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## Algebra Level 1 Test 2

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*Algebra Level 1 Test 2*.You scored %%SCORE%% out of %%TOTAL%%.You correct answer percentage: %%PERCENTAGE%% .Your performance has been rated as %%RATING%% Your answers are highlighted below.

Question 1 |

If x + 1/x = 13, what is the value of x

^{2}+ 1/x^{2}?169 | |

167 | |

143 | |

26 |

Question 1 Explanation:

x + 1/x = 13

(Squaring both sides)

x

x

(Squaring both sides)

x

^{2}+ 1/x^{2}+ 2.x.1/x = 169.x

^{2}+ 1/x^{2}= 169 â€“ 2 = 167.Question 2 |

Find the value of X & Y in Â 3X + 4Y = 25, 4X â€“ 2Y = 4

X = 4 Y = 3 | |

X = 3 Y = 4
â€ƒ | |

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 â€“ 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 â€“ 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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