Aptitude

Percentage

Logarithms

  • if \(log_a(x)=y\) then, x=?

    \(\boxed{x = a^y}\)

    \(log_a(x) \ne x \cdot log(a)\)

  • \(log_a(m \times n)\)

    \(= log_a(m) + log_a(n)\)

    \(\neq log_a(m + n)\)

  • \(log_a(\frac{m}{n})\)

    \(= log_a(m)- log_a(n)\)

    \(\neq log_a(m-n)\)

  • \(log_a(x^m)\)

    \(= m\cdot log_a(x)\)

  • \(log_{a^n}(x)\)

    \(=\frac{1}{n}log_a(x)\)

  • \(log_a(1)\)

    \(=0\)

  • \(log_a(a)\)

    \(=1\)

  • \(a^{log_a(x)}\)

    \(=x\)

  • \(a^{\frac{1}{n}}\)

    \(a^{\frac{1}{n}} = \sqrt[n]{a}\) = \(n^{th} \text{ root of a}\)

    \(a^{\frac{1}{2}} = \sqrt[2]{a}\)

    \(a^{\frac{1}{3}} = \sqrt[3]{a}\)

SI-CI

  • CI

    \(\text{amount}=P(1+\frac{r}{100} )^n\)

    \(\text{CI}=A-I= P(1+\frac{r}{100} )^n -P\)

    \(\text{CI}\ne P(1+\frac{r}{100} )^n\)

    CI = CI in this year - CI of previous Year

  • Difference CI-SI for 2 years

    \(CI-SI=\color {green}P(\frac{r}{100})^2\)

  • Difference CI-SI for 3 years

    \(CI-SI=\color {green}P(\frac{r}{100})^3 + 3P(\frac{r}{100})^2\)

Permutation And Combination

  • Permutation

    \(^nP_r=^nC_r\times R!\)

  • \(^nC_{r-1}+^nC_r=^{n+1}C_r\)

  • \(\binom{n}{r}\)

    \(=\frac{n!}{(n-r)!r!}\)

  • properties of \(\binom{n}{r}\)

    \(\binom{n}{r}=\binom{n}{n-r}\)

    \(\binom{n}{0}=1\)

    \(\binom{n}{1}=n\)

  • \((a + b)^n\)

    \(= \binom{n}{0}a^{n}b^{0} +\binom{n}{1}a^{n-1}b^{1}+\binom{n}{2}a^{n-2}b^{2}+...+\binom{n}{n-1}a^{1}b^{n-1}+\binom{n}{n}a^{0}b^{n}\)

Number System

  • \(N = a^p b^q c^r\)

    • number of factors of N

      =(p + 1) (q + 1) (r + 1)

    • can be expressed as the product of two factors in

      1/2 {(p + 1) (q + 1) (r + 1)} ways

    • if N is a perfect square, it can be expressed

      • as a product of two DIFFERENT factors in

        1/2 {(p + 1) (q + 1) (r + 1) - 1 } ways

      • as a product of two factors in

        1/2 {(p + 1) (q + 1) (r + 1) +1} ways

  • LCM × HCF

    = product of two numbers

  • \(\sum\limits_{k=1}^{n} k=1^1 + 2^1 + 3^1 + … + n^1\)

    \(= \dfrac{n(n + 1)}{2}\)

  • \(\sum\limits_{k=1}^{n} k^2=1² + 2² + 3² + … + n²\)

    \(= \dfrac{n ( n + 1 ) (2n + 1)} { 6}\)

  • \(1³ + 2³ + 3³ + …+ n³\)

    \(= \Big(\dfrac{n(n + 1)}{ 2}\Big)^2\)