Aptitude Test

Aptitude test : percentage shortcuts and tricks

In this post I describe some of the tricks of percentage calculation . by a certain percent , we means that many hundredths thus , x percent means x hundredths written as x%

fraction value of percentage : 1/2 = 50 % 1/3 = 33.33% 1/6 =16.67% 1/7 = 14.28%

1/4 = 25% 1/8 =12.5%

1/5 = 20 % 1/9 = 11.11

Note :these fraction value helps to minimize the calculation

Formula: 1. if the price of a commodity increases by R% , then the reduction in consumption so as not to increase the expenditure is 100*R/(100+R)%. 2. if the price of a commodity decreases by R% , then the increases n consumption so as not to decreases the expenditure is 100*R/(100-R)%. 3. value of machine after n years = p(1-R/100)power n . 4. if A is R% more than B , then b is less than A by 100*R/(100+R)%. 5. if A is R% less than B , then b is more than A by 100*R/(100-R)%. Example: 1. if 1/5 of a number is 25 less then 1/4 of a same number . what will be 3/2 of number . ans. as you know 1/5 = 20% and 1/4 = 25% so difference b/w these two is 5% which is equal to 25 then 5% *20 ---- 25*20 100% ----500 and 100% percent is the real value you can now calculate 3/2 so ans is 750 2. if the numerator of fraction is increase by 20% and denominator is decrease by 80% then fraction becomes 18/15. what is the actual fraction . ans .

120% N/20% D = 18/15 then N/D = (18/15 )*20/120 N/D = 1/5 answer

3 . if A's salary 20% less then B's salary then B's salary is how much more then A. ans . suppose B's salary is 100 then A's salary is 80 now you can calculate b's salary in respect to A ( 20/80)* 100 = 25% is answer 4. A number is increased by 12% and again deceased by 18% then how much change in that value. formula for this problem % change = (x)+(y)+x*y/100 Note take the value with symbol , if increase then take '"plus" if decease in value then take "minus" this formula is applicable for circle , square , triangle etc 12- 18 - 216/100 = -8.16 decrease is value by 8.16%

11:01 PM Posted by manoj singh at 11:01 PM No comments: Category : shortcuts for quantitive, TIME AND DISTANCE SHORTCUTS FOR QUANTITATIVE APTITUDE

10 quantitative aptitude tricks for Govt. exam

Quantitative aptitude is the most common part of every exam . In this post I want to motivate you to solve the quantitative problem with the help of great videos. These videos are made by different institute and math teachers. Don't Just watch the videos but try to do same practice. Read more » Rest of the post

11:45 AM Posted by manoj singh at 11:45 AM Category : shortcuts for quantitive

TIME AND DISTANCE SHORTCUTS FOR QUANTITATIVE APTITUDE TIME AND DISTANCE SHORTCUTS FOR QUANTITATIVE APTITUDE

Formula: Distance = Speed x Time Time = Distance / Speed

Speed = Distance / Time

Point to care: To convert speed in kmph to m/sec, multiply it with 5/18 To convert speed in m/sec to kmph , multiply it with 18/5. Example 1. If a men travels from point A to point B with a speed of 'a' and back to point A (from point B) with a speed of b, then the average speed of the body is:

2ab/(a+b). A car covers a certain distance at a speed of 90 km/hr while going and returns to the starting point at a speed of 60 km/hr. Find the average speed of the car for the whole journey? Ans: Average speed = (2 x 90 x 60)/ (60+90) = 72 km/hr . 2. If a car does a journey in 'T' hrs, the first half at 'a' km/hr and the second half at 'b' km/hr. The total distance covered by the car:

(2 x Time x a x b ) / (a + b). A motorcar does a journey in 10 hrs, the first half at 21 kmph and the second half at 24 kmph. Find the distance? Ans: Distance = (2 x 10 x 21 x 24) / (21+24) = 10080 / 45 = 224 km. 3.If a person goes from 'A' to 'B' at a speed of 'a' kmph and returns at a speed of 'b' kmph and takes 'T' hours in all, then the distance between the A and B:

Total time taken x (Product of the two Speeds / Addition of the two speeds) A boy goes to school at a speed of 3 kmph and returns to the village at a speed of 2 kmph. If he takes 5 hrs in all, what is the distance between the village and the school? Ans: Let the required distance be x km. Then time taken during the first journey = x/3 hr. and time taken during the second journey = x/2 hr. x/3 + x/2 = 5 => (2x + 3x) / 6 = 5 => 5x = 30. => x = 6

Required distance = 6 km. 4: Walking ¾ of his speed, a person is 10 min late to his office. Find his usual time to cover the distance? Ans: Usual time = Late time / {1/ (3/4) - 1) = 10 / (4/3 -1 ) = 10 / (1/3) = 30 minutes. 5: Running 4/3 of his usual speed, a person improves his timing by 10 minutes. Find his usual timing by 10 minutes. Find his usual time to cover the distance? Ans: Usual time = Improved time / { 1 - (1/ (3/4)} = 10 / { 1- (3/4) } = 40 minutes. 6 A train travelling 25 kmph leaves Delhi at 9 a.m. and another train travelling 35 kmph starts at 2 p.m. in the same direction. How many km from will they be together ? Ans: Meeting point's distance from the starting point = [25 x 35 x (2p.m. - 9 a.m)] / (35 -25) = (25 x 35x 5) / 10 = 4375 / 10 = 437.5 km .