Simplification: Quantitative Aptitude MCQ Question and Answer

Solution:  2.002 + 7.9 {(2.8 – 6.3 (3.6 – 1.5) + 15.6)}

= 2.002 + 7.9 (2.8 – 6.3 × 2.1 + 15.6)

= 2.002 + 7.9 (2.8 – 13.23 + 15.6)

= 2.002 + 7.9 × 15.17

= 2.002 + 40.843 = 42.845

Solution:

$3\div \left[ \left( 8-5 \right) \div \left\{ \left( 4-2 \right) \div \left( 2+\frac{8}{13} \right) \right\} \right] $

$=3\div \left[ 3\div \left\{ 2\div \frac{34}{13} \right\} \right] $

$=3\div \left[ 3\div \frac{13}{17} \right] =3\div \frac{3\times 17}{13}$

$=3\times \frac{13}{3\times 17}=\frac{13}{17}$

Solution:

$9-1\frac{2}{9}\mathrm{of} 3\frac{3}{11}\div 5\frac{1}{7}\,\,\mathrm{of} \frac{7}{9}\,\,$

$=\,\,9-\frac{11}{9}\,\,\mathrm{of} \frac{36}{11}\,\,\times \frac{1}{\frac{36}{7}\times \frac{7}{9}}$

$=9-4 \times \frac{1}{4}\,\,=8$

Solution:

Solution: $999\frac{1}{7}+999\frac{2}{7}+999\frac{3}{7}+999\frac{4}{7}+999\frac{5}{7}+999\frac{6}{7}$

$=\,\,999+\frac{1}{7}\,\,+999+\frac{2}{7}+999+\frac{3}{7}+999+\frac{4}{7}+999+\frac{5}{7}+999+\frac{6}{7}$

$6\times 999 +\,\,\frac{1+2+3+4+5+6}{7}\,\, =\,\,5994 +\,\,3 =5997$

Solution: $\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}$

$=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}$

$=\frac{1}{2}\left( 1-\frac{1}{11} \right) \,\,=\frac{5}{11}$

Triks: $\frac{1}{\mathrm{n}\left( \mathrm{n}+2 \right)}+\frac{1}{\left( \mathrm{n}+2 \right) \left( \mathrm{n}+4 \right)}+…. +\frac{1}{\left\{ \mathrm{n}\left( \mathrm{m}-2 \right) \right\} \left( \mathrm{n}+\mathrm{m} \right)}$

$=\,\,\frac{1}{2}\left( \frac{1}{\mathrm{n}}-\frac{1}{\mathrm{n}+\mathrm{m}} \right) $

Solution: $0.\bar{3}+0.\bar{6}+0.\bar{7}+0.\bar{8}$

$=\frac{3}{9}+\frac{6}{9}+\frac{7}{9}+\frac{8}{9}=\frac{3+6+7+8}{9}=\frac{24}{9}=2\frac{2}{3}$

Solution:  total number of students participated = $\frac{\mathrm{B}}{4}+\frac{3\mathrm{G}}{8}$

Required percentage = $\frac{\frac{\mathrm{B}}{4}+\frac{3\mathrm{G}}{8}}{\left( \mathrm{B}+\mathrm{G} \right)}\times 100 %$

Since, B and G are unknown ; Clearly given Data is  inadequate.

Solution: 1^st ⇒ $1+\frac{4}{5}=\frac{9}{5}$

2^nd ⇒ $2+\frac{3\times 5}{9}=\frac{11}{3}$

3^rd ⇒ $1+\frac{2\times 3}{11}=\frac{17}{11}$

4^th ⇒$1+\frac{11}{17}=1\frac{11}{17}$

Solution: 1^st ⇒ $1+\frac{1}{2}=\frac{3}{2}$

2^nd ⇒ $1+\frac{2}{3}=\frac{5}{3}$

3^rd ⇒ $1+\frac{3}{5}=\frac{8}{5}$

4^th ⇒ $1+\frac{5}{8}=\frac{13}{8}$

∴ $2\mathrm{x}+\frac{7}{4}=2\times \frac{13}{8}+\frac{7}{4}=5$