Maths-
General
Easy

Question

In a school, 5 over 8𝑡ℎ of the students are boys, if there are 240 girls, find the number of boys in the school?

Hint:

Find the total number of students and then subtract number of girls from it to
get the total number of boys.

The correct answer is: 400


    Let the total number of students = x
    Number of girls = 240
    So, number of boys = x - 240…(i)
    Given that 5 over 8 to the power of t h end exponent of the total students are boys. That is,5 over 8 x …(ii)
    On equating (i) and (ii) , we get 5 over 8 x equals x minus 240
    Multiply by 8 on both sides.
    We have 5 x equals 8 left parenthesis x minus 240 right parenthesis
    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 not stretchy rightwards double arrow 5 x equals 8 x minus 1920 end cell row cell not stretchy rightwards double arrow 8 x minus 5 x equals 1920 end cell end table
    not stretchy rightwards double arrow 3 x equals 1920
    not stretchy rightwards double arrow x equals 1920 over 3 equals 640
    Hence total number of students = 640
    From (ii), we get total number of boys equals 5 over 8 x equals 5 over 8 left parenthesis 640 right parenthesis equals 400
    So, number of boys in the school = 400

    Related Questions to study

    General
    Maths-

    If 24 pairs of trousers of equal size can be prepared with 54 m of cloth, what length of cloth is required for each pair of trousers?

    Let the cloth required for one trouser = x meters
    Then the cloth required for 24 equal size trousers =  54 m
    That is, 24x = 54
    On dividing both the sides by 24, we get fraction numerator 24 x over denominator 24 end fraction equals 54 over 24
    not stretchy rightwards double arrow 1 x equals 9 over 4 equals 2.25 straight m
    Hence, cloth required for one trouser x equals 9 over 4

    If 24 pairs of trousers of equal size can be prepared with 54 m of cloth, what length of cloth is required for each pair of trousers?

    Maths-General
    Let the cloth required for one trouser = x meters
    Then the cloth required for 24 equal size trousers =  54 m
    That is, 24x = 54
    On dividing both the sides by 24, we get fraction numerator 24 x over denominator 24 end fraction equals 54 over 24
    not stretchy rightwards double arrow 1 x equals 9 over 4 equals 2.25 straight m
    Hence, cloth required for one trouser x equals 9 over 4
    General
    Maths-

    Convert into decimal and identify whether it is a non-terminating and repeating decimal or a terminating decimal:24 over 11

    Let’s say x =24 over 11
    Upon dividing 24 by 11 we got
    x equals 24 over 11 equals 2.181818 horizontal ellipsis
    The resultant decimal is a non-terminating and repeating decimal.
    Final Answer:
    As the resultant decimal has an infinite number of digits following the decimal and these digits are repeated after a fixed interval. Hence, the decimal value of 24 over 11 is 2.18 with bottom parenthesis below and it is a non-terminating and repeating decimal.

    Convert into decimal and identify whether it is a non-terminating and repeating decimal or a terminating decimal:24 over 11

    Maths-General
    Let’s say x =24 over 11
    Upon dividing 24 by 11 we got
    x equals 24 over 11 equals 2.181818 horizontal ellipsis
    The resultant decimal is a non-terminating and repeating decimal.
    Final Answer:
    As the resultant decimal has an infinite number of digits following the decimal and these digits are repeated after a fixed interval. Hence, the decimal value of 24 over 11 is 2.18 with bottom parenthesis below and it is a non-terminating and repeating decimal.
    General
    Maths-

    By what rational number fraction numerator negative 8 over denominator 39 end fraction should  be multiplied to obtain 5 over 26 ?

    Complete step by step solution:
    Given rational number = fraction numerator negative 8 over denominator 39 end fraction and let the other rational number be x
    On multiplication, we get the result to be 5 over 26
    That is, we have fraction numerator negative 8 over denominator 39 end fraction x equals 5 over 26
    On dividing both sides by fraction numerator negative 8 over denominator 39 end fraction,

    we get x equals 5 over 26 divided by fraction numerator negative 8 over denominator 39 end fraction

    not stretchy rightwards double arrow x equals 5 over 26 cross times fraction numerator 39 over denominator negative 8 end fraction equals fraction numerator negative 15 over denominator 16 end fraction
    not stretchy rightwards double arrow x equals fraction numerator negative 15 over denominator 16 end fraction
    So, fraction numerator negative 15 over denominator 16 end fraction must be multiplied with fraction numerator negative 8 over denominator 39 end fraction  to get 5 over 26.

    By what rational number fraction numerator negative 8 over denominator 39 end fraction should  be multiplied to obtain 5 over 26 ?

    Maths-General
    Complete step by step solution:
    Given rational number = fraction numerator negative 8 over denominator 39 end fraction and let the other rational number be x
    On multiplication, we get the result to be 5 over 26
    That is, we have fraction numerator negative 8 over denominator 39 end fraction x equals 5 over 26
    On dividing both sides by fraction numerator negative 8 over denominator 39 end fraction,

    we get x equals 5 over 26 divided by fraction numerator negative 8 over denominator 39 end fraction

    not stretchy rightwards double arrow x equals 5 over 26 cross times fraction numerator 39 over denominator negative 8 end fraction equals fraction numerator negative 15 over denominator 16 end fraction
    not stretchy rightwards double arrow x equals fraction numerator negative 15 over denominator 16 end fraction
    So, fraction numerator negative 15 over denominator 16 end fraction must be multiplied with fraction numerator negative 8 over denominator 39 end fraction  to get 5 over 26.

    parallel
    General
    Maths-

    By what rational number fraction numerator negative 8 over denominator 39 end fraction should  be multiplied to obtain 5 over 26 ?

    Complete step by step solution:
    Given rational number = fraction numerator negative 8 over denominator 39 end fraction and let the other rational number be x
    On multiplication, we get the result to be 5 over 26
    That is, we have fraction numerator negative 8 over denominator 39 end fraction x equals 5 over 26
    On dividing both sides by fraction numerator negative 8 over denominator 39 end fraction,

    we get x equals 5 over 26 divided by fraction numerator negative 8 over denominator 39 end fraction

    not stretchy rightwards double arrow x equals 5 over 26 cross times fraction numerator 39 over denominator negative 8 end fraction equals fraction numerator negative 15 over denominator 16 end fraction
    not stretchy rightwards double arrow x equals fraction numerator negative 15 over denominator 16 end fraction
    So, fraction numerator negative 15 over denominator 16 end fraction must be multiplied with fraction numerator negative 8 over denominator 39 end fraction  to get 5 over 26.

    By what rational number fraction numerator negative 8 over denominator 39 end fraction should  be multiplied to obtain 5 over 26 ?

    Maths-General
    Complete step by step solution:
    Given rational number = fraction numerator negative 8 over denominator 39 end fraction and let the other rational number be x
    On multiplication, we get the result to be 5 over 26
    That is, we have fraction numerator negative 8 over denominator 39 end fraction x equals 5 over 26
    On dividing both sides by fraction numerator negative 8 over denominator 39 end fraction,

    we get x equals 5 over 26 divided by fraction numerator negative 8 over denominator 39 end fraction

    not stretchy rightwards double arrow x equals 5 over 26 cross times fraction numerator 39 over denominator negative 8 end fraction equals fraction numerator negative 15 over denominator 16 end fraction
    not stretchy rightwards double arrow x equals fraction numerator negative 15 over denominator 16 end fraction
    So, fraction numerator negative 15 over denominator 16 end fraction must be multiplied with fraction numerator negative 8 over denominator 39 end fraction  to get 5 over 26.

    General
    Maths-

    Convert into decimal and identify whether it is a non-terminating and repeating decimal or a terminating decimal:59 over 8

    Let’s say x =59 over 8
    Upon dividing 59 by 8 we got
    x equals 59 over 8 equals 7.375
    The resultant decimal is a terminating decimal
    Final Answer:
    As the resultant decimal has a finite number of digits following the decimal. Hence, the decimal value of 59 over 8  is 7.375 and it is a terminating decimal.

    Convert into decimal and identify whether it is a non-terminating and repeating decimal or a terminating decimal:59 over 8

    Maths-General
    Let’s say x =59 over 8
    Upon dividing 59 by 8 we got
    x equals 59 over 8 equals 7.375
    The resultant decimal is a terminating decimal
    Final Answer:
    As the resultant decimal has a finite number of digits following the decimal. Hence, the decimal value of 59 over 8  is 7.375 and it is a terminating decimal.
    General
    Maths-

    A sim earns Rs. 40000 per month. He spends 2 over 5 of his income on food, 5 over 12 of the remaining on education of children and 3 over 7 of the remainder on the house rent. How much money is still left with him?

    Total money earned by Asim = 40000 Rs
    Money spent on food from his income = 2 over 5 cross times 40000 equals 16000 Rs
    Amount left with him = 40000 - 16000 = 24000 Rs
    Money spent on education of children from the remaining money  = 5 over 12 cross times 24000 equals 10000 Rs
    Amount left with him = 24000 - 10000 = 14000 Rs
    Money spent on house rent from the remaining money  = 3 over 7 cross times 14000 equals 6000 Rs
    Amount left with him = 14000 - 6000 = 8000 Rs
    Money still left with him = 8000 Rs

    A sim earns Rs. 40000 per month. He spends 2 over 5 of his income on food, 5 over 12 of the remaining on education of children and 3 over 7 of the remainder on the house rent. How much money is still left with him?

    Maths-General
    Total money earned by Asim = 40000 Rs
    Money spent on food from his income = 2 over 5 cross times 40000 equals 16000 Rs
    Amount left with him = 40000 - 16000 = 24000 Rs
    Money spent on education of children from the remaining money  = 5 over 12 cross times 24000 equals 10000 Rs
    Amount left with him = 24000 - 10000 = 14000 Rs
    Money spent on house rent from the remaining money  = 3 over 7 cross times 14000 equals 6000 Rs
    Amount left with him = 14000 - 6000 = 8000 Rs
    Money still left with him = 8000 Rs
    parallel
    General
    Maths-

    Find the cost of 32 over 5 m of cloth at Rs 363 over 4.  per meter.
     

    Given that the cost of 1 m of cloth equals 36 3 over 4 equals 147 over 4 Rs
    ∴ Cost of 3 2 over 5 m equals 17 over 5 m equals 147 over 4 cross times 17 over 5
    equals fraction numerator 147 cross times 17 over denominator 4 cross times 5 end fraction
    equals 2499 over 20
    = 124.93 Rs
    Hence cost of 17 over 5 meter = 2499 over 20 Rs

    Find the cost of 32 over 5 m of cloth at Rs 363 over 4.  per meter.
     

    Maths-General
    Given that the cost of 1 m of cloth equals 36 3 over 4 equals 147 over 4 Rs
    ∴ Cost of 3 2 over 5 m equals 17 over 5 m equals 147 over 4 cross times 17 over 5
    equals fraction numerator 147 cross times 17 over denominator 4 cross times 5 end fraction
    equals 2499 over 20
    = 124.93 Rs
    Hence cost of 17 over 5 meter = 2499 over 20 Rs
    General
    Maths-

    Find the area of circular ring whose external and internal radii are 10cm and 3 cm respectively

    It is given that External radii, r1 = 10 cm
    Internal radii, r2 = 3 cm
    Area of circular ring = Area of outer ring – Area of inner ring
    straight pi (r1)2  - straight pi r2
    22 over 7 × (r1)2  -  22 over 7 × (r2)2
    22 over 7 × (102 – 32)
    22 over 7 × 91  =  286 cm2

    Find the area of circular ring whose external and internal radii are 10cm and 3 cm respectively

    Maths-General
    It is given that External radii, r1 = 10 cm
    Internal radii, r2 = 3 cm
    Area of circular ring = Area of outer ring – Area of inner ring
    straight pi (r1)2  - straight pi r2
    22 over 7 × (r1)2  -  22 over 7 × (r2)2
    22 over 7 × (102 – 32)
    22 over 7 × 91  =  286 cm2
    General
    Maths-

    The circumference of a circle is 123.2 cm. Calculate (i) the radius of the circle in cm. (ii) the area of the circle, correct to nearest cm2

    It is given that Circumference of the circle = 123.2 cm
    (i). Circumference of the circle = 2 straight pi r
    123.2  =  2 ×  22 over 7 × r
    19.6  = r
    (ii) Area of circle = straight pi r2
    = 22 over 7 × (19.6)2
    = 1207.36 cm2

    The circumference of a circle is 123.2 cm. Calculate (i) the radius of the circle in cm. (ii) the area of the circle, correct to nearest cm2

    Maths-General
    It is given that Circumference of the circle = 123.2 cm
    (i). Circumference of the circle = 2 straight pi r
    123.2  =  2 ×  22 over 7 × r
    19.6  = r
    (ii) Area of circle = straight pi r2
    = 22 over 7 × (19.6)2
    = 1207.36 cm2
    parallel
    General
    Maths-

    The area enclosed by the circumference of two concentric circles is 364.5 cm2. If the circumference of the inner circle is 88cm, calculate the radius of the outer circle.

    It is given that Circumference of inner circle = 88 cm
    straight pi r =  88
    2 × 22 over 7 ×  r  =  8
    r  =  14 cm
    Let radius of outer circle = r1
    Area enclosed by the circumference of two concentric circles = 364.5 cm2
    i.e. Area of outer circle – Area of inner circle = 364.5 cm2
    straight pi (r1)2  - straight pi r2     = 364.5
    22 over 7 × (r1)2  - 22 over 7 × (14)2  = 364.5
    22 over 7((r1)2 – 142) = 364.5
    ((r1)2 – 142   = fraction numerator 2551.5 over denominator 22 end fraction
    (r1)2 = 311.977
    r1  = 17.66 cm

    The area enclosed by the circumference of two concentric circles is 364.5 cm2. If the circumference of the inner circle is 88cm, calculate the radius of the outer circle.

    Maths-General
    It is given that Circumference of inner circle = 88 cm
    straight pi r =  88
    2 × 22 over 7 ×  r  =  8
    r  =  14 cm
    Let radius of outer circle = r1
    Area enclosed by the circumference of two concentric circles = 364.5 cm2
    i.e. Area of outer circle – Area of inner circle = 364.5 cm2
    straight pi (r1)2  - straight pi r2     = 364.5
    22 over 7 × (r1)2  - 22 over 7 × (14)2  = 364.5
    22 over 7((r1)2 – 142) = 364.5
    ((r1)2 – 142   = fraction numerator 2551.5 over denominator 22 end fraction
    (r1)2 = 311.977
    r1  = 17.66 cm
    General
    Maths-

    A track is in the form of a ring whose inner circumference is 352 m, and the outer circumference is 396 m. Find the width of the track

    It is given that Circumference of inner circle = 352 m
    straight pi r = 352
    22 over 7 r  = 352
    r  =  56 m
    Circumference of outer circle = 396 m
    straight pi r =   396
    2 22 over 7 ×  r  = 3
    r =  63 m
    Width of the track
    = Radius of outer circle – Radius of inner circle
    =  63 – 56
    =  7 m

    A track is in the form of a ring whose inner circumference is 352 m, and the outer circumference is 396 m. Find the width of the track

    Maths-General
    It is given that Circumference of inner circle = 352 m
    straight pi r = 352
    22 over 7 r  = 352
    r  =  56 m
    Circumference of outer circle = 396 m
    straight pi r =   396
    2 22 over 7 ×  r  = 3
    r =  63 m
    Width of the track
    = Radius of outer circle – Radius of inner circle
    =  63 – 56
    =  7 m
    General
    Maths-

    The area of a circular ring is 352 m2. If the diameter of its outer circle is 32 m, find the diameter of its inner circle.

    It is given that Area of circular ring = 352 m2
    Diameter of outer circle = 32 m
    rightwards double arrow Radius of outer circle,  r1 = fraction numerator text  diameter  end text over denominator 2 end fraction = 32 over 2 = 16 m
    Now, we know that area of the ring is difference of the areas of two circle
    Let radius of outer circle be r
    i.e. Area of ring = Area of outer circle – Area of inner circle
    352   = straight pi (r1)2  - straight pi r2
    352   = 22 over 7 ×  (16)2  - 22 over 7 × (r)2
    352 = 22 over 7 ×  (162 – r2)
    112 = 256 – r2
    r2 = 256 – 112 = 144
    r = square root of 144
    r = 12
    Diameter = 2  radius = 2  12 = 24 m

    The area of a circular ring is 352 m2. If the diameter of its outer circle is 32 m, find the diameter of its inner circle.

    Maths-General
    It is given that Area of circular ring = 352 m2
    Diameter of outer circle = 32 m
    rightwards double arrow Radius of outer circle,  r1 = fraction numerator text  diameter  end text over denominator 2 end fraction = 32 over 2 = 16 m
    Now, we know that area of the ring is difference of the areas of two circle
    Let radius of outer circle be r
    i.e. Area of ring = Area of outer circle – Area of inner circle
    352   = straight pi (r1)2  - straight pi r2
    352   = 22 over 7 ×  (16)2  - 22 over 7 × (r)2
    352 = 22 over 7 ×  (162 – r2)
    112 = 256 – r2
    r2 = 256 – 112 = 144
    r = square root of 144
    r = 12
    Diameter = 2  radius = 2  12 = 24 m
    parallel
    General
    Maths-

    The diameter of a circular park is 94.4 m. A 2.8 m wide road is constructed outside this circular park, i.e., around it. Find the cost of paving this road at the rate of Rs. 50 square meter

    Since the road of width 2.8 m is on outside , it means
    Diameter of inner circle = 94.4 m
    rightwards double arrow Radius =  fraction numerator text  diameter  end text over denominator 2 end fractionfraction numerator 94.4 over denominator 2 end fraction  = 47.2 m
    We know that area of a circle = p r2
    Radius of outer circle = Radius of inner circle + Width of road
    = 47.2 + 2.8 = 50 m
    Area of the road = Area of outer circle – Area of inner circle
    22 over 7 × (50)2  - 22 over 7 ×  (47.2)2
    22 over 7 × ( 50 – 47.2)(50 + 47.2)
    (therefore a2 – b2) = (a – b)(a + b)
    = 22 over 7  2.8 × 97.2
    =  855.36 m2
    Now, Cost of constructing 1 m2 road = Rs. 5
    Cost of constructing 855.36 m2 road = Rs. 50  855.36
    = Rs. 42768

    The diameter of a circular park is 94.4 m. A 2.8 m wide road is constructed outside this circular park, i.e., around it. Find the cost of paving this road at the rate of Rs. 50 square meter

    Maths-General
    Since the road of width 2.8 m is on outside , it means
    Diameter of inner circle = 94.4 m
    rightwards double arrow Radius =  fraction numerator text  diameter  end text over denominator 2 end fractionfraction numerator 94.4 over denominator 2 end fraction  = 47.2 m
    We know that area of a circle = p r2
    Radius of outer circle = Radius of inner circle + Width of road
    = 47.2 + 2.8 = 50 m
    Area of the road = Area of outer circle – Area of inner circle
    22 over 7 × (50)2  - 22 over 7 ×  (47.2)2
    22 over 7 × ( 50 – 47.2)(50 + 47.2)
    (therefore a2 – b2) = (a – b)(a + b)
    = 22 over 7  2.8 × 97.2
    =  855.36 m2
    Now, Cost of constructing 1 m2 road = Rs. 5
    Cost of constructing 855.36 m2 road = Rs. 50  855.36
    = Rs. 42768
    General
    Maths-

    The diameter of a circular park is 104 m. A road of width 3.5 m is on the outside around it. Find the cost of constructing the road at Rs. 15 per square meter.

    Since the road of width 3.5 m is on outside , it means
    Diameter of inner circle = 104 m
    rightwards double arrow Radius = fraction numerator text  diameter  end text over denominator 2 end fraction  = 104 over 2 = 52 m
    Radius of outer circle = Radius of inner circle + Width of road
    =  52 + 3.5 = 55.5 m
    We know that area of a circle = straight pi r2
    Area of the road = Area of outer circle – Area of inner circle
    22 over 7 × (55.5)2  - 22 over 7 ×  52
    22 over 7 ×  ( 55.5 – 52)(55.5 + 52)
    (therefore a2 – b2) = (a – b)(a + b)
    =  22 over 7 3.5 × 107.5
    = 1182.5 m2
    Now, Cost of constructing 1 m2 road = Rs. 15
    Cost of constructing 1182.5 m2 road = Rs. 15 × 1182.5
    = Rs. 17737.5

    The diameter of a circular park is 104 m. A road of width 3.5 m is on the outside around it. Find the cost of constructing the road at Rs. 15 per square meter.

    Maths-General
    Since the road of width 3.5 m is on outside , it means
    Diameter of inner circle = 104 m
    rightwards double arrow Radius = fraction numerator text  diameter  end text over denominator 2 end fraction  = 104 over 2 = 52 m
    Radius of outer circle = Radius of inner circle + Width of road
    =  52 + 3.5 = 55.5 m
    We know that area of a circle = straight pi r2
    Area of the road = Area of outer circle – Area of inner circle
    22 over 7 × (55.5)2  - 22 over 7 ×  52
    22 over 7 ×  ( 55.5 – 52)(55.5 + 52)
    (therefore a2 – b2) = (a – b)(a + b)
    =  22 over 7 3.5 × 107.5
    = 1182.5 m2
    Now, Cost of constructing 1 m2 road = Rs. 15
    Cost of constructing 1182.5 m2 road = Rs. 15 × 1182.5
    = Rs. 17737.5
    General
    Maths-

    A wire is in a circular shape of radius 35 cm. if it is bent in the form of square, find the area of square so formed.

    It is given that radius of the circle = 35 cm
    Circumference of the circle = 2 straight pi r
    = 2 × 22 over 7  ×35
    = 220 cm
    Since same wire is used, circumference/perimeter of both the shapes will be equal.
    rightwards double arrow Perimeter of square = Circumference of the circle
    4 × (side) =  220 cm
    Side = 55 cm
    Now, Area of square = (side)2
    =  552 = 3025 cm2

    A wire is in a circular shape of radius 35 cm. if it is bent in the form of square, find the area of square so formed.

    Maths-General
    It is given that radius of the circle = 35 cm
    Circumference of the circle = 2 straight pi r
    = 2 × 22 over 7  ×35
    = 220 cm
    Since same wire is used, circumference/perimeter of both the shapes will be equal.
    rightwards double arrow Perimeter of square = Circumference of the circle
    4 × (side) =  220 cm
    Side = 55 cm
    Now, Area of square = (side)2
    =  552 = 3025 cm2
    parallel

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