Aptitude and Reasoning Questions

In all other pairs, 2nd number = (1st number × 3) + 1.

P(X) = 15, P(Y) =14 , P(Z) = 13

Required probability:

= [ P(A)P(B){1−P(C)} ] + [ {1−P(A)}P(B)P(C) ] + [ P(A)P(C){1−P(B)} ] + P(A)P(B)P(C)

=14*13*45+34*13*15+23*14*15+14*13*15

= 460+360+260+160= 1060= 16

The given figure can be labelled as :

The simplest triangles are AJF, FBG, HDI, GCH and JEI i.e 5 in number.
The triangles composed of the three components each are AIC, FCE, ADG, EBH and DJB i.e 5 in number.
Thus, there are 5 + 5 = 10 triangles in the given figure.

Assume x soldiers join the fort. 1200 soldiers have provision for 1200 (days for which provisions last them)(rate of consumption of each soldier)
= (1200)(30)(3) kg.

Also provisions available for (1200 + x) soldiers is (1200 + x)(25)(2.5) k

As the same provisions are available
=> (1200)(30)(3) = (1200 + x)(25)(2.5)

x = ([(1200)(30)(3)] / (25)(2.5)) - 1200 => x = 528.

when a cube is formed by folding the sheet shown in fig. (A), then   is one of the faces of the cube and this face lies opposite to a blank face. Also, a face bearing a square lies opposite to another blank face. The remaining two blank faces lie opposite to each other. Clearly, in the cube shown in fig. (1), the face consisting of the four symbols is not the same as that formed (as shown above). Hence, the cube in fig. (1) cannot be formed

The amount of petrol left after 4 operations = 200×1-402004=81.92 

Hence the amount of kerosene = 200 - 81.92 = 118.08 liters