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## Algebra: Quadratic Equations Test-5

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Question 1 |

The product of the present ages of Sarita and Gauri is 320. Eight years from now, Sarita's age will be three times the age of Gauri. What was the age of Sarita when Gauri was born?

40 yr | |

32 yr | |

48 yr | |

36 yr |

Question 1 Explanation:

Let the age of Sarita be s years and Gauri be g years.

sg=320,

s+8=3(g+8)

s+8=3g+24

=>s-3g=16,sg=320

Solving the equation we can get ,

g=8 and s=40

Thus the age of Sarita when Gauri was born is 40-8=32 years.

sg=320,

s+8=3(g+8)

s+8=3g+24

=>s-3g=16,sg=320

Solving the equation we can get ,

g=8 and s=40

Thus the age of Sarita when Gauri was born is 40-8=32 years.

Question 2 |

In a class, the number of girls is one less than the number of the boys. If the product of the number of boys and that of girls is 272, then the number of girls in the class is

15 | |

14 | |

16 | |

17 |

Question 2 Explanation:

Let the number of girls be g and the number of boys be b.

g=b-1

gb=272

Solving the equation or by using options we can find that

g=16 and b=17.

Correct options is (c).

g=b-1

gb=272

Solving the equation or by using options we can find that

g=16 and b=17.

Correct options is (c).

Question 3 |

If you subtract Rs.1 from the money Bholu has, take its reciprocal, then add it to the square of the money and subtract the money Bholu has, you get two rupees more than the reciprocal of money after subtracting 1 from it. Find the money Bholu has.

Rs.7 | |

Rs.5 | |

Rs.10 | |

None of these |

Question 3 Explanation:

Let the amount with Bholu be Rs. b.

$ \begin{array}{l}\frac{1}{b-1}+{{b}^{2}}-b=2+\frac{1}{b-1}\\=>b(b-1)=2\\=>b=2\end{array}$

$ \begin{array}{l}\frac{1}{b-1}+{{b}^{2}}-b=2+\frac{1}{b-1}\\=>b(b-1)=2\\=>b=2\end{array}$

Question 4 |

A class decided to have a party for their class at a total cost of Rs.720. Four students decided to stay out of the party. To meet the expenses the remaining students have to increase their share by Rs.9. What is the original cost per student?

Rs.18, | |

Rs.24 | |

Rs.36 | |

Rs.20 |

Question 4 Explanation:

Let the number of students = s and the amount paid by each be r.

sr=720

and (s-4)(y+9)=720

Solving the two equations,

s=20 and r=36.

Thus the original cost per student is Rs. 36.

The correct option is (c).

sr=720

and (s-4)(y+9)=720

Solving the two equations,

s=20 and r=36.

Thus the original cost per student is Rs. 36.

The correct option is (c).

Question 5 |

$ \displaystyle Given\,\,\,\,\frac{\left( \sqrt{x+4}+\sqrt{x-10} \right)}{\sqrt{x+4}-\sqrt{x-10}}=\frac{5}{2},$

The value of x is

The value of x is

1 | |

331/5 | |

263/20 | |

17/21 |

Question 5 Explanation:

$\displaystyle Given\,\,\,\,\frac{\left( \sqrt{x+4}+\sqrt{x-10} \right)}{\sqrt{x+4}-\sqrt{x-10}}=\frac{5}{2},$

$ \begin{array}{l}\frac{\left( \sqrt{x+4}+\sqrt{x-10} \right)}{\sqrt{x+4}-\sqrt{x-10}}=\frac{5}{2},\\=>2\left( \sqrt{x+4}+\sqrt{x-10} \right)=5\left( \sqrt{x+4}-\sqrt{x-10} \right)\\=>2\sqrt{x+4}+2\sqrt{x-10}=5\sqrt{x+4}-5\sqrt{x-10}\\=>7\sqrt{x-10}=3\sqrt{x+4}\\=>49(x-10)=9(x+4)\\=>49x-490=9x+36\\=>40x=526\\=>x=\frac{526}{40}=\frac{263}{20}\end{array}$

$ \begin{array}{l}\frac{\left( \sqrt{x+4}+\sqrt{x-10} \right)}{\sqrt{x+4}-\sqrt{x-10}}=\frac{5}{2},\\=>2\left( \sqrt{x+4}+\sqrt{x-10} \right)=5\left( \sqrt{x+4}-\sqrt{x-10} \right)\\=>2\sqrt{x+4}+2\sqrt{x-10}=5\sqrt{x+4}-5\sqrt{x-10}\\=>7\sqrt{x-10}=3\sqrt{x+4}\\=>49(x-10)=9(x+4)\\=>49x-490=9x+36\\=>40x=526\\=>x=\frac{526}{40}=\frac{263}{20}\end{array}$

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