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Numerical Ability
Number System
Three times the first of three consecutive odd integers is 3 more than twice the third. The third integer is:
A. 9 B. 11
C. 13 D. 15
Read Solution (Total 7)
-
- Let numbers are n,n+2,n+4
now 3n=2(n+4)+3
3n-2n=11
so
n=11 then consecutive odd numbers are 11,13,15 - 12 years agoHelpfull: Yes(10) No(7)
- ANS. D
from the given answers the possible three consecutive odd integers to consider to satisfy the condition are
1) 5, 7, 9
2) 7, 9, 11
3) 9, 11, 13
4) 11, 13, 15
Only the option 4 satisfies Three times the first odd integers is 3 more than twice the third.
Hence the third integer is 15 - 13 years agoHelpfull: Yes(7) No(5)
- ans 9
2x+1, 2x+3, 2x+5 are the tree nos.
3* 2x+1 + 3= 2 * 2x+5
x=2
ans 9 - 11 years agoHelpfull: Yes(6) No(3)
- Answer: Option D
Explanation:
Let the three integers be x, x + 2 and x + 4.
Then, 3x = 2(x + 4) + 3 x = 11.
Third integer = x + 4 = 15. - 9 years agoHelpfull: Yes(3) No(1)
- Three integers x,x+2,x+4
3x=3+2(x+4)
3x=3+2x+8
x=11
Therefore 3rd integer is: x+4=11+4=15 - 8 years agoHelpfull: Yes(3) No(0)
- Three consecutive odd numbers
2x+1,2x+3,2x+5
3(2x+1) = 2(2x+5) + 3
Slove this,
X=5
Third num is 2(5)+5=15 - 6 years agoHelpfull: Yes(1) No(0)
- let numbers are x+1,x+3,x+5
3(x+1)=3+2(X+5)
X=10
SO,
11,13,15 are the numbers. - 5 years agoHelpfull: Yes(0) No(0)
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