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19 Apr 2016, 06:05
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70% (02:33) correct
30%
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John wants to put the 78 pencils in the 6 boxes. The number of the each pencil in the boxes is different one another and there is no empty boxes. The second largest box of the number of the pencils is named to be B, is the pencil's number of the box B more than 9?
1) The largest number box has 40 pencils and the smallest number box has 5 pencils
2) The box B has the pencils less than the number 14
* A solution will be posted in two days.
1) The largest number box has 40 pencils and the smallest number box has 5 pencils
2) The box B has the pencils less than the number 14
* A solution will be posted in two days.
19 Apr 2016, 16:55
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Since the no. of pencils in each box is supposed to be different and not zero, let the equation be stated as
a+b+c+d+e+f= 78
Statement 1 says the largest box has 40 pencils. The rest of them should add up to 38.
The box with least number of pencil contains 5 pencils.
So , 40+b+c+d+e+5= 78
b+c+d+e= 33
If we take the least values possible for the rest of the boxes apart from box b, we get
b+8+7+6= 33
b=12
Separately, when you take other values for c,d,e , you will find the least value for box B , the box with 2nd highest no. of pencils is 10
Statement 1 is Sufficient
Statement 2 say box B has less than 14 pencils which can either be greater than or less than 9.
Hence, not sufficient
Correct Option : A
a+b+c+d+e+f= 78
Statement 1 says the largest box has 40 pencils. The rest of them should add up to 38.
The box with least number of pencil contains 5 pencils.
So , 40+b+c+d+e+5= 78
b+c+d+e= 33
If we take the least values possible for the rest of the boxes apart from box b, we get
b+8+7+6= 33
b=12
Separately, when you take other values for c,d,e , you will find the least value for box B , the box with 2nd highest no. of pencils is 10
Statement 1 is Sufficient
Statement 2 say box B has less than 14 pencils which can either be greater than or less than 9.
Hence, not sufficient
Correct Option : A
21 Apr 2016, 01:23
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.
John wants to put the 78 pencils in the 6 boxes. The number of the each pencil in the boxes is different one another and there is no empty boxes. The second largest box of the number of the pencils is named to be B, is the pencil's number of the box B more than 9?
1) The largest number box has 40 pencils and the smallest number box has 5 pencils
2) The box B has the pencils less than the number 14
First of all, when it comes to inequality questions, it is important that if range of que includes range of con, that con is sufficient. The question is B>9?, and ,in 2) B<14, the range of que doesn’t include the range of con, which is not sufficient. Also, if the question refers to “more than”, you need to figure out a minimum value. For 1), 5+x+y+z+B+40=78 -> x+y+z+B=33. In this case, (x,y,z,B)=(6,7,8,9) is impossible and (6,8,9,10) is possible, which makes B=10>9. When it is minimum, it is always yes and sufficient.
Thus, the answer is A.
John wants to put the 78 pencils in the 6 boxes. The number of the each pencil in the boxes is different one another and there is no empty boxes. The second largest box of the number of the pencils is named to be B, is the pencil's number of the box B more than 9?
1) The largest number box has 40 pencils and the smallest number box has 5 pencils
2) The box B has the pencils less than the number 14
First of all, when it comes to inequality questions, it is important that if range of que includes range of con, that con is sufficient. The question is B>9?, and ,in 2) B<14, the range of que doesn’t include the range of con, which is not sufficient. Also, if the question refers to “more than”, you need to figure out a minimum value. For 1), 5+x+y+z+B+40=78 -> x+y+z+B=33. In this case, (x,y,z,B)=(6,7,8,9) is impossible and (6,8,9,10) is possible, which makes B=10>9. When it is minimum, it is always yes and sufficient.
Thus, the answer is A.
