# 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) Friday | B) Saturday |

C) Sunday | D) Thursday |

Explanation:

**15 Aug, 1947** = (1946 years + Period from 1.1.1947 to 15.8.1947)

**Odd days** in 1600 years = 0

**Odd days** in 300 years = 1

**46 years** = (35 ordinary years + 11 leap years) = (35 x 1 + 11 x 2)= 57 (8 weeks + 1 day) = **1 odd day **

**Jan. Feb. Mar. Apr. May. Jun. Jul. Aug **

( **31 + 28 + 31 + 30 + 31 + 30 + 31 + 15** ) = 227 days = (32 weeks + 3 days) = **3 odd days.**

**Total number of odd days = (0 + 1 + 1 + 3) = 5 odd days. **

Hence, as the number of **odd days = 5 , given day is Friday.**

A) Tuesday | B) Monday |

C) Sunday | D) Saturday |

Explanation:

Each day of the week is repeated after 7 days. So, after 63 days, it will be Monday.

After 61 days, it will be Saturday.

A) 2500 | B) 2700 |

C) 2900 | D) 3100 |

Explanation:

Total number of votes = 7500

Given that 20% of Percentage votes were invalid

=> Valid votes = 80%

Total valid votes = 7500*(80/100)

1st candidate got 55% of the total valid votes.

Hence the 2nd candidate should have got 45% of the total valid votes

=> Valid votes that 2nd candidate got = total valid votes x (45/100)

7500*(80/100)*(45/100) = 2700

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) 500 | B) 1000 |

C) 1500 | D) 2000 |

Explanation:

Let the total profit be Rs. 100.

After paying to charity, A's share = (95*3/5) = Rs. 57.

If A's share is Rs. 57, total profit = Rs. 100.

If A's share is Rs. 855, total profit = (100/57*855) = 1500.

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) 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) 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