Set B Solutions

B1. Answer: (d) 29.16 L P=40, x=4, n=3. Milk left =40(1-4/40)^3=40(0.9)3=40(0.729)=29.16 L.

B2. Answer: (c) 80 L Let milk =5x, water =3x. After adding 16 L water: (5x)/(3x+16)=(5)/(4) ⇒ 20x=15x+80 ⇒ 5x=80 ⇒ x=16. Milk =5x=80 L. (Check: water =3(16)+16=64; 80:64=5:4 ✓.)

B3. Answer: (b) 23:22 Take each glass =15 units (LCM of 3,5,5=15). Glass 1 (1:2, sum 3): acid =5, water =10. Glass 2 (2:3, sum 5): acid =6, water =9. Glass 3 (4:1, sum 5): acid =12, water =3. Total acid =5+6+12=23; total water =10+9+3=22. Ratio =23:22.

B4. Answer: (b) 1/5 Initial: water =(3)/(8), syrup =(5)/(8) (fraction of whole). Let fraction x of the mixture be drawn off and replaced with water. Syrup remaining (as fraction of total) =(5)/(8)(1-x). We need syrup = water =(1)/(2) each: (5)/(8)(1-x)=(1)/(2) ⇒ 1-x=(4)/(5) ⇒ x=(1)/(5)

B5. Answer: (c) 540 Alligation on average salary: ratio Officers : Non-officers = (12000-10000):(30000-12000) = 2000:18000=1:9. Officers =60 ⇒ 1 part =60 ⇒ 9 parts (non-officers) =540.

B6. Answer: (c) 600 kg Ratio (sold at 8%) : (sold at 18%) = (18-14):(14-8)=4:6=2:3. Total =1000 kg ⇒ sold at 18% =1000×(3)/(5)=600 kg.

B7. Answer: (b) 4:1 Mean CP =(20)/(1.25)=16. Milk (Rs 20), Water (Rs 0), Mean =16. Ratio =(0-16):(16-20)→ magnitudes 16:4=4:1.

B8. Answer: (b) 2/3 Fraction replaced =(40-26)/(40-19)=(14)/(21)=(2)/(3). (General rule: if a solution of strength a% has fraction f replaced by a solution of strength b% resulting in strength c%, then f=(a-c)/(a-b), which comes directly from a(1-f)+bf=c.)

B9. Answer: (c) 60 g Sugar =120 g; other content =180 g (unchanged). Let sugar added =y: (120+y)/(300+y)=0.5 ⇒ 240+2y=300+y ⇒ y=60 g.

B10. Answer: (a) 7:9 Spirit fractions: A =(5)/(7), B =(7)/(13), target =(8)/(13). Ratio A:B = ((8)/(13)-(7)/(13)):((5)/(7)-(8)/(13)) = (1)/(13):(65-56)/(91)=(1)/(13):(9)/(91). Multiply both by 91: 7:9. (Verification: Let A =7 units (spirit 5, water 2, since 7 units = (7)/(7)× scale where sum of A’s ratio is 7); let B =9 units, so with sum 13, m=(9)/(13): spirit =7m=(63)/(13), water =6m=(54)/(13). Total spirit =5+(63)/(13)=(128)/(13), total water =2+(54)/(13)=(80)/(13), ratio =128:80=8:5 ✓.)

B11. Answer: (c) 70 L Let total =7x (milk =4x, water =3x). Removing 21 L of mixture removes it in the same 4:3 ratio: milk removed =21×(4)/(7)=12; water removed =21×(3)/(7)=9. Remaining milk =4x-12; remaining water =3x-9+21=3x+12 (21 L water added back). (4x-12)/(3x+12)=(2)/(3) ⇒ 3(4x-12)=2(3x+12) ⇒ 12x-36=6x+24 ⇒ 6x=60 ⇒ x=10 Initial total =7x=70 L.

B12. Answer: (b) 21 L Let total =12x (A =7x, B =5x). Removing 9 L mixture removes A and B in ratio 7:5: A removed =9×(7)/(12)=5.25; B removed =9×(5)/(12)=3.75. Remaining A =7x-5.25; remaining B =5x-3.75+9=5x+5.25 (9 L of B added back). (7x-5.25)/(5x+5.25)=(7)/(9) ⇒ 9(7x-5.25)=7(5x+5.25) ⇒ 63x-47.25=35x+36.75 ⇒ 28x=84 ⇒ x=3 Initial A =7x=21 L.

B13. Answer: (b) 30 kg Mean CP =(20)/(1.25)=16. C_1=15 (90 kg), C_2=19 (unknown y kg). Ratio (Rs15 : Rs19) =(19-16):(16-15)=3:1. So 90:y=3:1 ⇒ y=30 kg.

B14. Answer: (a) 48 km/hr This is an equal-distance average speed case, not equal-time — so alligation/arithmetic mean does NOT apply here; the correct tool is the harmonic mean: s=(2 s_1 s_2)/(s_1+s_2)=(2(40)(60))/(40+60)=(4800)/(100)=48 km/hr (A student who wrongly applies simple alligation/average here would get 50 km/hr, choice (b) — a deliberately placed distractor testing this exact confusion, as flagged in Section 4.)

B15. Answer: (c) 25 L Let total =8x (P =5x, Q =3x). Removing 16 L of mixture removes P and Q in ratio 5:3: P removed =16×(5)/(8)=10; Q removed =16×(3)/(8)=6. Remaining P =5x-10; remaining Q =3x-6+16=3x+10 (16 L of Q added back). (5x-10)/(3x+10)=(3)/(5) ⇒ 5(5x-10)=3(3x+10) ⇒ 25x-50=9x+30 ⇒ 16x=80 ⇒ x=5 Initial P =5x=25 L.

B16. Answer: (b) 6:6:13 Let quantities of Rs 40 and Rs 60 varieties each be x, and Rs 100 variety be y. (40x+60x+100y)/(2x+y)=76 ⇒ 100x+100y=152x+76y ⇒ 24y=52x ⇒ y=(13)/(6)x Taking x=6 gives y=13, so the ratio is 6:6:13. Check: quantities 6, 6, 13; cost =40(6)+60(6)+100(13)=240+360+1300=1900; total quantity =25; mean =(1900)/(25)=76 ✓.

B17. Answer: (b) 4 Removing one-fourth each time gives a multiplying factor of (3)/(4) per operation. 256((3)/(4))^n = 81 ⇒ ((3)/(4))^n = (81)/(256) = ((3)/(4))^4 ⇒ n=4 (Since 3^4=81 and 4^4=256 exactly.)

B18. Answer: (c) 32 L 128(1-(x)/(128))^3 = 54 ⇒ (1-(x)/(128))^3 = (54)/(128)=(27)/(64)=((3)/(4))^3 ⇒ 1-(x)/(128)=(3)/(4) ⇒ (x)/(128)=(1)/(4) ⇒ x=32 Check: 128(3/4)^3 = 128×(27)/(64)=54 ✓.

B19. Answer: (c) 24 L Milk =40×(3)/(4)=30 L, water =10 L. Let y L of mixture (ratio (3)/(4):(1)/(4)) be removed and replaced with pure milk. Milk after =30-(3y)/(4)+y=30+(y)/(4); water after =10-(y)/(4). (30+(y)/(4))/(10-(y)/(4))=(9)/(1) ⇒ 30+(y)/(4)=90-(9y)/(4) ⇒ (10y)/(4)=60 ⇒ y=24 Check: removing 24 L takes out 18 L milk and 6 L water, leaving milk =12, water =4; adding back 24 L pure milk gives milk =36, water =4, ratio =9:1 ✓.

B20. Answer: (d) 56 years Total age of 40 students =40×15=600. New total with teacher included =41×16=656. Teacher’s age =656-600=56 years. (Alligation check: ratio students : teacher =(T-16):(16-15)=40:1 ⇒ T-16=40 ⇒ T=56 ✓.)

B21. Answer: (a) 1:2 Mean CP =(20)/(1.25)=16. Ratio (Rs 12 : Rs 18) =(18-16):(16-12)=2:4=1:2.

B22. Answer: (b) 14:11 Vessel A (15 L, 2:1): milk =10 L, water =5 L. Combined mixture (40 L, 3:2): total milk =40×(3)/(5)=24 L, total water =16 L. Vessel B’s milk =24-10=14 L; vessel B’s water =16-5=11 L (sums correctly to B’s 25 L). Ratio =14:11.

B23. Answer: (a) 30 km/hr For n equal-distance legs, average speed is the harmonic mean: s = (n)/((1)/(s_1)+(1)/(s_2)+(1)/(s_3)). (1)/(30)+(1)/(60)+(1)/(20) = (2)/(60)+(1)/(60)+(3)/(60)=(6)/(60)=(1)/(10) s = (3)/(1/10) = 30 km/hr

B24. Answer: (a) 8.33% Gain =9(1)/(11)%=(100)/(11)%. Mean CP =(1)/(1+(100/11)/(100))=(1)/(1+(1)/(11))=(1)/((12)/(11))=(11)/(12). Milk:Water =(0-(11)/(12)):((11)/(12)-1) → magnitudes (11)/(12):(1)/(12)=11:1. Water’s share =(1)/(12)≈ 8.33%.

B25. Answer: (a) 41.47% Replacing 20% of the solution each time gives a multiplying factor of 0.8 per operation (this applies directly to the milk percentage since total volume is constant): 81% × (0.8)^3 = 81×0.512 = 41.472% ≈ 41.47%

B26. Answer: (c) 20% Mean CP =(20(15)+30(25))/(50)=(300+750)/(50)=(1050)/(50)=21 per kg. Profit per kg =25.20-21=4.20. Profit% = (4.20)/(21)×100 = 20%

B27. Answer: (b) 30 L Let total =5x (acid =3x, water =2x). Removing 10 L of mixture removes acid and water in ratio 3:2: acid removed =10×(3)/(5)=6; water removed =10×(2)/(5)=4. Remaining acid =3x-6; remaining water =2x-4+10=2x+6 (10 L water added back). (3x-6)/(2x+6)=(2)/(3) ⇒ 3(3x-6)=2(2x+6) ⇒ 9x-18=4x+12 ⇒ 5x=30 ⇒ x=6 Initial total =5x=30 L.

B28. Answer: (c) 28(8)/(9)% Container 1 (20 L, 3:1): milk =15, water =5. Container 2 (30 L, 4:1): milk =24, water =6. Container 3 (40 L, 5:3, sum 8): milk =40×(5)/(8)=25, water =40×(3)/(8)=15. Total milk =15+24+25=64; total water =5+6+15=26; total volume =90 (check: 64+26=90 ✓). Water% = (26)/(90)×100 = (260)/(9)% = 28(8)/(9)%

B29. Answer: (b) 2:1 Copper fraction in A =(5)/(8)=0.625; in B =(1)/(4)=0.25; target =0.5. Ratio A:B =(0.5-0.25):(0.625-0.5)=0.25:0.125=2:1. Check: mixing 2 parts A and 1 part B: copper =2(0.625)+1(0.25)=1.25+0.25=1.5; total =3; fraction =0.5 ✓.

B30. Answer: (a) 3:2, Rs 66.70 Ratio (Rs 50 : Rs 70) =(70-58):(58-50)=12:8=3:2. SP per kg for 15% profit =58×1.15=66.70.

End of Chapter 8. In the next chapter, we build on these ratio and average techniques to tackle Time, Speed and Distance, where alligation reappears in average-speed problems exactly as flagged in Trick 7 and Question B14 above — revisit this chapter’s Section 2.4/2.7 methods if that connection feels unfamiliar.