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## Number System: Level 1 Test - 6

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Question 1 |

The number of two-digit numbers exactly divisible 3 is

33 | |

32 | |

31 | |

30 |

Question 1 Explanation:

Required numbers are 12, 15, 18, ………. , 99

This is an AP with

Tn=

99 = 12+ (n -1) x 3

n -1 = 99 -12 /3

n = 29 + 1 = 30

This is an AP with

*a*= 12 and*d*= 3Tn=

*a*+ (n - 1)*d*99 = 12+ (n -1) x 3

n -1 = 99 -12 /3

n = 29 + 1 = 30

Question 2 |

The number which when increased by 20 is equal to six times of the old number. Find the old number?

7 | |

5 | |

3 | |

4 |

Question 2 Explanation:

Let the whole number be

x =1/6 (x + 20)

6x=x+ 20

5x=20 and x =4

*x.*x =1/6 (x + 20)

6x=x+ 20

5x=20 and x =4

Question 3 |

A number when divided by 765 leaves a remainder 42. Find the remainder if the same number is divided by 17?

8 | |

5 | |

7 | |

6 |

Question 3 Explanation:

Let the number be

When this number is divided by 17,

then quotient will be

*(765x*+ 42).When this number is divided by 17,

then quotient will be

*(45x*+ 2) and remainder will be 8.Question 4 |

A company manufactured 6000 scooters in the third year of after establishment and 7000 scooters in the seventh year. Assuming that the production increases uniformly by in the given years , Find the production of the company in the tenth year?

7850 | |

7650 | |

7750 | |

7950 |

Question 4 Explanation:

Production in third year =6000

Production in seventh year =7000

Production in fourth year = 1000 ie

Production increases @ 250 scooters every year.

Production in tenth year = (7000 + 250 x 3) = 7750

Production in seventh year =7000

Production in fourth year = 1000 ie

*,*Production increases @ 250 scooters every year.

Production in tenth year = (7000 + 250 x 3) = 7750

Question 5 |

The digits of a three-digit number A are written in the reverse order to form another three-digit number B. If B > A and B-A is perfectly divisible by 7, then which of the following is necessarily true?

100 < A < 299 | |

106 < A < 305 | |

112 < A < 311 | |

118< A < 317 |

Question 5 Explanation:

Let A = abc, then B = cba

Given, B > A which implies c > a ...(1)

As B – A = (100c + 10b + a) – (100a + 10b + 1)

B – A = 100 (c – a) + (a – c)

B – A = 99 (c – a) and (B – A) is divisible by 7;

As 99 is not divisible by 7 (no factor like 7 or 7^2)

Therefore (c – a) must be divisible by 7 {i.e., (c – a)

Must be 7, 7^2 etc.} as c & a are single digits. (c – a)

Must be 7 only, the possible values (c, a) {with c > a}

Are (9, 2) & (8, 1), with this we can write A as A : abc ≡ 1b8 or 2b9

As b can take values from 0 to 9, the smallest & largest possible value of are:

A min = 108 & A max = 299

only option b satisfies the given condition, so (b) is the ans

Given, B > A which implies c > a ...(1)

As B – A = (100c + 10b + a) – (100a + 10b + 1)

B – A = 100 (c – a) + (a – c)

B – A = 99 (c – a) and (B – A) is divisible by 7;

As 99 is not divisible by 7 (no factor like 7 or 7^2)

Therefore (c – a) must be divisible by 7 {i.e., (c – a)

Must be 7, 7^2 etc.} as c & a are single digits. (c – a)

Must be 7 only, the possible values (c, a) {with c > a}

Are (9, 2) & (8, 1), with this we can write A as A : abc ≡ 1b8 or 2b9

As b can take values from 0 to 9, the smallest & largest possible value of are:

A min = 108 & A max = 299

only option b satisfies the given condition, so (b) is the ans

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