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Short and Long Question: Exponents And Powers | Short & Long Answer Questions for Class 7 PDF Download

Q1: Can you find two integers m,n such that 2m+n = 2mn ?
Ans:
Equating the powers of 2m+n =2mn  that is m+n = mn.
Now we have to find such integers so that the sum of the integers is equal to their product and those integers are 0 and 2:
Let us first take m=2 and n=2, we get
2 + 2 = 2 × 2
⇒ 4 = 4
which is true
Now take m=0 and n=0, we get
0 + 0 = 0 × 0
⇒ 0 = 0
which is true
Hence the integers are m = 0, 2 and n = 0, 2.

Q2: Simplify the following using laws of indices:
Short and Long Question: Exponents And Powers | Short & Long Answer Questions for Class 7
Ans: 
The given expression Short and Long Question: Exponents And Powers | Short & Long Answer Questions for Class 7 can be simplified as follows:
Short and Long Question: Exponents And Powers | Short & Long Answer Questions for Class 7

Short and Long Question: Exponents And Powers | Short & Long Answer Questions for Class 7

Hence Short and Long Question: Exponents And Powers | Short & Long Answer Questions for Class 7 = 121/4

Q3: Evaluate
(i) 2-2
(ii) (-2)-2
(iii) (3/2)
-5
Ans: As we know that
b-n = 1/bn
(i) 2-2 = 1/ 22 = 1/4
(ii) (-2)-2 = 1/(-2)= 1/4
(iii) (3/2)-5 = 3-5/ 2-5 =25/35 = 32/243

Q4: Simplify and express the result in power notation with positive exponent.
(i) (-2)5 ÷ (-2)4
(ii) (1/2)× (2/5)2
(iii) (-5)2 × (3/5)
Ans:
 
(i) (-2)5 ÷ (-2)4
= (-2)/ (-2)4
= (-2)5-4
=-2
(ii) (1/2)2 × (2/5)2
= (1/4) X (4/25)
= 1/25
(iii) (-5)2 × (3/5)
=25 × (3/5) =15

Q5: Express the following numbers in standard form.
(i) 0.0000000015
(ii) 0.00000001425
Ans:

(i) 0.0000000015
=1.5 ×10-9
(ii) 0.00000001425
=1.425×10-8
(iii) 102000000000000000
=1.02×1017

Q6: Simplify the following:
(i) 103 × 90 + 33 × 2 + 70
(ii) 63 × 70 + (-3)4 – 90
Ans:

(i) 103 × 90 + 33 × 2 + 70
= 1000 + 54 + 1
= 1055
(ii) 6× 7+ (-3)4 – 90
= 216 × 1 + 81 – 1
= 216 + 80
= 296

Q7: Write the following in expanded form:
(i) 70,824
(ii) 1,69,835
Ans:

(i) 70,824
= 7 × 10000 + 0 × 1000 + 8 × 100 + 2 × 10 + 4 × 100
= 7 × 104 + 8 × 10+ 2 × 101 + 4 × 100
(ii) 1,69,835= 1 × 100000 + 6 × 10000 + 9 × 1000 + 8 × 100 + 3 × 10 + 5 × 100
= 1 × 105 + 6 × 10+ 9 × 103 + 8 × 102 + 3 × 101 + 5 × 100

Q8: Find the number from each of the expanded form:
(i) 7 × 108 + 3 × 105 + 7 × 102 + 6 × 101 + 9
(ii) 4 × 107 + 6 × 103 + 5
Ans:

(i) 7 × 108 + 3 × 105 + 7 × 102 + 6 × 101 + 9
= 7 × 100000000 + 3 × 100000 + 7 × 100 + 6 × 10 + 9
= 700000000 + 300000 + 700 + 60 + 9
= 700300769
(ii) 4 × 107 + 6 × 103 + 5
= 4 × 10000000 + 6 × 1000 + 5
= 40000000 + 6000 + 5
= 40006005

Q9: Express the following in standard form:
(i) 8,19,00,000
(ii) 5,94,00,00,00,000
(iii) 6892.25
Ans:

(i) 8,19,00,000 = 8.19 × 107
(ii) 5,94,00,00,00,000 = 5.94 × 1011
(iii) 6892.25 = 6.89225 × 103


Q10: Express the following in exponential form:
(i) 5 × 5 × 5 × 5 × 5
(ii) 4 × 4 × 4 × 5 × 5 × 5
(iii) (-1) × (-1) × (-1) × (-1) × (-1)
(iv) a × a × a × b × c × c × c × d × d
Ans: :

(i) 5 × 5 × 5 × 5 × 5 = (5)5
(ii) 4 × 4 × 4 × 5 × 5 × 5 = 43 × 53
(iii) (-1) × (-1) × (-1) × (-1) × (-1) = (-1)5
(iv) a × a × a × b × c × c × c × d × d = a3b1c3d2

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