Maths-
General
Easy

Question

Show the conjecture is false by finding a counterexample.
Two adjacent angles always form a linear pair

The correct answer is: the given conjecture is false


    We have given the statement
    Two adjacent angles always form a linear pair.
    We have to prove the statement is false.
    The linear pair of angles is a pair of angles which is created by the intersection of two straight lines and the sum of a linear pair of angles is 180°.
    Now, if we take an angle which is less than 180° and bisect that angle with the help of a straight line, then we will get two adjacent angles. And, the sum of those two adjacent angles isn't 180°.
    In this way, we can come to a conclusion that every pair of adjacent angles will be not considered as the linear pair of angles.
    So, the given conjecture is false.

    Related Questions to study

    General
    Maths-

    A cubical tank 50 cm in length and 36 cm in breadth contain water. A cube of x cm edge is dropped into it and fully immersed. If the rise in water level is 15 cm, solve for x. and hence find the T.S.A of the cubical tank

    We have given the dimensions of cubical tank
    Length l = 50cm
    Breadth b= 36cm
    The dimensions of cube dropped are
    Side = x cm
    We have to find the value of x and TSA of cubical tank.
    It is given that rise in water level is 15 cm , so it will be the height of the water in the cubical tank

    h = 15 cm
    Therefore the volume of the cube dropped will be

    Volume = l cross timescross times h

    x3 = 50 cross times 36 cross times 15

    = 27000

    As 27000 is a cube of 30

    x3 = (30)3

    therefore, x = 30
    The TSA of the cubical tank = 2[lb + bh + hl]

    = 2[(50)(36) + (36)(15) + (15)(50)l]

    = 2[1800 + 540 + 750]

    = 2[3090]

    = 6180
    Which is approximately equal to 6200.
    Therefore, the correct option is a) 30 , 6200.

    A cubical tank 50 cm in length and 36 cm in breadth contain water. A cube of x cm edge is dropped into it and fully immersed. If the rise in water level is 15 cm, solve for x. and hence find the T.S.A of the cubical tank

    Maths-General
    We have given the dimensions of cubical tank
    Length l = 50cm
    Breadth b= 36cm
    The dimensions of cube dropped are
    Side = x cm
    We have to find the value of x and TSA of cubical tank.
    It is given that rise in water level is 15 cm , so it will be the height of the water in the cubical tank

    h = 15 cm
    Therefore the volume of the cube dropped will be

    Volume = l cross timescross times h

    x3 = 50 cross times 36 cross times 15

    = 27000

    As 27000 is a cube of 30

    x3 = (30)3

    therefore, x = 30
    The TSA of the cubical tank = 2[lb + bh + hl]

    = 2[(50)(36) + (36)(15) + (15)(50)l]

    = 2[1800 + 540 + 750]

    = 2[3090]

    = 6180
    Which is approximately equal to 6200.
    Therefore, the correct option is a) 30 , 6200.

    General
    Maths-

    Show the conjecture is false by finding a counterexample.
    If the product of two numbers is even, then the two numbers must be even.

    Solution:-
    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.

    Show the conjecture is false by finding a counterexample.
    If the product of two numbers is even, then the two numbers must be even.

    Maths-General
    Solution:-
    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.
    General
    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.

    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.
    parallel
    General
    Maths-

    calculate the surface area of a cube and volume of a cube with the given diagonal length of 15square root of 3cm

    We have given the diagonal of the cube
    D= 15square root of 3 cm
    We have to find the Surface area and volume of the cube.
    We know that Diagonal of the cube = square root of 3a
    Where 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 15square root of 3cm

    Maths-General
    We have given the diagonal of the cube
    D= 15square root of 3 cm
    We have to find the Surface area and volume of the cube.
    We know that Diagonal of the cube = square root of 3a
    Where 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.

    General
    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.

    General
    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

    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

    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

    parallel
    General
    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.

    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.
    General
    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.
    • a = 1
    • d = 2
    • tn = (2n – 1)
    The formula of sum of an A.P series
    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.
    • a = 1
    • d = 2
    • tn = (2n – 1)
    The formula of sum of an A.P series
    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.
    General
    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 cubesquare root of 3 cross times side equals square root of 3 a unit
    Solution:- 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,
    Area of the square=72,075 cm²
    Side of square=Diagonal of cube
    Let the edge of the cube be 'a' cm.
    As we know,
    length of the diagonal of cube= square root of 3×edge of the cube=square root of 3a 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)
    table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell equals left parenthesis square root of 3 a right parenthesis squared cm squared end cell row cell equals 3 a squared cm squared end cell end table
    • We know that, area of square drawn on the diagonal of a cube= 72,075 cm²
    Therefore,
    table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell 3 a squared equals 72075 cm squared end cell row cell => a squared equals 72075 over 3 end cell row cell => a squared equals 24025 end cell row cell => a equals square root of 24025 end cell row cell => a equals 155 cm end cell end table

    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 cubesquare root of 3 cross times side equals square root of 3 a unit
    Solution:- 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,
    Area of the square=72,075 cm²
    Side of square=Diagonal of cube
    Let the edge of the cube be 'a' cm.
    As we know,
    length of the diagonal of cube= square root of 3×edge of the cube=square root of 3a 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)
    table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell equals left parenthesis square root of 3 a right parenthesis squared cm squared end cell row cell equals 3 a squared cm squared end cell end table
    • We know that, area of square drawn on the diagonal of a cube= 72,075 cm²
    Therefore,
    table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell 3 a squared equals 72075 cm squared end cell row cell => a squared equals 72075 over 3 end cell row cell => a squared equals 24025 end cell row cell => a equals square root of 24025 end cell row cell => a equals 155 cm end cell end table

    Therefore , the side of the cube is 155 cm.
    parallel
    General
    Maths-

    What are all the possible values of b for which 3 x squared plus b x minus 8 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, 3 x squared plus b x minus 8 is it factorable  so be must be sum of factors of (-8)(3)
    Factors of -24 are
    (1,-24) not stretchy rightwards double arrow b can be -24 + 1 = -23
    (2,-12) not stretchy rightwards double arrow b can be -12 + 2 = -10
    (3,-8) not stretchy rightwards double arrow b can be -8 + 3 = -5
    (4,-6) not stretchy rightwards double arrow b can be 4 - 6 = -2
    (6,-4) not stretchy rightwards double arrow b can be 6 - 4 = 2
    (8,-3) not stretchy rightwards double arrow b can be 8 - 3 = 5
    (12,-2) not stretchy rightwards double arrow b can be 12 - 2 = 10
    (24,-1) not stretchy rightwards double arrow b can be 24 - 1 = 23
    So, 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 3 x squared plus b x minus 8 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, 3 x squared plus b x minus 8 is it factorable  so be must be sum of factors of (-8)(3)
    Factors of -24 are
    (1,-24) not stretchy rightwards double arrow b can be -24 + 1 = -23
    (2,-12) not stretchy rightwards double arrow b can be -12 + 2 = -10
    (3,-8) not stretchy rightwards double arrow b can be -8 + 3 = -5
    (4,-6) not stretchy rightwards double arrow b can be 4 - 6 = -2
    (6,-4) not stretchy rightwards double arrow b can be 6 - 4 = 2
    (8,-3) not stretchy rightwards double arrow b can be 8 - 3 = 5
    (12,-2) not stretchy rightwards double arrow b can be 12 - 2 = 10
    (24,-1) not stretchy rightwards double arrow b can be 24 - 1 = 23
    So, there are 8 possible values of b they are {-23, -10, -5, -2, 2, 5, 10, 23}
    General
    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.

    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.
    General
    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

    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
    parallel
    General
    Maths-

    The area of a playground is 36 x squared minus 16 y squared. 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 a squared minus b squared equals left parenthesis a plus b right parenthesis left parenthesis a minus b right parenthesis 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, A equals 36 x squared minus 16 y squared
    Write 16 y squared text  as  end text left parenthesis 4 y right parenthesis squared text  and  end text 36 x squared text  as  end text left parenthesis 6 x right parenthesis squared text  then we get  end text left parenthesis 6 x right parenthesis squared minus left parenthesis 4 y right parenthesis squared
    As a squared minus b squared equals left parenthesis a plus b right parenthesis left parenthesis a minus b right parenthesis
    Here a equals 6 x semicolon b equals 4 y
    We get A equals 36 x squared minus 16 y squared equals left parenthesis 6 x plus 4 y right parenthesis left parenthesis 6 x minus 4 y right parenthesis
    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 36 x squared minus 16 y squared. 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 a squared minus b squared equals left parenthesis a plus b right parenthesis left parenthesis a minus b right parenthesis 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, A equals 36 x squared minus 16 y squared
    Write 16 y squared text  as  end text left parenthesis 4 y right parenthesis squared text  and  end text 36 x squared text  as  end text left parenthesis 6 x right parenthesis squared text  then we get  end text left parenthesis 6 x right parenthesis squared minus left parenthesis 4 y right parenthesis squared
    As a squared minus b squared equals left parenthesis a plus b right parenthesis left parenthesis a minus b right parenthesis
    Here a equals 6 x semicolon b equals 4 y
    We get A equals 36 x squared minus 16 y squared equals left parenthesis 6 x plus 4 y right parenthesis left parenthesis 6 x minus 4 y right parenthesis
    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
    General
    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 ,…

    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 ,…
    General
    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 cross times 3cm cross times 5cm
    Size of box = 6cm cross times 3cm cross times 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
                                                                  table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell equals fraction numerator 6 cross times 3 cross times 15 over denominator 2 cross times 3 cross times 5 end fraction end cell row cell equals 270 over 30 end cell row cell equals 9 end cell end table                            
    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 cross times 3cm cross times 5cm
    Size of box = 6cm cross times 3cm cross times 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
                                                                  table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell equals fraction numerator 6 cross times 3 cross times 15 over denominator 2 cross times 3 cross times 5 end fraction end cell row cell equals 270 over 30 end cell row cell equals 9 end cell end table                            
    Therefore, the correct option is a) 9.
    parallel

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