# 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) 125% | B) 150% |

C) 175% | D) 110% |

Explanation:

Let the edge = a cm

So increase by 50 % = a + a/2 = 3a/2

Total surface Area of original cube = $6{a}^{2}$

TSA of new cube = $6{\left(\frac{3a}{2}\right)}^{2}$ =$6\left(\frac{9{a}^{2}}{4}\right)$= $13.5{a}^{2}$

Increase in area = $13.5{a}^{2}-6{a}^{2}$ =$7.5{a}^{2}$

$7.5{a}^{2}$ Increase % =$\frac{7.5{a}^{2}}{6{a}^{2}}\times 100$ = 125%

A) 20 | B) 30 |

C) 40 | D) 50 |

Explanation:

The fruit content in both the fresh fruit and dry fruit is the same.

Given, fresh fruit has 68% water.so remaining 32% is fruit content. weight of fresh fruits is 100kg

Dry fruit has 20% water.so remaining 80% is fruit content.let weight if dry fruit be y kg.

Fruit % in freshfruit = Fruit% in dryfruit

Therefore, (32/100) x 100 = (80/100 ) x y

we get, y = 40 kg.

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

A) 35 | B) 38 |

C) 40 | D) 42 |

Explanation:

Let the number of correct answers be X.

Number of incorrect answers = (60 – X).

4x – (60 – x) = 130

=> 5x = 190

=> x = 38

A) Rs. 169.50 | B) Rs.1700 |

C) Rs. 175.50 | D) Rs. 180 |

Explanation:

Since first second varieties are mixed in equal proportions, so their average price = Rs.(126+135)/2= Rs.130.50

So, Now the mixture is formed by mixing two varieties, one at Rs. 130.50 per kg and the other at say Rs. 'x' per kg in the ratio 2 : 2, i.e., 1 : 1. We have to find 'x'.

Cost of 1 kg tea of 1st kind Cost of 1 kg tea of 2nd kind

x-153/22.50 = 1 => x - 153 = 22.50 => x=175.50.

Hence, price of the third variety = Rs.175.50 per kg.

A) 500 | B) 600 |

C) 800 | D) 1000 |

Explanation:

Given that the student got 125 marks and still he failed by 40 marks

=> The minimum pass mark = 125 + 40 = 165

Given that minimum pass mark = 33% of the total mark

=> total mark =33/100 =165

=> total mark = 16500/33 = 500

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) 200, 250, 300 | B) 300, 200, 250 |

C) 200, 300, 400 | D) None of these |

Explanation:

A's 5 days work = 50%

B's 5 days work = 33.33%

C's 2 days work = 16.66% [100- (50+33.33)]

Ratio of contribution of work of A, B and C = $50:33\frac{1}{3}:16\frac{2}{3}$ = 3 : 2 : 1

A's total share = Rs. 1500

B's total share = Rs. 1000

C's total share = Rs. 500

A's one day's earning = Rs.300

B's one day's earning = Rs.200

C's one day's earning = Rs.250