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## Algebra Level 2 Test 1

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*Algebra Level 2 Test 1*. 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 the ages of P and R are added to twice the age of Q, the total becomes 59. If the ages of Q and Rare added to thrice the age of P, the total becomes 68 and if the age of P is added to thrice the age of Q and thrice the age of R, the total becomes 108. What is the age of P?

17 yr | |

19 yr | |

15 yr | |

12 yr |

Question 1 Explanation:

From the given conditions we can say that
2Q +P +R = 59 …………………………………1
Q + R + 3P = 68…………………………………2
P +3Q +3R = 108……………………………….3
From 2 and 3
3Q + 3R + 9P = 204
P +3Q +3R = 108
So therefore 8P= 96 = P = 12

Question 2 |

A father's age is three times the sum of the ages of his two children, but 20 yr hence his age will be equal to the sum of their ages. Then, the father's age is

30 yr | |

40 yr | |

35 yr | |

45 yr |

Question 2 Explanation:

Let the age of father = p years
Let the age of the children = q and r years
From the question
(q + r) = p/3
And also
(q +r )+ 20 +20 = p + 20
On solving for p
p = 30 years

Question 3 |

In a family, the father took ¼ of the cake and he had 3 times as much as others had. The total number of family members is

7 | |

3 | |

10 | |

12 |

Question 3 Explanation:

Let the number of members other than the father = p
Therefore 3/4cake owned by these members
Therefore 3/4p x 3 =1/4
p = 9
Therefore the number of members in the family = 10

Question 4 |

Five engines consume 6 metric tonne of coal when each is running 9 h a day. How much coal (in metric tonnes) will be needed for 8 engines, each running 10 h a day, it being given that 3 engines of the former type consume as much as 4 engines of the latter type?

6.48 | |

3 ^{1}/_{8} | |

8 | |

8 ^{5}/_{9} |

Question 4 Explanation:

Let the quantity to be needed= p
So from question we can say that
(5 x 9)/6 = (8 x 10)/p x ¾
p = {(8 x 10 x 3) / (4 x 5 x 9)} x 6 = 8

Question 5 |

A body of 7300 troops is formed of 4 battalions so that 1/2 of the first, 2/3 of the second, 3/4 of the third and 4/5 of the fourth are all composed of the same number of men. Find the same number.

1250 | |

1200 | |

1300 | |

1350 |

Question 5 Explanation:

Number of troops = 7300
Number of battalions = 4
Let battalions are p, q, r, s, respectively
So therefore from the question we can say that
p/2 = 2q/3 = 3r/4 = 4s/5
Therefore the values will be
q = 3/4p
r = 2/3p
s = 5/8p
And also we are given by p + q + r + s = 7300
By putting these all above values
p would be = 2400
Therefore same number of men are = 2400/2 = 1200

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