Maths-
General
Easy

Question

Determine if the given conjecture is true or not. Give a counterexample if it is false.
Given: ∠𝑃 𝑎𝑛𝑑 ∠𝑄 are complementary. ∠𝑄 𝑎𝑛𝑑 ∠𝑅 are complementary.
Conjecture: ∠𝑃 ≅ ∠𝑅

The correct answer is: the given conjecture is true


    We have given that

                ∠𝑃 𝑎𝑛𝑑 ∠𝑄 are complementary.

     ∠𝑄 𝑎𝑛𝑑 ∠𝑅 are complementary.
    Which means ∠𝑃 + ∠𝑄 = 90
    And  ∠𝑄 + ∠𝑅 = 90
    If we subtract ∠𝑄 from both sides of both equations we get,

    ∠P  =90 - ∠𝑄

    ∠R = 90 - ∠𝑄
    If we compare RHS of the both obtained equations
    We can say that

    ∠𝑃 ≅ ∠𝑅
    Therefore, the given conjecture is true.

    Related Questions to study

    General
    Maths-

    A plate of 3 cm thick, 9 cm broad, and 27 m long is melted into a cube. find the difference in surface area of the two solids correct to the nearest whole number.

    We are given the dimensions of the plate
    Thickness h = 3 cm
    Breadth b= 9cm
    Length l = 27cm
    This plate is melted into cube , let x be the length of side of cube
    So, as the plate is melted into cube the volume of both the solids will be same
    Therefore, l x b x h = x3

    27 x 9 x 3 = x3

    729 = x3
    As we know, 729 is the cube of 9
    Therefore, x = 9
    Surface area of the plate = 2[lb + bh + hl]

    = 2[(27)(9) + (9)(3) + (3)(27)]

    = 2[243 + 27 + 81]

    = 2(351)

    = 702

    Surface area of the cube = 6x2

    = 6(9)2

    = 6 (81)

    = 486
    Difference between both the areas= 702 – 486

    = 216
    Therefore the difference between the volumes is surface area .

    A plate of 3 cm thick, 9 cm broad, and 27 m long is melted into a cube. find the difference in surface area of the two solids correct to the nearest whole number.

    Maths-General
    We are given the dimensions of the plate
    Thickness h = 3 cm
    Breadth b= 9cm
    Length l = 27cm
    This plate is melted into cube , let x be the length of side of cube
    So, as the plate is melted into cube the volume of both the solids will be same
    Therefore, l x b x h = x3

    27 x 9 x 3 = x3

    729 = x3
    As we know, 729 is the cube of 9
    Therefore, x = 9
    Surface area of the plate = 2[lb + bh + hl]

    = 2[(27)(9) + (9)(3) + (3)(27)]

    = 2[243 + 27 + 81]

    = 2(351)

    = 702

    Surface area of the cube = 6x2

    = 6(9)2

    = 6 (81)

    = 486
    Difference between both the areas= 702 – 486

    = 216
    Therefore the difference between the volumes is surface area .

    General
    Maths-

    Which number is next in the sequence?
    35, 85, 135, 185, __

    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 the pattern of numbers

    35, 85, 135, 185, __
    We have to find the next number.
    We can see clearly that it is an Arithmetic progression With Common difference = 85 - 35 = 50
    = 135 - 85 = 50
    = 185 - 135 = 50
    Therefore next term will be
    185 + Common difference
    185 + 50 = 235
    Therefore, the pattern will be
    35, 85, 135, 185, 235

    Which number is next in the sequence?
    35, 85, 135, 185, __

    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 the pattern of numbers

    35, 85, 135, 185, __
    We have to find the next number.
    We can see clearly that it is an Arithmetic progression With Common difference = 85 - 35 = 50
    = 135 - 85 = 50
    = 185 - 135 = 50
    Therefore next term will be
    185 + Common difference
    185 + 50 = 235
    Therefore, the pattern will be
    35, 85, 135, 185, 235

    General
    General

    A cuboidal water tank is 6m long , 5m wide and 4.5 m deep.
    How many litres of water can it hold ?

    Step-by-step solution:
    The given cuboidal water tank has-
    • Length = 6m,
    • Width = 5m and
    • Height = 4.5m
    Now, we know that-
    Volume of a cuboid = Length * Width * Height
    ∴ Volume of the cuboidal tank = 6 * 5 * 4.5
    ∴ Volume of the cuboidal tank = 135 m3.
    Now, we know that-
    1m cubed = 1,000 L
    ∴ 135 m cubed = 1,000 * 135
    ∴ 135 m cubed = 1,35,000 L.

    A cuboidal water tank is 6m long , 5m wide and 4.5 m deep.
    How many litres of water can it hold ?

    GeneralGeneral
    Step-by-step solution:
    The given cuboidal water tank has-
    • Length = 6m,
    • Width = 5m and
    • Height = 4.5m
    Now, we know that-
    Volume of a cuboid = Length * Width * Height
    ∴ Volume of the cuboidal tank = 6 * 5 * 4.5
    ∴ Volume of the cuboidal tank = 135 m3.
    Now, we know that-
    1m cubed = 1,000 L
    ∴ 135 m cubed = 1,000 * 135
    ∴ 135 m cubed = 1,35,000 L.
    parallel
    General
    Maths-

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

    Solution:-
    Given, two supplementary angles always form a liner pair 


    We have to determine if the given statement is true or false.
    When the sum of measures of two angles is 180 degrees, then the angles are called supplementary angles.
    When two lines intersect each other at a single point, linear pairs of angles are formed.

    From the figure,
    ∠1 + ∠2 = 180°
    Linear pair of angles occurs in a straight line.
    So, two supplementary angles do not always form a linear pair.
    Therefore, the given statement is false.

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

    Maths-General
    Solution:-
    Given, two supplementary angles always form a liner pair 

    We have to determine if the given statement is true or false.
    When the sum of measures of two angles is 180 degrees, then the angles are called supplementary angles.
    When two lines intersect each other at a single point, linear pairs of angles are formed.

    From the figure,
    ∠1 + ∠2 = 180°
    Linear pair of angles occurs in a straight line.
    So, two supplementary angles do not always form a linear pair.
    Therefore, the given statement is false.
    General
    Maths-

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

    Solution :- 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.

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

    Maths-General
    Solution :- 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.
    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.

    parallel
    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.
    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.

    parallel
    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

    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.
    parallel
    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.
    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}
    parallel

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