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Two quantities I and II are given in the following questions. You have to find the values of both I and II by using your knowledge of mathematics and choose the most suitable relation between quantity I and II from the given options. Quantity I: A and B working alone can complete a piece of work in ‘x – 2’ days and ‘x + 6’ days respectively and the ratio of efficiencies of B and C is 3 : 4. If B and C together can finish the work in 48/7 days, then find the value of x. Quantity II: When the sum of the first 50 factorials is divided by 7, the remainder will be R. Find the value of 2R. 

Two quantities I and II are given in the following questions. You have to find the values of both I and II by using your knowledge of mathematics and choose the most suitable relation between quantity I and II from the given options. Quantity I: A and B working alone can complete a piece of work in ‘x – 2’ days and ‘x + 6’ days respectively and the ratio of efficiencies of B and C is 3 : 4. If B and C together can finish the work in 48/7 days, then find the value of x. Quantity II: When the sum of the first 50 factorials is divided by 7, the remainder will be R. Find the value of 2R.  Correct Answer Quantity I = Quantity II or No relation can be established

Quantity I:

Time taken by A alone = ‘x – 2’ days

Time taken by B alone = ‘x + 6’ days

Time taken by C alone = (x + 6) × (3/4) = 3(x + 6)/4 days

According to the question:

+ = (7/48)

⇒ 1/(x + 6) = 1/16

⇒ 16 = x + 6

⇒ x = 10

Quantity II:

We know that every term greater than 6! Will contain 7 and thus will not give any remainder.

⇒ We will have to check only for the sum of the first 6 factorials.

So,

Sum = 1! + 2! + 3! + 4! + 5! + 6! = 1 + 2 + 6 + 24 + 120 + 720 = 873

When 873 is divided by 7 it will give a remainder of 5.

⇒ 2R = 10

∴ Quantity I = Quantity II

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