If you are writing afcat exam in near future, there are few simple afcat exam numerical ability maths tips & formulas you want to keep on your finger tips to solves the mathematics question of afcat question paper. In this article we are sharing 6 Must Know Tips To Crack AFCAT 2 2017 Entry Numerical Ability Section.
6 Must Know Tips To Crack AFCAT 2 2017 Entry Numerical Ability Section
- In case of ratio, remember important derivatives like if a:b,b:c given then value of a:b:c and similar forms.
- In case of profit & loss, learn formulas for sp, c.p, m.p, successive discount, error% etc.
- For eg successive discount= a+b+ab/100 note here the sign for digits will be negative since a discount refers to deduction in amount.
- In involution, learn derivatives for cubic terms like A cube + B cube etc.
- In simple interest and C.I general formulas are needed. Same is for work force. Remember practice questions based on canisters/buckets/pipes.
Decimal Fraction:
- A decimal fraction is a fraction in which denominator is an integer power of ten. (The term decimals are commonly used to refer decimal fractions). Generally, a decimal fraction is expressed using decimal notation and its denominator is not mentioned explicitly
Examples: Â 1/10 = .1 , 1/100 = .01
- Conversion of a Decimal into Common Fraction: Put 1 in the denominator under the decimal point and annex with it as many zeros as is the number of digits after the decimal point. Now, remove the decimal point and reduce the fraction to its lowest terms.
Examples: 0.5 = 5/10=1/2
- Some Basic Formulas:
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- (a + b)(a – b) = (a2 – b2)
- (a + b)2 = (a2 + b2 + 2ab)
- (a – b)2 = (a2 + b2 – 2ab)
- (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
- (a3 + b3) = (a + b)(a2 – ab + b2)
- (a3 – b3) = (a – b)(a2 + ab + b2)
- (a3 + b3 + c3 – 3abc) = (a + b + c)(a2 + b2 + c2 – ab – bc – ac)
- When a + b + c = 0, then a3 + b3 + c3 = 3abc.
Simplification:
- Rule of ‘BODMAS’: This BODMAS rule depicts the correct sequence in which the operations are to be executed, so as to find out the value of given expression. Full form of BODMAS is B – Bracket, O – of, D – Division, M – Multiplication, A – Addition and S – Subtraction. Thus, while solving or simplifying a problem, first remove all brackets, strictly in the order (), {} and ||. After removing the brackets, we will use the following operations strictly in the following order: (i) of (ii) Division (iii) Multiplication (iv) Addition (v) Subtraction.
Average:
- Average = (Sum of observations/Number of observations)
- Suppose a train covers a certain distance at x kmph and an equal distance at y kmph. Then, the average speed of train during the whole journey is kmph (2xy/x+y)kmph.
Percentage:
- By a certain percent, we mean that many hundredths. Thus, x percent means x hundredths, written as x%. To express x% as a fraction: We have, x% = x/100
- To express a/b as a percent: We have, a/b = (a/b x 100)%
- Percentage Increase/Decrease: If the price of a commodity increases by R%, then the reduction in consumption so as not to increase the expenditure is: [(R/(100+R)) x 100]%. If the price of a commodity decreases by R%, then the increase in consumption so as not to decrease the expenditure is: [(R/(100-R)) x 100]%.
- Result on Population: Let the population of a town be P now and suppose it increases at the rate of R% per annum, then: population after n years = P (1+(R/100))n Population n years ago= P/ (1+(R/100))n
- Result on Depreciation: Let the present value of a machine be P. Suppose it depreciates at the rate of R% per annum. Then: Value of the machine after n years = P (1-(R/100)n , Value of the machine n years ago = P/ (1-(R/100)n, If A is R% more than B, then B is less than A by [(R/(100+R)) x 100] %, If A is R% less than B, then B is more than A by [(R/(100-R)) x 100] %
