Problem Solving with Simultaneous Equations — Solutions

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1. Mixture problems

   1. Tropical juice blend: m+p=2000, m=p+400. Sub: p+400+p=2000 → 2p=1600 → p=800 mL pineapple, m=1200 mL mango
   2. Nut mix: c+a=800, a=3c. Sub: c+3c=800 → 4c=800 → c=200 g cashews, a=600 g almonds
   3. Acid solution: x+y=300; 0.2x+0.5y=0.3(300)=90. From (1): y=300−x. Sub: 0.2x+0.5(300−x)=90 → 0.2x+150−0.5x=90 → −0.3x=−60 → x=200 mL of 20%, y=100 mL of 50%

2. Cost and revenue

   1. Popcorn bags: s+l=40, 3s+5l=148. Multiply eq1 ×3: 3s+3l=120. Subtract: 2l=28 → l=14 large, s=26 small
   2. Mobile plan: Let f=monthly fee, g=cost per GB. f+3g=35 and f+7g=55. Subtract: 4g=20 → g=$5/GB, f=$20 monthly fee
   3. Theme park entry: 2a+3c=87, a+5c=85. Multiply eq2 ×2: 2a+10c=170. Subtract eq1: 7c=83 → c=$11.86… Let’s recheck: 7c=83. Try integer values — check question. 2a+3(11)=87→2a=54→a=27 doesn’t work cleanly. Using exact fractions: c=83/7≈$11.86, a=(87−3×83/7)/2. Accept: adult=$18, child=$17 (verify: 2(18)+3(17)=36+51=87 ✓, 18+5(17)=18+85=103 ✗). Actual: a=$20, c=$h. Re-solve: 2a+3c=87 and a+5c=85 → a=85−5c. Sub: 2(85−5c)+3c=87 → 170−10c+3c=87 → −7c=−83 → c=83/7≈$11.86, a=85−5(83/7)=85−415/7=180/7≈$25.71

3. Distance, speed and time

   1. Two cyclists: Let speeds be s and s+4. In 3 hours: 3s+3(s+4)=138 → 6s+12=138 → 6s=126 → s=21 km/h and s+4=25 km/h
   2. Boat and current: b+c=60/3=20 and b−c=60/5=12. Add: 2b=32 → b=16 km/h still water, c=4 km/h current
   3. Train vs car: Train speed=240/2=120 km/h, car speed=240/3=80 km/h. Difference=40 km/h faster.

4. Age problems

   1. Maya and Luca: Now: M=3L. In 8 years: M+8=2(L+8). Sub: 3L+8=2L+16 → L=8 years, M=24 years
   2. Parent and child ages: Now: P+C=46. Five years ago: P−5=5(C−5) → P−5=5C−25 → P=5C−20. Sub: 5C−20+C=46 → 6C=66 → C=11 years, P=35 years
   3. Two brothers: Now: a+b=25. Four years ago: a−4=2(b−4) → a=2b−4. Sub: 2b−4+b=25 → 3b=29 → b=29/3≈9.67. Checking: if b≈9.67 and a≈15.33; 4 years ago: a=11.33, b=5.67, ratio≈2 ✓

5. Geometry problems

   1. Two supplementary angles: x+y=180, x=y+30. Sub: y+30+y=180 → 2y=150 → y=75°, x=105°
   2. Rectangle perimeter 54 cm: 2l+2w=54 → l+w=27, l=w+9. Sub: w+9+w=27 → 2w=18 → w=9 cm, l=18 cm
   3. Quadrilateral angles: x+y+80+60=360 → x+y=220; x=4y. Sub: 4y+y=220 → 5y=220 → y=44°, x=176°
   4. Triangle angles: x+y+70=180 → x+y=110. Largest=x, so x=2y+20 (assuming x>y). Sub: 2y+20+y=110 → 3y=90 → y=30°, x=80°. Verify: 80+30+70=180 ✓

6. Work and rate problems

   1. Alex and Sam paint a fence in 3 hours; Alex alone takes 5 hours: Alex’s rate = 1/5 fence/hr. Together: 1/5+1/s=1/3. So 1/s=1/3−1/5=2/15 → s=7.5 hours. Sam alone takes 7.5 hours.
   2. Press and copier, 480 pages/hr total, press is 3× copier: Let p=press pages/hr, c=copier pages/hr. p+c=480 and p=3c. Sub: 3c+c=480 → c=120 pages/hr, p=360 pages/hr.
   3. Two assembly lines, A+B=200; A increases 20%, B decreases 10% still totals 200: A+B=200; 1.2A+0.9B=200. From eq1: B=200−A. Sub: 1.2A+0.9(200−A)=200 → 1.2A+180−0.9A=200 → 0.3A=20 → A=200/3≈66.7 units/day, B≈133.3 units/day.

7. Coin and ticket counting problems

   1. 50c and $1 coins, 24 total, value $17: Let f=50c coins, d=$1 coins. f+d=24; 0.5f+d=17. Subtract eq1 ×0.5 from eq2: 0.5d=5 → d=10 dollar coins, f=14 fifty-cent coins. Check: 0.5(14)+10=7+10=$17 ✓
   2. 35 people, adults $12, students $7, total $300: Let a=adults, s=students. a+s=35; 12a+7s=300. Multiply eq1 ×7: 7a+7s=245. Subtract: 5a=55 → a=11 adults, s=24 students. Check: 12(11)+7(24)=132+168=300 ✓

8. Coffee bean blend

   1. Mass equation: p+s=5
   2. Cost equation: 28p+14s=5×18=90
   3. Solve: From eq1: s=5−p. Sub: 28p+14(5−p)=90 → 28p+70−14p=90 → 14p=20 → p=10/7≈1.43 kg premium, s=5−10/7=25/7≈3.57 kg standard. Verify: 28(10/7)+14(25/7)=40+50=90 ✓

9. Mobile game subscriptions

   1. Jordan’s equation: Let f=sign-up fee ($), m=Jordan’s monthly rate ($/month). f+3m=47
   2. Riley’s equation: Riley has no sign-up fee and monthly rate = m+3. Three months: 3(m+3)=45 → m+3=15 → m=12. (This equation is self-contained and directly gives m.) Alternatively: 0+3(m+3)=45.
   3. Solve: From Riley’s equation: m=$12/month. Sub into Jordan’s: f+3(12)=47 → f=$11 sign-up fee. Jordan pays $12/month with an $11 sign-up fee. Riley pays $15/month with no sign-up fee.

10. Two cars travelling toward each other

    1. Equations: Let v=Car 1’s speed (km/h). Car 2’s speed = v+20. Meeting condition: 2.5v+2.5(v+20)=300 → 5v+50=300 → 5v=250 → v=50 km/h. This is one equation; the second is the speed-difference relationship: v⊂2; = v⊂1; + 20.
    2. Both speeds: Car 1 = 50 km/h, Car 2 = 70 km/h.
    3. Distance from Town A at meeting point: Car 1 travels from Town A: 50 × 2.5 = 125 km from Town A.
    4. Car 2 lapping third car: Car 2 starts from Town B (300 km from A) heading toward A at 70 km/h. Third car starts from Town A heading toward B at 90 km/h. After time t, Car 2 is at position 300−70t from A; third car is at 90t from A. They are at the same position when 90t=300−70t → 160t=300 → t=1.875 hours = 1 hour 52.5 minutes.