Given:
A. The value of (1)/(4) + (1)/(4×5) + (1)/(4×5×6) is equal to (37)/(120).
B. 1+2÷{1+2÷(1+(1)/(3))} is equal to (4)/(5).
C. If the average of 6 numbers is 7, the average of 3 numbers of them is 5, then the average of remaining numbers is 9.
D. If 75% of a number is added to 75, then the result is the number itself. Then the number is 300.
E. Largest fraction from (9)/(31), (6)/(23), (3)/(17), (4)/(11) is (9)/(31).
Formula used:
For A: Add fractions directly with LCM.
For B: Solve using BODMAS rule.
For C: Average formula = (Sum of observations) ÷ (Number of observations).
For D: Solve percentage equation.
For E: Compare fractions by cross-multiplication.
Calculations:
For A:
(1)/(4) + (1)/(4×5) + (1)/(4×5×6)
⇒ (1/4) + (1/20) + (1/120)
LCM of 4, 20, 120 = 120
⇒ (30/120) + (6/120) + (1/120)
⇒ (37/120)
Statement A is correct.
For B:
1+2÷{1+2÷(1+(1)/(3))}
⇒ 1+2÷{1+2÷(1+0.333)}
⇒ 1+2÷{1+2÷1.333}
⇒ 1+2÷{1+1.5}
⇒ 1+2÷2.5
⇒ 1+0.8
⇒ 1.8 or (4/5)
Statement B is correct.
For C:
Average of 6 numbers = 7
⇒ Total = 6×7 = 42
Average of 3 numbers = 5
⇒ Total of 3 numbers = 3×5 = 15
Remaining total = 42 - 15 = 27
Average of remaining numbers = 27 ÷ 3 = 9
Statement C is correct.
For D:
Let the number be x.
75% of x + 75 = x
⇒ (75/100)x + 75 = x
⇒ (3/4)x + 75 = x
⇒ x - (3/4)x = 75
⇒ (1/4)x = 75
⇒ x = 75×4 = 300
Statement D is correct.
For E:
Compare fractions using cross-multiplication:
(9/31) vs (6/23): 9×23 = 207, 6×31 = 186, so (9/31) > (6/23).
(9/31) vs (3/17): 9×17 = 153, 3×31 = 93, so (9/31) > (3/17).
(9/31) vs (4/11): 9×11 = 99, 4×31 = 124, so (4/11) > (9/31).
Largest fraction is (4/11), not (9/31).
Statement E is incorrect.
Correct Answer: A, C, and D only