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13 Jun 2025, 06:50
Kudos
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
44% (02:57) correct
56%
(03:27)
wrong
based on 54
sessions
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A bag contains red balls that weigh 100 grams each, blue balls that weigh 50 grams each and green balls that weigh 50 grams each. If the total weight of the balls in the bag is between 1.05 kilogram and 1.50 kilogram, and the number of green balls is 9 more than the number of red balls, at most, how many balls are there in the bag?
(A) 17
(B) 23
(C) 27
(D) 28
(E) 30
(A) 17
(B) 23
(C) 27
(D) 28
(E) 30
13 Jun 2025, 13:58
Kudos
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siddhantvarma
The total weight of the balls is between 1050 grams and 1500 grams => 1050 < w < 1500
g = r + 9
Since red balls are the heaviest as compared to green and blue balls, and we need to find the maximum number of balls the bag can have, we will minimize the number of red balls. So, r = 1
=> g = 10
1 r = 100 grams, 10 g = 10 * 50 = 500 grams, so these 11 balls weigh 600 grams. Now, since blue balls weigh 50 grams each, the maximum weight of total no. of balls has to be 1450 grams. Thus, weight of blue balls is 1450 - 600 = 850 grams.
No. of blue balls = 850 / 50 = 17
Maximum no. of balls = 1 + 10 + 17 = 28
Answer D.
13 Jun 2025, 15:31
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siddhantvarma
\(1050<100r+50y+50g <1500 \)
\(1050<100r+50b+50(r+9)<1500\) ... \(g=r+9\)
\(600 < 150r+50b <1050\)
We need most number of balls, so lets concentrate on the higher limit.
\(150r+50b <1050\)
\(3r+b<21\)
\(3r+b=20 \) ..since \(r\) and \(b\) are integers, \(20\) is the max value of the above inequality.
After some tests we can see when \(b=17 \) and \( r=1\) gives max balls for the above eqn.
since \(g = r+9\) therefore \(g=10\)
Thus total balls \(17+1+10= 28\)
Ans D
Hope it helped.
