Question

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Answer 1

Answer:

Q. 1.

Answer is 131 because

1+3+1 = 5. &

If digits at one and hundreds places are interchanged that are 1 the new number will be same

Q. 2.

Here arswer is 24

x = 3y

48 = 2*2*2*2*3

From these factors of 48 we get different values for x & y e.g.

If x =2 y = 6

If x =4 y =12 and so on....

But according to given options

If x = 8 y =24

Answer 2

$1.$

Lets write down all 3 digits number who's sum is '5'

1. .104
2. 113
3. 122
4. 131
5. 140
6. 203
7. 212
8. 221
9. 230
10. 302
11. 311
12. 320
13. 401
14. 410
15. 500

Out of these numbers let's pick prime numbers.

Prime numbers are those numbers which are divisible by only 1 & itself.

• 113
• 131
• 311
• 401

Now lets flip the 'hundreds' and 'ones' place

• 131
• 113
• 311
• 410

Definitely '410' is not a prime number after interchange digits,so we eliminate it aswell.Remaing:

• 131
• 113
• 311

Therefore, 131,113,311 are the required numbers.Since 131 is same as 113 when we interchange the digits.So,we can consider either of them.Final numbers are 113 & 311.

$2.$

'x' is 3 times of 'y'

$x=3y.....(1)$

48 is common multiple(lcm) of 'x' and 'y'

But,we know that

$\blue{\boxed{Product\:of\: two\:no.=lcm×HCF}}$

where,

• HCF(GCD)- is highest common factor or greatest common divisor.
• Here,let it be 'k'

$x×y=48×HCF$

From (1)

$3y×y=48×k$

$3y^{2}=48×k$

$y^{2}=16×k$

$\boxed{y=4\sqrt{k}}$

$x=3y$

$x=3(4\sqrt{k})$

$\boxed{x=12\sqrt{k})}$

By trial and error,

$\bf If\: k=1$

$y=4\sqrt{1}$

$y=4$

$x=12(\sqrt{1})$

$x=12$

This is not true at k=2 because LCM of 'x' & 'y' is not 48.(in this case LCM is 12)

$\bf If\: k=4$

$y=4\sqrt{4}$

$y=4×2$

$y=8$

$x=12(\sqrt{4})$

$x=12×2$

$x=24$

It is ture at k=4.Since LCM of 'x' & 'y' is 48.

Therefore,

$\orange{x=24}$

VERIFICATION:

$\blue{\boxed{Product\:of\: two\:no.=lcm×HCF}}$$8×24=LCM×4$

$192=4LCM$

$LCM=\frac{192}{4}$

$LCM=48$

HENCE VERIFIED And,

x=24

y=8

HOPE THIS HELPS!!