#### SBI PO Exam Quantitative Aptitude Study Material

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b) 335 m/sec

c) 330 m/sec

d) 300 m/sec

=Distance/speed

=((1.34*1000)/4)metre/second

=(1340/4) metre/second

=335 metre/second

b) 6.5 km

c) 7 km

d) 6 km

=xkm(let)

Difference of time

=7+5=12minutes =12/60 hour =1/5 hour

Time =distance/speed

Therefore (x/5)-(x/6)=1/5

⇒((6x-5x)/30)=1/5

⇒x/30=1/5

x=30/5 =6Km

b) 9 km/hr

c) 12 km/hr

d) 10 km/hr

30/(x-y) + 21/(x+y) = 6.5

Taking 1/(x-y) as X and 1/(x+y) as Y

24X + 28Y = 6

30X + 21Y = 6.5

solving for X and Y=⇒

X = 1/6 and Y = 1/14

so x-y = 6 and x+y = 14 hence

x = 10 kmph and y = 4kmph

b) 8.30 AM

c) 9:00:00 AM

d) 10:00:00 AM

Relative speed = 4 + 6 =10 kmph

Time required to cover 20 km = 20/10 = 2 Hr

They started at 7 AM, so they will meet at 9 AM.

b) 25 : 26

c) 28 : 25

d) 56 : 25

⇒ 1 step of Policeman = 7/5 steps of thief

⇒ 8 Steps of Policeman = 8 x 7/5 steps of thief = 56/5 steps of thief. ? Ratio of speeds of policeman and thief = ratio of distance covered by policeman and thief in same time

= In same time policeman moves 8 steps and thief moves 10 steps

= 56/5 : 10 (because as we calculated earlier 8 steps of policeman = 56/5 steps of thief )

= 56 : 50 = 28 : 25

b) 700 km/hr

c) 680 km/hr

d) 1020 km/hr

Speed = Distance/Time

Given time is 2 2/3 hours = 2 + 2/3 = (2*3) + 2/3 = 8/3 hours

Therefore, speed = 2720/(8/3)

Required speed = 2720 * 3 / 8 = 340 * 3 = 1020 km/hr

b) 4.5 hrs

c) 5 hrs

d) 4.8 hrs

Relative speed = 33+39 = 72 km/hr

= (72×10)/36 m/s

Time = distance/speed

= 240 / (72×10)/36

= 240×36 / 720

= 12 s

b) 10

c) 12

d) 15

676 = P (1 + R/100)

(1 + R/100) = 26/25…… (iii)

Substituting iii in I, we get

650 = P x 26/25

26P = 16250

P = 625

b) 4 : 1

c) 1 : 4

d) 2 : 1

b) 24

c) 16.4

d) 12.8

2x/15+9x/20+10=x

⇒x-2x/15-9x/20=10

⇒25x/60=10

X=24 km