Maths-
General
Easy

Question

What is the number that is to be subtracted from fraction numerator negative 7 over denominator 8 end fraction 𝑡𝑜 𝑔𝑒𝑡 fraction numerator negative 13 over denominator 12 end fraction

Hint:

Frame the equation according to the question and then get the number to be
subtracted

The correct answer is: 5/24


    Given rational number = fraction numerator negative 7 over denominator 8 end fraction and let the other rational number to be subtracted from
    fraction numerator negative 7 over denominator 8 end fraction =  x
    On subtraction, we get the result to be fraction numerator negative 13 over denominator 12 end fraction
    That is, we have fraction numerator negative 7 over denominator 8 end fraction minus x equals fraction numerator negative 13 over denominator 12 end fraction
    On rearranging the equation, we get  x equals 13 over 12 plus fraction numerator negative 7 over denominator 8 end fraction
    On taking LCM of 12 and 8 , we have 24 as the LCM.
    not stretchy rightwards double arrow x equals fraction numerator 13 cross times 2 over denominator 12 cross times 2 end fraction plus fraction numerator negative 7 cross times 3 over denominator 8 cross times 3 end fraction
    not stretchy rightwards double arrow x equals 26 over 24 minus 21 over 24
    not stretchy rightwards double arrow x equals 5 over 24

    Related Questions to study

    General
    Maths-

    The circle touch each other internally. The sum of their area is 116 pie cm2 and the distance between their centres is 6 cm. Find the radii of the circles.

    It is given that the distance between their centres is 6 cm
    rightwards double arrow r2 - r1 = 6 rightwards double arrow r2 = 6 + r1
    The sum of their area of two circles =  116 straight pi cm2
    rightwards double arrowstraight pi (r1)2straight pi (r2)2 = 116 straight pi
    rightwards double arrow straight pi [ (r1)2 + (6 + r1)2 ] = 116 straight pi
    rightwards double arrow  (r1)2 + 36 +  (r1)2 + 12 r = 116
    rightwards double arrow 2 (r1)2 + 12 r + 36 – 116 = 0
    rightwards double arrow 2 (r1)2 + 12 r – 80 = 0 rightwards double arrow (r1)2 + 6 r – 40 = 0
    Using splitting the middle term,
    We get, (r1)2 + 10 r – 4r – 40 = 0
    (r1 + 10 )( r1 – 4) = 0 rightwards double arrow r1 = 4 , - 10
    Since radius is always positive, rightwards double arrow r1 = 4 cm
    rightwards double arrowr2 = 4 + 6 = 10 cm

    The circle touch each other internally. The sum of their area is 116 pie cm2 and the distance between their centres is 6 cm. Find the radii of the circles.

    Maths-General
    It is given that the distance between their centres is 6 cm
    rightwards double arrow r2 - r1 = 6 rightwards double arrow r2 = 6 + r1
    The sum of their area of two circles =  116 straight pi cm2
    rightwards double arrowstraight pi (r1)2straight pi (r2)2 = 116 straight pi
    rightwards double arrow straight pi [ (r1)2 + (6 + r1)2 ] = 116 straight pi
    rightwards double arrow  (r1)2 + 36 +  (r1)2 + 12 r = 116
    rightwards double arrow 2 (r1)2 + 12 r + 36 – 116 = 0
    rightwards double arrow 2 (r1)2 + 12 r – 80 = 0 rightwards double arrow (r1)2 + 6 r – 40 = 0
    Using splitting the middle term,
    We get, (r1)2 + 10 r – 4r – 40 = 0
    (r1 + 10 )( r1 – 4) = 0 rightwards double arrow r1 = 4 , - 10
    Since radius is always positive, rightwards double arrow r1 = 4 cm
    rightwards double arrowr2 = 4 + 6 = 10 cm
    General
    Maths-

    A bucket is raised from a well by means of a rope which is wound round a wheel of diameter 77cm. Given that the bucket ascends in 1 minute 28 seconds with a uniform speed of 1.1 m/s, calculate the number of complete revolutions of the wheel makes in raising the bucket.

    It is given that diameter of the wheel = 77 cm
    rightwards double arrow Radius = fraction numerator text  diameter  end text over denominator 2 end fraction = 38.5
    Circumference of the wheel = 2 straight pi r
    = 2 × 22 over 7 × 38.5 = 242 cm
    It is given that Speed of the bucket = 1.1 m/s
    Time taken = 1 min 28 sec = 88 sec
    Distance covered by the wheel = speed × time
    = 1.1 × 88 = 96.8 m
    = 96.8 × 100 cm
    = 9680 cm
    No of revolutions =  fraction numerator text  distance covered by wheel  end text over denominator text  circumference of wheel  end text end fraction
    =  9680 over 242 = 40 revolutions

    A bucket is raised from a well by means of a rope which is wound round a wheel of diameter 77cm. Given that the bucket ascends in 1 minute 28 seconds with a uniform speed of 1.1 m/s, calculate the number of complete revolutions of the wheel makes in raising the bucket.

    Maths-General
    It is given that diameter of the wheel = 77 cm
    rightwards double arrow Radius = fraction numerator text  diameter  end text over denominator 2 end fraction = 38.5
    Circumference of the wheel = 2 straight pi r
    = 2 × 22 over 7 × 38.5 = 242 cm
    It is given that Speed of the bucket = 1.1 m/s
    Time taken = 1 min 28 sec = 88 sec
    Distance covered by the wheel = speed × time
    = 1.1 × 88 = 96.8 m
    = 96.8 × 100 cm
    = 9680 cm
    No of revolutions =  fraction numerator text  distance covered by wheel  end text over denominator text  circumference of wheel  end text end fraction
    =  9680 over 242 = 40 revolutions
    General
    Maths-

    A rectangle with one side 4 cm inscribed in a circle of radius 2.5 cm. Find the area of the rectangle.

    It is given that radius of the circle = 2.5 cm
    rightwards double arrow AC = 2 × radius = 5 cm
    Length of the rectangle, AB = 4 cm ( given)
    We need to find breadth of the rectangle i.e. BC
    Using Pythagoras Theorem
    AC2 = AB2 + BC2
    52 = 42 + BC2
    25 – 16 = BC2
    9 = BCrightwards double arrow BC = 3 cm
    Area of rectangle = length  breadth
    = 4 × 3 = 12 cm2

    A rectangle with one side 4 cm inscribed in a circle of radius 2.5 cm. Find the area of the rectangle.

    Maths-General
    It is given that radius of the circle = 2.5 cm
    rightwards double arrow AC = 2 × radius = 5 cm
    Length of the rectangle, AB = 4 cm ( given)
    We need to find breadth of the rectangle i.e. BC
    Using Pythagoras Theorem
    AC2 = AB2 + BC2
    52 = 42 + BC2
    25 – 16 = BC2
    9 = BCrightwards double arrow BC = 3 cm
    Area of rectangle = length  breadth
    = 4 × 3 = 12 cm2
    parallel
    General
    Maths-

    The area of a circular ring enclosed between two concentric circles is 286 cm2. Find the radii of the two circles, given that their difference is 7 cm.

    Let radius of inner circle be r
    It is given that difference between the radii of both the circles is 7 cm.
    rightwards double arrow Radius of  outer circle – Radius of inner circle = 7 cm
    rightwards double arrow Radius of  outer circle = 7 + r
    Further, it is given that Area of a circular ring enclosed between two concentric circles = 286 cm2
    We know that area enclosed between two concentric circles is the difference of the areas of two circles
    rightwards double arrow Area of outer circle – Area of inner circle = 286 cm2.
    straight pi (r + 7)2  - p r2   =  286
    22 over 7 × ((r + 7)2 – r2 ) = 286
    r2 + 72 + 14 r – r2 = 91
    49 + 14 r = 91
    14 r = 91 – 49 = 42
    r = 3 cm
    So, radius of inner circle = 3 cm
    Radius of outer circle = 3 + 7 = 10 cm

    The area of a circular ring enclosed between two concentric circles is 286 cm2. Find the radii of the two circles, given that their difference is 7 cm.

    Maths-General
    Let radius of inner circle be r
    It is given that difference between the radii of both the circles is 7 cm.
    rightwards double arrow Radius of  outer circle – Radius of inner circle = 7 cm
    rightwards double arrow Radius of  outer circle = 7 + r
    Further, it is given that Area of a circular ring enclosed between two concentric circles = 286 cm2
    We know that area enclosed between two concentric circles is the difference of the areas of two circles
    rightwards double arrow Area of outer circle – Area of inner circle = 286 cm2.
    straight pi (r + 7)2  - p r2   =  286
    22 over 7 × ((r + 7)2 – r2 ) = 286
    r2 + 72 + 14 r – r2 = 91
    49 + 14 r = 91
    14 r = 91 – 49 = 42
    r = 3 cm
    So, radius of inner circle = 3 cm
    Radius of outer circle = 3 + 7 = 10 cm
    General
    Maths-

    A circular track has an inside circumference of 1320 m. If the width of the track is 7m, what is the outside circumference

    It is given that Circumference of inner circle = 1320 m
    straight pi r = 1320
    2 × 22 over 7 × r  = 1320
    r  =  210 m
    Radius of outer circle = Radius of inner circle +Width of track
    = 210 + 7 = 217 m
    Circumference of outer circle = 2 straight pi r
    = 2 × 22 over 7 × 217
    = 1364 m

    A circular track has an inside circumference of 1320 m. If the width of the track is 7m, what is the outside circumference

    Maths-General
    It is given that Circumference of inner circle = 1320 m
    straight pi r = 1320
    2 × 22 over 7 × r  = 1320
    r  =  210 m
    Radius of outer circle = Radius of inner circle +Width of track
    = 210 + 7 = 217 m
    Circumference of outer circle = 2 straight pi r
    = 2 × 22 over 7 × 217
    = 1364 m
    General
    Maths-

    In 2015 the populations of City X and City Y were equal. From 2010 to 2015 , the population of City X increased by 20% and the population of City  decreased by 10%. If the population of City  was 120,000 in 2010 , what was the population of City Y in 2010?

    Explanation:
    • We have given In 2015 the populations of City X and City Y were equal. From 2010 to 2015, the population of City X increased by 20% and the population of City Y decreased by 10%
    • We have to find what was the population of City Y in 2010.
    Step 1 of 1:
    1: population of city x in 2010 = 120000
    it increases by 20 % in 5 years . so in 2015 it will be
    ∣ 20 divided by 100 cross times 120000 equals 24000 equals 120000 plus 24000 equals 144000
    Thus population of city x in 2015 is 144000.
    2: the population of city x and y are same in year 2015.
    3: to find= population of city y in 2010 .if it reduces by 10% in 5 years.it will be 10 % more in year 2010
    4: let population of city y in year 2015 be 100.
    it gets increased by 10% =90
    thus-
    100 divided by 90 equals 144000 divided by x
    x equals 144000 cross times 100 over 90
    X =160000
    population of city x is .
    Hence, Option C is correct.

    In 2015 the populations of City X and City Y were equal. From 2010 to 2015 , the population of City X increased by 20% and the population of City  decreased by 10%. If the population of City  was 120,000 in 2010 , what was the population of City Y in 2010?

    Maths-General
    Explanation:
    • We have given In 2015 the populations of City X and City Y were equal. From 2010 to 2015, the population of City X increased by 20% and the population of City Y decreased by 10%
    • We have to find what was the population of City Y in 2010.
    Step 1 of 1:
    1: population of city x in 2010 = 120000
    it increases by 20 % in 5 years . so in 2015 it will be
    ∣ 20 divided by 100 cross times 120000 equals 24000 equals 120000 plus 24000 equals 144000
    Thus population of city x in 2015 is 144000.
    2: the population of city x and y are same in year 2015.
    3: to find= population of city y in 2010 .if it reduces by 10% in 5 years.it will be 10 % more in year 2010
    4: let population of city y in year 2015 be 100.
    it gets increased by 10% =90
    thus-
    100 divided by 90 equals 144000 divided by x
    x equals 144000 cross times 100 over 90
    X =160000
    population of city x is .
    Hence, Option C is correct.
    parallel
    General
    Maths-

    The area enclosed between two concentric circles is 770 cm2. If the radius of the outer circle is 21cm, calculate the radius of inner circle

    It is given that radius of outer circle, r1 = 21 cm
    Let radius of inner circle = r
    Area enclosed between two concentric circles = 770 cm2.
    rightwards double arrow Area of outer circle – Area of inner circle = 770 cm2.
    straight pi (r1)2  - straight pi r2   =  770
    22 over 7 × (21)2  - 22 over 7 ×  (r)2  = 770
    22 over 7 × (212 – r2 ) = 770
    441 – r2 = 245
    r2 = 441 – 245 = 196
    r = 14 cm
    Hence, radius of inner circle is 14 cm.

    The area enclosed between two concentric circles is 770 cm2. If the radius of the outer circle is 21cm, calculate the radius of inner circle

    Maths-General
    It is given that radius of outer circle, r1 = 21 cm
    Let radius of inner circle = r
    Area enclosed between two concentric circles = 770 cm2.
    rightwards double arrow Area of outer circle – Area of inner circle = 770 cm2.
    straight pi (r1)2  - straight pi r2   =  770
    22 over 7 × (21)2  - 22 over 7 ×  (r)2  = 770
    22 over 7 × (212 – r2 ) = 770
    441 – r2 = 245
    r2 = 441 – 245 = 196
    r = 14 cm
    Hence, radius of inner circle is 14 cm.
    General
    Maths-


    A researcher surveyed a random sample of students from a large university about how often they see movies. Using the sample data, the researcher estimated that 23% of the students in the population saw a movie at least once per month. The margin of error for this estimation is 4 straight percent sign . Which of the following is the most appropriate conclusion about all students at the university, based on the given estimate and margin of error?

    Explanation:
    • We have given the researcher estimated that 23% of the students in the population saw a movie at least once per month. The margin of error for this estimation is 4%.
    • We have to find which of the option is the most appropriate conclusion about all students at the university.
    • Step 1 of 1:
      There was a  4% error estimation error. But it didn’t specify whether it was caused by over estimation or under estimation.
      So, the researcher is between (23 - 24) % and (23+24)%
    • left parenthesis 23 minus 4 right parenthesis equals 19 straight percent sign
    • left parenthesis 23 plus 4 right parenthesis equals 27 straight percent sign
    • Therefore, It is plausible that the percentage of students who see a movie at least once per month is between 19% and 27%.
    • Hence, Option D is correct.


    A researcher surveyed a random sample of students from a large university about how often they see movies. Using the sample data, the researcher estimated that 23% of the students in the population saw a movie at least once per month. The margin of error for this estimation is 4 straight percent sign . Which of the following is the most appropriate conclusion about all students at the university, based on the given estimate and margin of error?

    Maths-General
    Explanation:
    • We have given the researcher estimated that 23% of the students in the population saw a movie at least once per month. The margin of error for this estimation is 4%.
    • We have to find which of the option is the most appropriate conclusion about all students at the university.
    • Step 1 of 1:
      There was a  4% error estimation error. But it didn’t specify whether it was caused by over estimation or under estimation.
      So, the researcher is between (23 - 24) % and (23+24)%
    • left parenthesis 23 minus 4 right parenthesis equals 19 straight percent sign
    • left parenthesis 23 plus 4 right parenthesis equals 27 straight percent sign
    • Therefore, It is plausible that the percentage of students who see a movie at least once per month is between 19% and 27%.
    • Hence, Option D is correct.
    General
    Maths-

    If x equals 5 plus 2 square root of 6, then find the value of x squared plus 1 over x squared

    Given x equals 5 plus 2 square root of 6
    Step 1 of 2:
    Find the value of x2
    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 squared equals left parenthesis 5 plus 2 square root of 6 right parenthesis squared end cell row cell x squared equals 25 plus 24 plus 2 cross times 5 cross times 2 square root of 6 end cell row cell x squared equals 49 plus 20 square root of 6 end cell end table
    Step 2 of 2:
    Put the value of x2 in  x squared plus 1 over x squared
    x squared plus 1 over x squared equals 49 plus 20 square root of 6 plus fraction numerator 1 over denominator 49 plus 20 square root of 6 end fraction
    Rationalizing fraction numerator 1 over denominator 49 plus 20 square root of 6 end fraction   in the upper equation
    table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell equals 49 plus 20 square root of 6 plus fraction numerator 1 over denominator 49 plus 20 square root of 6 end fraction cross times fraction numerator 49 minus 20 square root of 6 over denominator 49 minus 20 square root of 6 end fraction end cell row cell equals 49 plus 20 square root of 6 plus fraction numerator 49 minus 20 square root of 6 over denominator 49 squared minus left parenthesis 20 square root of 6 right parenthesis squared end fraction end cell row cell equals 49 plus 20 square root of 6 plus fraction numerator 49 minus 20 square root of 6 over denominator 2401 minus 2400 end fraction end cell row cell equals 49 plus 20 square root of 6 plus 49 minus 20 square root of 6 end cell row cell x squared plus 1 over x squared equals 98 end cell end table
    Final Answer:
    Hence, the value of x squared plus 1 over x squared  is 98.

    If x equals 5 plus 2 square root of 6, then find the value of x squared plus 1 over x squared

    Maths-General
    Given x equals 5 plus 2 square root of 6
    Step 1 of 2:
    Find the value of x2
    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 squared equals left parenthesis 5 plus 2 square root of 6 right parenthesis squared end cell row cell x squared equals 25 plus 24 plus 2 cross times 5 cross times 2 square root of 6 end cell row cell x squared equals 49 plus 20 square root of 6 end cell end table
    Step 2 of 2:
    Put the value of x2 in  x squared plus 1 over x squared
    x squared plus 1 over x squared equals 49 plus 20 square root of 6 plus fraction numerator 1 over denominator 49 plus 20 square root of 6 end fraction
    Rationalizing fraction numerator 1 over denominator 49 plus 20 square root of 6 end fraction   in the upper equation
    table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell equals 49 plus 20 square root of 6 plus fraction numerator 1 over denominator 49 plus 20 square root of 6 end fraction cross times fraction numerator 49 minus 20 square root of 6 over denominator 49 minus 20 square root of 6 end fraction end cell row cell equals 49 plus 20 square root of 6 plus fraction numerator 49 minus 20 square root of 6 over denominator 49 squared minus left parenthesis 20 square root of 6 right parenthesis squared end fraction end cell row cell equals 49 plus 20 square root of 6 plus fraction numerator 49 minus 20 square root of 6 over denominator 2401 minus 2400 end fraction end cell row cell equals 49 plus 20 square root of 6 plus 49 minus 20 square root of 6 end cell row cell x squared plus 1 over x squared equals 98 end cell end table
    Final Answer:
    Hence, the value of x squared plus 1 over x squared  is 98.
    parallel
    General
    Maths-

    A right circular cone has a volume of 24 pi cubic inches. If the height of the cone is 2 inches, what is the radius, in inches, of the base of the cone?

    Explanation:
    • We have given a right circular cone with volume 24 pi, if the height of the cone istext  2 inches  end text .
    • We have to find the radius of the cone 
    • We know that the volume of a cone is given by v equals 1 third pi r squared h.
    Step 1 of 1:
    We know that the volume of a cone is given by v equals 1 third pi r squared h.
    Height is given 2 inches
    So,
    v equals 1 third pi r squared h
    24 pi equals 1 third pi r squared cross times 2
    24 equals 1 third r squared cross times 2
    r squared = 36
    r = 6
    Hence, Option B is correct.

    A right circular cone has a volume of 24 pi cubic inches. If the height of the cone is 2 inches, what is the radius, in inches, of the base of the cone?

    Maths-General
    Explanation:
    • We have given a right circular cone with volume 24 pi, if the height of the cone istext  2 inches  end text .
    • We have to find the radius of the cone 
    • We know that the volume of a cone is given by v equals 1 third pi r squared h.
    Step 1 of 1:
    We know that the volume of a cone is given by v equals 1 third pi r squared h.
    Height is given 2 inches
    So,
    v equals 1 third pi r squared h
    24 pi equals 1 third pi r squared cross times 2
    24 equals 1 third r squared cross times 2
    r squared = 36
    r = 6
    Hence, Option B is correct.
    General
    Maths-

    In the given figure OACB is a quadrant of a circle. The radius OA = 3.5 cm, OD = 2cm. Calculate the area of shaded portion

    It is given that radius of the circle = 3.5 cm
    Area of the circle = straight pi r2
    = 22 over 7 × (3.5)2 = 38.5 cm2
    Area of quadrant of a circle = fraction numerator text  area of circle  end text over denominator 4 end fraction = fraction numerator 38.5 over denominator 4 end fraction = 9.625 cm2
    Now, Area of triangle OAD = 1 half ×  base × height
    1 half ×  3.5 × 2 = 3.5 cm2
    Area of shaded region
    = Area of quadrant of the circle – Area of triangle
    = 9.625 – 3.5 = 6.125 cm2

    In the given figure OACB is a quadrant of a circle. The radius OA = 3.5 cm, OD = 2cm. Calculate the area of shaded portion

    Maths-General
    It is given that radius of the circle = 3.5 cm
    Area of the circle = straight pi r2
    = 22 over 7 × (3.5)2 = 38.5 cm2
    Area of quadrant of a circle = fraction numerator text  area of circle  end text over denominator 4 end fraction = fraction numerator 38.5 over denominator 4 end fraction = 9.625 cm2
    Now, Area of triangle OAD = 1 half ×  base × height
    1 half ×  3.5 × 2 = 3.5 cm2
    Area of shaded region
    = Area of quadrant of the circle – Area of triangle
    = 9.625 – 3.5 = 6.125 cm2
    General
    Maths-

    The inner circumference of a circular track is 220 m. The track is 7 m wide everywhere. Calculate the cost of putting up a fence along the circumference of outer circle at the rate of Rs 2 per meter

    It is given that Circumference of inner circle = 220 m
    straight pi r  =   220
    2 × 22 over 7 × r  =   220
    r  =  35 m
    Radius of outer circle = Radius of inner circle +Width of track
    =  35 + 7 = 42 m
    Circumference of outer circle = 2 straight pi r
    = 2 × 22 over 7 × 42
    = 264 m
    Cost of fencing per meter = Rs. 2
    Cost of fencing outer circle = Rs. 2 × 264
    = Rs. 528

    The inner circumference of a circular track is 220 m. The track is 7 m wide everywhere. Calculate the cost of putting up a fence along the circumference of outer circle at the rate of Rs 2 per meter

    Maths-General
    It is given that Circumference of inner circle = 220 m
    straight pi r  =   220
    2 × 22 over 7 × r  =   220
    r  =  35 m
    Radius of outer circle = Radius of inner circle +Width of track
    =  35 + 7 = 42 m
    Circumference of outer circle = 2 straight pi r
    = 2 × 22 over 7 × 42
    = 264 m
    Cost of fencing per meter = Rs. 2
    Cost of fencing outer circle = Rs. 2 × 264
    = Rs. 528
    parallel
    General
    Maths-

    Is 5 over 7  rational or irrational? Explain.

    5 over 7 is a rational number because it is in the form of p/q with p = 5 and q = 7 and both p and q are integers
    Final Answer:
    Hence 5 over 7 is a rational number.

     

    Is 5 over 7  rational or irrational? Explain.

    Maths-General
    5 over 7 is a rational number because it is in the form of p/q with p = 5 and q = 7 and both p and q are integers
    Final Answer:
    Hence 5 over 7 is a rational number.

     
    General
    Maths-

    How many times will the wheel of a car rotate in a journey of 88 km if its known that the diameter of the wheel is 56 cm?

    It is given that diameter of the wheel = 56 cm
    rightwards double arrow Radius = fraction numerator text  diameter  end text over denominator 2 end fraction = 56 over 2 = 28 cm
    Circumference of the circle = 2 straight pi r
    = 2 × 22 over 7 × 28
    = 176 cm
    Total distance = 88 km = 8800000 cm
    No of rotations = fraction numerator text  total distance  end text over denominator text  circumference  end text end fraction  =  8800000 over 176 = 50000 rotations

    How many times will the wheel of a car rotate in a journey of 88 km if its known that the diameter of the wheel is 56 cm?

    Maths-General
    It is given that diameter of the wheel = 56 cm
    rightwards double arrow Radius = fraction numerator text  diameter  end text over denominator 2 end fraction = 56 over 2 = 28 cm
    Circumference of the circle = 2 straight pi r
    = 2 × 22 over 7 × 28
    = 176 cm
    Total distance = 88 km = 8800000 cm
    No of rotations = fraction numerator text  total distance  end text over denominator text  circumference  end text end fraction  =  8800000 over 176 = 50000 rotations
    General
    Maths-

    Express the following recurring decimal as a vulgar fraction
    1.3454545454545…………

    Hint:
    A repeating decimal is a decimal representation of a number whose digits are periodic (repeating its values at regular intervals) and the infinitely repeated portion is not zero. These repeating can be expressed in p/q form using some simple methods.
    Solution
    The given number is  1.3454545454545…… which can be written as 1.3 bottom enclose 45
    Let’s say X=1.3 bottom enclose 45
    Multiplying both sides by 10
    10 x equals 1.3 bottom enclose 45 ................(1)
    Step 2 of 3:
    Subtracting equation (2) by equation (1)
    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 1000 x minus 10 x equals 1345.45 with _ below minus 13.45 with _ below end cell row cell 990 x equals 1332 end cell row cell x equals 1332 over 990 end cell end table
    Simplifying the fraction:
    Final Answer:
    Hence, the p/q form of 1.3454545454545…… is .74 over 55
    Note: This question can also be solved by using the concept of the sum of infinite terms which are in geometric progression.

     
     

    Express the following recurring decimal as a vulgar fraction
    1.3454545454545…………

    Maths-General
    Hint:
    A repeating decimal is a decimal representation of a number whose digits are periodic (repeating its values at regular intervals) and the infinitely repeated portion is not zero. These repeating can be expressed in p/q form using some simple methods.
    Solution
    The given number is  1.3454545454545…… which can be written as 1.3 bottom enclose 45
    Let’s say X=1.3 bottom enclose 45
    Multiplying both sides by 10
    10 x equals 1.3 bottom enclose 45 ................(1)
    Step 2 of 3:
    Subtracting equation (2) by equation (1)
    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 1000 x minus 10 x equals 1345.45 with _ below minus 13.45 with _ below end cell row cell 990 x equals 1332 end cell row cell x equals 1332 over 990 end cell end table
    Simplifying the fraction:
    Final Answer:
    Hence, the p/q form of 1.3454545454545…… is .74 over 55
    Note: This question can also be solved by using the concept of the sum of infinite terms which are in geometric progression.

     
     
    parallel

    card img

    With Turito Academy.

    card img

    With Turito Foundation.

    card img

    Get an Expert Advice From Turito.

    Turito Academy

    card img

    With Turito Academy.

    Test Prep

    card img

    With Turito Foundation.