Simplification: Quantitative Aptitude MCQ Question and Answer

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Simplification MCQ Question with Details Answer

 

Q1. 2.002 + 7.9 {(2.8 – 6.3 (3.6 – 1.5) + 15.6)}

(a) 2.002

(b) 4.2845

(c) 4.0843

(d) 42.845

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

Q2. The value of $3\div \left[ \left( 8-5 \right) \div \left\{ \left( 4-2 \right) +\left( 2+\frac{8}{13} \right) \right\} \right] $

(a) $\frac{15}{17}$

(b) $\frac{13}{17}$

(c) $\frac{15}{19}$

(d) $\frac{13}{19}$

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] $

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

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

Q3.  $9-1\frac{2}{9}\mathrm{of} 3\frac{3}{11}\div 5\frac{1}{7}\,\,\mathrm{of} \frac{7}{9}\,\,=?$

(a) $\frac{5}{4}$

(b) 8

(c) $8\frac{32}{81}$

(d) 9

Solution:

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

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

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

 

Q4.  777.07 + 77.77 + 0.77 + 7.07 + 7 + 77 = ?

(a) 946.78

(b) 946.68

(c) 964.68

(d) 946.86

Solution:

s-quiz1-a4

Q5. Find the value of $999\frac{1}{7}+999\frac{2}{7}+999\frac{3}{7}+999\frac{4}{7}+999\frac{5}{7}+999\frac{6}{7}$

(a) 5997

(b) 5979

(c) 5994

(d) 6997

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

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

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

Q6. Find the value of $\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}$

(a) $\frac{10}{11}$

(b) $\frac{9}{11}$

(c) $\frac{5}{11}$

(d) $\frac{7}{11}$

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) $

Q7. Find the value of  $0.\bar{3}+0.\bar{6}+0.\bar{7}+0.\bar{8}$

(a) $2\frac{3}{10}$

(b) $2\frac{2}{3}$

(c) $20.\overline{35}$

(d) $5\frac{3}{10}$

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

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

Q8. 1/4th of number of boys and 3/8th of number of girls participated in annual sports of the school. What fractional part of total number of students participated?

(a) 3

(b) 2

(c) 3

(d) Data inadequate

(e) None of the above

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.

Q9. Simplify : $1+\frac{1}{1+\frac{2}{2+\frac{3}{1+\frac{4}{5}}}}$

(a) $1\frac{11}{17}$

(b) $1\frac{5}{7}$

(c) $1\frac{6}{17}$

(d) $1\frac{21}{17}$

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

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

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

4th ⇒$1+\frac{11}{17}=1\frac{11}{17}$

Q10. . If x =  $1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{2}}}}$ then, the value of $2\mathrm{x}+\frac{7}{4}$  is

(a) 3

(b) 4

(c) 5

(d) 6

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

2nd ⇒ $1+\frac{2}{3}=\frac{5}{3}$

3rd ⇒ $1+\frac{3}{5}=\frac{8}{5}$

4th ⇒ $1+\frac{5}{8}=\frac{13}{8}$

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

 

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