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## Algebra Level 2 Test 10

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*Algebra Level 2 Test 10*.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 |

What is the value of m which satisfies 3m

^{2}– 21m + 30 < 0?A | m < 2 or m > 5 |

B | m > 2 c. |

C | 2 < m < 5 |

D | Both a and c |

Question 1 Explanation:

3m

Dividing by 3

Therefore m

Now on solving this quadratic equation

(m – 5)(m – 2) < 0.

Now there are two cases

Either (m – 5) < 0 and (m – 2) > 0 or (m – 2) < 0 and (m –5) > 0.

Hence, either m < 5 and m > 2, i.e. 2< m < 5 or m < 2 and m > 5.

^{2}– 21m + 30 < 0Dividing by 3

Therefore m

^{2}– 7m + 10 < 0,Now on solving this quadratic equation

(m – 5)(m – 2) < 0.

Now there are two cases

Either (m – 5) < 0 and (m – 2) > 0 or (m – 2) < 0 and (m –5) > 0.

Hence, either m < 5 and m > 2, i.e. 2< m < 5 or m < 2 and m > 5.

Question 2 |

The value of [{(55)

^{3}+ (45)^{3}}/ {(55)^{3 }– 55 x 45 + (45)^{3}}]A | 100 |

B | 105 |

C | 125 |

D | 75 |

Question 2 Explanation:

As we know that

[{(p)

Compare the question with this and get the answer as the option number (a)

[{(p)

^{3}+ (q)^{3}}/ {(p)^{2 }– p x q + (q)^{2}}] = (p + q)Compare the question with this and get the answer as the option number (a)

Question 3 |

One root of x

^{2 }+ kx – 8 = 0 is square of the other. Then the value of k isA | 2 |

B | 8 |

C | -8 |

D | –2 |

Question 3 Explanation:

Let us suppose the general equation is ax

Since we know that the product of roots is ‘ca’

And the sum of roots is –b/a

If the roots are a and a

Therefore a = –2.

Hence, sum of the roots

= -k/1 = –(a + a

= –(–2 + 4) = – 2.

^{2}+ bx +cSince we know that the product of roots is ‘ca’

And the sum of roots is –b/a

If the roots are a and a

^{2}, the product of roots = a^{3 }= –8.Therefore a = –2.

Hence, sum of the roots

= -k/1 = –(a + a

^{2})= –(–2 + 4) = – 2.

Question 4 |

Once I had been to the post office to buy five-rupee, two-rupee and one-rupee stamps. I paid the clerk Rs. 20, and since he had no change, he gave me three more one-rupee stamps. If the number of stamps of each type that I had ordered initially was more than one, what was the total number of stamps that I bought?10

A | 10 |

B | 9 |

C | 12 |

D | 8 |

Question 4 Explanation:

We will do this question with logic

I have discussed that I paid the clerk Rs. 20, and since he had no change,

he gave me three more one-rupee stamps.

That means I had ordered the stamps for Rs.17,

therefore this can only happen when I order for 2 stamps of each type.

So the total expenditure was 2 x 5 + 2 x 3 + 2 x 1= Rs.16

but the total expenditure was 17 so I must have bought one more stamp of Re. 1.

So I ordered for 7 stamps and the clerk gave me three more stamps of 1 Re. as change,

thus, the total number of stamps I had was 10.

I have discussed that I paid the clerk Rs. 20, and since he had no change,

he gave me three more one-rupee stamps.

That means I had ordered the stamps for Rs.17,

therefore this can only happen when I order for 2 stamps of each type.

So the total expenditure was 2 x 5 + 2 x 3 + 2 x 1= Rs.16

but the total expenditure was 17 so I must have bought one more stamp of Re. 1.

So I ordered for 7 stamps and the clerk gave me three more stamps of 1 Re. as change,

thus, the total number of stamps I had was 10.

Question 5 |

Which of the following values of x do not satisfy the inequality (x

^{2 }– 3x + 2 > 0) at all?A | 1 ≤ x ≤ 2 |

B | –1 ≥ x ≥ –2 |

C | 0 ≤ x ≤ 2 |

D | 0 ≥ x ≥ –2 |

Question 5 Explanation:

x

= (x – 1)(x – 2) > 0

For this product to be greater than zero, either both the factors should be greater than zero or both of them should be less than zero.

Now there will be two cases

(x – 1) > 0 and (x – 2) > 0 or (x – 1) < 0 and (x – 2) < 0.

Hence, x > 1 and x > 2 or x < 1 and x < 2.

Again if we combine this either x > 2 or x < 1.

So for any value of x equal to or between 1 and 2, the above equation is not satisfied, hence, the right answer is option number 1.

^{2}– 3x + 2 > 0,= (x – 1)(x – 2) > 0

For this product to be greater than zero, either both the factors should be greater than zero or both of them should be less than zero.

Now there will be two cases

(x – 1) > 0 and (x – 2) > 0 or (x – 1) < 0 and (x – 2) < 0.

Hence, x > 1 and x > 2 or x < 1 and x < 2.

Again if we combine this either x > 2 or x < 1.

So for any value of x equal to or between 1 and 2, the above equation is not satisfied, hence, the right answer is option number 1.

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