Free Exponents and Powers Subjective Test 02 Practice Test - 7th grade
Question 1
Express 472 as a power of its prime factors. [1 MARK]
SOLUTION
Solution :
472=(2)×(2)×(2)×(59)
⇒472=23×59
Question 2
A man has many chocolates and he takes 37 chocolates and distributes it to 34 students. How many chocolates do each student get if each student should get an equal number of chocolates? [2 MARKS]
SOLUTION
Solution :Concept: 1 Mark
Answer: 1 Mark
Given that,
A man has many chocolates and he takes 37 chocolates.
He has to distribute them equally between 34 students.
The number of chocolates a student gets is:
=3734
=37−4 [∵aman=am−n]
=33
=27
So, each student will get 27 chocolates.
Question 3
Simplify: [2 MARKS]
25×3×42×5
SOLUTION
Solution : Steps: 1 Mark
Answer: 1 Mark
Given equation is:
25×3×42×5
=25×3×(22)2×5
=25×3×24×5
∵(am)n=amn
=25+4×3×5
∵am×an=am+n
=29×3×5
=29×3×5
=7680
Question 4
What is the difference between 43 and 32? [2 MARKS]
SOLUTION
Solution :Steps: 1 Mark
Answer: 1 Mark
43=(4)×(4)×(4)
⇒43=64
32=(3)×(3)
⇒32=9
The difference between given numbers is:
=64−9=55
The difference between 43 and 32 is 55.
Question 5
The distance between two cities is 3245678 km. Convert it into meters and express in standard form. [2 MARKS]
SOLUTION
Solution :Conversion: 1 Mark
Standard form: 1 Mark
Given distance that the distance between two cities 3245678 km.
Now, 1 km = 1000 m
3245678 km = (3245678 × 1000) m = 3245678000 m.
Standard form is,
3245678000m=3.245678×109
Question 6
In a stck there are 5 books each of thickness 20mm and 5 paper sheets of thickness 0.016mm. What is the total thickness of the stack? (1mm = 0.001m) [3 MARKS]
SOLUTION
Solution :Steps: 1 Mark
Application: 1 Mark
Answer: 1 Mark
Thickness of each book =20mm
Hence, thickness of 5 books =5×20=100mm
Thickness of each paper sheet =0.016mm
Hence, thickness of 5 paper sheets =5×0.016=0.080mm
Total thickness of the stack
= Thickness of 5 books + Thickness of 5 paper sheets
=(100+0.080)mm
=100.08mm
=1.0008×102mm
Question 7
Simplify : [3 MARKS]
38×45×32×54×7539×44×75
SOLUTION
Solution :Concept: 1 Mark
Steps: 1 Mark
Answer: 1 Mark
Given,
38×45×32×54×7539×44×75
=38+2×45×54×7539×44×75
[ ∵am×an=am+n]
=310−9×45−4×54×75−5
[ ∵aman=am−n]
=31×41×54×70
=3×4×54 [ ∵a0=1]
=7500
Question 8
Write the following numbers in the expanded form:
(i) 279404
(ii) 3006194
(iii) 2806196
[3 Marks]
SOLUTION
Solution : Each option: 1 Mark
i) 279404 = 2,00,000 + 70,000 + 9,000 + 400 + 00 +4
=2×100000+7×10000+9×1000+4×100+0×10+4×1
= 2×105+7×104+9×103+4×102+0×101+4×100
ii) 3006194 = 30,00,000 + 0 + 0 + 6,000 + 100 + 90 + 4
=3×1000000+0×100000+0×10000+6×1000+1×100+9×10+4×1
=3×106+0×105+0×104+6×103+1×102+9×101+4×100
iii) 2806196 = 20,00,000 + 8,00,000 + 0 + 6,000 + 100 + 90 + 6
= 2×1000000+8×100000+0×10000+6×1000+1×100+9×10+6×1
=2×106+8×105+0×104+6×103+1×102+9×101+6×100
Question 9
Simplify and express in exponential form:
[(52)3×54]÷57 [3 MARKS]
SOLUTION
Solution : Steps: 2 Marks
Answer: 1 Mark
=[(52)3×54]÷57
=[56×54]÷57 [∵(am)n=am×n]
=[56+4]÷57 [∵am×an=am+n]
=510÷57
=510−7 [∵am÷an=am−n]
=53
Question 10
The area of a certain number of triangles is equal to the sum of the exponents of the prime factors of the number 1628, and each prime factor represents a triangle. Find the sum of areas of the triangles and find the number of the triangles. [4 MARKS]
SOLUTION
Solution :Prime factorization: 1 Marks
Number of Triangles: 1 Mark
Sum of the area: 1 Mark
Steps: 1 Mark
Given that,
The area of a certain number of triangles is equal to the exponents of the prime factors of the number 1628 and each prime factor represents a triangle.
Prime factors of 1628 are:
1628=22×3×7×19.
Since there are 5 prime factors,
⇒ The number of given triangles are = 5
The area of the triangles is the sum of powers of the prime factors.
⇒The sum of areas of the triangle = 2 + 1 + 1 + 1
= 5 square units
The number of triangles is 5 and the sum of areas of the triangle is 5 square units.
Question 11
Find the value of 'n' if (am)2n=a2m. Find the numerical value of a2m if a=2 and m=4. [4 MARKS]
SOLUTION
Solution :Formula: 1 Mark
Value of n: 1 Mark
Numerical value: 1 Mark
Steps: 1 Mark
Given that,
(am)2n=a2m
a2mn=a2m [∵(am)n=amn]
Since bases are same, their exponents also should be equal.
⇒2mn=2m
⇒n=2m2m
⇒n=1
As per question,
a=2 and m=4
On substituting the values, we get:
a2m=22×4
=28
=2×2×2×2×2×2×2×2
=256
Hence, the required value is 256.
Question 12
The value of a man's property is greater than the multiplication of cube and square of properties of A and B respectively.
What is the minimum value of his property, if the value of properties A and B are Rs 20 and Rs 400 respectively?
The actual value of his property was Rs 2,00,00,00,000 but due to recession, the value of the property decreased by 30%.
Find the present value of the property. [4 MARKS]
SOLUTION
Solution :Steps: 2 Marks
Minimum Value: 1 Mark
Present Value: 1 Mark
Given that,
A's property = Rs 20
⇒ cube of A's property =203
B's property = Rs 400
⇒ Square of B's property =4002
The minimum value of the man's property should be
=Rs 203×4002
=Rs 128×107
=Rs 1.28×109
Given that
The actual value of the property is Rs2,00,00,00,000.
The value of the property depreciated by 30%.
The present value of the property=2000000000−2000000000×30100
=2000000000−600000000
=1400000000
Hence, the present value of the property is Rs 1400000000
Question 13
Find the value of 'a' satisfying the equation.
a) (42)a=(7a)2.
b) (2a)5=24×43
[4 MARKS]
SOLUTION
Solution :Each option: 2 Marks
a) Given that,
(42)a=(7a)2
42a=72a [amn=amn]
Given, 42a=72a
Since 4≠7 the only case where the above equation is true is when both the exponents are zero.
⇒ a=0
⇒42a=72a=1 [∵40=1, 70=1]
b) The given equation is
(2a)5=24×43
⇒2a×5=24×(22)3
∵(am)n=am×n
⇒2a×5=24×(22×3)
⇒2a×5=24×(26)
⇒2a×5=24+6=210
∵am×an=am+n
Since their bases are same and they are equal, threfore their powers must be same,
So, 5×a=10
⇒a=105
⇒a=2
So, the value of a = 2.
Question 14
Simplify:
(i)0.005×102−5×10−1
(ii)(25)2×7383×7
[4 MARKS]
SOLUTION
Solution :Each question: 2 Marks
i) It's easy to subtract powers when we convert them into normal form.
0.005×102=0.5 (move the decimal point 2 places to the right)
5×10−1=0.5 (move the decimal point 1 place to the left)
Now,
0.005×102−5×10−1
=0.5−0.5
=0
ii)
(25)2×7383×7
= (25)2×7383×7
= (25×2)×73(23)3×7
= 210×7329×7
= 210−9×73−1=2×72
= 2×49=98
Question 15
Find the value of m and n, if 6−5×5−2×63×5m×6n=1. [4 MARKS]
SOLUTION
Solution :Steps: 2 Marks
Application: 1 Mark
Answer: 1 Mark
6−5×5−2×63×5m×6n=1
6−5×63×6n×5−2×5m=1
6−5+3+n×5−2+m=1
6−2+n×5−2+m=1
The bases of both the numbers are not equal.
So, the value will be equal to 1, when the exponents of these bases will be equal to zero.
Any non - zero number raised to the power zero is 1.
So, the exponents of both 6 and 5 are zero.
Therefore,
−2+m=0 ⇒m=2
−2+n=0 ⇒n=2
∴m=n