Tables MCQs for UPSC Prelims

Per capita income = (income in crores / population in lakhs) x 100 rupees: 1992: 1010/20 x 100 = Rs. 5,050; 1993: 1111/21 x 100 = about Rs. 5,290; 1994: 1225/22 x 100 = about Rs. 5,568; 1995: 1345/23 x 100 = about Rs. 5,848. All four exceed Rs. 5,000. Population growth is 5%, 4.76% and 4.55%, so not 5% or more every year. Income growth is 10%, 10.26% and 9.8%, so not 10% or more every year. Per capita income peaks in 1995, not 1994.

First use the English column to find the ratio of girls to boys. Let g = number of girls, b = number of boys. The overall English average is 8.8, between girls' 9 and boys' 8: (9g + 8b) / (g + b) = 8.8 9g + 8b = 8.8g + 8.8b 0.2g = 0.8b, so g = 4b (four times as many girls as boys). Now apply this ratio to Hindi (girls 8, boys 7): x = (8g + 7b) / (g + b) = (8(4b) + 7b) / (4b + b) = (32b + 7b) / 5b = 39b / 5b = 39/5 = 7.8. So x = 7.8.

The total payment for each car is 6 per km hired plus the diesel cost, which equals 40 multiplied by (distance divided by mileage). For each car, set total payment = 6d + 40(d / mileage) and solve for distance d, then divide by hours to get average speed. Car A: 6d + 40(d/8) = 6d + 5d = 11d = 2120, so d = 192.7 km; speed = 192.7 / 20 = about 9.6 km per hour. Car B: 6d + 40(d/10) = 6d + 4d = 10d = 1950, so d = 195 km; speed = 195 / 25 = 7.8 km per hour. Car C: 6d + 40(d/9) = 10.44d = 2064, so d = about 197.7 km; speed = 197.7 / 24 = about 8.2 km per hour. Car D: 6d + 40(d/11) = 9.636d = 1812, so d = about 188 km; speed = 188 / 22 = about 8.5 km per hour. Car A has the highest average speed, so the answer is Car A.

The fastest run scorer has the highest runs per ball over both innings. A: (61 + 14) / (99 + 76) = 75/175 = 0.43. B: (5 + 50) / (12 + 85) = 55/97 = 0.57. C: (15 + 20) / (75 + 50) = 35/125 = 0.28. D: (13 + 12) / (55 + 50) = 25/105 = 0.24. B has the best scoring rate. A scored the most runs (75) but used far more balls, so A loses out once the rate, not the total, is compared.

Natural growth rate = birth rate - death rate. Computing for the offered options: 1911-1921: 48.1 - 35.5 = 12.6; 1941-1951: 39.9 - 27.4 = 12.5; 1961-1971: 43.1 - 18.9 = 24.2; 1971-1981: 37.1 - 14.8 = 22.3. The maximum, 24.2, occurs in 1961-1971.

Aggregate percentage = total marks obtained divided by total maximum marks, times 100. Student A: obtained = 60 + 70 + 50 + 30 = 210; maximum = 100 + 100 + 100 + 50 = 350; percentage = 210/350 = 60%. Student B: obtained = 80 + 70 + 60 + 15 = 225; maximum = 150 + 100 + 100 + 25 = 375; percentage = 225/375 = 60%. Both aggregate to 60%, so the difference is Zero.

Row totals: P = 40+30+20 = 90; Q = 50+20+30 = 100; R = 30+40+10 = 80; S = 60+10+40 = 110. S has the highest total (110), so S is correct. Q (100) is the nearest competitor but still lower; P and R are clearly less.

Wednesday = 36. A 25 percent increase gives 36 x 1.25 = 45, which is Thursday's value, so Thursday is correct. Monday (48) is 36 x 1.333, i.e. 33.3 percent more. Tuesday (60) is 36 x 1.667, i.e. 66.7 percent more. Wednesday is the base day itself (0 percent), so it cannot be 25 percent more than itself.