# 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) 5.33 | B) 6 |

C) 5.74 | D) 5.68 |

A) 42 | B) 36 |

C) 30 | D) 24 |

Explanation:

Let present age of Pranay = 15p

That of Sai = 17p

ATQ,

15p + 6/17p + 6 = 9/10

=> 150p + 60 = 153p + 54

=> p = 2

Therefore, Age of Pranay after 6 yrs = 15 x 2 + 6 = 36 years.

A) 19 kmph | B) 13 kmph |

C) 7 kmph | D) 9 kmph |

Explanation:

Let the speed of the slower train = p kmph

ATQ,

Speed of the faster train = (p + 6) kmph

Then,

(p + p + 6) x 5 = 160

10p + 30 = 160

10p = 130

p = 13 kmph

Then, speed of the faster train = p + 6 = 13 + 6 = **19 kmph**.

A) Rs. 14,400 | B) Rs. 15,600 |

C) Rs. 14,850 | D) Rs. 15,220 |

Explanation:

Let the required Sum = Rs.S

From the given data,

1008 = [(S x 11 x 5)/100] - [(S x 8 x 6)/100]

=> S = Rs. 14,400.

A) 60 | B) 55 |

C) 45 | D) 40 |

Explanation:

From the given data,

let the length, breadth and height of the cuboid are m, n, r

m x n = 12

n x r = 20

r x m = 15

Hence, m x n x n x r x r x m = 12 x 20 x 15

mnr = sqrt of (12x20x15) = **60 cub.cm.**

A) 6 min | B) 8 min |

C) 12 min | D) 10 min |

Explanation:

Let pipe A takes p min to fill

Then,

pipe B takes 3p min to fill

=> 3p - p = 32

=> p = 16 min => 3p = 48 min

Required, both pipes to fill = (48 x 16)/(48 + 16) min = 12 min.

A) 9 | B) 16 |

C) 22 | D) 36 |

Explanation:

In the given series 1 4 9 16 22 36

1 = 1 x 1

4 = 2 x 2

9 = 3 x 3

16 = 4 x 4

25 = 5 x 5 (Not 22)

36 = 6 x 6

Hence, the odd man in the series is 22.

A) 4 | B) 5 |

C) 7 | D) 9 |

Explanation:

Given equation 5x-17 = -x+7

Add 1x to each side of the equation

5x-17+x = -x+7+x

6x=24

x=4