Free Exponents and Powers Subjective Test 01 Practice Test - 7th grade
Question 1
Express 625 as a power of 5. [1 MARK]
SOLUTION
Solution :
625=(5)×(5)×(5)×(5)
⇒625=54
Question 2
Which of the numbers 726 and 938, when expressed as factors has greater power of 2?
[2 MARKS]
SOLUTION
Solution :Steps: 1 Mark
Answer: 1 Mark
The prime factorisation of 726 and 938 are given below.
726=(2)×(3)×(11)×(11)
⇒726=21×31×112
938=(2)×(7)×(67)
⇒938=21×71×67
Both have the same powers of 2 which is 1.
Question 3
Find the difference between 45 and 54? [2 MARKS]
SOLUTION
Solution :Steps: 1 Mark
Answer: 1 Mark
45=(4)×(4)×(4)×(4)×(4)
⇒45=1024
54=(5)×(5)×(5)×(5)
⇒54=625
The difference between 45 and 54 is
45−54=1024−625
⇒45−54=399
The difference between 45 and 54 is 399.
Question 4
Write (8)×(8)×(8)×(8)×(8) in the exponential form, taking 2 as base. [2 MARKS]
SOLUTION
Solution :Concept: 1 Mark
Answer: 1 Mark
Here we know that for any non zero number if the base is same and the number is multiplied, then the exponents are added.
(8)×(8)×(8)×(8)×(8)
=23×23×23×23×23
=23+3+3+3+3
∵ab×ac = ab+c
=215
Question 5
Is 30=(1000)0 true or false? Justify your answer. [2 MARKS]
SOLUTION
Solution : Concept: 1 Mark
Answer: 1 Mark
Consider, 30=(1000)0
L.H.S. is 30=1 [ ∵a0 = 1]
and R.H.S. (1000)0=1 [ ∵a0 = 1]
⇒ L.H.S. = R.H.S.
⇒30=(1000)0.
∴ It is true.
We should remember that any non-zero number raised to the power 0 is always equal to 1.
Question 6
Simplify: 84×143×5×3445×73×62 [3 MARKS]
SOLUTION
Solution :Application: 1 Mark
Steps: 1 Mark
Answer: 1 Mark
Consider,
84×143×5×3445×73×62
84=(23)4=212
143=(2×7)3=23×73
45=(22)5=210
62=(2×3)2=22×32
On substituting the values we get:
212×23×73×5×34210×73×22×32
=212+3−10−2×73−3×5×34−2 [∵ aman=am−n]
=23×5×32 [∵a0=1]
=8×5×9
=360
Question 7
When a number is multiplied with 25 it gives 78125. Find the number and express it as a power of 5. What is the product if the number is multiplied by (−5)×(5)? [3 MARKS]
SOLUTION
Solution :Concept: 1 Mark
Application: 1 Mark
Answer: 1 Mark
Let the number be x.
Given that,
(x)×(25)=78125
x=3125
⇒ The required number is 3125
Now,
3125=5×5×5×5×5
⇒3125=55
Now, if the number was multiplied by (−5)×(5) then the product=(−5)×(5)×3125 = −78125
Question 8
A shopkeeper A has Rs 34265 in his counter and shopkeeper B has 432500 paise in his counter. Express the difference in rupees in standard form. [ 3 MARKS]
SOLUTION
Solution :Conversion: 1 Mark
Difference: 1 Mark
Standard form: 1 Mark
As per the question
The amount shopkeeper A has = Rs 34265.
The amount shop keeper B has = 432500 paise ⇒ Rs 4325.
Difference = Rs 34265 - Rs 4325 = Rs 29940.
The standard form of Rs 29940.
29940=2.994×104
The difference in rupees is 2.994×104
Question 9
Express each of the following numbers using the exponential notation:
(i) 512
(ii) 343
(iii) 729
SOLUTION
Solution : Each option: 1 Mark
i)512=2×2×2×2×2×2×2×2×2=29
ii)343=7×7×7=73
iii)729=3×3×3×3×3×3=36
Question 10
A ball is released from a height of x meters. Given that it bounces back to half of its original height. Find the product of heights after eight bounces. [4 MARKS]
SOLUTION
Solution :Formula: 1 Mark
Application: 1 Mark
Answer: 2 Marks
Given that
The ball is released from a height of x m
Height reached after first bounce =x2
Height reached after second bounce =x4=x22
Height reached after third bounce =x8=x23
Height reached after fourth bounce =x16=x24
Height reached after fifth bounce =x32=x25
Height reached after sixth bounce =x64=x26
Height reached after seventh bounce =x128=x27
Height reached after eight bounce =x256=x28
∴ Product of first 9 heights =x×x2×x22×x23×x24×x25×x26×x27×x28
=x(1+1+1+1+1+1+1+1+1)2(1+2+3+4+5+6+7+8)
(∵ab×ac = ab+c)
=x9236
The product of heights after 8 bounces =x9236
Question 11
A man X has a land whose area is 34 meter square and a man Y has 43 meter square of land. If the land costs Rs 50 per meter square, who has higher property value? [4 MARKS]
SOLUTION
Solution :Step: 2 Marks
Application: 1 Mark
Answer: 1 Mark
The area of land owned by X is
=34 m2=81 m2
The value of given property is
=(81)×(50)=Rs.4050
The area of land owned by Y is
43 m2=64 m2
The value of given property is
=(64)×(50)=Rs.3200
⇒ Value of property owned by X is greater than property owned by Y.
Question 12
Simplify: [4 MARKS]
i)(58)−7×(85)−5
ii)(25)2×7383×7
SOLUTION
Solution : Each: [2 Marks]
i)(58)−7×(85)−5
=5−78−7×8−55−5
=5−75−5×8−58−7
=5(−7)−(−5)×8(−5)−(−7)
(∵ aman=am−n)
=5−2×82=8252=6425
ii)(25)2×7383×7
(25)2×7383×7
= (25×2)×73(23)3×7
(∵(am)n=am×n)
= 210×7329×7
=210−9×73−1=2×72
=2×49=98
Question 13
If abc = 1, then find the value of:
11+a+b−1+11+b+c−1+11+c+a−1 [4 MARKS]
SOLUTION
Solution : Formula: 1 Mark
Steps: 2 Marks
Answer: 1 Mark
Given that,
abc=1
So, c=1÷ab=(ab)−1
Similarly,
a=(bc)−1
b=(ac)−1
And,
ab=c−1
ac=b−1
bc=a−1
11+a+b−1+11+b+c−1+11+c+a−1
=11+a+b−1+b−1b−1+bb−1+b−1c−1+aa+ac+aa−1
[Multiplying 2nd term by b−1b−1 and 3rd term by aa]
=11+a+b−1+b−1b−1+1+a+aa+b−1+1
∵ aman=am−n
=1+b−1+a1+a+b−1
=1
Question 14
If x=ya, y=zb, z=xc then find abc. If a = 12, b = 1288, then find the value of c. [4 MARKS]
SOLUTION
Solution : Steps: 1 Marks
Application: 1 Mark
Value of abc: 1 Mark
Value of c: 1 Mark
It is given that
x=ya, y=zb, z=xc
x=ya
=(zb)a (∵y=zb)
=zab
=(xc)ab (∵xc=z)
x=xabc ∵((am)n=amn)
⇒x1=xabc
As bases are equal, exponents are equal.
⇒abc=1
Now, it is also given that,
a = 12, b = 1288
a×b×c = 1
c = 1a×b
On substituting the values we get:
c = 112×1288
c = 28812
c = 24
Hence, the value of c is 24.
Question 15
In a race Karan covers 1m in 1st second, in the next second he covers twice the distance he covered in 1st second, in 3rd second he covers twice the distance which he covered in 2nd second and he continues to run like this till six seconds. After that he slows down, as in 7th second he covers only half the distance which he covered in 6th second and in 8thsecond he covers half the distance which he covered in 7thsecond and he continuous in that manner. If Karan completes the race in 9 seconds, find the length of given track. [4 MARKS]
SOLUTION
Solution :Application: 1 Mark
Steps: 2 Marks
Answer: 1 Mark
Here are the distances covered by Karan in each second
1st second = 1m
2nd second =1×2=2m
Hence, the length of the track is 2 m.
3rd second =2×2= 4m
4th second =4×2= 8m
5th second = 8×2=16m
6th second = 16×2=32m
7th second =32÷2= 16m
8th second =16÷2= 8m
9th second =8÷2= 4m
Length of the given track = (1+2+4+8+16+32+16+8+4) m
⇒Length of the given track = 91 m
Hence, the length of the track is 91 m.