Quantitative Aptitude - Arithmetic Ability Questions


What is Quantitative Aptitude- Arithmetic Ability?


Quantitative aptitude test helps measure one's numerical ability, problem solving and mathematical skills. Quantitative aptitude is found in almost all the entrance exams, competitive exams and placement exams. Quantitative aptitude 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 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 problems.


Wide range of Quantitative aptitude 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 Competitive Exams, ICET, UPSC,SNAP Test, KPSC, XAT, GRE, Defence Competitive exams, LIC/G IC Competitive exams, Railway Competitive exam ,TNPSC , University Grants Commission (UGC), Career Aptitude test (IT companies), Government Exams and etc.


Q:

What was the day on 15th august 1947 ?

A) Friday B) Saturday
C) Sunday D) Thursday
 
Answer & Explanation Answer: A) Friday

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 + 1 ) = 213 days = (30 weeks + 3 days) = 3 odd days.

 

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

 

Given day is Friday.

Report Error

View Answer Workspace Report Error Discuss

184 33204
Q:

Today is Monday. After 61 days, it will be :

A) Tuesday B) Monday
C) Sunday D) Saturday
 
Answer & Explanation Answer: 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.

Report Error

View Answer Workspace Report Error Discuss

248 23873
Q:

A bag contains 50 P, 25 P and 10 P coins in the ratio 5: 9: 4, amounting to Rs. 206. Find the number of coins of each type respectively.

A) 360, 160, 200 B) 160, 360, 200
C) 200, 360,160 D) 200,160,300
 
Answer & Explanation Answer: C) 200, 360,160

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

Report Error

View Answer Workspace Report Error Discuss

176 20122
Q:

In an election between two candidates, one got 55% of the total valid votes, 20% of the votes were invalid. If the total number of votes was 7500, the number of valid votes that the other candidate got, was :

A) 2500 B) 2700
C) 2900 D) 3100
 
Answer & Explanation Answer: B) 2700

Explanation:

Total number of votes = 7500

Given that 20% of Percentage votes were invalid

=> Valid votes = 80%

Total valid votes = 

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 

Report Error

View Answer Workspace Report Error Discuss

117 19086
Q:

If each side of a square is increased by 25%, find the percentage change in its area?

A) 65.25 B) 56.25
C) 65 D) 56
 
Answer & Explanation Answer: B) 56.25

Explanation:

let each side of the square be a , then area = a x a

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

New area =(5a x 5a) / (4 x 4) = (25a²/16) 

increased area== (25a²/16) - a²

Increase %= [(9a²/16 ) x (1/a² ) x 100]% = 56.25%

Report Error

View Answer Workspace Report Error Discuss

96 17279