LCM and HCF

LCM and HCF

Easy Questions
Moderate Questions
Difficult Questions
LCM and HCF Tutorial

Aptitude

Simplification
Number System
Fractions
Elementary Algebra
LCM and HCF
Approximation
Unitary Method
Linear Equation
Quadratic Equation
Discount
Average
Surds and Indices
Percentage
Square root and cube root
Order of Magnitude
Work and Wages
Odd Man Out and Series
Algebra
Profit and Loss
Stocks and Shares
True Discount
Ratio, Proportion
Partnership
Alligation or Mixture
Time and Work
Pipes and Cistern
Speed, Time and Distance
Problem on Trains
Height and Distance
Banker's Discount
Boats and Streams
Races and games
Problems on Ages
Clocks and Calendars
Simple interest
Compound Interest
Sets and Functions
Area and Perimeter
Volume and Surface Area of Solid Figures
Sequences and Series
Plane Geometry
Logarithm
Probability
Permutation and Combination
Statistics
Mensuration
Trigonometry
Aptitude miscellaneous
  1. The LCM and the HCF of the numbers 28 and 42 are in the ratio :








  1. View Hint View Answer Discuss in Forum

    L.C.M. of 28 and 42

    L.C.M. of 28 and 42 = 2 × 2 × 7 × 3 = 84
    H.C. F. of 28 and 42

    ∴ H.C. F = 14

    Correct Option: A

    L.C.M. of 28 and 42

    L.C.M. of 28 and 42 = 2 × 2 × 7 × 3 = 84
    H.C. F. of 28 and 42

    ∴ H.C. F = 14

    Required ratio =
    84
    		 = 6 : 1
    14

  1. Two pipes of length 1.5 m and 1.2 m are to be cut into equal pieces without leaving any extra length of pipes. The greatest length of the pipe pieces of same size which can be cut from these two lengths will be








  1. View Hint View Answer Discuss in Forum

    Maximum length of each piece = HCF of 1.5 metre and 1.2 metre
    First of all we find HCF of 1.5 metre and 1.2 metre

    ∴ HCF of 1.5 and 1.2 metre = 0.3 metre

    Correct Option: C

    Maximum length of each piece = HCF of 1.5 metre and 1.2 metre
    First of all we find HCF of 1.5 metre and 1.2 metre

    ∴ HCF of 1.5 and 1.2 metre = 0.3 metre
    Required Maximum length of each piece = 0.3 metre

  1. In a school, 391 boys and 323 girls have been divided into the largest possible equal classes, so that each class of boys numbers the same as each class of girls. What is the number of classes ?








  1. View Hint View Answer Discuss in Forum

    First of all we find HCF of 391 and 323.

    Correct Option: D

    First of all we find HCF of 391 and 323.

    ∴ Number of classes = 17

  1. Find the greatest number that will divide 390, 495 and 300 without leaving a remainder.








  1. View Hint View Answer Discuss in Forum

    Required number = HCF of 390, 495 and 300
    Firstly , we find HCF of 390, 495 and 300

    Correct Option: B

    Required number = HCF of 390, 495 and 300
    Firstly , we find HCF of 390, 495 and 300


    HCF of 15 and 300 = 15
    Hence , Required number = 15

  1. The product of two 2–digit numbers is 2160 and their H.C.F. is 12. The numbers are








  1. View Hint View Answer Discuss in Forum

    HCF of numbers = 12
    Let the numbers be 12p and 12q where p and q are co–prime.
    According to the question,
    12p × 12q = 2160

    ⇒ pq =
    2160
    		 = 15
    12 × 12

    pq = 3 × 5 or 1 × 15
    Here , p = 3 , q = 5

    Correct Option: C

    HCF of numbers = 12
    Let the numbers be 12p and 12q where p and q are co–prime.
    According to the question,
    12p × 12q = 2160

    ⇒ pq =
    2160
    		 = 15
    12 × 12

    pq = 3 × 5 or 1 × 15
    Here , p = 3 , q = 5
    ∴ Required numbers = 12 × 3 = 36 and 12 × 5 = 60