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.


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)

15th_August_1947_day1534219857.jpg image

 

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.

Report Error

View Answer Report Error Discuss

3734 392042
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 Report Error Discuss

2448 276888
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 = 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

Report Error

View Answer Report Error Discuss

2406 271715
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 Report Error Discuss

1777 218580
Q:

If 20% of a = b, then b% of 20 is the same as :

A) 4% of a B) 6% of a
C) 8% of a D) 10% of a
 
Answer & Explanation Answer: A) 4% 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.

Report Error

View Answer Report Error Discuss

1261 145171
Q:

Fresh fruit contains 68% water and dry fruit contains 20% water. How much dry fruit can be obtained from 100 kg of fresh fruits ?

A) 20 B) 30
C) 40 D) 50
 
Answer & Explanation Answer: C) 40

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.

Report Error

View Answer Report Error Discuss

1160 93781
Q:

A trader mixes 26 kg of rice at Rs. 20 per kg with 30 kg of rice of other variety at Rs. 36 per kg and sells the mixture at Rs. 30 per kg. His profit percent is:

A) No profit, no loss B) 5%
C) 8% D) 10%
 
Answer & Explanation Answer: B) 5%

Explanation:

C.P. of 56 kg rice = Rs. (26 x 20 + 30 x 36) = Rs. (520 + 1080) = Rs. 1600.

 

S.P. of 56 kg rice = Rs. (56 x 30) = Rs. 1680.

 

Gain =(80/1600*100) % = 5%

Report Error

View Answer Report Error Discuss

988 129448
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 = a2 

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

New side = 125a/100 = 5a/4 

 

New area = 5a42  

 

Increased area= 25a216-a2 

 

Increase %=9a2/16a2*100  % = 56.25%

Report Error

View Answer Report Error Discuss

941 137184