NCERT Exemplar Problems Class 7 Maths – Exponents and Powers

NCERT Exemplar Problems Class 7 Maths – Exponents and Powers

Multiple Choice Questions (MCQs)

Question 1:
[(- 3)²]³ is equal to
(a) (-3)⁸               (b) (- 3)⁶
(c) (-3)⁵                (d) (-3)²³
Solution:
(b) We know that, if ‘a’ is a rational number, m and n are natural numbers, then (a^m)ⁿ = a^{m x n}
So, [(- 3)² ]³ = (- 3)^{2 x 3} = (- 3)⁶

Question 2:
For a non-zero rational number x, x⁸ ÷ x² is equal to
(a) x⁴                     (b) x⁶                    (c) x¹⁰                  (d) x¹⁶
Solution:
(b) We know that, if ‘a’ is a rational number, m and n are natural numbers such that m> n, then
a^m ÷ aⁿ = a^m-n
So, x⁸ ÷ x² = x⁸/x² = x^8-2 = x⁶

Question 3:
x is a non-zero rational number. Product of the square of x with the cube of x is equal to the
(a) second power of x                 (b) third power of x
(c) fifth power of x                      (d) sixth power of x
Solution:
(c) Square of x = x²
Cube of x = x³
Product of square with the cube of x = x² x x³ = x²⁺³ [∴ a^m x aⁿ =a^m+n]
i.e. fifth power of x.

Question 4:
For any two non-zero rational numbers x and y, x⁵ ÷ y⁵ is equal to
(a) (x ÷ y)¹                      (b) (x ÷ y)⁰
(c) (x ÷ y)⁵                     (d) (x ÷ y)¹⁰
Solution:

Question 5:
a^m x aⁿ  is equal to
(a) (a²)^mn             (b) a^m-n
(c) a^m+n                (d) a^mn
Solution:
(c) We know that, if ‘a’ is a rational number, m and n are natural numbers, then
a^m x aⁿ =a^m+n

Question 6:
(1° + 2° + 3°) is equal to
(a) 0                       (b) 1                       (c) 3                          (d) 6
Solution:
(c) As we know, a0 = 1
10 + 2° + 3° =1 + 1 + 1 = 3.

Question 7:

Solution:

Question 8:
The standard form of the number 12345 is
(a) 1234.5 x 10¹                      (b) 123.45 x 10²
(c) 12.345 x 10³                     (d) 1.2345 x 10⁴
Solution:
(d) A number in standard form is written as a x 10^k, where a is a terminating decimal such that 1 ≤ a ≤ 10 and k is any integer.
So, 12345 = 1.2345 x 10⁴

Question 9:

Solution:

Question 10:
Which of the following is equal to 1?
(a) 2° + 3° + 4°                     (b) 2° x 3° x 4°
(c) (3° -2°) x 4°                     (d) (3° – 2°) x (3° + 2°)
Solution:
(b) Let us solve all the expressions,
For option (a),
2° + 3° + 4° = 1 + 1 + 1 [∴ a° = 1 ]
= 3
For option (b),
2° x 3° x 4° = 1 x 1 x 1 [∴ a° = 1 ]
= 1
Hence, option (b) is the answer.

Question 11:
In standard form, the number 72105.4 is written as 7.21054 x 10ⁿ, where n is equal to
(a) 2                (b) 3                       (c) 4                        (d) 5
Solution:
(c) We know that, if the given number is greater than or equal to 10, then the power of 10 (i.e. n) is a positive integer equal to the number of places the decimal point has been shifted.
Hence, 72105.4 = 7.21054 x 10⁴

Question 12:
Square of [-2/3] is

Solution:

Question 13:
The cube [-1/4] is

Solution:

Question 14:

Solution:

Question 15:

Solution:

Question 16:

Solution:

Question 17:
In standard form, the number 829030000 is written as K x 10⁸, where K is equal to
(a) 82903                    (b) 829.03                      (c) 82.903                  (d) 8.2903
Solution:
(d) We have,
A number in a standard form is written as K x 10⁸, then K will be a terminating decimal such that 1 ≤ K ≤ 10.
So, there is only one option, where K = 8.2903 < 10.

Question 18:

Solution:

Question 19:
In standard form 72 crore is written as
(a) 72 x 10⁷                    (b) 72 x 10⁸                      (c)7.2 x 10⁸                      (d)7.2 x 10⁷
Solution:
(c) We know that,
A number in standard form is written as a x 10^k, where a is the terminating decimal such that 1 ≤ a ≤ 10 and k is any integer.
So, 72 crore = 720000000 =7.2 x10⁸
Note Here, power of 10 (i.e. k) is a positive integer equal to the number of places the decimal point has been shifted.

Question 20:

Solution:

Question 21:
Which of the following is not true?
(a) 3² >2³                     (b) 4³ =2⁶                 (c)3³=9                   (d) 2⁵ > 5²
Solution:

Question 22:
Which power of 8 is equal to 2⁶?
(a) 3                 (b) 2                     (c) 1                    (d) 4
Solution:

Fill in the Blank

In questions 23 to 39, fill in the blanks to make the  statements True .
Question 23:

Solution:

Question 24:

Solution:

Question 25:

Solution:

Question 26:

Solution:

Question 27:

Solution:

Question 28:

Solution:

Question 29:

Solution:

Question 30:

Solution:

Question 31:
a⁶ x a⁵ x a° = a^—-
Solution:
Since, a⁶ x a⁵ x a°=a⁶⁺⁵⁺⁰ = a¹¹
∴ a⁶ x a⁵ x a°=a¹¹

Question 32:
1 lakh = 10^—-
Solution:
Here, 1 lakh = 100000= 10⁵.
∴ 1 lakh = 10⁵

Question 33:
1 million = 10^—-
Solution:
Here, 1 million = 1000000 = 10⁶
∴ 1 million = 10⁶

Question 34:
729 = 3^—-
Solution:
Here, firstly we find out the factors of given expression.

So, 729 = 3 x 3 x 3 x 3 x 3 x 3=3⁶
∴ 729 = 3⁶

Question 35:
432 = 2⁴ x 3^—-
Solution:
Firstly, we find out the factors of given expression.

So, 432 = 2 x 2 x 2 x 2 x 3 x 3 x 3= 2⁴ x 3³
∴ 432 =2⁴ x 3³

Question 36:
53700000 = ___  x 10⁷
Solution:
Given, 53700000
For standard form, 53700000 = 537 x 10⁵ Also, 537 = 5.37 x 10²
So, 5.37 x 10² x 10⁵ = 5.37 x 10⁷
53700000 = 5.37 x 10⁷

Question 37:
88880000000 = ___ x 10¹⁰ .
Solution:
Given, 88880000000
For standard form, 88880000000 = 8888 x 10⁷ Also, 8888 = 8.888 x 10³
So, 8.888 x 10³ x 10⁷ = 8.888 x 10¹⁰
88880000000 = 8.888 x 10¹⁰

Question 38:
27500000 = 2.75 x 10^—-
Solution:
Given, 27500000
For standard form, 27500000 = 275 x 10⁵ Also, 275 = 2.75 x 10²
So, 2.75 x 10² x 10⁵= 2.75 x 10⁷ [∴ a^m x aⁿ = a^m+n]
27500000 = 2.75 x 10⁷

Question 39:
340900000 = 3.409 x 10^—-
Solution:
For standard form, 340900000 = 3409 x 10⁵
Also, 3409= 3.409 x 10³
So, 3.409 x 10³ x 10⁵ = 3.409 x 10⁸ [∴ a^m x aⁿ = a^m+n]
340900000 = 3.409 x 10⁸

Question 40:

Solution:

True/ False

In questions 41 to 65, state whether the given statements are True or False.
Question 41:
One million = 10⁷
Solution:
False
One million = 10 lakhs = 1000000 = 10⁶
Hence, 10⁶ ≠ 10⁷

Question 42:
One hour = 60² seconds
Solution:
True
1 h = 60 min = 60 x 60 s = 60² s

Question 43:
1° x 0¹ =1
Solution:
False
∴1° x 0¹ = 1 x 0 = 0 ≠ 1.

Question 44:
(- 3)⁴ = – 12
Solution:
False
(- 3)⁴ = (- 3) x (- 3) x (- 3) x (- 3) = 81 ≠ -12

Question 45:
3⁴ > 4³
Solution:
True
∴ 3⁴ = 3 x 3 x 3 x 3=81
and 4³ = 4 x 4 x 4 = 64
81 > 64
Hence, 3⁴ > 4³

Question 46:

Solution:

Question 47:
(10 + 10)¹⁰ = 10¹⁰ + 10¹⁰
Solution:
False
We know that, (a x b)^m = a^m x b^m
So, (10 x 10)¹⁰ = 10¹⁰ x 10¹⁰

Question 48:
x° x x° =x° + x° is true for all non-zero values of x.
Solution:
True
∴ x° x x° = 1 x 1 = 1 [∴ a° = 1]
and x°+x° = 1+1 = 1 [∴ a°=1]
Hence, x°x x°=x°+ x°

Question 49:
In the standard form, a large number can be expressed as a decimal number between 0 and 1, multiplied by a power of 10.
Solution:
False
A number in standard form is written as a x 10^k, where 1 ≤ a ≤ 10 and k is any integer.

Question 50:
4² is greater than 2⁴.
Solution:
False
4²=4 x 4 = 16 [∴ a^m = a x a x a x… x a (m times)]
and 2⁴ =2 x 2 x 2 x 2 = 16
So, 4² = 2⁴

Question 51:
x^m + x^m = x^2m, where x is a non-zero rational number and m is a positive integer.
Solution:
False

a^m x aⁿ =a^m+n
x^m x x^m =x^m+m =x^2m
Also, a^k + a^k =2 a^k
So, x^m + x^m =2 x^m

Question 52:
x^m x y^m = (x x y)^2m, where x and y are non-zero rational numbers and m is a positive integer.
Solution:
False
If a and b are rational numbers, then
x^m x y^m = (ab)^m
x^m x y^m = (xy)^m = (x x y)^m
Hence, x^m x y^m ≠ (x x y)^2m

Question 53:
x^m ÷ y^m = (x ÷ y)^m, where x and y are non-zero rational numbers and m is a positive integer.
Solution:
True
If x and y are rational numbers, then (x/y)^m = (x^m/y^m)   or  x^m ÷ y^m = (x ÷ y)^m

Question 54:
a^m x aⁿ = a^m+n, where x is a non-zero rational number and m, n are positive integers.
Solution:
True
If x is a rational number and m and n are positive integers, then
a^m x aⁿ = a^m+n

Question 55:
4⁹ is greater than 16³.
Solution:
True
∴ 16³ =(4²)³ [∴ 16 = 4 x 4 = 4²]
= 4⁶
Now, in 4⁹ and 4⁶, 4⁹ > 4⁶ as powers 9 > 6

Question 56:

Solution:

Question 57:

Solution:

Question 58:

Solution:

Question 59:

Solution:

Question 60:
5° x 25° x 125° =(50)6
Solution:
True
Here,5° x 25° x 125°=5° x(5 x 5)° x (5 x 5 x 5)° [∴ 25 = 5 x 5 and 125 = 5 x 5 x 5]
= 5° x 5° x 5° x 5° x 5° x 5° [∴ a^m x b^m = a^mb^m]
= (5° )⁶
Hence, 5° x 25° x 125° =(5⁰)⁶

Question 61:
876543 = 8 x 10⁵ + 7 x 10⁴ + 6 x 10³ +5 x 10² + 4 x 10¹ + 3 x 10°
Solution:
True
Take RHS = 8 x 10⁵ + 7 x 10⁴ + 6 x 10³ +5 x 10² + 4 x 10¹ + 3 x 10°
= 8 x 100000 + 7 x 10000 + 6 x 1000 + 5 x 100 + 4 x 10 + 3 x 1 [∴ a° = 1]
= 800000 + 70000 + 6000 + 500+ 40+3 = 876543 = LHS
Hence, RHS = LHS

Question 62:
600060 = 6 x 10⁵ + 6 x 10²
Solution:
False
Take RHS =6 x 10⁵ + 6 x 10² = 6 x 100000 + 6 x 100= 600000 + 600 = 600600 ≠ LHS
Hence, RHS ≠ LHS

Question 63:
4 x 10⁵+ 3 x 10⁴ + 2 x 10³ + 1 x 10° = 432010
Solution:
False
Take LHS
= 4 x 10⁵+ 3 x 10⁴ + 2 x 10³ + 1 x 10°
= 4 x 100000+ 3 x 10000 + 2 x 1000+ 1 x 1 [∴ a° = 1]
= 400000 + 30000 + 2000 + 1 = 432001 ≠ RHS
Hence, LHS ≠ RHS

Question 64:
8 x 10⁶ + 2 x 10⁴ + 5 x 10² + 9 x 10° = 8020509
Solution:
True
Take LHS
= 8 x 10⁶ + 2 x 10⁴ + 5 x 10² + 9 x 10°
= 8 x 1000000 + 2 x 10000 + 5 x 100 + 9 x 1 [∴ a° = 1]
= 8000000 + 2000 + 500+9 = 8020509 = RHS
Hence, LHS = RHS

Question 65:
4° +5° +6° = (4 + 5 + 6)°
Solution:
False
Here, 4° + 5° + 6° = 1 + 1 + 1= 3 [∴ a°=1]
and (4+ 5+ 6)° =(15)° =1
Hence, 4° + 5° + 6° ≠ (4 + 5 + 6)°

Question 66:
Arrange in ascending order.
2⁵, 3³, 2³ x 2, (3³)², 3⁵, 4°, 2³ x 3¹
Solution:
In ascending order, the numbers are arranged from smallest to largest.
We have, 2⁵=2 x 2 x 2 x 2 x 2 = 32.
3³ =3 x 3 x 3=27.
2³ x 2=2 x 2 x 2 x 2 = 16.
(3³)² = 3^{3 x 2}  [∴ (a^m)ⁿ = a^mn]
3⁶ = 3 x 3 x 3 x 3 x 3 x 3 = 729.
3⁵ = 3 x 3 x 3 x 3 x 3 = 243.
4° = 1 [∴ a⁰ = 1]
and 2³ x 3¹=2 x 2 x 2 x 3=24 Thus, the required ascending order will be
4° < 2³ x 2 < 23 x 31 < 33 < 25 < 35 < (33)2.

Question 67:
Arrange the following exponents in descending order. 2²⁺³ , (2²)³ , (2 x 2²) , 3⁵/3² , (3² x 3⁰ ) , (2² x 5²)   .
Solution:
In descending order, the numbers are arranged from largest to smallest. We have, 22 + 3 = 25=2 x 2 x 2 x 2 x 2 = 32.
(2²)³ => 2⁶= 2 x 2 x 2 x 2 x 2 x 2=64.
2 x 2² = 2¹⁺² => 2³ = 8.
3⁵/3²  = 3^5-2 => 3³ = 27.
3² x 3⁰ = 3²⁺⁰ => 3² = 9.
2³ x 5² =2 x 2 x 2 x 5 x 5 => 8 x 25 = 200.
Thus, the required descending order will be
(2² x 5²) > (2²)³  > 2²⁺³  > 3⁵/3²  > (3² x 3⁰ )  > (2 x 2²).

Question 68:
By what number should (-4)⁵ be divided so that the quotient may be equal to (- 4)³ ?
Solution:
In order to find the number, which should divide (- 4)⁵ to get the quotient (- 4)³, we will
divide (- 4)⁵ by (- 4)³.
Hence, required number = = (- 4)^5-3 = (-4)²

Question 69:

Solution:

Question 70:

Solution:

Question 71:

Solution:

Question 72:
Find the value of

Solution:

Question 73:

Solution:

Question 74:
Express the following in usual form.
(a) 8.01 x 10⁷                       (b) 1.75 x 10^-3
Solution:

Question 75:
Find the value of
(a) 2⁵                         (b)(-3⁵)                        (c) – (- 4⁴)
Solution:

Question 76:
Express the following in exponential form.
(a)3 x 3 x 3 x a x a x a x a                     (b ) a x a x b x b x b x c x c x c x c                           (c) s x s x t x t x s x s x t
Solution:
We know that,
a x a x … x a (n times) = aⁿ
(a) 3 x 3 x 3 x a x a x a x a=3³ x a⁴ = 27 a⁴
(b) a x a x b x b x b x c x c x c x c = a² x b³ x c⁴
(c) s x s x t x t x s x s x t = s x s x s x s x t x t x t = s⁴ x t³

Question 77:
How many times of 30 must be added together to get a sum equal to 30⁷.
Solution:

Question 78:
Express each of the following numbers using exponential notations,

Using prime factorization of 1024, we have
Solution:

Question 79:
Identify the greater number, in each of the following.
(a) 2⁶ or 6²                (b) 2⁹ or 9²                   (c) 7.9 x 10⁴ or 5.28 x 10⁵
Solution:
(a) We have, 2⁶=2 x 2 x 2 x 2 x 2 x 2=64 and 6² = 6 x 6 = 36 So, 2⁶ > 6²
(b) We have, 2⁹ = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 512 and 9² = 9 x 9 = 81
So, 2⁹ > 9²
(c) We have, 7.9 x 10⁴ = 7.9 x 10000 = 79000 and 5.28 x 10⁵ = 5.28 x 100000 = 528000 So, 5.28 x 10⁵ > 7.9 x 10⁴

Question 80:
Express each of following as a product of powers of their prime factors,
(a) 9000                      (b) 2025                               (c) 800
Solution:
(a)

Question 81:
Express each of the following in single exponential form,
(a) 2³ x 3³
(b) 2⁴ x 4²
(c) 5² x 7²
(d) (- 5)⁵ x (- 5)
(e) (- 3)³ x (- 10)³
(f) (- 11)² x (- 2)²
Solution:

Question 82:
Express the following numbers in standard form.
(a) 76,47,000                            (b) 8,19,00,000
(c) 5,83,00,00,00,000           (d) 24 billion
Solution:
(a) We have, 76,47,000 = 7647000,00
A number in standard form is written as a x 10^k, where a is the terminating decimal such that 1 < a < 10 and k is any integer.
So,7647000 = 7647 x 10³
= 7.647 x 10³ x 10³ = 7.647 x 10⁶
Similarly,
(b) 8,19,00,000 = 81900000.00= 819 x 10 s = 8.19 x 10² x 10⁵ = 8.19 x 10⁷
(c) 5,83,00,00,00,000 = 583000000000.00= 583 x 10⁹ = 5.83 x 10² x 10⁹ = 5.83 x 10¹¹
(d) 24 billion = 24,00,00,00,000 = 24 x 10⁹ = 2.4 x 10¹ x 10⁹ = 2.4 x 10¹⁰

Question 83:
The speed of light in vacuum is 3 x 10⁸ m/s. Sunlight takes about 8 minutes to reach the Earth. Express distance of Sun from Earth in standard form.
Solution:
It is given that,
Speed of light = 3 x 10⁸ m/s
Time taken by light to reach the Earth = 8 min= 8 x 60 s= 480 s [∴ 1 min = 60s]
We know that,
Distance = Speed x Time=3 x 10⁸ x 480 = 1440 x 10⁸ = 1.440 x 10³ x 10⁸
= 1.44 x 10¹¹ [∴ 10³ x 10⁸ = 10¹¹]
Hence, the distance of Sun from the Earth is 1.44 x 10¹¹ m.

Question 84:
Simplify and express each of the following in exponential form.

Solution:

Question 85:
Evaluate

Solution:

Question 86:
Express the given information in Scientific notation (standard form) and then arrange them in ascending order of their size.

Solution:
1. Area of Kalahari, South Africa = 932,400 = 932400.00 [∴ standard form a x 10^k]
= 9324 x 10² = 9.324 x 10³ x 10² = 9.324 x 10⁶
2. Area of Thar, India = 199,430 = 199430.00 = 19943 x 10¹ = 1.9943 x 10⁴ x 10¹ = 1.9943 x 10⁵
3. Area of Gibson, Australia = 155,400= 155400.00= 1554 x 10² = 1.554 x 10³ x 10² = 1.554 x 10⁵
4. Area of Great Victoria, Australia = 647,500= 647500.00
= 6475 x 10² = 6.475 x 10³ x 10⁵ = 6475 x 10⁵
5. Area of Sahara, North-Africa = 8,598,800= 8598800.00
= 85988 x 10² = 8.5988 x 10⁴ x 10² = 85988 x 10⁶
Two numbers written in scientific notation can be compared. The number with the larger power of 10 is greater than the number with the smaller power of 10. If the powers of ten are the same, then the number with larger factor is the larger number.
Hence, required ascending order of the size will be
Gibson, Australia < Thar, India < Great Victoria, Australia < Kalahari, South-Africa < Sahara, North-Africa.

Question 87:
Express the given information in scientific notation and then arrange them in descending order of their size.

Solution:
A number is written in standard form as a x 10^k, where a is terminating decimal and k is an integer.

Two numbers written in scientific notation can be compared. The number with the larger power of 10 is greater than the number with the smaller power of 10. If the powers of ten are the same, then the number with larger factor is the larger number.
Hence, the required descending order of the size will be
Jupiter > Saturn > Neptune > Uranus > Earth > Venus > Mars > Mercury > Pluto

Question 88:
Write the number of seconds in scientific notation.

Solution:
1. 1 min = 60 s = 60 x 10¹ s = 6 x 101 s
2. 1 h = 3,600 s = 36 x 10² s = 3.6 x 10 x 10² s = 3.6 x 10³ s
3. 1 day = 86,400 s = 8.64 x 10² s = 8.64 x 10² x 10² s = 8.64 x 10⁴ s = 8.6 x 10⁴ s
4. 1 month = 2,600,000 s=26 x 10⁵ s =2.6 x 10 x 10⁵ s = 2.6 x 10⁶ s
5. 1 yr = 32,000,000 s = 32 x 10⁶ s = 32 x 10⁶ s = 3.2 x 10 x 10⁶ s = 32 x 10⁷ s
6. 10 yr = 320,000,000 s = 32 x 10⁷ S = 32 x 10 x 10⁷ s = 3.2 x 10⁸ s

Question 89:
In our own planet Earth, 361,419,000 square kilometre of area is covered with water and 148,647,000 square kilometre of area is covered by land. Find the approximate ratio of area covered with water to area covered by land converting these numbers into scientific notation.
Solution:

Question 90:
If 2ⁿ⁺² +2ⁿ⁺¹ + 2n = c x 2ⁿ, then find c.
Solution:
We have, 2ⁿ⁺² +2ⁿ⁺¹ + 2n = c x 2ⁿ
=> 2ⁿ 2² +2ⁿ 2¹ + 2ⁿ = c x 2ⁿ [∴ a^m+n = a^m x aⁿ]
=> 2ⁿ [2² – 2¹ + 1] = c x 2ⁿ [taking common 2ⁿ in LHS]
=> 2ⁿ[4-2 + 1]=c x 2ⁿ
=> 3  x 2ⁿ  = c x 2ⁿ
3  x 2ⁿ x 2^-n =c  x 3ⁿ x c^-n [multiplying both sides by 2^-n]
=> 3 x 2^n-1 = c x 2^n-n   [∴ a^m+n = a^m x aⁿ]
=> 3 x 2° =c x 2°
=> 3x 1 =c x 1    [∴ a°=1]
∴  3 = c

Question 91:
A light year is the distance that light can travel in one year.
1 light year = 9,460,000,000,000 km.
(a) Express one light year in scientific notation.
(b) The average distance between Earth and Sun is 1.496 x 10⁸ km. Is the distance between Earth and the Sun greater than, less than or equal to one light year?

Solution:

Question 92:
Geometry Application
The number of diagonals of an n-sided figure is  ½(n² – 3n). Use the formula to find the number of diagonals for a 6-sided figure (hexagon).

Solution:

Question 93:
Life Science
Bacteria can divide in every 20 minutes. So, 1 bacterium can multiply to 2 in 20 minutes, 4 in 40 minutes, and so on. How many bacteria will there be in 6 hours? Write your answer using exponents, then evaluate.

Most Bacteria reproduce by a type of simple cell division known as binary fission.
Each species reproduce best at a specific temperature and moisture level.
Solution:
We know that, 1 h = 60 min
6 h = 60 x 6 min = 360 min Given, a bacteria doubles itself in every 20 min.
Number of times it will double itself = 360 min/20 min = 18
∴ Bacteria will there in 6 h = 2 x 2 x 2 x … x 2 (18 times) =2¹⁸

Question 94:
Blubber makes up 27 per cent of a blue whale’s body weight. Deepak found the average weight of blue whales and used it to calculate the average weight of their blubber. He wrote the amount as 22 x 32 x 5 x 17 kg. Evaluate this amount.

Solution:
Weight calculated by Deepak = 22 x 32 x 5 x 17 kg
= 2 x 2 x 3 x 3 x 5 x 17 = 4 x 9 x 5 x 17
= 36 x 5 x 17= 180 x 17 = 3060 kg Hence, weight calculated by Deepak was 3060 kg.

Question 95:
Life Science Application
The major components of human blood are red blood cells, white blood cells, platelets and plasma. A typical red blood cell has a diameter of approx 7 x 10^-6 metre. A typical platelet has a diameter of approximately 2.33 x 10^-6 metre.
Which has a greater diameter, a red blood cell or a platelet?
Solution:
Given, diameter of red blood cell = 7 x 10^-6 m
and diameter of platelet = 2.33 x 10^-6 m
We know that, two numbers written in scientific notation can be compared. The number with the larger power of 10 is greater than the number with the smaller power of 10. If the powers of ten are the same, then the number with the larger factor is the larger number. Therefore, red blood cell has a greater diameter than a platelet.

Question 96:
A googol is the number 1 followed by 100 zeroes.
(a) How is a googol written as a power?
(b) How is a googol times a googol written as a power?
Solution:

Question 97:
What’s the Error?
A student said that 3⁵/5⁵ is the same as 1/3.  What mistake has the student  made?
Solution:

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