Category: Class 7

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) 5x² + 3y + 7
Answer:
Given expression is 5x² + 3y + 7 Number of terms = 3

(ii) 5x²y + 3
Answer:
Given expression is 5x²y + 3
Number of terms = 2

(iii) 3x²y
Answer:
Given expression is 3x²y
Number of terms = 1

(iv) 5x – 7
Answer:
Given expression is 5x – 7
Number of terms = 2

(v) 7x³ – 2x
Answer:
Given expression is 7x³ – 2x
Number of terms = 2

Question 2.
Write numeric and algebraic terms in ‘ the above expressions.
(i) 5x² + 3y + 7
Answer:
Given expression is 5x² + 3y + 7
Numerical terms = 7
Algebraic terms = 5x², 3y

(ii) 5x²y + 3
Answer:
Given expression is 5x²y + 3
Numerical terms = 3
Algebraic terms = 5x²y

(iii) 3x²y
Answer:
Given expression is 3x²y
Numerical terms = No
Algebraic terms = 3x²y

(iv) 5x – 7
Answer:
Given expression is 5x – 7
Numerical terms = – 7
Algebraic terms = 5x

(v) 7x³ – 2x
Answer:
Given expression is 7x³ – 2x
Numerical terms = No
Algebraic terms = 7x³, – 2x

Question 3.
Write the terms in the following expressions.
– 3x + 4, 2x – 3y, \(\frac{4}{3}\)a² + \(\frac{5}{2}\)b,
1.2ab + 5.1b – 3.2a
Answer:

Expressions	Terms
– 3x+4	– 3x, 4
2x – 3y	2x, – 3y
\( \frac{4}{3} \)a2 + \( \frac{5}{2} \)b	\( \frac{4}{3} \)a2 + \( \frac{5}{2} \)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 m² and write the coefficients of m².
(i) mn² + m²n
Answer:
Given expression is mn² + m²n

m2 term	Coefficeint
m2n	n

(ii) 7m² – 5m – 3
Answer:
Given expression is 7m² – 5m – 3

m2 term	Coefficeint
7m2	7

(iii) 11 – 5m² + n + 8 mn
Answer:
Given expression is 11 – 5m² + n + 8 mn

m2 term	Coefficeint
– 5m2	-5

Check Your Progress [Page No. 52]

Question 1.
Write like terms from the following :
– xy², – 4yx, 8x, 2xy², 7y, – 11x², – 100x, – 11yx, 20x²y, – 6x², y, 2xy, 3x
Answer:
Given – xy², – 4yx, 8x, 2xy², 7y, – 11x², – 100x, – 11yx, 20x²y, – 6x², y, 2xy, 3x

• – xy², 2xy² are like terms because they contain same algebraic factor xy².
• – 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.
• – 11x², – 6x². are like terms because they contain same algebraic factor x².

Question 2.
Write 3 like terms for
(i) 3x²y
Answer:
Like terms of 3x²y are – 8x²y, \(\frac{5}{3}\) x²y, 2.9 x²y

(ii) – ab²c
Answer:
Like terms of – ab²c are 3ab²c, 5.8 ab²c, \(\frac{-1}{4}\) ab²c.

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, b2c, xy2z, ………………..
Binomial	x + 2y, 4b – 3c, x2y – 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) 10x², – 3x², 5x²
Answer:
Sum of 10x², – 3x², 5x²
= 10x² + (- 3x²) + 5x²
= 10x² – 3x² + 5x²
= (10 – 3 + 5) x²
= 12x²

(iii) – y², 5y², 8y², – 14y²
Answer:
Sum of – y², 5y², 8y², – 14y²
= – y² + 5y² + 8y² + (- 14y²)
= – 1y² + 5y² + 8y² – 14y²
= (- 1 + 5 + 8 – 14) y²
= – 2y²

(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 + 2l² + 4
Answer:
2l² – 5l + A

(ii) 4b² + 5 – 3b
Answer:
4b² – 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) 4m² – 7n² + 5mn, 3n² + 5m² – 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) 4a², 10a²
Answer:
10a² – 4a² = (10 – 4)a² = 6a²

(iii) 15p, 3p
Answer:
3p – 15p = (3 – 15)p = – 12p

(iv) 6m²n, – 20m²n
Answer:
– 20 m²n – 6m²n = (- 20 – 6) m²n
= – 26m²n

(v) a²b², – a²b²
Answer:
– a²b² – a²b² = (- 1 – 1)a²b² = – 2a²b²

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) 3t²
Answer:
3t² ……….. 1 term

(iii) 9p³ + 10q – 15
Answer:
9p³ + 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, – a²b², – 7ba, – 105b, 9b²a², – 5a², 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’.
(- a²b², 9b²a²) are like terms because they contain same algebraic factor ‘a²b²‘.

Question 5.
State with reasons, classify the following expressions into monomials, binomials, trinomials.
a + 4b, 3x²y, px² + qx + 2, qz², x² + 2y, 7xyz, 7x² + 9y³ – 10z⁴, 3l² – m², x, – abc.
Answer:

Expressions	Type of the Expression	Reason
x, 7xyz, 3x2y, qz2, – abc	Monomial	One term
a + 4b, x2 + 2y, 3l2 – m2	Binomial	Two unlike terms
px2 + qx + 2, 7x2 + 9y3 – 10z4	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) 5p²q, 2p²q
Answer:
Sum of 5p²q, 2p²q = 5p²q + 2p²q
= (5 – 2) p²q = 7p²q

(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:
6a² + 3ab + 5b² – 2ab – b² + 2a² + 4ab + 2b² – a²
Answer:
6a² + 3ab + 5b² – 2ab – b² + 2a² + 4ab + 2b² – a²
= (6a² + 2a² – a²) + (3ab – 2ab + 4ab) + (5b² – b² + 2b²)
= [(6 + 2 – 1) a²] + [(3 – 2 + 4)ab] + [(5 – 1 + 2)b²]
= 7a² + 5ab + 6b²

Question 9.
Add 2x² – 3x + 5 and 9 + 6x² 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) 6x² – 4x + 5
Answer:
Additive inverse of 6x² – 4x + 5
= (6x² – 4x + 5) = – 6x² + 4x – 5

Question 11.
Subtract 2p² – 3 from 9p² – 8.
Answer:
9p² – 8 – (2p² – 3) = 9p² – 8 – 2p² + 3
= (9 – 2) p² – 8 + 3
= 7p² – 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 3m³ + 4 from 6m³ + 4m² + 7m – 3 in stepwise method.
Answer:
Let us solve this in stepwise.
Step 1: 6m³ + 4m² + 7m – 3 – (3m³ + 4)
Step 2: 6m³ + 4m² + 7m – 3 – 3m³ – 4
Step 3: 6m³ – 3m³ + 4m² + 7m – 3 – 4 (rearranging like terms)
Step 4: (6 – 3)m³ + 4m² + 7m – 7 (distributive law)
Thus, the required answer = 3m³ + 4m² + 7m – 7

Question 14.
Subtract 4m² – 7n² + 5mn from 3n² + 5m² – 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

SCERT AP 7th Class Maths Solutions Pdf Chapter 6 Algebraic Expressions Unit Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 6th Lesson Algebraic Expressions Unit Exercise

Question 1.
Fill in the blanks:
(i) The constant term in the expression a + b + 1 is ………………..
Answer:
1.

(ii) The variable in the expression 3x – 8 is ………………..
Answer:
x

(iii) The algebraic term in the expression 2d – 5 is ………………..
Answer:
2d

(iv) The number of terms in the expression ……………….. p² – 3pq + q is
Answer:
3

(v) The numerical coefficient of the term – ab is …………………..
Answer:
– 1

Question 2.
Write below statements are True or False:
(i) \(\frac{3 x}{9 y}\) is a binomial.
Answer:
False

(ii) The coefficient of b in – 6abc is – 6a.
Answer:
False

(iii) 5pq and – 9qp are like terms.
Answer:
True

(iv) The sum of a + b and 2a + 7 is 3a + 7b.
Answer:
False

(v) When x = – 2, then the value of x + 2 is 0.
Answer:
True.

Question 3.
Identify like terms among the following:
3a, 6b, 5c, – 8a, 7c, 9c, – a, \(\frac{2}{3}\)b, \(\frac{7 c}{9}\), \(\frac{a}{2}\).
Answer:
Given terms are
3a, 6b, 5c, – 8a, 7c, 9c, – a, \(\frac{2}{3}\)b, \(\frac{7 c}{9}\), \(\frac{a}{2}\)
Like terms: 3a, – 8a, – a, \(\frac{a}{2}\)
6b, \(\frac{2}{3}\)b
5c, 7c, 9c, \(\frac{7 c}{9}\)

Question 4.
Arjun and his friend George went to a stationary shop. Arjun bought 3 pens and 2 pencils whereas George bought one pen and 4 pencils. If the price of each pen and pencil is ₹ x and ₹ y respectively, then find the total bill amount in x and y.
Answer:
Given cost of each pen is ₹ x and cost of each pencil is ₹ y .
Arjun bought 3 pens and 2 pencils. Cost of 3 pens = 3 × ₹ x = ₹ 3x
Cost of 2 pencils = 2 × ₹y = ₹ 2y
Amount paid by Arjun = 3x + 2y = ₹ (3x + 2v)
George bougth one pen and 4 pencils.
Cost of 1 pen = 1 × ₹ x = ₹ x
Cost of 4 pencils = 4 × ₹ y = ₹ 4y
Amount paid by George = x + 4y = ₹(x + 4y)
Total bill amount
= Arjun amout + George amount
= (3x + 2y) + (x + 4y)
= 3x + 2y + x + 4y
= 3x + x + 2y + 4y
= (3 + 1)x + (2 +4)y
∴ Total bill amount = ₹ (4x + 6y)

Question 5.
Find the errors and correct the following :
(i) 7x + 4y = 11xy
Answer:
7x and 4y are not like terms and different variables x, y.
So, we should not add the coefficients.

(ii) 8a² + 6ac = 14a³c
Answer:
8a² and 6ac are not like terms. So, we should not add the coefficients.

(iii) 6pq² – 9pq² = 3pq2
Answer:
6pq² – 9pq² = (6 – 9)pq²
= – 3pq²

(iv) 15mn – mn = 15
Answer:
15mn – mn = (15 – 1) mn
= 14 mn

(v) 7 – 3a = 4a
Answer:
7 and 3a are not like terms.
So, we should not subtract the coefficient 7 and 3.

Question 6.
Add the expressions
(i) 9a + 4, 2 – 3a
Answer:
Given expressions are 9a + 4; 2 – 3a
Write the given expressions in standard form.
9a + 4, – 3a + 2.
The sum = (9a + 4) + (- 3a + 2)
= 9a + 4 – 3a + 2
= (9a – 3a) + (4 + 2)
= (9 – 3)a + 6
= 6a + 6

(ii) 2m – 7n, 3n + 8m, m + n
Answer:
Given expressions are
2m – 7n, 3n + 8m, m + n
Write the given expressions in standard form.
2m – 7n, 8m + 3n, m + n
The sum
= (2m 7n) + (8m + 3n) + (m + n)
= 2m – 7n + 8m + 3n + m + n
= (2m + 8m + m) + (- 7n + 3n + n)
= (2 + 8 + 1)m + (- 7 + 3 + 1)n
= 11 m + (- 3)n
= 11 m – 3n

Question 7.
Subtract:
(i) – y from y
Answer:
– y from y = y – (-y) = y + y = 2y

(ii) 18 pq from 25pq
Answer:
18pq from 2 5pq
= 25 pq – 18 pq
= (25 – 18) pq = 7pq

(iii) 6t + 5 from 1 – 9t
Answer:
Given expressions are (6t + 5), (1 – 9t)
Write the given expressions in the standard form (6t + 5); (- 9t + 1)
(6t + 5) from (- 9t + 1)
= (- 9t + 1) – (6t + 5)
= – 9t + 1 – 6t – 5
= (- 9t – 6t) + (1 – 5)
= (- 9 – 6) t + (- 4) = – 15t – 4

Question 8.
Simplify the following :
(i) t + 2 + t + 3 + t + 6- t- 6 + t
Answer:
Given t + 2 + t + 3+ t + 6 – t – 6 + t

= 3t + 5

(ii) (a + b + c) + (2a + 3b – c) – (4a + b – 2c)
Answer:
Given (a + b + c) + (2a + 3b – c) – (4a + b – 2c)
= a + b + c + 2a + 3b – c – 4a – b + 2c
= (a + 2a – 4a) + (b + 3b – b) + (c – c + 2c)
= (1 + 2 – 4)a + (1 + 3 – 1)b + (1 – 1 + 2)c
= (- 1) a + 3b + 2c
= – a + 3b + 2c

(iii) x + (y + 1) + (x + 2) + (y + 3) + (x + 4) + (y + 5)
Answer:
Given x + (y + 1) + (x + 2) + (y + 3) + (x + 4) + (y + 5)
= x + y + 1 + x + 2 + y + 3 + x + 4 + y + 5
= (x + x + x) + (y + y + y) + (1 + 2 + 3 + 4 + 5)
= 3x + 3y + 15

Question 9.
The perimeter of a triangle is 8x² + 7x – 9 and two of its sides are x² – 3x + 4, 2x² + x – 9 respectively, then find third side.
Answer:
Let the sides of triangle are A, B, C.
A = x² – 3x + 4; B = 2x² + x – 9 ; C = ?
Perimeter = 8x² + 7x – 9
Perimeter of the triangle = A + B + C
To get the third side (C). subtract sum of A and B from the perimeter.
∴ C = Perimeter – (A + B)
So,
A + B = (x² – 3x + 4) + (2x² + x – 9)
= x² – 3x + 4 + 2x² + x – 9
= x² + 2x² – 3x + x +4 – 9
= (1 + 2) x² + (- 3 + 1) x – 5
A + B = 3x² – 2x – 5
Additive inverse of A + B is – (A + B)
– (A + B) = – (3x² – 2x – 5) .
– (A + B) = – 3x² + 2x + 5
C = perimeter + [- (A + B)]
= (8x² + 7x – 9) + (- 3x² + 2x + 5)
= 8x² + 7x – 9 – 3x² + 2x + 5
= 8x² – 3x² + 7x + 2x – 9 + 5
= (8 – 3) x² + (7 + 2) x – 4
C = 5x² + 9x – 4
∴ Third side is 5x² + 9x – 4.

Question 10.
The perimeter of a rectangle is 2a³ – 4a² – 12a + 10, if length is 3a² – 4, find its breadth.
Answer:
Given length of rectangle l = 3a² – 4 and breadth b = ?
Perimeter of a rectangle = 2a³ – 4a² – 12a + 10
Perimeter of a rectangle
= 2(l + b)
= 2a³ – 4a² – 12a + 10

= \(\frac{1}{2}\) × 2(a³ – 2a² – 6a + 5)
⇒ l + b = (a³ – 2a² – 6a + 5)
⇒ l + b – l = (a³ – 2a² – 6a + 5) – l
∴ b = (a³ – 2a² – 6a + 5) – l
Additive inverse of 1 is – 1 = – (3a² – 4)
∴ – l = – 3a² + 4 .
b = (a³ – 2a² – 6a + 5) + (- 1)
= (a³ – 2a² – 6a + 5) + (- 3a² + 4)
= a³ – 2a² – 6a + 5 – 3a² + 4
= a³ – 2a² – 3a² – 6a + 5 + 4
= a³ + (- 2 – 3) a² – 6a + 9
= a³ + (- 5) a² – 6a + 9
∴ Breadth of rectangle
= a³ – 5a² – 6a + 9

SCERT AP 7th Class Maths Solutions Pdf Chapter 11 Area of Plane Figures Ex 11.1 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 11th Lesson Area of Plane Figures Ex 11.1

Question 1.
Calculate the area of the following triangles given below :
(a)

Answer:
From the figure, base = 6 cm
Height = 3 cm
Area of the triangle = \(\frac{1}{2}\) ∙ b ∙ h

= 9 sq.cm

(b)

Answer:
From the figure, base = 4.2 cm
Height = 3.2 cm
Area of the triangle = \(\frac{1}{2}\) ∙ b ∙ h

= 6.72 sq.cm

(a)

Answer:
From the figure, base = 3 cm
Height = 4 cm
Area of the triangle = \(\frac{1}{2}\) ∙ b ∙ h

= 6 sq.cm

(a)

Answer:
From the figure, base = 5 cm
Height = 2 cm
Area of the triangle = \(\frac{1}{2}\) ∙ b ∙ h

= 5 sq.cm

Question 2.
Find the area of triangle with base 3.8 cm and height 4.6 cm.
Answer:
Given base of the triangle b = 3.8 cm
Height of the triangle h = 4.6 cm
Area of the triangle = \(\frac{1}{2}\) ∙ b ∙ h

∴ Area of the triangle = 8.74 sq.cm.

Question 3.
The surface area of a triangular shape window is 24 sq.m, and height is 6m., then find the base of the triangle, if the cost of glass fitting per sqm is ₹ 250, then find total cost of glass fitting for window.

Answer:
Let,
base of the triangular window = b m
Height of the triangular window h= 6m
Area of triangular window = \(\frac{1}{2}\) ∙ b ∙ h

= 24 sq.cm
⇒ 3b = 24
⇒ b = \(\frac{24}{3}\)
= 8
∴ Base of the triangular window = 8m
Cost of glass fitting per sq.m = ₹ 250
Cost of glass fitting per 24 sq.m = 250 × 24
∴ Total cost of glass fitting for window = ₹ 6000/-

Question 4.
A traffic signal plate in the shape of triangle is having base 20 cm, height 15 cm. Find the area of the triangle. If the cost of painting is ₹ 2 per Sq.cm, then find the total cost for painting the signal board on one side.

Answer:
Given the base of the signal plate b = 20 cm
Height of the signal plate h = 15 cm
Area of the signal plate = \(\frac{1}{2}\) ∙ b ∙ h

= 150 sq.cm
∴ Area of the signal plate = 150 sq.cm
Cost of painting per sq.cm = ₹ 2
Cost of painting for 150 sq.cm = 150 × 2
∴ Total cost for painting the signal board = ₹ 300

Question 5.
find the area of triangular shaped wall painting whose base is 24m, height is 38m. Find the area of the triangle, if the cost of painting is ₹ 50 per Sq. m. and also find the total cost of painting.

Answer:
Given base of the triangular wall painting = 24 m Height = 38 m
Area of wall painting = \(\frac{1}{2}\) ∙ b ∙ h

= 456 sq.m.
Cost of painting per 1 sq.m = ₹ 50
Cost of painting for 456 sq.m = 456 × 50
∴ Total cost for painting per 456 sq.m = ₹ 22,800

Question 6.
The area of triangle shaped elevation of a house is 195 Sq.m. Its base is 26m. Find the height of elevation. Find the total cost of cementing, if the cost of cementing per Sq.m, is ₹ 250.

Answer:
Given, base of triangle shaped elevation of house b = 26 m height h = ?
Area of the triangular elevation = 195 sq.m
= \(\frac{1}{2}\) ∙ b ∙ h

⇒ 13 h = 195

Height of triangular elevation = 15 m
Cost of cementing per sq.m = ₹ 250
Cost of cementing per 195 sq.m
= 195 × 250
∴ Total cost of cementing of triangular elevation = ₹ 48,750

SCERT AP 7th Class Maths Solutions Pdf Chapter 10 Construction of Triangles Unit Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 10th Lesson Construction of Triangles Unit Exercise

Construct triangles for the following :

Question 1.
Construct ∆PQR with measurements PQ = 5.8 cm, QR = 6.5 cm and PR = 4.5 cm.
Answer:
Given measurements of ∆PQR are PQ = 5.8 cm, QR = 6.5 cm and PR = 4.5 cm

Steps of Construction:

1. Draw a rough sketch of triangle arid label it with given measurements.
2. Draw a line segment with PQ = 5.8 cm.,
3. Draw ati arc with centre P and radius 4.5 cm.
4. Draw another arc with centre Q and radius 6.5 cm. to intersect the previous arc at R.
5. Join PR and QR.
   Thus, required ∆PQR is constructed with the given measurements.

Question 2.
Construct an isosceles triangle LMN with measurements LM = LN = 6.5 cm and MN=8cm.
Answer:
Given measurements of ∆LMN are LM = LN = 6.5 cm and MN = 8 cm.

Steps of Construction :

1. Draw a rough sketch of triangle and label it with given measurements.
2. Draw a line segment with MN = 8 cm.
3. Draw an arc with centre M and radius 6.5 cm.
4. Draw another arc with centre N and same radius (6.5 cm) to intersect the previous are at L
5. Join ML and NL.
   Hence, required ∆LMN is constructed with the given measurements.

Question 3.
Construct ∆ABC with measurements ∠A = 60°, ∠B = 706 and AB = 7 cm.
Answer:
Given measurements of ∆ABC are ∠A = 60°, ∠B = 70° and AB = 7 cm.

Steps of Construction :

1. Draw a rough sketch of triangle and label it with given measurements.
2. Draw a line segment AB = 7 cm.
3. Draw a ray AX such that ∠BAX = 60°.
4. Draw another ray BY such that ∠ABY = 70°.
5. Name the intersecting point of AX and BY as C.
   Hence, required ∆ABC is constructed with the given measurements.

Question 4.
Construct a right angled triangle XYZ in which ∠Y = 90°, XY = 5 cm and YZ = 7 cm.
Answer:
Given measurements of ∆XYZ are ∠Y = 90°, XY = 5 cm and YZ = 7 Cm.

Steps of Construction:

1. Draw a rough sketch of triangle and label it with given measurements.
2. Draw a line segment with XY = 5 cm.
3. Draw a ray YP such that ∠XYP = 90°
4. Draw an arc with centre Y and radius 7 cm to intersect YP at point Z.
5. Join XZ.
   Hence, required ∆XYZ is constructed with the given measurements.

Question 5.
Construct an equilateral triangle DEF in which DE = EF = FD = 5 cm.
Answer:
Given measurements of ∆DEF are DE = EF = FD = 5 cm.

Steps of Construction:

1. Draw a rough sketch of the triangle and label it with the given measurements.
2. Draw a line segment DE of length 5 cm.
3. Draw an arc with centre D and radius 5 cm.
4. Draw another arc with centre E and same radius (5 cm) to intersect the previous arc at F.
5. Join DF and EF.
   Thus, required triangle ∆DEF is constructed with the given measurements.

Question 6.
Construct a triangle with non-included angle for the sides of ST and SU of lengths 6 v and 7 cm. respectively and ∠T = 80°.
Answer:
Given measurements of ∆STU are ST = 6 cm, SU = 7 cm and ∠T = 80°.

Steps of Construction:

1. Draw a rough sketch of triangle and label it with given measurements.
2. Draw a line segment with ST = 6 cm.
3. Draw a ray TX such that ∠STX = 80°.
4. Draw an arc with centre S and radius 7 cm to intersect TX at point U.
5. Join SU.
   Hence, required triangle ∆STU is constructed with the given measurements.

Question 7.
Can you construct ∆DEF with DE = 7 cm, EF = 14 cm and FD = 5 cm. ? If not give reasons. .
Answer:
No, we can’t construct the triangle. Because the given sides of ∆DEF are DE = 7 cm, EF = 14 cm, FD = 5 cm.
In any triangle sum of any. two sides is always greater than the third side.
DE + FD = 7 + 5 = 12 cm < 14 cm
Sum of DE + FD < EF.
So, with the given measurements construction of ADEF is not possible.

SCERT AP 7th Class Maths Solutions Pdf Chapter 10 Construction of Triangles Ex 10.3 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 10th Lesson Construction of Triangles Ex 10.3

Question 1.
Construct ∆DEF with measurements ∠D = 60°, ∠F = 50° and DF = 4 cm.
Answer:
Given measurements of ∆DEF are ∠D = 60°, ∠F = 50° and DF =. 4 cm.

Steps of Construction :

1. Draw a rough sketch of triangle and label it with given measurements.
2. Draw a line segment with DF = 4 cm.
3. Draw a ray DX such that ∠FDX = 60°.
4. Draw another ray FY such that ∠DFY = 50° which intersects the previous ray at E. Thus, required ∆DEF is constructed with the given measurements.

Question 2.
Construct the triangle with the measurements XY = 7.2 cm, ∠Y = 30° and ∠Z = 100°.
Answer:
Given measurements of ∆XYZ are XY = 7.2 cm, ∠Y = 30° and ∠Z = 100°.
We know stun of angles of triangle is 180°
So, ∠X + ∠Y + ∠Z = 180°
⇒ ∠X + 30° + 100° = 180°
⇒ ∠X= 180°-130°
∴ ∠X = 50°

Steps of Construction:

1. Draw a rough sketch of triangle and label it with given measurements.
2. Draw a line segment with XY = 7.2 cm.
3. Draw a ray \(\overrightarrow{\mathrm{XP}}\) such that ∠YXP = 50°
4. Draw another ray \(\overrightarrow{\mathrm{YQ}}\) such that ∠XYQ = 30° which intersects the previous ray at Z. Thus, required ∆XYZ is constructed with the given measurements.

Question 3.
Construct ∆PQR with the measurements ∠P = ∠Q = 60° and PQ = 7 cm.
Answer:
Given measurements of ∆PQR are ∠P = ∠Q = 60° and PQ = 7 cm.

Steps of Construction:

1. Draw a rough sketch of the triangle and label it with given measurements.
2. Draw a line segment with PQ = 7 cm.
3. Draw a ray PX such that ∠QPX — 60°
4. Draw another ray QY such that ∠PQY = 60° which intersects the previous ray at R. Thus, required ∆PQR is constructed with the given measurements.

SCERT AP 7th Class Maths Solutions Pdf Chapter 10 Construction of Triangles Ex 10.2 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 10th Lesson Construction of Triangles Ex 10.2

Question 1.
Construct ∆ABC with measurements AB = 4.5 cm, BC = 6 cm and ∠B=75°
Answer:
Given measurements of ∆ABC are AB = 4.5 cm, BC = 6 cm and ∠B = 75°.

Steps of Construction :

1. Draw a rough sketch of triangle and label it with given measurements.
2. Draw a line segment with BC = 6 cm.
3. Draw a ray \(\overrightarrow{\mathrm{BX}}\) such that ∠CBX = 15°.
4. Draw an arc with centre B and radius 4.5 cm, to intersect \(\overrightarrow{\mathrm{BX}}\) at point A.
   Thus, required ∆ABC is constructed with the given measurements.

Question 2.
Construct an isosceles triangle with measurements DE = 7 cm, EF = 7 cm and ∠E = 60°
Answer:
Given measurements of ∆DEF are DE = 7 cm, EF = 7 cm and ∠E = 60°.

Steps of Construction:

1. Draw a rough sketch of triangle and label it with given measurements.
2. Draw a line segment with DE = 7 cm.
3. Draw a ray \(\overrightarrow{\mathrm{EX}}\) such that ∠DEX = 60°.
4. Draw an arc with centre E and radius 7 cm to intersect \(\overrightarrow{\mathrm{EX}}\) at point F.
5. Join DF.
   Hence, required ∆DEF is constructed with the given measurements.

Question 3.
Draw a triangle with,measurements ∠B = 50°, AB = 3 cm and AC = 4 cm.
Answer:
Given measurements of a triangle are ∠B = 50°, AB = 3 cm and AC = 4cm..

Steps of Construction:

1. Draw a rough sketch of triangle and label it with given measurements.
2. Draw a line segment with AB = 3 cm.
3. Draw a ray \(\overrightarrow{\mathrm{BX}}\) such that ∠ABX = 50°.
4. Draw an arc with centre A and radius 4 cm to intersect \(\overrightarrow{\mathrm{BX}}\) at point C.
5. Join AC.
   Hence, required ∆ABC is constructed with the given measurements.

Question 4.
Construct a right angled triangle in which, XY = 5 cm, XZ = 6 cm and right tingle at X.
Answer:
Given measurements of ∆XYZ are XY = 5 cm, XZ = 6 cm and ZX = 90° (∠YXZ = 90°)

Steps of Construction:

1. Draw a rough sketch of triangle arid label it with given measurements.
2. Draw a line segment with XY = 5 cm.
3. Draw a ray \(\overrightarrow{\mathrm{XP}}\) such that ∠YXP = 90°.
4. Draw an arc with centre X and radius 6 cm to intersect XP at point Z.
5. Join YZ.
   Hence, required triangle ∆XYZ is constructed with the given measurements.

Question 5.
Construct a right angled with measurements ABC, ∠B = 90°, AB = 8 cm and AC = 10 cm.
Answer:
Given measurements of ∆ABC are ∠B = 90°, AB = 8 cm and AC = 10 cm.

Steps of Construction :

1. Draw a rough sketch of triangle and label it with given measurements.
2. Draw a line segment with AB = 8 cm.
3. Draw a ray \(\overrightarrow{\mathrm{BX}}\) such that ∠ABX = 90°.
4. Draw an arc with centre A and radius 10 cm to intersect \(\overrightarrow{\mathrm{BX}}\) at point C.
5. Join AC.
   Thus, the required ∆ABC is constructed with the given measurements.

SCERT AP 7th Class Maths Solutions Pdf Chapter 10 Construction of Triangles Ex 10.1 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 10th Lesson Construction of Triangles Ex 10.1

Question 1.
Construct ∆ABC with measurements such that AB = 3.5 cm, BC = 4 cm and AC = 4.5 cm.
Answer:
Given sides of ∆ABC are AB = 3.5 cm, BC = 4 cm and AC = 4.5 cm.

Steps of Construction:

1. Draw a rough sketch of the triangle and label it with the given measurements.
2. Draw a line segment BC of length 4 cm.
3. Draw an arc with centre B and radius 3.5 cm.
4. Draw another arc with centre C and radius 4.5 cm to intersect the previous arc at A.
5. Join AB and AC.
   Thus, required ∆ABC is constructed with the given measurements.

Question 2.
Construct an equilateral triangle PQR with side PQ = 5.5 cm
Answer:
Given sides of an equilateral triangle ∆PQR is PQ = 5.5 cm.
In an equilateral triangle all sides are equal in length.
∴ PQ = QR = PR = 5.5 cm

Steps of Construction:

1. Draw a rough sketch of the triangle and label it with the given measurements.
2. Draw a line segment PQ of length 5.5 cm.
3. Draw an arc with centre P arid radius 5.5 cm.
4. Draw another arc with centre Q and the same radius (5.5 cm) to intersect the previous arc at R.
5. Join PR and QR.
   Thus, required ∆PQR is constructed with the given measurements.

Question 3.
Construct ∆XYZ with measurements XY = 3.5 cm, YZ = 5 cm and ZX = 3.5 cm. Which type of triangle is this ? . ‘
Answer:
Given sides of the ∆XYZ are XY = 3.5 cm, YZ 5 cm and ZX = 3.5 cm. In ∆XYZ two sides (XY = ZX = 3.5 cm) are equal.
So, ∆XYZ is an isosceles triangle.

Steps of Construction:

1. Draw a rough sketch of the triangle and label it with the given measurements.
2. Draw a line segment YZ of length 5 cm.
3. Draw an arc with centre Y and radius 3.5 cm.
4. Draw another arc with centre Z and same radius (3.5 cm) to intersect the previous arc at X.
5. Join XY and XZ.
   Thus, required ∆XYZ is constructed with the given measurements.

Question 4.
Construct a triangle with measurements of 5 cm, 3 cm and 4.5 cm.
Answer:
Given sides of the ∆ADI are AD = 5 cm, AI = 3 cm, DI = 4.5 cm.

Steps of Construction:

1. Draw a rough sketch of the triangle, and label it with the given measurements.
2. Draw a line segment with AD = 5 cm.
3. Draw an arc with centre A and radius 3 cm.
4. Draw another arc with centre D and radius 4.5 cm to intersect the previous arc at I.
5. Join AI and DI.
   Thus, required AADI is constructed with the given measurements.

SCERT AP 7th Class Maths Solutions Pdf Chapter 10 Construction of Triangles Review Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 10th Lesson Construction of Triangles Review Exercise

Question 1.
Draw the angles 70° and 110° by using protractor,
(i) ∠AOB = 70°
Answer:
∠AOB = 70°

1. Draw a ray \(\overrightarrow{\mathrm{OA}}\)
2. At ‘O’ erect a ray OB such that ∠AOB = 70°

(ii) ∠AOB = 110°
Answer:
∠AOB = 110°

1. Draw a ray \(\overrightarrow{\mathrm{OA}}\).
2. At ‘O’ erect a ray \(\overrightarrow{\mathrm{OB}}\) such that ∠AOB = 110°

Question 2.
Construct the angles 60° and 120° by using ruler and compass.
(i) ∠AOB = 60°
Answer:
∠AOB = 60°

1. Draw a ray \(\overrightarrow{\mathrm{OA}}\).
2. Draw an arc with any radius (convenient) taking ‘O’ as a centre and cutting \(\overrightarrow{\mathrm{OA}}\) at X.
3. Draw an arc with the same radius – taking X as a centre and meeting the previous arc at Y.
4. Join OY and produce it to B.
5. ∠AOB is formed.
6. ∠AOB = 60°

(ii) ∠ADI = 120°
Answer:
∠ADI = 120°

1. Draw a ray \(\overrightarrow{\mathrm{DA}}\).
2. Draw an arc with any radius taking ‘D’ as centre intersecting \(\overrightarrow{\mathrm{DA}}\) at X.
3. Now draw two successive arcs from X intersecting the previous arc at Y and Z.
4. Join D and Z and produce DZ to I.
5. Now the angle formed ∠ADI is equal to 120°.

Question 3.
Draw a line segment PQ = 4.5 cm and construct its perpendicular bisector by using ruler and compass.
Answer:

∠PMY = ∠QMY = 90°
So, PQ ⊥ XY

1. Draw a line segment PQ of length 4.5 cm.
2. Draw two arcs with centre P and radius > \(\frac{1}{2}\)PQ on both sides of PQ.
3. Repeat step 2 with centre ‘Q’ intersecting the previous arc at X and Y.
4. Join X and Y and produce it on either sides to form \(\overrightarrow{\mathrm{XY}}\).
5. XY is the perpendicular bisector of PQ.

Question 4.
Construct ∠DEF = 60° and its angular bisector by using ruler and compass.
Answer:

∠DEF = 60°
∠DEK = ∠FEK = \(\frac{\angle \mathrm{DEF}}{2}=\frac{60^{\circ}}{2}\) = 30°

1. Construct ∠DEF = 60°
2. Draw an arc with a convenient radius taking E as centre intersecting ED at X and EF at Y.
3. Draw two arcs with same radius taking X and Y as centres intersect¬ing at Z.
4. Join E, Z and produce it to K.
5. ∠DEK = ∠KEF = 30°.

Question 5.
Construct an angle 90° without using a protractor.
Answer:
∠SRI = 90°

1. Draw a ray \(\overrightarrow{R S}\).
2. Draw an arc of convenient radius taking R as a centre and intersecting \(\overrightarrow{R S}\) at X.
3. With the same radius draw two successive arcs from X intersecting the previous arc at Y & Z.
4. Draw bisector to \(\widehat{\mathrm{YZ}}\) intersecting at I.
5. Join R, I.

SCERT AP 7th Class Maths Solutions Pdf Chapter 78 Exponents and Powers InText Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 8th Lesson Exponents and Powers InText Questions

Check Your Progress [Page No. 28]

Question 1.
Write the following in exponential form by using 10 as the base number :
(i) 10,00,00,000
(ii) 100,00,00,000
Answer:

Number	Expanded form	Exponential
(i) 10,00,00,000	10 × 10 × 10 × 10 × 10 × 10 × 10 × 10	108
(ii) 100,00,00,000	10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10	109

Let’s Explore [Page No. 29]

Question 1.
Observe and complete the following table. First one is done for you.

Answer:

Question 2.
Write the following numbers in exponential form. Also state the base, exponent and how to read. .
(i) 16
(ii) 49
(iii) 512
(iv) 243
Answer:

Question 3.
Compute the following and write the greater one.
(i) 4³ or 3⁴
Answer:
4³ = 4 × 4 × 4 = 64
3⁴ = 3 × 3 × 3 × 3 = 81 is greater.
64 < 81 (or) 81 > 64
So, 4³ < 3⁴ (or) 3⁴ > 4³

(ii) 5³ or 3⁵
Answer:
5³ = 5 × 5 × 5 = 125
3⁵ = 3 × 3 × 3 × 3 × 3 = 243 is greater.
125 < 243 (or) 243 > 125
So, 5³ < 3⁵ (or) 3⁵ > 5³

Question 4.
Is 32 equal to 23 ? Justify your answer.
Answer:
3² = 3 × 3 = 9
23 = 2 × 2 × 2 = 8
9 ≠ 8
3² ≠ 2³
So, a^b ≠ b^a unless a = b

Check Your Progress [Page No. 30]

Express the following number in exponential form using prime factorisation:
(i) 432
Answer:

432 = 2 × 216
= 2 × 2 × 108
= 2 × 2 × 2 × 54
= 2 × 2 × 2 × 2 × 27
= 2 × 2 × 2 × 2 × 3 × 9
= 2 × 2 × 2 × 2 × 3 × 3 × 3
∴ 432 = 2⁴ × 3³

(ii) 1296
Answer:
1296 = 2 × 648
= 2 × 2 × 324
= 2 × 2 × 2 × 162
= 2 × 2 × 2 × 2 × 81
= 2 × 2 × 2 × 2 × 3 × 27
= 2 × 2 × 2 × 2 × 3 × 3 × 9
= 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3

∴ 1296 = 2⁴ × 3⁴

(iii) 729
Answer:

729 = 3 × 243
= 3 × 3 × 81
= 3 × 3 × 3 × 27
= 3 × 3 × 3 × 3 × 9
= 3 × 3 × 3 × 3 × 3 × 3
729 = 3⁶

(iv) 1600
Answer:

1600 = 2 × 800
= 2 × 2 × 400
= 2 × 2 × 2 × 200
= 2 × 2 × 2 × 2 × 100
= 2 × 2 × 2 × 2 × 2 × 50
= 2 × 2 × 2 × 2 × 2 × 2 × 25
= 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5
1600 = 2⁶ × 5²

Let’s Explore [Page No. 32]

Question 1.
Write the appropriate number in place of ▢ in the following.
Let ‘b’ be any non-zero integer.

Answer:

= b² × b³
= b × b × b × b × b = b⁵

Answer:

then b¹⁰ × b¹ = b¹⁴
b¹⁰⁺¹ = b¹⁴
b¹¹ ≠ b¹⁴

then b¹⁰ × b² = b¹⁴
b¹⁰⁺² = b¹⁴
b¹² ≠ b¹⁴

then b¹⁰ × b³ = b¹⁴
b¹⁰⁺³ = b¹⁴
b¹³ ≠ b¹⁴

then b¹⁰ × b⁴ = b¹⁴
b¹⁰⁺⁴ = b¹⁴
∴ b¹⁴ = b¹⁴

Question 2.
Simplify the following using the formula a^m × aⁿ = a^m+n
(i) 5⁷ × 5⁴
Answer:
5⁷ × 5⁴
We know that a^m × aⁿ = a^m+n;
5⁷ × 5⁴ = 5⁷⁺⁴ = 5¹¹
∴ 5⁷ × 5⁴ = 5¹¹

(ii) p³ × p²
Answer:
p³ × p²
We know that a^m × aⁿ = a^m+n
p³ × p² = p³⁺² = p⁵
∴ p³ × p² = p⁵

(iii) (-4)¹⁰ × (-4)³ × (-4)²
Answer:
(-4)¹⁰ × (-4)³ × (-4)²
We know that a^m × aⁿ = a^m+n
(- 4)¹⁰⁺³⁺² = (- 4)¹⁵
∴ (-4)¹⁰ × (-4)³ × (-4)² = (- 4)¹⁵

Let’s Explore [Page No. 33]

Question 1.
Write the following in exponential form using the formula (a^m)ⁿ = a^mn.
(i) (6²)⁴
Answer:
(6²)⁴
We know (a^m)ⁿ = a^mn
(6²)⁴ = 6^2×4 = 6⁸
∴ (6²)⁴ = 6⁸

(ii) (2²)¹⁰⁰
Answer:
(2²)¹⁰⁰
We know (a^m)ⁿ = a^mn
(2²)¹⁰⁰ = 2^2×100 = 2²⁰⁰
∴ (2²)¹⁰⁰ = 2²⁰⁰

(iii) (20⁶)²
Answer:
(20⁶)²
We know (a^m)ⁿ = a^mn
(20⁶)² = 20^6×2 = 20¹²
∴ (20⁶)² = 20¹²

(iv) [(-10)³]⁵
Answer:
[(-10)³]⁵
We know (a^m)ⁿ = a^mn
[(-10)³]⁵ =(-10)^3×5 = (-10)¹⁵
∴ [(- 10)³]⁵ = (- 10)¹⁵

Check Your Progress [Page No. 34]

Simplify the following by using the law , a^m × b^m = (ab)^m.
(i) 7⁶ × 3⁶
Answer:
7⁶ × 3⁶
We know, a^m × b^m = (ab)^m
7⁶ × 3⁶ = (7 × 3)⁶ = (21)⁶
∴ 7⁶ × 3⁶ = 21⁶

(ii) (3 × 5)⁴
Answer:
(3 × 5)⁴
We know, (ab)^m = a^m × b^m
(3 × 5)⁴ = 3⁴ × 5⁴
∴ (3 × 5)⁴ = 3⁴ × 5⁴

(iii) a⁴ × b⁴
Answer:
a⁴ × b⁴
We know, a^m × b^m = (ab)^m
a⁴ × b⁴ = (a . b)⁴
∴ a⁴ × b⁴ = (a . b)⁴

(iv) 3² × a²
Answer:
3² × a²
We know, a^m × b^m = (ab)^m
3² × a² = (3 × a)² = (3a)²
∴ 3² × a² = (3a)²

Let’s Explore [Page No. 37]

Question 1.
Simplify and write In the form of a^m-n Or \(\frac{1}{\mathbf{a}^{\mathbf{n}-\mathbf{m}}}\)
(i) \(\frac{10^{8}}{10^{4}}\)
Answer:
We Know \(\frac{10^{8}}{10^{4}}\) = a^m-n
\(\frac{10^{8}}{10^{4}}\) = 10^8-4 = 10⁴
∴\(\frac{10^{8}}{10^{4}}\) = 10⁴

(ii) \(\frac{(-7)^{13}}{(-7)^{10}}\)
Answer:
\(\frac{(-7)^{13}}{(-7)^{10}}\)
We Know \(\frac{a^{m}}{a^{n}}\) = a^m-n (m > n)
\(\frac{(-7)^{13}}{(-7)^{10}}\) = (-7)^13-10
= (-7)³ = -7 × -7 × -7 = – 343
∴\(\frac{(-7)^{13}}{(-7)^{10}}\) = – 343

(iii) \(\frac{12^{5}}{12^{8}}\)
Answer:

(iv) \(\frac{3^{4}}{3^{7}}\)
Answer:

Question 2.
Fill the appropriate number In the box


Answer:

We Know \(\frac{a^{m}}{a^{n}}\) = a^m-n (m > n)
\(\frac{7^{12}}{7^{7}}\) = 7^12-7 = 7⁵
∴ \(\frac{7^{12}}{7^{7}}\) = 7⁵

Answer:

Answer:

Answer:

Question 3.
Simplify the following :
(i) \(\frac{6^{8}}{6^{8}}\)
Answer:
\(\frac{6^{8}}{6^{8}}\)
We Know \(\frac{a^{m}}{a^{n}}\) = a^m-n
\(\frac{6^{8}}{6^{8}}\) = 6^8-8 = 6⁰ = 1 (∵ a⁰ = 1)

(ii) \(\frac{t^{10}}{t^{10}}\)
Answer:
\(\frac{t^{10}}{t^{10}}\)
We Know \(\frac{a^{m}}{a^{n}}\) = a^m-n
\(\frac{t^{10}}{t^{10}}\) = t^10-10 = t⁰ = 1 (∵ a⁰ = 1)

(iii) \(\frac{12^{7}}{12^{7}}\)
Answer:
\(\frac{12^{7}}{12^{7}}\)
We Know \(\frac{a^{m}}{a^{n}}\) = a^m-n
\(\frac{12^{7}}{12^{7}}\) = 12^7-7 = 12⁰ = 1 (∵ a⁰ = 1)

(iv) \(\frac{p^{5}}{p^{5}}\)
Answer:
\(\frac{p^{5}}{p^{5}}\)
We Know \(\frac{a^{m}}{a^{n}}\) = a^m-n
\(\frac{p^{5}}{p^{5}}\) = p^10-10 = p⁰ = 1 (∵ a⁰ = 1)

Check Your Progress [Page No. 38]

Question 1.
Complete the following boxes .

Answer:

Answer:

Answer:

Answer:

Check Your Progress [Page No. 39]

Question 1.
Express the following in exponential form.
(i) \(\frac{-27}{125}\)
Answer:

(ii) \(\frac{-32}{243}\)
Answer:

(iii) \(\frac{-125}{1000}\)
Answer:

(iv) \(\frac{-1}{625}\)
Answer:

Let’s Think [Page No. 39]

Question 1.
Deekshltha and Harsha computed 4(3)² in different ways.
Deeksbitha did it like this
4(3)² = (4 × 3)²
= 12²
= 144

Harsha did it like this
4(3)² = 4 × (3 × 3)
= 4 × 9
= 36
Who has done the problem Incorrectly?
Discuss the reason for the mistake with your friends.
Answer:
In 4(3)² square only belongs to 3, but not 4.
So, Deekshitha did wrong and Harsha did correct.

Let’s Do Activity [Page No. 40]

Finding the pair : Divide the classroom into two groups. Each group has a set of cards. Each student of group 1 has to pair with one suitable student of group 2 by stating in the reason.

Answer:

Note : This activity can be extended till all the children in the class are familiarised with the laws of exponents.

Project Work [Page No. 42]

Collect the annual income of 5 families in your location by observing their ration card and rounded into the nearest thousand / Lakh and express in the exponential form. One done for you.

Answer:

Examples:

Question 1.
Which one is greater 8² or 2⁸? Justify.
Answer:
8² = 8 × 8 = 64
2⁸ = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256
256 >64
Therefore, 2⁸ > 8².

Question 2.
Simplify the following using the formula a^m × aⁿ = a^m+n
(i) (- 5)⁷ × (- 5)⁴
Answer:
(- 5)⁷ × (- 5)⁴ = (- 5)⁷⁺⁴
(∵ a^m × aⁿ = a^m+n)
= (- 5)¹¹
∴ (- 5)⁷ × (- 5)⁴ = (- 5)¹¹

(ii) 3³ × 3² × 3⁴
Answer:
3³ × 3² × 3⁴ = 3³⁺²⁺⁴
(∵ a^m × aⁿ = a^m+n)
= 3⁹
∴ 3³ × 3² × 3⁴ = 3⁹

Question 3.
Simplify the following using the formula (a^m)ⁿ = a^mn
(i) (8³)⁴
Answer:
(8³)⁴ = 8^3×4
= 8¹²
∴ (8³)⁴ = 8¹²

(ii) [(-11)⁵]²
Answer:
[(-11)⁵]² = (- 11)^5×2
= (- 11)¹⁰
∴ [(-11)⁵]² = (- 11)¹⁰

(iii) (7⁵⁰)²
Answer:
(7⁵⁰)²= 7^50×2 = 7¹⁰⁰
∴ (7⁵⁰)² = 7¹⁰⁰

Question 4.
Simplify the following using the expo-nential law a^m × b^m = (ab)^m
(i) 5² × 3²
Answer:
5² × 3² = (5 × 3)² [∵ a^m × b^m = (ab)^m]

(ii) p³ × q³
Answer:
p³ × q³ = (p × q)³

(iii) (7 × 8)⁴
Answer:
(7 × 8)⁴ = 7⁴ × 8⁴ [∵ (ab)^m = a^m × b^m]

Question 5.
Simplify the following and write in the form of \(\frac{a^{m}}{a^{n}}\) = a^m-n or \(\frac{a^{m}}{a^{n}}=\frac{1}{a^{n-m}}\)
(i) \(\frac{2^{9}}{2^{3}}\)
Answer:
\(\frac{2^{9}}{2^{3}}\) = 2^9-3 [∵ \(\frac{a^{m}}{a^{n}}\) = a^m-n]
= 2⁶

(ii) \(\frac{(-9)^{11}}{(-9)^{7}}\)
Answer:
\(\frac{(-9)^{11}}{(-9)^{7}}\) = (- 9)^11-7 = (-9)⁴

(iii) \(\frac{7^{10}}{7^{13}}\)
Answer:
\(\frac{7^{10}}{7^{13}}=\frac{1}{7^{13-10}}\)
[∵ \(\frac{a^{m}}{a^{n}}=\frac{1}{a^{n-m}}\)]
= \(\frac{1}{7^{3}}\)

(iv) \(\frac{6^{2}}{6^{5}}\)
Answer:
\(\frac{6^{2}}{6^{5}}=\frac{1}{6^{5-2}}=\frac{1}{6^{3}}\)

Question 6.
Simplify the following by using formula \(\frac{a^{m}}{b^{m}}=\left(\frac{a}{b}\right)^{m}\).
(i) \(\frac{5^{3}}{2^{3}}\)
Answer:
\(\frac{5^{3}}{2^{3}}=\left(\frac{5}{2}\right)^{3}\) [∵ \(\frac{a^{m}}{b^{m}}=\left(\frac{a}{b}\right)^{m}\)]

(ii) \(\left(\frac{8}{5}\right)^{4}\)
Answer:
\(\left(\frac{8}{5}\right)^{4}=\frac{8^{4}}{5^{4}}\)
[∵ \(\frac{a^{m}}{b^{m}}=\left(\frac{a}{b}\right)^{m}\)]

Question 7.
Evaluate:
(1)⁴, (1)⁵, (1)⁷, (- 1)², (- 1)³, (- 1)⁴, (- 1)⁵
Answer:
(1)⁴ = 1 × 1 × 1 × 1 = 1
(1)⁵ = 1 × 1 × 1 × 1 × 1 = 1
(1)⁷ = 1 × 1 × 1 × 1 × 1 × 1 × 1 = 1
(- 1)²= (-1) × (-1) = 1
(- 1)³ = (-1) × (1) × (-1) = – 1
(- 1)⁴ =(-1) × (-1) × (-1) × (-1) = 1
(- 1)⁵ = (-1) × (-1) × (-1) × (-1) × (- 1) = – 1

From the above illustrations,
(i) It raised. to any power is 1.
(ii) (-1) raised to even power is (- 1) and
(-1) raised to an odd power is (-1).

Thus (- 1)^m = 1 if ’m’ is even
(- 1)^m 1 if ‘m’ is odd

Question 8.
Express \(\) In exponential form.
Answer:
-8 = (-2) × (-2) × (-2) = (-2)³
27 = 3 × 3 × 3 = (3)³
∴ \(\frac{-8}{27}: \frac{(-2)^{3}}{3^{3}}=\left(\frac{-2}{3}\right)^{3}\)

Question 9.
Abhllash computed a³. a² as a⁶. Is it correct?
Answer:
Abhilash has done it incorrectly.
Bcause a³. a² = a³⁺² = a⁵ [∵ a^m . aⁿ = a^mn]
Therefore, a³. a² = a⁵ is correct answer.

Question 10.
Riyaz computed \(\frac{a^{8}}{a^{2}}\) as a⁴. Has he done It correctly? Justify your answer.
Answer:
Riyaz has done it incorrectly.
Because \(\frac{a^{8}}{a^{2}}\) = a^8-2
= a⁶ [∵ \(\frac{a^{m}}{a^{n}}\) = a^m-n]
∴ \(\frac{a^{8}}{a^{2}}\) = a⁶ is correct answer.

Question 11.
Write the following into standard form.
(i) 7465
Answer:
7465 = 7.465 × 1000 (Decimal is shifted three places to the left)
= 7.465 × 10³

(ii) The height of Mount Everest is 8848 m.
Answer:
The height of Mount Everest
= 8848 m
= 8.848 × 1000 m (Decimal is shifted three places to the left)
= 8.848 × 10³m

(iii) The distance from the Sun to Earth is 149,600,000,000 m.
Answer:
The distance from the Sun to Earth
= 149,600,000,000 m
= 1.496 × 100000000000 m
= 1.496 × 10¹¹ m

Reasoning Corner [Page No. 45: Odd one out in numbers]

In each of the following questions, there are 4 numbers. Three of them are similar in a certain way but one is not like the other three. One has to identify the similarity and then strike the odd one out as answer option.
The number can be odd/ even /consecutive, prime numbers, multiple of some number, single, square or cubes of different numbers, plus/minus of some other number or combinations of any mathematical calculation.
Question 1.
(a) 12
(b) 25
(c) 37
(d) 49
Answer:
(c) 37

Hint:
Prime number

Question 2.
(a) 13
(b) 63
(c) 83
(d) 43
Answer:
(b) 63

Hint:
Not a prime number

Question 3.
(a) 21
(b) 49
(c) 56
(d) 36
Answer:
(d) 36

Hint:
Not divisible by 7

Question 4.
(a) 112
(b) 256
(c) 118
(d) 214
Answer:
(b) 256

Hint:
Square number

Question 5.
(a) 42
(b) 21
(c) 84
(d) 35
Answer:
(d) 35

Hint:
Not divisible by 3

Question 6.
(a) 11
(b) 13
(c) 15
(d) 17
Answer:
(c) 15

Hint:
Not a prime number

Question 7.
(a) 10
(b) 11
(c) 15
(d) 16
Answer:
(b) 11

Hint:
Prime number

Question 8.
(a) 49
(b) 63
(c) 77
(d) 81
Answer:
(d) 81

Hint:
Not divisible by 7

Question 9.
(a) 28
(b) 65
(c) 129
(d) 215
Answer:
(a) 28

Hint:
Even number

Question 10.
(a) 51
(b) 144
(c) 64
(d) 121
Answer:
(a) 51

Hint:
Not square number

Practice Questions [Page No. 46]

Question 1.
(a) 3
(b) 9
(c) 5
(d) 7
Answer:
(b) 9

Explanation:
9 is composite number. The remaining numbers are primes.

Question 2.
(a) 6450
(b) 1776
(c) 2392
(d) 3815
Answer:
(d) 3815

Explanation:
All others are even numbers.

Question 3.
(a) 24
(b) 48
(c) 42
(d) 12
Answer:
(c) 42

Explanation:
12, 24 and 48 are multiples of 12.

Question 4.
(a) 616
(b) 252
(c) 311
(d) 707
Answer:
(c) 311

Explanation:
616, 252 and 707 are Palindromes.

Question 5.
(a) 18
(b) 12
(c) 30
(d) 20
Answer:
(d) 20

Explanation:
12, 18 and 30 are 3 (or) 6 multiples. But, 20 is not 3 (or) 6 multiple.

Question 6.
Find the odd one from the given
(a) 3730
(b) 6820
(c) 5568
(d) 4604
Answer:
(a) 3730

Explanation:
3730, All others are divisible by 4.

Question 7.
(a) 2587
(b) 7628
(c) 8726
(d) 2867
Answer:
(a) 2587

Explanation:
All others are formed by 2, 6, 7 and 8.

Question 8.
(a) 63
(b) 29
(c) 27
(d) 25
Answer:
(d) 25

Explanation:
63, 29 and 27 are not squares. 25 only the square.

Question 9.
(a) 23
(b) 37
(c) 21
(d) 31
Answer:
(c) 21

Explanation:
23,37 and 31 are prime numbers.
21 only the composite number.

Question 10.
(a) 18
(b) 9
(c) 21
(d) 7
Answer:
(d) 7

Explanation:
18, 9, 21 are composite numbers.
7 only the prime number.

SCERT AP 7th Class Maths Solutions Pdf Chapter 8 Exponents and Powers Unit Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 8th Lesson Exponents and Powers Unit Exercise

Question 1.
Answer the following.
(i) The exponential form 14⁹ should read as
Answer:
14 is raised to the power of 9.

(ii) When base is 12 and exponent is 17, it’s exponential form is _________
Answer:
12¹⁷.

(iii) The value of (14 × 21)⁰ is
Answer:
We know a⁰ = 1
So, (14 × 21)⁰ = 1

Question 2.
Express the following numbers as a product of powers of prime factors :
(i) 648
Answer:

Given 648 = 2 × 324
= 2 × 2 × 162
= 2 × 2 × 2 × 81
= 2 × 2 × 2 × 3 × 27
= 2 × 2 × 2 × 3 × 3 × 9
= 2 × 2 × 2 × 3 × 3 × 3 × 3
∴ 648 = 2³ × 3⁴

(ii) 1600
Answer:
Given 1600 = 2 × 800
= 2 × 2 × 400
= 2 × 2 × 2 × 200
= 2 × 2 × 2 × 2 × 100
= 2 × 2 × 2 × 2 × 2 × 2 × 25
= 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5

∴ 1600 = 2⁶ × 5²

(iii) 3600
Answer:

Given 3600 = 2 × 1800
= 2 × 2 × 900
= 2 × 2 × 2 × 450
= 2 × 2 × 2 × 2 × 225
= 2 × 2 × 2 × 2 × 3 × 75
= 2 × 2 × 2 × 2 × 3 × 3 × 25
= 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5
∴ 3600 = 2⁴ × 3² × 5²

Question 3.
Simplify the following using laws of exponents.
(i) a⁴ × a¹⁰
Answer:
a⁴ × a¹⁰
We know a^m × aⁿ = a^m+n
= a⁴⁺¹⁰
∴ a⁴ × a¹⁰ = a¹⁴

(ii) 18¹⁸ ÷ 18¹⁴
Answer:
18¹⁸ ÷ 18¹⁴
We Know a^m ÷ aⁿ = a^m-n
= 18^18-14
∴ 18¹⁸ ÷ 18¹⁴ = 18⁴

(iii) (x^m)⁰
Answer:
(x^m)⁰
We Know (a^m)ⁿ = a^m.n
= x^m×n = x⁰(∵ a⁰ = 1)
∴ (x^m)⁰ = 1

(iv) (6² × 6⁴) ÷ 6³
Answer:
(6² X 6⁴) ÷ 6³
We Know a^m × aⁿ = a^m+n
= (6²⁺⁴) ÷ 6³
= 6⁶ ÷ 6³
We Know a^m ÷ aⁿ = a^m-n
= 6^6-3
∴ (6² × 6⁴) ÷ 6³ = 6³

(v) \(\left(\frac{2}{3}\right)^{p}\)
Answer:
\(\left(\frac{2}{3}\right)^{p}\)
We Know \(\left(\frac{a}{b}\right)^{m}=\frac{a^{m}}{b^{m}}=\frac{2^{p}}{3^{p}}\)
∴ \(\left(\frac{2}{3}\right)^{\mathrm{p}}=\frac{2^{\mathrm{p}}}{3^{\mathrm{p}}}\)

Question 4.
Identify the greater number in each of the following andjustify your answer.
(i) 2¹⁰ or 10²
Answer:
2¹⁰ = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
2¹⁰ = 1024
10² = 10 × 10
10² = 100
1024 > 100
So, 2¹⁰ > 10²
∴ 2¹⁰ is greater.

(ii) 5⁴ or 4⁵
Answer:
5⁴ = 5 × 5 × 5 × 5 = 625
4⁵ =4 × 4 × 4 × 4 × 4 = 1024
1024 > 625
So, 4⁵ > 5⁴
∴ 4⁵ is greater number.

Question 5.
If \(\left(\frac{4}{5}\right)^{2} \times\left(\frac{4}{5}\right)^{5}=\left(\frac{4}{5}\right)^{k}\), then find the value of ‘k’
Answer:
Given \(\left(\frac{4}{5}\right)^{2} \times\left(\frac{4}{5}\right)^{5}=\left(\frac{4}{5}\right)^{k}\)
We Know a^m × aⁿ = a^m+n
⇒ \(\left(\frac{4}{5}\right)^{2+5}=\left(\frac{4}{5}\right)^{\mathrm{k}}\)
⇒ \(\left(\frac{4}{5}\right)^{7}=\left(\frac{4}{5}\right)^{\mathrm{k}}\)
If the bases are equal powers should be equal
⇒ 7 = k
∴ k = 7

Question 6.
If 5^2p+1 ÷ 5² = 125, then find the value of ‘p’.
Answer:
Given 5^2p+1 ÷ 5² = 125
We Know a^m ÷ aⁿ = a^m-n
⇒ 5^2p+1-2 = 5 × 5 × 5
⇒ 5^2p-1 = 5³
If the bases are equal, powers should be equal.
⇒ 2p – 1 =3
⇒ 2p = 3 + 1
⇒ 2p = 4
⇒ \(\frac{2 \mathrm{p}}{2}=\frac{4}{2}\)
∴ p = 2

Question 7.
Prove that \(\left(\frac{x^{b}}{x^{c}}\right)^{a} \times\left(\frac{x^{c}}{x^{a}}\right)^{b} \times\left(\frac{x^{a}}{x^{b}}\right)^{c}\) = 1
Answer:
Given \(\left(\frac{x^{b}}{x^{c}}\right)^{a} \times\left(\frac{x^{c}}{x^{a}}\right)^{b} \times\left(\frac{x^{a}}{x^{b}}\right)^{c}\) = 1

Question 8.
Express the following numbers in the expanded form.
(i) 20068
Answer:
20068 = (2 × 10,000) + (0 × 1000) + (0 × 100) + (6 × 10) + (8 × 1)
∴ 20068 = (2 × 104) + (6 × 101) + (8 × 1)

(ii) 120718
Answer:
120718 = (1 × 1,00,000) + (2 × 10,000) + (0 × 1000) + (7 × 100) + (1 × 10) + (8 × 1)
∴ 120718 = (1 × 105) + (2 × 104) + (7 × 102) + (1 × 101) + (8 × 1)

Question 9.
Express the number appearing in the following statements in standard form :
(i) The Moon is 384467000 meters away from the Earth approximately.
Answer:
Distance of Moon from the Earth = 384467000 metres
= 3.84467000 × 100000000

Decimal is shifted eight places to the left.
= 3.84 4 67 × 10⁸ m

Distance of Moon from the earth = 3.84 4 67 × 10⁸ m

(ii) Mass of the Sun is 1 ,989,000,000,000,000,000,00000,000000 kg.
Answer:
Mass of Sun = 1 ,989,000,000,000,000,000,00000,000000 kg
= 1.989 × 1 ,000,000,000,000,000,00000,000000
= 1.989 × 10³⁰ kg
∴ Mass of Sun = 1.989 × 10³⁰ kg.

Question 10.
Lasya solved some problems of exponents and powers in the following way. Do you agree with the solution ? If not why? Justify your answer.
Answer:
(i) x³ × x² = x⁶
Answer:
No. I won’t agree with this solution.
Given, x³ × x²
We know a^m × aⁿ = a^m+n
= a³⁺²
= x⁵ which is ≠ x⁶
so, x³ × x² ≠ x⁶
We have to add to powers. But, Lasya multiplied the powers. That’s why Lasya’s solution is wrong.

(ii) (6³)¹⁰ = 6¹³
Answer:
No, (6³)¹⁰ is not equal to 6.
We know (a^m)ⁿ = a^mn
= 6^3×10
= 6³⁰ which is ≠ 6¹³
so, (6³)¹⁰ ≠ 6¹³
We have to multiply. the powers. But, Lasya added the powers. That’s why Lasya’s solution is wrong.

(iii) \(\frac{4 x^{6}}{2 x^{2}}\) = 2x³
Answer:
No, I won’t agree with this solution.
\(\frac{4 x^{6}}{2 x^{2}}=\frac{2^{2}}{2^{1}} \times \frac{x^{6}}{x^{2}}\)
We know a^m ÷ aⁿ = a^m-n
= 2^2-1 × x^6-2
= 2¹.x⁴
= 2x⁴ which is ≠ 2x³
so, \(\frac{4 x^{6}}{2 x^{2}}\) ≠ 2x³
We have to subtract the powers. But, Lasya divided the powers. That’s why Lasya’s solution is wrong.

(iv) \(\frac{3^{5}}{9^{5}}=\frac{1}{3}\)
Answer:
No. I won’t agree with this solution.
\(\frac{3^{5}}{9^{5}}=\frac{3^{5}}{\left(3^{2}\right)^{5}}\)

We Know (a^m)ⁿ = a^mn

We have to subtract the powers. But, Lasya divided the powers. That’s why Lasya’s solution is wrong.

Question 11.
Is – 2² is equal to 4? Justify your answer.
Answer:
– 2² = – (2 × 2) = – 4 ≠ 4
∴ – 2² = – 4

Question 12.
Beulah computed 2⁵ × 2¹⁰ = 2⁵⁰. Has she done it correctly? Give the reason.
Answer:
Given 2⁵ × 2¹⁰
We know a^m × aⁿ = a^m+n
= 2⁵⁺¹⁰ = 2¹⁵ ≠ 2⁵⁰
∴ 2⁵ × 2¹⁰ ≠ 2⁵⁰
Beulah did wrong.
Here we have to add the powers.
But, he multiplied the powers.

Question 13.
Rafi computed \(\frac{3^{9}}{3^{3}}\) as 3³. Has he done
Answer:
Given \(\frac{3^{9}}{3^{3}}\)
We know \(\frac{a^{m}}{a^{n}}\) = a^m-n
= 3^9-3 = 3⁶ ≠ 3³
So, \(\frac{3^{9}}{3^{3}}\) ≠ 3³
Rafi computed wrong.
Here we have to subtract the powers. But, he divided the powers.

Question 14.
Is (a²)³ equal to a⁸? Give the reason.
Answer:
Given (a²)³ equal to a⁸
We Know (a^m)ⁿ = a^mn
= (a²)³ = a^2×3 = a⁶ ≠ a⁸
∴ (a²)³ = a⁶

We have to do 2 × 3 = 6, But not
2³ = 2 × 2 × 2 = 8

SCERT AP 7th Class Maths Solutions Pdf Chapter 8 Exponents and Powers Ex 8.3 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 8th Lesson Exponents and Powers Ex 8.3

Question 1.
Write the following numbers in expanded form.
(i) 23468
Answer:
23468 = (2 × 10,000) + (3 × 1000) + (4 × 100) + (6 × 10) + (8 × 1)
= (2 × 10⁴) + (3 × 10³) + (4 × 10²) + (6 × 10¹) + (8 × 1)

(ii) 120718
Answer:
120718 = (1 × 1,00,000) + (2 × 10,000) + (0 × 1000) + (7 × 100) + (1 × 10) + (8 × 1)
= (1 × 10⁵) + (2 × 10⁴) + (7 × 10²) + (1 × 10¹) + (8 × 1)

(iii) 806190
Answer:
806190 = (8 × 1,00,000) + (0 × 10,000) + (6 × 1000) + (1 × 100) + (9 × 10) + (0 × 1)
= (8 × 10⁵) + (6 × 10³) + (1 × 10²) + (9 × 10¹)

(iv) 3006194
Answer:
3006194 = (3 × 10,00,000) + (0 × 1,00,000) + (0 × 10,000) + (6 × 1000) + (1 × 100) + (9 × 10) + (4 × 1)
= (3 × 10⁶) + (6 × 10³) + (1 × 10²) + (9 × 10¹) + (4 × 1)

Question 2.
Write the following numbers in standard fortti:
(i) 5,00,000
Answer:
5,00,000 = 5 × 1,00,000 = 5 × 10⁵

(ii) 48,30,000
Answer:
48,30,000 = 4.830000 × 10,00,000 (Decimal is shifted six places to the left).
= 4.830000 × 10⁶
∴ 48,30,000 = 4.83 × 10⁶

(iii) 3,94,00,00,00,000
Answer:
3,94,00,00,00,000 = 3.94000000000 × 100000000000 = 3.94 × 10¹¹

(iv) 30000000
Answer:
30000000 = 3 × 10000000 = 3 × 10⁷
∴ 30000000 = 3 × 10⁷

(v) 180000
Answer:
180000 = 1.80000 × 100000 = 1.8 × 10⁵

Question 3.
Express the number appearing in the following statements in standard form,
(i) The Universe is estimated to be about 12,000,000,000 years old.
Answer:
The Universe is 12,000,000,000 years old = 12 × 1,000,000,000
= 12 × 10⁹ years.
= 1.2 × 10¹⁰ years.

(ii) Earth circumference is about 402000000 km.
Answer:
Earth circumference = 402000000 km
= 4.02000000 × 100000000 (Decimal is shifted eight places to left)
= 4.02 × 10⁸ km.
∴ Earth circumference = 4.02 × 10⁸ km.

SCERT AP 7th Class Maths Solutions Pdf Chapter 8 Exponents and Powers Ex 8.2 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 8th Lesson Exponents and Powers Ex 8.2

Question 1.
Simplify the following by using Laws of Exponents.
(i) 3⁷ × 3⁸
Answer:
3⁷ × 3⁸
We know a^m × aⁿ = a^m+n
3⁷ × 3⁸ = 3⁷ + 8 = 3¹⁵
∴ 3⁷ × 3⁸ =3¹⁵

(ii) 9² × 9⁰ × 9³
Answer:
9² × 9⁰ × 9³
We know a^p × a^q × a^r = p + q + r
9² × 9⁰ × 9³ = 9^{2 + 0 + 3} = 9⁵
∴ 9² × 9⁰ × 9³ = 9⁵

(iii) (2⁸)³
Answer:
(2⁸)³
We know (a^m)ⁿ = a^mn
(2⁸)³ = 2^8×3 =2²⁴
∴ (2⁸)³ = 2²⁴

(iv) (a⁵)⁴
Answer:
(a⁵)⁴
We Know (a^m)ⁿ = a^mn
(a⁵)⁴ = a^5×4 = a²⁰
(a⁵)⁴ = a²⁰

(v) \(\left(\frac{2}{5}\right)^{4} \times\left(\frac{2}{5}\right)^{3} \times\left(\frac{2}{5}\right)^{8}\)
Answer:

(vi) 7⁵ ÷ 7⁸
Answer:
7⁵ ÷ 7⁸
We know a^m = \(\frac{1}{a^{n-m}}\) (n>m)
7⁵ ÷ 7⁸ = \(\frac{1}{7^{8-5}}=\frac{1}{7^{3}}\)
∴ 7⁵ ÷ 7⁸ = \(\frac{1}{7^{3}}\)

(vii) \(\frac{(-6)^{9}}{(-6)^{5}}\)
Answer:
\(\frac{(-6)^{9}}{(-6)^{5}}\)
We Know a^m ÷ aⁿ = a^m-n(m > n)
\(\frac{(-6)^{9}}{(-6)^{5}}\) = (-6)^9-5 = (-6)⁴
∴ \(\frac{(-6)^{9}}{(-6)^{5}}\) = (-6)⁴

(viii) (6⁴ × 6²) ÷ 6⁵
Answer:
(6⁴ × 6²) ÷ 6⁵
We know atm X a = a”
= (6⁴⁺²) ÷ 6⁵
= 6⁶ ÷ 6⁵
We know a^m ÷ aⁿ = a^m-n (m> n)
= 6^6-5 = 6¹ = 6
∴ (6⁴ × 6²) ÷ 6⁵ = 6

(ix) \(\frac{5^{3}}{2^{3}}\)
Answer:
\(\frac{5^{3}}{2^{3}}\)
We know \(\frac{a^{m}}{b^{m}}=\left(\frac{a}{b}\right)^{m}\)
∴ \(\frac{5^{3}}{2^{3}}=\left(\frac{5}{2}\right)^{3}\)

(x) (-3)³ × (-3)¹⁰ × (-3)⁷
Answer:
(-3)³ × (-3)¹⁰ × (-3)⁷
we know a^p . a^q. a^r = a^p+q+r
(-3)³ × (-3)⁰ × (-3)⁷ = (-3)³⁺¹⁰⁺⁷
∴ (- 3)³ × (- 3)¹⁰ × (- 3)⁷ = (- 3)²⁰

Question 2.
Simplify and express each of the following in Exponential form.
(i) \(\left(\frac{a^{5}}{a^{3}}\right)\) × a⁸
Answer:
\(\left(\frac{a^{5}}{a^{3}}\right)\) × a⁸
We Know \(\frac{a^{m}}{a^{n}}\) = a^m-n(m > n)
= (a^5-3) × a⁸
= a² × a⁸

We know a^m × aⁿ = a^m+n
= a²⁺⁸ = a¹⁰
∴ \(\left(\frac{a^{5}}{a^{3}}\right)\) × a⁸ = a¹⁰

(ii) 2⁰ + 3⁰ – 4⁰
Answer:
2⁰ + 3⁰ – 4⁰
We know a⁰ = 1

∴ 2⁰ + 3⁰ – 4⁰ = 1

(iii) (2³ × 2)²
Answer:
(2³ × 2)²
We know a^m.aⁿ = a^m+n
(2³ × 2)² = (2³⁺¹)² = (2⁴)²

We know (a^m)ⁿ = a^m.n
= 2^4×2 = 2⁸
∴(2³ × 2)² = 2⁸

(iv) [(5²)³ × 5⁴] ÷ 5⁷
Answer:
[(5²)³ × 5⁴] ÷ 5⁷

We know (a^m) = a^m.n
= [5^2×3 × 5⁴] ÷ 5⁷
= [5⁶ < 5⁴] ÷ 5⁷

We know a^m .aⁿ = a^m+n
= [5⁶⁺⁴] ÷ 5⁷
= 5¹⁰ ÷ 5⁷
We know a^m ÷ aⁿ = a^m-n (m > n)
= 5^10-7 = 5³
∴ [(5²)³ × 5⁴] ÷ 5⁷ = 5³

Question 3.
Simplify \(\left(\frac{x^{a}}{x^{b}}\right) \times\left(\frac{x^{b}}{x^{c}}\right) \times\left(\frac{x^{c}}{x^{a}}\right)\)
Answer:
\(\left(\frac{x^{a}}{x^{b}}\right) \times\left(\frac{x^{b}}{x^{c}}\right) \times\left(\frac{x^{c}}{x^{a}}\right)\)

Question 4.
Find the value of the following.
(i) (-1)¹⁰⁰⁰
Answer:
(-1)¹⁰⁰⁰
(-1)¹ = – 1
(-1)² = – 1 × – 1 = + 1
(-1)³ = – 1 × -1 × -1 = – 1
(-1)⁴ = -1 × -1 × – 1 × – 1 = + 1
(-1)⁵ = -1 × -1 × -1 × -1 × -1 = -1
(-1)¹⁰⁰ = -1 × -1 × -1 × ………… 100 times = + 1

So, (-1)^{odd number} = -1 and
(-1)^{even number} = + 1

Here 1000 is even number
So, (-1)¹⁰⁰⁰ = + 1

(ii) (1)²⁵⁰
Answer:
1²⁵⁰ = 1 × 1 × 1 ……… 250 times = 1
∴ 1²⁵⁰ = 1

(iii) (1)¹²¹
Answer:
1¹²¹ = -1 × -1 × -1 ……..(121 times)
Here 121 is an odd number
So, (1)¹²¹ = – 1

(iv) (10000)⁰
Answer:
(10000)⁰
we know a⁰ = 1
So, (10000)⁰ = 1

Question 5.
If 7⁵ × 7^3x = 7²⁰ then find the value of ’x’.
Answer:
Given, 7⁵ × 7^3x = 7²⁰
W know a^m X aⁿ = a^m+n
7^{5 + 3x} = 7²⁰
If the bases are equal, then the powers should also be equal.
⇒ 5 + 3x = 20
⇒ 5 + 3x – 5 = 20 – 5
⇒ 3x = 15
⇒ \(\frac{3 x}{3}=\frac{15}{3}\) = 5
∴ x = 5

Question 6.
If 10^y = 10000 then 5y =?
Answer:
Given, 10^y = 10000
10^y = 10 × 10 × 10 × 10
10^y = 10⁴
If the bases are equal, then the powers should also be equal.
⇒ y = 4
Multiply by 5 on bothsides,
5 × y = 5 × 4
∴ 5y = 20

Question 7.
If 5^x = 100 then find the following values.
(i) 5^x+y
Answer:
Given 5^x+y = 100
5^x = 10²
Multiply by 5²
5^x × 5² = 100 × 5²
5^x+2 = 100 × 25 (∵ a^m × aⁿ = a^m+n)
∴ 5^x+2 = 2500

(ii) 5^x-2
Answer:
Given 5^x = 100
5^x = 10²
Divide by 5² on both sides
5^x ÷ 5² = 100 ÷ 5²

∴ 5^x-2 = 4

Question 8.
By what number should 3⁴ be multiplied so that the product is 243?
Answer:
Given number is 3⁴
If we multiply 3⁴ by ‘a’ the product is 243.
i.e. 3⁴ × a = 243
= 3 × 3 × 3 × 3 × 3
3⁴ × a = 35
Divide by 3⁴ on both sides

If we multiply 3⁴ by 3 we get 243.

Question 9.
Arushi computed (5²)⁴ as 5¹⁶ Has she done it correctly or not’? Justify your answer.
Answer:
Given, (5²)⁴
We know) (a^m)ⁿ = a^m.n
= 5^2×4 = 5⁸
∴ (5²)⁴ = 5⁸
But Arushi got 5¹⁶ So, (5²)⁴ ≠ 5¹⁶
Therefore Arushi did wrong.

Question 10.
Is 3⁵ × 4⁵ equal to 12²⁵? If not why? Justify your answer.
Answer:
Given, 3⁵ × 4⁵
We know a^m × b^m = (a × b)^m
So, 3⁵ × 4⁵ = 12 but not 12^5×5
= (3 × 4)⁵
= (12)⁵
3⁵ × 4⁵= 12⁵
But given 3⁵ × 4⁵ = 12²⁵.
So, this is wrong.
So, 3⁵ × 4⁵ ≠ 12²⁵