Practice LCM & HCF

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Find the HCF of \[336, \ 240\] and \[96\]

\[40\]

\[42\]

\[46\]

\[48\]

\[\therefore\] HCF of \[336, 240, 96 = 48\]

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Find the LCM of \[15, \ 20, \ 36\] and \[48\]

\[480\]

\[540\]

\[680\]

\[720\]

LCM \[= 2 × 2 × 3 × 5 × 3 × 4\]

\[ = 720\]

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HCF of \[\frac{2}{3},\ \frac{3}{4}, \ \frac{6}{7}\] is :

\[\frac{1}{100}\]

\[\frac{1}{150}\]

\[\frac{2}{100}\]

\[\frac{2}{150}\]

HCF of \[\frac{2}{3},\ \frac{3}{4}, \ \frac{6}{7}\]

\[=\frac{HCF\ \ of\ \ 2,4,6}{LCM\ \ of\ \ 3,5,7}\]

\[=\frac{2}{150}\]

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LCM of \[\frac{2}{3},\frac{4}{9},\frac{5}{6}\] is

\[\frac{10}{3}\]

\[\frac{20}{3}\]

\[\frac{3}{10}\]

\[\frac{3}{20}\]

LCM of \[\frac{2}{3},\ \frac{4}{9}, \ \frac{5}{6}\]

\[=\frac{LCM\ \ of\ \ 2,4,5}{HCF\ \ of\ \ 3,9,6}\]

\[=\frac{20}{3}\]

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The LCM of two numbers is \[864\] and their HCF is \[144\]. If one of the number is \[288\], the other number is :

\[430\]

\[432\]

\[434\]

\[436\]

Required number

\[=\frac{LCM\times HCF}{First\ Number}\]

\[=\frac{864\times144}{288}\]

\[=432\]

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Four bells ring at the intervals of \[5 , 6 , 8\] and \[9\] seconds. All the bells ring simultaneously at some time. They will again ring simultaneously after

6 minutes

12 minutes

18 minutes

24 minutes

The LCM of \[5, 6, 8\] and \[9\]

\[= 360\] seconds

\[= 6\] minutes

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The traffic lights at three different road crossings change after \[24\] seconds, \[36\] seconds and \[54\] seconds respectively. If they all change simultaneously at \[10 : 15 : 00\] AM, then at what time will they again change simultaneously?

\[10 : 16 : 30\] a.m.

\[10 : 18 : 32\] a.m.

\[10 : 16 : 30\] a.m.

\[10 : 18 : 36\] a.m.

LCM of \[24, 36\] and \[54\] seconds

\[= 216\] seconds

\[= 3\] minutes \[36\] seconds

\[\therefore\] Required time

\[= 10 : 15 : 00 + 3\] minutes \[36\] seconds

\[= 10 : 18 : 36\] a.m.

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The smallest perfect square divisible by each of \[6, 12\] and \[18\] is

\[196\]

\[144\]

\[108\]

\[36\]

The LCM of \[6, 12\] and \[18\]

\[= 36\]

\[= 6^2\]

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The least number which when divided by \[4, 6, 8, 12\] and \[16\] leaves a remainder of \[2\] in each case is

L.C.M. of \[4, 6, 8, 12\] and \[16 = 48\]

\[\therefore\] Required number

\[= 48 + 2 = 50\]

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Find the greatest number of five digits which when divided by \[3, 5, 8, 12\] have \[2\] as remainder :

99999

99980

99960

99962

The greatest number of five digits is \[99999\].

LCM of \[3, 5, 8\] and \[12\].

LCM \[= 2 × 2 × 3 × 5 × 2 = 120\]

After dividing \[99999\] by \[120\], we get \[39\] as remainder

\[99999 – 39 = 99960\]

\[= (833 × 120) = 99960\]

\[99960\] is the greatest five digit number divisible by the given divisors.

In order to get \[2\] as remainder in each case we will simply add \[2\] to \[99960\].

\[\therefore\] Greatest number

\[= 99960 + 2\]

\[= 99962\]

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Which is the least number which when doubled will be exactly divisible by \[12, 18, 21\] and \[30\] ?

\[320\]

\[430\]

\[520\]

\[630\]

The LCM of \[12, 18, 21, 30\]

LCM \[= 2 × 3 × 2 × 3 × 7 × 5 \]

\[= 1260\]

\[\therefore\] The required number

\[=\frac{1260}{2}=630\]

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The product of two numbers is \[2160\] and their HCF is \[12\]. Number of such possible pairs is

\[1\]

\[2\]

\[3\]

\[4\]

HCF \[= 12\]

Numbers \[= 12 x\] and \[12 y\]

where \[x\] and \[y\] are prime to each other.

\[\therefore 12 x \times 12 y = 2160\]

\[\Rightarrow xy=\frac{2160}{12\times12}\]

\[=15 = 3 × 5, 1 × 15\]

Possible pairs \[= (36, 60)\] and \[(12, 180)\]

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The maximum number of students among whom \[1001\] pens and \[910\] pencils can be distributed in such a way that each student gets same number of pens and same number of pencils, is

910

1001

1911

Maximum number of students

\[=\] The greatest common divisor

\[=\] HCF of \[1001\] and \[910\]

\[=91\]

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84 Maths books, 90 Physics books and 120 Chemistry books have to be stacked topicwise. How many books will be there in each stack so that each stack will have the same height too ?

As the height of each stack is same, the required number of books in each stack

\[=\] HCF of \[84, 90\] and \[120 \]

\[84 = 2 × 2 × 3 × 7\]

\[90 = 2 × 3 × 3 × 5\]

\[120 = 2 × 2 × 2 × 3 × 5\]

HCF \[ = 2 × 3 = 6\]

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Two numbers are in the ratio \[3 : 4\]. If their HCF is \[4\], then their LCM is

Numbers \[= 3 x\] and \[4 x\]

HCF \[= x = 4\]

LCM \[= 12 x = 12 × 4 = 48\]

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The ratio of two numbers is \[4 : 5\] and their H.C.F. is \[8\]. Then the larger number is

\[10\]

\[20\]

\[30\]

\[40\]

Let the numbers be \[4 x\] and \[5 x\] .

H.C.F. \[= x = 8\]

\[\therefore\] Smaller number \[= 4x = 4 \times 8 = 32\]

and Larger number \[= 5x = 5 \times 8 = 40\]

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The ratio of two numbers is \[4 : 5\] and their L.C.M. is \[120\]. The numbers are

\[10, 12\]

\[12, 36\]

\[24, 30\]

\[32, 36\]

Suppose the numbers are \[4x\] and \[5x\] respectively

According to question

\[x × 4 × 5 = 120\]

\[x = 6\]

Required numbers

\[= 4 × 6 = 24\]

and

\[5 × 6 = 30\]

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The product of the LCM and HCF of two numbers is \[24\]. The difference of the two numbers is \[2\]. Find the numbers ?

\[1\] and \[2]\

\[2\] and \[4]\

\[4\] and \[6]\

\[6\] and \[8]\

Let the numbers be \[x\] and \[( x + 2)\].

Product of numbers \[= HCF × LCM \]

\[\implies x ( x + 2) = 24 \]

\[\implies x^2 + 2 x – 24=0 \]

\[\implies x^2 + 6 x – 4 x – 24=0 \]

\[\implies x ( x + 6) – 4 ( x + 6)=0 \]

\[\implies ( x – 4) ( x + 6)=0 \]

\[\implies x = 4\], as \[x \neq – 6=0 \]

Numbers are \[4\] and \[6\].

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The least number to be subtracted from \[36798\] to get a number which is exactly divisible by \[78\] is

When \[36798\] is divided by \[78\], remainder \[= 60\]

\[\therefore\] The least number to be subtracted \[= 60\]

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Find the greatest number which will divide \[42, 49\] and \[56\] and leave remainders \[6, 7\] and \[8\] respectively.

Take the HCF of :

\[(42 - 6) = 36\]
\[(49 - 7) = 42\]
\[(56 - 8) = 48 \]

And, HCF of \[6\] and \[48\] is also \[6\].

So, the required greatest number will be \[6\].