A container of milk was 4/5 full. When 12 bottles of milk were taken out and 8 bottles of milk were poured into it, it was 3/4 full. How many bottles of milk can the container contain?

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

Solution 1.
Let the question solved correctly by Mohan = X ;
and number of question solved wrongly by Mohan = 30 - X ;
According to the question,
3X - (30 - X) x 2 = 40
? 3x - 60 + 2x = 40
? 5x - 60 = 40
? 5x = 40 + 60 ? x = 100/5 = 20
? Mohan attempted 20 questions correctly .

Solution 2.
Let the question solved correctly by Mohan = X ;
and number of question solved wrongly by Mohan = Y ;
According to given question;
X + Y = 30 ; -----------i
3X - 2Y = 40 ;-------ii

Now multiply the first equation by 2;
we will get
2X + 2Y = 60 ;--------iii
After Add the equation ii and iii , we will get
3X - 2Y + 2X + 2Y = 40 + 60 ;
? 5X = 100 ;
? X = 20;

? Mohan attempted 20 questions correctly .

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

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

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

Hint 1.
Let consider the monthly income = x ;
Amount spend on House rent = 2x/5;
Amount spend on food = 3x/10;
Amount spend on conveyance = x/8;
After Expenditure , The left amount = 2800;

According to Question;
x - ( 2x/5 + 3x/10 + x/8 ) = 2800 ;
x- ((16x + 12x + 5x)/40) = 2800 ;
x - 33x/40 = 2800 ;
(40x - 33x)40 = 2800 ;
7x/40 = 2800 ;
7x = 2800 x 40 ;
x = 2800 x 40/7 ;
x = 400 x 40 ;
x = 16000 ;
Expenditure on food = 3x/10 = 3 x 16000 /10= ? 4800
Expenditure on conveyance = x/8 = 1 x 16000/8 = ? 2000
? Expenditure on food and conveyance together
=? (4800 + 2000) = ? 6800



Solution 2 .

part of income left
= 1 - (2/5 + 3/10 + 1/8 )
=1- (16 + 12 + 5/40)
= 1- 33/40 = 40 - 33/40 = 7 /40
Let the monthly income = x
According to the question,
7/40 x = 2800
? x = 2800 x 40/7
= 400 x 40 = ? 16000
Expenditure on food = 3/10 x 16000 = ? 4800
Expenditure on conveyance = 1/8 x 16000 = ? 2000
? Expenditure on food and conveyance together
=? (4800 + 2000) = ? 6800

Let share of each friend = a.
Then, share of each daughter = 4a.
and share of each son = 5a.
According to the question,
5 x 5a + 4 x 4a + 2 x a = 17200
? 25x + 16x + 2x = 17200
? 43x = 17200
? x = 17200/43 = 400
? Share of each daughter
= 4x = 4 x 400
= ? 1600

Let at present, there are c men, then number of women = 5c
Before the men had left
number of men = c + 45 and
number of women = 5c
? c + 45 = 2 x 5c
? 9c = 45
? c = 45/9 = 5
Hence, number of women in the beginning
= 5c + 15 = 5 x 5 + 15
= 25 + 15 = 40
? Double of the women = 40 x 2 = 80