Table Chart — Part 1 MCQs with Explanations

Total Students = 800 No. of students in secondary = 80% of 800 = 640 Rest Students = 800 - 640 = 160 Rest students are divided equally into class 12 and 11. So, No. of students in class 12 = $$\frac{160}{2}$$ = 80 Now, total vegetarian = 53% No. of Total vegetarian = 53% of 800 = 424 55% of secondary students are vegetarian. No. of vegetarian in secondary = 55% of 640 = 352 No. vegetarian in 11 = 50% of 80 = 40 Thus, no. of vegetarian in 12, = 424 - 352 - 40 = 32 Thus, In class 12 total vegetarian = 32. So, % of vegetarian = $$\frac{32}{80}$$ × 100 = 40%

Male vegetarian = 32 - 8 = 24 Difference = 24 - 8 = 16

% of male students in secondary section, = $$\frac{288}{640}$$ × 100 [∵ 800 × 80% = 640] = 45%

$$\eqalign{ & \text{Average in four subjects,} \cr & = \frac{56+68+68+48}{4} \cr & = \frac{240}{4} \cr & = 60 \cr} $$

Total marks = 40 × 68 = 2720

Difference in 2007 is 34 which is maximum.

Difference in 28 is minimum in 2008 i.e. 8.

Required difference, = (200 + 195 + 245 + 200 + 225) - (150 + 200 + 250 + 230 + 200) = 1065 - 1030 = 35

$$\eqalign{ & = \left(\frac{175}{900}\times100\right)\% \cr & = \frac{175}{9}\% \cr & = 19\frac{4}{9}\% \cr} $$

Here, A = 820, B = 830, C = 880, D = 700 and E = 905 Hence, The school, which has the maximum number of players is E.

$$\eqalign{ & x\% \text{ of } 130 = 65 \cr & \Rightarrow \frac{x\times130}{100} = 65 \cr & \Rightarrow x = 50 \cr} $$

The required diet can be formed by mixing P and S or Q and S but the lowest cost per unit can be obtained by Q and S only.

To make diet with at lest 60% carbohydrate we can use option (B) and (E) only, but the lowest cost per unit can be given by when P, Q and S are mixed in the ratio 4 : 1 : 1

As the ingredients are mixed in equal amounts, so we can take the average of the constituent % of the element used. Only O and S satisfies all the conditions.

As the diet should contain 10% minerals and only two ingredient contain 10 minerals i.e. O and Q. hence, only be mixing O and Q a diet with 10% minerals can be formed. Hence, there is only one way.

% change in the cost incurred in the fifth month, $$\eqalign{ & = \frac{100000-40000}{40000}\times100 \cr & = 150\% \cr} $$

In new technique 5 people working in 5 th month, 4 in 6 th , 5 in 7 th and 5 in 8 th . Total people working in this period = 19 Each get paid Rs. 10000, Then total cost = Rs. 1,90,000

Difference between old and new technique, = 190000 - 130000 = Rs. 60,000

Cost incurred in specification stage, = (80000 + 120000) = Rs. 200000 Which is maximum cost.

Average cost is lowest for 11 - 15

Average number of candidates appeared from T all over the years $$\eqalign{ & = \frac{\left(6.2+6.2+6.4+7.8+9.9+11.8\right)\times100}{6} \cr & = \frac{48.3\times100}{6} \cr & = \frac{4830}{6} \cr & = 805 \cr} $$

In zone S, the difference between the appeared candidates and the qualified candidates in various years is given below: 2005 → (4.2 - 2.4) = 1.8 2006 → (7.4 - 3.3) = 4.1 2007 → (8.3 - 5.6) = 2.7 2008 → (9.3 - 6.4) = 2.9 2009 → (11.4 - 5.2) = 6.2 2010 → (14.2 - 11.4) = 2.8 If was lowest in 2005 and second lowest in 2007

Total number of candidates who qualified the test in various zones are: Total number of qualified students in P = 2.5 + 3.4 + 4.9 + 5.7 + 4.8 + 5.6 ⇒ 26.9 Total number of qualified students in Q = 1.4 + 4.2 + 4.5 + 3.4 + 5.2 + 6.2 ⇒ 25.1 Total number of qualified students in R = 2.2 + 4.2 +4.8 + 3.2 + 6.8 + 7.4 ⇒ 28.8 Total number of qualified students in S = 2.4 + 3.3 + 5.6 + 6.4 + 5.2 + 11.4 ⇒ 34.3 Total number of qualified students in T = 2.6 +4.8 +4.2 + 6.2 + 9.4 + 6.9 ⇒ 34.1 34.1 is the second highest ∴ Zone T is the second highest in total number of qualified students.

$$\eqalign{ & \text{Required percentage} \cr & = \left(\frac{7.4}{5.4}\times100\right)\% \cr & = \frac{3700}{27}\% \cr & = 137\% \cr} $$

$$\eqalign{ & \text{Required ratio} \cr & = \frac{3.2}{5.6} \cr & = \frac{32}{56} \cr & = \frac{4}{7} \cr & = 4:7 \cr} $$

Total number of employees in all the departments = 840 + 220 + 900 + 360 + 450 + 540 = 3310

Number of females in Production : Number of females in Marketing $$\eqalign{ & = \left(900\times\frac{23}{100}\right):\left(450\times\frac{44}{100}\right) \cr & = 207:198 \cr & = 23:22 \cr} $$

Total number of males in IT and Customer Service $$\eqalign{ & = \left(840\times\frac{55}{100}\right)+\left(540\times\frac{60}{100}\right) \cr & = 462 + 324 \cr & = 786 \cr} $$

(Number of females in HR and Accounts) : (Number of males in HR and Accounts) = $$ \left\{\left(\frac{65}{100}\times360\right)+\left(\frac{35}{100}\times220\right)\right\}$$ $$: \left\{\left(\frac{35}{100}\times360\right)+\left(\frac{65}{100}\times220\right)\right\}$$ = (234 + 77) : (126 + 143) = 311 : 269

$$\eqalign{ & \text{Required percentage} \cr & = \left(\frac{360}{220}\times100\right)\% \cr & = 163.6\% \cr & \approx164\% \cr} $$

Average semester fee for M. Sc. course over the give years. $$=\frac{\left(5.8+6.4+10.2+14.6+17.7+20.9 \right)\times1000}{6}$$ $$\eqalign{ & = \frac{75.6\times1000}{6} \cr & = \frac{75600}{6} \cr & = \text{Rs. }12600 \cr} $$

Total semester fee for all the courses in 2006 = Rs. (14.5 + 6.4 + 11.6 + 5.8 + 3.2) × 1000 = Rs. (41.5 × 1000) = Rs. 41500

Percent increase in semester fees of B. Ed. in the year 2007 as compared to 2006. $$\eqalign{ & = \left\{ \frac{\left(13.9-11.6\right)}{11.6}\times100 \right\} \% \cr & = \left(\frac{2.3}{11.6}\times100\right)\% \cr & = \frac{2300}{116}\% \cr & = \frac{575}{29}\% \cr & = 19.8\% \cr & \approx 20\% \cr} $$

Required difference = Rs. {(1.8 + 3.2 + 4.8 + 5.6 + 12.5 + 14.9) × 1000 - (35.8 × 1000)} = Rs. {(42.8 × 1000) - (35.8 × 1000)} = Rs. (7 × 1000) = Rs. 7000

$$\eqalign{ & \text{Required Percentage} \cr & = \left\{\frac{\left(12.7\times1000\right)}{\left(17.7\times1000\right)}\times100\right\} \%\cr & = \left(\frac{127}{177}\times100\right)\% \cr & = \frac{12700}{177}\% \cr & = 71.75\% \cr & \approx 72\% \left(\text{Approx}\right) \cr} $$

$$\eqalign{ & \text{No. of candidates from city C} \cr & = 1.08\times100000 \cr & = 108000 \cr & \text{No. of candidates from city B} \cr & = 3.14\times100000 \cr & = 314000 \cr & \therefore \text{Required percentage} \cr & = \left(\frac{108000}{314000}\times100\right)\% \cr & = \left(\frac{108}{314}\times100\right)\% \cr & = \left(\frac{54}{157}\times100\right)\% \cr & = \frac{5400}{157}\% \cr & = 34.39\% \cr & \approx 34\% \cr} $$

Number of candidates passing the exam from City E. $$\eqalign{ & = 1.85\times100000\times\frac{3}{5} \cr & = 185000\times\frac{3}{5} \cr & = 37000\times3 \cr & = 111000 \cr} $$

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 $$\eqalign{ & = \frac{7}{12}\times2.73\times100000 \cr & = \frac{7\times273\times1000}{12} \cr & = 7\times91\times250 \cr & = 159250 \cr & \therefore \text{Required percentage} \cr & = \left(\frac{159250}{1232000}\times100\right)\% \cr & = \frac{15925}{1232}\% \cr & = 12.926\% \cr & \approx 12.93\% \cr} $$

$$\eqalign{ & \text{No. of candidates failing from city D} \cr & = \frac{3}{4}\times2.27\times100000 \cr & = \frac{3}{4}\times\frac{227}{100}\times100000 \cr & = 227\times750 \cr & \text{No. of candidates failing from city A} \cr & = \frac{3}{10}\times1.25\times100000 \cr & = \frac{3}{10}\times\frac{125}{100}\times100000 \cr & = 300\times125 \cr & \therefore \text{Required ratio} \cr & = \frac{227\times750}{300\times125} \cr & = \frac{227}{50} \cr & = 227:50 \cr} $$

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.