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## Arithmetic : Level 3 Test -5

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*Arithmetic : Level 3 Test -5*.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 |

At the end of year 1998, Sheppard bought nine dozen goats. Henceforth, every year he added p% of the goats at the beginning of the year and sold q% of the goats at the end of the year where p > 0 and q > 0. If Sheppard had nine dozen goats at the end of year 2002, after making the sales for that year, which of the following is true?

p = q. | |

p < q | |

p > q | |

p = q/2 |

Question 1 Explanation:

Since the number of goats remains unchanged that means the percentage that is added every time is equal to the percentage that is sold, therefore there should be a net decrease. The same will be the case if the percentage added is less than the percentage sold.
The only way, the number of goats will remain the same is if p > q

Question 2 |

Instead of walking along two adjacent sides of a rectangular field, a boy took a short cut along the diagonal and saved a distance equal to half the longer side. Then the ratio of the shorter side to the longer side is

Â½ | |

2/3 | |

1/4 | |

Â¾ |

Question 2 Explanation:

We will do this question with the help of options

Letâ€™s take 1/2

Diagonal will be =âˆš 5

Distance saved by the boy = 3 â€“âˆš 5 = approx.0.75

which is not equal to the half of the longer side Therefore this is incorrect option

Now letâ€™s take Â¾

â‡’ Diagonal would be = 5

Distance saved by the boy = (4 + 3) â€“ 5 = 2 = Which is equal to the half the larger side.

Therefore d is the right option

Letâ€™s take 1/2

Diagonal will be =âˆš 5

Distance saved by the boy = 3 â€“âˆš 5 = approx.0.75

which is not equal to the half of the longer side Therefore this is incorrect option

Now letâ€™s take Â¾

â‡’ Diagonal would be = 5

Distance saved by the boy = (4 + 3) â€“ 5 = 2 = Which is equal to the half the larger side.

Therefore d is the right option

Question 3 |

Mayank, Mirza, Little and Jaspal bought a motorbike for $60. Mayank paid one-half of the sum of the amounts paid by the other boys. Mirza paid one-third of the sum of the amounts paid by the other boys. Little paid one-fourth of the sum of the amounts paid by the other boys. How much did Jaspal have to pay?

$15 | |

$13 | |

$17 | |

None of these |

Question 3 Explanation:

Given that Â Mayank paid Â½Â of what others paid.

Let the money paid by others R

Therefore the money paid by Mayank = R/2

Total amount = 60

R + R/2 = 60

R= 40

Therefore the money that paid by Mayank is $. 20

Similarly,

Mirza paid $15 and little paid $12.

Remaining amount of $60 â€“ $20 â€“ $15 â€“ $12 = $13 is paid by Jaspal.

Let the money paid by others R

Therefore the money paid by Mayank = R/2

Total amount = 60

R + R/2 = 60

R= 40

Therefore the money that paid by Mayank is $. 20

Similarly,

Mirza paid $15 and little paid $12.

Remaining amount of $60 â€“ $20 â€“ $15 â€“ $12 = $13 is paid by Jaspal.

Question 4 |

A piece of string is 40 cm long. It is cut into three pieces. The longest piece is three times as long as the middle-sized and the shortest piece is 23 cm shorter than the longest piece. Find the length of the shortest piece.

27 | |

5 | |

4 | |

9 |

Question 4 Explanation:

Let the length of the largest piece = 3l

Therefore the length of the Middle piece = l

Therefore the length of the shortest piece= 3l â€“ 23 or 3l + l + (3l â€“ 23) = 40

or l = 9 or the shortest piece = 3(9) â€“ 23 = 4

Therefore the length of the Middle piece = l

Therefore the length of the shortest piece= 3l â€“ 23 or 3l + l + (3l â€“ 23) = 40

or l = 9 or the shortest piece = 3(9) â€“ 23 = 4

Question 5 |

Fresh grapes contain 90% water by weight while dried grapes contain 20% water by weight. What is the weight of dry grapes available from 20 kg of fresh grapes?

2 kg | |

2.4 kg | |

2.5 kg | |

None of these |

Question 5 Explanation:

This is very simple one

Quantity of pulp in fresh grapes = 10%

Therefore the quantity of grapes in 20 kg od grapes = 2 kg

Quantity of pulp in dry grapes = 80%

Therefore grapes contain 2 kg of pulp = 2/0.8 = 20/8 = 2.5Â kg dry grapes

Quantity of pulp in fresh grapes = 10%

Therefore the quantity of grapes in 20 kg od grapes = 2 kg

Quantity of pulp in dry grapes = 80%

Therefore grapes contain 2 kg of pulp = 2/0.8 = 20/8 = 2.5Â kg dry grapes

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