Coded Directions Sense Questions, type of which were asked in IBPS PO Mains 2017 Exam
Directions (1-5): Read the following information carefully to answer the questions that follow. The questions are based on following coding formats:
# – North
@ – South
% – East
$ – West
! – Either 4 or 6 m
& – Either 3 or 10 m
Examples: P@Q means P is South of Q, P#$Q means P is North-West of Q, P$!Q means P is West of Q at a distance of either 4 or 6 m.
Conditions given are as:
- Given H$G, HG= 3 m, B#H. Find distance between B and C.
4 m5 m6 m8 mCannot be determinedOption B
To make G south-west of C, length of CD should be smaller than FE, which is possible when CD = 4 m, and FE = 6 m. Also FG should be smaller than DE, which is possible with FG = 4 or 6 and DE = 10 m
Assume point P – north of G and west of C. So PC = 2 m (FE-CD). Since B is north of H, so BP = HG = 3 m
So BC = BP+PC = 3+2 = 5 m
- If B#G, then which of the following is true regarding points B and G?
G#&BB#&GB@!GG@!BNone of theseOption D
FG = 4 or 6 m,
DE = 10 m as explained in Q1.
B is west of D. So FB should be 10 m
Hence, if FG = 4, then GB = 6, and if FG = 6, then GB = 4, G!B
- A man starts walking from point A, walks for 5 m in ‘%’ direction, reaches point J, turns ‘@’ direction, and now walks 7 m to reach point G. Given J#G. What is the distance between points C and A. [Also take data from Question No. 1]
√38 m4√5 m√34 m4√6 m√42 mOption C
J#G – J is north of G. and JG = 7 m, which is only possible when AB = 3 m
from Q1, BC = 5m
So AC = √(32 + 52) = √34 m
- If distance between points G and E is 6√2 m, the
2√5 m3√5 m5√6 m4√6 m5√2 mOption A
GE= 6√2 m
So FG = √(GE2 – FE2) = √(6√2)2 – 62) = √36 = 6 m
Assume point S – east of G and south of C
So GC = √(GS2 + CS2)
GS = FE – CD = 2 m, CS = DE – FG = 10-6 = 4m, now find GC = 2√5 m
- If K@!D and G$K, what is the maximum area of the quadrilateral formed by joining points K, G, F and E?
24 m230 m218 m236 m2None of theseOption D
K@!D – K is south of D and at a distance of 4 or 6 m.
G is west of K. So figure formed by joining given points is a rectangle.
So if DK = 4, KE = FG = 6, and if DK – 6, then KE = FG = 4. Area will be maximum in 1st case. So Ar of rectangle KEFG = FE*De = 6*6 = 36 m2
Directions (6-10): Read the following information carefully to answer the questions that follow. The questions are based on following coding formats:
$ – North
@ – South
# – East
% – West
! – Either 3 or 8 m
& – Either 4 or 9 m
© – Either 1 or 6 m
Examples: P@Q means P is South of Q, P$%Q means P is North-West of Q, P#!Q means P is East of Q at a distance of either 3 or 8 m.
Conditions given are as:
- If distance between points P and D is 8 m, P$D, P%B, find the distance between points P and B.
4 m5 m3 m8 mNone of theseOption A
Since E is east of F, so AB should be greater than CD, which is possible with CD = 4, AB = 8
Point P is 8 m north of point D, point P is west of point B
So PB = AB-CD = 8-4 = 4 m
- Given that L@!A, and L$C, find the distance between points L and C.
2 m7 m8 m3 m5 mOption E
For L to be north of C and A to be north of L, length AL should be 3, and AC be 8. so LC = AC-AL = 8-3 = 5 m
- If K@C, then which of the following is definitely false?
We have to find the one which is definitely false, for 1st and 2nd options, there is a possibility. In D, B is north-west of K, but B will be definitely north-east of K.
- If distance between points D and F is greater than distance between points B and E, then which of the given conditions above will contradict given that E#F is definitely true?
For Q, conditions will be DF = 6, and BE = 4. Now keeping F west of E as same, we get
So in this, A is south of C.
- If distance between points B and E is 4 m, find the distance and direction of point A with respect to point D?
8 m10 m11 m5 m4 mOption D
If BE = 4, then DF = 1 and AC = 3, so
AD = √(AC2+CD2) = √(9+16) = 5 m