1. Option B
Solution: Let distance be x km and speed be y km/hr
x/(y-10)-x/y=40==> x=4y(y-10) ——(1)
x/y-x/(y+5)=10==>x=2y(y+5) ——-(2)
Equate 1 and 2 4y(y-10)=2y(y+5) 2y-20=y+5==>y=25km/hr
Then x=2*25(25+5)=50*30=1500km

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2. Option C
Solution: Let the original speed be 6x.
Travelling 60km at 5/6th of original speed cost  12 mins etc.
60/5x = 60/6x +12/60
==>x=10 Original speed 6x=60km/hr.

3. Option A
Solution: If the distance be x km, then

4. Option D
Solution: Let the duration of the flight be x hrs.
Then, 600/x-600/(x+1/2)=200
2x^2+x-3=0 X=1hr.

5. Option B
Solution: Let second racer takes x hr with speed s2
First racer takes x-5/6 hr with speed s1
Total distance = 50km
S1 = 50/(x-(5/6))
S2= 50/x
As they cross each other in 1hr…
Total speed = s1 + s2
Now, T = D / S
50/(s1+s2) = 1
x = 5/2, 1/3
Put x= 5/2 in s2 –> 20km/hr

6. Option A
Solution: Time taken by P to complete one round 800/40=20
Time taken by Q to complete one round 800/20=40
LCM of 20 40=40
Every 40sec they would be together at the starting point

7. Option D
Solution: If the 1st 3hr Ram covers 90km
So the rest 650-90=560km
Now they both travel together towards each other
So, the time is 560/70=8hr
Then ram travel total 3+8=11hrs
Thus the distance travelled by Ram 11*30=330km

8. Option C
Solution: Speed of bus = 480/8 = 60km/ h
Speed of car = (60*4)/3=80 km / h
Speed of car : Speed of bike = 16 : 15
Speed of bike =80/16 * 15 = 75 km/ h
Distance covered by bike in 6 hr = 75 × 6 = 450 km

9. Option B
Solution: Relative speed =42-30=12km/hr
Time= (50+80)*18/12*5
=130*18/12*5==>13*3 =39sec.

10. Option C Solution:
Average speed=total distance/total time
Total time=20/8 + 45/10
Avg speed=(20+45)/(20/8 + 45/10)
= 65/((200+360)/80)
=65*80/560 = 65/7 =9.3km/hr

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