## Arithmetic Aptitude 6

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Question 1 |

Which digits should come in place of * and $ if the number 4675*2$ is divisible by both 5 & 8?

4, 0 | |

4, 5 | |

1, 0 | |

8, 0 |

**Arithmetic Aptitude 6**

**Number Divisibility**

**Discuss it**

Question 1 Explanation:

Since the given number is divisible by 5, 0 or 5 must come in place of $. But a number ending with 5 is never divisible with 8 because a divisible number must be even. Therefore 0 will replace $. If the number is divisible by 8, the number formed by last three digits must be divisible by 8. The number *20 must be divisible by 8. Among all 4 options, only placing 1 in place of * makes it divisible by 8.

Question 2 |

What least number must be added to 4000 to obtain a number exactly divisible by 17?

12 | |

14 | |

15 | |

13 |

**Arithmetic Aptitude 6**

**Number Divisibility**

**Discuss it**

Question 2 Explanation:

Let us divide 4000 by 17. 17) 4000 (235 34 …. 60 51 ….. 90 85 ….. 5 We get 5 as remainder. Number to be added = 17-5 = 12

Question 3 |

What least number must be subtracted from 5000 to get a number exactly divisible by 23?

10 | |

2 | |

7 | |

9 |

**Arithmetic Aptitude 6**

**Number Divisibility**

**Discuss it**

Question 3 Explanation:

Let us divide 5000 by 23. 23) 5000 (217 46 …. 40 23 …. 170 161 ….. 9 We get 9 as remainder. Therefore required number to be subtracted = 9

Question 4 |

The largest 4 digit number exactly divisible by 5, 6 and 7 is:

9980 | |

9870 | |

9540 | |

9640 |

**Arithmetic Aptitude 6**

**Number Divisibility**

**Discuss it**

Question 4 Explanation:

The required number must be divisible by L.C.M. of 5,6 and 7. L.C.M. of 5, 6 and 7 = 5 x 6 x 7 = 210 Let us divide 9999 by 210. 210) 9999 (47 840 ---- 1599 1470 ---- 129 Required number = 9999 – 129 = 9870

Question 5 |

34. What should be the next number in below series?
3, 8, 15, 24, 35……..

44 | |

47 | |

48 | |

49 |

**Arithmetic Aptitude 6**

**Number Divisibility**

**Discuss it**

Question 5 Explanation:

Pattern is (2^{2}- 1), (3^{2}- – 1), (4^{2}- – 1), (5^{2}- – 1), .... Next number will be = 7^{2}- – 1 = 49 – 1 = 48

Question 6 |

The average of 21 results is 20. Average of 1

^{st}10 of them is 24 that of last 10 is 14. the result of 11'th is :42 | |

44 | |

46 | |

40 |

**Arithmetic Aptitude 6**

**Numbers**

**Discuss it**

Question 6 Explanation:

11'th result = sum of 21 results – sum of 20 results = 21 x 20 – (24 x 10 + 14 x 10) = 420 – (240 + 140) = 420- 380 = 40

There are 6 questions to complete.