SBI Associate PO 2010

Average distance travelled by truck S = $$\frac{325+314+312+278+292+274}{6} = 299\frac{1}{6}$$ km

Distance travelled by P on Monday = 240 km
Speed = 19.2 kmph
Speed = Distance/Time
Time = $$\frac{240}{19.2} = 12.5$$ hr
$$\therefore$$ P took 12 hr 30 minutes to cover the distance.

Distance travelled by truck Q on Friday = 302 km
Time taken = 8 hrs
Speed = $$\frac{302}{8}$$ = 37.75 kmph

Distance travelled by all the trucks on Saturday = 292+284+260+274+280+242 = 1632 km

If the speeds are same, the ratio of time taken is same as that of the distance travelled.
$$\frac{\text{Time taken by truck R}}{\text{Time taken by truck T}} = \frac{308}{318} = \frac{154}{159}$$

The persons appeared in interview are 84%of 25780 is 3094 Aprox

,Persons who selected for the interview from engineering are 11% Of 7390 Is= 813 hence,B

candidates appeared from management = (12/100)*25780 = 3093.6

from others = (12/100)*25780 = 3093.6

from eng. = (16/100)*25780 = 4124.8

reqd ans = (3093.6+3093.6)/4124.8 = 6187.2/4124.8 = 1.5

%age = 1.5*100 = 150

(shortcut)

total % from mgmt and others = 12 + 12 =24%

from eng = 16%

ans 24/16 = 1.5

=> 150%

Engineering discipline students are 11%, agriculture discipline students are 7%, Difference is 4% Of 7390 So approx 296. We can calculate 11% of 7390 and 7% of 7390 can subtract them individually but the earler method is the easiest

For the difference to be max we need higher upper limit,

hence, consider 28% of 25780

we have 5421 And 32% of 7390,

We have The dif beyond 3000, All other values are less

Commerce & agriculture together constitutes 23% .

also It is23% Of 7390 &

(23/100)*7390

Equal to 1700 Approx