Rs Aggarwal 2017 Solutions for Class 7 Math Chapter 5 Exponents are provided here with simple step-by-step explanations. These solutions for Exponents are extremely popular among Class 7 students for Math Exponents Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2017 Book of Class 7 Math Chapter 5 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2017 Solutions. All Rs Aggarwal 2017 Solutions for class Class 7 Math are prepared by experts and are 100% accurate.
Page No 90:
Question 1:
Write each of the following in power notation:
(i)
(ii)
(iii)
(iv) (−8) × (−8) × (−8) × (−8) × (−8)
Answer:
(i)
(ii)
(iii)
(iv)
Page No 91:
Question 2:
Express each of the following in power notation:
(i)
(ii)
(iii)
(iv)
Answer:
(i) [since 25 = 52 and 36 = 62]
(ii) [since −27 = (−3)3 and 64 = 43]
(iii) [since −32 = (−2)5 and 243 = 35]
(iv) [since (−1)7 = −1 and 128 = 27]
Page No 91:
Question 3:
Express each of the following as a rational number:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii) (−1)9
Answer:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii) [Since (-1) an odd natural number = -1]
Page No 91:
Question 4:
Express each of the following as a rational number:
(i) (4)−1
(ii) (−6)−1
(iii)
(iv)
Answer:
(i) [since ]
(ii) [since ]
(iii) [since ]
(iv) [since ]
Page No 91:
Question 5:
Find the reciprocal of each of the following:
(i)
(ii)
(iii) 67
(iv) (−4)3
Answer:
We know that the reciprocal of is .
(i) Reciprocal of
(ii) Reciprocal of
(iii) Reciprocal of 67 = Reciprocal of =
(iv) Reciprocal of (− 4)3 = Reciprocal of =
Page No 91:
Question 6:
Find the value of each of the following:
(i) 80
(ii) (−3)0
(iii) 40 + 50
(iv) 60 × 70
Answer:
(i) 80 = 1
(ii) (−3)0 = 1
(iii) 40 + 50 = 1 + 1 = 2
(iv) 60 × 70 = 1 × 1 = 1
Note: a0 = 1
Page No 91:
Question 7:
Simplify each of the following and express each as a rational number:
(i)
(ii)
(iii)
(iv)
(v)
Answer:
(i)
(ii)
=
(iii)
=
(iv)
(v)
Page No 91:
Question 8:
Simplify and express each as a rational number:
(i)
(ii)
(iii)
Answer:
(i)
= =
(ii)
=
=
=
(iii)
=
=
=
Page No 91:
Question 9:
Express each of the following as a rational number:
(i) 5−3
(ii) (−2)−5
(iii)
(iv)
(v)
(vi)
(vii) (5−1−7−1)−1
(viii)
(ix)
(x)
Answer:
Note:
(i) 5−3 =
(ii) (−2)−5 =
(iii)
(iv) =
(v)
(vi)
(vii)
=
(viii) =
(ix)
(x) [since a0 = 1 for every integer a]
Page No 91:
Question 10:
Simplify:
(i)
(ii)
(iii)
(iv)
Answer:
(i)
(ii)
=
= [since (−2)6 = 64 and (3)6 = 729]
(iii)
=
(iv)
=
Page No 91:
Question 11:
By what number should (−5)−1 be multiplied so that the product is (8)−1?
Answer:
Let the required number be x.
(−5)-1 x = (8)-1
⇒
∴ x = =
Hence, the required number is .
Page No 91:
Question 12:
By what number should 3−3 be multiplied to obtain 4?
Answer:
Let the required number be x.
(3)-3 x x = 4
⇒
⇒
∴ x = 4 x 27 = 108
Hence, the required number is 108.
Page No 91:
Question 13:
By what number should (−30)−1 be divided to get 6−1?
Answer:
Let the required number be x.
(-30)-1 ÷ x = 6-1
⇒
⇒
∴ x =
=
Hence, the required number is .
Page No 91:
Question 14:
Find x such that .
Answer:
⇒
⇒
On equating the exponents:
−3 = 2x − 1
⇒ 2x = −3 + 1
⇒ 2x = −2
∴ x =
Page No 91:
Question 15:
Simplify: .
Answer:
Page No 92:
Question 16:
Simplify: .
Answer:
⇒
⇒
⇒
⇒
Page No 92:
Question 17:
Find the value of n when:
(i) 52n × 53 = 59
(ii) 8 × 2n+2 = 32
(iii) 62n+1 ÷ 36 = 63
Answer:
(i) 52n × 53 = 59
52n+3 = 59 [since an × am = am+n]
On equating the coefficients:
2n + 3 = 9
⇒ 2n = 9 − 3
⇒ 2n = 6
∴ n =
(ii) 8 × 2n+2 = 32
⇒ (2)3 × 2n+2 = (2)5 [since 23 = 8 and 25 = 32]
⇒ (2)3+ (n+2) = (2)5
On equating the coefficients:
3 + n + 2 = 5
⇒ n + 5 = 5
⇒ n = 5 − 5
∴ n = 0
(iii) 62n+1 ÷ 36 = 63
⇒ 62n+1 ÷ 62 = 63 [since 36 = 62]
⇒
⇒ [since ]
⇒ 62n-1 = 63
On equating the coefficients:
2n - 1 = 3
⇒ 2n = 3 + 1
⇒ 2n = 4
∴ n =
Page No 92:
Question 18:
If 2n−7 × 5n−4 = 1250, find the value on n.
Answer:
⇒ [since 1250 = 2 × 54]
⇒
⇒ [using cross multiplication]
⇒ [since am × an = am+n ]
⇒
⇒ [since an × bn = (a × b)n ]
⇒
⇒ n = 8
Page No 92:
Question 1:
Express each of the following numbers in standard form:
(i) 538
(ii) 6428000
(iii) 82934000000
(iv) 940000000000
(v) 23000000
Answer:
(i) 538 = 5.38 102 [since the decimal point is moved 2 places to the left]
(ii) 6428000 = 6.428 106 [since the decimal point is moved 6 places to the left]
(iii) 82934000000 = 8.2934 1010 [since the decimal point is moved 10 places to the left]
(iv) 940000000000 = 9.4 1011 [since the decimal point is moved 11 places to the left]
(v) 23000000 = 2.3 107 [since the decimal point is moved 7 places to the left]
Page No 92:
Question 2:
Express each of the following numbers in standard form:
(i) Diameter of Earth = 12756000 m.
(ii) Distance between Earth and Moon = 384000000 m.
(iii) Population of India in March 2001 = 1027000000.
(iv) Number of stars in a galaxy = 100000000000.
(v) The present age of universe = 12000000000 years
Answer:
(i) Diameter of the Earth = 1.2756 107 m
[since the decimal point is moved 7 places to the left]
(ii) Distance between the Earth and the Moon = 3.84 108 m
[since the decimal point is moved 8 places to the left]
(iii) Population of India in March 2001 = 1.027 109
[since the decimal point is moved 9 places to the left]
(iv) Number of stars in a galaxy = 1.0 1011
[since the decimal point is moved 11 places to the left]
(v) Present age of the universe = 1.2 1010 years
[since the decimal point is moved 10 places to the left]
Page No 93:
Question 3:
Write the following numbers in expanded form:
(i) 684502
(ii) 4007185
(iii) 5807294
(iv) 50074
Answer:
(i) 684502 = 6 x 105 + 8 x 104 + 4 x 103 + 5 x 102 + 0 x 101 + 2 x 100
(ii) 4007185 = 4 x 106 + 0 x 105 + 0 x 104 + 7 x 103 + 1 x 102 + 8 x 101 + 5 x 100
(iii) 5807294 = 5 x 106 + 8 x 105 + 0 x 104 + 7 x 103 + 2 x 102 + 9 x 101 + 4 x 100
(iv) 50074 = 5 x 104 + 0 x 103 + 0 x 102 + 7 x 101 + 4 x 100
Note: a0 = 1
Page No 93:
Question 4:
Write the numeral whose expanded form is given below:
(i) 6 × 104 + 3 × 103 + 0 × 102 + 7 × 101 + 8 × 100
(ii) 9 × 106 + 7 × 105 + 0 × 104 + 3 × 103 + 4 × 102 + 6 × 101 + 2 × 100
(iii) 8 × 105 + 6 × 104 + 4 × 103 + 2 × 102 + 9 × 101 + 6 × 100
Answer:
(i) 6 × 104 + 3 × 103 + 0 × 102 + 7 × 101 + 8 × 100
= 6 x 10000 + 3 x 1000 + 0 x 100 + 7 x 10 + 8 x 1 = 63078
(ii) 9 × 106 + 7 × 105 + 0 × 104 + 3 × 103 + 4 × 102 + 6 × 101 + 2 × 100
= 9 x 1000000 + 7 x 100000 + 0 x 10000 + 3 x 1000 + 4 x 100 + 6 x 10 + 2 x 1 = 9703462
(iii) 8 × 105 + 6 × 104 + 4 × 103 + 2 × 102 + 9 × 101 + 6 × 100
= 8 x 100000 + 6 x 10000 + 4 x 1000 + 2 x 100 + 9 x 10 + 6 x 1 = 864296
Page No 93:
Question 1:
Mark (✓) against the correct answer
(6−1 − 8−1)−1=?
(a)
(b) −2
(c)
(d) 24
Answer:
(d) 24
= [since L.C.M. of 6 and 8 is 24]
=
=
Page No 93:
Question 2:
Mark (✓) against the correct answer
(5−1 × 3−1)−1=?
(a)
(b)
(c) 15
(d) −15
Answer:
(c) 15
We have:
=
=
Page No 93:
Question 3:
Mark (✓) against the correct answer
(2−1 − 4−1)2 = ?
(a) 4
(b) −4
(c)
(d)
Answer:
(c)
We have:
= [since L.C.M. of 2 and 4 is 4]
=
=
Page No 93:
Question 4:
Mark (✓) against the correct answer
(a)
(b) 29
(c)
(d) none of these
Answer:
(b) 29
We have:
= (22 + 32 + 42)
= (4 + 9 + 16)
= 29
Page No 94:
Question 5:
Mark (✓) against the correct answer
(a)
(b)
(c)
(d) none of these
Answer:
(c)
We have:
= [since L.C.M. of 3 and 6 is 6]
=
=
Page No 94:
Question 6:
Mark (✓) against the correct answer
(a) −64
(b) 64
(c)
(d)
Answer:
(b) 64
We have:
Page No 94:
Question 7:
Mark (✓) against the correct answer
(a)
(b)
(c)
(d)
Answer:
(b)
= [ since L.C.M. of 1 and 3 is 3]
=
=
=
Page No 94:
Question 8:
Mark (✓) against the correct answer
(a)
(b) 16
(c)
(d) −16
Answer:
(a)
Page No 94:
Question 9:
Mark (✓) against the correct answer
(a)
(b) 0
(c) 1
(d) none of these
Answer:
(c) 1
(a)0 = 1
∴
Page No 94:
Question 10:
Mark (✓) against the correct answer
(a)
(b)
(c)
(d)
Answer:
(b)
=
Page No 94:
Question 11:
Mark (✓) against the correct answer
(a)
(b)
(c)
(d)
Answer:
(b)
Page No 94:
Question 12:
Mark (✓) against the correct answer
(a)
(b)
(c)
(d) none of these
Answer:
(b)
We have:
=
Page No 94:
Question 13:
Mark (✓) against the correct answer
(a)
(b)
(c)
(d)
Answer:
(d)
= =
Page No 94:
Question 14:
Mark (✓) against the correct answer
(a)
(b)
(c)
(d) none of these
Answer:
(a)
We have:
=
=
=
=
=
Page No 95:
Question 15:
Mark (✓) against the correct answer
(a)
(b)
(c) (−5)5
(d)
Answer:
(c) (-5)5
We have:
=
Page No 95:
Question 16:
Mark (✓) against the correct answer
(a)
(b)
(c)
(d)
Answer:
(a)
=
Page No 95:
Question 17:
Mark (✓) against the correct answer
(a)
(b)
(c)
(d)
Answer:
(c)
Page No 95:
Question 18:
Mark (✓) against the correct answer
(a)
(b)
(c)
(d) none of these
Answer:
(b)
We have:
Page No 95:
Question 19:
Mark (✓) against the correct answer
If , then x = ?
(a)
(b)
(c)
(d)
Answer:
(c)
⇒ [ since ]
⇒
On equating the coefficients:
6 = 8x
∴ x =
Page No 95:
Question 20:
Mark (✓) against the correct answer
By what number should (−8)−1 be multiplied to get 10−1?
(a)
(b)
(c)
(d) none of these
Answer:
(c)
Let the required number be x.
(−8)-1 x x = (10)-1
⇒
∴ x = =
Hence, the required number is .
Page No 95:
Question 21:
Mark (✓) against the correct answer
Which of the following numbers is in standard form?
(a) 21.56 × 105
(b) 215.6 × 104
(c) 2.156 × 106
(d) none of these
Answer:
(c) 2.156 × 106
A given number is said to be in standard form if it can be expressed as k x 10n, where k is a real number such that 1 ≤ k < 10 and n is a positive integer.
For example: 2.156 × 106
Page No 96:
Question 1:
Write the reciprocal of:
(i)
(ii)
(iii) 25
(iv) (−5)6
Answer:
We know that the reciprocal of is .
(i) Reciprocal of =
(ii) Reciprocal of
(iii) Reciprocal of 25 = Reciprocal of
(iv) Reciprocal of (−5)6 = Reciprocal of
Page No 96:
Question 2:
By what number should we multiply (−6)−1 to obtain a product equal to 9−1?
Answer:
Let the required number be x.
(−6)-1 × x = (9)-1
⇒
∴ x =
Hence, the required number is .
Page No 96:
Question 3:
By what number should (−20)−1 be divided to obtain (−10)−1?
Answer:
Let the required number be x.
(−20)-1 ÷ x = (−10)-1
⇒
⇒
∴ x =
Hence, the required number is 2-1.
Page No 96:
Question 4:
(i) Express 2000000 in standard form.
(ii) Express 6.4 × 105 in usual form.
Answer:
(i) 2000000 = 2.000000 × 106 [since the decimal point is moved 6 places to the left]
= 2 × 106
(ii) 6.4 × 105 = 6.4 × 100000
= 640000
Page No 96:
Question 5:
Simplify:
Answer:
We have:
⇒
⇒
⇒
⇒
Page No 96:
Question 6:
If 2n-7 × 5n-4 = 1250, find the value of n.
Answer:
We have:
⇒ [since 1250 = 2 × 54]
⇒
⇒ [using cross multiplication]
⇒ [since am × an = am+n ]
⇒
⇒ [since an × bn = (a × b)n ]
⇒
⇒ n = 8
Page No 96:
Question 7:
Mark (✓) against the correct answer
(a) 0
(b)
(c) 1
(d) none of these
Answer:
(c) 1
We know:
(a)0 = 1
∴
Page No 96:
Question 8:
Mark (✓) against the correct answer
(a)
(b)
(c)
(d)
Answer:
(d)
=
=
=
Page No 96:
Question 9:
Mark (✓) against the correct answer
(a)
(b)
(c)
(d)
Answer:
(b)
=
Page No 96:
Question 10:
Mark (✓) against the correct answer
(a) 19
(b)
(c) −19
(d)
Answer:
(a) 19
=
= (27 − 8) = 19
Page No 96:
Question 11:
Mark (✓) against the correct answer
(a)
(b)
(c)
(d) none of these
Answer:
(a)
=
=
=
Page No 96:
Question 12:
Which of the following numbers is in standard form?
(a) 32.63 × 104
(b) 326.3 × 103
(c) 3.263 × 105
(d) none of these
Answer:
(c) 3.263 x 105
A given number is said to be in standard form if it can be expressed as k x 10n, where k is a real number such that 1 ≤ k < 10 and n is a positive integer.
For example: 3.263 x 105
Page No 96:
Question 13:
Fill in the blanks.
(i) If 9 × 3n = 36, then n = ...... .
(ii) (8)0 = ?
(iii)
(iv) (−2)−5 = ......
Answer:
(i) If 9 × 3n = 36, then n = 4.
Explanation:
If 9 × 3n = 36
⇒ 32 × 3n = 36
⇒ 3(2 + n) = 36
Equating the powers:
⇒ ( 2 + n) = 6
⇒ n = (6 - 2) = 4
(ii) (8)0 = 1
Explanation:
By definition, we have a0 = 1 for every integer a.
∴ (8)0 = 1
(iii) =
Explanation:
We know:
(iv) (−2)−5 =
Explanation:
(−2)−5 =
=
Page No 96:
Question 14:
Write 'T' for true and 'F' for false for each of the following.
(i) 645 in standard form is 6.45 × 102.
(ii) 27000 in standard form is 27 × 103.
(iii) (30 + 40 + 50) = 12.
(iv) Reciprocal of 56 is 65.
(v) If 5−1 × x = 8−1, then x = .
Answer:
(i) True
645 = 6.45 x 102 [since the decimal point is moved 2 places to the left]
(ii) False
27000 = 2.7 x 104 [since the decimal point is moved 4 places to the left]
(iii) False
(30 + 40 + 50) = 1 [since a0 = 1 for every integer a]
(iv) False
Reciprocal of 56 = Reciprocal of
(v) False
5−1 × x = 8−1
⇒
⇒ x =
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