Maths-
General
Easy

Question

Venu has as many sisters as he has brothers. If Karuna, Venu’s sister has thrice as many brothers as she has sisters, then Venu has how many sisters?

Hint:

let no. of sisters for venu be x
By condition no. of brothers of venu is also x
If Karuna, Venu’s sister has thrice as many brothers as she has sisters
Find no. of karuna’s brothers and no. of karuna’s sisters  and equate them as per above condition.

The correct answer is: 2 sisters


    Ans :- Venu has 2 sisters
    Explanation :-
    let no. of sisters for venu be x
    By condition no. of brothers of venu is also x
    Step 1:-Find no. of karuna’s brothers and no. of karuna’s sisters
    No. of sisters of karuna is no. of sister of venu except herself
    No. of karunas sisters = no. of sister for venu - 1 = x -1
    No. of brothers of karuna are no. of brothers of venu including venu.
    No. of brothers of karuna = x + 1
    Step 2:-equate the condition
    If Karuna, Venu’s sister has thrice as many brothers as she has sisters
    Now ,left parenthesis x plus 1 right parenthesis equals 3 left parenthesis x minus 1 right parenthesis not stretchy rightwards double arrow x plus 1 equals 3 x minus 3
    not stretchy rightwards double arrow 1 plus 3 equals 3 x minus x not stretchy rightwards double arrow 4 equals 2 x
    not stretchy rightwards double arrow x equals 2
    ∴venu has 2 sisters.

    Related Questions to study

    General
    Maths-

    A pit 10 m long and 2 m wide is filled with 4.8 cubic m. of sand. What is the depth of the pit?

    Hint:
    A pit is a rectangular cylinder. So, we use l b h as the formula for volume of the pit where l comma b comma h are the length, width and depth of the pit, respectively and solve the problem.
    Explanations:
    Step 1 of 1:
    The parameters of the pit is given by, l = 10; b = 2 and volume = 4.8
    Therefore,
    l b h equals 4.8
    not stretchy rightwards double arrow 10 cross times 2 cross times h equals 4.8
    not stretchy rightwards double arrow h equals 48 over 10 cross times 1 over 20
    not stretchy rightwards double arrow h equals 0.24
    Final Answer:
    The depth of the pit is 0.24 m

    A pit 10 m long and 2 m wide is filled with 4.8 cubic m. of sand. What is the depth of the pit?

    Maths-General
    Hint:
    A pit is a rectangular cylinder. So, we use l b h as the formula for volume of the pit where l comma b comma h are the length, width and depth of the pit, respectively and solve the problem.
    Explanations:
    Step 1 of 1:
    The parameters of the pit is given by, l = 10; b = 2 and volume = 4.8
    Therefore,
    l b h equals 4.8
    not stretchy rightwards double arrow 10 cross times 2 cross times h equals 4.8
    not stretchy rightwards double arrow h equals 48 over 10 cross times 1 over 20
    not stretchy rightwards double arrow h equals 0.24
    Final Answer:
    The depth of the pit is 0.24 m
    General
    Maths-

    The circumference of Earth is estimated to be 40,030 kilometers at the equator. Which of the following best approximates the diameter, in miles, of Earth's equator? (1 kilometer almost equal to 0.62137  miles)

    Given,
    Circumference of the earth at the equator = 40,030 kilomteres
    The equator can be considered as a circle,
    And the circumference of a circle = 2 pi r , where  r is the radius of the circle.
    As, diameter of a circle is twice the radius, that is, diameter = 2r ,
    So we get
    cicum ference equals pi cross times diameter
    Thus,
    text  diameter  end text equals fraction numerator text  circumference  end text over denominator pi end fraction
    Putting the value of the circumference in the above equation
    text  diameter  end text equals 40030 over pi
    Using the calculator, we get

    text  diameter  end text equals 12741.94
    Thus, we get
    Diameter of the earth  12741.94 kilometres
    It is given that,
    1 text  kilometer  end text almost equal to 0.62137 text  miles  end text
    This is the unit conversion factor.
    Thus, we have
    12741.94 text  kilometres  end text almost equal to 12741.94 cross times 0.62137 text  miles  end text
    = 7917.45
    almost equal to 7917 text  miles  end text
    Thus, the correct option is C)

    The circumference of Earth is estimated to be 40,030 kilometers at the equator. Which of the following best approximates the diameter, in miles, of Earth's equator? (1 kilometer almost equal to 0.62137  miles)

    Maths-General
    Given,
    Circumference of the earth at the equator = 40,030 kilomteres
    The equator can be considered as a circle,
    And the circumference of a circle = 2 pi r , where  r is the radius of the circle.
    As, diameter of a circle is twice the radius, that is, diameter = 2r ,
    So we get
    cicum ference equals pi cross times diameter
    Thus,
    text  diameter  end text equals fraction numerator text  circumference  end text over denominator pi end fraction
    Putting the value of the circumference in the above equation
    text  diameter  end text equals 40030 over pi
    Using the calculator, we get

    text  diameter  end text equals 12741.94
    Thus, we get
    Diameter of the earth  12741.94 kilometres
    It is given that,
    1 text  kilometer  end text almost equal to 0.62137 text  miles  end text
    This is the unit conversion factor.
    Thus, we have
    12741.94 text  kilometres  end text almost equal to 12741.94 cross times 0.62137 text  miles  end text
    = 7917.45
    almost equal to 7917 text  miles  end text
    Thus, the correct option is C)
    General
    Maths-

    Length of the fence of a trapezium shaped field ABCD is 120m. If BC = 48m , CD = 17m and AD = 40m , Find the area of this field. Side AB is perpendicular to the parallel sides AD and BC.

    Ans :-660 m2.
     Explanation :-
    Step 1:- Find the length of AB from perimeter and other given side lengths.
    Perimeter of trapezium = 120 m
    Given BC = 48 m ;CD = 17 m and AD = 40 m.
    Side AB is perpendicular to the parallel sides AD and BC
    Perimeter of trapezium = AB+BC+CD+DA
    not stretchy rightwards double arrow 120 m equals A B plus 48 m plus 17 m plus 40 m
    not stretchy rightwards double arrow 120 m equals A B plus 105 m not stretchy rightwards double arrow A B equals 120 minus 105 equals 15 m

    Step 2:- find the area of trapezium
    text  Area of trapezium  end text equals 1 half left parenthesis text  height  end text right parenthesis text  (sum of lengths of parallel sides)  end text
    text  Area of trapezium  end text equals 1 half left parenthesis 15 right parenthesis left parenthesis 40 plus 48 right parenthesis equals left parenthesis 15 cross times 88 right parenthesis divided by 2 equals 15 cross times 44
    = 660 m2
    Therefore, Area of trapezium shaped field ABCD  = 660 m2.

    Length of the fence of a trapezium shaped field ABCD is 120m. If BC = 48m , CD = 17m and AD = 40m , Find the area of this field. Side AB is perpendicular to the parallel sides AD and BC.

    Maths-General
    Ans :-660 m2.
     Explanation :-
    Step 1:- Find the length of AB from perimeter and other given side lengths.
    Perimeter of trapezium = 120 m
    Given BC = 48 m ;CD = 17 m and AD = 40 m.
    Side AB is perpendicular to the parallel sides AD and BC
    Perimeter of trapezium = AB+BC+CD+DA
    not stretchy rightwards double arrow 120 m equals A B plus 48 m plus 17 m plus 40 m
    not stretchy rightwards double arrow 120 m equals A B plus 105 m not stretchy rightwards double arrow A B equals 120 minus 105 equals 15 m

    Step 2:- find the area of trapezium
    text  Area of trapezium  end text equals 1 half left parenthesis text  height  end text right parenthesis text  (sum of lengths of parallel sides)  end text
    text  Area of trapezium  end text equals 1 half left parenthesis 15 right parenthesis left parenthesis 40 plus 48 right parenthesis equals left parenthesis 15 cross times 88 right parenthesis divided by 2 equals 15 cross times 44
    = 660 m2
    Therefore, Area of trapezium shaped field ABCD  = 660 m2.

    parallel
    General
    Maths-

    Mukesh has some goats and hens in his shed. Upon counting, Mukesh found that the total number of legs is 112 and the total number of heads is 40. Find the number of hens in his shed.

    Ans :- There are 24 hens in the shed
    Explanation :-
    let no. of hens be x  and no. of goats be y .
    No. of legs for each hen = 2 ;No. of legs for each goat = 4
    No. of heads for each hen = 1 ;No. of heads for each goat = 1
    Step 1:- Form the equation based on given conditions
    Total no. of legs = No. of legs for x hens + No. of legs for y goats = 112
    xcross times No. of legs for each hen + ycross times No. of legs for each goat = 112
    not stretchy rightwards double arrow 2 x plus 4 y equals 112 not stretchy rightwards double arrow x plus 2 y equals 56 — Eq1
    Total no. of heads = No. of heads for x hens + No. of head for y goats = 40
    xcross times No. of heads for each hen + ycross times No. of heads for each goat = 40
    not stretchy rightwards double arrow x plus y equals 40 —Eq2
    Step 2:- eliminate y to find x
    Doing 2(Eq2) -(Eq1)
    2 left parenthesis x plus y right parenthesis minus left parenthesis x plus 2 y right parenthesis equals 2 left parenthesis 40 right parenthesis minus 56
    not stretchy rightwards double arrow 2 x minus x equals 80 minus 56
    ∴x = 24
    ∴There are 24 hens in the shed

    Mukesh has some goats and hens in his shed. Upon counting, Mukesh found that the total number of legs is 112 and the total number of heads is 40. Find the number of hens in his shed.

    Maths-General
    Ans :- There are 24 hens in the shed
    Explanation :-
    let no. of hens be x  and no. of goats be y .
    No. of legs for each hen = 2 ;No. of legs for each goat = 4
    No. of heads for each hen = 1 ;No. of heads for each goat = 1
    Step 1:- Form the equation based on given conditions
    Total no. of legs = No. of legs for x hens + No. of legs for y goats = 112
    xcross times No. of legs for each hen + ycross times No. of legs for each goat = 112
    not stretchy rightwards double arrow 2 x plus 4 y equals 112 not stretchy rightwards double arrow x plus 2 y equals 56 — Eq1
    Total no. of heads = No. of heads for x hens + No. of head for y goats = 40
    xcross times No. of heads for each hen + ycross times No. of heads for each goat = 40
    not stretchy rightwards double arrow x plus y equals 40 —Eq2
    Step 2:- eliminate y to find x
    Doing 2(Eq2) -(Eq1)
    2 left parenthesis x plus y right parenthesis minus left parenthesis x plus 2 y right parenthesis equals 2 left parenthesis 40 right parenthesis minus 56
    not stretchy rightwards double arrow 2 x minus x equals 80 minus 56
    ∴x = 24
    ∴There are 24 hens in the shed
    General
    Maths-

    The capacity of a cup is 300 ml. The capacity of a jug is twice that of a cup. What is the capacity of 3 cups and 3 jugs?

    Hint:
    We simply find out the capacity of a jug and solve the problem using multiplication.
    Explanations:
    Step 1 of 3:
    The capacity of a cup is 300 ml.
    The capacity of a jug is twice that of a cup, hence the capacity of a jug is300 cross times 2 equals 600 ml.
    Step 2 of 3:
    The capacity of 3 jugs is 300 cross times 3 equals 900 ml
    The capacity of 3 jugs is 600 cross times 3 equals 1800 ml
    Step 3 of 3:
    The capacity of 3 cups and 3 jugs is 900+1800 = 2700 ml, i.e., 2.7 ltr (since 1 ltr = 1000 ml)
    Final Answer:
    The capacity of 3 cups and 3 jugs is 2700 ml, i.e., 2.7 ltr

    The capacity of a cup is 300 ml. The capacity of a jug is twice that of a cup. What is the capacity of 3 cups and 3 jugs?

    Maths-General
    Hint:
    We simply find out the capacity of a jug and solve the problem using multiplication.
    Explanations:
    Step 1 of 3:
    The capacity of a cup is 300 ml.
    The capacity of a jug is twice that of a cup, hence the capacity of a jug is300 cross times 2 equals 600 ml.
    Step 2 of 3:
    The capacity of 3 jugs is 300 cross times 3 equals 900 ml
    The capacity of 3 jugs is 600 cross times 3 equals 1800 ml
    Step 3 of 3:
    The capacity of 3 cups and 3 jugs is 900+1800 = 2700 ml, i.e., 2.7 ltr (since 1 ltr = 1000 ml)
    Final Answer:
    The capacity of 3 cups and 3 jugs is 2700 ml, i.e., 2.7 ltr
    General
    Maths-

    A mother said to her son, ‘the sum of our present ages is twice my age 12 years ago and nine years hence, the sum of our ages will be thrice my age 14 years ago’. What is her son’s present age?

    Ans :- present age of son is 12 years
    Explanation :-
    Step 1:- find the equation by 1st condition.
    let present ages of son is x years and present age of mom is y years
    The sum of our present ages is twice my age 12 years ago.
    x + y =2( y-12) [ the age of 12 years ago is present age -12 years]
    not stretchy rightwards double arrow x plus y equals 2 y minus 24 not stretchy rightwards double arrow x minus y equals negative 24 — Eq1
    Step 2:- find the equation by applying second condition
    After 9 years the sons age = x+9
    After 9 years the mothers age = y+9
    Mother age 14 years ago = present age - 14 years = y - 14
    After 9 years ,the sum of our ages will be thrice my age 14 years ago
    left parenthesis x plus 9 right parenthesis plus left parenthesis y plus 9 right parenthesis equals 3 left parenthesis y minus 14 right parenthesis
    not stretchy rightwards double arrow x plus y plus 18 equals 3 y minus 42
    not stretchy rightwards double arrow x minus 2 y equals negative 60— Eq2
    Step 3:- find value of x by eliminating y
    Doing 2(Eq1)-Eq2 to eliminate y
    not stretchy rightwards double arrow 2 left parenthesis x minus y right parenthesis minus left parenthesis x minus 2 y right parenthesis equals 2 left parenthesis negative 24 right parenthesis minus left parenthesis negative 60 right parenthesis
    not stretchy rightwards double arrow 2 left parenthesis x minus y right parenthesis minus left parenthesis x minus 2 y right parenthesis equals 2 left parenthesis negative 24 right parenthesis minus left parenthesis negative 60 right parenthesis
    therefore space x space equals space 12
    ∴ present age of son = x= 12 years.

    A mother said to her son, ‘the sum of our present ages is twice my age 12 years ago and nine years hence, the sum of our ages will be thrice my age 14 years ago’. What is her son’s present age?

    Maths-General
    Ans :- present age of son is 12 years
    Explanation :-
    Step 1:- find the equation by 1st condition.
    let present ages of son is x years and present age of mom is y years
    The sum of our present ages is twice my age 12 years ago.
    x + y =2( y-12) [ the age of 12 years ago is present age -12 years]
    not stretchy rightwards double arrow x plus y equals 2 y minus 24 not stretchy rightwards double arrow x minus y equals negative 24 — Eq1
    Step 2:- find the equation by applying second condition
    After 9 years the sons age = x+9
    After 9 years the mothers age = y+9
    Mother age 14 years ago = present age - 14 years = y - 14
    After 9 years ,the sum of our ages will be thrice my age 14 years ago
    left parenthesis x plus 9 right parenthesis plus left parenthesis y plus 9 right parenthesis equals 3 left parenthesis y minus 14 right parenthesis
    not stretchy rightwards double arrow x plus y plus 18 equals 3 y minus 42
    not stretchy rightwards double arrow x minus 2 y equals negative 60— Eq2
    Step 3:- find value of x by eliminating y
    Doing 2(Eq1)-Eq2 to eliminate y
    not stretchy rightwards double arrow 2 left parenthesis x minus y right parenthesis minus left parenthesis x minus 2 y right parenthesis equals 2 left parenthesis negative 24 right parenthesis minus left parenthesis negative 60 right parenthesis
    not stretchy rightwards double arrow 2 left parenthesis x minus y right parenthesis minus left parenthesis x minus 2 y right parenthesis equals 2 left parenthesis negative 24 right parenthesis minus left parenthesis negative 60 right parenthesis
    therefore space x space equals space 12
    ∴ present age of son = x= 12 years.
    parallel
    General
    Maths-

    The cross section of a canal is a trapezium is shape. If the canal is 10m wide at the top and 6m wide at the bottom and the area of cross section is 840sq.m. Find the depth of the canal?

    Ans :- 105 m.
    Explanation :-
    Given the parallel sides of trapezium are 10 cm and 6 cm.
    Sum of lengths of parallel sides =10+6 =16 cm
    Area of trapezium = 840 sq. m
    Let height or depth of cross section  be h m
    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
    840 equals 1 half left parenthesis text  height  end text right parenthesis left parenthesis 16 right parenthesis not stretchy rightwards double arrow text  height  end text equals 840 divided by 8 equals 105 straight m

    Therefore ,the depth of the canal is 105 m .

    The cross section of a canal is a trapezium is shape. If the canal is 10m wide at the top and 6m wide at the bottom and the area of cross section is 840sq.m. Find the depth of the canal?

    Maths-General
    Ans :- 105 m.
    Explanation :-
    Given the parallel sides of trapezium are 10 cm and 6 cm.
    Sum of lengths of parallel sides =10+6 =16 cm
    Area of trapezium = 840 sq. m
    Let height or depth of cross section  be h m
    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
    840 equals 1 half left parenthesis text  height  end text right parenthesis left parenthesis 16 right parenthesis not stretchy rightwards double arrow text  height  end text equals 840 divided by 8 equals 105 straight m

    Therefore ,the depth of the canal is 105 m .
    General
    Maths-

    A certain colony of bacteria began with one cell, and the population doubled every 20 minutes. What was the population of the colony after 2 hours?

    The relation between hours and minutes is given by
    1 hour = 60 minutes
    This can be written as
    1 hour = 3 to the power of cross times 20 text  minutes  end text
    So, we get
    2 hours = 2. open parentheses 3 to the power of cross times 20 close text  minutes)  end text
    equals 6 to the power of cross times 20 text  minutes  end text
    That is , 2 hours have 6 sessions of 20 minutes.
    Given, the initial population is 1.
    Also, the population doubles every 20 minutes.
    And this happens 6 times,
    So the population of the colony after 2 hours equals 1 to the power of cross times 2 to the power of cross times 2 to the power of cross times 2 to the power of cross times 2 to the power of cross times 2 to the power of cross times 2 left parenthesis equalsopen 2 to the power of 6 close parentheses
    64
    Thus, the correct option is D)

    A certain colony of bacteria began with one cell, and the population doubled every 20 minutes. What was the population of the colony after 2 hours?

    Maths-General
    The relation between hours and minutes is given by
    1 hour = 60 minutes
    This can be written as
    1 hour = 3 to the power of cross times 20 text  minutes  end text
    So, we get
    2 hours = 2. open parentheses 3 to the power of cross times 20 close text  minutes)  end text
    equals 6 to the power of cross times 20 text  minutes  end text
    That is , 2 hours have 6 sessions of 20 minutes.
    Given, the initial population is 1.
    Also, the population doubles every 20 minutes.
    And this happens 6 times,
    So the population of the colony after 2 hours equals 1 to the power of cross times 2 to the power of cross times 2 to the power of cross times 2 to the power of cross times 2 to the power of cross times 2 to the power of cross times 2 left parenthesis equalsopen 2 to the power of 6 close parentheses
    64
    Thus, the correct option is D)
    General
    Maths-

    A Rectangular sheet of paper, 36 cm x 22 cm, is rolled along its length to form a cylinder.  Find the volume of the cylinder so formed?

    Hint:
    If a rectangular sheet of paper is rolled along its length to form a cylinder, then its length becomes the circumference of the cylinder base, and its width becomes the height of the cylinder.
    Step 1 of 3:
    The length and width of a rectangular sheet of paper are given by, 36 cm and 22 cm, respectively.
    As per the hint, if the paper is rolled along its length to form a cylinder, then the height of the cylinder would be 22 cm(= h), and the circumference of the cylinder base would be 36 cm.
    Step 2 of 3:
    Therefore,  2 pi r equals 36 not stretchy rightwards double arrow r equals 36 cross times 7 over 22 cross times 1 half equals 5.73 cm
    The radius of the cylinder formed is 5.73 cm(= r).
    Step 3 of 3:
    The volume of the formed cylinder is pi r squared h equals 22 over 7 cross times 5.73 squared cross times 22 equals 2770.16 cm3
    Final Answer:
    The volume of the cylinder so formed is 2770.16 cm3.

    A Rectangular sheet of paper, 36 cm x 22 cm, is rolled along its length to form a cylinder.  Find the volume of the cylinder so formed?

    Maths-General
    Hint:
    If a rectangular sheet of paper is rolled along its length to form a cylinder, then its length becomes the circumference of the cylinder base, and its width becomes the height of the cylinder.
    Step 1 of 3:
    The length and width of a rectangular sheet of paper are given by, 36 cm and 22 cm, respectively.
    As per the hint, if the paper is rolled along its length to form a cylinder, then the height of the cylinder would be 22 cm(= h), and the circumference of the cylinder base would be 36 cm.
    Step 2 of 3:
    Therefore,  2 pi r equals 36 not stretchy rightwards double arrow r equals 36 cross times 7 over 22 cross times 1 half equals 5.73 cm
    The radius of the cylinder formed is 5.73 cm(= r).
    Step 3 of 3:
    The volume of the formed cylinder is pi r squared h equals 22 over 7 cross times 5.73 squared cross times 22 equals 2770.16 cm3
    Final Answer:
    The volume of the cylinder so formed is 2770.16 cm3.
    parallel
    General
    Maths-

    The parallel sides of a trapezium are 20cm and 10cm. Its non parallel sides are both equal. Each being 13cm. Find the area of the trapezium ?

    Ans :- 180 cm2.
    Explanation :-
    Step 1:- Given lengths of parallel sides is 10 and 20 cm .
    Length of non parallel sides is 13 and 13 cm
    We get a triangle and parallelogram ABED
    we get BE = 13 cm ; DE = 10 cm (opposites sides of parallelogram )
    CE = CD-DE = 20 - 10 = 10 cm .


    Step 2:- Equate the areas and find value of height h
    5 h equals 60 not stretchy rightwards double arrow h equals 60 over 5 equals 12 cm
    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 12 right parenthesis left parenthesis 10 plus 20 right parenthesis equals 6 cross times 30 equals 180 cm squared
    Therefore, Area of trapezium ABCD  = 180 cm2.

    The parallel sides of a trapezium are 20cm and 10cm. Its non parallel sides are both equal. Each being 13cm. Find the area of the trapezium ?

    Maths-General
    Ans :- 180 cm2.
    Explanation :-
    Step 1:- Given lengths of parallel sides is 10 and 20 cm .
    Length of non parallel sides is 13 and 13 cm
    We get a triangle and parallelogram ABED
    we get BE = 13 cm ; DE = 10 cm (opposites sides of parallelogram )
    CE = CD-DE = 20 - 10 = 10 cm .


    Step 2:- Equate the areas and find value of height h
    5 h equals 60 not stretchy rightwards double arrow h equals 60 over 5 equals 12 cm
    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 12 right parenthesis left parenthesis 10 plus 20 right parenthesis equals 6 cross times 30 equals 180 cm squared
    Therefore, Area of trapezium ABCD  = 180 cm2.

    General
    Maths-

    In a fraction, if the numerator is increased by 2 and the denominator is decreased by 3, then the fraction becomes 1. Instead, if the numerator is decreased by 2 and denominator is increased by 3, then the fraction becomes 3/ 8. Find the fraction.

    Ans :- 8/13 is the value of fraction .
    Explanation :-
    Step 1:- Frame the equation based on given conditions.
    let the  numerator of fraction be x and denominator of fraction is y
    if the numerator is increased by 2 and the denominator is decreased by 3, then the fraction becomes 1
    We get fraction numerator x plus 2 over denominator y minus 3 end fraction equals 1 not stretchy rightwards double arrow x plus 2 equals y minus 3 not stretchy rightwards double arrow x plus 5 equals y — Eq1
    If the numerator is decreased by 2 and denominator is increased by 3, then the fraction becomes 3/ 8.
    We get fraction numerator x minus 2 over denominator y plus 3 end fraction equals 3 over 8 not stretchy rightwards double arrow 8 left parenthesis x minus 2 right parenthesis equals 3 left parenthesis y plus 3 right parenthesis not stretchy rightwards double arrow 8 x minus 16 equals 3 y plus 9
    8 x minus 16 minus 9 equals 3 y not stretchy rightwards double arrow 8 x minus 25 equals 3 y — Eq2
    Step 2:- eliminate the y and find x
    Doing Eq2 -3(Eq1) to eliminate y .
    8 x minus 25 minus 3 left parenthesis x plus 5 right parenthesis equals 3 y minus 3 left parenthesis y right parenthesis not stretchy rightwards double arrow 8 x minus 3 x equals 25 plus 15
    not stretchy rightwards double arrow 5 x equals 40
    ∴x = 8
    Step 3:- find y by substituting the value of x in eq1.
    x plus 5 equals y not stretchy rightwards double arrow 8 plus 5 equals y
    not stretchy rightwards double arrow y equals 13
    ∴y = 13
    We know the fraction is x/y = 8 over 13

    In a fraction, if the numerator is increased by 2 and the denominator is decreased by 3, then the fraction becomes 1. Instead, if the numerator is decreased by 2 and denominator is increased by 3, then the fraction becomes 3/ 8. Find the fraction.

    Maths-General
    Ans :- 8/13 is the value of fraction .
    Explanation :-
    Step 1:- Frame the equation based on given conditions.
    let the  numerator of fraction be x and denominator of fraction is y
    if the numerator is increased by 2 and the denominator is decreased by 3, then the fraction becomes 1
    We get fraction numerator x plus 2 over denominator y minus 3 end fraction equals 1 not stretchy rightwards double arrow x plus 2 equals y minus 3 not stretchy rightwards double arrow x plus 5 equals y — Eq1
    If the numerator is decreased by 2 and denominator is increased by 3, then the fraction becomes 3/ 8.
    We get fraction numerator x minus 2 over denominator y plus 3 end fraction equals 3 over 8 not stretchy rightwards double arrow 8 left parenthesis x minus 2 right parenthesis equals 3 left parenthesis y plus 3 right parenthesis not stretchy rightwards double arrow 8 x minus 16 equals 3 y plus 9
    8 x minus 16 minus 9 equals 3 y not stretchy rightwards double arrow 8 x minus 25 equals 3 y — Eq2
    Step 2:- eliminate the y and find x
    Doing Eq2 -3(Eq1) to eliminate y .
    8 x minus 25 minus 3 left parenthesis x plus 5 right parenthesis equals 3 y minus 3 left parenthesis y right parenthesis not stretchy rightwards double arrow 8 x minus 3 x equals 25 plus 15
    not stretchy rightwards double arrow 5 x equals 40
    ∴x = 8
    Step 3:- find y by substituting the value of x in eq1.
    x plus 5 equals y not stretchy rightwards double arrow 8 plus 5 equals y
    not stretchy rightwards double arrow y equals 13
    ∴y = 13
    We know the fraction is x/y = 8 over 13
    General
    Maths-

    What is the volume of a cylindrical container with base radius 5 cm and height 20 cm?
     

    Hint:
    The volume of a cylindrical container with base radius r and height h, is pi r squared h cubic units.
    Explanations:
    Step 1 of 1:
    Given, r = 5 and   h = 20.
    Therefore, the volume of the cylindrical container is given by
    pi r squared h
    equals 22 over 7 cross times 5 squared cross times 20
    equals 1571.428
    almost equal to 1571.43 cm cubed
    Final Answer:
    The required volume of the cylindrical container is 1571.43 cm3

    What is the volume of a cylindrical container with base radius 5 cm and height 20 cm?
     

    Maths-General
    Hint:
    The volume of a cylindrical container with base radius r and height h, is pi r squared h cubic units.
    Explanations:
    Step 1 of 1:
    Given, r = 5 and   h = 20.
    Therefore, the volume of the cylindrical container is given by
    pi r squared h
    equals 22 over 7 cross times 5 squared cross times 20
    equals 1571.428
    almost equal to 1571.43 cm cubed
    Final Answer:
    The required volume of the cylindrical container is 1571.43 cm3
    parallel
    General
    Maths-

    In the x - y plane, the graph of line ell  has slope 3 . Line k  is parallel to line  ell and contains the point  ( 3, 10 ). Which of the following is an equation of line k ?

    Given,
    Slope of line  l = 3
    As, line  is parallel to l , their slopes are equal.
    So, slope of line  k (m) = 3
    Also,  passes through the point (3, 10)
    The equation of a line in point-slope form is given by
    y minus b equals m left parenthesis x minus a right parenthesis
    where, m  is the slope of the line and
    ( a, b ) is any point lying on the line.
    Using the above formula, the equation of line  is given by
    y minus 10 equals 3 left parenthesis x minus 3 right parenthesis
    Simplifying the above equation, we have
    y minus 10 equals 3 x minus 9
    Adding 10 both sides, we get
    y equals 3 x minus 9 plus 10
    Finally, we have
    y equals 3 x plus 1
    Hence, the equation of line  k is y = 3x + 1
    Thus, the correct option is D)

    In the x - y plane, the graph of line ell  has slope 3 . Line k  is parallel to line  ell and contains the point  ( 3, 10 ). Which of the following is an equation of line k ?

    Maths-General
    Given,
    Slope of line  l = 3
    As, line  is parallel to l , their slopes are equal.
    So, slope of line  k (m) = 3
    Also,  passes through the point (3, 10)
    The equation of a line in point-slope form is given by
    y minus b equals m left parenthesis x minus a right parenthesis
    where, m  is the slope of the line and
    ( a, b ) is any point lying on the line.
    Using the above formula, the equation of line  is given by
    y minus 10 equals 3 left parenthesis x minus 3 right parenthesis
    Simplifying the above equation, we have
    y minus 10 equals 3 x minus 9
    Adding 10 both sides, we get
    y equals 3 x minus 9 plus 10
    Finally, we have
    y equals 3 x plus 1
    Hence, the equation of line  k is y = 3x + 1
    Thus, the correct option is D)
    General
    Maths-

    The ratio of the lengths of the parallel sides of a trapezium is 3:2. The shortest distance between them is 15cm. If the area of the trapezium is 450 sq. cm. Find the sum of the lengths of the parallel sides ?

    Ans :- 60 cm
    Explanation :-
    Step 1:-  Find the equation between x and y  and find the sum of lengths of parallel sides in terms of y.
    let the length of parallel sides be x and y (x>y).
    Given ratio of x:y = 3:2 not stretchy rightwards double arrow x over y equals 3 over 2 not stretchy rightwards double arrow y equals 2 x divided by 3
    Sum of lengths of parallel sides = x plus y equals x plus 2 x divided by 3 equals 5 x divided by 3
    Step 2:-substitute the values in the area formula find x
    Given area of trapezium = 450 sq.cm
    Altitude (or) height = 15 cm
    area of trapezium = ½ × (height)(sum of lengths of parallel side)
    not stretchy rightwards double arrow 450 equals 1 half cross times left parenthesis 15 right parenthesis cross times open parentheses fraction numerator 5 x over denominator 3 end fraction close parentheses not stretchy rightwards double arrow 30 cross times 2 equals fraction numerator 5 x over denominator 3 end fraction not stretchy rightwards double arrow fraction numerator 5 x over denominator 3 end fraction equals 60
    We know Sum of lengths of parallel sides = x plus y equals x plus 2 x divided by 3 equals 5 x divided by 3
    So, Sum of lengths of parallel sides = 60 cm.
    Therefore, Sum of lengths of parallel sides is 60 cm.

    The ratio of the lengths of the parallel sides of a trapezium is 3:2. The shortest distance between them is 15cm. If the area of the trapezium is 450 sq. cm. Find the sum of the lengths of the parallel sides ?

    Maths-General
    Ans :- 60 cm
    Explanation :-
    Step 1:-  Find the equation between x and y  and find the sum of lengths of parallel sides in terms of y.
    let the length of parallel sides be x and y (x>y).
    Given ratio of x:y = 3:2 not stretchy rightwards double arrow x over y equals 3 over 2 not stretchy rightwards double arrow y equals 2 x divided by 3
    Sum of lengths of parallel sides = x plus y equals x plus 2 x divided by 3 equals 5 x divided by 3
    Step 2:-substitute the values in the area formula find x
    Given area of trapezium = 450 sq.cm
    Altitude (or) height = 15 cm
    area of trapezium = ½ × (height)(sum of lengths of parallel side)
    not stretchy rightwards double arrow 450 equals 1 half cross times left parenthesis 15 right parenthesis cross times open parentheses fraction numerator 5 x over denominator 3 end fraction close parentheses not stretchy rightwards double arrow 30 cross times 2 equals fraction numerator 5 x over denominator 3 end fraction not stretchy rightwards double arrow fraction numerator 5 x over denominator 3 end fraction equals 60
    We know Sum of lengths of parallel sides = x plus y equals x plus 2 x divided by 3 equals 5 x divided by 3
    So, Sum of lengths of parallel sides = 60 cm.
    Therefore, Sum of lengths of parallel sides is 60 cm.

    General
    Maths-

    The sum of two numbers is 52 and their difference is 20. find the numbers?

    Ans :- 16 and 36 are the numbers
    Explanation :-
    Step 1:-Frame the system of linear equations based on the
    let the  two number be x and y where y > x
    Given sum of = 52
    x + y =52 — Eq1
    Given difference = 20
    Difference is bigger number y - smaller number x = 20
    rightwards double arrowy = x = 20 — Eq2
    Step 2:- eliminate the x and find y
    Doing Eq2 +Eq1 to eliminate x .
    x plus y plus y minus x equals 52 plus 20 not stretchy rightwards double arrow 2 y equals 72
    not stretchy rightwards double arrow y equals 36
    ∴y = 36
    Step 3:- find x by substituting the value of y in eq1.
    x plus y equals 52 not stretchy rightwards double arrow x plus 36 equals 52
    not stretchy rightwards double arrow x equals 52 minus 36
    ∴x = 16
    ∴16 and 36 are the numbers which satisfy the given conditions.

    The sum of two numbers is 52 and their difference is 20. find the numbers?

    Maths-General
    Ans :- 16 and 36 are the numbers
    Explanation :-
    Step 1:-Frame the system of linear equations based on the
    let the  two number be x and y where y > x
    Given sum of = 52
    x + y =52 — Eq1
    Given difference = 20
    Difference is bigger number y - smaller number x = 20
    rightwards double arrowy = x = 20 — Eq2
    Step 2:- eliminate the x and find y
    Doing Eq2 +Eq1 to eliminate x .
    x plus y plus y minus x equals 52 plus 20 not stretchy rightwards double arrow 2 y equals 72
    not stretchy rightwards double arrow y equals 36
    ∴y = 36
    Step 3:- find x by substituting the value of y in eq1.
    x plus y equals 52 not stretchy rightwards double arrow x plus 36 equals 52
    not stretchy rightwards double arrow x equals 52 minus 36
    ∴x = 16
    ∴16 and 36 are the numbers which satisfy the given conditions.
    parallel

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