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02 Jul 2025, 08:59
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initial solution= 2% salt which is 98% water
let, initial weight= 100g (2g salt + 98g water)
after adding water
let, water= X g
total salt remains= 2g
2/100+x=0.01
x=100g
final water: 98g + 100g= 198g
final ration( final water: inital water)
198/98= 99/49
ans: D (99:49)
let, initial weight= 100g (2g salt + 98g water)
after adding water
let, water= X g
total salt remains= 2g
2/100+x=0.01
x=100g
final water: 98g + 100g= 198g
final ration( final water: inital water)
198/98= 99/49
ans: D (99:49)
02 Jul 2025, 09:00
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Given conditions,
Initial total weight of the solution = W1
Initial weight of salt = S
Initial weight of water = H1
Given that the saline solution is 2% salt by weight initially, S=0.02×W1
The rest is water, so:
H1 =W1−S=0.98W1
Now, more water is added.
Final total weight of the solution = W2
Final weight of salt = S (amount of salt remains the same, only water is added)
Final weight of water = H2
After water is added, the salt concentration decreases to 1% by weight i.e.,S=0.01×W2
We have two expressions for the weight of salt S: 0.02W1 =0.01W2
W1 =2W1
the final weight of water i.e., H2 = W2 −S
Substitute S=0.02W1 and W2 =2W1
H2=2W1 −0.02W1 =(2−0.02)W1 =1.98W1
ratio of the final weight of water / the initial weight of water:
H2/H1 =1.98/0.98 =198/98
So, the ratio is 99/49
.
Initial total weight of the solution = W1
Initial weight of salt = S
Initial weight of water = H1
Given that the saline solution is 2% salt by weight initially, S=0.02×W1
The rest is water, so:
H1 =W1−S=0.98W1
Now, more water is added.
Final total weight of the solution = W2
Final weight of salt = S (amount of salt remains the same, only water is added)
Final weight of water = H2
After water is added, the salt concentration decreases to 1% by weight i.e.,S=0.01×W2
We have two expressions for the weight of salt S: 0.02W1 =0.01W2
W1 =2W1
the final weight of water i.e., H2 = W2 −S
Substitute S=0.02W1 and W2 =2W1
H2=2W1 −0.02W1 =(2−0.02)W1 =1.98W1
ratio of the final weight of water / the initial weight of water:
H2/H1 =1.98/0.98 =198/98
So, the ratio is 99/49
.
02 Jul 2025, 09:02
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Bookmarks
From the details given,
Salt content does not change; Water is added
1.)Before adding water:
Weight of salt = sw = 2g (assume)
Let Initial weight of water = wi = 98g (assume)
Let Initial weight of water = wf
We are to find out wf/wi
2.)After adding water:
Since we know salt is not added, sw = 2g(as per assumption)
It's mentioned that the salt content becomes 1% by weight of the saline solution
To find final weight of solution,
Let total weights before and after adding water be ti & tf
.01 * tf = 2
∴tf = 200; ∴ wf = tf -2 = 198
∴wf/wi = 198/98
∴Ration = 99:49
Correct option is Option D
Salt content does not change; Water is added
1.)Before adding water:
Weight of salt = sw = 2g (assume)
Let Initial weight of water = wi = 98g (assume)
Let Initial weight of water = wf
We are to find out wf/wi
2.)After adding water:
Since we know salt is not added, sw = 2g(as per assumption)
It's mentioned that the salt content becomes 1% by weight of the saline solution
To find final weight of solution,
Let total weights before and after adding water be ti & tf
.01 * tf = 2
∴tf = 200; ∴ wf = tf -2 = 198
∴wf/wi = 198/98
∴Ration = 99:49
Correct option is Option D
