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 the two consecutive numbers are N and N+2
According to the question
N (N + 2) = 19043
? N2 + 2N - 19043 = 0
? N2 + 139N - 137N - 19043= 0
? N( N + 139) - 137 ( N - 137 ) = 0
? N( N + 139) ( N - 137) = 0
? N = 137 and N = - 139
? N = 137
A will reach at starting point in 5 * 2 / 5 = 2 hours ;
B will reach at starting point in 5 / 3 hours ;
C will reach at starting point in 5 / 2 hours ;
Then, on the starting point all three will meet after the L.C.M. of 2, 5 / 3, 5 / 2, 10 / 1 = 10 hours.
Circumference = 2?r
= 2 x (22 / 7) x 70 cm
= 440 cm
Distance travelled in 10 revolutions = 440 x 10 cm
= 4400 cm
= 44 m
? Speed = distance / time
= 44 / 5 m/sec
= (44 / 5) x (18 / 5) km/hr
= 31.68 km/hr
Regarding all copies of the same book as one book, we have only 5 books. These 5 books can be arranged in 5! ways. But all copies of the same book being identical can be arranged in only one way.
? Required number = 5! x 1! x 1! x 1! x 1! = 120
We known that,
LCM of fractions = (LCM of numerators) / (HCF of denominators)
? Required LCM = (LCM of 1, 2, 5, and 4) / (HCF of 3, 9, 6 and 27)
? = 20/3
The speed of A and B are in the ratio 11 : 8.
Let, speeds be 11s and 8s (in m/sec).
Let, race be of P meter.
Then, time taken by A to run P meter is same as that of B to run (P - 120) meter.
? P / 11 = (P - 120) / 8
? P / 11 = (P - 120 / 8
? 8P = 11 x (P - 120 )
? 8P = 11P - 120 x 11
? 11P - 8P = 120 x 11
? 3P = 11 × 120
? P = 440.
? = 124.35% of 8096
= (8096 x 124.35)/100 = 1006737.6/100
= 10067.376 = 10000
log105 = log10(10/2)
= log1010 - log102
= 1-0.3010
= 0.6990
Friday will fall on 3, 10, 17, 24, 31
So, it will be 5th Friday on 31 st
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