Question

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SOLVE this attachment question

NOTE >Solve both questions

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Answer 1

$<b > here is your ans$

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$<b> question 1$ ⤵

Let 2x & 3x be the numbers their L.C.M will be 6x

But their L.C.M is given = 48

⇒6x = 48

⇒x = 8

So, the numbers are 2 × 8 i.e 16 & 3 × 8 i.e 24

Hence the sum of numbers = 16 + 24 = 40

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$<b> question 2$⤵

First of all we have to take the L.C.M. of 16, 18 and 20

Prime factorization of 16 = 2 × 2 × 2 × 2

Prime factorization of 18 = 2 × 3 × 3

Prime factorization = 20 = 2 × 2 × 5

= 2 × 2 × 2 × 2 × 3 × 3 × 5

L.C.M. of 16 , 18 and 20 = 720

Now , the L.C.M. of 16 , 18 and 20 is 720

So, the required number will be in the form of (720 × x) + 4

Now , we have to apply the hit and trial method to find the least value of x for

which (720 × x) + 4 is divisible by 7 ... by putting x = 1, 2, 3, 4.........n.

First by putting x = 1

⇒ (720 × 1) + 4

⇒ 720 + 4

⇒ 724

724 is not divisible by 7.

Now, putting x = 2

⇒ (720 × 2) + 4

⇒ 1440 + 4

⇒ 1444

1444 is also not divisible by 7.

Now, putting x = 3

⇒ (720 × 3) + 4

⇒ 2160 + 4

⇒ 2164

2164 is also not divisible by 7

Now, putting x = 4

⇒ (720 × 4) + 4

⇒ 2880 + 4

⇒ 2884

2884 is exactly divisible by 7

So , for the value of x = 4 , the required number comes 2884

2884 is the least number which when divided by 16 , 18 and 20 leaves a remainder 4 in each case but exactly divisible 7

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$<marquee > hope it helps u$