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Let the total work = 96 units (LCM of 24, 48 and 32) Amount of work done by A alone in one day = 96/24 = 4 units Amount of work done by B alone in one day = 96/48 = 2 units Amount of work done by C alone in one day = 96/32 = 3 units Amount of work done by A, B and C together in 2 days = 2 × (4 + 2 + 3) = 18 units Amount of work done by B and C together in 3 days = 3 × (2 + 3) = 15 units Remaining work = 96 – 18 – 15 = 63 units So, the time taken by C alone to complete 63 units work = 63/3 = 21 days So, C worked for 21 + 2 + 3 = 26 days
If x 2 – 15x + 51 = 0, then determine the value of (x – 5) + {1/(x – 5)}.
√(92×8 ×52+700) = ?
{(21/20) + (20/21)}2 - {(21/20) - (20/21)}2 = ?
In a college, the proportion of boys to girls is 7:9, and the proportion of graduate students to postgraduate students is 4:5. Determine the total numbe...
The average of three numbers a, b and c is 2 more than c. The average of a and b is 48. If d is 10 less than c, then the average of c and d is:
If a3 + b3 = 6240 and a + b = 30, then find the value of {(a + b)2 – 3ab}
If 15a2 + 1 = 20a, then find the value of {(9a2 + 1)/25a2}
1/3 + 1/15 + 1/35 + 1/63 + 1/99 = ?
In a class of 70 students and 25 teachers, each student got gifts that were 20% of total number of students and each teacher got gifts that were 10% of ...
