Maths-
General
Easy

Question

Ram's mother is four times as old as Ram now. After sixteen years, she will be twice as old as Ram. Find their present ages.

Hint:

let the rams present age be x ; find the ram’s mothers age now
Calculate ram’s and ram’s mothers age after 16 yr by adding 16 to their present ages . After sixteen years, she will be twice as old as Ram.
Apply the condition and calculate the present ages .

The correct answer is: 32 years


    Ans :- The present age of ram is 8  and the present age of ram’s mother 32 (in years) .
    Explanation :-
    Step 1:- calculate present age ram’s mother
    let the rams present age be x
    Ram's mother is four times as old as Ram now
    Now , Ram's mother = 4x
    Step 2:-After 16 year calculate the ram’s and ram’s mother age
    After 16 yr ram’s age  = x+16
    After 16 yr ram’s mother age  = 4x+16
    Step 3:-Apply the given condition and find their respective ages
    But After sixteen years, she will be twice as old as Ram
    So, 4x+16 = 2(x+16)
    4x +16 = 2x + 32
    4x-2x = 32 -16
    2x = 16
    x =8
    ∴The present age of ram = x = 8 yrs  and the present age of ram’s mother = 4x = 32 yrs.

    Related Questions to study

    General
    Maths-

    Tanay obtained 98 marks in a mathematics test. His score is the highest in the class and it is also 8 more than three times the lowest score. Create the equation and calculate the lowest score.

    Ans :- 30 is the lowest score of the class
    Explanation :-
    let the lowest score be x,
    Given tanay's score is 8 more than three times the lowest score.
    Tanay’s score = 8 + 3(x) .
    Where tanay’s score = 98 (given)
    So, 3x +8 = 98 3x = 98 -8
    rightwards double arrow3x = 90
    therefore space x space equals space 30
    therefore 30 is the lowest score in the class.

    Tanay obtained 98 marks in a mathematics test. His score is the highest in the class and it is also 8 more than three times the lowest score. Create the equation and calculate the lowest score.

    Maths-General
    Ans :- 30 is the lowest score of the class
    Explanation :-
    let the lowest score be x,
    Given tanay's score is 8 more than three times the lowest score.
    Tanay’s score = 8 + 3(x) .
    Where tanay’s score = 98 (given)
    So, 3x +8 = 98 3x = 98 -8
    rightwards double arrow3x = 90
    therefore space x space equals space 30
    therefore 30 is the lowest score in the class.
    General
    Maths-

    Find three consecutive odd numbers whose sum is 159.

    Ans :- The three consecutive odd numbers which satisfy the given condition are 51,53 and 55
    Explanation :-
    let the three consecutive odd numbers be 2n+1;2n+3 and 2n+5 .
    Sum of three consecutive odd numbers  = 159.
    (2n+1) + (2n+3) +(2n+5) = 159
    rightwards double arrow2n + 2n + 2n +1+3+5 =159
    rightwards double arrow6n + 9 =159
    rightwards double arrow6n = 150
    rightwards double arrow3(2n) = 150
    rightwards double arrow2n = 50
    We get 2n = 50 So, 2n+1 = 50+ 1 =51; 2n +3 = 50+3 =53; 2n + 5 = 55
    ∴ The three consecutive odd numbers which satisfy the given condition is 51,53,55.

    Find three consecutive odd numbers whose sum is 159.

    Maths-General
    Ans :- The three consecutive odd numbers which satisfy the given condition are 51,53 and 55
    Explanation :-
    let the three consecutive odd numbers be 2n+1;2n+3 and 2n+5 .
    Sum of three consecutive odd numbers  = 159.
    (2n+1) + (2n+3) +(2n+5) = 159
    rightwards double arrow2n + 2n + 2n +1+3+5 =159
    rightwards double arrow6n + 9 =159
    rightwards double arrow6n = 150
    rightwards double arrow3(2n) = 150
    rightwards double arrow2n = 50
    We get 2n = 50 So, 2n+1 = 50+ 1 =51; 2n +3 = 50+3 =53; 2n + 5 = 55
    ∴ The three consecutive odd numbers which satisfy the given condition is 51,53,55.
    General
    Maths-

    Solve 3x - 2y = 6 and x over 3 minus y over 6 equals 1 half

    Ans :- x = 0 ; y = -3
    Explanation :-
    x over 3 minus y over 6 equals 1 half not stretchy rightwards double arrow x over 3 equals y over 6 plus 1 half not stretchy rightwards double arrow x equals y over 2 plus 3 over 2 minus text  eq  end text 1
    3 x minus 2 y equals 6 minus negative text  eq  end text 2
    Step 1 :- find x by substituting x equals y over 2 plus 3 over 2 in eq 2.
    3 open parentheses y over 2 plus 3 over 2 close parentheses minus 2 y equals 6 not stretchy rightwards double arrow fraction numerator 3 y over denominator 2 end fraction plus 9 over 2 minus 2 y equals 6
    not stretchy rightwards double arrow fraction numerator negative y over denominator 2 end fraction equals 6 minus 9 over 2 not stretchy rightwards double arrow fraction numerator negative y over denominator 2 end fraction equals 3 over 2 not stretchy rightwards double arrow negative y equals 3
     rightwards double arrow y space equals space minus 3
    Step 2 :- substitute value of y and find x
    x equals y over 2 plus 3 over 2 not stretchy rightwards double arrow x equals fraction numerator negative 3 over denominator 2 end fraction plus 3 over 2
    not stretchy rightwards double arrow x equals 0
     therefore space x space equals space 0
    therefore spacex = 0 and y = -3 is the solution of the given pair of equations.

    Solve 3x - 2y = 6 and x over 3 minus y over 6 equals 1 half

    Maths-General
    Ans :- x = 0 ; y = -3
    Explanation :-
    x over 3 minus y over 6 equals 1 half not stretchy rightwards double arrow x over 3 equals y over 6 plus 1 half not stretchy rightwards double arrow x equals y over 2 plus 3 over 2 minus text  eq  end text 1
    3 x minus 2 y equals 6 minus negative text  eq  end text 2
    Step 1 :- find x by substituting x equals y over 2 plus 3 over 2 in eq 2.
    3 open parentheses y over 2 plus 3 over 2 close parentheses minus 2 y equals 6 not stretchy rightwards double arrow fraction numerator 3 y over denominator 2 end fraction plus 9 over 2 minus 2 y equals 6
    not stretchy rightwards double arrow fraction numerator negative y over denominator 2 end fraction equals 6 minus 9 over 2 not stretchy rightwards double arrow fraction numerator negative y over denominator 2 end fraction equals 3 over 2 not stretchy rightwards double arrow negative y equals 3
     rightwards double arrow y space equals space minus 3
    Step 2 :- substitute value of y and find x
    x equals y over 2 plus 3 over 2 not stretchy rightwards double arrow x equals fraction numerator negative 3 over denominator 2 end fraction plus 3 over 2
    not stretchy rightwards double arrow x equals 0
     therefore space x space equals space 0
    therefore spacex = 0 and y = -3 is the solution of the given pair of equations.
    parallel
    General
    Maths-

    Solve the following by substitution method 2x - 3y = 7 and x + 6y = 11.

    Ans :- x = 5 ; y = 1
    Explanation :-
    x plus 6 y equals 11 not stretchy rightwards double arrow x equals 11 minus 6 y minus text  eq  end text 1
    2 x minus 3 y equals 7 minus text  eq  end text 2
    Step 1 :- find x by substituting x = 11-6y in eq 2.
    2 left parenthesis 11 minus 6 y right parenthesis minus 3 y equals 7 not stretchy rightwards double arrow 22 minus 12 y minus 3 y equals 7
    not stretchy rightwards double arrow 22 minus 15 y equals 7 not stretchy rightwards double arrow 15 y equals 22 minus 7 not stretchy rightwards double arrow 15 y equals 15
     rightwards double arrow y space equals space 1
    Step 2 :- substitute value of y and find x
    x equals 11 minus 6 y not stretchy rightwards double arrow x equals 11 minus 6 left parenthesis 1 right parenthesis
    not stretchy rightwards double arrow x equals 11 minus 6 not stretchy rightwards double arrow x equals 5
     therefore space x space equals space 5
    therefore x = 5 and y = 1 is the solution of the given pair of equations.

    Solve the following by substitution method 2x - 3y = 7 and x + 6y = 11.

    Maths-General
    Ans :- x = 5 ; y = 1
    Explanation :-
    x plus 6 y equals 11 not stretchy rightwards double arrow x equals 11 minus 6 y minus text  eq  end text 1
    2 x minus 3 y equals 7 minus text  eq  end text 2
    Step 1 :- find x by substituting x = 11-6y in eq 2.
    2 left parenthesis 11 minus 6 y right parenthesis minus 3 y equals 7 not stretchy rightwards double arrow 22 minus 12 y minus 3 y equals 7
    not stretchy rightwards double arrow 22 minus 15 y equals 7 not stretchy rightwards double arrow 15 y equals 22 minus 7 not stretchy rightwards double arrow 15 y equals 15
     rightwards double arrow y space equals space 1
    Step 2 :- substitute value of y and find x
    x equals 11 minus 6 y not stretchy rightwards double arrow x equals 11 minus 6 left parenthesis 1 right parenthesis
    not stretchy rightwards double arrow x equals 11 minus 6 not stretchy rightwards double arrow x equals 5
     therefore space x space equals space 5
    therefore x = 5 and y = 1 is the solution of the given pair of equations.
    General
    Maths-

    Solve the following by using elimination method: 2x + y = 6 , 3y = 8 + 4x

    Ans :- x = 1 ; y = 4
    Explanation :-
    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 2 x plus y equals 6 minus text  eq  end text 1 end cell row blank end table
    3 y equals 8 plus 4 x not stretchy rightwards double arrow 4 x plus 8 minus 3 y equals 0 text  - eq  end text 2
    Step 1 :- find y by eliminating x
    Eliminating x by doing eq 2- 2(eq1) then
    4 x plus 8 minus 3 y minus 2 left parenthesis 2 x plus y right parenthesis equals 0 minus 2 left parenthesis 6 right parenthesis
    not stretchy rightwards double arrow 4 x minus 4 x minus 3 y minus 2 y plus 8 equals negative 12 not stretchy rightwards double arrow negative 5 y equals negative 20
     rightwards double arrow y space equals space 4
    Step 2 :- substitute value of y and find x
    not stretchy rightwards double arrow 2 x plus y equals 6 not stretchy rightwards double arrow 2 x plus left parenthesis 4 right parenthesis equals 6
    not stretchy rightwards double arrow 2 x equals 6 minus 4 not stretchy rightwards double arrow 2 x equals 2
     therefore space x space equals space 1
    therefore x = 1 and y = 4 is the solution of the given pair of equations.

    Solve the following by using elimination method: 2x + y = 6 , 3y = 8 + 4x

    Maths-General
    Ans :- x = 1 ; y = 4
    Explanation :-
    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 2 x plus y equals 6 minus text  eq  end text 1 end cell row blank end table
    3 y equals 8 plus 4 x not stretchy rightwards double arrow 4 x plus 8 minus 3 y equals 0 text  - eq  end text 2
    Step 1 :- find y by eliminating x
    Eliminating x by doing eq 2- 2(eq1) then
    4 x plus 8 minus 3 y minus 2 left parenthesis 2 x plus y right parenthesis equals 0 minus 2 left parenthesis 6 right parenthesis
    not stretchy rightwards double arrow 4 x minus 4 x minus 3 y minus 2 y plus 8 equals negative 12 not stretchy rightwards double arrow negative 5 y equals negative 20
     rightwards double arrow y space equals space 4
    Step 2 :- substitute value of y and find x
    not stretchy rightwards double arrow 2 x plus y equals 6 not stretchy rightwards double arrow 2 x plus left parenthesis 4 right parenthesis equals 6
    not stretchy rightwards double arrow 2 x equals 6 minus 4 not stretchy rightwards double arrow 2 x equals 2
     therefore space x space equals space 1
    therefore x = 1 and y = 4 is the solution of the given pair of equations.
    General
    Maths-

    Solve 2a – 3/b = 12 and 5a – 7/b = 1

    Ans :- a = -81 ; b = -1/58
    Explanation :-
    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 2 a minus 3 divided by b equals 12 minus eq 1 end cell row cell 5 a minus 7 divided by b equals 1 minus eq 2 end cell end table
    Step 1:- find the value of 1/b by eliminating a
    By 5cross timeseq1 - 2cross timeseq2 we get
    5 left parenthesis 2 a minus 3 divided by b right parenthesis minus 2 left parenthesis 5 a minus 7 divided by b right parenthesis equals 5 left parenthesis 12 right parenthesis minus 2
    not stretchy rightwards double arrow 10 a minus 10 a minus 15 divided by b plus 14 divided by b equals 60 minus 2
    not stretchy rightwards double arrow negative 1 divided by b equals 58
     therefore space b space equals space minus 1 divided by 58
    Step 2:- find the value of a by substitution -1/b =  58 in eq 1
    2 a plus 3 left parenthesis negative 1 divided by b right parenthesis equals 12 not stretchy rightwards double arrow 2 a plus 3 left parenthesis 58 right parenthesis equals 12
    not stretchy rightwards double arrow 2 a equals negative 162

     therefore space a space equals space minus 81
    therefore a = -81 and b = -1/58 are the solution of the given equation.

    Solve 2a – 3/b = 12 and 5a – 7/b = 1

    Maths-General
    Ans :- a = -81 ; b = -1/58
    Explanation :-
    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 2 a minus 3 divided by b equals 12 minus eq 1 end cell row cell 5 a minus 7 divided by b equals 1 minus eq 2 end cell end table
    Step 1:- find the value of 1/b by eliminating a
    By 5cross timeseq1 - 2cross timeseq2 we get
    5 left parenthesis 2 a minus 3 divided by b right parenthesis minus 2 left parenthesis 5 a minus 7 divided by b right parenthesis equals 5 left parenthesis 12 right parenthesis minus 2
    not stretchy rightwards double arrow 10 a minus 10 a minus 15 divided by b plus 14 divided by b equals 60 minus 2
    not stretchy rightwards double arrow negative 1 divided by b equals 58
     therefore space b space equals space minus 1 divided by 58
    Step 2:- find the value of a by substitution -1/b =  58 in eq 1
    2 a plus 3 left parenthesis negative 1 divided by b right parenthesis equals 12 not stretchy rightwards double arrow 2 a plus 3 left parenthesis 58 right parenthesis equals 12
    not stretchy rightwards double arrow 2 a equals negative 162

     therefore space a space equals space minus 81
    therefore a = -81 and b = -1/58 are the solution of the given equation.
    parallel
    General
    Maths-

    Solve the following equations by using the elimination method: x - y = 1 , 3x - y = 9

    Ans :- x = 4 ; y = 3
    Explanation :-
    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 x minus y equals 1 minus text  eq  end text 1 end cell row cell 3 x minus y equals 9 minus text  eq  end text 2 end cell end table
    Step 1 :- find x by eliminating y
    Eliminating x by subtracting  eq 2 - eq1 then
    left parenthesis 3 x minus y right parenthesis minus left parenthesis x minus y right parenthesis equals 9 minus 1
    not stretchy rightwards double arrow 3 x minus x minus y plus y equals 8 not stretchy rightwards double arrow 2 x equals 8
     rightwards double arrow x space equals space 4
    Step 2 :- substitute value of x and find y
    not stretchy rightwards double arrow x minus y equals 1 not stretchy rightwards double arrow 4 minus y equals 1
    not stretchy rightwards double arrow y equals 4 minus 1 not stretchy rightwards double arrow y equals 3
     therefore space y space equals space 3
    therefore x = 4 and y = 3 is the solution of the given pair of equations.

    Solve the following equations by using the elimination method: x - y = 1 , 3x - y = 9

    Maths-General
    Ans :- x = 4 ; y = 3
    Explanation :-
    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 x minus y equals 1 minus text  eq  end text 1 end cell row cell 3 x minus y equals 9 minus text  eq  end text 2 end cell end table
    Step 1 :- find x by eliminating y
    Eliminating x by subtracting  eq 2 - eq1 then
    left parenthesis 3 x minus y right parenthesis minus left parenthesis x minus y right parenthesis equals 9 minus 1
    not stretchy rightwards double arrow 3 x minus x minus y plus y equals 8 not stretchy rightwards double arrow 2 x equals 8
     rightwards double arrow x space equals space 4
    Step 2 :- substitute value of x and find y
    not stretchy rightwards double arrow x minus y equals 1 not stretchy rightwards double arrow 4 minus y equals 1
    not stretchy rightwards double arrow y equals 4 minus 1 not stretchy rightwards double arrow y equals 3
     therefore space y space equals space 3
    therefore x = 4 and y = 3 is the solution of the given pair of equations.
    General
    Maths-

    Solve the following system of linear equations: 2x - 4y = 6 , x - 3y = 12

    Ans :- x = -15 ; y = -9
    Explanation :-
    2x - 4y = 6 — eq 1
    x - 3y = 12 — eq 2
    Step 1 :- find y by eliminating x
    Eliminating x by subtracting e2q 2- eq1 then
    2 left parenthesis x minus 3 y right parenthesis minus left parenthesis 2 x minus 4 y right parenthesis equals 2 left parenthesis 12 right parenthesis minus 6
    not stretchy rightwards double arrow 2 x minus 2 x minus 6 y plus 4 y equals 24 minus 6 not stretchy rightwards double arrow negative 2 y equals 18
     rightwards double arrow y space equals space minus 9
    Step 2 :- substitute value of y and find x
    not stretchy rightwards double arrow 2 x minus 4 y equals 6 not stretchy rightwards double arrow 2 x minus 4 left parenthesis negative 9 right parenthesis equals 6
    not stretchy rightwards double arrow 2 x equals 6 minus 36 not stretchy rightwards double arrow x equals negative 30 divided by 2
     therefore space x space equals space minus 15
    therefore x = -15 and y = -9 is the solution of the given pair of equations.

    Solve the following system of linear equations: 2x - 4y = 6 , x - 3y = 12

    Maths-General
    Ans :- x = -15 ; y = -9
    Explanation :-
    2x - 4y = 6 — eq 1
    x - 3y = 12 — eq 2
    Step 1 :- find y by eliminating x
    Eliminating x by subtracting e2q 2- eq1 then
    2 left parenthesis x minus 3 y right parenthesis minus left parenthesis 2 x minus 4 y right parenthesis equals 2 left parenthesis 12 right parenthesis minus 6
    not stretchy rightwards double arrow 2 x minus 2 x minus 6 y plus 4 y equals 24 minus 6 not stretchy rightwards double arrow negative 2 y equals 18
     rightwards double arrow y space equals space minus 9
    Step 2 :- substitute value of y and find x
    not stretchy rightwards double arrow 2 x minus 4 y equals 6 not stretchy rightwards double arrow 2 x minus 4 left parenthesis negative 9 right parenthesis equals 6
    not stretchy rightwards double arrow 2 x equals 6 minus 36 not stretchy rightwards double arrow x equals negative 30 divided by 2
     therefore space x space equals space minus 15
    therefore x = -15 and y = -9 is the solution of the given pair of equations.
    General
    Maths-

    Solve: x - 2y = 8 , 4x + 2y = 7 by using elimination method.

    Ans :- x = 3 ; y = -5/2
    Explanation :-
    x - 2y = 8 — eq 1
    4x = 2y = 7 — eq 2
    Step 1 :- find x by eliminating y
    Eliminating x by adding eq 2 and eq1 then
    left parenthesis x minus 2 y right parenthesis plus left parenthesis 4 x plus 2 y right parenthesis equals 8 plus 7
    not stretchy rightwards double arrow x plus 4 x plus 2 y minus 2 y equals 15 not stretchy rightwards double arrow 5 x equals 15
     rightwards double arrowx space equals space 3
    Step 2 :- substitute value of x and find y
    not stretchy rightwards double arrow x minus 2 y equals 8 not stretchy rightwards double arrow left parenthesis 3 right parenthesis minus 2 y equals 8
    not stretchy rightwards double arrow negative 2 y equals 8 minus 3 not stretchy rightwards double arrow negative 2 y equals 5
     therefore space y space equals space minus 5 divided by 2
    therefore space x space equals space 3 and y = -5/2 is the solution of the given pair of equations.

    Solve: x - 2y = 8 , 4x + 2y = 7 by using elimination method.

    Maths-General
    Ans :- x = 3 ; y = -5/2
    Explanation :-
    x - 2y = 8 — eq 1
    4x = 2y = 7 — eq 2
    Step 1 :- find x by eliminating y
    Eliminating x by adding eq 2 and eq1 then
    left parenthesis x minus 2 y right parenthesis plus left parenthesis 4 x plus 2 y right parenthesis equals 8 plus 7
    not stretchy rightwards double arrow x plus 4 x plus 2 y minus 2 y equals 15 not stretchy rightwards double arrow 5 x equals 15
     rightwards double arrowx space equals space 3
    Step 2 :- substitute value of x and find y
    not stretchy rightwards double arrow x minus 2 y equals 8 not stretchy rightwards double arrow left parenthesis 3 right parenthesis minus 2 y equals 8
    not stretchy rightwards double arrow negative 2 y equals 8 minus 3 not stretchy rightwards double arrow negative 2 y equals 5
     therefore space y space equals space minus 5 divided by 2
    therefore space x space equals space 3 and y = -5/2 is the solution of the given pair of equations.
    parallel
    General
    Maths-

    2 x minus y equals negative 4
    2 x plus y equals 4
    For the solution of the system of equations above, what is the value of  ?

    The given equations are
    2 x minus y equals negative 4
    2 x plus y equals 4
    First, we eliminate y by adding the equations together
    left parenthesis 2 x minus y right parenthesis plus left parenthesis 2 x plus y right parenthesis equals negative 4 plus 4
    Simplifying, we get
    2 x minus y plus 2 x plus y equals 0
    We have,
    2 x plus 2 x equals 0
    Thus, we get
    4 x equals 0
    As 4 not equal to 0 , we must have x =0 .
    Thus, the correct option is C).

    2 x minus y equals negative 4
    2 x plus y equals 4
    For the solution of the system of equations above, what is the value of  ?

    Maths-General
    The given equations are
    2 x minus y equals negative 4
    2 x plus y equals 4
    First, we eliminate y by adding the equations together
    left parenthesis 2 x minus y right parenthesis plus left parenthesis 2 x plus y right parenthesis equals negative 4 plus 4
    Simplifying, we get
    2 x minus y plus 2 x plus y equals 0
    We have,
    2 x plus 2 x equals 0
    Thus, we get
    4 x equals 0
    As 4 not equal to 0 , we must have x =0 .
    Thus, the correct option is C).
    General
    Maths-

    If E, F, G and H are respectively the mid points of the sides of a parallelogram ABCD and ar(EFGH) = 40 sq.cm , find the ar (parallelogram ABCD).

    Ans :- 80 cm2
    Explanation :-
    Step 1:- Find area of parallelogram ABCD
    Let  height Dk = h and base AB = b
    We get area of parallelogram ABCD  = base × height = bh
    Step 2:- Find height GI and EJ
    In parallelogram line joining midpoints of opposites sides is parallel to the base and
    equal to the base length i.e HF = AB = b
    Let us considerstraight triangle ADK ,
    Here  HF // AB and passing through mid point of side AD of straight triangle ADK
    So, the line intersects the side DK at midpoint L i.e DL = LK = DK/2 = h/2.
    As the are the perpendicular to HF so , the are the perpendicular length between lines DC , HF and AB
    As we know perpendicular length between DC and HF is h/2 then GI = h/2
    As we know perpendicular length between AB and HF is h/2 then EJ = h/2

    Step 3:- Find the areas of EFH and HGF
    As the area of triangle = ½ base × height
    Area of ΔEFH = ½ base length of hf × height EJ = 1 half cross times b cross times h over 2 equals fraction numerator h b over denominator 4 end fraction
    Area of ΔHGF = ½ base length of hf × height GI = 1 half cross times b cross times h over 2 equals fraction numerator h b over denominator 4 end fraction
    Area of EFGH = Area of ΔEFH + Area of ΔHGF = fraction numerator h b over denominator 4 end fraction plus fraction numerator h b over denominator 4 end fraction equals fraction numerator h b to the power of blank over denominator 2 end fraction ϵ
    Step 4:- Equate the areas  of EFGH
    fraction numerator h b over denominator 2 end fraction equals 40 sq. cm not stretchy rightwards double arrow h b equals 80 sq. cm
    We Know that area of parallelogram ABCD = bh  = 80 sq.cm.

    If E, F, G and H are respectively the mid points of the sides of a parallelogram ABCD and ar(EFGH) = 40 sq.cm , find the ar (parallelogram ABCD).

    Maths-General
    Ans :- 80 cm2
    Explanation :-
    Step 1:- Find area of parallelogram ABCD
    Let  height Dk = h and base AB = b
    We get area of parallelogram ABCD  = base × height = bh
    Step 2:- Find height GI and EJ
    In parallelogram line joining midpoints of opposites sides is parallel to the base and
    equal to the base length i.e HF = AB = b
    Let us considerstraight triangle ADK ,
    Here  HF // AB and passing through mid point of side AD of straight triangle ADK
    So, the line intersects the side DK at midpoint L i.e DL = LK = DK/2 = h/2.
    As the are the perpendicular to HF so , the are the perpendicular length between lines DC , HF and AB
    As we know perpendicular length between DC and HF is h/2 then GI = h/2
    As we know perpendicular length between AB and HF is h/2 then EJ = h/2

    Step 3:- Find the areas of EFH and HGF
    As the area of triangle = ½ base × height
    Area of ΔEFH = ½ base length of hf × height EJ = 1 half cross times b cross times h over 2 equals fraction numerator h b over denominator 4 end fraction
    Area of ΔHGF = ½ base length of hf × height GI = 1 half cross times b cross times h over 2 equals fraction numerator h b over denominator 4 end fraction
    Area of EFGH = Area of ΔEFH + Area of ΔHGF = fraction numerator h b over denominator 4 end fraction plus fraction numerator h b over denominator 4 end fraction equals fraction numerator h b to the power of blank over denominator 2 end fraction ϵ
    Step 4:- Equate the areas  of EFGH
    fraction numerator h b over denominator 2 end fraction equals 40 sq. cm not stretchy rightwards double arrow h b equals 80 sq. cm
    We Know that area of parallelogram ABCD = bh  = 80 sq.cm.

    General
    Maths-

    A soft drink is available in two packs. i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?

    Hint:
    The volume of a cylinder with any kind of base (rectangular or circular) is the area of its base multiplied by the height of the cylinder, i.e., if the height of a cylinder is h , then its volume is given by, (area x h) cubic units.
    Explanations:
    Step 1 of 3:
    The volume of a cylinder, with a rectangular base of length l and b width  and height of the cylinder be h ,   is l cross times b cross times h , (since the area of a rectangle is length  x width).
    So, the capacity (or volume) of the tin can with a rectangular base of length ( l) 5 cm and width (b) 4 cm, having a height (h) of 15 cm, is given by,
    l cross times b cross times h equals 5 cross times 4 cross times 15 equals 300 text  cubic  end text cm
    Step 2 of 3:
    We know that, if the radius of a cylinder is r and height be h , then the volume is pi r squared h .
    So, the capacity of the plastic cylinder with circular base of diameter 7 cm (so radius (r) is 7/2 cm) and height (h) 10 cm, is given by,
    pi r squared h equals 22 over 7 cross times open parentheses 7 over 2 close parentheses squared cross times 10 equals 22 over 7 cross times 7 over 2 cross times 7 over 2 cross times 10 equals 385 cm3
    Step 3 of 3:
    The tin can have a capacity of 300 cm3 and the plastic cylinder has of 385 cm3
    So, clearly 385 > 300, hence the plastic cylinder has greater capacity, by (385 – 300) = 85 cm3 more.
    Final Answer:
    The plastic cylinder has greater capacity by 85 cm3 more than the tin can.

    A soft drink is available in two packs. i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?

    Maths-General
    Hint:
    The volume of a cylinder with any kind of base (rectangular or circular) is the area of its base multiplied by the height of the cylinder, i.e., if the height of a cylinder is h , then its volume is given by, (area x h) cubic units.
    Explanations:
    Step 1 of 3:
    The volume of a cylinder, with a rectangular base of length l and b width  and height of the cylinder be h ,   is l cross times b cross times h , (since the area of a rectangle is length  x width).
    So, the capacity (or volume) of the tin can with a rectangular base of length ( l) 5 cm and width (b) 4 cm, having a height (h) of 15 cm, is given by,
    l cross times b cross times h equals 5 cross times 4 cross times 15 equals 300 text  cubic  end text cm
    Step 2 of 3:
    We know that, if the radius of a cylinder is r and height be h , then the volume is pi r squared h .
    So, the capacity of the plastic cylinder with circular base of diameter 7 cm (so radius (r) is 7/2 cm) and height (h) 10 cm, is given by,
    pi r squared h equals 22 over 7 cross times open parentheses 7 over 2 close parentheses squared cross times 10 equals 22 over 7 cross times 7 over 2 cross times 7 over 2 cross times 10 equals 385 cm3
    Step 3 of 3:
    The tin can have a capacity of 300 cm3 and the plastic cylinder has of 385 cm3
    So, clearly 385 > 300, hence the plastic cylinder has greater capacity, by (385 – 300) = 85 cm3 more.
    Final Answer:
    The plastic cylinder has greater capacity by 85 cm3 more than the tin can.
    parallel
    General
    Maths-

    Two transversals are cutting two parallel lines as shown in the diagram. Find the missing values: x, y and

    ANS :-  x = 90 ,the triangle is an obtuse angled triangle.
    Explanation :- Given the two lines are parallel and two are transversals .
    As alternate interior angles are equal we get, z° = 20°
    ∴z = 20
    Using sum of angles in triangle is 180° we get ,
    20°+ 30°+ 2x°=180° (sum of angles in a triangle)
    2x°+50° = 180°
    2x° = 130°  x° = 65°
    ∴ x = 65
    2x° = y° ( vertically opposite angles are equal)
    130° = y°
    ∴ y = 130

    Two transversals are cutting two parallel lines as shown in the diagram. Find the missing values: x, y and

    Maths-General
    ANS :-  x = 90 ,the triangle is an obtuse angled triangle.
    Explanation :- Given the two lines are parallel and two are transversals .
    As alternate interior angles are equal we get, z° = 20°
    ∴z = 20
    Using sum of angles in triangle is 180° we get ,
    20°+ 30°+ 2x°=180° (sum of angles in a triangle)
    2x°+50° = 180°
    2x° = 130°  x° = 65°
    ∴ x = 65
    2x° = y° ( vertically opposite angles are equal)
    130° = y°
    ∴ y = 130
    General
    Maths-

    Find the value of x and classify the triangle by its angles.

    ANS :-  x = 90 ,the triangle is an obtuse angled triangle.
    Explanation :-
    Given , the angles of the triangle are 50°,30°and (x+10)°.
    50°+ 30°+ x+10°=180° (sum of angles in a triangle)
    x°+90° = 180°
    x° = 90°
    ∴ x = 90
    We get angle x + 10 = 90 + 10 = 100° (> 90° so, obtuse angle)
    One of the angles of the triangle is obtuse. so, the triangle is an obtuse angled triangle

    Find the value of x and classify the triangle by its angles.

    Maths-General
    ANS :-  x = 90 ,the triangle is an obtuse angled triangle.
    Explanation :-
    Given , the angles of the triangle are 50°,30°and (x+10)°.
    50°+ 30°+ x+10°=180° (sum of angles in a triangle)
    x°+90° = 180°
    x° = 90°
    ∴ x = 90
    We get angle x + 10 = 90 + 10 = 100° (> 90° so, obtuse angle)
    One of the angles of the triangle is obtuse. so, the triangle is an obtuse angled triangle
    General
    Maths-

    Let A be the event that the house selected at random is a single storey house with slate roof.
    From the table given, it can be seen that
    The number of houses which are single storey having slate roof =4.
    Total number of houses =48
    So,
    The number of outcomes favourable to A (m)=4
    Total number of outcomes (n)= 48
    Hence, the probability of selecting a single storey house with slate roof,
    P left parenthesis A right parenthesis equals m over n equals 4 over 48
    Thus, the correct option is A)

    Maths-General
    Let A be the event that the house selected at random is a single storey house with slate roof.
    From the table given, it can be seen that
    The number of houses which are single storey having slate roof =4.
    Total number of houses =48
    So,
    The number of outcomes favourable to A (m)=4
    Total number of outcomes (n)= 48
    Hence, the probability of selecting a single storey house with slate roof,
    P left parenthesis A right parenthesis equals m over n equals 4 over 48
    Thus, the correct option is A)
    parallel

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