Maths-
General
Easy

Question

Given a rectangular room whose width is 4/7 of its length y and perimeter z. Write down an equation connecting y and z. Calculate the length of the rectangular room when the given perimeter is 4400 cm.

Hint:

let the length of the rectangular room is y and width is 4/7 of length(i.e4y/7)
Perimeter of rectangular room z = 4400cm
Find perimeter in terms of lengths and width and solve to get the equation
Now substitute the value of z in the equation to get y.

The correct answer is: 1400 cm.


    Ans :- length of rectangle is 1400 cm .
    Explanation :-
    Step 1:- frame the equation from the given set of conditions .
    let the length of the rectangular room is y
    width is 4/7 of length  width = 4y/7
    Perimeter of rectangle = 2(length + width) = 2(y + fraction numerator 4 y over denominator 7 end fraction)
    Z = equals 2 open parentheses fraction numerator 11 y over denominator 7 end fraction close parentheses not stretchy rightwards double arrow 22 y equals 7 z
    The equation connecting y and z is 22y = 7z
    Step 2:- find the y by substituting z=4400cm(given)
    22 y equals 7 left parenthesis 4400 right parenthesis not stretchy rightwards double arrow y equals 7 left parenthesis 200 right parenthesis
    ∴ y = 1400 cm.
    ∴ length of the rectangle is 1400 cm .

    Related Questions to study

    General
    Maths-

    A drainage tile is a cylindrical shell 21 cm long. The inside and outside diameters are 4.5 cm and 5.1 cm respectively. What is the volume of the clay required for the tile?

    Hint:
    Subtracting the volume of the shell with inner radius from the volume with outer radius would give us the volume of clay required for the tile.
    Explanations:
    Step 1 of 2:
    Given, h = 21cm; r = 4.5/2 cm (inner radius) and R  = 5.1/2 cm (outer radius)
    The volume with inner radiusequals pi r squared h equals pi open parentheses fraction numerator 4.5 over denominator 2 end fraction close parentheses squared 21 cm3
    The volume with outer radius = pi R squared h equals pi open parentheses fraction numerator 5.1 over denominator 2 end fraction close parentheses squared 21  cm3
    Step 2 of 2:
    The volume of the clay required for the tile
    pi R squared h minus pi r squared h
    equals pi open parentheses R squared minus r squared close parentheses h
    equals pi left parenthesis R plus r right parenthesis left parenthesis R minus r right parenthesis h
    equals 22 over 7 cross times open parentheses fraction numerator 5.1 over denominator 2 end fraction plus fraction numerator 4.5 over denominator 2 end fraction close parentheses cross times open parentheses fraction numerator 5.1 over denominator 2 end fraction minus fraction numerator 4.5 over denominator 2 end fraction close parentheses cross times 21
    equals 22 over 7 cross times 1 half left parenthesis 5.1 plus 4.5 right parenthesis cross times 1 half left parenthesis 5.1 minus 4.5 right parenthesis cross times 21
    equals 95.04 cm cubed
    Final Answer:
    The volume of the clay required for the tile is 95.04 cm3
     

    A drainage tile is a cylindrical shell 21 cm long. The inside and outside diameters are 4.5 cm and 5.1 cm respectively. What is the volume of the clay required for the tile?

    Maths-General
    Hint:
    Subtracting the volume of the shell with inner radius from the volume with outer radius would give us the volume of clay required for the tile.
    Explanations:
    Step 1 of 2:
    Given, h = 21cm; r = 4.5/2 cm (inner radius) and R  = 5.1/2 cm (outer radius)
    The volume with inner radiusequals pi r squared h equals pi open parentheses fraction numerator 4.5 over denominator 2 end fraction close parentheses squared 21 cm3
    The volume with outer radius = pi R squared h equals pi open parentheses fraction numerator 5.1 over denominator 2 end fraction close parentheses squared 21  cm3
    Step 2 of 2:
    The volume of the clay required for the tile
    pi R squared h minus pi r squared h
    equals pi open parentheses R squared minus r squared close parentheses h
    equals pi left parenthesis R plus r right parenthesis left parenthesis R minus r right parenthesis h
    equals 22 over 7 cross times open parentheses fraction numerator 5.1 over denominator 2 end fraction plus fraction numerator 4.5 over denominator 2 end fraction close parentheses cross times open parentheses fraction numerator 5.1 over denominator 2 end fraction minus fraction numerator 4.5 over denominator 2 end fraction close parentheses cross times 21
    equals 22 over 7 cross times 1 half left parenthesis 5.1 plus 4.5 right parenthesis cross times 1 half left parenthesis 5.1 minus 4.5 right parenthesis cross times 21
    equals 95.04 cm cubed
    Final Answer:
    The volume of the clay required for the tile is 95.04 cm3
     
    General
    Maths-

    Find the area of the given figure.

    Ans :- 40.5cm2.
    Explanation :-
    Step 1:- Find the area of square
    Given side of square = 3 cm .
    Area of square = left parenthesis text  side  end text right parenthesis squared equals 3 squared equals 9 cm squared
    Step 2:- Find the area of trapezium
    We get the parallel sides of trapezium as 3 cm and 6 cm from the given diagram .
    Height (or) distance between parallel lines = 10 - 3 = 7 cm
    text  Area of trapezium  end text equals 1 half left parenthesis text  height  end text right parenthesis left parenthesis text  sum of lengths of parallel sides  end text right parenthesis
    text  Area of trapezium  end text equals 1 half left parenthesis 7 right parenthesis left parenthesis 3 plus 6 right parenthesis equals 1 half left parenthesis 7 right parenthesis left parenthesis 9 right parenthesis equals 63 divided by 2 equals 31.5 cm squared
    Step 3:- Find the Area of Figure
    Area of polygon  = area of  square + area of trapezium
    Area of polygon  = 9 + 31.5 = 40.5 cm2.
    Therefore, the area of the given figure is 40.5cm2.

    Find the area of the given figure.

    Maths-General
    Ans :- 40.5cm2.
    Explanation :-
    Step 1:- Find the area of square
    Given side of square = 3 cm .
    Area of square = left parenthesis text  side  end text right parenthesis squared equals 3 squared equals 9 cm squared
    Step 2:- Find the area of trapezium
    We get the parallel sides of trapezium as 3 cm and 6 cm from the given diagram .
    Height (or) distance between parallel lines = 10 - 3 = 7 cm
    text  Area of trapezium  end text equals 1 half left parenthesis text  height  end text right parenthesis left parenthesis text  sum of lengths of parallel sides  end text right parenthesis
    text  Area of trapezium  end text equals 1 half left parenthesis 7 right parenthesis left parenthesis 3 plus 6 right parenthesis equals 1 half left parenthesis 7 right parenthesis left parenthesis 9 right parenthesis equals 63 divided by 2 equals 31.5 cm squared
    Step 3:- Find the Area of Figure
    Area of polygon  = area of  square + area of trapezium
    Area of polygon  = 9 + 31.5 = 40.5 cm2.
    Therefore, the area of the given figure is 40.5cm2.
    General
    Maths-

    Four years back in time, a father was 3 times as old as his son then was. 8 years later father will be 2 times as old as his son will then be. Find the ages of son and father.

    Ans :- Age of Son is 16 years and Age of father is 40 years .
    Explanation :-
    Step 1:- frame the equations from the given set of conditions .
    let present ages of son is x years and present age of father is y years
    Fours years back, age of son = x-4
    Fours years back, age of father  = y-4
    Four years back in time, a father was 3 times as old as his son then was
    left parenthesis y minus 4 right parenthesis equals 3 space cross times space left parenthesis x minus 4 right parenthesis
    not stretchy rightwards double arrow y equals 3 x minus 8 —Eq1
    After 8 years, age of son = x+8
    After 8 years, age of father = y+8
    8 years later father will be 2 times as old as his son will then be
    not stretchy rightwards double arrow left parenthesis y plus 8 right parenthesis equals 2 left parenthesis x plus 8 right parenthesis not stretchy rightwards double arrow y plus 8 equals 2 x plus 16
    not stretchy rightwards double arrow y equals 2 x plus 8 — Eq2
    Step 2:- find x by eliminating y
    Doing Eq1-Eq2 to eliminate x
    left parenthesis 3 x minus 8 right parenthesis minus left parenthesis 2 x plus 8 right parenthesis equals y minus y not stretchy rightwards double arrow x minus 16 equals 0
    ∴ x= 16
    Step 3:- substitute x=16 in Eq2.
    y equals 2 left parenthesis 16 right parenthesis plus 8 not stretchy rightwards double arrow y equals 32 plus 8
    ∴ y = 40
    ∴ The present age of son =x=16 years and present age of father = 40 years

    Four years back in time, a father was 3 times as old as his son then was. 8 years later father will be 2 times as old as his son will then be. Find the ages of son and father.

    Maths-General
    Ans :- Age of Son is 16 years and Age of father is 40 years .
    Explanation :-
    Step 1:- frame the equations from the given set of conditions .
    let present ages of son is x years and present age of father is y years
    Fours years back, age of son = x-4
    Fours years back, age of father  = y-4
    Four years back in time, a father was 3 times as old as his son then was
    left parenthesis y minus 4 right parenthesis equals 3 space cross times space left parenthesis x minus 4 right parenthesis
    not stretchy rightwards double arrow y equals 3 x minus 8 —Eq1
    After 8 years, age of son = x+8
    After 8 years, age of father = y+8
    8 years later father will be 2 times as old as his son will then be
    not stretchy rightwards double arrow left parenthesis y plus 8 right parenthesis equals 2 left parenthesis x plus 8 right parenthesis not stretchy rightwards double arrow y plus 8 equals 2 x plus 16
    not stretchy rightwards double arrow y equals 2 x plus 8 — Eq2
    Step 2:- find x by eliminating y
    Doing Eq1-Eq2 to eliminate x
    left parenthesis 3 x minus 8 right parenthesis minus left parenthesis 2 x plus 8 right parenthesis equals y minus y not stretchy rightwards double arrow x minus 16 equals 0
    ∴ x= 16
    Step 3:- substitute x=16 in Eq2.
    y equals 2 left parenthesis 16 right parenthesis plus 8 not stretchy rightwards double arrow y equals 32 plus 8
    ∴ y = 40
    ∴ The present age of son =x=16 years and present age of father = 40 years
    parallel
    General
    Maths-

    A number which is formed by two digits is 3 times the sum of the digits. While interchanging the digits it is 45 more than the actual number. Find the number.

    Ans :- 27 is the number which satisfies the given condition.
    Explanation :-
    Step 1:- frame the equations from the given set of conditions .
    Given The number formed is 3 times the sum of digits
    10 x plus y equals 3 left parenthesis x plus y right parenthesis not stretchy rightwards double arrow 7 x equals 2 y not stretchy rightwards double arrow y equals 7 over 2 x— Eq1
    If digits are reversed the number is reduced by 45
    y x minus x y equals left parenthesis 10 y plus x right parenthesis minus left parenthesis 10 x plus y right parenthesis equals 45
    not stretchy rightwards double arrow 9 y minus 9 x equals 45
    not stretchy rightwards double arrow y minus x equals 5—- Eq2
    Step 2:- Substitute Eq1 in Eq2
    We get not stretchy rightwards double arrow 7 over 2 x minus x equals 5
    not stretchy rightwards double arrow 5 over 2 x equals 5
    ∴ x = 2
    Step 3:- find the value of y by substituting x=2 in Eq2
    y minus x equals 5 not stretchy rightwards double arrow y minus 2 equals 5 not stretchy rightwards double arrow y equals 7
    ∴ y = 7
    ∴ 27 is the number which satisfy the given conditions.

    A number which is formed by two digits is 3 times the sum of the digits. While interchanging the digits it is 45 more than the actual number. Find the number.

    Maths-General
    Ans :- 27 is the number which satisfies the given condition.
    Explanation :-
    Step 1:- frame the equations from the given set of conditions .
    Given The number formed is 3 times the sum of digits
    10 x plus y equals 3 left parenthesis x plus y right parenthesis not stretchy rightwards double arrow 7 x equals 2 y not stretchy rightwards double arrow y equals 7 over 2 x— Eq1
    If digits are reversed the number is reduced by 45
    y x minus x y equals left parenthesis 10 y plus x right parenthesis minus left parenthesis 10 x plus y right parenthesis equals 45
    not stretchy rightwards double arrow 9 y minus 9 x equals 45
    not stretchy rightwards double arrow y minus x equals 5—- Eq2
    Step 2:- Substitute Eq1 in Eq2
    We get not stretchy rightwards double arrow 7 over 2 x minus x equals 5
    not stretchy rightwards double arrow 5 over 2 x equals 5
    ∴ x = 2
    Step 3:- find the value of y by substituting x=2 in Eq2
    y minus x equals 5 not stretchy rightwards double arrow y minus 2 equals 5 not stretchy rightwards double arrow y equals 7
    ∴ y = 7
    ∴ 27 is the number which satisfy the given conditions.
    General
    Maths-

    The gas mileage M (s), in miles per gallon, of a car traveling s miles per hour is modeled by the function below, where  20 less or equal than s less or equal than 75.
    M left parenthesis s right parenthesis equals negative 1 over 24 s squared plus 4 s minus 50
    According to the model, at what speed, in miles per hour, does the car obtain its greatest gas mileage?

    Given,
    Gas mileage is denoted by M(s), in miles per gallon,
    Speed of the car is denoted by s miles per hour.
    The relation between gas mileage and speed is given by
    M left parenthesis s right parenthesis equals negative 1 over 24 s squared plus 4 s minus 50
    We need to find the value of  for which M(s) is maximum.
    We use the second derivative test.
    First, we find M to the power of straight prime left parenthesis s right parenthesis
    M to the power of straight prime left parenthesis s right parenthesis equals negative 1 over 24 times 2 s plus 4 plus 0
    That is
    M to the power of straight prime left parenthesis s right parenthesis equals negative s over 12 plus 4
    TO find critical points,
    M to the power of straight prime left parenthesis s right parenthesis equals 0
    Which gives
    negative s over 12 plus 4 equals 0
    Solving for , we have
    s over 12 equals 4 not stretchy rightwards double arrow s equals 48
    The only critical point, we get is s = 48 . So, we check if M (s)  has a maximum at s = 48
    Now
    M to the power of ′′ left parenthesis s right parenthesis equals negative 1 over 12 text  for all  end text s
    So M to the power of straight prime left parenthesis s right parenthesis less than 0 text  for  end text s equals 48
    Thus, M (s)  has a maximum at s = 48
    The correct option is B)

    The gas mileage M (s), in miles per gallon, of a car traveling s miles per hour is modeled by the function below, where  20 less or equal than s less or equal than 75.
    M left parenthesis s right parenthesis equals negative 1 over 24 s squared plus 4 s minus 50
    According to the model, at what speed, in miles per hour, does the car obtain its greatest gas mileage?

    Maths-General
    Given,
    Gas mileage is denoted by M(s), in miles per gallon,
    Speed of the car is denoted by s miles per hour.
    The relation between gas mileage and speed is given by
    M left parenthesis s right parenthesis equals negative 1 over 24 s squared plus 4 s minus 50
    We need to find the value of  for which M(s) is maximum.
    We use the second derivative test.
    First, we find M to the power of straight prime left parenthesis s right parenthesis
    M to the power of straight prime left parenthesis s right parenthesis equals negative 1 over 24 times 2 s plus 4 plus 0
    That is
    M to the power of straight prime left parenthesis s right parenthesis equals negative s over 12 plus 4
    TO find critical points,
    M to the power of straight prime left parenthesis s right parenthesis equals 0
    Which gives
    negative s over 12 plus 4 equals 0
    Solving for , we have
    s over 12 equals 4 not stretchy rightwards double arrow s equals 48
    The only critical point, we get is s = 48 . So, we check if M (s)  has a maximum at s = 48
    Now
    M to the power of ′′ left parenthesis s right parenthesis equals negative 1 over 12 text  for all  end text s
    So M to the power of straight prime left parenthesis s right parenthesis less than 0 text  for  end text s equals 48
    Thus, M (s)  has a maximum at s = 48
    The correct option is B)
    General
    Maths-

    Given a rectangle ABCD, where AB = (4y-z) cm, BC = (y+4) cm, CD = (2y+z+8) cm, DA = 2z cm. Find y and z

    Ans :- y = 12 and z = 8
    Explanation :-
    Given a rectangle ABCD, where AB = (4y-z) cm, BC = (y+4) cm, CD = (2y+z+8) cm, DA = 2z cm.
    Step 1:- form the system of linear equations using the given condition
    AB = CD (opposite sides of the rectangle are equal.)
    4 y minus z equals 2 y plus z plus 8 not stretchy rightwards double arrow 2 y equals 2 z plus 8
    not stretchy rightwards double arrow y equals z plus 4 — Eq1
    BC = DA(opposite sides of the rectangle are equal.)
    not stretchy rightwards double arrow y plus 4 equals 2 z — Eq2
    Step 2:- substitute the Eq1 in Eq2.
    not stretchy rightwards double arrow y plus 4 equals 2 z not stretchy rightwards double arrow left parenthesis z plus 4 right parenthesis plus 4 equals 2 z
    not stretchy rightwards double arrow z plus 8 equals 2 z
    not stretchy rightwards double arrow 8 equals z
    ∴ z = 8
    Step 3:- the value of z in Eq1.
    not stretchy rightwards double arrow y equals z plus 4 not stretchy rightwards double arrow y equals 8 plus 4
    not stretchy rightwards double arrow y equals 12
    ∴ y = 12

    Given a rectangle ABCD, where AB = (4y-z) cm, BC = (y+4) cm, CD = (2y+z+8) cm, DA = 2z cm. Find y and z

    Maths-General
    Ans :- y = 12 and z = 8
    Explanation :-
    Given a rectangle ABCD, where AB = (4y-z) cm, BC = (y+4) cm, CD = (2y+z+8) cm, DA = 2z cm.
    Step 1:- form the system of linear equations using the given condition
    AB = CD (opposite sides of the rectangle are equal.)
    4 y minus z equals 2 y plus z plus 8 not stretchy rightwards double arrow 2 y equals 2 z plus 8
    not stretchy rightwards double arrow y equals z plus 4 — Eq1
    BC = DA(opposite sides of the rectangle are equal.)
    not stretchy rightwards double arrow y plus 4 equals 2 z — Eq2
    Step 2:- substitute the Eq1 in Eq2.
    not stretchy rightwards double arrow y plus 4 equals 2 z not stretchy rightwards double arrow left parenthesis z plus 4 right parenthesis plus 4 equals 2 z
    not stretchy rightwards double arrow z plus 8 equals 2 z
    not stretchy rightwards double arrow 8 equals z
    ∴ z = 8
    Step 3:- the value of z in Eq1.
    not stretchy rightwards double arrow y equals z plus 4 not stretchy rightwards double arrow y equals 8 plus 4
    not stretchy rightwards double arrow y equals 12
    ∴ y = 12
    parallel
    General
    Maths-

    Find the volume of a right circular cylinder of length 80 cm and diameter of the base 14 cm.
     

    Hint:
    A right circular cylinder is a cylinder that has a closed circular surface. The volume of a right circular cylinder with base radius r and  height h, is pi r squared h cubic units.
    Therefore, the volume of
    Explanations:
    Step 1 of 1:
    The diameter of the base of a right circular cylinder is given by, 14cm. Then the base radius
    r = 14/2 = 7cm.
    The height is given by, h = 80 cm
    the given right circular cylinder
    pi 7 squared cross times 80
    equals 22 over 7 cross times 7 squared cross times 80
    equals 12320cm3
    Final Answer:
    The volume of the given right circular cylinder is 12320 cm3.

    Find the volume of a right circular cylinder of length 80 cm and diameter of the base 14 cm.
     

    Maths-General
    Hint:
    A right circular cylinder is a cylinder that has a closed circular surface. The volume of a right circular cylinder with base radius r and  height h, is pi r squared h cubic units.
    Therefore, the volume of
    Explanations:
    Step 1 of 1:
    The diameter of the base of a right circular cylinder is given by, 14cm. Then the base radius
    r = 14/2 = 7cm.
    The height is given by, h = 80 cm
    the given right circular cylinder
    pi 7 squared cross times 80
    equals 22 over 7 cross times 7 squared cross times 80
    equals 12320cm3
    Final Answer:
    The volume of the given right circular cylinder is 12320 cm3.
    General
    Maths-

    Find the area of the polygon in the given picture, text  if  end text MP equals 9 cm comma MD equals 7 cm comma MC equals 6cm, MB equals 4 cm text  and  end text MA equals 2 cm

    Ans :- 31.75cm2.
    Explanation :-
    Given ,MP = 9 cm, MD = 7 cm, MC = 6 cm, MB = 4 cm and MA = 2 cm.
    OD =3 cm , NA = 2.5 cm , RB = 2.5 cm and QC = 2cm
    Step 1:- Find the area of  ΔMAN
    Consider ΔMAN ,
    we have MA = 2 cm and NA = 2.5 cm with right angle at A
    Area of triangle ΔMAN  = ½ × b × h = ½  × 2 × 2.5 = 2.5 cm2
    Step 2:- Find the area of  ANOD
    Consider ANOD,
    As NA and DO are perpendicular to AD , NA is parallel to DO ; AD is perpendicular to
    both. Here ANOD is a Trapezium .parallel sides are  DO and AD
    Here AD is height ; AD = MD - MA = 7-2 = 5 cm
    Area of trapezium ANOD  equals 1 half left parenthesis text  height  end text right parenthesis left parenthesis text  sum of lengths of parallel sides  end text right parenthesis
    equals 1 half left parenthesis 5 right parenthesis left parenthesis 2.5 plus 3 right parenthesis equals 1 half left parenthesis 5 right parenthesis left parenthesis 5.5 right parenthesis equals 13.75 cm squared
    Step 3 :- Find the area of ΔDOP
    Consider ΔDOP ,
    We have OD = 3 cm and DP = MP-MD =9-7 = 2 cm with right angle at D
    Area of triangle ΔDOP   = ½ × b × h = ½ × 2 × 3 = 3 cm2
    Step 4 :- Find the area of ΔMBR
    Consider ΔMBR ,
    we have MB = 4 cm and RB  = 2.5 cm with right angle at B
    Area of triangle ΔMAN  = ½ × b × h = ½ × 4 × 2.5 = 5 cm2
    Step 5 :- Find the area of RBCQ
    Consider RBCQ,
    As RB and CQ are perpendicular to BC, RB is parallel to CQ ; BC is perpendicular to
    both. Here RBCQ is a Trapezium .parallel sides are  RB and CQ
    Here BC is height ; BC = MC - MB = 6 -4 = 2 cm
    Area of trapezium ANOD  equals space 1 half left parenthesis text  height  end text right parenthesis left parenthesis text  sum of lengths of parallel sides)  end text
    equals 1 half left parenthesis 2 right parenthesis left parenthesis 2.5 plus 2 right parenthesis equals 1 half left parenthesis 2 right parenthesis left parenthesis 4.5 right parenthesis equals 4.5 cm squared
    Step 6 :- Find the area of ΔQCP
    Consider ΔQCP ,
    We have QC = 2 cm and CP = MP-MC =9-6 = 3 cm with right angle at C
    Area of triangle ΔDOP   = ½ × b × h = ½ × 2 × 3 = 3 cm2
    Step 7:-Find the area of the polygon.
    Area of polygon  = area of  ΔMAN + area of  ANOD+ area of ΔDOP +
    area of ΔMBR + area of RBCQ +  area of ΔQCP
    Area of polygon  = 2.5 +13.75 + 3 +5 + 4.5 + 3 = 31.75 cm2
    Therefore, area of polygon is 31.75 cm2

    Find the area of the polygon in the given picture, text  if  end text MP equals 9 cm comma MD equals 7 cm comma MC equals 6cm, MB equals 4 cm text  and  end text MA equals 2 cm

    Maths-General
    Ans :- 31.75cm2.
    Explanation :-
    Given ,MP = 9 cm, MD = 7 cm, MC = 6 cm, MB = 4 cm and MA = 2 cm.
    OD =3 cm , NA = 2.5 cm , RB = 2.5 cm and QC = 2cm
    Step 1:- Find the area of  ΔMAN
    Consider ΔMAN ,
    we have MA = 2 cm and NA = 2.5 cm with right angle at A
    Area of triangle ΔMAN  = ½ × b × h = ½  × 2 × 2.5 = 2.5 cm2
    Step 2:- Find the area of  ANOD
    Consider ANOD,
    As NA and DO are perpendicular to AD , NA is parallel to DO ; AD is perpendicular to
    both. Here ANOD is a Trapezium .parallel sides are  DO and AD
    Here AD is height ; AD = MD - MA = 7-2 = 5 cm
    Area of trapezium ANOD  equals 1 half left parenthesis text  height  end text right parenthesis left parenthesis text  sum of lengths of parallel sides  end text right parenthesis
    equals 1 half left parenthesis 5 right parenthesis left parenthesis 2.5 plus 3 right parenthesis equals 1 half left parenthesis 5 right parenthesis left parenthesis 5.5 right parenthesis equals 13.75 cm squared
    Step 3 :- Find the area of ΔDOP
    Consider ΔDOP ,
    We have OD = 3 cm and DP = MP-MD =9-7 = 2 cm with right angle at D
    Area of triangle ΔDOP   = ½ × b × h = ½ × 2 × 3 = 3 cm2
    Step 4 :- Find the area of ΔMBR
    Consider ΔMBR ,
    we have MB = 4 cm and RB  = 2.5 cm with right angle at B
    Area of triangle ΔMAN  = ½ × b × h = ½ × 4 × 2.5 = 5 cm2
    Step 5 :- Find the area of RBCQ
    Consider RBCQ,
    As RB and CQ are perpendicular to BC, RB is parallel to CQ ; BC is perpendicular to
    both. Here RBCQ is a Trapezium .parallel sides are  RB and CQ
    Here BC is height ; BC = MC - MB = 6 -4 = 2 cm
    Area of trapezium ANOD  equals space 1 half left parenthesis text  height  end text right parenthesis left parenthesis text  sum of lengths of parallel sides)  end text
    equals 1 half left parenthesis 2 right parenthesis left parenthesis 2.5 plus 2 right parenthesis equals 1 half left parenthesis 2 right parenthesis left parenthesis 4.5 right parenthesis equals 4.5 cm squared
    Step 6 :- Find the area of ΔQCP
    Consider ΔQCP ,
    We have QC = 2 cm and CP = MP-MC =9-6 = 3 cm with right angle at C
    Area of triangle ΔDOP   = ½ × b × h = ½ × 2 × 3 = 3 cm2
    Step 7:-Find the area of the polygon.
    Area of polygon  = area of  ΔMAN + area of  ANOD+ area of ΔDOP +
    area of ΔMBR + area of RBCQ +  area of ΔQCP
    Area of polygon  = 2.5 +13.75 + 3 +5 + 4.5 + 3 = 31.75 cm2
    Therefore, area of polygon is 31.75 cm2
    General
    Maths-

    A man bought 3 paise and 5 paise denomination stamps for Rs 1.00. In total he bought 22 stamps. Find out total number of 3 paise stamps bought by him.

    Ans :- There are 5 stamps of 3 paise.
    Explanation :-
    let the no. of 3 paise stamps bought be x and no. of 5 paise stamps bought be y
    Step 1:-Frame the equations from the given conditions.
    Total no. of stamps = no. of 3 paise stamps +no. of 5 paise stamps = 22
    x plus y equals 22— Eq1
    Total cost of stamps = 3 cross times no. of 3 paise stamps + 5cross times no. of 5 paise stamps = 100 paise
    3 x plus 5 y equals 100 — Eq2
    Step 2:- Eliminate y to find x
    Doing 5(Eq1) -(Eq2) to eliminate y
    5 left parenthesis x plus y right parenthesis minus left parenthesis 3 x plus 5 y right parenthesis equals 5 left parenthesis 22 right parenthesis minus 100
    not stretchy rightwards double arrow 5 x minus 3 x plus 5 y minus 5 y equals 110 minus 100
    not stretchy rightwards double arrow 2 x equals 10
    ∴ x = 5
    ∴ The no. of 3 paise stamps = x = 5 stamps

    A man bought 3 paise and 5 paise denomination stamps for Rs 1.00. In total he bought 22 stamps. Find out total number of 3 paise stamps bought by him.

    Maths-General
    Ans :- There are 5 stamps of 3 paise.
    Explanation :-
    let the no. of 3 paise stamps bought be x and no. of 5 paise stamps bought be y
    Step 1:-Frame the equations from the given conditions.
    Total no. of stamps = no. of 3 paise stamps +no. of 5 paise stamps = 22
    x plus y equals 22— Eq1
    Total cost of stamps = 3 cross times no. of 3 paise stamps + 5cross times no. of 5 paise stamps = 100 paise
    3 x plus 5 y equals 100 — Eq2
    Step 2:- Eliminate y to find x
    Doing 5(Eq1) -(Eq2) to eliminate y
    5 left parenthesis x plus y right parenthesis minus left parenthesis 3 x plus 5 y right parenthesis equals 5 left parenthesis 22 right parenthesis minus 100
    not stretchy rightwards double arrow 5 x minus 3 x plus 5 y minus 5 y equals 110 minus 100
    not stretchy rightwards double arrow 2 x equals 10
    ∴ x = 5
    ∴ The no. of 3 paise stamps = x = 5 stamps
    parallel
    General
    Maths-

    A ball pen costs Rs 3.50 more than pencil. Adding 3 ball pens and 2 pencils sum up to Rs 13. Taking y and z as costs of ball pen and pencil respectively. Write down 2 simultaneous equations in terms of y and z which satisfy the above statement. Find out the values of y and z.

    Ans :- cost ball pen = 4.00 rs and cost of pencil  = 0.50 rs
    Explanation :-
    Taking y and z as costs of ball pen and pencil respectively
    L.e Cost of ball pen = y and cost of pencil = z
    Step 1:- form the system of linear equations using the given condition
    A ball pen costs Rs 3.50 more than pencil
    so, y = z + 3.50 —- Eq1
    Adding 3 ball pens and 2 pencils sum up to Rs 13.
    So, 3y + 2z = 13 —- Eq2
    We got Eq1 and Eq2 are simultaneous equations .
    Step 2:-  substitute the Eq1 in Eq2.
    3 y plus 2 z equals 13 not stretchy rightwards double arrow 3 left parenthesis z plus 3.50 right parenthesis plus 2 z equals 13
    not stretchy rightwards double arrow 3 z plus 2 z plus 10.50 equals 13
    not stretchy rightwards double arrow 5 z equals 2.50
    ∴ z = 0.50
    Step 3:-substitute the value of z in Eq1 to get the value of y
    y equals z plus 3.50 not stretchy rightwards double arrow y equals 0.50 plus 3.50 equals 4.00
    ∴ y = 4.00
    ∴ z = 0.50 rs is the cost of pencil and y = 4.00 rs is the cost of ball pen.

    A ball pen costs Rs 3.50 more than pencil. Adding 3 ball pens and 2 pencils sum up to Rs 13. Taking y and z as costs of ball pen and pencil respectively. Write down 2 simultaneous equations in terms of y and z which satisfy the above statement. Find out the values of y and z.

    Maths-General
    Ans :- cost ball pen = 4.00 rs and cost of pencil  = 0.50 rs
    Explanation :-
    Taking y and z as costs of ball pen and pencil respectively
    L.e Cost of ball pen = y and cost of pencil = z
    Step 1:- form the system of linear equations using the given condition
    A ball pen costs Rs 3.50 more than pencil
    so, y = z + 3.50 —- Eq1
    Adding 3 ball pens and 2 pencils sum up to Rs 13.
    So, 3y + 2z = 13 —- Eq2
    We got Eq1 and Eq2 are simultaneous equations .
    Step 2:-  substitute the Eq1 in Eq2.
    3 y plus 2 z equals 13 not stretchy rightwards double arrow 3 left parenthesis z plus 3.50 right parenthesis plus 2 z equals 13
    not stretchy rightwards double arrow 3 z plus 2 z plus 10.50 equals 13
    not stretchy rightwards double arrow 5 z equals 2.50
    ∴ z = 0.50
    Step 3:-substitute the value of z in Eq1 to get the value of y
    y equals z plus 3.50 not stretchy rightwards double arrow y equals 0.50 plus 3.50 equals 4.00
    ∴ y = 4.00
    ∴ z = 0.50 rs is the cost of pencil and y = 4.00 rs is the cost of ball pen.
    General
    Maths-

    In a town there are 2 brothers namely Anand and Suresh. Adding 1/3 rd anand’s age and suresh age would sum up to 10. Also adding Anand’s age together with 1⁄2 of suresh age would sum up to 10. Find Anand and Suresh ages.

    Ans :- age of anand = 6 yrs ; age of suresh  = 8 yrs
    Explanation :-
    let the age of Anand be x years and age of suresh be y years
    Step 1:- Frame  the equations using the given  conditions.
    Adding 1/3 rd anand’s age and suresh age would sum up to 10.
    So, x over 3 plus y equals 10 — Eq1
    Adding  Anand’s age together with 1⁄2 of suresh age would sum up to 10
    So, x plus y over 2 equals 10 — Eq2
    Step 2:- Eliminate x and find y
    Doing, 3(Eq1)-Eq2 to eliminate x.
    3 open parentheses bold italic x over 3 plus bold italic y close parentheses minus open parentheses x plus y over 2 close parentheses equals 3 left parenthesis 10 right parenthesis minus 10
    not stretchy rightwards double arrow x minus x plus 3 y minus y over 2 equals 30 minus 10
    not stretchy rightwards double arrow fraction numerator 5 y over denominator 2 end fraction equals 20 not stretchy rightwards double arrow y equals 20 open parentheses 2 over 5 close parentheses equals 4 space cross times space 2 equals 8
    ∴ y = 8
    Step 3:- Find x by substituting y= 8 in Eq2
    x plus y over 2 equals 10 not stretchy rightwards double arrow x plus 4 equals 10
    not stretchy rightwards double arrow x equals 10 minus 4 equals 6
    ∴ x = 6
    ∴ age of Anand is 6 years and age of suresh is 8 years .

    In a town there are 2 brothers namely Anand and Suresh. Adding 1/3 rd anand’s age and suresh age would sum up to 10. Also adding Anand’s age together with 1⁄2 of suresh age would sum up to 10. Find Anand and Suresh ages.

    Maths-General
    Ans :- age of anand = 6 yrs ; age of suresh  = 8 yrs
    Explanation :-
    let the age of Anand be x years and age of suresh be y years
    Step 1:- Frame  the equations using the given  conditions.
    Adding 1/3 rd anand’s age and suresh age would sum up to 10.
    So, x over 3 plus y equals 10 — Eq1
    Adding  Anand’s age together with 1⁄2 of suresh age would sum up to 10
    So, x plus y over 2 equals 10 — Eq2
    Step 2:- Eliminate x and find y
    Doing, 3(Eq1)-Eq2 to eliminate x.
    3 open parentheses bold italic x over 3 plus bold italic y close parentheses minus open parentheses x plus y over 2 close parentheses equals 3 left parenthesis 10 right parenthesis minus 10
    not stretchy rightwards double arrow x minus x plus 3 y minus y over 2 equals 30 minus 10
    not stretchy rightwards double arrow fraction numerator 5 y over denominator 2 end fraction equals 20 not stretchy rightwards double arrow y equals 20 open parentheses 2 over 5 close parentheses equals 4 space cross times space 2 equals 8
    ∴ y = 8
    Step 3:- Find x by substituting y= 8 in Eq2
    x plus y over 2 equals 10 not stretchy rightwards double arrow x plus 4 equals 10
    not stretchy rightwards double arrow x equals 10 minus 4 equals 6
    ∴ x = 6
    ∴ age of Anand is 6 years and age of suresh is 8 years .
    General
    Maths-

    From a cylindrical wooden log of length 30 cm and base radius 72 cm, biggest cuboid of square base is  made. Find the volume of wood wasted?

    Hint:
    Look at the diagram. Observe that AC is the diameter of the base of the wooden log. With this hint, we find a and subtract the volume of square cuboid from the volume of log to solve the problem.

    Explanations:
    Step 1 of 3:
    In straight capital delta A B C comma a squared plus a squared equals A C squared,
    AC is the diameter of the wooden log base, therefore AC = 2 x 72 = 144 cm
    therefore a squared plus a squared equals 144 squared
    not stretchy rightwards double arrow 2 a squared equals 144 squared
    not stretchy rightwards double arrow a squared equals 144 cross times 72
    not stretchy rightwards double arrow a equals 12 cross times 6 square root of 2 equals 72 square root of 2
    Step 2 of 3:
    For wooden log, base radius r  = 72cm and height h = 30cm
    So, the volume of the wooden log = pi cross times 72 squared cross times 30 cm3
    For square cuboid, side a = 72 square root of 2 cm text  and height  end text h equals 30cm
    Then, the volume of the cuboid = left parenthesis 72 square root of 2 right parenthesis squared cross times 30 cm3
    Step 3 of 3:
    The
    is 30 cross times 72 squared left parenthesis pi minus 2 right parenthesiscm3. volume of the wood wasted is given by,
    Log volume – cuboid volume  equals pi 72 squared cross times 30 minus left parenthesis 72 square root of 2 right parenthesis squared cross times 30 equals 30 cross times 72 squared left parenthesis pi minus 2 right parenthesis cm cubed
    Final Answer:
    The volume of wood wasted

    From a cylindrical wooden log of length 30 cm and base radius 72 cm, biggest cuboid of square base is  made. Find the volume of wood wasted?

    Maths-General
    Hint:
    Look at the diagram. Observe that AC is the diameter of the base of the wooden log. With this hint, we find a and subtract the volume of square cuboid from the volume of log to solve the problem.

    Explanations:
    Step 1 of 3:
    In straight capital delta A B C comma a squared plus a squared equals A C squared,
    AC is the diameter of the wooden log base, therefore AC = 2 x 72 = 144 cm
    therefore a squared plus a squared equals 144 squared
    not stretchy rightwards double arrow 2 a squared equals 144 squared
    not stretchy rightwards double arrow a squared equals 144 cross times 72
    not stretchy rightwards double arrow a equals 12 cross times 6 square root of 2 equals 72 square root of 2
    Step 2 of 3:
    For wooden log, base radius r  = 72cm and height h = 30cm
    So, the volume of the wooden log = pi cross times 72 squared cross times 30 cm3
    For square cuboid, side a = 72 square root of 2 cm text  and height  end text h equals 30cm
    Then, the volume of the cuboid = left parenthesis 72 square root of 2 right parenthesis squared cross times 30 cm3
    Step 3 of 3:
    The
    is 30 cross times 72 squared left parenthesis pi minus 2 right parenthesiscm3. volume of the wood wasted is given by,
    Log volume – cuboid volume  equals pi 72 squared cross times 30 minus left parenthesis 72 square root of 2 right parenthesis squared cross times 30 equals 30 cross times 72 squared left parenthesis pi minus 2 right parenthesis cm cubed
    Final Answer:
    The volume of wood wasted
    parallel
    General
    Maths-

    Find the area of the figure.GE equals 14 cm comma HF equals 10 cm comma AD equals 24 cm straight & BC equals 12
    .

    Ans :- 340 cm2.
    Explanation :-
    By observing the diagram EFGH is a quadrilateral where all sides are equal so
    EFGH is a rhombus .
    ABCD is quadrilateral with AD // BC .
    so, ABCD is trapezium with AD and BC as parallel sides.
    STEP 1:- Find the area of rhombus EFGH
    GE = 14 cm and HF = 10 cm ( diagonals of rhombus EFGH)
    Area of rhombus = 1 half cross times d subscript 1 cross times d subscript 2 text  where  end text d subscript 1 text  and  end text d subscript 2 text  are lengths of diagonals.  end text
    Area of rhombus =1 half cross times 14 cross times 10 equals 7 cross times 10 equals 70 cm squared
    We get area of rhombus EFGH is 70cm2.
    STEP 2:- Find the area of trapezium ABCD
    AD =  24 cm ; BC = 12 cm ;distance between parallel lines AD and BC = 15 cm
    text  Area of trapezium  end text equals 1 half left parenthesis text  height  end text right parenthesis left parenthesis text  sum of lengths of parallel sides  end text right parenthesis
    text  Area of trapezium  end text equals 1 half left parenthesis 15 right parenthesis left parenthesis 12 plus 24 right parenthesis equals 1 half left parenthesis 15 right parenthesis left parenthesis 36 right parenthesis equals 15 cross times 18
    = 270 cm2
    Area of trapezium ABCD is 270 cm2
    Step 3:- Find the area of figure
    Area of figure = area of rhombus EFGH + area of trapezium ABCD .
    Area of figure = 70+270 = 340 cm2.
    Therefore , the area of the entire figure is 340cm2.

    Find the area of the figure.GE equals 14 cm comma HF equals 10 cm comma AD equals 24 cm straight & BC equals 12
    .

    Maths-General
    Ans :- 340 cm2.
    Explanation :-
    By observing the diagram EFGH is a quadrilateral where all sides are equal so
    EFGH is a rhombus .
    ABCD is quadrilateral with AD // BC .
    so, ABCD is trapezium with AD and BC as parallel sides.
    STEP 1:- Find the area of rhombus EFGH
    GE = 14 cm and HF = 10 cm ( diagonals of rhombus EFGH)
    Area of rhombus = 1 half cross times d subscript 1 cross times d subscript 2 text  where  end text d subscript 1 text  and  end text d subscript 2 text  are lengths of diagonals.  end text
    Area of rhombus =1 half cross times 14 cross times 10 equals 7 cross times 10 equals 70 cm squared
    We get area of rhombus EFGH is 70cm2.
    STEP 2:- Find the area of trapezium ABCD
    AD =  24 cm ; BC = 12 cm ;distance between parallel lines AD and BC = 15 cm
    text  Area of trapezium  end text equals 1 half left parenthesis text  height  end text right parenthesis left parenthesis text  sum of lengths of parallel sides  end text right parenthesis
    text  Area of trapezium  end text equals 1 half left parenthesis 15 right parenthesis left parenthesis 12 plus 24 right parenthesis equals 1 half left parenthesis 15 right parenthesis left parenthesis 36 right parenthesis equals 15 cross times 18
    = 270 cm2
    Area of trapezium ABCD is 270 cm2
    Step 3:- Find the area of figure
    Area of figure = area of rhombus EFGH + area of trapezium ABCD .
    Area of figure = 70+270 = 340 cm2.
    Therefore , the area of the entire figure is 340cm2.
    General
    Maths-

    y equals 2 x plus 4
    y equals left parenthesis x minus 3 right parenthesis left parenthesis x plus 2 right parenthesis
    The system of equations above is graphed in the xy -plane. At which of the following points do the graphs of the equations intersect?

    The given equations are:
    y equals 2 x plus 4
    y equals left parenthesis x minus 3 right parenthesis left parenthesis x plus 2 right parenthesis
    We will use the method of substitution to solve these equations.
    From the first equation, we use the value of  in the second equation,
    2 x plus 4 equals left parenthesis x minus 3 right parenthesis left parenthesis x plus 2 right parenthesis
    We get an equation in one variable.
    Simplifying this, we get
    2 x plus 4 equals x left parenthesis x plus 2 right parenthesis minus 3 left parenthesis x plus 2 right parenthesis
    2 x plus 4 equals x squared plus 2 x minus 3 x minus 6
    2 x plus 4 equals x squared minus x minus 6
    Taking all the terms on one side, we have
    x squared minus x minus 2 x minus 6 minus 4 equals 0
    Thus, we get a quadratic equation,
    x squared minus 3 x minus 10 equals 0
    factorization to solve We use middle term the above equation.
    x squared minus 5 x plus 2 x minus 10 equals 0
    factorization to solve We use middle term the above equation.
    x squared minus 5 x plus 2 x minus 10 equals 0
    Now, taking  common from the first two terms and 2 from the last two terms, we have
    x left parenthesis x minus 5 right parenthesis plus 2 left parenthesis x minus 5 right parenthesis equals 0
    left parenthesis x minus 5 right parenthesis left parenthesis x plus 2 right parenthesis equals 0
    So, we get two solutions for  given by
    x minus 5 equals 0 not stretchy rightwards double arrow x equals 5 space of 1em text  and  end text space of 1em x plus 2 equals 0 not stretchy rightwards double arrow x equals negative 2
    Using these values of x in the first equation, we get two values of y
    text  For  end text x equals 5 text , we have  end text y equals 2.5 plus 4 equals 10 plus 4 equals 14
    text  For  end text x equals negative 2 text , we have  end text y equals 2 left parenthesis negative 2 right parenthesis plus 4 equals negative 4 plus 4 equals 0
    Thus, we get two solutions for the given system of equation
    left parenthesis 5 comma 14 right parenthesis text  and  end text left parenthesis negative 2 comma 0 right parenthesis
    We can see that ( 5, 14 )  is in the options.
    The correct option is D).

    y equals 2 x plus 4
    y equals left parenthesis x minus 3 right parenthesis left parenthesis x plus 2 right parenthesis
    The system of equations above is graphed in the xy -plane. At which of the following points do the graphs of the equations intersect?

    Maths-General
    The given equations are:
    y equals 2 x plus 4
    y equals left parenthesis x minus 3 right parenthesis left parenthesis x plus 2 right parenthesis
    We will use the method of substitution to solve these equations.
    From the first equation, we use the value of  in the second equation,
    2 x plus 4 equals left parenthesis x minus 3 right parenthesis left parenthesis x plus 2 right parenthesis
    We get an equation in one variable.
    Simplifying this, we get
    2 x plus 4 equals x left parenthesis x plus 2 right parenthesis minus 3 left parenthesis x plus 2 right parenthesis
    2 x plus 4 equals x squared plus 2 x minus 3 x minus 6
    2 x plus 4 equals x squared minus x minus 6
    Taking all the terms on one side, we have
    x squared minus x minus 2 x minus 6 minus 4 equals 0
    Thus, we get a quadratic equation,
    x squared minus 3 x minus 10 equals 0
    factorization to solve We use middle term the above equation.
    x squared minus 5 x plus 2 x minus 10 equals 0
    factorization to solve We use middle term the above equation.
    x squared minus 5 x plus 2 x minus 10 equals 0
    Now, taking  common from the first two terms and 2 from the last two terms, we have
    x left parenthesis x minus 5 right parenthesis plus 2 left parenthesis x minus 5 right parenthesis equals 0
    left parenthesis x minus 5 right parenthesis left parenthesis x plus 2 right parenthesis equals 0
    So, we get two solutions for  given by
    x minus 5 equals 0 not stretchy rightwards double arrow x equals 5 space of 1em text  and  end text space of 1em x plus 2 equals 0 not stretchy rightwards double arrow x equals negative 2
    Using these values of x in the first equation, we get two values of y
    text  For  end text x equals 5 text , we have  end text y equals 2.5 plus 4 equals 10 plus 4 equals 14
    text  For  end text x equals negative 2 text , we have  end text y equals 2 left parenthesis negative 2 right parenthesis plus 4 equals negative 4 plus 4 equals 0
    Thus, we get two solutions for the given system of equation
    left parenthesis 5 comma 14 right parenthesis text  and  end text left parenthesis negative 2 comma 0 right parenthesis
    We can see that ( 5, 14 )  is in the options.
    The correct option is D).
    General
    Maths-

    Find the area of an isosceles trapezium ABCD, if the parallel sides are 6cm and 14 cm, its non parallel sides is 5cm each.

    Ans :- 30 cm2.
    Explanation :-
    Step 1:- Given lengths of parallel sides is 6 and 14 cm .
    Length of non parallel sides is 5 and 5 cm
    We get a triangle and parallelogram ABED
    we get CE = 5 cm ; AE = 6 cm (opposites sides of parallelogram )
    BE = BA-AE = 14- 6 = 8 cm .


    Step 2:- Equate the areas and find value of height h
    bold 4 bold italic h equals 12 not stretchy rightwards double arrow bold italic h equals 3 bold italic c bold italic m
    Step 3:-Find the area of trapezium
    text  Areaoftrapezium  end text equals 1 half cross times left parenthesis text  height  end text right parenthesis cross times left parenthesis text  sumoflengthsofparallelsides  end text right parenthesis
    text  Area of trapezium  end text equals 1 half left parenthesis 3 right parenthesis left parenthesis 6 plus 14 right parenthesis equals 3 cross times 10 equals 30 cm squared
    Therefore, Area of trapezium ABCD  =  30 cm2.

    Find the area of an isosceles trapezium ABCD, if the parallel sides are 6cm and 14 cm, its non parallel sides is 5cm each.

    Maths-General
    Ans :- 30 cm2.
    Explanation :-
    Step 1:- Given lengths of parallel sides is 6 and 14 cm .
    Length of non parallel sides is 5 and 5 cm
    We get a triangle and parallelogram ABED
    we get CE = 5 cm ; AE = 6 cm (opposites sides of parallelogram )
    BE = BA-AE = 14- 6 = 8 cm .


    Step 2:- Equate the areas and find value of height h
    bold 4 bold italic h equals 12 not stretchy rightwards double arrow bold italic h equals 3 bold italic c bold italic m
    Step 3:-Find the area of trapezium
    text  Areaoftrapezium  end text equals 1 half cross times left parenthesis text  height  end text right parenthesis cross times left parenthesis text  sumoflengthsofparallelsides  end text right parenthesis
    text  Area of trapezium  end text equals 1 half left parenthesis 3 right parenthesis left parenthesis 6 plus 14 right parenthesis equals 3 cross times 10 equals 30 cm squared
    Therefore, Area of trapezium ABCD  =  30 cm2.
    parallel

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