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Updated on: 01 Oct 2023, 13:48
Originally posted by Sajjad1994 on 02 May 2020, 10:07.
Last edited by BottomJee on 01 Oct 2023, 13:48, edited 3 times in total.
Last edited by BottomJee on 01 Oct 2023, 13:48, edited 3 times in total.
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\(x\): 3
\(z\): 6
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
5%
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Question Stats:
91% (01:01) correct
9%
(01:28)
wrong
based on 475
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Given the equation \(x^2−x−z=0\), which of the following could be a value of \(z\) and a corresponding value of \(x\)?
| \(x\) | \(z\) | Statements |
| 1 | ||
| 3 | ||
| 5 | ||
| 6 | ||
| 10 | ||
| 15 |
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Official Answer
\(x\): 3
\(z\): 6
bM22
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02 May 2020, 11:25
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Given equation: x^2−x−z=0
Now we will mostly use elimination technique. Going through options:
1 - rejected , since if either x = 1, or z = 1, then eq. wont be = 0.
3 - could be for x, when z = 6, but z cannot be 3 as again eq. wont be equal to 0.
So we have one pair, when x = 3, z=6 => 3^2 - 3 - 6 = 0.
5 - if x = 5 we need z = 20, not given so rejected.
6 - if x - 6, we need z = 30, not given so rejected.
Similarly we can reject 10 and 15 also.
Now lets also check vice versa for last 4 values.
if z = 15, we need x^2 - x = 15, which is not possible with integer values.
if z = 10, we need x^2 - x = 10, which is again not possible with integer values.
similarly, if we go reverse, for z = 6, we get x = 3. which is given the table.
So answer is:
x = 3,
z = 6.
Now we will mostly use elimination technique. Going through options:
1 - rejected , since if either x = 1, or z = 1, then eq. wont be = 0.
3 - could be for x, when z = 6, but z cannot be 3 as again eq. wont be equal to 0.
So we have one pair, when x = 3, z=6 => 3^2 - 3 - 6 = 0.
5 - if x = 5 we need z = 20, not given so rejected.
6 - if x - 6, we need z = 30, not given so rejected.
Similarly we can reject 10 and 15 also.
Now lets also check vice versa for last 4 values.
if z = 15, we need x^2 - x = 15, which is not possible with integer values.
if z = 10, we need x^2 - x = 10, which is again not possible with integer values.
similarly, if we go reverse, for z = 6, we get x = 3. which is given the table.
So answer is:
x = 3,
z = 6.
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05 May 2020, 14:37
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x= 3 and z=6
First of all, the question has a small typo. Instead of value of Z and X, it says Z and a
The equation: x^2 - x = z.
Using the options:
1 : If x=1 , z= 0. If z=1, x<>R
3 If x=3, z=6. Both listed as options. If z=3, x<>R
5 if x=5, z=20.
6 If x=6, z= 30. However, inverse is true as explained in x=3
10 If x= 10, z= 90. If z= 10, X<>R
15 Same, need not solve.
First of all, the question has a small typo. Instead of value of Z and X, it says Z and a
The equation: x^2 - x = z.
Using the options:
1 : If x=1 , z= 0. If z=1, x<>R
3 If x=3, z=6. Both listed as options. If z=3, x<>R
5 if x=5, z=20.
6 If x=6, z= 30. However, inverse is true as explained in x=3
10 If x= 10, z= 90. If z= 10, X<>R
15 Same, need not solve.
