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21. If HCF of two numbers is 11 and the product of these numbers is 363, what is the the greater number? | |

A. 22 | B. 11 |

C. 33 | D. 9 |

Discuss |

answer with explanation

Answer: Option C

Explanation:

Let the numbers be 11a and 11b

11a × 11b = 363

=> ab = 3

co-primes with product 3 are (1, 3)

(Reference: Co-prime Numbers)

Hence the numbers with HCF 11 and product 363

= (11 × 1, 11 × 3)

= (11, 33)

Hence, numbers are 11 and 33

The greater number = 33

22. What is the greatest number which on dividing 1223 and 2351 leaves remainders 90 and 85 respectively? | |

A. 42 | B. 1100 |

C. 127 | D. 1133 |

Discuss |

answer with explanation

Answer: Option D

Explanation:

Required number

= HCF of (1223 - 90) and (2351 - 85)

= HCF of 1133 and 2266

= 1133

23. What is the least multiple of 7 which leaves a remainder of 4 when divided by 6, 9, 15 and 18 ? | |

A. 343 | B. 350 |

C. 371 | D. 364 |

Discuss |

answer with explanation

Answer: Option D

Explanation:

LCM of 6, 9, 15 and 18 = 90

Required Number = (90k + 4) which is a multiple of 7

Put k = 1. We get number as (90 × 1) + 4 = 94. But this is not a multiple of 7

Put k = 2. We get number as (90 × 2) + 4 = 184. But this is not a multiple of 7

Put k = 3. We get number as (90 × 3) + 4 = 274. But this is not a multiple of 7

Put k = 4. We get number as (90 × 4) + 4 = 364. This is a multiple of 7

Hence 364 is the answer.

24. Three numbers which are co-prime to each other are such that the product of the first two is 119 and that of the last two is 391. What is the sum of the three numbers? | |

A. 43 | B. 51 |

C. 47 | D. 53 |

Discuss |

answer with explanation

Answer: Option C

Explanation:

Since the numbers are co-prime, their HCF = 1

Product of first two numbers = 119

Product of last two numbers = 391

The middle number is common in both of these products. Hence, if we take HCF of 119 and 391, we get the common middle number.

HCF of 119 and 391 = 17

=> Middle Number = 17

First Number $=\dfrac{119}{17}=7$

Last Number $=\dfrac{391}{17}=23$

Sum of the three numbers = 7 + 17 + 23 = 47

25. Reduce $\dfrac{4329}{4662}$ to its lowest terms | |

A. $\dfrac{13}{14}$ | B. $\dfrac{7}{13}$ |

C. $\dfrac{7}{12}$ | D. $\dfrac{13}{17}$ |

Discuss |

answer with explanation

Answer: Option A

Explanation:

We need to find out HCF of 4329 and 4662

Hence, HCF of 4329 and 4662 = 333

4329 ÷ 333 = 13

4662 ÷ 333 = 14

Hence, $\dfrac{4329}{4662}=\dfrac{13}{14}$

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