Important formulas for Data Interpretation

Data Interpretation Formulas

We use some aptitude formulas in solving DI questions. Most Common questions are asking in this format from Percentages

1) X is what % of y that means x/y * 100.

Eg: 72 is what % of 360 = [72/360]×100 =20%

2) What percentage is x of y? I.e. x/y * 100.

3) X is what percent more than y = x-y/y * 100

4) X is what percent less than y = y-x/y * 100

5) Percentage change between two particular values = [final – initial/ initial] * 100

Memorize some fractional values and percentage values.

      • 1/2 = 50%
      • 1/3 = 33.33%
      • ¼ = 25%
      • 1/5 = 20%
      • 1/6 =16.66%
      • 1/7 = 14.28%
      • 1/8 = 12.5%
      • 1/9 = 11.11%
      • 1/10 = 10%
      • 1/11 =9.09 %
      • 1/12 = 8.33%
      • 1/13 = 7.69%
      • 1/14 = 7%
      • 1/15 = 6.66%
      • 1/16 = 6.25%
      • 1/17 = 5.88%
      • 1/18 =5.55%
      • 1/19 = 5.26%
      • 1/20= 5%
      • 5%= 1/20 = 0.05
      • 10% =1/10 =0.1
      • 15% =3/20
      • 20% = 1/5
      • 25% = ¼
      • 30% =3/10
      • 40% =2/5
      • 50% =1/2
      • 55%=11/20
      • 60% = 3/5
      • 70% = 7/10
      • 75% = 3/4
      • 80% = 4/5
      • 90% = 9/10
      • 100%=1
      • 6 ¼ % = 1/6
      • 12 ½ % = 1/8
      • 16 2/3 % = 1/6
      • 33 1/3 % = 1/3
      • 66 2/3 % =2/3
      • 125% =5/4
      • 150% = 3/2

Comparison of two fractions:

Eg: Compare, 3/4, and 1/4.

By seeing itself, we can say 3/4 is big. Implement here our logic of cross multiplication of two fractions.

¾ ¼ we get, 12 and 4 we know 12 > 4 so we can say ¾ >1/4

Apply the same logic big fractions it’s very simple and takes less time.

Formulas on Averages:

  • Average = sum of elements/ no. of elements  (or) (a1 + a2 + a3 + a4 +………… an)/n
  • Sum of first n odd numbers:   1 + 3 + 5 + ………… + (2n-1)= n2
  • Sum of first n even numbers:  2 + 4 + 6 + …………+ 2n = n (n+1).
  • Sum of first n natural numbers: 1+2+3 +4 + ……..+n = n(n+1)/2.
  • Sum of squares of first n natural numbers: 12 + 22 + 32 + 42 + 52 + 62 + …………+n[n (n+1)(2n+1)]/6.
  • Sum of cubes of first n natural numbers:  13 + 23+33 + 43 + …… + n3  [n(n+1)/2]2

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