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IBPS PO Mains 2018 Paper for online practice
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© examsnet.com
Question : 101
Total: 155
Uday can cover 'X' distance with 'Y' speed in 'Z' time. He can cover same distance with 'Y + 10' speed in (Z – 2) time. He can cover same distance 'X' with 'Y'–15 speed in (Z + 6) time. What can be found from the given data.
(i) time to cover 200 km with speed 'Y + 10'
(ii) distance covered in (Z + 6) time with (Y + 10) speed
(iii) speed by which a tunnel can be crossed in
Z
2
hour
(iv) Ratio between time to cover distance 'D' with speed 'Y' to time to cover distance (X – 5) with speed (Y + 10)
Only (ii)
Only (ii) and (iii)
Only (i) and (iii) (d)
all of the above
Only (i), (ii) and (iv)
Validate
Solution:
ATQ,
Y =
X
Z
... (i)
(Y + 10) =
X
Z
−
2
... (ii)
(Y - 15) =
X
Z
+
6
... (iii)
On solving (i), (ii) & (iii)
X = 400 km, Y = 40 km/hr. Z = 10 hour
Statement 1, 2 and 4 can be found out from the given data but statement 3 can’t be solved as length of tunnel is not given.
© examsnet.com
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