Quantitative Aptitude - Arithmetic Ability Questions

A triangle is formed by joining any three non-collinear points in pairs.

There are 7 non-collinear points

The number of triangles formed = 7C3 = 35

First,we count the number of odd days for the left over days in the given period.

Here,given period is 12.2.1986 to 1.1.1987

Feb Mar Apr May June July  Aug Sept Oct Nov Dec Jan

16   31    30  31   30    31   31   30   31  30   31   1 (left days)

2 + 3 + 2 + 3 + 2 + 3 + 3 + 2 + 3 + 2 + 3 + 1(odd days) = 1 odd day

So,given day Wednesday + 1 = Thursday is the required result.

Average speed = (2xy) /(x + y) km/hr

                      = (2 * 50 * 30) / (50 + 30) km/hr.

                         37.5 km/hr.

Let A, B, C be the events of getting a white ball in first, second and third draw respectively, then 

 Required probability = PA∩B∩C 

= PA PBA PCA∩B

 Now, P(A) = Probability of drawing a white ball in first draw = 4/18 = 2/9

When  a white ball is drawn in the first draw there are 17 balls left in the urn, out of which 3 are white

 ∴PBA=317 

Since the ball drawn is not replaced, therefore after drawing a white ball in the second draw there are 16 balls left in the urn, out of which 2 are white.

 ∴PCA∩B =216=18

 Hence the required probability = 29×317×18=1204

Sum of digits = (5 + 1 + 7 + x + 3 + 2 + 4) = ( 22 + x ), which must be divisible by 3.

When x=2. [(22 + 2) = 24 is divisible by 3]

So, the answer is 2

area of cross roads = (55 x 4) + (35 x 4)- (4 x 4) = 344sq m

cost of graveling = 344 x  (75/100) = Rs. 258

We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys). 

Required number of ways = 6C1*4C3+6C2*4C2+6C3*4C1+6C4  

= 6C1*4C1+6C2*4C2+6C3*4C1+6C2 = 209.