Maths-
General
Easy
Question
Show the conjecture is false by finding a counterexample.
If the product of two numbers is even, then the two numbers must be even.
The correct answer is: the given conjecture is false
We have given a statement
If the product of two numbers is even, then the two numbers must be even.
We have to show the given conjecture is false.
Let us take two even numbers 6 and 8
Product of considered numbers = 6 x 8 = 48
Product we get is an even number , so the given statement is true for considered values.
If we consider one odd and one even number , 5 and 6
Product = 5 x 6 = 30
So, the product obtained is an even which contradicts the given statement
So, We can say that if the product of two numbers is even, then the two numbers need be even.
So, the given conjecture is false.
Related Questions to study
Maths-
Show the conjecture is false by finding a counterexample.
If a + b = 0, then a = b = 0.
Solution:-
We have given a statement
If a + b = 0, then a = b = 0
We have to show the given conjecture is false.
Let us take a = 0 and b = 0
Then a + b = 0 + 0 = 0
Given statement is true for considered values of a and b.
Let us take a = 3 and b = - 3
Then a + b = 3 + (- 3) = 3 – 3 = 0
From this example we can say that it is not necessary that and b should be 0 to get sum of a and b to be 0 .
Therefore, the given conjecture is false.
We have given a statement
If a + b = 0, then a = b = 0
We have to show the given conjecture is false.
Let us take a = 0 and b = 0
Then a + b = 0 + 0 = 0
Given statement is true for considered values of a and b.
Let us take a = 3 and b = - 3
Then a + b = 3 + (- 3) = 3 – 3 = 0
From this example we can say that it is not necessary that and b should be 0 to get sum of a and b to be 0 .
Therefore, the given conjecture is false.
Show the conjecture is false by finding a counterexample.
If a + b = 0, then a = b = 0.
Maths-General
Solution:-
We have given a statement
If a + b = 0, then a = b = 0
We have to show the given conjecture is false.
Let us take a = 0 and b = 0
Then a + b = 0 + 0 = 0
Given statement is true for considered values of a and b.
Let us take a = 3 and b = - 3
Then a + b = 3 + (- 3) = 3 – 3 = 0
From this example we can say that it is not necessary that and b should be 0 to get sum of a and b to be 0 .
Therefore, the given conjecture is false.
We have given a statement
If a + b = 0, then a = b = 0
We have to show the given conjecture is false.
Let us take a = 0 and b = 0
Then a + b = 0 + 0 = 0
Given statement is true for considered values of a and b.
Let us take a = 3 and b = - 3
Then a + b = 3 + (- 3) = 3 – 3 = 0
From this example we can say that it is not necessary that and b should be 0 to get sum of a and b to be 0 .
Therefore, the given conjecture is false.
Maths-
calculate the surface area of a cube and volume of a cube with the given diagonal length of 15
cm
We have given the diagonal of the cube
D= 15 cm
We have to find the Surface area and volume of the cube.
We know that Diagonal of the cube =a
Where a is side of the cube
Comparing with the given value we get
D= 15
cmWe have to find the Surface area and volume of the cube.
We know that Diagonal of the cube =
aWhere a is side of the cube
Comparing with the given value we get
a = 15 cm
Therefore , Surface area of cube = 6a2
= 6(15)2
= 6(225)
= 1350
And, volume of the given cube = a3
= (15)3
= 3375
Therefore the correct option is c)1350, 3375.
calculate the surface area of a cube and volume of a cube with the given diagonal length of 15cm
Maths-General
We have given the diagonal of the cube
D= 15 cm
We have to find the Surface area and volume of the cube.
We know that Diagonal of the cube =a
Where a is side of the cube
Comparing with the given value we get
D= 15
cmWe have to find the Surface area and volume of the cube.
We know that Diagonal of the cube =
aWhere a is side of the cube
Comparing with the given value we get
a = 15 cm
Therefore , Surface area of cube = 6a2
= 6(15)2
= 6(225)
= 1350
And, volume of the given cube = a3
= (15)3
= 3375
Therefore the correct option is c)1350, 3375.
Maths-
Show the conjecture is false by finding a counterexample.
If the product of two numbers is positive, then the two numbers must both be positive.
Solution:-
We have given a statement If the product of two numbers is positive, then the two numbers must both be positive.
If we consider two positive numbers 7 and 9
Then , their product, 7 x 9 = 63
Hence, the statement is correct for positive integers
Suppose we take two negative integers - 7 and - 9
Then, their product, - 7 x - 9 = 63
So,The product of any two positive numbers is positive, and the product of any two negative numbers is also positive.
Therefore, the given statement is false.
Show the conjecture is false by finding a counterexample.
If the product of two numbers is positive, then the two numbers must both be positive.
Maths-General
Solution:-
We have given a statement If the product of two numbers is positive, then the two numbers must both be positive.
If we consider two positive numbers 7 and 9
Then , their product, 7 x 9 = 63
Hence, the statement is correct for positive integers
Suppose we take two negative integers - 7 and - 9
Then, their product, - 7 x - 9 = 63
So,The product of any two positive numbers is positive, and the product of any two negative numbers is also positive.
Therefore, the given statement is false.
Maths-
Show the conjecture is false by finding a counterexample.
The square root of any positive integer x is always less than x.
Solution: - We have given the statement
The square root of any positive integer x is always less than
Let us take any positive integer x
We consider that the square root of any interger x is always less than x
The square root of any positive integer x is always less than
Let us take any positive integer x
We consider that the square root of any interger x is always less than x
But if we consider integer 1 then,
The square root of 1 is 1.
Which is not less than 1
So, the given statement is false
Show the conjecture is false by finding a counterexample.
The square root of any positive integer x is always less than x.
Maths-General
Solution: - We have given the statement
The square root of any positive integer x is always less than
Let us take any positive integer x
We consider that the square root of any interger x is always less than x
The square root of any positive integer x is always less than
Let us take any positive integer x
We consider that the square root of any interger x is always less than x
But if we consider integer 1 then,
The square root of 1 is 1.
Which is not less than 1
So, the given statement is false
Maths-
Show the conjecture is false by finding a counterexample.
All prime numbers are odd.
Solution :- We have given a statement that
All prime numbers are odd
Before deciding we will know the definition of prime numbers
Definition - a prime number has only 2 factors - itself and 1.
Hence the smallest natural prime number is 2, and the only on that is even. All other prime numbers are odd, and there are infinitely many prime numbers.
Therefore, it shows that All prime numbers are not odd.
All prime numbers are odd
Before deciding we will know the definition of prime numbers
Definition - a prime number has only 2 factors - itself and 1.
Hence the smallest natural prime number is 2, and the only on that is even. All other prime numbers are odd, and there are infinitely many prime numbers.
Therefore, it shows that All prime numbers are not odd.
Show the conjecture is false by finding a counterexample.
All prime numbers are odd.
Maths-General
Solution :- We have given a statement that
All prime numbers are odd
Before deciding we will know the definition of prime numbers
Definition - a prime number has only 2 factors - itself and 1.
Hence the smallest natural prime number is 2, and the only on that is even. All other prime numbers are odd, and there are infinitely many prime numbers.
Therefore, it shows that All prime numbers are not odd.
All prime numbers are odd
Before deciding we will know the definition of prime numbers
Definition - a prime number has only 2 factors - itself and 1.
Hence the smallest natural prime number is 2, and the only on that is even. All other prime numbers are odd, and there are infinitely many prime numbers.
Therefore, it shows that All prime numbers are not odd.
Maths-
Complete the conjecture.
The sum of the first n odd positive integers is ____.
Solution :- A sequence of numbers which has a common difference between any two consecutive numbers is called an arithmetic progression (A.P.). The arithmetic progression general form is given by a, a + d, a + 2d, a + 3d, . . .. Hence, the formula to find the nth term is:
an = a + (n – 1) × d
To find the sum of arithmetic progression, we have to know the first term, the number of terms and the common difference between each term. Then use the formula given below:
S = n/2[2a + (n − 1) × d]
To find: Sum of first n odd natural numbers
The first n odd natural numbers are given by 1,3,5,7,9…… (2n – 1) and this forms an AP.
S = n/2[2a + (n − 1) × d]
S = n /2 [2 + 2n – 2]
S = n /2 [2n]
S = n2
The sum of first n odd natural numbers is n2.
an = a + (n – 1) × d
To find the sum of arithmetic progression, we have to know the first term, the number of terms and the common difference between each term. Then use the formula given below:
S = n/2[2a + (n − 1) × d]
To find: Sum of first n odd natural numbers
The first n odd natural numbers are given by 1,3,5,7,9…… (2n – 1) and this forms an AP.
- a = 1
- d = 2
- tn = (2n – 1)
S = n/2[2a + (n − 1) × d]
S = n /2 [2 + 2n – 2]
S = n /2 [2n]
S = n2
The sum of first n odd natural numbers is n2.
Complete the conjecture.
The sum of the first n odd positive integers is ____.
Maths-General
Solution :- A sequence of numbers which has a common difference between any two consecutive numbers is called an arithmetic progression (A.P.). The arithmetic progression general form is given by a, a + d, a + 2d, a + 3d, . . .. Hence, the formula to find the nth term is:
an = a + (n – 1) × d
To find the sum of arithmetic progression, we have to know the first term, the number of terms and the common difference between each term. Then use the formula given below:
S = n/2[2a + (n − 1) × d]
To find: Sum of first n odd natural numbers
The first n odd natural numbers are given by 1,3,5,7,9…… (2n – 1) and this forms an AP.
S = n/2[2a + (n − 1) × d]
S = n /2 [2 + 2n – 2]
S = n /2 [2n]
S = n2
The sum of first n odd natural numbers is n2.
an = a + (n – 1) × d
To find the sum of arithmetic progression, we have to know the first term, the number of terms and the common difference between each term. Then use the formula given below:
S = n/2[2a + (n − 1) × d]
To find: Sum of first n odd natural numbers
The first n odd natural numbers are given by 1,3,5,7,9…… (2n – 1) and this forms an AP.
- a = 1
- d = 2
- tn = (2n – 1)
S = n/2[2a + (n − 1) × d]
S = n /2 [2 + 2n – 2]
S = n /2 [2n]
S = n2
The sum of first n odd natural numbers is n2.
Maths-
A square is drawn with the length of the side equal to the diagonal of the cube. if the square area is 72075 cm2, then find the side of the cube?
Hint:-
Area of square=side × side=a²
Diagonal of cube=
unit
Solution:- The length of the square and the diagonal of the cube are coincided. Therefore, the value of the two will be equal.
Side of square=Diagonal of cube
Let the edge of the cube be 'a' cm.
As we know,
length of the diagonal of cube=×edge of the cube=a cm
Therefore , the side of the cube is 155 cm.
Area of square=side × side=a²
Diagonal of cube=
unitSolution:- The length of the square and the diagonal of the cube are coincided. Therefore, the value of the two will be equal.
- We are given that,
Side of square=Diagonal of cube
Let the edge of the cube be 'a' cm.
As we know,
length of the diagonal of cube=
×edge of the cube=a cm- The length of the side of the square is equal to the length of the diagonal of the cube.
- Area of square(drawn on diagonal of cube)
- We know that, area of square drawn on the diagonal of a cube= 72,075 cm²
Therefore , the side of the cube is 155 cm.
A square is drawn with the length of the side equal to the diagonal of the cube. if the square area is 72075 cm2, then find the side of the cube?
Maths-General
Hint:-
Area of square=side × side=a²
Diagonal of cube= unit
Solution:- The length of the square and the diagonal of the cube are coincided. Therefore, the value of the two will be equal.
Side of square=Diagonal of cube
Let the edge of the cube be 'a' cm.
As we know,
length of the diagonal of cube=×edge of the cube=a cm
Therefore , the side of the cube is 155 cm.
Area of square=side × side=a²
Diagonal of cube=
unitSolution:- The length of the square and the diagonal of the cube are coincided. Therefore, the value of the two will be equal.
- We are given that,
Side of square=Diagonal of cube
Let the edge of the cube be 'a' cm.
As we know,
length of the diagonal of cube=
×edge of the cube=a cm- The length of the side of the square is equal to the length of the diagonal of the cube.
- Area of square(drawn on diagonal of cube)
- We know that, area of square drawn on the diagonal of a cube= 72,075 cm²
Therefore , the side of the cube is 155 cm.
Maths-
What are all the possible values of b for which 
is it factorable using only integer coefficients and constant?
Ans:- {-23,-10,-5,-2,2,5,10,23} are all possible values of b.
Given, is it factorable so be must be sum of factors of (-8)(3)
Factors of -24 are
(1,-24)
b can be -24 + 1 = -23
(2,-12) b can be -12 + 2 = -10
(3,-8) b can be -8 + 3 = -5
(4,-6) b can be 4 - 6 = -2
(6,-4) b can be 6 - 4 = 2
(8,-3) b can be 8 - 3 = 5
(12,-2) b can be 12 - 2 = 10
(24,-1) b can be 24 - 1 = 23
So, there are 8 possible values of b they are {-23, -10, -5, -2, 2, 5, 10, 23}
Given,
is it factorable so be must be sum of factors of (-8)(3)Factors of -24 are
(1,-24)
b can be -24 + 1 = -23(2,-12)
b can be -12 + 2 = -10(3,-8)
b can be -8 + 3 = -5(4,-6)
b can be 4 - 6 = -2(6,-4)
b can be 6 - 4 = 2(8,-3)
b can be 8 - 3 = 5(12,-2)
b can be 12 - 2 = 10(24,-1)
b can be 24 - 1 = 23So, there are 8 possible values of b they are {-23, -10, -5, -2, 2, 5, 10, 23}
What are all the possible values of b for which is it factorable using only integer coefficients and constant?
Maths-General
Ans:- {-23,-10,-5,-2,2,5,10,23} are all possible values of b.
Given, is it factorable so be must be sum of factors of (-8)(3)
Factors of -24 are
(1,-24) b can be -24 + 1 = -23
(2,-12) b can be -12 + 2 = -10
(3,-8) b can be -8 + 3 = -5
(4,-6) b can be 4 - 6 = -2
(6,-4) b can be 6 - 4 = 2
(8,-3) b can be 8 - 3 = 5
(12,-2) b can be 12 - 2 = 10
(24,-1) b can be 24 - 1 = 23
So, there are 8 possible values of b they are {-23, -10, -5, -2, 2, 5, 10, 23}
Given,
is it factorable so be must be sum of factors of (-8)(3)Factors of -24 are
(1,-24)
b can be -24 + 1 = -23(2,-12)
b can be -12 + 2 = -10(3,-8)
b can be -8 + 3 = -5(4,-6)
b can be 4 - 6 = -2(6,-4)
b can be 6 - 4 = 2(8,-3)
b can be 8 - 3 = 5(12,-2)
b can be 12 - 2 = 10(24,-1)
b can be 24 - 1 = 23So, there are 8 possible values of b they are {-23, -10, -5, -2, 2, 5, 10, 23}
Maths-
Find a counterexample to show that the following conjecture is false.
Conjecture: The square of any integer is always greater than the integer.
We have given, the square of every integer is always greater than the integer.
We have to prove that given statement false.
Number obtained when a number is multiplied by itself three times is called a cube number.
If m = n², then m is a perfect square where m and n are natural numbers.
Example: consider 1
Square of 1 = 1
Example: consider 2
Square of 2 = 4
Therefore, the square of every natural number is not always greater than the number itself.
Therefore, given statement is false.
We have to prove that given statement false.
Number obtained when a number is multiplied by itself three times is called a cube number.
If m = n², then m is a perfect square where m and n are natural numbers.
Example: consider 1
Square of 1 = 1
Example: consider 2
Square of 2 = 4
Therefore, the square of every natural number is not always greater than the number itself.
Therefore, given statement is false.
Find a counterexample to show that the following conjecture is false.
Conjecture: The square of any integer is always greater than the integer.
Maths-General
We have given, the square of every integer is always greater than the integer.
We have to prove that given statement false.
Number obtained when a number is multiplied by itself three times is called a cube number.
If m = n², then m is a perfect square where m and n are natural numbers.
Example: consider 1
Square of 1 = 1
Example: consider 2
Square of 2 = 4
Therefore, the square of every natural number is not always greater than the number itself.
Therefore, given statement is false.
We have to prove that given statement false.
Number obtained when a number is multiplied by itself three times is called a cube number.
If m = n², then m is a perfect square where m and n are natural numbers.
Example: consider 1
Square of 1 = 1
Example: consider 2
Square of 2 = 4
Therefore, the square of every natural number is not always greater than the number itself.
Therefore, given statement is false.
Maths-
Three blocks that are in the shape of a cube with each side 3.2 cm are attached end to end. Calculate the T.S.A of the resulting cuboid?
We are given the side of the cube as 3.2 cm.
When three cubes of side 3.2 cm are joined, the resulting shape is a cuboid of
length l = 3.2×3=9.6 cm;
breadth b = 3.2 cm
and height h= 3.2 cm
So, total surface area of a cuboid = 2(lb + bh + lh)
=2(9.6×3.2+3.2×3.2+9.6×3.2)
=143.36 cm2
Therefore the total surface area of the cuboid obtained is 143.36 cm2
Therefore the correct option is a)143.36
When three cubes of side 3.2 cm are joined, the resulting shape is a cuboid of
length l = 3.2×3=9.6 cm;
breadth b = 3.2 cm
and height h= 3.2 cm
So, total surface area of a cuboid = 2(lb + bh + lh)
=2(9.6×3.2+3.2×3.2+9.6×3.2)
=143.36 cm2
Therefore the total surface area of the cuboid obtained is 143.36 cm2
Therefore the correct option is a)143.36
Three blocks that are in the shape of a cube with each side 3.2 cm are attached end to end. Calculate the T.S.A of the resulting cuboid?
Maths-General
We are given the side of the cube as 3.2 cm.
When three cubes of side 3.2 cm are joined, the resulting shape is a cuboid of
length l = 3.2×3=9.6 cm;
breadth b = 3.2 cm
and height h= 3.2 cm
So, total surface area of a cuboid = 2(lb + bh + lh)
=2(9.6×3.2+3.2×3.2+9.6×3.2)
=143.36 cm2
Therefore the total surface area of the cuboid obtained is 143.36 cm2
Therefore the correct option is a)143.36
When three cubes of side 3.2 cm are joined, the resulting shape is a cuboid of
length l = 3.2×3=9.6 cm;
breadth b = 3.2 cm
and height h= 3.2 cm
So, total surface area of a cuboid = 2(lb + bh + lh)
=2(9.6×3.2+3.2×3.2+9.6×3.2)
=143.36 cm2
Therefore the total surface area of the cuboid obtained is 143.36 cm2
Therefore the correct option is a)143.36
Maths-
The area of a playground is 
. Without removing common factors, factor to possible dimensions of the playground. How are the side-lengths related? What value would you need to subtract from the longer side and add to the shorter side for the playground to be a square?
Hint :- factorize the given area using 
and find the possible values of length and breadth of the rectangle .Find the value that is needed to subtract from the longer side and add to the shorter side for the rug to be a square.
Ans :- 6x + 4y and 6x - 4y are the dimensions of a rectangular playground . 4y needs to be subtracted from the longer side(6x + 4y) and adding it to the shorter side(6x - 4y) gives the playground to be a square .
Explanation :-
Given,
Write
As
Here
We get
As the garden is rectangular , we can write area A = length × breadth
The possibility is length = 6x + 4y and breadth = 6x - 4y (or) viceversa
4y needs to be subtracted from the longer side(6x + 4y) and adding it to the shorter side(6x - 4y) gives the playground to be a square
and find the possible values of length and breadth of the rectangle .Find the value that is needed to subtract from the longer side and add to the shorter side for the rug to be a square.Ans :- 6x + 4y and 6x - 4y are the dimensions of a rectangular playground . 4y needs to be subtracted from the longer side(6x + 4y) and adding it to the shorter side(6x - 4y) gives the playground to be a square .
Explanation :-
Given,
Write
As
Here
We get
As the garden is rectangular , we can write area A = length × breadth
The possibility is length = 6x + 4y and breadth = 6x - 4y (or) viceversa
4y needs to be subtracted from the longer side(6x + 4y) and adding it to the shorter side(6x - 4y) gives the playground to be a square
The area of a playground is . Without removing common factors, factor to possible dimensions of the playground. How are the side-lengths related? What value would you need to subtract from the longer side and add to the shorter side for the playground to be a square?
Maths-General
Hint :- factorize the given area using and find the possible values of length and breadth of the rectangle .Find the value that is needed to subtract from the longer side and add to the shorter side for the rug to be a square.
Ans :- 6x + 4y and 6x - 4y are the dimensions of a rectangular playground . 4y needs to be subtracted from the longer side(6x + 4y) and adding it to the shorter side(6x - 4y) gives the playground to be a square .
Explanation :-
Given,
Write
As
Here
We get
As the garden is rectangular , we can write area A = length × breadth
The possibility is length = 6x + 4y and breadth = 6x - 4y (or) viceversa
4y needs to be subtracted from the longer side(6x + 4y) and adding it to the shorter side(6x - 4y) gives the playground to be a square
and find the possible values of length and breadth of the rectangle .Find the value that is needed to subtract from the longer side and add to the shorter side for the rug to be a square.Ans :- 6x + 4y and 6x - 4y are the dimensions of a rectangular playground . 4y needs to be subtracted from the longer side(6x + 4y) and adding it to the shorter side(6x - 4y) gives the playground to be a square .
Explanation :-
Given,
Write
As
Here
We get
As the garden is rectangular , we can write area A = length × breadth
The possibility is length = 6x + 4y and breadth = 6x - 4y (or) viceversa
4y needs to be subtracted from the longer side(6x + 4y) and adding it to the shorter side(6x - 4y) gives the playground to be a square
Maths-
What is the first three numbers in the pattern?
−, −, −, 64, 128, 256, …
Hint:- A sequence of numbers which has a common difference between any two consecutive numbers is called an arithmetic progression (A.P.). The arithmetic progression general form is given by a, a + d, a + 2d, a + 3d, . . .. Hence, the formula to find the nth term is:
an = a + (n – 1) × d
Solution:-
We have given a pattern
−, −, −, 64, 128, 256, …
We have to find the first 3 terms of the given pattern
Fourth term is 64 and fifth is 128 and sixth is 256
On analysis we get that , the next term is double of the previous term
Therefore the first three terms will be
Third term = 64 / 2 = 32
Second term = 32 / 2 = 16
First term = 16/2 = 8
Therefore, the pattern will be 8 , 16 , 32 , 64, 128 , 256 ,…
an = a + (n – 1) × d
Solution:-
We have given a pattern
−, −, −, 64, 128, 256, …
We have to find the first 3 terms of the given pattern
Fourth term is 64 and fifth is 128 and sixth is 256
On analysis we get that , the next term is double of the previous term
Therefore the first three terms will be
Third term = 64 / 2 = 32
Second term = 32 / 2 = 16
First term = 16/2 = 8
Therefore, the pattern will be 8 , 16 , 32 , 64, 128 , 256 ,…
What is the first three numbers in the pattern?
−, −, −, 64, 128, 256, …
Maths-General
Hint:- A sequence of numbers which has a common difference between any two consecutive numbers is called an arithmetic progression (A.P.). The arithmetic progression general form is given by a, a + d, a + 2d, a + 3d, . . .. Hence, the formula to find the nth term is:
an = a + (n – 1) × d
Solution:-
We have given a pattern
−, −, −, 64, 128, 256, …
We have to find the first 3 terms of the given pattern
Fourth term is 64 and fifth is 128 and sixth is 256
On analysis we get that , the next term is double of the previous term
Therefore the first three terms will be
Third term = 64 / 2 = 32
Second term = 32 / 2 = 16
First term = 16/2 = 8
Therefore, the pattern will be 8 , 16 , 32 , 64, 128 , 256 ,…
an = a + (n – 1) × d
Solution:-
We have given a pattern
−, −, −, 64, 128, 256, …
We have to find the first 3 terms of the given pattern
Fourth term is 64 and fifth is 128 and sixth is 256
On analysis we get that , the next term is double of the previous term
Therefore the first three terms will be
Third term = 64 / 2 = 32
Second term = 32 / 2 = 16
First term = 16/2 = 8
Therefore, the pattern will be 8 , 16 , 32 , 64, 128 , 256 ,…
Maths-
Calculate the maximum number of chocolates of size 2cmx3 cmx5 cm that can be kept in a rectangular box of dimensions 6cmx3cmx15 cm
We are given that
Size of chocolate = 2cm
3cm 5cm
Size of box = 6cm 3cm 15cm
We have to find out the maximum number of chocolates that can be kept in the given box,
So, we can get our required answer by dividing Volume of box by volume of one soap.
Maximum number of chocolates = Volume of box / volume of one soap
Therefore, the correct option is a) 9.
Size of chocolate = 2cm
3cm 5cmSize of box = 6cm
3cm 15cmWe have to find out the maximum number of chocolates that can be kept in the given box,
So, we can get our required answer by dividing Volume of box by volume of one soap.
Maximum number of chocolates = Volume of box / volume of one soap
Therefore, the correct option is a) 9.
Calculate the maximum number of chocolates of size 2cmx3 cmx5 cm that can be kept in a rectangular box of dimensions 6cmx3cmx15 cm
Maths-General
We are given that
Size of chocolate = 2cm 3cm 5cm
Size of box = 6cm 3cm 15cm
We have to find out the maximum number of chocolates that can be kept in the given box,
So, we can get our required answer by dividing Volume of box by volume of one soap.
Maximum number of chocolates = Volume of box / volume of one soap
Therefore, the correct option is a) 9.
Size of chocolate = 2cm
3cm 5cmSize of box = 6cm
3cm 15cmWe have to find out the maximum number of chocolates that can be kept in the given box,
So, we can get our required answer by dividing Volume of box by volume of one soap.
Maximum number of chocolates = Volume of box / volume of one soap
Therefore, the correct option is a) 9.
Maths-
Make and test a conjecture about the sign of the cube of negative integers
Solution :-
The cube of a number is the number multiplied by itself thrice. For the negative numbers, cubes are negative since the negative numbers are multiplied thrice. When two negative numbers are squared, the result is positive. However, when the negative numbers are cubed, the result is negative
(- 3)3
= (-3) × (- 3) = 9
= 9 × (- 3)
= - 27
Hence, the cube of (-3) is -27.
Therefore, the sign of cube of negative intergers is negative.
The cube of a number is the number multiplied by itself thrice. For the negative numbers, cubes are negative since the negative numbers are multiplied thrice. When two negative numbers are squared, the result is positive. However, when the negative numbers are cubed, the result is negative
(- 3)3
= (-3) × (- 3) = 9
= 9 × (- 3)
= - 27
Hence, the cube of (-3) is -27.
Therefore, the sign of cube of negative intergers is negative.
Make and test a conjecture about the sign of the cube of negative integers
Maths-General
Solution :-
The cube of a number is the number multiplied by itself thrice. For the negative numbers, cubes are negative since the negative numbers are multiplied thrice. When two negative numbers are squared, the result is positive. However, when the negative numbers are cubed, the result is negative
(- 3)3
= (-3) × (- 3) = 9
= 9 × (- 3)
= - 27
Hence, the cube of (-3) is -27.
Therefore, the sign of cube of negative intergers is negative.
The cube of a number is the number multiplied by itself thrice. For the negative numbers, cubes are negative since the negative numbers are multiplied thrice. When two negative numbers are squared, the result is positive. However, when the negative numbers are cubed, the result is negative
(- 3)3
= (-3) × (- 3) = 9
= 9 × (- 3)
= - 27
Hence, the cube of (-3) is -27.
Therefore, the sign of cube of negative intergers is negative.
Maths-
Given the area of a square, factor it to find the side length.
Hint :- Using the formula area = side2. find the side of the square by factoring the area .
Ans:- 12x-1 is the side length of the square.
Explanation:-
Given , Area A =
Write
we get A =
Applying
We get A =
But, area of square A =
∴ length of side is 12x - 1 .
Ans:- 12x-1 is the side length of the square.
Explanation:-
Given , Area A =
Write
we get A =
Applying
We get A =
But, area of square A =
∴ length of side is 12x - 1 .
Given the area of a square, factor it to find the side length.
Maths-General
Hint :- Using the formula area = side2. find the side of the square by factoring the area .
Ans:- 12x-1 is the side length of the square.
Explanation:-
Given , Area A =
Write
we get A =
Applying
We get A =
But, area of square A =
∴ length of side is 12x - 1 .
Ans:- 12x-1 is the side length of the square.
Explanation:-
Given , Area A =
Write
we get A =
Applying
We get A =
But, area of square A =
∴ length of side is 12x - 1 .
