A can contains a mixture of two liquids A and B in proportion 7:5. When 9 litre of mixture are drawn off and the can is filled with B, the proportion of A and B becomes 7:9. How many litres of liquid A was contained by the can initially.

M4maths Previous Puzzles

19 Aug 2015 Mixture A can contains a mixture of two liquids A and B in proportion 7:5. When 9 litre of mixture are drawn off and the can is filled with B, the proportion of A and B becomes 7:9. How many litres of liquid A was contained by the can initially.: OPtion
1) 25
2) 16
3) 15
4) 20
5) 22
6) 18
7) 10
8) 24
9) 21
10)None of these; Correct Option: 9 Best Solution (81)

sudiptaa 9 years AGO

IR = 7:5
FR= 7:9

Cross Product = 63-35 = 28
Common factor of first ratio = 9/12 +9*7/28=3
A = 7*3 =21 litres
B= 5*3= 15 litres

Like? Yes (94) | No (37)

sanjeev kumar 9 years AGO

Initial
A=7x ,B=5x , Mix.=12x
After removing 9 lit. n filled by B mix.=12x-9
A=7/12(12x-9) ,B=5/12(12x-9)+9
A:B=7:9
solving these x=3
Initial A=7x=7*3=21 Liter

Like? Yes (45) | No (7)

vivek kr. chatterjee 9 years AGO

since initially ratio was 7:5, so just for hit and trial ,take any option which is multiple of 7, such option is only 21

take that value for now, do the calculation and u will it satisfies the conditions.

Like? Yes (43) | No (19)

vimal tiwari 9 years AGO

initially 7/12 prts is A

after addition become 7/16

let M be total amount than...

(7/12)/(7/16)=(1-(9/M))
M = 36 litres

so A initially was. 7/12 of 36 litre i.e. 21 litres

Like? Yes (28) | No (12)

Chandrajeet Yadav 9 years AGO

let A = 7x & B = 5x
total mixture = 7x+5x=12x

after 9 litre of mixture are drawn off remaining misture = 12x-9

after filling can with B:
A = (12x-9) * 7/12= 28x-21
B = {(12x-9)* 5/12}+9= 20x+21

So, (28x-21)/(20x+21)= 7/9
by solving above equation, we get
x=3

So liquid A was contained by the can initially = 7x = 7*3 = 21

So Ans = option (9)

Like? Yes (27) | No (12)

ABHINAV PUSHPAM 9 years AGO

Suppose the can initially contains 7x and 5x litres of mixtures A and B resp.
Quantity of A in mixture left
= (7x - 7/12 x 9) litres = (7x - 21/4) litres.
Quantity of B in mixture left
= (5x - 5/12 x 9) litres = (5x - 15/4) litres.
(7x - 21/4) / [(5x - 15/4)+9] = 7/9 = › 28x - 21/20x + 21 = 7/9 =› 252x - 189 = 140x + 147
=› 112x = 336 =’ x = 3.
So, the can contained 21 litres of A.
option 9)

Like? Yes (9) | No

Abhishek Anand 9 years AGO

Suppose the can initially contains 7x and 5x of mixtures A and B respectively
Quantity of A in mixture left= (7x - 7/12 * 9) = (7x - 21/4)ltr
Quantity of B in mixture left = (5x - 5/12 *9) = (5x - 15/4)ltr

so, (7x-21/4) / (5x -15/4) + 9 = 7/9

28x-21/20x+21 = 7/9
on solving for x
we get 112x=336
x=3
So, the can contained 21 litres of A.

Like? Yes (6) | No

bhaskar 9 years AGO

(7x-(7*9/12))/(5x-(5*9/12)+9)=7/9
x=3
7x=21

Like? Yes (7) | No (1)

shreya saklecha 9 years AGO

7x/(5x+9)=7/9
X=9/4
=>[[9/4*12]+9]*7/12=21

Like? Yes (25) | No (20)

kondapuram tirumalreddy 9 years AGO

X/Y=7/5 - RATIO OF A to B
5X-7Y=0......(1)
9 litres would have : 7/12*9=5.25 of A and 9-5.25=3.75 of B

New ratio: (X-5.25)/(Y-3.75+9)=7/9
9X-7Y=16*5.25 ....... (2)

Solve eq 1 and 2 : answer is 21 for x

Like? Yes (4) | No

Tinku 9 years AGO

Suppose the can initially contains 7x and 5x of mixtures A and B respectively.

Quantity of A in mixture left = (7x - 7/12 x 9 litres
=( 7x - 21/4) litres.

Quantity of B in mixture left = (5x - 5/12 x 9 ) litres
=( 5x - 15/4 ) litres.

Since ( 7x - 21/4) / ( 5x - 15/4 ) + 9 = 7/ 9

= 28x -21 / 20x + 21 = 7 / 9

=> 252x -189 = 140x + 147

=> 112x 336

=> x =3

So, can contained 21 litres of A.

Like? Yes (4) | No

sahil arora 9 years AGO

simple but tricky question just chck the options which 1 is dividible by 7

Like? Yes (10) | No (6)

guru teja 9 years AGO

consider 7x and 5x of mixtures A and B.
A's quantity left 7x-21/4 liters
B's quantity left 5x-15/4 liters
x = 3.
So, the can contained 21 litres of A.

Like? Yes (3) | No

ritu yadav 9 years AGO

let liquid A be 7x and B be 5x
when 9 litres are drawn off and liquid B is added
then , remaining liquid A = 7x-(7/12*9)=7x-21/4
remaing liquid B = 5x-(5/12*9)+9=5X-21/4
now (7x-21/4):(5x-21/4)= 7:9
then x=3
but liquid A is 7x so it will be 21 litres

Like? Yes (4) | No (1)

vinusha 9 years AGO

Suppose the can initially contains 7x and 5x of mixtures A and B respectively.

Quantity of A in mixture left = 7x - 7 x 9 litres = 7x - 21 litres.
12 4
Quantity of B in mixture left = 5x - 5 x 9 litres = 5x - 15 litres.
12 4

7x - 21
4
= 7
5x - 15 + 9
4
9
28x - 21 = 7
20x + 21 9
252x - 189 = 140x + 147

112x = 336

x = 3.

So, the can contained 21 litres of A.

Like? Yes (2) | No

Sangeetha G 9 years AGO

(7x-7*9/12)/(5x-5*9/12)
x=3
A=21 litres

Like? Yes (3) | No (1)

bishal 9 years AGO

first A is 7x and B is 5x total mixtuer 12x now 9 liter drawn so 12x-9+z=16x so z=4x+9 now cheeking answer it came 21

Like? Yes (2) | No

Nanthini 9 years AGO

A=(7x-97/12)*9)=(7x-21/4)
B=(5x-(5/12)*9)=(5x-15/4)
A/B+9=7/9
=>252x-189=140x+147
x=3
therefore 7x=21

Like? Yes (2) | No

Mugesh sivagnanam Krishnapillai 9 years AGO

Let the initial mixture be in ratio 7x and 5x
After taking 9 litres
A mixture left=(7x-((7/12)*9)
B mixture left=(5x-((5/12)*9)
then 9 litre of b mixture is added so B=(5x-((5/12)*9)+9
A/B=(7x-((7/12)*9)/(5x-((5/12)*9)+9=7/9
solving we get x=3
Therefore 7x=21 litres.

Like? Yes (2) | No

Koteswararao 9 years AGO

let total volume (A+B) be=x ltrs (intial)
then A's concentration = 7*x/12
After removal of 9 ltrs A's concentration=(7*x/12)-(63/12)
therfore 7x-63/12=7x/16
x=36;
intial volume of A=7*36/12=21

Like? Yes (2) | No

shahida 9 years AGO

let initially a and b have 7x and 5x litres..
when 9ltrs are drawn off,quantity of A in mixture left:
7x-(7/12*9)=7x-21/4ltrs
quantity of b left:
5x-(15/4)ltr
therefore the ratio will be:-
7x-(21/4)/5x-(15/4)+9=7/9
after solving,we get the value of x as 3.
which means the answer is 7*3=21

Like? Yes (2) | No

Nabarupa Das 9 years AGO

{7x-(7/12)*9}/[{5x-(5/12)*9}+9]=7/9
hence x=3.
so A was=3*7=21

Like? Yes (2) | No

Surit Datta 9 years AGO

Suppose the can initially contains 7x and 5x litres of mixtures A and B respectively
Quantity of A in mixture left
= (7x - 7/12 x 9) litres = (7x - 21/4) litres.
Quantity of B in mixture left
= (5x - 5/12 x 9) litres = (5x - 15/4) litres.
(7x - 21/4) / [(5x - 15/4)+9] = 7/9 = › 28x - 21/20x + 21 = 7/9 =› 252x - 189 = 140x + 147
=› 112x = 336 =’ x = 3.
So, the can contained 21 litres of A.
so option 9
that's it.

Like? Yes (6) | No (5)

kumar 9 years AGO

Like? Yes (1) | No

Narendra Gopi 9 years AGO

Initially ratio is 7:5
9 liters drawn of from can A is 7x-7/12*9=7x-21/4
9 liters drawn from can B is 5x-5/12*9=5x-15/4
their ratio is and adding 9 liters to B is (7x-21/4)/(5x-15/4)+9
=(28x-21)/(20x+19)
their ratio is equal to 7:9
==>(28x-21)/(20x+19)=7/9
==>252x-189=140x+147
==>112x=336
==>x=336/112
==>x=3
A filled by initially is 7x=7*3=21

Like? Yes (1) | No

SAGNIK DATTA 9 years AGO

(7x-(63/12))/(5x+(63/12))=7/9
Solve x. x=3
Ans : 7x = 21

Like? Yes (2) | No (1)

ishan chourishi 9 years AGO

let a be 21 so b becomes 15 by ratio 7:5 nd total becomes 36 litres.
when 9 litre of mixture is drawn, total mixture remains 27 litre.
which is 15.75 nd 11.25 by ratio 7:5.
9 litre is added to 11.25 so it becomes 20.25.
so the new mixture has A & B 15.75 & 20.25 litres respectively which is in the ratio 7:9.
hence option 9.

Like? Yes (2) | No (1)

Vivek 9 years AGO

The original ratio = 7/5
9 litres of mixture is removed
The amount of A solution = 7x - 7/12*9
similarly
The amount of B solution = 5x - 5/12*9
9 litres of B is added hence solution B becomes
5x - 5/12*9+9
Equating the ratio of A and B solution to the new ratio we have
(7x - 7/12*9)/(5x - 5/12*9+9) = 7/9
solving we get
x=3
Hence the amount of original solution =7*3=21
therefore answer is option 9)

Like? Yes (1) | No

Akhilesh Pandey 9 years AGO

A:B=7:5
9LT drawn.
A=9*7/12=21/4
B=9*5/12=15/4

so,
(7x-21/4)/(5x+21/4)=7/9
x=3
A=7*3=21
option 9

Like? Yes (1) | No

jayaram 9 years AGO

Suppose the can initially contains 7x and 5x of mixtures A and B respectively.

Quantity of A in mixture left = 7x - 7 x 9 litres = 7x - 21 litres.
12 4

Quantity of B in mixture left = 5x - 5 x 9 litres = 5x - 15 litres.
12 4

7x - 21
4
= 7
5x - 15 + 9
4
9

28x - 21 = 7
20x + 21 9

252x - 189 = 140x + 147

112x = 336

x = 3.

So, the can contained 21 litres of A.

Like? Yes (1) | No

Arpan Chattopadhyay 9 years AGO

if A=21l,the B=15l. if 9 litres is removed then A becomes 15.75l and B 11.25l. now adding 9litres to B gives 20.25l. So 15.75:20.25 gives 7:9.

Like? Yes (2) | No (1)

Venkatesh kadali 9 years AGO

Suppose the can initially contains 7x and 5x of mixtures A and B respectively.

Quantity of A in mixture left=

(7x - 7/12)x 9 litres =(7x -21/4)litres.

Quantity of B in mixture left =

(5x - 5/12)x 9 litres = (5x - 15/4) litres.

---> (7x-21/4) / (5x-15/4)+9 = 7/9

-----> (28x - 21) /(20x + 21)= 7/9

------> 112x=336

----> X=3
So, the A can contained initially 7*x=7*3= 21 litres of A

Like? Yes (1) | No

Arun Karthick 9 years AGO

Quantity of A in mixture left = ((7x) – ((7/12) * 9)) => 7x – (21 /4) liters
Quantity of B in mixture left = ((5x)-((5/12) * 9)) => 5x – (15 / 4) liters
ratio of A and B in mixture is 7 : 9,
(7x – (21/4)) / ((5x) – (15/4)) = 7 / 9 => x = 3
initially can contains 7x liters of A = 21 liters

Like? Yes (1) | No

ramya 9 years AGO

Let the can initially contains 7x and 5x litres of mixtures A and B respectively
Quantity of A in mixture left
= (7x - 7/12 x 9) litres = (7x - 21/4) litres.
Quantity of B in mixture left
= (5x - 5/12 x 9) litres = (5x - 15/4) litres.
(7x - 21/4) / [(5x - 15/4)+9] = 7/9 = › 28x - 21/20x + 21 = 7/9 =› 252x - 189 = 140x + 147
=› 112x = 336 =’ x = 3.
So, the can contained 21 litres of A.

Like? Yes (1) | No

Ajay Kumar Gupta 9 years AGO

Suppose the can initially contains 7x and 5x of mixtures A and B respectively.

Quantity of A in mixture left =(7x -7/12*9) litres =(7x -21/4) litres.
Quantity of B in mixture left =(5x -5/12x 9)litres =(5x -15/4) litres.
(7x -21/4)/(5x-15/4)+9=7/9
so, (28x - 21)/(20x + 21)=7/9
so, 252x - 189 = 140x + 147
so,112x = 336
i.e,x = 3.
So, the can contained 7x=21 litres of A.

Like? Yes (1) | No

Swati 9 years AGO

Like? Yes (1) | No

Vishal 9 years AGO

Like? Yes (1) | No

Dinni 9 years AGO

Like? Yes (1) | No

sourav pal 9 years AGO

Suppose the can initially contains 7x and 5x of mixtures A and B respectively.

Quantity of A in mixture left = 7x - 7 x 9 litres = 7x - 21 litres.
12 4

Quantity of B in mixture left = 5x - 5 x 9 litres = 5x - 15 litres.
12 4

7x - 21
4
= 7
5x - 15 + 9
4
9

28x - 21 = 7
20x + 21 9

252x - 189 = 140x + 147

112x = 336

x = 3.

So, the can contained 21 litres of A.

Like? Yes (1) | No

Sanjeeb Chatterjee 9 years AGO

X/Y=7/5 - RATIO OF A to B
5X-7Y=0......(1)

9 litres would have : 7/12*9=5.25 of A and 9-5.25=3.75 of B

New ratio: (X-5.25)/(Y-3.75+9)=7/9
9X-7Y=16*5.25 ....... (2)

Solve eq 1 and 2 : answer is 21 for x

Like? Yes (1) | No

Palak 9 years AGO

Like? Yes (1) | No

SHARMILA BISWAS 9 years AGO

let initially qty of A=7x and B=5x
in 9 l of mixture,qty of A=(7/12)*9=21/4 and B=(5/12)*9=15/4
after 9 l is drawn out,A=7x-(21/4) and B=5x-(15/4)
after the can is filled with B again,B=5x-(15/4)+9
and A/B=7/9
solving we get x=3
so initially A was 7x therefore A=21.

Like? Yes (1) | No

Santosh Kumar Kasaudhan 9 years AGO

let total mixture=x
then
x*7/12-63/12 : x*5/12-45/12+9 = 7:9
so
x*7/12
=
21

Like? Yes (1) | No

Kethineni Vinodh 9 years AGO

Suppose the can initially contains 7x and 5x of mixtures A and B respectively.

Quantity of A in mixture left = 7x - 7 x 9 litres = 7x - 21 litres.
12 4

Quantity of B in mixture left = 5x - 5 x 9 litres = 5x - 15 litres.
12 4

7x - 21
4
= 7
5x - 15 + 9
4
9

28x - 21 = 7
20x + 21 9

252x - 189 = 140x + 147

112x = 336

x = 3.

So, the can contained 21 litres of A.

Like? Yes (1) | No

JAHAGIRDAR RUPALI MANOJ 9 years AGO

Suppose the can initially contains 7x and 5x of mixtures A and B respectively.

Quantity of A in mixture left = 7x - (7 *9)/12litres = 7x - (21 /4) litres.

Quantity of B in mixture left = 5x -( 5 * 9)/12 litres = 5x - (15/4) litres.

[7x - (21 /4) ] /[(5x - (15/4) )+9]=7/9

solving this ,
112x = 336
x = 3.

So, the can contained 21 litres of A.

Like? Yes (1) | No

pranay jaroli 9 years AGO

let x litre is contained initially
amount of a=(7/12)*x
amount of b=(5/12)*x

when 9L is taken
amount of a=21/4
amount of b=15/4

remaining a 7x/12 -21/4
remaining b 5x/12 - 15/4 + 9(after adding 9L)
final ratio is given
(7x-63/5x+63 ) = 7/9
solving x=36

part of a contained initially (7/12)*36=21 ans

Like? Yes (1) | No

raj 9 years AGO

let 7x and 5x are the initial value.
now, ( 7x- (7/12)*9 ) / 12x =7/16
so x=3
therefore, initial concentration = 7 * 3=21

Like? Yes (1) | No

siddhi datri 9 years AGO

let initial mixtures of A & B be 7x & 5x.
new proportion of A=7x-7*9/12
new proportion of B=5x-5*9/12+9
therefore,( 7x-7*9/12 )/(5x-5*9/12+9)=7/9
7x=21

Like? Yes (1) | No

Gaurav Verma 9 years AGO

initially
______________
a=7x, b=5x
after decreasing 9 liter
a=7x- 9(7/12)x b=5x- 9(5/12)x

Adding 9 l in B
b=5x - 45x/12 +9

Dividing a/b and equating it with 7/9
we get 7x=a=21L

Like? Yes (1) | No

ezhilane 9 years AGO

A:B = 7:5
Total = 5+7=12 units

When 9 litres of mixture is withdrawn , ((7/12) * 9) litres of A is withdrawn, and ((5/12) * 9) litres of B is withdrawn
Then 9 litres of B is added
so that new ratio is 7:9

In new mixture, Total volume of A is 7x - ((7/12) * 9) = 7x - 21/4
And total volume of B is 5x - ((5/12) * 9) + 9 = 5x - 3 3/4 + 9 = 5x + 21/4

So, (7x - 21/4)/(5x + 21/4) = 7/9

Solving for x:
x=3
Original Volume of A = 7x = 21 .
so answer is 21.

Like? Yes (1) | No

KAMARAJAR 9 years AGO

X/Y=7/5 - RATIO OF A to B
5X-7Y=0......(1)

9 litres would have : 7/12*9=5.25 of A and 9-5.25=3.75 of B

New ratio: (X-5.25)/(Y-3.75+9)=7/9
9X-7Y=16*5.25 ....... (2)

Solve eq 1 and 2 : answer is 21 for x

Like? Yes (1) | No

Manikanteswar 9 years AGO

Assume that the can initially contains 7x and 5x of mixtures A and B respectively
quantity of A in the mixture = 7x-21/4
quantity of b in the miixture = 5x-15/4
but the given ratio after the alteration is 7:9
so
(7x-21/4)/(5x-(15/4)+9) = 7/9
Solve the equation and you will get the answer
and also its apparent that there is no other option which is a factor of 7.

Like? Yes (1) | No

lavanya 9 years AGO

Let can contains 7x and 5x litres of mixtures A and B respectively.
Quantity of A in mix. left =(7x-7/12*9) = 7x-21/4
Quantity of B in mix. left =(5x-5/12*9) = 5x-15/4
Therefore
(7x-21/4)/( 5x-15/4) = 7/9
x=3
therefore A contains 21 litres

Like? Yes (1) | No

tamilmani 9 years AGO

Like? Yes (1) | No

Gomathi P 9 years AGO

initially the A and B is 7x and 5x respectively

quantity of A mixtures is left=7x-(7/12*9)=7x-(21/4)liters
quantity of B mixtures is left=5x-(5/12*9)=5x-(15/4)liters

Therefore,

(7x-(21/4))/(5x-15/4+9)=7:9

28x-21/20x+21=7:9
9(28x-21)=7(20X+21)
252x-189=140x+147
112x=336
x=3

so A contains 7*3=21liters

Like? Yes (1) | No

apoorv gupta 9 years AGO

Initially can has 7x litre of liquid A, 5x litre of liquid B.
After removing 9 litres of mixture and adding 9 litres of B -
A = 7x - (7/12)*9
B = 5x - (5/12)*9 + 9

(7x - (7/12)*9) : (5x - (5/12)*9 + 9) = 7:9
x = 3
So, initially A = 7x = 7*3 = 21

Like? Yes (1) | No

kajal singh 9 years AGO

Like? Yes (1) | No

gaurav shukla 9 years AGO

let x be total volume in liter.
then A=7x/12 B=5x/12
(7(x-9)/12)/(5(x-9)/12 + 9)=7/9
x=36
A=21

Like? Yes (1) | No

Karan Somjhani 9 years AGO

let the initial quantity was-
A = 7x, B= 5x ;
from this 9l of liquid was taken out and replaced by B,
so, new quantities are-
A = 7x-(7/12)9 and
B= 5x -(5/12)9 +9
now, above A:B = 7/9 (i.e the new ratio)
solving this we get x=3
so, initially A was 7x=7*3=21 (ans).

Like? Yes (1) | No

sweta jha 9 years AGO

(7x-21/4)/(5x-15/4)+9=7/9
=>x=3
so A=7*3=21

Like? Yes (1) | No

NAGENDRA P 9 years AGO

Like? Yes (1) | No

prakiti maji 9 years AGO

initially can contains 7x and 5x litres of mixtures..
quantity of a in mixture left 7x-21/4.. .............................(1)
quantity of b in mixture left 5x-15/4.. .............................(2)
.....
.....
dividing 1 by 2, we get x =3 and after getting the value of x we put the value of x in 7x so the answer is 21.

Like? Yes (1) | No

Sireesha Tumati 9 years AGO

//7x-(21/4)/(5x-(15/4)+9)=7/9
by solving
x=3
then A's content is 7x=7*3=21

Like? Yes (1) | No

Swapnil Gupta 9 years AGO

let,initially can contains x ltrs. of liquid.
so, A will be 7x/12 and B will be 5x/12.
now when 9 ltrs. of liquid is taken out so new composition becomes :-
A= 7(x-9)/12 and B= 5(x-9)/12.
now 9 ltrs. of liquid B is added. So, new ratio becomes:-
A:B=7(x-9)/12:[5(x-9)/12]+9=7:9.
now after solving we get x=36 ltrs.
so A=7x/12=21 ltrs.
hence, option (9) is the correct ans.

Like? Yes (1) | No

praveen Kumar 9 years AGO

a:b=7:5 A:B=7:9
let no. of litres=x
a=7/12,b=5/12 A=7/16 B=9/16
((7/12)x-((7/12)*9))/((5/12)x-((5/12)*9))=7/9
x=36
a=(7/12)*36
a =21

Like? Yes (1) | No

Sagar Wadjikar 9 years AGO

7K-7/12 (9) =7a

5k - 5/12 (9) + 9 = 5a

Solve for k
k = 3
7k = 21

Like? Yes (1) | No

KAVITHA 9 years AGO

the can initially contains 7x and 5x of mixtures A and B respectively.

Quantity of A in mixture left = 7x - 7 x 9 litres = 7x - 21 litres.
12 4

Quantity of B in mixture left = 5x - 5 x 9 litres = 5x - 15 litres.
12 4

7x - 21
4
= 7
5x - 15 + 9
4
9

28x - 21 = 7
20x + 21 9

252x - 189 = 140x + 147

112x = 336

x = 3.

So, the can contained 21 litres of A.

Like? Yes (1) | No

Archie 9 years AGO

Fraction of B in original mixture =5/12
Feaction of B in 2nd mixture= 1
Fraction of B in resulting =9/16
Using alligations we get the mixture in which 2 liquids are mixed as 3:1 (Ratio of ((9/16)-(5/12)) & ((1-9/16))
Hence for 9 ltrs of liquid B i s added after taking out 9ltrs , 27ltrs will be qty. of liquid left.
Hence, total mixture contained=27+9=36ltrs
qty. of A contained =(36*7)/12 =21ltrs
option 9 :)

Like? Yes (1) | No

anjali singh 9 years AGO

Like? Yes (1) | No

rahul shaw 9 years AGO

Suppose the can initially contains 7x�and 5x�of mixtures A and B respectively.
so acc to question
7x-(7x*9/12)/5x-(5x*9/12)
after solving we get
x=3
so A=21 INITIALLY

Like? Yes (2) | No (1)

Sushmita Ghanty 9 years AGO

let, initial amount of A and B are 7x and 5x litre
now, A =7x-7/12*9=7x-21/4 litre
B=5x-5/12*9=5x-15/4 litre
(7x-7/12*9)/((5x-5/12*9)+9)=7/9
or, x=3;
so A contained 7*3=21litre

Like? Yes (1) | No (1)

anand vijay 9 years AGO

Let the can initially contains 7x and 5x of mixtures A and B respectively.

quantity of A left in the mixture=7x-7*9/12 litres= 7x-21/4 litres
quantity of B left in the mixture=5x-5*9/12 litres=5x-15/4
litres
therefore, (7x-21/4)/(5x-15/4)+9=7/9
on solving we get, x=3
hence can contains liquid A 7*3= 21 litres initially
option 9)

Like? Yes (1) | No (1)

PRABIR 9 years AGO

A-7x
B-5x
In 9L A is 21/4 L, B is 15/4
So (7x-21/4)/(5x-15/4 +9)=7/9
7x=21L

Like? Yes (1) | No (1)

Kunal nemu 9 years AGO

let ,the can initially contains A=7x,B=5x
quantity of A=(7x-7/12*9)
quantity of B=(5x-5/12*9)

so,(7x-21/4)/(5x-15/4)+9=7/9
or (28x-21)/(20x+21)=7/9
or 252x-189=140x+147
or 112x=336
or x=3
So,the can contained 21lit. of A

Like? Yes (1) | No (1)

Ayush Kumar Singh 9 years AGO

let a+b=x
and A/B=7/5
then A=7/12 of x initially
final will be,A= 7/12(x-9)
and same procedure with B

Like? Yes (2) | No (2)

Meganathan.I 9 years AGO

X/Y=7/5 - RATIO OF A to B
5X-7Y=0......(1)

9 litres would have : 7/12*9=5.25 of A and 9-5.25=3.75 of B

New ratio: (X-5.25)/(Y-3.75+9)=7/9
9X-7Y=16*5.25 ....... (2)

Solve eq 1 and 2 : answer is 21 for x

Like? Yes (1) | No (1)

santosh kumar patel 9 years AGO

Suppose the can initially contains 7x and 5x litres of mixtures A and B respectively.
Quantity of A in mixture left
= (7x - 7/12 x 9) litres = (7x - 21/4) litres.
Quantity of B in mixture left
= (5x - 5/12 x 9) litres = (5x - 15/4) litres.
(7x - 21/4) / [(5x - 15/4)+9] = 7/9 = › 28x - 21/20x + 21 = 7/9 =› 252x - 189 = 140x + 147
=› 112x = 336 =’ x = 3.
So, the can contained 21 litres of A.

Like? Yes (1) | No (1)

vikas kulshrestha 9 years AGO

Solution:

Suppose the can initially contains 7x and 5x litres of mixtures A and B respectively.

When 9 litres of mixture are drawn off, quantity of A in mixture left:
7x–(
7
12
)×9=7x−
21
4
litres

Similarly quantity of B in mixture left:
5x−(
5
12
)×9=5x−(
15
4
) litres

Therefore ratio becomes:
7x–(
21
4
)
5x−(
15
4
)+9
=
7
9

⇒
(28x−21)
(20x+21)
=
7
9

⇒(252x−189)=140x+147
⇒112x=336
⇒ x=3

So the can contained: 7×x=7×3= 21 litres of A

Like? Yes (1) | No (1)

Thirupathi Reddy 9 years AGO

A should be a multiple of 7
only one option is divisible by 7 i.e., 21

Like? Yes (3) | No (3)

Pranav 9 years AGO

considering the amount to be 7x and 5x the the question can be solved.

Like? Yes (1) | No (1)

Naveenhaa B 9 years AGO

initially can contains 7x and 5x of mixtures A and B.

Quantity of A in mixture left = (7x-(2/4))
Quantity of B in mixture left = (5x-(15/4))

(7x - (21/4))
divided by (5x-(15?4)) and which is equal to 7 : 9
on solving
x = 3.

So, the can contained 21 litres of A.

Like? Yes (1) | No (1)

Dimple

India

Punjab

Time: 00:01:33

Points

Maths Quotes

amcat ques , guys help me what is the value of these log0.2 base 5

amcat

"Obvious" is the most dangerous word in mathematics.

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A can contains a mixture of two liquids A and B in proportion 7:5. When 9 litre of mixture are drawn

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