Mohan gets 3 marks for each correct sum and loses 2 marks for each wrong sum. He attempts 30 sums and obtains 40 marks. The number of sums solved correctly is

? = 683.46 - 227.39 - 341.85
= 683.46 - 569.24 = 114.22

8648 - 7652 = ? x 40
996 = ? x 40 ? ? = 996/40 = 24.9

Multiply the given in question.
The answer would be
? = 10.8 x 5.5 x 8.4 = 498.96

Given expression
= 6 - [9 - {18 - (15 -12 - 9)}]
= 6 - [9 - {18 - (15 - 12 + 9)}]
= 6 - [9 - {18 - 12}]
= 6 - [9 - 6] = 6 - 3 = 3

We know that,
a³ + b³ + c³ - 3abc
= (a + b + c) 1/2 [(a - b) ² + (b - c) ² + (c - a) ²]
= (225 + 226 + 227) 1/2 [1 + 1 + 4]
= 678 x 3 = 2034

Let the maximum marks = x.
According to the question,
40x/100 = 200 + 8
? x = 208 x 10/4 = 520

Let Rahul's marks in Mathematics = x
and in Hindi = y
According to the question,
1/3 x - 1/2y = 30
? 2x - 3y = 180 ...(i)
Also, given
x + y = 480 ...(ii)
By solving Eqs. (i) and (ii), we get
x = 324
and y =156

Let x bottles can fill the container completely.
According to the questions,
4/5 x - 3/4 x = (12 - 8)
? 16x - 15x/20 = (12 - 8)
? x/20 = 4
? x = 80
? Required number of bottles = 80.

Let the ten's place digit = x
and unit's place digit = y
? The original number = 10x + y
After interchanging the digits
Number will become 10y + x
? according to the given question,
y + x = 13 and
(10y + x ) - (10x + y) = 27
? 9y - 9x = 27
? y - x = 3 ................(i)
and y + x = 13 .................(ii)
On solving Eqs.(i) and (ii) , we get
y = 8 and x = 5
? Required number is
10x + y = 10 x 5 + 8 = 58

Let first, second and third numbers be x , y and z, respectively.
Then , xy = 42.................(i)
yz = 56 ..................(ii)
xz = 48 ....................(iii)
Multiplying Eqs ( i), (ii) and (iii), we get
(xyz)² = 42 x 56 x 48
? (xyz)² = 112896
? xyz = 336 ................(iv)
Dividing Eq. (iv) by Eq. (i), we get
xyz/xy = 336/42 ? z = 8