Question
Ankit began a task but left it
after working for 32 days. The remaining 60% of the task was then completed by Bittu and Sittu working together over 45 days. Sittu is 120% more efficient than Bittu. Determine how long Sittu would take to complete 22% of the task on his own.Solution
ATQ, Total time taken by ‘Bittu’ and ‘Sittu’ to complete the whole task = 45/0.6 = 75 days Time taken by ‘Ankit’ to complete total task = 32/0.40 = 80 days Let total work = 1200 units (LCM of 75 and 80) Efficiency of ‘Ankit’ = 1200/80 = 15 units/day Efficiency of (Bittu + Sittu) = 1200/75 = 16 units/day Let the efficiency of ‘Bittu’ be ‘b’ units/day Efficiency of ‘Sittu’ = 2.20 × b = 2.2b units/day According to the question, 2.2b + b = 16 Or, 3.2b = 16 Or, b = 5 Therefore, efficiency of ‘Bittu’ = b = 5 units/day Efficiency of ‘Sittu’ = 16 – 5 = 11 units/day Time taken by ‘Sittu’ to complete 22% of the task = (0.22 × 1200)/11 = 24 days
If cos(90° − θ) = 3/5 for an acute angle θ, find:
Simplify:
15cos 41° cosec 49° - 6tan 55° tan 35°What is the value of [(sin x + sin y) (sin x – sin y)]/[(cosx + cosy) (cosy – cosx)]?
- If cos2B = sin(1.5B - 36 o ) , then find the measure of 'B'.
If 6sin²x + 2cos²x − 3 = 0, then find the value of sinx, given that 0° < x < 90°.
- Find the simplified value of the expression:sin 2 45 o  + sin 2 60 o  - (1/3) X tan 2 60 o
If x+  1/x = 2cosθ, then the value of x³+  1/x³ is
- If 4cos²A + 5sin²A = 4.5, then find the value of (sec²A - 1)

The Value of (sin38Ëš)/(cos52Ëš) + (cos12Ëš)/(sin78Ëš) - 4cos²60Ëš is