# Quantitative Aptitude - Arithmetic Ability Questions

## What is Quantitative Aptitude - Arithmetic Ability?

Quantitative Aptitude - Arithmetic Ability test helps measure one's numerical ability, problem solving and mathematical skills. Quantitative aptitude - arithmetic ability is found in almost all the entrance exams, competitive exams and placement exams. Quantitative aptitude questions includes questions ranging from pure numeric calculations to critical arithmetic reasoning. Questions on graph and table reading, percentage analysis, categorization, simple interests and compound interests, clocks, calendars, Areas and volumes, permutations and combinations, logarithms, numbers, percentages, partnerships, odd series, problems on ages, profit and loss, ratio & proportions, stocks &shares, time & distance, time & work and more .

Every aspirant giving Quantitative Aptitude Aptitude test tries to solve maximum number of problems with maximum accuracy and speed. In order to solve maximum problems in time one should be thorough with formulas, theorems, squares and cubes, tables and many short cut techniques and most important is to practice as many problems as possible to find yourself some tips and tricks in solving quantitative aptitude - arithmetic ability questions.

Wide range of Quantitative Aptitude - Arithmetic Ability questions given here are useful for all kinds of competitive exams like Common Aptitude Test(CAT), MAT, GMAT, IBPS and all bank competitive exams, CSAT, CLAT, SSC Exams, ICET, UPSC, SNAP Test, KPSC, XAT, GRE, Defence, LIC/G IC, Railway exams,TNPSC, University Grants Commission (UGC), Career Aptitude test (IT companies), Government Exams and etc.

A) 153 | B) 165 |

C) 142 | D) 127 |

Explanation:

From the beginning, the next term comes by adding prime numbers in a sequence of 2, 3, 5, 7, 9, 11, 13... to its previous term. But 165 will not be in the series as it must be replaced by 166 since 153+13 = 166.

A) 1/2 | B) 3/5 |

C) 9/20 | D) 8/15 |

Explanation:

Here, S = {1, 2, 3, 4, ...., 19, 20}.

Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.

P(E) = n(E)/n(S) = 9/20.

A) 100% | B) 200% |

C) 300% | D) 400% |

Explanation:

Let the C.P be Rs.100 and S.P be Rs.x, Then

The profit is (x-100)

Now the S.P is doubled, then the new S.P is 2x

New profit is (2x-100)

Now as per the given condition;

=> 3(x-100) = 2x-100

By solving, we get

x = 200

Then the Profit percent = (200-100)/100 = 100

Hence the profit percentage is 100%

A) 142 | B) 181 |

C) 156 | D) 179 |

Explanation:

Here the series follow the rule that

${1}^{2}+{2}^{2}=5$

${3}^{2}+{4}^{2}=25$

${5}^{2}+{6}^{2}=61$

${7}^{2}+{8}^{2}=113$

SO NEXT IS, ${9}^{2}+{10}^{2}=181$

A) 391 | B) 421 |

C) 481 | D) 511 |

Explanation:

Numbers are (2^{3} - 1), (3^{3} - 1), (4^{3} - 1), (5^{3} - 1), (6^{3} - 1), (7^{3} - 1) etc.

So, the next number is (8^{3} - 1) = (512 - 1) = 511.

A) 54 % | B) 64 % |

C) 74 % | D) 84 % |

Explanation:

Let the number be x.

Then, ideally he should have multiplied by x by 5/3. Hence Correct result was x * (5/3)= 5x/3.

By mistake he multiplied x by 3/5 . Hence the result with error = 3x/5

Then, error = (5x/3 - 3x/5) = 16x/15

Error % = (error/True vaue) * 100 = [(16/15) * x/(5/3) * x] * 100 = 64 %

A) 71 | B) 72 |

C) 73 | D) 69 |

Explanation:

With close observation, you will note that each number in the list is in the middle of two prime numbers. Thus:

4 is in the middle of 3 and 5, 6 is in the middle of 5 and 7, 12 is in the middle of 11 and 13, 18 is in the middle of 17 and 19, 30 is in the middle of 29 and 31. 42 is in the middle of 41 and 43, 60 is in the middle of 59 and 61.

Therefore, the next number would be the one that is in the middle of the next two prime numbers, which is 72 (which is in the middle of 71 and 73).

A) 32 | B) 36 |

C) 30 | D) 34 |

Explanation:

The given series follows the pattern that,

18, 97, 26, 91, ?, 83

=> 18 + 8 = 26

=> 97 - 8 = 91

=> 26 + 8 = 34

=> 91 - 8 = 83

Hence, the missing number is **34**.