(a) 23
(b) 24
(c) 14
(d) 28
(e) none of the these
(a) 3/7
(b) 5/14
(c) 1/12
(d) 6/11
(e)none of these
(a) 2
( b) 3
(c) 13
(d) 27
(e) None of these
(a) 57/216
(b) 67/216
(c) 77/216
(d) 87/216
(e) None of these
Solutions & Answers
Solution (1)
This can be solved by trial and error method
You know well the probability of getting black balls is 2/7 and of red balls is 1/2 , also Number of blue balls is 6.
Now the total number of balls will be a multiple of both 7 and 2 i.e. it can be 14 , 28 , etc. Suppose if we take total number of balls to be 14 and try distributing accordingly, we get
Suppose if we take total number of balls to be 28 and try distributing accordingly, we get
(1) So 28 is the total number of balls (D)
You know well the probability of getting black balls is 2/7 and of red balls is 1/2 , also Number of blue balls is 6.
Solution (2)
21 is the only number which is multiple of both 3 and 7, so while calculating probability separately for 3 and 7, this number 21 will be taken twice so to avoid confusion we take it once here and will exclude 21 while calculating probability separately. Also, number 21 can only be taken once.
➡ Probability of getting number multiple of both 3 and 7 (number 21) =1/28
➡ Probability of getting numbers which are multiple of 3 ( 3,6,9,12,15,18,24,27) = 9 / 28
➡ Probability of getting numbers which are multiple of 7 (7,14,28) =3/28
➡ Probability of getting a number either multiple of 3 or 7 = 3/28 + 9/28 + 1/ 28 = 12/28 = 3/7
Hence 3/7 (a) is the right answer.
➡ Probability of getting number multiple of both 3 and 7 (number 21) =1/28
➡ Probability of getting numbers which are multiple of 3 ( 3,6,9,12,15,18,24,27) = 9 / 28
➡ Probability of getting numbers which are multiple of 7 (7,14,28) =3/28
➡ Probability of getting a number either multiple of 3 or 7 = 3/28 + 9/28 + 1/ 28 = 12/28 = 3/7
Hence 3/7 (a) is the right answer.
Solution 3
This can also be done on trial and error method
Given
Given
Suppose total number balls is 27 then red balls are 15, yellow balls will be 6 and green balls will be 6.the ratio of green and yellow balls must be 9:4 , this condition is not satisfied. Same way even for 36 , 45 …. Not agreeable.
- If the total number of balls taken to be 54.
- Then Red balls = 15
- Yellow balls = 12
- Green balls = 27 (green and yellow ratio 27:12 = 9:4 satisifed )
- Therefore the total number of balls is 54 in bag A.
- Now using this total number balls calculate the values for bag B i.e. given 2/3 times green balls in bag A = 2/3 *27 = 18 . Also total number of balls in Bag B is 5/6 times of toal balls of A = 54*5/6 = 45
- Now red balls are 3 less the yellow balls .let yellow balls be X and red be x-3
- 45- 18= 27
- 27 is both red plus yellow balls i.e. x + ( x-30) = 27
- X=15
- Also red balls in bag C 1/3 ( total red balls in A and B) = 1/3 ( 27)= 9
- Also 20% more green balls than B’s green bag i.e. 15*120/100 = 18
- And using probability 9/16 remaining data is found
if p yellow balls are removed from bag A and transferred to bag C then
18+ p / 48+ p = 2/5
Solving this we get 2. Hence (a)
Solution (4)
Only yellow balls have to be chosen from any of the bags
Firstly in bag A you have to choose yellow ball out of 3colors so 1/3, then out of 12/54
Similarly for other two bags
1/3 ( 12/54) + 1/3 ( 15/45) + 1/3 ( 18/48)
1/3( 12/54 + 15/45 + 18/48) = 1/3(67/216)
Firstly in bag A you have to choose yellow ball out of 3colors so 1/3, then out of 12/54
Similarly for other two bags
1/3 ( 12/54) + 1/3 ( 15/45) + 1/3 ( 18/48)
1/3( 12/54 + 15/45 + 18/48) = 1/3(67/216)
Hence (b)
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