Reasoning Aptitude > Data Interpretation
TABULATION MCQs
Table Charts
Total Questions : 185
| Page 19 of 19 pages
Answer: Option B. -> 34%
$$\eqalign{
& \text{No. of candidates from city C} \cr
& = 1.08\times100000 \cr
& = 108000 \cr
& \text{No. of candidates from city B} \cr
& = 3.14\times100000 \cr
& = 314000 \cr
& \therefore \text{Required percentage} \cr
& = \left(\frac{108000}{314000}\times100\right)\% \cr
& = \left(\frac{108}{314}\times100\right)\% \cr
& = \left(\frac{54}{157}\times100\right)\% \cr
& = \frac{5400}{157}\% \cr
& = 34.39\% \cr
& \approx 34\% \cr} $$
$$\eqalign{
& \text{No. of candidates from city C} \cr
& = 1.08\times100000 \cr
& = 108000 \cr
& \text{No. of candidates from city B} \cr
& = 3.14\times100000 \cr
& = 314000 \cr
& \therefore \text{Required percentage} \cr
& = \left(\frac{108000}{314000}\times100\right)\% \cr
& = \left(\frac{108}{314}\times100\right)\% \cr
& = \left(\frac{54}{157}\times100\right)\% \cr
& = \frac{5400}{157}\% \cr
& = 34.39\% \cr
& \approx 34\% \cr} $$
Answer: Option C. -> 227 : 50
$$\eqalign{
& \text{No. of candidates failing from city D} \cr
& = \frac{3}{4}\times2.27\times100000 \cr
& = \frac{3}{4}\times\frac{227}{100}\times100000 \cr
& = 227\times750 \cr
& \text{No. of candidates failing from city A} \cr
& = \frac{3}{10}\times1.25\times100000 \cr
& = \frac{3}{10}\times\frac{125}{100}\times100000 \cr
& = 300\times125 \cr
& \therefore \text{Required ratio} \cr
& = \frac{227\times750}{300\times125} \cr
& = \frac{227}{50} \cr
& = 227:50 \cr} $$
$$\eqalign{
& \text{No. of candidates failing from city D} \cr
& = \frac{3}{4}\times2.27\times100000 \cr
& = \frac{3}{4}\times\frac{227}{100}\times100000 \cr
& = 227\times750 \cr
& \text{No. of candidates failing from city A} \cr
& = \frac{3}{10}\times1.25\times100000 \cr
& = \frac{3}{10}\times\frac{125}{100}\times100000 \cr
& = 300\times125 \cr
& \therefore \text{Required ratio} \cr
& = \frac{227\times750}{300\times125} \cr
& = \frac{227}{50} \cr
& = 227:50 \cr} $$
Question 183. Directions (1 - 5): Study the following table carefully and answer the question that follow:![The Number Of Candidates Passing The Examination From City F...]( "Click to Enlarge Image")
The number of candidates passing the examination from City F is what per cent of the total number of candidates appearing from all the cities together? ( rounded off to two digits after the decimal)
The number of candidates passing the examination from City F is what per cent of the total number of candidates appearing from all the cities together? ( rounded off to two digits after the decimal)
Answer: Option A. -> 12.93%
Total number of candidates from al the 5 cities
= {(1.25 + 3.14 + 1.08 + 2.27 + 1.85 + 2.73) × 100000}
= 12.32 × 100000
= 1232000
Number of candidates passing the exam from city F
$$\eqalign{
& = \frac{7}{12}\times2.73\times100000 \cr
& = \frac{7\times273\times1000}{12} \cr
& = 7\times91\times250 \cr
& = 159250 \cr
& \therefore \text{Required percentage} \cr
& = \left(\frac{159250}{1232000}\times100\right)\% \cr
& = \frac{15925}{1232}\% \cr
& = 12.926\% \cr
& \approx 12.93\% \cr} $$
Total number of candidates from al the 5 cities
= {(1.25 + 3.14 + 1.08 + 2.27 + 1.85 + 2.73) × 100000}
= 12.32 × 100000
= 1232000
Number of candidates passing the exam from city F
$$\eqalign{
& = \frac{7}{12}\times2.73\times100000 \cr
& = \frac{7\times273\times1000}{12} \cr
& = 7\times91\times250 \cr
& = 159250 \cr
& \therefore \text{Required percentage} \cr
& = \left(\frac{159250}{1232000}\times100\right)\% \cr
& = \frac{15925}{1232}\% \cr
& = 12.926\% \cr
& \approx 12.93\% \cr} $$
Answer: Option A. -> 13000
Number of candidates passing the exam from City E.
$$\eqalign{
& = 1.85\times100000\times\frac{3}{5} \cr
& = 185000\times\frac{3}{5} \cr
& = 37000\times3 \cr
& = 111000 \cr} $$
Number of candidates passing the exam from City E.
$$\eqalign{
& = 1.85\times100000\times\frac{3}{5} \cr
& = 185000\times\frac{3}{5} \cr
& = 37000\times3 \cr
& = 111000 \cr} $$
Answer: Option D. -> D
Number of failures from different cities:
$$\eqalign{
& \text{A}→ 1.25\times100000\times\frac{3}{10} \cr
& \,\,\,\,\,\, = 37500 \cr
& \text{B}→ 3.14\times100000\times\frac{3}{8} \cr
& \,\,\,\,\,\, = \frac{314000\times3}{8} \cr
& \,\,\,\,\,\, = 117750 \cr
& \text{C}→ 1.08\times100000\times\frac{5}{9} \cr
& \,\,\,\,\,\, = 108000\times\frac{5}{9} \cr
& \,\,\,\,\,\, = 60000 \cr
& \text{D}→ 2.27\times100000\times\frac{3}{4} \cr
& \,\,\,\,\,\, = 227000\times\frac{3}{4} \cr
& \,\,\,\,\,\, = 170250 \cr
& \text{E}→ 1.85\times100000\times\frac{2}{5}\cr
& \,\,\,\,\,\, = 185000\times\frac{2}{5} \cr
& \,\,\,\,\,\, = 74000 \cr
& \text{F}→ 2.73\times100000\times\frac{5}{12} \cr
& \,\,\,\,\,\, = 273000\times\frac{5}{12} \cr
& \,\,\,\,\,\, = 113750 \cr } $$
So,
The maximum number failures are from city D.
Number of failures from different cities:
$$\eqalign{
& \text{A}→ 1.25\times100000\times\frac{3}{10} \cr
& \,\,\,\,\,\, = 37500 \cr
& \text{B}→ 3.14\times100000\times\frac{3}{8} \cr
& \,\,\,\,\,\, = \frac{314000\times3}{8} \cr
& \,\,\,\,\,\, = 117750 \cr
& \text{C}→ 1.08\times100000\times\frac{5}{9} \cr
& \,\,\,\,\,\, = 108000\times\frac{5}{9} \cr
& \,\,\,\,\,\, = 60000 \cr
& \text{D}→ 2.27\times100000\times\frac{3}{4} \cr
& \,\,\,\,\,\, = 227000\times\frac{3}{4} \cr
& \,\,\,\,\,\, = 170250 \cr
& \text{E}→ 1.85\times100000\times\frac{2}{5}\cr
& \,\,\,\,\,\, = 185000\times\frac{2}{5} \cr
& \,\,\,\,\,\, = 74000 \cr
& \text{F}→ 2.73\times100000\times\frac{5}{12} \cr
& \,\,\,\,\,\, = 273000\times\frac{5}{12} \cr
& \,\,\,\,\,\, = 113750 \cr } $$
So,
The maximum number failures are from city D.
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