# 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) 4% of a | B) 6% of a |

C) 8% of a | D) 10% of a |

Explanation:

20% of a = b

=> (20/100) * a = b

b% of 20 =(b/100) x 20 = [(20a/100) / 100] x 20= 4a/100 = 4% of a.

A) 65.25 | B) 56.25 |

C) 65 | D) 56 |

Explanation:

let each side of the square be a , then area = ${a}^{2}$

As given that The side is increased by 25%, then

New side = 125a/100 = 5a/4

New area = ${\left(\frac{5a}{4}\right)}^{2}$

Increased area= $\frac{25{a}^{2}}{16}-{a}^{2}$

Increase %=$\frac{\left[9{a}^{2}/16\right]}{{a}^{2}}*100$ % = 56.25%

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) 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) 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) 48 min. past 12. | B) 46 min. past 12. |

C) 45 min. past 12. | D) 47 min. past 12. |

Explanation:

Time from 8 a.m. on a day to 1 p.m. on the following day = 29 hours.

24 hours 10 min. of this clock = 24 hours of the correct clock.

$\frac{145}{6}$ hrs of this clock = 24 hours of the correct clock.

29 hours of this clock = $24*\frac{6}{145}*29$ hrs of the correct clock

= 28 hrs 48 min of the correct clock.

Therefore, the correct time is 28 hrs 48 min. after 8 a.m.

This is 48 min. past 12.

A) 18 litres | B) 24 litres |

C) 32 litres | D) 42 litres |

Explanation:

Let the quantity of the wine in the cask originally be x litres

Then, quantity of wine left in cask after 4 operations =$\left[x{\left(1-\frac{8}{x}\right)}^{4}\right]$litres

$\therefore \left[\frac{x{\left(1-{\displaystyle \frac{8}{x}}\right)}^{4}}{x}\right]=\frac{16}{81}$

$\Rightarrow {\left[1-\frac{8}{x}\right]}^{4}={\left(\frac{2}{3}\right)}^{4}$

$\Rightarrow x=24$

A) 1/2 | B) 7/15 |

C) 8/15 | D) 1/9 |

Explanation:

Let S be the sample space

Then n(S) = no of ways of drawing 2 balls out of (6+4) =$10{C}_{2}$ 10 =$\frac{10*9}{2*1}$ =45

Let E = event of getting both balls of same colour

Then,n(E) = no of ways (2 balls out of six) or (2 balls out of 4)

=$6{C}_{2}+4{C}_{2}$ = $\frac{6*5}{2*1}+\frac{4*3}{2*1}$= 15+6 = 21

Therefore, P(E) = n(E)/n(S) = 21/45 = 7/15