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In a zoo, the ratio of elephants to lions is same as lions to deer. If
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In a zoo, the ratio of elephants to lions is same as lions to deer. If there are at least 5 deer and all three  deer, lion and elephant are different in number, what is the number of elephants? (1) The ratio of elephant to lion is 1:5 (2) The number of lions is a Prime number. source self made
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In a zoo, the ratio of elephants to lions is same as lions to deer. If
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chetan2u wrote: In a zoo, the ratio of elephants to lions is same as lions to deer. If there are at least 5 deer, what is the number of elephants?
(1) The ratio of elephant to lion is 1:5 (2) The number of lions is a Prime number.
source self made It should be B. From stem, we know E : L is same as L : D and know there are at least 5 deer but we do not have exact ratios. Statement 1: We know \(\frac{E}{L} = \frac{L}{D} = \frac{1}{5}\). If there is just 1 Elephant, E : L tells us there must be at least 5 Lions and if there are at least 5 Lions, there must be at least 25 Deer since L : D = 1 : 5. However, this is not sufficient to provide us actual numbers. Statement 2: If number of lions is a prime number, then for the ratio to work and for Lions to be prime, one of either Elephants or Deer needs to be 1, however we know there are at least 5 deer, then there can only be 1 elephant since for a greater value, Lions would not remain a prime number. Lets look at these cases to understand this better. Case 1: Suppose E : L is 1:5 then we know E : L : D would be 1:5:25 and this does not violate the condition of L being prime. Case 2: If there are more than 1 Elephants say 2 and we have the ratio E:L = 2:5 , then since E: L = L : D, combined ratio E : L : D would be 4 : 10 : 25 but we know Lions are a prime number and this violates the condition. Case 3: Say Elephants are larger in number as in 25 and there are 5 lions. Then by the ratio E : L = L : D, there must be only 1 Deer, but this violates the conditions given in stem about deer being at least 5. If we need to make Deer at least 5, we would have to multiply Elephants and Lions to the common ratio but that would again make Lions nonprime. This is a great question and it took me a while to grasp this and I am still hoping to learn more from it.
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Originally posted by jedit on 16 Jul 2017, 05:49.
Last edited by jedit on 16 Jul 2017, 06:22, edited 5 times in total.



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Re: In a zoo, the ratio of elephants to lions is same as lions to deer. If
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16 Jul 2017, 11:04
chetan2u wrote: In a zoo, the ratio of elephants to lions is same as lions to deer. If there are at least 5 deer and all three  deer, lion and elephant are different in number, what is the number of elephants?
(1) The ratio of elephant to lion is 1:5 (2) The number of lions is a Prime number.
source self made My answer: "C"Ques > E/L = L/D and D>=5 Stmt 1: E/L =1/5; Insufficient Stmt 2: L = Prime; Insufficient Since, L=3 E=1 then E/L = 1/3 Therefore D can be 9 so that L/D = 3/9=1/3 Similarly we can come up with infinitely many such cases Stmt 1&2 combined: Sufficient E/L=1/5 and L has to be prime. Then the only feasible way is for the common factor to be "1". So that L=5; Thus E=1.
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Re: In a zoo, the ratio of elephants to lions is same as lions to deer. If
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16 Jul 2017, 11:07
gmatexam439 wrote: chetan2u wrote: In a zoo, the ratio of elephants to lions is same as lions to deer. If there are at least 5 deer and all three  deer, lion and elephant are different in number, what is the number of elephants?
(1) The ratio of elephant to lion is 1:5 (2) The number of lions is a Prime number.
source self made My answer: "C"Ques > E/L = L/D and D>=5 Stmt 1: E/L =1/5; Insufficient Stmt 2: L = Prime; Insufficient Since, L=3 E=1 then E/L = 1/3 Therefore D can be 9 so that L/D = 3/9=1/3 Similarly we can come up with infinitely many such cases Stmt 1&2 combined: Sufficient E/L=1/5 and L has to be prime. Then the only feasible way is for the common factor to be "1". So that L=5; Thus E=1. in your statement 2 explanation, you proved elephant has to be 1 for any prime number of lions... Sent from my Redmi Note 4 using GMAT Club Forum mobile app
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Re: In a zoo, the ratio of elephants to lions is same as lions to deer. If
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16 Jul 2017, 22:02
jedit wrote: chetan2u wrote: In a zoo, the ratio of elephants to lions is same as lions to deer. If there are at least 5 deer, what is the number of elephants?
(1) The ratio of elephant to lion is 1:5 (2) The number of lions is a Prime number.
source self made It should be B. From stem, we know E : L is same as L : D and know there are at least 5 deer but we do not have exact ratios. Statement 1: We know \(\frac{E}{L} = \frac{L}{D} = \frac{1}{5}\). If there is just 1 Elephant, E : L tells us there must be at least 5 Lions and if there are at least 5 Lions, there must be at least 25 Deer since L : D = 1 : 5. However, this is not sufficient to provide us actual numbers. Statement 2: If number of lions is a prime number, then for the ratio to work and for Lions to be prime, one of either Elephants or Deer needs to be 1, however we know there are at least 5 deer, then there can only be 1 elephant since for a greater value, Lions would not remain a prime number. Lets look at these cases to understand this better. Case 1: Suppose E : L is 1:5 then we know E : L : D would be 1:5:25 and this does not violate the condition of L being prime. Case 2: If there are more than 1 Elephants say 2 and we have the ratio E:L = 2:5 , then since E: L = L : D, combined ratio E : L : D would be 4 : 10 : 25 but we know Lions are a prime number and this violates the condition. Case 3: Say Elephants are larger in number as in 25 and there are 5 lions. Then by the ratio E : L = L : D, there must be only 1 Deer, but this violates the conditions given in stem about deer being at least 5. If we need to make Deer at least 5, we would have to multiply Elephants and Lions to the common ratio but that would again make Lions nonprime. This is a great question and it took me a while to grasp this and I am still hoping to learn more from it. E  1 L  2 D  4 E  1 L  3 D  9 etc... Sent from my D5503 using GMAT Club Forum mobile app



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Re: In a zoo, the ratio of elephants to lions is same as lions to deer. If
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16 Jul 2017, 23:40
VachePBH wrote: jedit wrote: chetan2u wrote: In a zoo, the ratio of elephants to lions is same as lions to deer. If there are at least 5 deer, what is the number of elephants?
(1) The ratio of elephant to lion is 1:5 (2) The number of lions is a Prime number.
source self made It should be B. From stem, we know E : L is same as L : D and know there are at least 5 deer but we do not have exact ratios. Statement 1: We know \(\frac{E}{L} = \frac{L}{D} = \frac{1}{5}\). If there is just 1 Elephant, E : L tells us there must be at least 5 Lions and if there are at least 5 Lions, there must be at least 25 Deer since L : D = 1 : 5. However, this is not sufficient to provide us actual numbers. Statement 2: If number of lions is a prime number, then for the ratio to work and for Lions to be prime, one of either Elephants or Deer needs to be 1, however we know there are at least 5 deer, then there can only be 1 elephant since for a greater value, Lions would not remain a prime number. Lets look at these cases to understand this better. Case 1: Suppose E : L is 1:5 then we know E : L : D would be 1:5:25 and this does not violate the condition of L being prime. Case 2: If there are more than 1 Elephants say 2 and we have the ratio E:L = 2:5 , then since E: L = L : D, combined ratio E : L : D would be 4 : 10 : 25 but we know Lions are a prime number and this violates the condition. Case 3: Say Elephants are larger in number as in 25 and there are 5 lions. Then by the ratio E : L = L : D, there must be only 1 Deer, but this violates the conditions given in stem about deer being at least 5. If we need to make Deer at least 5, we would have to multiply Elephants and Lions to the common ratio but that would again make Lions nonprime. This is a great question and it took me a while to grasp this and I am still hoping to learn more from it. E  1 L  2 D  4 E  1 L  3 D  9 etc... Sent from my D5503 using GMAT Club Forum mobile appYes, number of elephants remains 1
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Re: In a zoo, the ratio of elephants to lions is same as lions to deer. If
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17 Jul 2017, 01:53
jedit wrote: gmatexam439 wrote: My answer: "C"
Ques > E/L = L/D and D>=5
Stmt 1: E/L =1/5; Insufficient Stmt 2: L = Prime; Insufficient Since, L=3 E=1 then E/L = 1/3 Therefore D can be 9 so that L/D = 3/9=1/3 Similarly we can come up with infinitely many such cases
Stmt 1&2 combined: Sufficient E/L=1/5 and L has to be prime. Then the only feasible way is for the common factor to be "1". So that L=5; Thus E=1. in your statement 2 explanation, you proved elephant has to be 1 for any prime number of lions... Sent from my Redmi Note 4 using GMAT Club Forum mobile appNo Bro, Read it again. I said if L=prime, then Case 1 > Let L=3. Now lets take E=1 then E/L=1/3. Thus, L/D=1/3 Hence D=9 > E=1 Case 2 > Let L=11. Now Lets take E=11 then E/L=11/11. Thus, L/D=1 Hence D=L=11 > E=11 Insufficient. Please let me know in case I have missed any point.  Kudos if it helps!
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Re: In a zoo, the ratio of elephants to lions is same as lions to deer. If
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17 Jul 2017, 02:03
jedit wrote: VachePBH wrote: jedit wrote: It should be B.
From stem, we know E : L is same as L : D and know there are at least 5 deer but we do not have exact ratios.
Statement 1: We know \(\frac{E}{L} = \frac{L}{D} = \frac{1}{5}\). If there is just 1 Elephant, E : L tells us there must be at least 5 Lions and if there are at least 5 Lions, there must be at least 25 Deer since L : D = 1 : 5. However, this is not sufficient to provide us actual numbers.
Statement 2: If number of lions is a prime number, then for the ratio to work and for Lions to be prime, one of either Elephants or Deer needs to be 1, however we know there are at least 5 deer, then there can only be 1 elephant since for a greater value, Lions would not remain a prime number. Lets look at these cases to understand this better.
Case 1: Suppose E : L is 1:5 then we know E : L : D would be 1:5:25 and this does not violate the condition of L being prime. Case 2: If there are more than 1 Elephants say 2 and we have the ratio E:L = 2:5 , then since E: L = L : D, combined ratio E : L : D would be 4 : 10 : 25 but we know Lions are a prime number and this violates the condition. Case 3: Say Elephants are larger in number as in 25 and there are 5 lions. Then by the ratio E : L = L : D, there must be only 1 Deer, but this violates the conditions given in stem about deer being at least 5. If we need to make Deer at least 5, we would have to multiply Elephants and Lions to the common ratio but that would again make Lions nonprime.
This is a great question and it took me a while to grasp this and I am still hoping to learn more from it.
E  1 L  2 D  4 E  1 L  3 D  9 etc... Sent from my D5503 using GMAT Club Forum mobile appYes, number of elephants remains 1 Hi, I think we are missing a point here: As per the given question stem: E/L = L/D => D=L^2/E Now if we take into account stmt 2: L=prime and since # of dears have to be positive integer, we have the following stmt: "E has to be a factor of prime number" This means that since L is prime, it can have 2 factors "1" and "L". Thus "E" can be either "1" or "L". Since we have 2 answers given by stmt 2, stmt 2 shouldn't be sufficient. chetan2u Please correct me if i am wrong.  Kudos if it helps!
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Re: In a zoo, the ratio of elephants to lions is same as lions to deer. If
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17 Jul 2017, 02:07
jedit wrote: VachePBH wrote: jedit wrote: [quote="chetan2u"]In a zoo, the ratio of elephants to lions is same as lions to deer. If there are at least 5 deer, what is the number of elephants?
(1) The ratio of elephant to lion is 1:5 (2) The number of lions is a Prime number.
source self made It should be B. From stem, we know E : L is same as L : D and know there are at least 5 deer but we do not have exact ratios. Statement 1: We know \(\frac{E}{L} = \frac{L}{D} = \frac{1}{5}\). If there is just 1 Elephant, E : L tells us there must be at least 5 Lions and if there are at least 5 Lions, there must be at least 25 Deer since L : D = 1 : 5. However, this is not sufficient to provide us actual numbers. Statement 2: If number of lions is a prime number, then for the ratio to work and for Lions to be prime, one of either Elephants or Deer needs to be 1, however we know there are at least 5 deer, then there can only be 1 elephant since for a greater value, Lions would not remain a prime number. Lets look at these cases to understand this better. Case 1: Suppose E : L is 1:5 then we know E : L : D would be 1:5:25 and this does not violate the condition of L being prime. Case 2: If there are more than 1 Elephants say 2 and we have the ratio E:L = 2:5 , then since E: L = L : D, combined ratio E : L : D would be 4 : 10 : 25 but we know Lions are a prime number and this violates the condition. Case 3: Say Elephants are larger in number as in 25 and there are 5 lions. Then by the ratio E : L = L : D, there must be only 1 Deer, but this violates the conditions given in stem about deer being at least 5. If we need to make Deer at least 5, we would have to multiply Elephants and Lions to the common ratio but that would again make Lions nonprime. This is a great question and it took me a while to grasp this and I am still hoping to learn more from it. E  1 L  2 D  4 E  1 L  3 D  9 etc... Sent from my D5503 using GMAT Club Forum mobile appYes, number of elephants remains 1[/quote] Yeah,we can prove it arithmetically E:L is the same as L:D The relationship between these will be as follows, If we have the fraction of E:L, then we can find the number of deers with the following formula D = L^2/E As we are talking about units of animals, then these figures should be integers. With the statement B, we are told that L is a prime number. So, if we want D to be an integer E should be a factor of prime number squared. The possible values for E 1. E = L^2, in this case D becomes 1, which is not acceptable as the problem told that the least number of dears is 5. 2. E = L, in this case D = L, which means that all numbers are the same, and this is also not acceptable, as the numbers are different. 3. E = 1, when D = L^2, the only possible solution. Hence the statement B is sufficient, and the number of Elephants is always 1. Sent from my D5503 using GMAT Club Forum mobile app



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Re: In a zoo, the ratio of elephants to lions is same as lions to deer. If
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17 Jul 2017, 02:17
VachePBH wrote: Yeah,we can prove it arithmetically E:L is the same as L:D The relationship between these will be as follows, If we have the fraction of E:L, then we can find the number of deers with the following formula D = L^2/E As we are talking about units of animals, then these figures should be integers. With the statement B, we are told that L is a prime number. So, if we want D to be an integer E should be a factor of prime number squared. The possible values for E 1. E = L^2, in this case D becomes 1, which is not acceptable as the problem told that the least number of dears is 5. 2. E = L, in this case D = L, which means that all numbers are the same, and this is also not acceptable, as the numbers are different. 3. E = 1, when D = L^2, the only possible solution. Hence the statement B is sufficient, and the number of Elephants is always 1. Sent from my D5503 using GMAT Club Forum mobile appHi, I think we can have more than 1 eligible values of E. Plz go through the explanation i posted. https://gmatclub.com/forum/inazoothe ... l#p1889701Please let me know if i missed out anything. I love healthy discussions  Kudos if it helps!
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Re: In a zoo, the ratio of elephants to lions is same as lions to deer. If
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17 Jul 2017, 02:22
gmatexam439 wrote: VachePBH wrote: Yeah,we can prove it arithmetically E:L is the same as L:D The relationship between these will be as follows, If we have the fraction of E:L, then we can find the number of deers with the following formula D = L^2/E As we are talking about units of animals, then these figures should be integers. With the statement B, we are told that L is a prime number. So, if we want D to be an integer E should be a factor of prime number squared. The possible values for E 1. E = L^2, in this case D becomes 1, which is not acceptable as the problem told that the least number of dears is 5. 2. E = L, in this case D = L, which means that all numbers are the same, and this is also not acceptable, as the numbers are different. 3. E = 1, when D = L^2, the only possible solution. Hence the statement B is sufficient, and the number of Elephants is always 1. Sent from my D5503 using GMAT Club Forum mobile appHi, I think we can have more than 1 eligible values of E. Plz go through the explanation i posted. https://gmatclub.com/forum/inazoothe ... l#p1889701Please let me know if i missed out anything. I love healthy discussions  Kudos if it helps! I guess you missed the point 2 of my explanation. The numbers should be different, it was given in the problem. So whenever you take E=L, then D=L, L=L. And as it was stated they should be different and this case is not possible. gmatexam439 wrote: jedit wrote: gmatexam439 wrote: My answer: "C"
Ques > E/L = L/D and D>=5
Stmt 1: E/L =1/5; Insufficient Stmt 2: L = Prime; Insufficient Since, L=3 E=1 then E/L = 1/3 Therefore D can be 9 so that L/D = 3/9=1/3 Similarly we can come up with infinitely many such cases
Stmt 1&2 combined: Sufficient E/L=1/5 and L has to be prime. Then the only feasible way is for the common factor to be "1". So that L=5; Thus E=1. in your statement 2 explanation, you proved elephant has to be 1 for any prime number of lions... Sent from my Redmi Note 4 using GMAT Club Forum mobile appNo Bro, Read it again. I said if L=prime, then Case 1 > Let L=3. Now lets take E=1 then E/L=1/3. Thus, L/D=1/3 Hence D=9 > E=1 Case 2 > Let L=11. Now Lets take E=11 then E/L=11/11. Thus, L/D=1 Hence D=L=11 > E=11 Insufficient. Please let me know in case I have missed any point.  Kudos if it helps! Sent from my D5503 using GMAT Club Forum mobile app



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Re: In a zoo, the ratio of elephants to lions is same as lions to deer. If
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17 Jul 2017, 02:34
Agreed. I found my mistake. Then option "B" is correct.
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Re: In a zoo, the ratio of elephants to lions is same as lions to deer. If
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17 Jul 2017, 02:35
gmatexam439 wrote: jedit wrote: It should be B.
From stem, we know E : L is same as L : D and know there are at least 5 deer but we do not have exact ratios.
Statement 1: We know \(\frac{E}{L} = \frac{L}{D} = \frac{1}{5}\). If there is just 1 Elephant, E : L tells us there must be at least 5 Lions and if there are at least 5 Lions, there must be at least 25 Deer since L : D = 1 : 5. However, this is not sufficient to provide us actual numbers.
Statement 2: If number of lions is a prime number, then for the ratio to work and for Lions to be prime, one of either Elephants or Deer needs to be 1, however we know there are at least 5 deer, then there can only be 1 elephant since for a greater value, Lions would not remain a prime number. Lets look at these cases to understand this better.
Case 1: Suppose E : L is 1:5 then we know E : L : D would be 1:5:25 and this does not violate the condition of L being prime. Case 2: If there are more than 1 Elephants say 2 and we have the ratio E:L = 2:5 , then since E: L = L : D, combined ratio E : L : D would be 4 : 10 : 25 but we know Lions are a prime number and this violates the condition. Case 3: Say Elephants are larger in number as in 25 and there are 5 lions. Then by the ratio E : L = L : D, there must be only 1 Deer, but this violates the conditions given in stem about deer being at least 5. If we need to make Deer at least 5, we would have to multiply Elephants and Lions to the common ratio but that would again make Lions nonprime.
This is a great question and it took me a while to grasp this and I am still hoping to learn more from it. E  1 L  2 D  4 E  1 L  3 D  9 etc... Sent from my D5503 using GMAT Club Forum mobile appYes, number of elephants remains 1 Hi, I think we are missing a point here: As per the given question stem: E/L = L/D => D=L^2/E Now if we take into account stmt 2: L=prime and since # of dears have to be positive integer, we have the following stmt: "E has to be a factor of prime number" This means that since L is prime, it can have 2 factors "1" and "L". Thus "E" can be either "1" or "L". Since we have 2 answers given by stmt 2, stmt 2 shouldn't be sufficient. chetan2u Please correct me if i am wrong.  Kudos if it helps![/quote] Hi, You are correct that E can take TWO values  1 and L... BUT there are ATLEAST 5 deers, so deer cannot be 1 and thus it will be L... ONLY remaining value of E is 1... So B is the answer
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Re: In a zoo, the ratio of elephants to lions is same as lions to deer. If
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17 Jul 2017, 04:15
chetan2u wrote: In a zoo, the ratio of elephants to lions is same as lions to deer. If there are at least 5 deer and all three  deer, lion and elephant are different in number, what is the number of elephants?
(1) The ratio of elephant to lion is 1:5 (2) The number of lions is a Prime number.
source self made Answer should be B E/L=L/D E=L*L/D Given that L is a prime number, So E can be integer only whenD=L*L D=L(Not possible since all the three numbers are different) or D=1 (Not possible as D>=5) So the only possible solution is D=L*L > E=1 Statement 2 is sufficient Statement 1 is insuff as many values for E are possible 5:25,25:125 etc




Re: In a zoo, the ratio of elephants to lions is same as lions to deer. If
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