Set B Solutions

B1. Sum of first 6 numbers = 10.5×6 = 63. Sum of last 6 numbers = 11.4×6 = 68.4. Total sum of all 11 numbers = 10.9×11 = 119.9. Since the 6th number is counted in both “first six” and “last six” groups, it is counted twice when we add 63+68.4. So: 6th number = (Sum of first 6 + Sum of last 6) − Total sum = 63+68.4−119.9 = 131.4−119.9 = 11.5. Answer: (c) 11.5

B2. New person’s weight = Old weight + n×d = 65 + (8×2.5) = 65+20 = 85 kg. Answer: (c) 85 kg

B3. Teacher’s age = New average×(n+1) − Old average×n. New average = 15+1 = 16 (since average increases by exactly 1 year). Teacher’s age = 16×41 − 15×40 = 656 − 600 = 56 years. Answer: (d) 56 years

B4. Equal distance → Average speed = 2xy/(x+y) = 2×50×75/(50+75) = 7500/125 = 60 km/h. Answer: (c) 60 km/h

B5. For 5 consecutive odd numbers, the average equals the middle (3rd) term. So the middle term = 61. The numbers are 57, 59, 61, 63, 65. Largest = 65. Answer: (b) 65

B6. Combined average = (30×45 + 20×40)/(30+20) = (1350+800)/50 = 2150/50 = 43 kg. Answer: (c) 43 kg

B7. Let the third number = x. Then second number = 2x, first number = 4x. Sum = 4x+2x+x = 7x = 42×3 = 126. So x = 18. First number = 4×18 = 72. Answer: (c) 72

B8. Five years ago, average age of 2 people = 20, so sum of their ages then = 40. Presently (5 years later), each has aged by 5 years, so sum increases by 2×5=10, new sum = 50, present average = 50/2 = 25 years. (Equivalently: average simply increases by 5, from 20 to 25.) Answer: (d) 25 years

B9. Let number of passed candidates = p, failed = (60−p). Total marks = 60×40 = 2400. 50p + 30(60−p) = 2400 50p + 1800 − 30p = 2400 20p = 600 p = 30. Answer: (c) 30 Verification by alligation: ratio of passed : failed = (Overall avg − Fail avg) : (Pass avg − Overall avg) = (40−30):(50−40) = 10:10 = 1:1. So passed = failed = 60/2 = 30 ✓.

B10. Original total (with errors) = 45×100 = 4500. Wrongly added sum = 32+47 = 79. Correct sum should have been = 23+74 = 97. Corrected total = 4500 − 79 + 97 = 4518. Correct average = 4518/100 = 45.18. Answer: (c) 45.18

B11. For n equal distances at speeds x_1, x_2, …, x_n: Average speed = n / (1/x₁+1/x₂+…+1/xₙ). Here n=4, speeds 10,20,30,60. Sum of reciprocals = 1/10+1/20+1/30+1/60 = 6/60+3/60+2/60+1/60 = 12/60 = 1/5. Average speed = 4/(1/5) = 4×5 = 20 km/h. Answer: (b) 20 km/h

B12. First 20 multiples of 7: 7,14,…,140. Average = (First+Last)/2 = (7+140)/2 = 147/2 = 73.5. Answer: (d) 73.5

B13. For 9 consecutive natural numbers, the average equals the middle (5th) term = 20. The numbers range from 16 to 24 (20−4 to 20+4). Largest = 20+4 = 24. Answer: (c) 24

B14. Teacher’s height = New average×(n+1) − Old average×n. New average = 150+1 = 151. Teacher’s height = 151×36 − 150×35 = 5436 − 5250 = 186 cm. Answer: (d) 186 cm

B15. New average = (Old average×n + New score)/(n+1) = (36×16 + 70)/17 = (576+70)/17 = 646/17 = 38 runs. Answer: (c) 38 runs Alternative (increase-per-innings) method: Let the average increase by d after the 17th innings. Using the join-formula from Section 2.7 (x = A + (n+1)×d): 70 = 36 + 17d → 17d = 34 → d = 2. New average = 36+2 = 38 ✓, matching the direct method exactly.

B16. Total increase in the group’s sum = n×d = 15×2 = 30. Sum of the two students replaced = 40+50 = 90. Sum of the two new students = 90+30 = 120. Average weight of the two new students = 120/2 = 60 kg. Answer: (c) 60 kg

B17. Old sum = 18×32 = 576. Removing the person aged 45: 576−45 = 531 (17 people remain). New sum (with new average 31, 18 people) = 31×18 = 558. Age of the new person = 558−531 = 27 years. Answer: (c) 27 years

B18. Let the onward speed = x, so the return speed = 1.5x. Since distance is equal both ways: Average speed = 2×x×1.5x/(x+1.5x) = 3x²/2.5x = 1.2x. Given average speed = 42: 1.2x = 42 → x = 35 km/h. Verification: 2×35×52.5/(35+52.5) = 3675/87.5 = 42 ✓. Answer: (c) 35 km/h

B19. Since the three legs cover unequal distances, use Total Distance ÷ Total Time. Time for leg 1 = 100/50 = 2 h. Time for leg 2 = 180/60 = 3 h. Time for leg 3 = 150/75 = 2 h. Total distance = 100+180+150 = 430 km. Total time = 2+3+2 = 7 h. Average speed = 430/7 = 61.43 km/h ≈ 61.4 km/h. Answer: (b) 61.4 km/h

B20. Combined average = (20×60 + 30×70 + 50×80)/(20+30+50) = (1200+2100+4000)/100 = 7300/100 = 73. Answer: (c) 73

B21. Average of squares of first n natural numbers = (n+1)(2n+1)/6 = (17)(33)/6 = 561/6 = 93.5. Answer: (c) 93.5

B22. Average of cubes of first n natural numbers = n(n+1)²/4 = 12×(13)²/4 = 12×169/4 = 2028/4 = 507. Answer: (c) 507

B23. Let boys = 3x, girls = 2x (ratio 3:2). Combined average = (3x×65 + 2x×55)/(3x+2x) = (195x+110x)/5x = 305x/5x = 61 kg. Answer: (c) 61 kg

B24. Assume A = 2000. Deviations: 1997−2000=−3, 2005−2000=+5, 1990−2000=−10, 2012−2000=+12, 1998−2000=−2, 2008−2000=+8. Sum of deviations = −3+5−10+12−2+8 = 10. Average = 2000 + 10/6 = 2000+1.67 = 2001.67. Direct check: Sum = 1997+2005+1990+2012+1998+2008 = 12010; 12010/6 = 2001.67 ✓. Answer: (c) 2001.67

B25. Original (incorrect) total = 38×50 = 1900. The number 45 was wrongly read as 54, so the total was inflated by 54−45=9. Correct total = 1900 − 9 = 1891. Correct average = 1891/50 = 37.82. Answer: (b) 37.82

B26. Old sum = 30×34 = 1020. Removing the two who retired (aged 50 and 55): 1020 − (50+55) = 1020−105 = 915 (28 employees remain). Adding the three new employees (average 26): 915 + 3×26 = 915+78 = 993 (31 employees). New average = 993/31 = 32.03 years (approx). Answer: (b) 32.03 years

B27. For 11 consecutive multiples of 4 (an odd count), the average equals the middle (6th) term, so the middle term = 84. Since consecutive multiples of 4 differ by 4, the smallest term = 84 − 5×4 = 84−20 = 64. Verify: terms run from 64 to 104; average = (64+104)/2 = 84 ✓. Answer: (c) 64

B28. Let his average after 20 innings be A, so his total after 20 innings = 20A. After scoring 30 in the 21st innings, his new average is (A−2), so: 20A + 30 = 21(A−2) → 20A+30 = 21A−42 → 30+42 = 21A−20A → A = 72. New average after 21st innings = A−2 = 72−2 = 70 runs. Verify: Total after 21 innings = 20×72+30 = 1440+30 = 1470; 1470/21 = 70 ✓. Answer: (c) 70 runs

B29. Sum of first 5 numbers = 42×5 = 210. Sum of last 4 numbers = 50×4 = 200. Total sum of all 8 numbers = 45×8 = 360. Since the 5th number is common to both the “first 5” and “last 4” groups (positions 1–5 and 5–8 overlap at position 5), it gets counted twice when 210 and 200 are added. So: 5th number = (210+200) − 360 = 410−360 = 50. Answer: (c) 50

B30. Present sum of the family’s 5 members = 26×5 = 130. The member who married in 3 years ago is presently 33 years old; subtracting this member: sum of the remaining 4 original members, presently = 130−33 = 97. Three years ago, each of these 4 members was 3 years younger, so their total was less by 4×3 = 12: sum 3 years ago = 97−12 = 85. Average age of these 4 members, three years ago = 85/4 = 21.25 years. Answer: (c) 21.25 years