Quantitative Aptitude Prep Tips for MBA
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3 months agoContributor-Level 9
In the first group, one question can be selected or can be rejected; so three questions can be dealt with in 2 * 2 * 2 ways, but this includes the case when all three questions have been left; so they can be selected in 23 – 1 = 7 ways. Similarly four questions of the second group can be selected in 24 – 1 = 15 ways. Thus all seven questions can be selected in 15 * 7 = 105 ways; but this includes the case when all questions have been solved; hence leaving that case, total number of ways required is 105 – 1 = 104
New answer posted
3 months agoContributor-Level 9
Every number between 1000 and 2000, which is divisible by five and which can be formed by the given digits, must contains 5 in unit's place and 1 in thousand's place. Thus we are left with four digits out of which we are to place two between 1 and 5, which can be done in 4P2 = 12 ways. Hence, 12 numbers can be formed.
New answer posted
3 months agoContributor-Level 9
Let L be the vertical distance between the centers of two adjacent balls.
Also, if the box can hold n sphere then (n – 1) L + 2r
≤ 9r [ the last ball height = 2r]
n ≤ 5.04
Since, n is a natural number.
n = 5
So, the box can hold maximum of 5 spheres.
New question posted
3 months agoNew answer posted
3 months agoContributor-Level 9
9B + 3 + 5G + 4 = 200 9B + 5G = 193
The positive integer solutions are (B, G) = (2, 35) (7, 25) (12, 17) (17, 8)
Only one of the (B, G) gives G > B and 9B + 3 > 100, i.e. (B, G) = (12, 17)
17 * 5 = 85
New answer posted
3 months agoContributor-Level 10
(a):Let, s be the speed of stream.
So, 8 + s = 2 (8 – s)
s = 8/3 mph
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