According to question,
Quantity of Milk/Quantity of Water = 3/2
Let us assume the product ratio = n.
Quantity of Milk = 3n and Quantity of Water = 2n liters
Quantity of Milk + Quantity of Water = 20 liters
3n + 2n = 20 liters
5n = 20 liters
n = 20/5
n = 4
Quantity of Milk = 3n liters
Put the value of n,
Quantity of Milk = 3 x 4 = 12 liters
Quantity of Water = 2n
Quantity of Water = 2 x 4 = 8 liters
If 10 liters of mixture are removed first time, we will find how much milk and water contain in mixture.
? 20 liters of mixture contains 12 liter of milk.
? 1 liters of mixture contains 12/20 liter of milk.
? 10 liters of mixture contains 10 x 12/20 liter of milk.
? 10 liters of mixture contains 6 liter of milk.
? 10 liters of mixture contains 4 liter of water.
If 10 liters of mixture are removed, then amount of milk removed = 6 liters and amount of water removed = 4 liters.
Remaining milk = 12 - 6 = 6 liters
Remaining water = 8 - 4 = 4 liters
10 liters of pure milk are added, therefore total milk = (6 + 10) = 16 liters and water = 4 liters.
If the process is repeated one more time and 10 liters of the mixture are removed second time, then
If 10 liters of mixture are removed, Again we will find how much milk and water contain in the mixture.
? 20 liters of mixture contains 16 liter of milk.
? 1 liters of mixture contains 16/20 liter of milk.
? 10 liters of mixture contains 10 x 16/20 liter of milk.
? 10 liters of mixture contains 8 liter of milk.
? 10 liters of mixture contains 2 liter of water.
If 10 liters of mixture are removed, then amount of milk removed = 8 liters and amount of water removed = 2 liters.
Remaining milk = (16 - 8) = 8 liters.
Remaining water = (4 - 2) = 2 liters.
Now 10 liters milk is added => total milk = 18 liters and water will be 2 liters.
The required ratio of milk and water in the final mixture obtained
Quantity of milk/Quantity of Water= 18:2 = 9:1
Let speed upstream = x km/hr
Then, speed downstream = 3x km/hr
? Speed in still water = (x + 3x)/2 km/hr = 2x km/hr
Speed of the current = (3x - x)/2 km/hr = x km/hr
&becaus 2x = 28/3
? x = 14/3 = 42/3 km/hr.
P --- 10 ----- 600
P --- 5 ----- 300
3P --- 5 ----- 900
__________
1200
(24 + 31 + 18) days = 73 days = 1/5 year .
P = Rs 3000 and R = 18 % p.a.
Principle amount = Rs. 29000
Interest = Rs. 10440
Let rate of interest = r%
=> So, time = r years
According to the question,
10440 = 29000 x r x r/100
290 x r x r = 10440
r x r = 1044/29 = 36
r = 6
Hence, the rate of interest = 6% and time = 6 yrs.
2500 in 5th year and 3000 in 7th year
So in between 2 years Rs. 500 is increased => for a year 500/2 = 250
So, per year it is increasing Rs.250 then in 5 years => 250 x 5 = 1250
Hence, the initial amount must be 2500 - 1250 = Rs. 1250
S.I. for 3 years = Rs. (12005 - 9800) = Rs. 2205.
S.I. for 5 years = Rs. = Rs.3675
Principle = Rs.(9800-3675) = Rs.6125
Hence, Rate = =12%
Let the sum invested be Rs. P
Let the rate of interest be R% per annum
=> Interest earned for 5 years = (P x 5 x R/100) = PR/20
Now, given that the interest earned increased by Rs. 600 if the Rate increased to (R+2)%
=> SI = (P x 5 x (R+2))/100 = PR/20 + 10P/100
Hence,
PR/20 + 10P/100 = PR/20 + 600
=> P = 6000
Therefore, the sum invested is Rs. 6000
Let x be the speed of the boat and y be the speed of the current.
Speed upstream = (x - y) km/h
Speed downstream = (x + y) km/h
According to the question,
20 /(x - y) + 20/(x + y) = 25/36
In this equation, there are two variables but only one equation. So, the value of x cannot be determined.
Let the principle amount be Rs. P
Interest rate = 12%
Total amount he paid after 5 years = Rs. 1280
ATQ,
Hence, the amount he borrowed = P = Rs. 800.
S.I. = Rs. (15500 - 12500) = Rs. 3000.
Rate = % = 6%
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