x = 10

x = 5

Let Sahil's monthly salary be Rs. x
Then, Suraj's monthly salary = Rs. 3x
After giving Rs.  6000 to Sahil, Sahil has money
= x + 6000 and Suraj has money = 3x - 6000
Then, according to the question,
2(x + 6000) = (3x -6000)
\(\Rightarrow\) 2x + 12000 = 3x - 6000
\(\Rightarrow\) 3x - 2x =12000+ 6000 =18000
x =18000
So, Sahil's monthly salary = Rs.18000
and Suraj's monthly salary = Rs. 54000.
(b) The value depicted by Suraj is their helpful nature. He helps his friend in the need

Let the number of gold coins initially be x 
Then, the number of non-gold coins be 3x. When,
10 more gold coins added
Then, the total number of gold coins = (10 + x)
Then, according to the question , \(\frac { (10+x) }{ 3x } =\frac { 1 }{ 2 } \)
\(\Rightarrow\) 2 (10 + x) = 3x  \(\Rightarrow\) 20 + 2x = 3x \(\Rightarrow\) x = 20
Then, total number of coins at last = 3x + 10 + x  = 4x + 10 = 4 x 20 + 10 = 90

Let box B's weight be x kg.
Since, box A weights \(2\frac { 1 }{ 2 } \) kg more than box Band  box C weights \(10\frac { 1 }{ 4 } \) kg more than box B.
Weight of box A = \(\left( x+2\frac { 1 }{ 2 } \right) kg=\left( x+\frac { 5 }{ 2 } \right) kg\)
Weight of box B = \(\left( x+10\frac { 1 }{ 4 } \right) kg=\left( x+\frac { 41 }{ 4 } \right) kg\)
Total weight of all the boxes
\(\left( x+\frac { 5 }{ 2 } +x+x+\frac { 41 }{ 4 } \right) kg\)
According to the question,
Total weight = \(48\frac { 3 }{ 4 } kg=\frac { 195 }{ 4 } kg\)
\(\therefore\)  \(x+\frac { 5 }{ 2 } +x+x+\frac { 41 }{ 4 } =\frac { 195 }{ 4 } \)
\(\Rightarrow\) 4x + 10+ 4x + 4x + 41 = 195  [multiplying both sides by 4]
\(\Rightarrow\) 12x + 51 = 195 \(\Rightarrow\) 12x = 144 \(\Rightarrow\) x = 12
\(\therefore\) weight box
A = \(\left( 12+\frac { 5 }{ 2 } \right) kg=\frac { 29 }{ 2 } kg=14\frac { 1 }{ 2 } kg\)

Let the numerator of the number be x. Then, denominator of the number be (x - 4)
So, fraction \(\frac { x }{ x-4 } \)
According to the question, if 6 is added to numerator, it becomes thrice the denominator
\(\therefore\)  \(\frac { x+6 }{ x-4 } =\frac { 3(x-4) }{ x-4 } \Rightarrow \frac { x+6 }{ x-4 } =3\)
\(\Rightarrow\)  3x -12 = x + 6 [by cross-multiplication]
\(\Rightarrow\)  \(2x18\quad \Rightarrow x=9\)
Put x = 9 in Eq. (i), we get
Fraction = \(\frac { x }{ x-4 } =\frac { 9 }{ 9-4 } =\frac { 9 }{ 5 } \)

Let Abdul buys 2x m of trouser material 
Then, the shirt material bought by him = 3x m
Sale price of 1 m of trouser material
Rs (90 + 12% of 90)
\(\left( 90+\frac { 12\times 90 }{ 100 } \right) =Rs.100.80\)
Sale price of 2x m of trouser material = Rs.(2xx 100.80) = Rs. 201.60x
Sale price of 1m of shirt material
Rs. 50 +20% of Rs 50 = \(\left( 50+\frac { 20\times 50 }{ 100 } \right) =Rs.60\)
Sale price of 3x m of shirt material = Rs 3x x 60
= Rs. 180x
Total sale = Rs. (201.60+180) x = Rs. 381.60x
\(\therefore\)  381.60 = 38160
\(x=\frac { 38160 }{ 38.160 } =100\)
So,Abdul bought 2 x 100 = 200 m of trouser material.

Value of first Prize = Rs.60, Value of second Prize = Rs. 50, value of third prize = Rs.40

28 yr

Let the number of Rs. 5 coins be x.
Then, the number of Rs. 2 coins = 3x
The total number of coins is 160 .
The number of coins of Rs. 1= 160 - (x + 3x)
= (160 - 4x)
The amount that I have from Rs.  5 coins = 5 x x = 5x
The amount that I have from Rs.  2 coins = 2 x 3x = 6x
The amount that I have from Rs.  1 coins
= 1 x (160 - 4x) = 160 - 4x
According to the question,
Total amount = 300
\(\Rightarrow\) 5x + 6x + (160 - 4x) = 300
\(\Rightarrow\) 5x +6x +160 - 4x = 300
\(\Rightarrow\) 7x + 160 = 300   [transposing 160 to RHS]
\(\Rightarrow\) 7x = 300 -160  [transposing 160 to RHS]
\(\Rightarrow\) 7x = 140
\(\Rightarrow\) x = \(\frac { 140 }{ 7 } \) = 20   [dividing both sides by 7]
Number of Rs. 5 coins = x = 20
Number of Rs. 2 coins = 3x = 3 x 20 = 60
and number of Rs.1 coins = 160 - 4x = 160 - 4 x 20
= 160 - 80 = SO
Hence, I have SO,60 and 20 coins of denomination Rs.1, Rs. 2 and Rs 5, respectively.