Quantitative Aptitude Prep Tips for MBA
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New answer posted
a month agoContributor-Level 10
Given quadratic equation x2 + ax + b = 0
Then product of roots αβ = b
Sum of roots α + β = − a
Next quadratic equation bx2 + ax + 1 = 0
Then product of roots =
Hence, clearly by visualising options
New roots we will be and
New answer posted
a month agoContributor-Level 10
We have (A – 20) = 0.4 (B + 20), i.e. A – 0.4B = 28
And (B – 40) = 0.4 (A + 40), i.e. B – 0.4A = 56
Solving we get, A = Rs. 60
New answer posted
a month agoContributor-Level 10
Let speeds of P, Q & R be P, Q & R km/hr respectively.
Thus 4P = 2R
= . (1)
= 5Q = 4R = . (2)
From (1) & (2),
=
=
New answer posted
a month agoContributor-Level 10
Since M is the midpoint of side PQ, the length of MQ is 2.
Hence, the area of? MQR = * 2 * 4 = 4.
Also area of? NSR = 4. Thus, the unshaded area of the figure = 4 + 4 = 8.
Hence, the area of quadrilateral PMRN
= Area of the square PQRS – The unshaded area of the figure
= 16 – 8 = 8
New answer posted
a month agoContributor-Level 10
Let n = the number of terms.
Then,
Let d be the common difference.
Then (= the eleventh term) = 17 + 10d
New answer posted
a month agoContributor-Level 10
Let oil in containers be A & B.
After 1st operation
Container A = 0.4 A
Container B = 0.6 A + B
After 2nd operation
Container A = 0.4 A + 0.3 A + 0.5 B
Container B = 0.3 A + 0.5 B
=
1.6A = 2B
Volume of A : B = 5 : 4.
New answer posted
a month agoContributor-Level 10
Let the distance be x km and speed be y km/hr and t be the time in hours. Then the equation will be x = yt …. (1),
Therefore,
Also,
On solving the above equations, we get speed is km/hr time is 6 hrs, and distance is 200 km.
New answer posted
a month agoContributor-Level 10
Let the speed of A be x km/hr and speed of B be y km/hr, then:
Also,
On solving both equation (1) & (2) we get x = 11 km/hr and y = 10 km/hr.
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