Quantitative Aptitude Prep Tips for MBA
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New answer posted
a month agoContributor-Level 10
f (x) = x – x2 + 1
g (x) = x2 + b + 3
f (2) g (1) < 0
(a – 4 + 1) (1 + 6 + 3) < 0
(a – 3) (b + 4) < 0
New answer posted
a month agoContributor-Level 10
100! has 24 zeroes.
100! + 200! = 100! [1 + 101 * 102 * … * 200]
Which will again give 24 zeroes at the end.
New answer posted
a month agoContributor-Level 10
Let 2P = 3Q = 4R = 5S = K
Then,
K=1200
Therefore. R = 300, S = 240
Difference=300 - 240 = 60
New answer posted
a month agoContributor-Level 10
400 students * 80 days = 240 students * d days
d =
The food was to last 80 more days but now it is lasting 53.33 days more.
New answer posted
a month agoContributor-Level 10
| A | B |
I | 5x | 3x |
II | 4y | 7y |
5x + 4y = 253 …. (1)
3x + 7y = 161 …. (2)
On solving equations (1) & (2), we get
x = 49 and y = 2.
Hence, type I Alloy = 392 kg, and Type II Alloy = 22 kg
New answer posted
a month agoContributor-Level 10
The distances covered by them are in the ratio 3:5 and difference is 30 m. This means 5x – 3x = 30 or 2x = 30 or x = 15 m
Therefore, the total distance covered is equal to 120 m. Their speeds are also in the ratio 3 : 5. Since the speed of the first person is 6 m/s, the speed of the second person will be 10 m/s and he has to cover a distance of 45 m.
Therefore time taken
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