Free Exponents and Powers Subjective Test 01 Practice Test - 7th grade

Question 1

Express 625 as a power of 5. [1 MARK]

SOLUTION

Solution :

$625 = (5) \times (5) \times (5) \times (5)$
$\Rightarrow 625 = 5^{4}$

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.
$726 = (2) \times (3) \times (11) \times (11)$

$\Rightarrow 726 = 2^{1} \times 3^{1} \times 11^{2}$

$938 = (2) \times (7) \times (67)$

$\Rightarrow 938 = 2^{1} \times 7^{1} \times 67$

Both have the same powers of 2 which is 1.

Question 3

Find the difference between $4^{5}$ and $5^{4}$ ? [2 MARKS]

SOLUTION

Solution :
Steps: 1 Mark
Answer: 1 Mark

$4^{5} = (4) \times (4) \times (4) \times (4) \times (4)$

$\Rightarrow 4^{5} = 1024$

$5^{4} = (5) \times (5) \times (5) \times (5)$

$\Rightarrow 5^{4} = 625$

The difference between $4^{5}$ and $5^{4}$ is

$4^{5} - 5^{4} = 1024 - 625$

$\Rightarrow 4^{5} - 5^{4} = 399$
The difference between $4^{5}$ and $5^{4}$ is $399$ .

Question 4

Write $(8) \times (8) \times (8) \times (8) \times (8)$ 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.
$(8) \times (8) \times (8) \times (8) \times (8)$

$= 2^{3} \times 2^{3} \times 2^{3} \times 2^{3} \times 2^{3}$

$= 2^{3 + 3 + 3 + 3 + 3}$

$∵ a^{b} \times a^{c} = a^{b + c}$

$= 2^{15}$

Question 5

Is $3^{0} = (1000)^{0}$ true or false? Justify your answer. [2 MARKS]

SOLUTION

Solution : Concept: 1 Mark
Answer: 1 Mark

Consider, $3^{0} = (1000)^{0}$

L.H.S. is $3^{0} = 1$ [ $∵ a^{0} = 1$ ]

and R.H.S. $(1000)^{0} = 1$ [ $∵ a^{0} = 1$ ]

$\Rightarrow$ L.H.S. = R.H.S.

$\Rightarrow 3^{0} = (1000)^{0}$ .

$∴$ It is true.
We should remember that any non-zero number raised to the power 0 is always equal to 1.

Question 6

Simplify: $\frac{8^{4} \times 14^{3} \times 5 \times 3^{4}}{4^{5} \times 7^{3} \times 6^{2}}$ [3 MARKS]

SOLUTION

Solution :
Application: 1 Mark
Steps: 1 Mark
Answer: 1 Mark

Consider,

$\frac{8^{4} \times 14^{3} \times 5 \times 3^{4}}{4^{5} \times 7^{3} \times 6^{2}}$

$8^{4} = (2^{3})^{4} = 2^{12}$

$14^{3} = (2 \times 7)^{3} = 2^{3} \times 7^{3}$

$4^{5} = (2^{2})^{5} = 2^{10}$

$6^{2} = (2 \times 3)^{2} = 2^{2} \times 3^{2}$

On substituting the values we get:

$\frac{2^{12} \times 2^{3} \times 7^{3} \times 5 \times 3^{4}}{2^{10} \times 7^{3} \times 2^{2} \times 3^{2}}$

$= 2^{12 + 3 - 10 - 2} \times 7^{3 - 3} \times 5 \times 3^{4 - 2}$ [ $∵ \frac{a^{m}}{a^{n}} = a^{m - n}$ ]

$= 2^{3} \times 5 \times 3^{2}$ [ $∵ a^{0} = 1$ ]

$= 8 \times 5 \times 9$

$= 360$

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 $(- 5) \times (5)$ ? [3 MARKS]

SOLUTION

Solution :
Concept: 1 Mark
Application: 1 Mark
Answer: 1 Mark

Let the number be $x$ .

Given that,

$(x) \times (25) = 78125$

$x = 3125$

$\Rightarrow$ The required number is 3125

Now,

$3125 = 5 \times 5 \times 5 \times 5 \times 5$

$\Rightarrow 3125 = 5^{5}$

Now, if the number was multiplied by $(- 5) \times (5)$ then the product $= (- 5) \times (5) \times 3125$ = $- 78125$

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 $\Rightarrow$ Rs 4325.

Difference = Rs 34265 - Rs 4325 = Rs 29940.

The standard form of Rs 29940.

$29940 = 2.994 \times 10^{4}$
The difference in rupees is $2.994 \times 10^{4}$

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) $512 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^{9}$

ii) $343 = 7 \times 7 \times 7 = 7^{3}$

iii) $729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 3^{6}$

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 $= \frac{x}{2}$

Height reached after second bounce $= \frac{x}{4} = \frac{x}{2^{2}}$

Height reached after third bounce $= \frac{x}{8} = \frac{x}{2^{3}}$

Height reached after fourth bounce $= \frac{x}{16} = \frac{x}{2^{4}}$

Height reached after fifth bounce $= \frac{x}{32} = \frac{x}{2^{5}}$

Height reached after sixth bounce $= \frac{x}{64} = \frac{x}{2^{6}}$

Height reached after seventh bounce $= \frac{x}{128} = \frac{x}{2^{7}}$

Height reached after eight bounce $= \frac{x}{256} = \frac{x}{2^{8}}$

$∴$ Product of first 9 heights $= x \times \frac{x}{2} \times \frac{x}{2^{2}} \times \frac{x}{2^{3}} \times \frac{x}{2^{4}} \times \frac{x}{2^{5}} \times \frac{x}{2^{6}} \times \frac{x}{2^{7}} \times \frac{x}{2^{8}}$
$= \frac{x^{(1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1)}}{2^{(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8)}}$
$(∵ a^{b} \times a^{c} = a^{b + c})$
$= \frac{x^{9}}{2^{36}}$

The product of heights after 8 bounces $= \frac{x^{9}}{2^{36}}$

Question 11

A man X has a land whose area is $3^{4}$ meter square and a man Y has $4^{3}$ 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

$= 3^{4} m^{2} = 81 m^{2}$

The value of given property is

$= (81) \times (50) = R s .4050$

The area of land owned by Y is

$4^{3} m^{2} = 64 m^{2}$

The value of given property is

$= (64) \times (50) = R s .3200$

$\Rightarrow$ Value of property owned by X is greater than property owned by Y.

Question 12

Simplify: [4 MARKS]

i) ${(\frac{5}{8})}^{- 7} \times {(\frac{8}{5})}^{- 5}$
ii) $\frac{(2^{5})^{2} \times 7^{3}}{8^{3} \times 7}$

SOLUTION

Solution : Each: [2 Marks]

i) ${(\frac{5}{8})}^{- 7} \times {(\frac{8}{5})}^{- 5}$

$= \frac{5^{- 7}}{8^{- 7}} \times \frac{8^{- 5}}{5^{- 5}}$

$= \frac{5^{- 7}}{5^{- 5}} \times \frac{8^{- 5}}{8^{- 7}}$

$= 5^{(- 7) - (- 5)} \times 8^{(- 5) - (- 7)}$

( $∵ \frac{a^{m}}{a^{n}} = a^{m - n}$ )

$= 5^{- 2} \times 8^{2} = \frac{8^{2}}{5^{2}} = \frac{64}{25}$

ii) $\frac{(2^{5})^{2} \times 7^{3}}{8^{3} \times 7}$

$\frac{(2^{5})^{2} \times 7^{3}}{8^{3} \times 7}$

= $\frac{(2^{5 \times 2}) \times 7^{3}}{(2^{3})^{3} \times 7}$

$(∵ (a^{m})^{n} = a^{m \times n})$

= $\frac{2^{10} \times 7^{3}}{2^{9} \times 7}$

= $2^{10 - 9} \times 7^{3 - 1} = 2 \times 7^{2}$

= $2 \times 49 = 98$

Question 13

If abc = 1, then find the value of:

$\frac{1}{1 + a + b^{- 1}} + \frac{1}{1 + b + c^{- 1}} + \frac{1}{1 + c + a^{- 1}}$ [4 MARKS]

SOLUTION

Solution : Formula: 1 Mark
Steps: 2 Marks
Answer: 1 Mark

Given that,
abc=1
So, $c = 1 \div a b = (a b)^{- 1}$
Similarly,
$a = (b c)^{- 1}$
$b = (a c)^{- 1}$
And,
$a b = c^{- 1}$
$a c = b^{- 1}$
$b c = a^{- 1}$

$\frac{1}{1 + a + b^{- 1}} + \frac{1}{1 + b + c^{- 1}} + \frac{1}{1 + c + a^{- 1}}$

$= \frac{1}{1 + a + b^{- 1}} + \frac{b^{- 1}}{b^{- 1} + b b^{- 1} + b^{- 1} c^{- 1}} + \frac{a}{a + a c + a a^{- 1}}$

[Multiplying 2nd term by $\frac{b^{- 1}}{b^{- 1}}$ and 3rd term by $\frac{a}{a}$ ]

$= \frac{1}{1 + a + b^{- 1}} + \frac{b^{- 1}}{b^{- 1} + 1 + a} + \frac{a}{a + b^{- 1} + 1}$

$∵ \frac{a^{m}}{a^{n}} = a^{m - n}$

$= \frac{1 + b^{- 1} + a}{1 + a + b^{- 1}}$

$= 1$

Question 14

If $x = y^{a}, y = z^{b}, z = x^{c}$ then find abc. If a = 12, b = $\frac{1}{288}$ , 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
$x = y^{a}, y = z^{b}, z = x^{c}$

$x = y^{a}$

$= (z^{b})^{a}$ $(∵ y = z^{b})$

$= z^{a b}$

$= (x^{c})^{a b}$ $(∵ x^{c} = z)$

$x = x^{a b c}$ $∵ ((a^{m})^{n} = a^{m n})$

$\Rightarrow x^{1} = x^{a b c}$

As bases are equal, exponents are equal.

$\Rightarrow a b c = 1$

Now, it is also given that,
a = 12, b = $\frac{1}{288}$
a $\times$ b $\times$ c = 1
c = $\frac{1}{a \times b}$

On substituting the values we get:
c = $\frac{1}{12 \times \frac{1}{288}}$

c = $\frac{288}{12}$

c = $24$
Hence, the value of c is $24$ .

Question 15

In a race Karan covers 1m in $1^{s t}$ second, in the next second he covers twice the distance he covered in $1^{s t}$ second, in $3^{r d}$ second he covers twice the distance which he covered in $2^{n d}$ second and he continues to run like this till six seconds. After that he slows down, as in $7^{t h}$ second he covers only half the distance which he covered in $6^{t h}$ second and in $8^{t h}$ second he covers half the distance which he covered in $7^{t h}$ second 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

$1^{s t}$ second = 1m

$2^{n d}$ second = $1 \times 2$ =2m

Hence, the length of the track is 2 m.

$3^{r d}$ second = $2 \times 2$ = 4m

$4^{t h}$ second = $4 \times 2$ = 8m

$5^{t h}$ second = $8 \times 2$ =16m

$6^{t h}$ second = $16 \times 2$ =32m

$7^{t h}$ second = $32 \div 2$ = 16m

$8^{t h}$ second = $16 \div 2$ = 8m

$9^{t h}$ second = $8 \div 2$ = 4m

Length of the given track = (1+2+4+8+16+32+16+8+4) m

$\Rightarrow$ Length of the given track = 91 m

Hence, the length of the track is 91 m.

Free Exponents and Powers Subjective Test 01 Practice Test - 7th grade

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

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