# 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) 76 | B) 79 |

C) 85 | D) 87 |

Explanation:

Average = total runs / no.of innings = 32

So, total = Average x no.of innings = 32 x 10 = 320.

Now increase in avg = 4runs. So, new avg = 32+4 = 36runs

Total runs = new avg x new no. of innings = 36 x 11 = 396

Runs made in the 11th inning = 396 - 320 = 76

A) Monday | B) Friday |

C) Sunday | D) Tuesday |

Explanation:

On 31st December, 2005 it was Saturday.

Number of odd days from 2006 to 2009 = (1 + 1 + 2 + 1) = 5 days.

On 31st December 2009, it was Thursday.

Thus, on 1st Jan, 2010 it is Friday

A) 38000 | B) 46800 |

C) 36700 | D) 50000 |

Explanation:

Let the original salaries of Ravi and Sumit be Rs. 2x and Rs. 3x respectively.

Then,

(2x+4000) / (3x+4000) = 40 / 57

⇒ 57 × (2x + 4000) = 40 × (3x+4000)

⇒ 6x = 68,000

⇒ 3x = 34,000

Sumit's present salary = (3x + 4000) = Rs.(34000 + 4000) = Rs. 38,000

A) 14 | B) 19 |

C) 33 | D) 38 |

Explanation:

Let the son's present age be x years .Then, (38 - x) = x => x= 19.

Son's age 5 years back = (19 - 5) = 14 years

A) 20% | B) 21% |

C) 22% | D) 23% |

Explanation:

Gain % = $\left(\frac{{\left(100+commongain\%\right)}^{2}}{100}-100\right)$%

=$\left(\frac{{\left(100+10\right)}^{2}}{100}-100\right)$

= 21%

A) 2/91 | B) 1/22 |

C) 3/22 | D) 2/77 |

Explanation:

Let S be the sample space.

Then, n(S) = number of ways of drawing 3 balls out of 15 = $15{C}_{3}$ =$\frac{15*14*13}{3*2*1}$= 455.

Let E = event of getting all the 3 red balls.

n(E) = $5{C}_{3}$ = $\frac{5*4}{2*1}$ = 10.

=> P(E) = n(E)/n(S) = 10/455 = 2/91.

A) 2/7 | B) 5/7 |

C) 1/5 | D) 1/2 |

Explanation:

Total number of outcomes possible, n(S) = 10 + 25 = 35

Total number of prizes, n(E) = 10

$P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{10}{35}=\frac{2}{7}$

A) 10000 | B) 12000 |

C) 14000 | D) 16000 |

Explanation:

Purchase price = $Rs.\left[\frac{8748}{{{}^{\left(1-{\displaystyle \frac{10}{100}}\right)}}^{3}}\right]$ = Rs. [8748 * (10/9) * (10/9 )* (10/9)] = Rs.12000