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11. What is the greatest possible length which can be used to measure exactly the lengths 8 m, 4 m 20 cm and 12 m 20 cm? | |

A. 30 cm | B. 25 cm |

C. 20 cm | D. 10 cm |

Discuss |

answer with explanation

Answer: Option C

Explanation:

Required length

= HCF of 800 cm, 420 cm, 1220 cm

= 20 cm

12. Which of the following fraction is the largest ? | |

A. $\dfrac{11}{12}$ | B. $\dfrac{21}{40}$ |

C. $\dfrac{5}{6}$ | D. $\dfrac{41}{50}$ |

Discuss |

answer with explanation

Answer: Option A

Explanation:

**Solution 1**

LCM of 6, 40, 50 and 12 = 600

$\dfrac{5}{6} = \dfrac{500}{600}\\~\\\dfrac{21}{40} = \dfrac{315}{600}\\~\\\dfrac{41}{50} = \dfrac{492}{600}\\~\\\dfrac{11}{12} = \dfrac{550}{600}$

Clearly $\dfrac{550}{600}=\dfrac{11}{12}$ is the largest among these fractions.

$\dfrac{5}{6} = .83\\~\\\dfrac{21}{40} = .52\\~\\\dfrac{41}{50} = .82\\~\\\dfrac{11}{12} = .92$

Clearly $.92=\dfrac{11}{12}$ is the largest among these.

13. The product of two 2 digit numbers is 2028 and their HCF is 13. What are the numbers ? | |

A. 13, 156 | B. 26, 78 |

C. 36, 68 | D. 39, 52 |

Discuss |

answer with explanation

Answer: Option D

Explanation:

Let the numbers be $13x$ and $13y~~$ (∵ HCF of the numbers = 13)

$13x × 13y = 2028\\\Rightarrow xy = 12$

co-primes with product 12 are (1, 12) and (3, 4) (∵ we need to take only co-primes with product 12. If we take two numbers with product 12, but not co-prime, the HCF will not remain as 13) (Reference: Co-prime Numbers)

Hence the numbers with HCF 13 and product 2028

= (13 × 1, 13 × 12) and (13 × 3, 13 × 4)

= (13, 156) and (39, 52)

Given that the numbers are 2 digit numbers.

Hence numbers are 39 and 52

14. The product of two numbers is 2028 and their HCF is 13. What are the number of such pairs? | |

A. 2 | B. 3 |

C. 4 | D. 1 |

Discuss |

answer with explanation

Answer: Option A

Explanation:

Let the numbers be $13x$ and $13y~~$ (∵ HCF of the numbers = 13)

$13x × 13y = 2028 \\\Rightarrow xy = 12$

co-primes with product 12 are (1, 12) and (3, 4) (∵ we need to take only co-primes with product 12. If we take two numbers with product 12, but not co-prime, the HCF will not remain as 13) (Reference: Co-prime Numbers)

Hence the numbers with HCF 13 and product 2028

= (13 × 1, 13 × 12) and (13 × 3, 13 × 4)

= (13, 156) and (39, 52)

So, there are 2 pairs of numbers with HCF 13 and product 2028

15. N is the greatest number which divides 1305, 4665 and 6905 and gives the same remainder in each case. What is the sum of the digits in N? | |

A. 5 | B. 4 |

C. 6 | D. 3 |

Discuss |

answer with explanation

Answer: Option B

Explanation:

If the remainder is same in each case and remainder is not given, HCF of the differences of the numbers is the required greatest number.

6905 - 1305 = 5600

6905 - 4665 = 2240

4665 - 1305 = 3360

Hence, the greatest number which divides 1305, 4665 and 6905 and gives the same remainder, N

= HCF of 5600, 2240, 3360

= 1120

Sum of digits in N

= Sum of digits in 1120

= 1 + 1 + 2 + 0

= 4

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