Elitmus
Exam
Numerical Ability
Number System
Let x = √4+√4-√4+√4-....infinity
then x equals:
A. 3
B. (√13 - 1)/2
✓C .(√13 + 1)/2
D.√13
Please Explain.
Read Solution (Total 14)
-
- To solve this question we have this formula.
[1+(sqrt 4n-3)] /2
so fir n=4, ans will be (1+sqrt 13)/2 - 10 years agoHelpfull: Yes(15) No(1)
- i think anwser is (sqrt13+1)/2
- 10 years agoHelpfull: Yes(8) No(0)
- after having equation x^4-8x^2+x+12=0;
then go through option .
1+sqrt13/2 will be root only - 9 years agoHelpfull: Yes(3) No(0)
- 2+2/(1-(-1))=3
- 10 years agoHelpfull: Yes(2) No(2)
- We know that √4 – anything will be less than 2. So we get √ 4 + (less than 2). Which should be more than 2 but less than 2.5. The only option that satisfied is c option with √13 is approximately 3.5. So (3.5 + 1)/2 = 2.25. The same sum can be solved by formulae but that will take time.We know that √4 – anything will be less than 2. So we get √ 4 + (less than 2). Which should be more than 2 but less than 2.5. The only option that satisfied is c option with √13 is approximately 3.5. So (3.5 + 1)/2 = 2.25. The same sum can be solved by formulae but that will take time.
- 9 years agoHelpfull: Yes(2) No(1)
- ans is c.. u can check it here http://testfunda.com/examprep/catcentre/cat-100-percentilers/article/prepare-for-cat-quant-section-with-less-than-1-month-to-go.htm?assetid=011475c1-4b2a-468f-8095-c22350e27520
- 9 years agoHelpfull: Yes(1) No(0)
- Ans. C
We know that √4 – anything will be less than 2. So we get √ 4 + (less than 2). Which should be more than 2 but less than 2.5. The only option that satisfied is c option with √13 is approximately 3.5. So (3.5 + 1)/2 = 2.25. The same sum can be solved by formula but that will take time. - 9 years agoHelpfull: Yes(1) No(1)
- correct answe is c
- 8 years agoHelpfull: Yes(1) No(1)
- the option a is 2 instead of 3
x=sqrt4+sqrt4-sqrt4+sqrt4-.......
x=sqrt4 + sqrt4 - x
2x=2(sqrt4)
x=sqrt4
x=2 - 7 years agoHelpfull: Yes(1) No(1)
- X=root4+root4-.................infinity
squaring both side
x2=4+x
so eqation becomes as--
x2-x-4=0
finding roots of x as--
-------------formula------------
roots=[b2-sqre (b2-4ac)]/2a
-------------------------
=[-(-1)+-sqre(-1)2-4*(1)(-4)]/2(-1)
=[1+sqre(1+16)]/2
correct answer is=(√17+1)/2
which is not given in the answer - 10 years agoHelpfull: Yes(0) No(27)
- (x^2-4)=root4-x
now put the values from option only option (c) satisfies the condition - 10 years agoHelpfull: Yes(0) No(1)
- @javed can u pls explain brother?
- 10 years agoHelpfull: Yes(0) No(1)
- 3 coz after expect fst term the total series is n gp.ie. 2-2+2.....and here the sum is 1 (2/1-(-1)) whre -1 is r.then add 1 to fst term ..and i.e.3
- 10 years agoHelpfull: Yes(0) No(1)
- If x=root 4+root 4-root 4+root 4 - root 4..........to infinity
x=(1+root 4n-3)/2 =1+root 13/2 - 7 years agoHelpfull: Yes(0) No(0)
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