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Algebra Level 1 Test 2
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Question 1 |
If x + 1/x = 13, what is the value of x2 + 1/x2?
169 | |
167 | |
143 | |
26 |
Question 1 Explanation:
x + 1/x = 13
(Squaring both sides)
x2 + 1/x2 + 2.x.1/x = 169.
x2 + 1/x2 = 169 – 2 = 167.
(Squaring both sides)
x2 + 1/x2 + 2.x.1/x = 169.
x2 + 1/x2 = 169 – 2 = 167.
Question 2 |
Find the value of X & Y in 3X + 4Y = 25 ,4X – 2Y = 4
X = 4 Y = 3 | |
X = 3 Y = 4
| |
X = 5 Y = 3 | |
X = 3 Y = 3 |
Question 2 Explanation:
Multiply the second equation with 2 and then add
both the equations, we get X= 3 and Y= 4
both the equations, we get X= 3 and Y= 4
Question 3 |
If product of the two numbers is 24, find the Maximum value of the sum of the numbers ?
25 | |
10 | |
14 | |
11 |
Question 3 Explanation:
The sum of the number will be maximum when the
difference of the number is maximum :
Hence axb= 24 so possible solution can be 1*24 = 24 (hence a=1 b =24 )
Maximum value of the sum = 1+24= 25
difference of the number is maximum :
Hence axb= 24 so possible solution can be 1*24 = 24 (hence a=1 b =24 )
Maximum value of the sum = 1+24= 25
Question 4 |
If (a + b) = 36 and ab = 323, find (a – b)
6 | |
2 | |
1 | |
7 |
Question 4 Explanation:
As a x b = 323, so possible pairs are 17 x 19 , 323 x 1
So only possible pair for a+b= 36 is when a=17 b=19
Thus, the difference is option b
So only possible pair for a+b= 36 is when a=17 b=19
Thus, the difference is option b
Question 5 |
If (a + b) = 70 and ab = 1221, find (a – b)
6 | |
2 | |
4 | |
7 |
Question 5 Explanation:
As a x b = 1221 = so possible pairs are 37 x 33 ,
111 x 11 , 1221 x 1
So only possible pair for a+b= 70 is when a=37 b=33
Thus, the difference is option 3
111 x 11 , 1221 x 1
So only possible pair for a+b= 70 is when a=37 b=33
Thus, the difference is option 3
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