Eureka!

If n=gold then mass of gold in water is n-(n/20)

if x=silver then mass of silver in water is x-(x/10)

When the crown was weighed under water it mass was

$(n-n/20)+(x-x/10)=9.25$ and you can simplify it to $19n+18x=185....(1)$

As king ordered that the sum mass of crown is made of 8kg of gold and 2kg of silver , gold smith replaced the gold mass with silver mass however the sum is 10kg. so..

$x+n=10....(2)$

Combine (1) and (2) by elimination hence you get x=5kg and n=5kg

King ordered 8kg of gold and 2kg of silver. So much of gold that stolen in kg was $8-5=3kg$

$((8+2)-(\frac{8}{20}+\frac{2}{10})-9.25)\cdot 20=3$

If x=gold weight and y=silver weight

The weight of the Crown must be 10kg, so x+y=10

Under the water, the weight is 9.25kg, so (19x/20) + (9y/10) = 9.25

Solving the system of equations x=5 y=5

Initially, the gold weight was 8kg, so 8 - 5= 3

Multiply 8kg to 1/20th this results in .4 use this answer and subtract it to 8kg and you will get 7.6 the weight of gold underwater Do the EXACT same thing to the other problem Multiply 2kg to 1/10th Result= .2 Subtract the product to 2kg Result after subtraction= 1.8 Add the end answers 1.8 plus 7.6 use this and subtract it to 9.25 you will get the lost gold amount which the difference between the two is .15 1 gold times 1/20= .5 using this number .5 I divided .15 and I get 3 it's a long method of trying to figure this out without using algebra

Gold= A Silver= B

A + B= 10

A/20+B/10= (10-9,25)

A= 10 -B

A + 2B = 0,75

10 - B +2B = 15

B= 5 A = 5 A has become 5 while at the beginning was 8. So the smith has stolen 3 kg of gold.

Let the weigh of Gold and Silver be G and S. We have 2 equations: G + S = 10 and 0.95G + 0.9S = 9.25 Therefore G = 5. So the result is 8 - 5 = 3

according to order gold to be taken 8 kg while silver 2kg.. as goldsmith replaced some amount of gold with silver so let GOLD=G and silver =S than S=2+s and G=8-s .....(1)

(1/20)G+(1/10)S=10kg-9.25kg

(1/20)(8-s)+(1/10)(2+s)=0.75kg

solving the above addition...

(12+s)/20=0.75kg

12+s=15kg

s=3kg

by putting the value of s in equation no. (1)

S=2+s and G=8-s

S=2+3 and G=8-3

S=5 and G=5

than 8kg-5kg= 3KG

(8-k)9/10+(2+k)19/20=9.25, hence the value of k is 3, said the sherlock holmes

Let $x$ be amount he stole.

$8-x$ becomes new amount of gold.

New amount of silver equals $2 + x$ .

Weight of gold under water is $1/20th$ less than its normal weight: $8-x - ((8-x)* 1/20)$ .

Weight of silver under water is $1/10th$ less than its normal weight: $2 + x - ((2+x) * 1/10)$ .

Adding up the weights, we get: $8-x - ((8-x) * 1/20) + 2 + x - ((2+x) * 1/10)$ . The combined weights are equal to $9.25$ .

$8-x-((8-x) * 1/20)+2+x-((2+x)*1/10) = 9.25.$

$8-x-((2/5) - (x/20))+2+x-((1/5) + (x/10)) = 9.25.$

$8-x-(2/5)+(x/20)+2+x-(1/5)-(x/10)=9.25.$

$(38/5)-(19x/20)+(9/5)+(9x/10)=9.25.$

$(47/5)-(x/20)=37/4.$

$-x/20 = (37/4)-(47/5).$

$-x/20 = (185/20)-(188/20).$

$-x/20 = -3/20.$

Comparing numerators,

$- x = -3$

$x = 3$

Let S be the weight of silver , and g the weight of gold. the crown lost 0.75kg of his weight when put under water , we know that gold loses 1/20 of its weight and silver loses 1/10 of it's weight (under water), the equation will be 1/20g +1/10s=0.75kg ( the amount lost under water ) and we know that g+s=10kg so, | 1/20g+1/10s=0,75 | g+s=10 ==> g=5 s=5. The king orderd a crown made with 8kg of gold and 2kg of silver so amount that golden smith has stolen is 8-5=3kg

Let $x$ be the amount of gold. Then the amount of weight lost is:

$\frac x{20}+\frac{10-x}{10}=\frac34$

Solution is $x=\boxed3$

kind of ugly but I solved it in this manner Crown = 10kg = 8-x+2+y, x=y

9.25 = (8-x) -((8-x)/20) + (2+y)-((2+y)/10)

y= (-3+21x)/22

10 - x -(3+21x)/22

x = -3, so 3 kg were stolen

Let: gold's weight = x and silver's weight = y

$x + y = 10, y = 10 - x$

$$

(gold's weight - $\frac{1}{20}$ gold's weight) + (silver's weight - $\frac{1}{10}$ silver's weight) = crown's wieght in the water

$x - 0.05x + (10 - x) - 0.1(10 - x) = 9,25$

$0.05x =0.25$

$x = 5$

$$

$x_0 = 8 kg$ and $x = 5 kg$ , so that: $\Delta x = 8 -5 = \boxed{3}$ $kg$

Let x be weight of gold in the crown made y be weight of silver Then x+y=10 Now in water (19x/20)+(18y/20)=9.25 Solving the two equations we get x= 5 And so gold taken away=8-3 = 5 kg Its a easy (basic) question of linear equation.

If 'x' is the weight(in Kg.) of the gold taken out and replaced by silver, then the weight of gold in water=1- weight of gold lost in water=1-(1/20)=19/20, similarly the weight of silver in water=1- weight of silver lost in water=1-(1/10)=9/10 [(19/20)(8-x)]+[(9/10)(2+x)]=9.25 On solving we get x=3

if "g" for gold and "s" for silver :

(g/20) + (s/10) = (10-9.25)=3/4 (eq.1)

g + s = 10 (eq.2)

summing equation 1 and 2, and we will get : g= 5 and s = 5 so, the gold that stolen by the gold smith is : 8 - 5 = 3 kg

king ordered that the crown should be made by 8kg of gold and 2 kg of silver bur the gold smith took away some gold and replaced it by an EQUAL number of silver so first the silver was 2kg but if we remove 3kg of gold from 8kg and add 3kg of silver to 2kg then we have equal amount of both.

let gold use was G kg and silver used was S kg initially, as crown was 10 kg, G+S=10,....................(EQ1) NOW, under the water, gold lost its 1/20th weight and silver lost 1/10th weight and hence the crown weighed 9.25 kg hence,
( G-G/20)+(S-S/10)=9.25...............(EQ2) ON SOLVING EQ1 and EQ2, we get G=5 and S=5.
SO 3Kg gold was stolen

Let gold stolen be x kg. Then new mass of gold in crown= (8-x)kg, and that of silver=(2+x)kg So mass of crown underwater= (8-x) + (2+x) - (8-x)/20 - (2+x)/10 = 9.25 Solving the above equation, we get x=3 kg, which is the amount of gold stolen.