2}, x, – abc.

Answer:

Expressions | Type of the Expression | Reason |

x, 7xyz, 3x^{2}y, qz^{2}, – abc |
Monomial | One term |

a + 4b, x^{2} + 2y, 3l^{2} – m^{2} |
Binomial | Two unlike terms |

px^{2} + qx + 2, 7x^{2} + 9y^{3} – 10z^{4} |
Trinomial | Three unlike terms |

Question 6.

Find the sum of the following like terms :

(i) 3a, 9a

Answer:

Sum of 3a, 9a = 3a + 9a

= (3 + 9)a = 12a

(ii) 5p^{2}q, 2p^{2}q

Answer:

Sum of 5p^{2}q, 2p^{2}q = 5p^{2}q + 2p^{2}q

= (5 – 2) p^{2}q = 7p^{2}q

(iii) 6m, – 15m, 2m

Answer:

Sum of 6m, – 15m, 2m

= 6m + (- 15m) + 2m

= 6m – 15m + 2m

= (6 – 15 + 2)m = – 7m

Question 7.

Write the perimeter of the given figure.

Answer:

The perimeter of the figure

= 10 + 4 + x + 3+ y

= x + y + (10 + 4 + 3)

= x + y + 17

Question 8.

Simplify:

6a^{2} + 3ab + 5b^{2} – 2ab – b^{2} + 2a^{2} + 4ab + 2b^{2} – a^{2}

Answer:

6a^{2} + 3ab + 5b^{2} – 2ab – b^{2} + 2a^{2} + 4ab + 2b^{2} – a^{2}

= (6a^{2} + 2a^{2} – a^{2}) + (3ab – 2ab + 4ab) + (5b^{2} – b^{2} + 2b^{2})

= [(6 + 2 – 1) a^{2}] + [(3 – 2 + 4)ab] + [(5 – 1 + 2)b^{2}]

= 7a^{2} + 5ab + 6b^{2}

Question 9.

Add 2x^{2} – 3x + 5 and 9 + 6x^{2} in vertical method.

Answer:

Question 10.

Find the additive inverse of the follow-ing expressions:

(i) 35

Answer:

Additive inverse of 35 = – 35

(ii) – 5a

Answer:

Additive inverse of – 5a = – (-5a) = 5a

(iii) 3p – 7

Answer:

Additive inverse of 3p – 7 = – (3p – 7) = – 3p + 7

(iv) 6x^{2} – 4x + 5

Answer:

Additive inverse of 6x^{2} – 4x + 5

= (6x^{2} – 4x + 5) = – 6x^{2} + 4x – 5

Question 11.

Subtract 2p^{2} – 3 from 9p^{2} – 8.

Answer:

9p^{2} – 8 – (2p^{2} – 3) = 9p^{2} – 8 – 2p^{2} + 3

= (9 – 2) p^{2} – 8 + 3

= 7p^{2} – 5

Question 12.

Subtract 3a + 4b – 2c from 6a – 2b + 3c in row method.

Answer:

Let A = 6a – 2b + 3c, B = 3a + 4b – 2c

Subtracting 3a + 4b – 2c from 6a – 2b + 3c is equal to adding additive inverse of 3a + 4b – 2c to 6a – 2b + 3c

i.e., A – B = A + (-B)

additive inverse of (3a + 4b – 2c) = – (3a + 4b – 2c) = – 3a – 4b + 2c

A – B = A + (-B)

= 6a – 2b + 3c + (- 3a – 4b + 2c)

= 6a – 2b + 3c – 3a – 4b +2c = (6 – 3)a – (2 + 4)b + (3 + 2)c

Thus, the required answer = 3a – 6b + 5c

Question 13.

Subtract 3m^{3} + 4 from 6m^{3} + 4m^{2} + 7m – 3 in stepwise method.

Answer:

Let us solve this in stepwise.

Step 1: 6m^{3} + 4m^{2} + 7m – 3 – (3m^{3} + 4)

Step 2: 6m^{3} + 4m^{2} + 7m – 3 – 3m^{3} – 4

Step 3: 6m^{3} – 3m^{3} + 4m^{2} + 7m – 3 – 4 (rearranging like terms)

Step 4: (6 – 3)m^{3} + 4m^{2} + 7m – 7 (distributive law)

Thus, the required answer = 3m^{3} + 4m^{2} + 7m – 7

Question 14.

Subtract 4m^{2} – 7n^{2} + 5mn from 3n^{2} + 5m^{2} – 2mn.

(For easy understanding, same colours are taken to like terms)

Answer:

Question 15.

Find the value of the following expressions, when x = 3.

(i) x + 6

Answer:

When x = 3, the value of x + 6

(substituting 3 in the place of x) is

(3) + 6 = 9

(ii) 8x – 1

Answer:

When x = 3,

the value of 8x – 1 = 8(3) – 1

= 24 – 1 = 23

(iii) 14 – 5x

Answer:

When x = 3,

the value of 14 – 5x = 14 – 5(3)

= 14 – 15 = – 1

Practice Questions [Page No: 65]

Question 1.

In a certain code BOARD: CNBQE, how ANGLE will be written in that code?

(a) BMHKF

(b) CNIJE

(c) BLGIF

(d) CMIKF

Answer:

(a) BMHKF

Explaination:

Each letter in a word is shifted to 1 position is alphabetical order and went 1 position is backward like this.

ANGLE = BMHKF

Question 2.

In a certain code, MOBILE: 56, how PHONE will be written in that code?

(a) 52

(b) 54

(c) 56

(d) 58

Answer:

(d) 58

Explaination:

Taking sum of positions of letters in for ward direction.

So, PHONE = 16 + 8 + 15 + 14 + 5 = 58

Question 3.

If BEAN: ABNE, then NEWS ?

(a) WSNE

(b) WSEN

(c) WNSE

(d) WNES

Answer:

(c) WNSE

Explaination:

Follow the same to decode the given word.

Question 4.

If ROSE : 6821, CHAIR : 73456, PREACH : 961473, then SEARCH ?

(a) 241673

(b) 214673

(c) 216473

(d) 216743

Answer:

(b) 214673

Explaination:

By comparing each letter with the symbol, by writing one below the other.

Question 5.

If COMPUTER: RFUVQNPC, then MEDICINE?

(a) EDJOJMEF

(b) EOJDJEFM

(c) EOJJDFEM

(d) EDJJOFME

Answer:

(b) EOJDJEFM

Explaination:

In the coded form the first & last letters have been interchanged while the rem¬aining letters are coded by taking their immediate next letters in the reverse order.

Question 6.

If LAKE = 7@$5, WALK = %@7$, then WAKE = ?

(a) @%75

(b) %@$5

(c) %5@7

(d) %@57

Answer:

(b) %@$5

Explaination:

By comparing each letter with symbol, by writing One below the other.

Question 7.

If MANY = OCPA, then LOOK = ?

(a) NQQM

(b) MQQN

(c) QMQN

(d) QNQM

Answer:

(a) NQQM

Explaination:

Each letter in a word is shifted to 2 position forward in alphabetical order.

∴ LOOK = NQQM

Question 8.

If SOME = PUB, then BODY ?‘

(a) LABY

(b)YBAL

(c) YLAV

(d) ABLY

Answer:

(c) YLAV

Explaination:

Bach letter in a word is shifted to 3 position backward in alphabetical order.

So, BODY = YLAV

Question 9.

If ARC = CVI, then RAY?

(a) TEU

(b) TEE

(c) TED

(d) TEF

Answer:

(b) TEE

Explaination:

Each letter in a word is forwarded alphabetical order as follows.

So, RAY = TEE

Question 10.

If MEAN = KGYI, then MODE = ?

(a) QBGK

(b) KBQC

(c) KGBQ

(d) kQBG

Answer:

(d) kQBG

Explaination:

Each letter in the word is shifted to 2 position backward and 2 position forward as following.

So, MODE = KQBG

Question 11.

If FIND = DNIF, then DONE ?

(a) ENOD

(b) ENDO

(c) NEOD

(d) ONED

Answer:

(a) ENOD

Explaination:

By reversing the word from left to right.

So, DONE = ENOD .

Question 12.

If BASE = SBEA, then AREA = ?

(a) AARE

(b) EAAR

(c) EARA

(d) REAA

Answer:

(b) EAAR

Explaination:

Follow the same to decode the given word.

Question 13.

If LESS = 55, then MORE = ?

(a)54

(b)50

(c) 51

(d) 52

Answer:

(c) 51

Explaination:

Taking sum of positions of letters in forward direction.

Question 14.

If BACK = 17, then CELL =?

(a) 33

(b) 30

(c) 31

(d) 32

Answer:

(d) 32

Explaination:

Taking sum of positions of letters in for ward direction.

3 + 5 + 12 + 12 = 32

Question 15.

If BIG = 63, then SMALL =?

(a) 76

(b) 78

(c) 74

(d) 72

Answer:

(b) 78

Explaination:

Taking sum of positions of letters in reverse direction.

8 + 14 + 26 + 15 + 15 = 78