Latest Infosys Aptitude Question SOLUTION: Q. If CROSS + ROADS = DANGER then D+A+N+G+E+R=? Option a)31 b)21 c)11 d)16

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(#M40032543) INFOSYS Logical Reasoning question Reasoning Keep an EYE

Q. If CROSS + ROADS = DANGER then D+A+N+G+E+R=?
Option
a)31
b)21
c)11
d)16

Asked In Infosys mounika (10 years ago)

Unsolved Read Solution (9)

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Read Solution (9) :

sandeep kumar  (10 years ago)

96233+62513=1+5+8+7+4+6=31

Is this solution Helpfull? Yes (14) | No (3)  

Divyanshu Gupta  (9 years ago)

C R O S S
+ R O A D S
D A N G E R

clearly d=1 becoz max. possible carry is 1. s+s=r means r is even, assume r=6 then s=3.
now s+d=e ==> e=4
now we are left with 0,2,5,6,7,8,9
assume c=9, as c+r=a ===>a=5
o+a=g , assume o=2 then g=7
at last n will be 8.

******All assumptions are made by hit and trial,i have written only the correct cases
so D+A+N+G+E+R=31

Is this solution Helpfull? Yes (7) | No (2)  

KARTIK MISHRA  (10 years ago)

plz explain clrly...

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mounika  (10 years ago)

please explain it...

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sajid  (10 years ago)

USA
USSR
------ hence p=1. & E=0 and since E not equal to U. so U=9 and adding a carry
PEACE 1 with it to 10 as PE. now U+s=10+A to produce a carry for the lastterm
i.e 9+S=10+A.....(1)
again E=0, so A+R=10, hense it produce a carry 1,which adds up with S+S to make C, i.e 2S+1=c, s9, but s=0,4 not possible as it gives C=1,9, which is not possible as P=1 & U=9. so S may be 1,2,3. but S=1 gives A=0 not possible, S=2 gives A=1 not possible. so only S=3 is possible. now S=3 gives A=2 & C=7. A+R=10, R=8. P+E+A+C+E=1+0+2+7+0= 10(Ans.)

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yashaswini kale  (10 years ago)

b (d+a+n+g+e+r=1+0+4+6+2+8=21)

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monika  (10 years ago)

pls help with the procedure to follow for solving these ques??

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priyamahalakshmi.c  (8 years ago)

can anyone explain these type of ques clearly??

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HARSHINI  (8 years ago)

ans:D+A+N+G+E+R=31

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