Percentage (āĻļāϤāĻžāĻ‚āĻļ) – Top 30 Standard Questions for Competitive Exams - 2


Q9. A vessel contains a mixture of milk and water in which milk is 80% of the mixture. By what percentage should water be added so that the milk becomes 60% of the new mixture ? (āĻāϟāĻž āĻĒāĻžāĻ¤ā§ā§°āϤ āĻĻ⧁āϧ āφ⧰⧁ āĻĒāĻžāύ⧀⧰ āĻŽāĻŋāĻļā§ā§°āĻŖāϤ āĻĻ⧁āϧ⧰ āĻĒā§°āĻŋāĻŽāĻžāĻŖ 80%āĨ¤ āĻŽāĻŋāĻļā§ā§°āĻŖāĻŸā§‹āϤ āĻ•āĻŋāĻŽāĻžāύ āĻļāϤāĻžāĻ‚āĻļ āĻĒāĻžāύ⧀ āϝ⧋āĻ— āϕ⧰āĻŋāϞ⧇ āĻĻ⧁āϧ⧰ āĻĒā§°āĻŋāĻŽāĻžāĻŖ āύāϤ⧁āύ āĻŽāĻŋāĻļā§ā§°āĻŖā§° 60% āĻš'āĻŦ ?)


Options: A) 20% B) 25% C) 33⅓% D) 40%


Ans: C) 33⅓%


Soln / āϏāĻŽāĻžāϧāĻžāύ:


Assume total mixture (āϧ⧰āĻž āϝāĻžāĻ“āĻ•, āĻŽā§āĻ  āĻŽāĻŋāĻļā§ā§°āĻŖ )= 100 units.


Milk (āĻĻ⧁āϧ) = 80 units, Water (āĻĒāĻžāύ⧀) = 20 units


After adding water, milk remains 80 units. (āĻĒāĻžāύ⧀ āϝ⧋āĻ— āϕ⧰āĻžā§° āĻĒāĻŋāĻ›āϤ⧋ āĻĻ⧁āϧ = 80 āĻāĻ•āĻ•āĨ¤)


Now, 80 units becomes 60% of the new mixture. (āĻāϤāĻŋāϝāĻŧāĻž, 80 āĻāĻ•āĻ• = āύāϤ⧁āύ āĻŽāĻŋāĻļā§ā§°āĻŖā§° 60%)


New mixture = (80 × 100) ÷ 60 = 133⅓ units


Water added = 133⅓ − 100 = 33⅓ units


Therefore, Percentage of water added = 33⅓%


Ans / āωāĻ¤ā§āϤ⧰: C) 33⅓%


Q10 / āĻĒā§ā§°āĻļā§āύ ā§§ā§Ļ: If 80% of A = 50% of B and B = x% of A, find the value of x. (A-ā§° 80% = B-ā§° 50% āφ⧰⧁ B = A-ā§° x% āĻšāϞ⧇, x-ā§° āĻŽāĻžāύ āύāĻŋā§°ā§āĻŖāϝāĻŧ āϕ⧰āĻžāĨ¤)


Options / āĻŦāĻŋāĻ•āĻ˛ā§āĻĒāϏāĻŽā§‚āĻš: A) 40  B) 80  C) 160  D) 150


Ans / āωāĻ¤ā§āϤ⧰: C) 160


Soln / āϏāĻŽāĻžāϧāĻžāύ:āύ


Assume (āϧ⧰āĻž āϝāĻžāĻ“āĻ•) A = 100.


Then 80% of A = 80. (āϤ⧇āĻ¨ā§āϤ⧇ A-ā§° 80% = 80āĨ¤)


Given, āĻĒā§ā§°āĻļā§āύāĻŽāϤ⧇, 50% of B = 80 (āĻ…ā§°ā§āĻĨāĻžā§Ž, B-ā§° 50% = 80)


Now, āϝāĻĻāĻŋ 50% = 80 Then, 100% = 160


āϤ⧇āĻ¨ā§āϤ⧇ 100% = 160, So, B = 160


Since A = 100, A = 100 āĻšā§‹ā§ąāĻž āĻŦāĻžāĻŦ⧇,


B = 160% of A


āĻ…ā§°ā§āĻĨāĻžā§Ž, B = A-ā§° 160%


Therefore, x = 160


 Ans / āωāĻ¤ā§āϤ⧰: C) 160