Percentage (āĻļāϤāĻžāĻāĻļ) â 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 ? (āĻāĻāĻž āĻĒāĻžāϤā§ā§°āϤ āĻĻā§āϧ āĻā§°ā§ āĻĒāĻžāύā§ā§° āĻŽāĻŋāĻļā§ā§°āĻŖāϤ āĻĻā§āϧ⧰ āĻĒā§°āĻŋāĻŽāĻžāĻŖ 80%āĨ¤ āĻŽāĻŋāĻļā§ā§°āĻŖāĻā§āϤ āĻāĻŋāĻŽāĻžāύ āĻļāϤāĻžāĻāĻļ āĻĒāĻžāύ⧠āϝā§āĻ āĻā§°āĻŋāϞ⧠āĻĻā§āϧ⧰ āĻĒā§°āĻŋāĻŽāĻžāĻŖ āύāϤā§āύ āĻŽāĻŋāĻļā§ā§°āĻŖā§° 60% āĻš'āĻŦ ?)
Options: A) 20% B) 25% C) 33â % D) 40%
Ans: C) 33â %
Soln / āϏāĻŽāĻžāϧāĻžāύ:
Assume total mixture (āϧ⧰āĻž āϝāĻžāĻāĻ, āĻŽā§āĻ āĻŽāĻŋāĻļā§ā§°āĻŖ )= 100 units.
Milk (āĻĻā§āϧ) = 80 units, Water (āĻĒāĻžāύā§) = 20 units
After adding water, milk remains 80 units. (āĻĒāĻžāύ⧠āϝā§āĻ āĻā§°āĻžā§° āĻĒāĻŋāĻāϤ⧠āĻĻā§āϧ = 80 āĻāĻāĻāĨ¤)
Now, 80 units becomes 60% of the new mixture. (āĻāϤāĻŋāϝāĻŧāĻž, 80 āĻāĻāĻ = āύāϤā§āύ āĻŽāĻŋāĻļā§ā§°āĻŖā§° 60%)
New mixture = (80 × 100) ÷ 60 = 133â units
Water added = 133â − 100 = 33â units
Therefore, Percentage of water added = 33â %
Ans / āĻāϤā§āϤ⧰: C) 33â %
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% āĻā§°ā§ B = A-ā§° x% āĻšāϞā§, x-ā§° āĻŽāĻžāύ āύāĻŋā§°ā§āĻŖāϝāĻŧ āĻā§°āĻžāĨ¤)
Options / āĻŦāĻŋāĻāϞā§āĻĒāϏāĻŽā§āĻš: A) 40 B) 80 C) 160 D) 150
Ans / āĻāϤā§āϤ⧰: C) 160
Soln / āϏāĻŽāĻžāϧāĻžāύ:āύ
Assume (āϧ⧰āĻž āϝāĻžāĻāĻ) A = 100.
Then 80% of A = 80. (āϤā§āύā§āϤ⧠A-ā§° 80% = 80āĨ¤)
Given, āĻĒā§ā§°āĻļā§āύāĻŽāϤā§, 50% of B = 80 (āĻ ā§°ā§āĻĨāĻžā§, B-ā§° 50% = 80)
Now, āϝāĻĻāĻŋ 50% = 80 Then, 100% = 160
āϤā§āύā§āϤ⧠100% = 160, So, B = 160
Since A = 100, A = 100 āĻšā§ā§ąāĻž āĻŦāĻžāĻŦā§,
B = 160% of A
āĻ ā§°ā§āĻĨāĻžā§, B = A-ā§° 160%
Therefore, x = 160
Ans / āĻāϤā§āϤ⧰: C) 160