# 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) 1/4 | B) 1/2 |

C) 3/4 | D) 7/12 |

Explanation:

Let A, B, C be the respective events of solving the problem and $\overline{)A},\overline{)B},\overline{)C}$ be the respective events of not solving the problem. Then A, B, C are independent event

$\therefore \overline{)A},\overline{)B},\overline{)C}$ are independent events

Now, P(A) = 1/2 , P(B) = 1/3 and P(C)=1/4

$P\left(\overline{)A}\right)=\frac{1}{2},P\left(\overline{)B}\right)=\frac{2}{3},P\left(\overline{)C}\right)=\frac{3}{4}$

$\therefore $ P( none solves the problem) = P(not A) and (not B) and (not C)

= $P\left(\overline{)A}\cap \overline{)B}\cap \overline{)C}\right)$

= $P\left(\overline{)A}\right)P\left(\overline{)B}\right)P\left(\overline{)C}\right)$ $\left[\because \overline{)A},\overline{)B},\overline{)C}areIndependent\right]$

= $\frac{1}{2}\times \frac{2}{3}\times \frac{3}{4}$

= $\frac{1}{4}$

Hence, P(the problem will be solved) = 1 - P(none solves the problem)

= $1-\frac{1}{4}$= **3/4**

A) 360, 160, 200 | B) 160, 360, 200 |

C) 200, 360,160 | D) 200,160,300 |

Explanation:

let ratio be x.

Hence no. of coins be 5x ,9x , 4x respectively

Now given total amount = Rs.206

=> (.50)(5x) + (.25)(9x) + (.10)(4x) = 206

we get x = 40

=> No. of 50p coins = 200

=> No. of 25p coins = 360

=> No. of 10p coins = 160

A) 4991 | B) 5467 |

C) 5987 | D) 6453 |

Explanation:

Total sale for 5 months = Rs. (6435 + 6927 + 6855 + 7230 + 6562) = Rs. 34009.

Required sale = Rs.[(6500 x 6) - 34009]

= Rs. (39000 - 34009)

= Rs. 4991.

A) 587 | B) 554 |

C) 489 | D) 499 |

Explanation:

The given **number series** follows a pattern that

5 **x 5 – 1 = 24**

24 **x 4 – 2 = 94**

94 **x 3 – 3 = 279**

279 **x 2 – 4 = 554**

A) 61 | B) 64 |

C) 72 | D) 70 |

Explanation:

The pattern is 1 x 2, 2 x 3, 3 x 4, 4 x 5, 5 x 6, 6 x 7, 7 x 8.

So, the next number is 8 x 9 = 72.

A) 72 | B) 110 |

C) 132 | D) 150 |

Explanation:

The numbers are 7 x 8, 8 x 9, 9 x 10, 10 x 11, 11 x 12, 12 x 13.

So, 150 is wrong.

A) 523 | B) 521 |

C) 613 | D) 721 |

Explanation:

Each number is twice the preceding one with 1 added or subtracted alternatively.

So, the next number is (2 x 261 + 1) = 523.

A) 18.25 | B) 28.375 |

C) 11.5 | D) 8 |

Explanation:

Given number series follows a pattern that,

**2**

**2 x 1.5 + 1 = 4**

**4 x 1.5 + 1 = 7 **(not equal to 8)

**7 x 1.5 + 1 = 11.5**

**11.5 x 1.5 + 1 = 18.25**

**18.25 x 1.5 + 1 = 28.375**

Hence, **the odd number in the given number series** is** 8.**