How to use this set

Ten ques­tions in all. The first five are straight­for­ward elim­i­na­tion; the rest are word prob­lems where you have to form the equa­tions first. For those, choose whichever method, sub­sti­tu­tion or elim­i­na­tion, suits the pair best, and say which one you used. In the exam either is accepted unless the ques­tion names one.

Sec­tion A, Equa­tions

  1. x+2y=8x + 2y = 8, x−2y=0x - 2y = 0.
  2. 3x+5y=263x + 5y = 26, 3x−2y=53x - 2y = 5.
  3. 2x+7y=312x + 7y = 31, 5x−3y=15x - 3y = 1.
  4. 4x−5y=−74x - 5y = -7, 6x+10y=286x + 10y = 28.
  5. 0.3x+0.5y=1.90.3x + 0.5y = 1.9, 0.5x−0.2y=0.40.5x - 0.2y = 0.4.

Sec­tion B, Word prob­lems

  1. Three pen­cils and four pens cost ₹ 5252. Four pen­cils and three pens cost ₹ 4646. Find the cost of one pen­cil and one pen.
  2. The sum of the dig­its of a two-digit num­ber is 1212. The num­ber exceeds the num­ber obtained by revers­ing its dig­its by 3636. Find the num­ber.
  3. A boat goes 2424 km upstream and 3636 km down­stream in 66 hours. It goes 3636 km upstream and 2424 km down­stream in 612\displaystyle 6\tfrac{1}{2} hours. Find the speed of the boat in still water and the speed of the stream.
  4. The ratio of incomes of two peo­ple is 8:58 : 5 and the ratio of their expen­di­tures is 5:35 : 3. Each saves ₹ 30003000 a month. Find their incomes.
  5. Places A and B are 120120 km apart. Two cars start from A and B at the same time. Trav­el­ling in the same direc­tion they meet in 66 hours; trav­el­ling towards each other they meet in 112\displaystyle 1\tfrac{1}{2} hours. Find their speeds.
Two sketches of places A and B 120 km apart. Top: both cars drive the same way and meet beyond B after 6 hours. Bottom: they drive towards each other and meet in between after 1.5 hours.
Ques­tion 10: the two cars in each sit­u­a­tion

Answers

Show answers
  1. Add: 2x=82x = 8, x=4x = 4; then y=2y = 2.
  2. Sub­tract: 7y=217y = 21, y=3y = 3; then 3x+15=263x + 15 = 26, x=113\displaystyle x = \frac{11}{3}. Check: 11+15=2611 + 15 = 26 ✓, 11−6=511 - 6 = 5 ✓.
  3. Mul­ti­ply first by 55, sec­ond by 22: 10x+35y=15510x + 35y = 155, 10x−6y=210x - 6y = 2. Sub­tract: 41y=15341y = 153 (check: 155−2=153155 - 2 = 153 and 35+6=4135 + 6 = 41), so y=15341\displaystyle y = \frac{153}{41}, which does not sim­plify. Then from 5x−3y=15x - 3y = 1: 5x=1+45941=50041\displaystyle 5x = 1 + \frac{459}{41} = \frac{500}{41}, x=10041\displaystyle x = \frac{100}{41}. Ver­ify in the first: 20041+107141=127141=31\displaystyle \frac{200}{41} + \frac{1071}{41} = \frac{1271}{41} = 31 ✓. The solu­tion is frac­tional, and that's per­fectly cor­rect as it stands; don't go hunt­ing for a mis­take that isn't there.
  4. Dou­ble the first: 8x−10y=−148x - 10y = -14. Add to the sec­ond: 14x=1414x = 14, x=1x = 1; then 4−5y=−74 - 5y = -7, y=115\displaystyle y = \frac{11}{5}. Check: 6+22=286 + 22 = 28 ✓.
  5. Mul­ti­ply both by 1010: 3x+5y=193x + 5y = 19, 5x−2y=45x - 2y = 4. Mul­ti­ply first by 22, sec­ond by 55: 6x+10y=386x + 10y = 38, 25x−10y=2025x - 10y = 20. Add: 31x=5831x = 58, x=5831\displaystyle x = \frac{58}{31}. Then y=19−3x5=19⋅31−174155=415155=8331\displaystyle y = \frac{19 - 3x}{5} = \frac{19 \cdot 31 - 174}{155} = \frac{415}{155} = \frac{83}{31}. Ver­ify: 0.5⋅5831−0.2⋅8331=29−16.631=12.431=0.4\displaystyle 0.5 \cdot \frac{58}{31} - 0.2 \cdot \frac{83}{31} = \frac{29 - 16.6}{31} = \frac{12.4}{31} = 0.4 ✓.
  6. Pen­cil ₹ xx, pen ₹ yy: 3x+4y=523x + 4y = 52, 4x+3y=464x + 3y = 46. Add: 7x+7y=987x + 7y = 98, so x+y=14x + y = 14. Sub­tract the sec­ond from the first: −x+y=6-x + y = 6, so y−x=6y - x = 6. Hence y=10y = 10, x=4x = 4. Pen­cil ₹ 44, pen ₹ 1010. Check: 12+40=5212 + 40 = 52 ✓, 16+30=4616 + 30 = 46 ✓.
  7. Dig­its xx (tens), yy (units): x+y=12x + y = 12 and (10x+y)−(10y+x)=36(10x + y) - (10y + x) = 36, i.e. 9x−9y=369x - 9y = 36, so x−y=4x - y = 4. Add: 2x=162x = 16, x=8x = 8, y=4y = 4. The num­ber is 8484. Check: 84−48=3684 - 48 = 36 ✓.
  8. Boat uu km/h, stream vv km/h. Upstream speed u−vu - v, down­stream u+vu + v. Let p=1u−v\displaystyle p = \frac{1}{u - v}, q=1u+v\displaystyle q = \frac{1}{u + v}: 24p+36q=624p + 36q = 6 and 36p+24q=132\displaystyle 36p + 24q = \frac{13}{2}. Mul­ti­ply first by 33, sec­ond by 22: 72p+108q=1872p + 108q = 18, 72p+48q=1372p + 48q = 13. Sub­tract: 60q=560q = 5, q=112\displaystyle q = \frac{1}{12}; then 24p+3=624p + 3 = 6, p=18\displaystyle p = \frac{1}{8}. So u−v=8u - v = 8 and u+v=12u + v = 12: u=10u = 10, v=2v = 2. Boat 1010 km/h, stream 22 km/h.
  9. Incomes 8k8k and 5k5k; expen­di­tures 5m5m and 3m3m. Sav­ings: 8k−5m=30008k - 5m = 3000 and 5k−3m=30005k - 3m = 3000. Mul­ti­ply first by 33, sec­ond by 55: 24k−15m=900024k - 15m = 9000, 25k−15m=1500025k - 15m = 15000. Sub­tract: k=6000k = 6000. Incomes ₹ 4800048000 and ₹ 3000030000. Check: m=9000m = 9000, and 48000−45000=3000=30000−2700048000 - 45000 = 3000 = 30000 - 27000 ✓.
  10. Speeds xx and yy km/h with x>yx > y. Same direc­tion: 6x−6y=1206x - 6y = 120, so x−y=20x - y = 20. Towards each other: 32(x+y)=120\displaystyle \frac{3}{2}(x + y) = 120, so x+y=80x + y = 80. Add: 2x=1002x = 100, x=50x = 50, y=30y = 30. Speeds 5050 km/h and 3030 km/h.