Bar Chart — Part 1 MCQs with Explanations

Increase in Population = 68.40 - 54.80 = 13.6 crores

Percentage increase in 2001 = 24.8

% increase in 1971 = [(36.14 × 31.85) × 100] = 13.47% least in all

Total increase = 68.40 - 27.90 = 40.5 crores Hence, annual increase $$\eqalign{ & = \frac{40.5}{50} \text{ crores} \cr & = \frac{40.5\times10000000}{50} \cr & = 8100000 \cr} $$

No. of other vehicles registered in January 1991 = 27 - 21 = 6 thousand No. of other vehicles registered in April 1991 = 35 - 20 = 15 thousand Hence, increased = 15 - 6 = 9 thousand

Registered increase = 28 - 21 = 7 thousand

No. of vehicles registered in May = 35 - 17 = 18 is maximum in all.

No. of vehicles registered in March 1991 = 25 - 15 = 10 thousand

Registered % increase = $$\frac{\left(15 - 15\right)\times100}{15}$$ = 0%

In the year 1998, (75 - 25) = 50 mixer-grinder purchased. Then, In 2000, 20% of 50 disposed off = 10. Similarly, number of mixer grinder purchased in 1997 is 25. Number of mixer-grinder disposed in 1999 = 20% of 25 = 5. Total number of mixer-grinder disposed off at end of 2000 = 10 + 5 = 15

It is given that 20% of the mixer-grinders are disposed off after 2 years. There for in 1999, 5 mixer-grinder disposed off (20% of 25). Also the number of mixer grinder in circulation in 1999 was 120 and 1998 was 75. The difference is 45. The total number of mixer-grinder purchased is (45 + 5) = 50

Since, information of Purana mixer-grinder is not given for the even-numbered years (1996, 1998, etc), the number of mixer-grinder disposed off in 2000 cannot determined.

$$\eqalign{ & \text{% increase in 2004,} \cr & = \frac{66.90-59.50}{59.50}\times100 \cr & = 12.44\% \cr & \cr & \text{% increase in 2003,} \cr & = \frac{59.50-48.60}{48.60}\times100 \cr & = 22.43\% \cr & \cr & \text{% increase in 2002,} \cr & = \frac{48.60-40.90}{40.90}\times100 \cr & = 18.83\% \cr & \cr & \text{% increase in 2001,} \cr & = \frac{40.90-35.80}{35.80}\times100 \cr & = 14.25\% \cr & \cr} $$ Thus, maximum increase in daily earning of men over preceding year is in 2003

Difference between earning of man and women in 2000, = 35.8 - 18.3 = 17.5 Difference between earning of man and women in 2001, = 40.90 - 21.10 = 19.8 Difference between earning of man and women in 2002, = 48.60 - 27.0 = 21.6 Difference between earning of man and women in 2003, = 59.50 - 34.20 = 25.3 Difference between earning of man and women in 2004, = 66.90 - 40.60 = 26.3 pattern show, its increasing.

Required ratio in 2000, = $$\frac{35.80}{18.30}$$ = 1.95 Required ration in 2001, = $$\frac{40.9}{21.1}$$ = 1.93 Required ration in 2002, = $$\frac{48.60}{27.0}$$ = 1.8 Required ration in 2003, = $$\frac{59.5}{34.20}$$ = 1.73 Required ration in 2004, = $$\frac{66.90}{40.60}$$ = 1.64 It is highest in 2000

Assume the total production of the first year as 10000, second year becomes 12000, third year 14400 and fourth year 17280. Then, 0.2 × 17280 - 0.2 × 12000 = 1056 But this difference is given as 2640. Hence, the value of production will be; 25000, 30000, 36000, and 43200 respectively for the 4 years. ∴ Revenue = 125 × 0.4 × 30000 = 1500000

Assume the total production of the first year as 10000, second year becomes 12000, third year 14400 and fourth year 17280. Then, 0.2 × 17280 - 0.2 × 12000 = 1056 But this difference is given as 2640. Hence, the value of production will be; 25000, 30000, 36000, and 43200 respectively for the 4 years. Ratio of the number of units C produced in four year, = 0.2 × 25000 : 0.2 × 30000 : 0.2 × 36000 : 0.2 × 43200 = 250 : 300 : 360 : 432 = 175 : 150 : 180 : 216

Assume the total production of the first year as 10000, second year becomes 12000, third year 14400 and fourth year 17280. Then, 0.2 × 17280 - 0.2 × 12000 = 1056 But this difference is given as 2640. Hence, the value of production will be; 25000, 30000, 36000, and 43200 respectively for the 4 years. Ratio of revenue generated in 2002, = 2 × 3 : 4 × 5 : 2 × 7 : 2 × 8 = 6 : 20 : 14 : 16 = 3 : 10 : 7 : 8

Assume the total production of the first year as 10000, second year becomes 12000, third year 14400 and fourth year 17280. Then, 0.2 × 17280 - 0.2 × 12000 = 1056 But this difference is given as 2640. Hence, the value of production will be; 25000, 30000, 36000, and 43200 respectively for the 4 years. Growth rate of production = $$\frac{\left(144-87.5\right)\times100}{87.5}$$ = 65% (approx)

From graph it is clear that southern region showed the highest growth in number of households in all the income categories for the period.

Region wise breakup is not available. hence, question cannot be answered.

The percentage increase in total number of households for the northern region for upper-middle income category is 200%.

It is clear from the table average income is Rs. 75,000.

$$\eqalign{ & \text{Number of students in:} \cr & \text{A → } \frac{12}{100}\times6000 = 720 \cr & \text{B → } \frac{9}{100}\times6000 = 540 \cr & \text{C → } \frac{26}{100}\times6000 = 1560 \cr & \text{D → } \frac{18}{100}\times6000 = 1080 \cr & \text{E → } \frac{29}{100}\times6000 = 1740 \cr & \text{F → } \frac{6}{100}\times6000 = 360 \cr} $$ Number of boys in: A → 500 B → 400 C → 900 D → 600 E → 1200 F → 100 Number of girls in: A → 720 - 500 = 220 B → 540 - 400 = 140 C → 1560 - 900 = 660 D → 1080 - 600 = 480 E → 1740 - 1200 = 540 F → 360 - 100 = 260 Required ratio = 900 : 140 : 1740 = 90 : 14 : 174 = 45 : 7 : 87

$$\eqalign{ & \text{Number of students in:} \cr & \text{A → } \frac{12}{100}\times6000 = 720 \cr & \text{B → } \frac{9}{100}\times6000 = 540 \cr & \text{C → } \frac{26}{100}\times6000 = 1560 \cr & \text{D → } \frac{18}{100}\times6000 = 1080 \cr & \text{E → } \frac{29}{100}\times6000 = 1740 \cr & \text{F → } \frac{6}{100}\times6000 = 360 \cr} $$ Number of boys in: A → 500 B → 400 C → 900 D → 600 E → 1200 F → 100 Number of girls in: A → 720 - 500 = 220 B → 540 - 400 = 140 C → 1560 - 900 = 660 D → 1080 - 600 = 480 E → 1740 - 1200 = 540 F → 360 - 100 = 260 Let (number of girls in A) be $$x$$% of number of students in B. Then, $$\eqalign{ & 220 = \frac{x}{100}\times540 \cr & \Rightarrow x = \frac{220\times100}{540} \cr & \Rightarrow x = \frac{1100}{27} \cr & \Rightarrow x = 40.7 \cr & \Rightarrow x \approx 40\% \cr} $$

$$\eqalign{ & \text{Number of students in:} \cr & \text{A → } \frac{12}{100}\times6000 = 720 \cr & \text{B → } \frac{9}{100}\times6000 = 540 \cr & \text{C → } \frac{26}{100}\times6000 = 1560 \cr & \text{D → } \frac{18}{100}\times6000 = 1080 \cr & \text{E → } \frac{29}{100}\times6000 = 1740 \cr & \text{F → } \frac{6}{100}\times6000 = 360 \cr} $$ Number of boys in: A → 500 B → 400 C → 900 D → 600 E → 1200 F → 100 Number of girls in: A → 720 - 500 = 220 B → 540 - 400 = 140 C → 1560 - 900 = 660 D → 1080 - 600 = 480 E → 1740 - 1200 = 540 F → 360 - 100 = 260 Required sum = 660 + 540 + 600 = 1800

$$\eqalign{ & \text{Number of students in:} \cr & \text{A → } \frac{12}{100}\times6000 = 720 \cr & \text{B → } \frac{9}{100}\times6000 = 540 \cr & \text{C → } \frac{26}{100}\times6000 = 1560 \cr & \text{D → } \frac{18}{100}\times6000 = 1080 \cr & \text{E → } \frac{29}{100}\times6000 = 1740 \cr & \text{F → } \frac{6}{100}\times6000 = 360 \cr} $$ Number of boys in: A → 500 B → 400 C → 900 D → 600 E → 1200 F → 100 Number of girls in: A → 720 - 500 = 220 B → 540 - 400 = 140 C → 1560 - 900 = 660 D → 1080 - 600 = 480 E → 1740 - 1200 = 540 F → 360 - 100 = 260 (Number of boys in E) - (Number of students in F) = 1200 - 360 = 840

$$\eqalign{ & \text{Number of students in:} \cr & \text{A → } \frac{12}{100}\times6000 = 720 \cr & \text{B → } \frac{9}{100}\times6000 = 540 \cr & \text{C → } \frac{26}{100}\times6000 = 1560 \cr & \text{D → } \frac{18}{100}\times6000 = 1080 \cr & \text{E → } \frac{29}{100}\times6000 = 1740 \cr & \text{F → } \frac{6}{100}\times6000 = 360 \cr} $$ Number of boys in: A → 500 B → 400 C → 900 D → 600 E → 1200 F → 100 Number of girls in: A → 720 - 500 = 220 B → 540 - 400 = 140 C → 1560 - 900 = 660 D → 1080 - 600 = 480 E → 1740 - 1200 = 540 F → 360 - 100 = 260 Number of girls in School E = 540 = Total number of students in B

(Number of students appeared from C) : (Number of students qualified from C) $$\eqalign{ & = \frac{3250}{2250} \cr & = \frac{13}{9} \cr & = 13:9 \cr} $$

Average number of students appeared from all the schools = $$\frac{1}{5}$$(3000 + 2250 + 3250 + 2500 + 3000) = $$\frac{14000}{5}$$ = 2800 Average number of students qualified from all the schools = $$\frac{1}{5}$$(1750 + 1250 + 2250 + 2000 + 2000) = $$\frac{9250}{5}$$ = 1850 ∴ Required difference = 2800 - 1850 = 950

(Number of students qualified from A) : (Number of students qualified from B) $$\eqalign{ & = \frac{1750}{1250} \cr & = \frac{7}{5} \cr & = 7:5 \cr} $$

Average number of students qualified from C and D $$\eqalign{ & = \frac{1}{2}\left(2250+2000\right) \cr & = \frac{4250}{2} \cr & = 2125 \cr} $$ Average number of students appeared from C and D $$\eqalign{ & = \frac{1}{2}\left(3250+2500\right) \cr & = \frac{5750}{2} \cr & = 2875 \cr & \therefore \text{ Required %} \cr & = \left(\frac{2125}{2875}\times100\right)\% \cr & = \left(\frac{85}{115}\times100\right)\% \cr & = \frac{1700}{23}\% \cr & = 73.91\% \cr} $$

Required % $$= \left(\frac{2500}{3000+2250+3250+2500+3000}\times100\right)\%$$ $$\eqalign{ & = \left(\frac{2500}{14000}\times100\right)\% \cr & = \frac{125}{7}\% \cr & \approx 18\% \cr} $$

Average export during the given period = Rs. 8.73 crores. If was above average in 2004, 2005, 2006 and 2007. i.e. in 4 years

Percentage increase in export from 2006 to 2007. $$\eqalign{ & = \left(\frac{11.4-9.5}{9.5}\times100\right)\% \cr & = \left(\frac{1.9}{9.5}\times100\right)\% \cr & = \left(\frac{19}{95}\times100\right)\% \cr & = 20\% \cr} $$

Average export of pearls over the period $$ = \left\{\frac{1}{7}\left(5.2+6.5+7.8+9.9+10.8+9.5+11.4\right)\right\}\text{crores}$$ $$\eqalign{ & = \frac{61.1}{7}\text{ crores} \cr & = 8.728 \text{ crores} \cr & = \text{Rs. }8.73 \text{ crores} \cr} $$

$$\eqalign{ & \text{Average export in 2003 and 2004} \cr & = \frac{7.8+9.9}{2} \text{ crores} \cr & = \frac{17.7}{2} \text{ crores} \cr & = \text{Rs. }8.85 \text{ crores} \cr & \approx \text{Rs. }9 \text{ crores} \cr} $$

$$\eqalign{ & \text{Increase % in various years} \cr & 2002 → \left\{\frac{\left(6.5-5.2\right)}{5.2}\times100\right\}\% \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \left(\frac{1.3}{5.2}\times100\right)\% \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \left(\frac{13}{52}\times100\right)\% \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 25\% \cr \cr & 2003 → \left\{\frac{\left(7.8-6.5\right)}{6.5}\times100\right\}\% \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \left(\frac{1.3}{6.5}\times100\right)\% \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \left(\frac{13}{65}\times100\right)\% \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 20\% \cr \cr & 2004 → \left\{\frac{\left(9.9-7.8\right)}{7.8}\times100\right\}\% \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \left(\frac{2.1}{7.8}\times100\right)\% \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \left(\frac{21}{78}\times100\right)\% \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 26.9\% \cr \cr & 2005 → \left\{\frac{\left(10.8-9.9\right)}{9.9}\times100\right\}\% \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \left(\frac{0.9}{9.9}\times100\right)\% \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \left(\frac{9}{99}\times100\right)\% \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{100}{11}\% \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 9.090\% \cr \cr & 2007 → \left\{\frac{\left(11.4-9.5\right)}{9.5}\times100\right\}\% \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \left(\frac{1.9}{9.5}\times100\right)\% \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \left(\frac{19}{95}\times100\right)\% \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 20\% \cr & \cr } $$ Clearly, the maximum increase was in 2004

$$\eqalign{ & \text{Speeds of vehicles on Day 1:} \cr & \text{A → } \frac{832}{16}\text{ km/hr}=\text{52 km/hr} \cr & \text{B → } \frac{516}{12}\text{ km/hr}=\text{43 km/hr} \cr & \text{C → } \frac{693}{11}\text{ km/hr}=\text{63 km/hr} \cr & \text{D → } \frac{552}{12}\text{ km/hr}=\text{46 km/hr} \cr & \text{E → } \frac{935}{17}\text{ km/hr}=\text{55 km/hr} \cr & \text{F → } \frac{703}{19}\text{ km/hr}=\text{37 km/hr} \cr & \cr & \text{Speed of vehicles on Day 2:} \cr & \text{A → } \frac{864}{16}\text{ km/hr}=\text{54 km/hr} \cr & \text{B → } \frac{774}{18}\text{ km/hr}=\text{43 km/hr} \cr & \text{C → } \frac{810}{18}\text{ km/hr}=\text{45 km/hr} \cr & \text{D → } \frac{765}{15}\text{ km/hr}=\text{51 km/hr} \cr & \text{E → } \frac{546}{14}\text{ km/hr}=\text{39 km/hr} \cr & \text{F → } \frac{636}{12}\text{ km/hr}=\text{53 km/hr} \cr} $$ Clearly B travelled at the same speed on both the days.

$$\eqalign{ & \text{Speeds of vehicles on Day 1:} \cr & \text{A → } \frac{832}{16}\text{ km/hr}=\text{52 km/hr} \cr & \text{B → } \frac{516}{12}\text{ km/hr}=\text{43 km/hr} \cr & \text{C → } \frac{693}{11}\text{ km/hr}=\text{63 km/hr} \cr & \text{D → } \frac{552}{12}\text{ km/hr}=\text{46 km/hr} \cr & \text{E → } \frac{935}{17}\text{ km/hr}=\text{55 km/hr} \cr & \text{F → } \frac{703}{19}\text{ km/hr}=\text{37 km/hr} \cr & \cr & \text{Speed of vehicles on Day 2:} \cr & \text{A → } \frac{864}{16}\text{ km/hr}=\text{54 km/hr} \cr & \text{B → } \frac{774}{18}\text{ km/hr}=\text{43 km/hr} \cr & \text{C → } \frac{810}{18}\text{ km/hr}=\text{45 km/hr} \cr & \text{D → } \frac{765}{15}\text{ km/hr}=\text{51 km/hr} \cr & \text{E → } \frac{546}{14}\text{ km/hr}=\text{39 km/hr} \cr & \text{F → } \frac{636}{12}\text{ km/hr}=\text{53 km/hr} \cr} $$ $$\eqalign{ & \text{Speed of C on Day 2} \cr & = 45 \text{ km/hr} \cr & = \left(45\times\frac{5}{18}\right) \text{m/sec} \cr & = \frac{25}{2} \text{ m/s} \cr & = 12.5 \text{ m/s} \cr} $$