Solved quantitative aptitude problems for competitive exam #2 ~ Arithmetic, Mensuration Formulas and competitive Math Tricks

Shortcut Math formula for competitive exams, Mensuration, Arithmetics of SSC, PSC, Railway Recruitment Banking exam math preparation page is New Math Tricks. Aptitude tricks are made with traditional and shortcuts formulas.

Friday, 15 December 2017

Solved quantitative aptitude problems for competitive exam #2

 This is the second part of solved quantitative aptitude problems for competitive exams. We are providing some important quantitative aptitudes for quicker math solution using maths tricks. Let's looks some previous years math problems and easy solution. 

Problem1: A boat travels 24 km in upstream in 6 hours and 20 km downstream in 4 hours. Then the speed of the boat in still water and the speed of the current are-
Solve: We knew the travelled distance of upstream and downstream.We need the speed of upstream and downstream of the boat to execute the answer by shortcut formula.
Upstream rate =24/6 =4 kmph.
Downstream =20/4 =5 kmph. 
solved-quantitative-aptitude-problems-for competitive exam-by-math-tricks
Problem 2: The single discount equivalent discount to two successive discounts of 40% and 10% is-
Answer:  Look at below math tricks, how solved it quickly. 

Problem 3: A single discount equivalent to consecutive discount of 20%, 20% and 10% is-
Quick solve: 
Problem 4: Average of three numbers is 40. The first number is twice the second and the second one is thrice the third number. The difference between the largest and the smallest number is-
Solution: here 1st no>2nd no>3rd no. If we assume 1st no as x, then 2nd and 3rd number will be fractional numbers. To avoid that we assume the lowest number (3rd no) as x. let's look at the quick solution. 
Problem 5: The average score of a cricketer has 60 in ten innings. How many runs to be scored in the eleventh innings to raise the average score to 65?  
Solution: The average score of 10 innings is 60, means the batsman scored (60*10) =600 runs. As it will be (65*11) =715 after eleventh innings. So the batsman must be the score in eleventh innings (715-60) =115 runs.
Problem 6: In an election, 80% of the voter has voted and three candidates contested. The first candidate got 30% votes and second got 46% votes. If the total number of registered voter were 50,000, then find the number of votes got by the 3rd candidate. 
Solution: Look at the question 
  • All voters are not voted, only 80% voters are voted and fifty thousand of voter are registered. 
  • The 3rd candidates got votes 100-(30+46) =24%. 

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