Mathematics
Grade-7
Easy

Question

Quotient of 0.11 divided by space fraction numerator negative 4 over denominator 6 end fraction  is _____.

  1. 11 over 100
  2. fraction numerator negative 33 over denominator 200 end fraction
  3. fraction numerator negative 100 over denominator 11 end fraction
  4. 200 over 33

Hint:

Solve using the simple BODMASS rule.

The correct answer is: fraction numerator negative 33 over denominator 200 end fraction


    STEP BY STEP SOLUTION
    0.11 space divided by space fraction numerator negative 4 over denominator 6 end fraction space
C o n v e r t i n g space d e c i m a l space i n t o space f r a c t i o n space t o space m a k e space i t space e a s i e r space
equals space 11 over 100 space divided by space 4 over 6 space
equals space 11 over 100 space cross times space fraction numerator negative 6 over denominator 4 end fraction
space equals space fraction numerator negative 66 over denominator 400 end fraction
space equals space fraction numerator negative 33 over denominator 200 end fraction
T h e space a n s w e r space equals space fraction numerator negative 33 over denominator 200 end fraction

    Related Questions to study

    Grade-7
    Mathematics

    If  straight g space cross times space fraction numerator negative 1 over denominator 2 end fraction space equals space fraction numerator negative 13 over denominator 4 end fraction , then value of g = _____.

    STEP BY STEP SOLUTION
    g space cross times space fraction numerator negative 1 over denominator 2 end fraction space equals space fraction numerator negative 13 over denominator 4 end fraction
T o space f i n d space t h e space v a l u e space o f space g space w e space w i l l space d i v i d e space fraction numerator negative 13 over denominator 4 end fraction space b y space fraction numerator negative 1 over denominator 2 end fraction
g space equals space fraction numerator negative 13 over denominator 4 end fraction space divided by space fraction numerator negative 1 over denominator 2 end fraction
g space equals space fraction numerator negative 13 over denominator 4 end fraction space cross times fraction numerator negative 2 over denominator 1 end fraction
g space equals space 26 over 4
g space equals space 13 over 2

    If  straight g space cross times space fraction numerator negative 1 over denominator 2 end fraction space equals space fraction numerator negative 13 over denominator 4 end fraction , then value of g = _____.

    MathematicsGrade-7
    STEP BY STEP SOLUTION
    g space cross times space fraction numerator negative 1 over denominator 2 end fraction space equals space fraction numerator negative 13 over denominator 4 end fraction
T o space f i n d space t h e space v a l u e space o f space g space w e space w i l l space d i v i d e space fraction numerator negative 13 over denominator 4 end fraction space b y space fraction numerator negative 1 over denominator 2 end fraction
g space equals space fraction numerator negative 13 over denominator 4 end fraction space divided by space fraction numerator negative 1 over denominator 2 end fraction
g space equals space fraction numerator negative 13 over denominator 4 end fraction space cross times fraction numerator negative 2 over denominator 1 end fraction
g space equals space 26 over 4
g space equals space 13 over 2
    Grade-7
    Mathematics

    The value of a, if a straight a space divided by space fraction numerator negative 3 over denominator 6 end fraction  is fraction numerator negative 4 over denominator 2 end fraction a = ___________.

    STEP BY STEP SOLUTION
    a space divided by space fraction numerator negative 3 over denominator 6 space end fraction space equals space fraction numerator negative 4 over denominator 2 end fraction
T o space f i n d space t h e space v a l u e space o f space a space
a space equals space fraction numerator negative 4 over denominator 2 end fraction space divided by space fraction numerator negative 6 over denominator 3 end fraction
a space equals space fraction numerator negative 4 over denominator 2 end fraction space cross times fraction numerator negative 3 over denominator 6 end fraction
a space equals space 12 over 12
a space equals space 1

    The value of a, if a straight a space divided by space fraction numerator negative 3 over denominator 6 end fraction  is fraction numerator negative 4 over denominator 2 end fraction a = ___________.

    MathematicsGrade-7
    STEP BY STEP SOLUTION
    a space divided by space fraction numerator negative 3 over denominator 6 space end fraction space equals space fraction numerator negative 4 over denominator 2 end fraction
T o space f i n d space t h e space v a l u e space o f space a space
a space equals space fraction numerator negative 4 over denominator 2 end fraction space divided by space fraction numerator negative 6 over denominator 3 end fraction
a space equals space fraction numerator negative 4 over denominator 2 end fraction space cross times fraction numerator negative 3 over denominator 6 end fraction
a space equals space 12 over 12
a space equals space 1
    Grade-7
    Mathematics

    The sum or difference of 9 over 13 plus open parentheses fraction numerator negative 7 over denominator 13 end fraction close parentheses is ___________.

    STEP BY STEP SOLUTION
    9 over 13 plus open parentheses fraction numerator negative 7 over denominator 13 end fraction close parentheses
O p e n i n g space b r a c k e t space
equals 9 over 13 minus 7 over 13
T a k i n g space L C M thin space
equals fraction numerator 9 minus 7 over denominator 13 end fraction
equals 2 over 13

    The sum or difference of 9 over 13 plus open parentheses fraction numerator negative 7 over denominator 13 end fraction close parentheses is ___________.

    MathematicsGrade-7
    STEP BY STEP SOLUTION
    9 over 13 plus open parentheses fraction numerator negative 7 over denominator 13 end fraction close parentheses
O p e n i n g space b r a c k e t space
equals 9 over 13 minus 7 over 13
T a k i n g space L C M thin space
equals fraction numerator 9 minus 7 over denominator 13 end fraction
equals 2 over 13
    parallel
    Grade-7
    Mathematics

    The sum or difference of  1 over 8 plus open parentheses fraction numerator negative 7 over denominator 8 end fraction close parentheses is __________.

    STEP BY STEP SOLUTION
    1 over 8 plus open parentheses fraction numerator negative 7 over denominator 8 end fraction close parentheses
O p e n space t h e space b r a c k e t space
equals 1 over 8 minus 7 over 8
T a k i n g space L C M space o r space c o n v e r t i n g space f r a c t i o n space i n t o space d e c i m a l
equals fraction numerator 1 minus 7 over denominator 8 end fraction space
equals fraction numerator negative 6 over denominator 8 end fraction
T h e space a n s w e r space o f space a b o v e space q u e s t i o n space i s space fraction numerator negative 6 over denominator 8 end fraction

    The sum or difference of  1 over 8 plus open parentheses fraction numerator negative 7 over denominator 8 end fraction close parentheses is __________.

    MathematicsGrade-7
    STEP BY STEP SOLUTION
    1 over 8 plus open parentheses fraction numerator negative 7 over denominator 8 end fraction close parentheses
O p e n space t h e space b r a c k e t space
equals 1 over 8 minus 7 over 8
T a k i n g space L C M space o r space c o n v e r t i n g space f r a c t i o n space i n t o space d e c i m a l
equals fraction numerator 1 minus 7 over denominator 8 end fraction space
equals fraction numerator negative 6 over denominator 8 end fraction
T h e space a n s w e r space o f space a b o v e space q u e s t i o n space i s space fraction numerator negative 6 over denominator 8 end fraction
    Grade-7
    Mathematics

    Jeff had 27lb can of mighty dog food. This dog eats 2 1 fourth lbs each day. Number of days the can lasts is_________

    STEP BY STEP SOLUTION
    Total capicity of can jeff have  = 27lbs
    Food dog eats per day = 2 1 fourth l b s
    So to find out for how many days the food will last we will divide the food that dog eats per day by the total capicity of can
    2 1 fourth space divided by space 27 space equals space 9 over 4 space divided by space 27 over 1 space equals space 9 over 4 space cross times space 1 over 27 space equals space 9 over 108 space equals space 3 over 36 space equals space 1 over 12

    Jeff had 27lb can of mighty dog food. This dog eats 2 1 fourth lbs each day. Number of days the can lasts is_________

    MathematicsGrade-7
    STEP BY STEP SOLUTION
    Total capicity of can jeff have  = 27lbs
    Food dog eats per day = 2 1 fourth l b s
    So to find out for how many days the food will last we will divide the food that dog eats per day by the total capicity of can
    2 1 fourth space divided by space 27 space equals space 9 over 4 space divided by space 27 over 1 space equals space 9 over 4 space cross times space 1 over 27 space equals space 9 over 108 space equals space 3 over 36 space equals space 1 over 12
    Grade-7
    Mathematics

    The sum or difference of 3 1 half plus left parenthesis negative 24.5 right parenthesis is ____________.

    STEP BY STEP SOLUTION
    3 1 half plus space left parenthesis negative 24.5 right parenthesis
    7 over 2 plus space left parenthesis negative 24.5 right parenthesis
3.5 space plus thin space left parenthesis negative 24.5 right parenthesis
3.5 space minus space 24.5
minus 21.5

    The sum or difference of 3 1 half plus left parenthesis negative 24.5 right parenthesis is ____________.

    MathematicsGrade-7
    STEP BY STEP SOLUTION
    3 1 half plus space left parenthesis negative 24.5 right parenthesis
    7 over 2 plus space left parenthesis negative 24.5 right parenthesis
3.5 space plus thin space left parenthesis negative 24.5 right parenthesis
3.5 space minus space 24.5
minus 21.5
    parallel
    Grade-7
    Mathematics

    Calculate the distance between 6.8 and 2.5 on the number line.

    We know that distance is the difference between the numbers.
    So, we will substract 2.5 from 6.8 to find the distance on the number line
    vertical line 6.8 minus 2.5 vertical line equals vertical line 4.3 vertical line equals 4.3 units.
    Hence, 4.3 is the distance between the above 2 units on the number line

    Calculate the distance between 6.8 and 2.5 on the number line.

    MathematicsGrade-7
    We know that distance is the difference between the numbers.
    So, we will substract 2.5 from 6.8 to find the distance on the number line
    vertical line 6.8 minus 2.5 vertical line equals vertical line 4.3 vertical line equals 4.3 units.
    Hence, 4.3 is the distance between the above 2 units on the number line
    Grade-7
    Mathematics

    The value of  7 open parentheses 3 over 4 close parentheses space plus space 7 open parentheses fraction numerator negative 4 over denominator 3 end fraction close parentheses is ___________

    STEP BY STEP SOLUTION
    7 open parentheses 3 over 4 close parentheses space plus space 7 open parentheses fraction numerator negative 4 over denominator 3 end fraction close parentheses space
M u l t i p l y i n g space 7 space i n space t h e space b r a c k e t
21 over 4 space plus space open parentheses fraction numerator negative 28 over denominator 3 end fraction close parentheses
T a k i n g space L C M space o f space 4 space & space 3 space left parenthesis i. e space 12 right parenthesis
fraction numerator 147 minus 196 over denominator 28 end fraction
space minus 49 over 12

    The value of  7 open parentheses 3 over 4 close parentheses space plus space 7 open parentheses fraction numerator negative 4 over denominator 3 end fraction close parentheses is ___________

    MathematicsGrade-7
    STEP BY STEP SOLUTION
    7 open parentheses 3 over 4 close parentheses space plus space 7 open parentheses fraction numerator negative 4 over denominator 3 end fraction close parentheses space
M u l t i p l y i n g space 7 space i n space t h e space b r a c k e t
21 over 4 space plus space open parentheses fraction numerator negative 28 over denominator 3 end fraction close parentheses
T a k i n g space L C M space o f space 4 space & space 3 space left parenthesis i. e space 12 right parenthesis
fraction numerator 147 minus 196 over denominator 28 end fraction
space minus 49 over 12
    Grade-7
    Mathematics

    Henry has 1 fourth acre of land. He plans to grow strawberries in 1 over 20 acre of land and tomatoes in 1 over 16 acre. The fraction of land he is left with is ____________.

    STEP BY STEP SOLUTION
    Total land = 1 fourth a c r e
    He plants strawberries = 1 over 20 a c r e
    He plants tomatoes =1 over 16 a c r e space
    To find the fraction of land left we will subrastract the land occupied by strawberries and tomatoes from the total land
    FRACTION OF LAND LEFT = TOTAL LAND - ( LAND OCCUPIED BY STRAWBERRIES + LAND OCCUPIED BY TOMATOES)
    1 fourth space minus space open parentheses 1 over 20 space plus space 1 over 16 close parentheses space equals space 1 fourth space minus space open parentheses fraction numerator 4 space plus space 5 over denominator 80 end fraction close parentheses space equals space fraction numerator 1 over denominator 4 space end fraction space minus space 9 over 80 space equals space fraction numerator 20 space minus space 9 over denominator 80 end fraction space equals space 11 over 80
    The fraction of land left = 11 over 80 a c r e s

    Henry has 1 fourth acre of land. He plans to grow strawberries in 1 over 20 acre of land and tomatoes in 1 over 16 acre. The fraction of land he is left with is ____________.

    MathematicsGrade-7
    STEP BY STEP SOLUTION
    Total land = 1 fourth a c r e
    He plants strawberries = 1 over 20 a c r e
    He plants tomatoes =1 over 16 a c r e space
    To find the fraction of land left we will subrastract the land occupied by strawberries and tomatoes from the total land
    FRACTION OF LAND LEFT = TOTAL LAND - ( LAND OCCUPIED BY STRAWBERRIES + LAND OCCUPIED BY TOMATOES)
    1 fourth space minus space open parentheses 1 over 20 space plus space 1 over 16 close parentheses space equals space 1 fourth space minus space open parentheses fraction numerator 4 space plus space 5 over denominator 80 end fraction close parentheses space equals space fraction numerator 1 over denominator 4 space end fraction space minus space 9 over 80 space equals space fraction numerator 20 space minus space 9 over denominator 80 end fraction space equals space 11 over 80
    The fraction of land left = 11 over 80 a c r e s
    parallel
    Grade-7
    Mathematics

    The value of  23 over 4 space minus space open parentheses negative 1 1 fourth close parentheses is __________.

    STEP BY STEP SOLUTION
    2 3 over 4 space minus space open parentheses negative 1 1 fourth close parentheses space
C o n v e r t i n g space m i x e d space f r a c t i o n space i n t o space s i m p l e space f r a c t i o n space
equals space 11 over 4 space plus space 5 over 4 space
T a k i n g space L C M space
equals space fraction numerator 11 space plus space 5 over denominator 4 end fraction space equals space 16 over 4 space equals space 4

    The value of  23 over 4 space minus space open parentheses negative 1 1 fourth close parentheses is __________.

    MathematicsGrade-7
    STEP BY STEP SOLUTION
    2 3 over 4 space minus space open parentheses negative 1 1 fourth close parentheses space
C o n v e r t i n g space m i x e d space f r a c t i o n space i n t o space s i m p l e space f r a c t i o n space
equals space 11 over 4 space plus space 5 over 4 space
T a k i n g space L C M space
equals space fraction numerator 11 space plus space 5 over denominator 4 end fraction space equals space 16 over 4 space equals space 4
    Grade-7
    Mathematics

    Simplify straight x space plus space open parentheses negative 8 over 9 close parentheses space equals space 13 over 4 space cross times space 1 third

    STEP BY STEP SOLUTION
    straight x space plus space open parentheses fraction numerator negative 8 over denominator 9 end fraction close parentheses space equals space 13 over 4 space cross times space 1 third
S o l v i n f space f o e space t h e space v a l u e space o f space x
straight x space equals space 13 over 4 space cross times space 1 third space plus space 8 over 9
F o l l o w i n g space t h e space B O D M A S S space r u l e space 1 s t space m u l t i p l y i n g space
equals space 13 over 12 space plus space 8 over 9
T a k i n g space L C M space o f space 12 space & space 9 space left parenthesis i. e space 36 right parenthesis
equals space fraction numerator 29 space plus space 32 over denominator 36 end fraction
equals space 71 over 36

    Simplify straight x space plus space open parentheses negative 8 over 9 close parentheses space equals space 13 over 4 space cross times space 1 third

    MathematicsGrade-7
    STEP BY STEP SOLUTION
    straight x space plus space open parentheses fraction numerator negative 8 over denominator 9 end fraction close parentheses space equals space 13 over 4 space cross times space 1 third
S o l v i n f space f o e space t h e space v a l u e space o f space x
straight x space equals space 13 over 4 space cross times space 1 third space plus space 8 over 9
F o l l o w i n g space t h e space B O D M A S S space r u l e space 1 s t space m u l t i p l y i n g space
equals space 13 over 12 space plus space 8 over 9
T a k i n g space L C M space o f space 12 space & space 9 space left parenthesis i. e space 36 right parenthesis
equals space fraction numerator 29 space plus space 32 over denominator 36 end fraction
equals space 71 over 36
    Grade-7
    Mathematics

    Calculate the distance between negative 4.2 and negative 2.2 on the number line.

    We know that distance is the difference between the numbers.
    So, we will substract the units mentioned in the question to find the distance
    vertical line minus 4.2 minus left parenthesis negative 2.2 right parenthesis vertical line equals vertical line minus 4.2 plus 2.2 vertical line equals space 2 spaceunits.
    Hence the distance between -4.2 and -2.2 is 2 units

    Calculate the distance between negative 4.2 and negative 2.2 on the number line.

    MathematicsGrade-7
    We know that distance is the difference between the numbers.
    So, we will substract the units mentioned in the question to find the distance
    vertical line minus 4.2 minus left parenthesis negative 2.2 right parenthesis vertical line equals vertical line minus 4.2 plus 2.2 vertical line equals space 2 spaceunits.
    Hence the distance between -4.2 and -2.2 is 2 units
    parallel
    Grade-7
    Mathematics

    Simplify 2 open parentheses negative 1 over 6 close parentheses space plus space 3 open parentheses negative 1 fifth close parentheses space plus space 4 open parentheses negative 1 third close parentheses

    STEP BY STEP SOLUTION
    2 open parentheses negative 1 over 6 close parentheses space plus space 3 open parentheses negative 1 fifth close parentheses space plus space 4 open parentheses negative 1 third close parentheses
    M u l t i p l y i n g space a n d space o p e n i n g space b r a c k e t s space
    fraction numerator negative 2 over denominator 6 end fraction space minus space 3 over 5 space minus space 4 over 3
T a k i n g space L C M space o f space 6 comma 5 space & 3 space left parenthesis i. e space 30 right parenthesis
space equals space fraction numerator negative 10 space minus space 15 space minus space 40 over denominator 30 end fraction space equals space fraction numerator negative 65 over denominator 30 end fraction space equals space fraction numerator negative 13 over denominator 6 end fraction space

    Simplify 2 open parentheses negative 1 over 6 close parentheses space plus space 3 open parentheses negative 1 fifth close parentheses space plus space 4 open parentheses negative 1 third close parentheses

    MathematicsGrade-7
    STEP BY STEP SOLUTION
    2 open parentheses negative 1 over 6 close parentheses space plus space 3 open parentheses negative 1 fifth close parentheses space plus space 4 open parentheses negative 1 third close parentheses
    M u l t i p l y i n g space a n d space o p e n i n g space b r a c k e t s space
    fraction numerator negative 2 over denominator 6 end fraction space minus space 3 over 5 space minus space 4 over 3
T a k i n g space L C M space o f space 6 comma 5 space & 3 space left parenthesis i. e space 30 right parenthesis
space equals space fraction numerator negative 10 space minus space 15 space minus space 40 over denominator 30 end fraction space equals space fraction numerator negative 65 over denominator 30 end fraction space equals space fraction numerator negative 13 over denominator 6 end fraction space
    Grade-7
    Mathematics

    An elevator descends into a mine shaft from 3 6 over 10 above the ground level. If it goes 8 6 over 10 m below the ground level, find the distance covered by the elevator.

    STEP BY STEP SOLUTION
    Lift distance above ground = 3 6 over 10
    Lift distance below the ground = -8 6 over 10
    To calculate the total distance covered by life we will simple add the the distance below the ground and distance above the ground
    Distance covered by lift = 3 6 over 10 plus open vertical bar negative 8 6 over 10 close vertical bar equals 3 6 over 10 plus 8 6 over 10 equals 3.6 plus 8.6 equals 12.2
    So the total distance covered by lift = 12.2 units

    An elevator descends into a mine shaft from 3 6 over 10 above the ground level. If it goes 8 6 over 10 m below the ground level, find the distance covered by the elevator.

    MathematicsGrade-7
    STEP BY STEP SOLUTION
    Lift distance above ground = 3 6 over 10
    Lift distance below the ground = -8 6 over 10
    To calculate the total distance covered by life we will simple add the the distance below the ground and distance above the ground
    Distance covered by lift = 3 6 over 10 plus open vertical bar negative 8 6 over 10 close vertical bar equals 3 6 over 10 plus 8 6 over 10 equals 3.6 plus 8.6 equals 12.2
    So the total distance covered by lift = 12.2 units
    Grade-7
    Mathematics

    The quotient of  negative 10 over 4 space divided by space minus 0.2 is _________.

    STEP BY STEP SOLUTION
    fraction numerator negative 10 over denominator 4 end fraction space divided by space minus 0.2
C o n v e r t i n g space d e c i m a l space i n t o space f r a c t i o n
space equals space fraction numerator negative 10 over denominator 4 end fraction space divided by space fraction numerator negative 2 over denominator 10 end fraction space
D i v i d i n g space
equals space fraction numerator negative 10 over denominator 4 end fraction space cross times space fraction numerator negative 10 over denominator 2 end fraction space equals space 100 over 8 space equals space 25 over 2

    The quotient of  negative 10 over 4 space divided by space minus 0.2 is _________.

    MathematicsGrade-7
    STEP BY STEP SOLUTION
    fraction numerator negative 10 over denominator 4 end fraction space divided by space minus 0.2
C o n v e r t i n g space d e c i m a l space i n t o space f r a c t i o n
space equals space fraction numerator negative 10 over denominator 4 end fraction space divided by space fraction numerator negative 2 over denominator 10 end fraction space
D i v i d i n g space
equals space fraction numerator negative 10 over denominator 4 end fraction space cross times space fraction numerator negative 10 over denominator 2 end fraction space equals space 100 over 8 space equals space 25 over 2
    parallel

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