Q1. Write the equation of direct variation when \(x\) varies directly as \(y\).

Ans: \(x = ky\) where \(k\) is constant of variation.

Q2. If cost of 12 notebooks is ₹240, what is cost of 1 notebook?

Ans: Cost per notebook \(= 240\div12 = ₹20.\)

Q3. Using the table, if cost per notebook is ₹20, cost of 3 notebooks = ?

Ans: \(3\times20=₹60.\)

Q4. If \(x\) varies inversely as \(y\), write the equation.

Ans: \(x = \dfrac{k}{y}\) or \(x\cdot y = k.\)

Q5. If \(x\propto y\) and when \(x=5\) then \(y=30\), find \(k\).

Ans: \(x=ky\Rightarrow 5=k\cdot30\Rightarrow k=\dfrac{1}{6}.\)

Q6. Write the direct variation equation for cost \(x\) (₹) and weight \(y\) (kg) if 5 kg costs ₹450.

Ans: \(k=450\div5=90\). So \(x=90y\).

Q7. If \(x\propto\sqrt{y}\) and \(x=24\) when \(y=16\), find \(k\) in \(x=k\sqrt{y}\).

Ans: \(\sqrt{16}=4\Rightarrow k=24/4=6.\)

Q8. If 4 labourers are paid ₹1000 in total, what is payment for 1 labourer (direct proportion)?

Ans: Per labourer = \(1000\div4=₹250.\)

Q9. If number of students in a row × number of rows = 240, what is rows when students/row = 40?

Ans: rows = \(240\div40=6.\)

Q10. If \(f \propto \dfrac{1}{d^2}\) and \(f=18\) when \(d=5\), find constant \(k\) in \(f d^2 = k.\)

Ans: \(k=18\times5^2=18\times25=450.\)

Q11. If speed \(s\) and time \(t\) are inversely proportional and \(s=48\) km/h when \(t=6\) h, find constant \(k\) in \(s\cdot t=k\).

Ans: \(k=48\times6=288.\)

Q12. If \(x = ky\) and \(k=\tfrac{1}{6}\), and \(y=12\), find \(x\).

Ans: \(x=\tfrac{1}{6}\times12=2.\)

Q13. If \(x\propto y\) and ratio \(x:y = 1:20\), what is \(k\) in \(x=ky\)?

Ans: \(x/y=1/20=k\Rightarrow k=\tfrac{1}{20}.\)

Q14. If 15 women finish a job in 8 days, how many women required to finish it in 6 days? (inverse proportion)

Ans: \(d\cdot n=k=15\times8=120\Rightarrow n=120/6=20.\)

Q15. If cost ₹90 per kg, cost of 100 kg = ?

Ans: \(90\times100=₹9000.\)

Q16. If \(x\propto y^2\) and \(x=18\) when \(y=3\), find \(k\) (so \(x=ky^2\)).

Ans: \(k=18/(3^2)=18/9=2.\)

Q17. If boxes need 27 when box holds 24 apples each, how many apples total? (direct)

Ans: Total apples = \(27\times24=648.\)

Q18. If \(x\cdot y=k\) and \(k=120\), find \(y\) when \(x=12.\)

Ans: \(y=120/12=10.\)

Q19. If \(f d^2 =450\) and \(f=50\), find \(d\).

Ans: \(50 d^2=450\Rightarrow d^2=9\Rightarrow d=±3\) (take positive if physical distance). So \(d=3.\)

Q20. If speed increases from 48 to 72 km/h, how does time change for same journey?

Ans: Time scales inversely: \(t_1/t_2 = s_2/s_1\Rightarrow t_2 = t_1\times (s_1/s_2) = 6\times (48/72)=6\times2/3=4\) hours.

Q1. Table: cost per dozen is ₹240. Fill costs for 3, 9, 24, 50 notebooks.

Ans: cost per notebook = 240/12 = ₹20. So:
3 → ₹60, 9 → ₹180, 24 → ₹480, 50 → ₹1000.

Q2. Write the statement: "Circumference \(c\) of a circle is directly proportional to radius \(r\)" using variation symbol.

Ans: \(c \propto r\) or \(c = k r\) (here \(k=2\pi\)).

Q3. Write "petrol consumption \(l\) and distance \(d\)" as variation statement.

Ans: Consumption \(l\propto d\) or \(l=kd\) (direct variation).

Q4. If \(m\propto n\) and \(m=154\) when \(n=7\), find \(m\) when \(n=14\).

Ans: \(k=m/n=154/7=22\), so when \(n=14\), \(m=22\times14=308.\)

Q5. Complete table: if \(n\propto m\) and given (m,n): (3,5), (6, ?), (12,20) etc. (use constant)

Ans: From (3,5) => \(k=5/3\). For m=6, n=6*k = 6*(5/3)=10. For m=12, n=20 given consistent. (Other blanks filled similarly.)

Q6. If \(y\propto\sqrt{x}\) and \(y=24\) when \(x=16\), find equation.

Ans: \(\sqrt{16}=4\Rightarrow k=24/4=6.\) So \(y=6\sqrt{x}.\)

Q7. Remuneration for harvesting: 4 labourers get ₹1000; find for 17 labourers (direct proportion in number).

Ans: cost per labourer = \(1000/4=₹250.\) For 17 → \(250\times17=₹4250.\)

Q8. If \(a\cdot b = 120\) and \(a=6\), find \(b\).

Ans: \(b=120/6=20.\)

Q9. If \(f\propto 1/d^2\), and \(f=18\) at \(d=5\), find \(f\) at \(d=10\).

Ans: \(k=450.\) Then \(f = 450/10^2 = 450/100 = 4.5.\)

Q10. Boxes & apples: 24 apples/box need 27 boxes. If 36 apples/box, how many boxes?

Ans: total apples = 27×24=648. Boxes when 36 per box = 648/36 = 18 boxes.

Q11. Workers-days table: Fill missing when workers 30→days6, 20→9, 10→12, ?→36. Find missing workers for 36 days.

Ans: Work constant k = workers×days = 30×6=180. For 36 days, workers = 180/36 = 5 workers.

Q12. If \(p\propto 1/q\) and \(p=15\) when \(q=4\), write equation and find \(p\) when \(q=2\).

Ans: \(p q = k = 15×4=60.\) So when \(q=2\), \(p=60/2=30.\)

Q13. If \(z\propto 1/w\) and \(z=2.5\) when \(w=24\), find equation and \(z\) when \(w=12\).

Ans: \(k=2.5×24=60.\) So \(z=60/w\). For \(w=12\), \(z=60/12=5.\)

Q14. If \(s\propto 1/(2t)\) and \(s=4\) when \(t=5\), find equation (interpretation: \(s\cdot(2t)=k\)).

Ans: \(s(2t)=k\Rightarrow k=4×(2×5)=40.\) So \(s=40/(2t)=20/t.\)

Q15. If \(x\propto 1/y\) and \(x=40\) when \(y=16\), find \(y\) when \(x=10\).

Ans: \(x y=k=40×16=640.\) When \(x=10\), \(y=640/10=64.\)

Q16. If \(x\) varies inversely as \(y\), with \(x=15,y=10\). Find \(y\) when \(x=20\).

Ans: \(k=15×10=150.\) When \(x=20\), \(y=150/20=7.5.\)

Q17. If 15 workers →48 hours, how many workers for 30 hours? (inverse)

Ans: k = 15×48 = 720. Workers = 720/30 = 24 workers.

Q18. A machine fills 120 half-litre bags in 3 minutes. Time to fill 1800 such bags?

Ans: Rate = 120 bags / 3 min = 40 bags/min. For 1800 bags → 1800/40 = 45 minutes.

Q19. Car speed 60 km/h takes 8 hours. To cover same distance in \(7\frac{1}{2}\) hours, what increase in speed required?

Ans: distance = 60×8=480 km. Required speed = 480 / 7.5 = 64 km/h. Increase = 4 km/h.

Q20. If speed and time inverse; speed increases by 50%, time becomes ?

Ans: speed ×1.5 ⇒ time divided by 1.5 ⇒ new time = old time × (2/3).

Q1. A shop sells notebooks at ₹240 per dozen. Find cost of 50 notebooks and show steps.

Ans: cost per notebook = \(240/12=20\). Cost of 50 = \(50\times20=₹1000.\)

Q2. If \(x\propto\sqrt{y}\) and \(x=24\) at \(y=16\), find \(x\) at \(y=81\).

Ans: \(k=6\). So \(x=6\sqrt{81}=6\times9=54.\)

Q3. Cost of groundnuts is directly proportional to weight. If 5 kg costs ₹450, find cost of 1 quintal (100 kg).

Ans: \(k=450/5=90\) ₹/kg. For 100 kg: \(90\times100=₹9000.\)

Q4. If \(a\propto\frac{1}{b}\) and \(a=6\) when \(b=20\), fill table for \(a\) when \(b=12,15,4\).

Ans: \(k=6\times20=120\).
When \(b=12\Rightarrow a=120/12=10.\)
When \(b=15\Rightarrow a=120/15=8.\)
When \(b=4\Rightarrow a=120/4=30.\)

Q5. If \(f\cdot d^2 =450\) and \(d=10\), find \(f\). Then find \(d\) when \(f=50\).

Ans: For \(d=10\): \(f=450/100=4.5.\) For \(f=50\): \(d^2=450/50=9\Rightarrow d=3\) (positive).

Q6. Time-work: 15 women finish in 8 days. How many women for 6 days? Explain formula used.

Ans: \(d\propto1/n\Rightarrow dn=k=15\times8=120.\) For \(d=6\): \(n=120/6=20\) women.

Q7. Speed-time: 48 km/h → 6 h. Find time at 72 km/h and show calculation.

Ans: \(s t = k=48\times6=288.\) For \(s=72\): \(t=288/72=4\) hours.

Q8. A box holds apples. If 24 apples/box → 27 boxes, find boxes when 12 apples/box and when 56 apples/box.

Ans: Total apples = \(27\times24=648.\) Boxes at 12 each: \(648/12=54.\) At 56 each: \(648/56=11.571...\) Not integer → need 12 boxes (if whole boxes) or 11 with a partial box. Exact = \(648/56=11\frac{32}{56}=11\frac{4}{7}.\)

Q9. If \(x\propto y\) and \(x:y=1:20\), find y when x=154.

Ans: \(x/y=1/20\Rightarrow y=20x=20\times154=3080.\) (But earlier example used reciprocal; ensure interpretation — common sense check.)

Q10. Given workers & days table, check consistency: 30 workers→6d, 20→9d, 10→12d. Show constant of work.

Ans: k = workers×days; 30×6=180, 20×9=180, 10×12=120? Wait 10×12 = 120 → inconsistent with others. If consistent, all should be 180. (Book’s original had consistent k=240 earlier; verify with given numbers.)

Q11. If \(x\) varies as \(y^2\) and \(x=18\) when \(y=3\), find \(x\) when \(y=6\).

Ans: \(k=18/9=2.\) So when \(y=6\): \(x=2\times36=72.\)

Q12. A car at 60 km/h takes 8 h. Find new speed to finish in 7½ h and percentage increase.

Ans: distance=480 km. New speed = 480/7.5=64 km/h. Increase = 4 km/h → percentage \(= 4/60\times100\%=6.666\%.\)

Q13. If \(p\propto 1/q\) and \(p=15\) at \(q=4\), find \(p\) at \(q=1.6\).

Ans: \(k=60.\) At \(q=1.6\): \(p=60/1.6=37.5.\)

Q14. A machine fills 120 half-litre bags in 3 minutes; find rate per minute and time for 1800 bags (full steps).

Ans: rate = 120/3 = 40 bags/min. Time = 1800/40 = 45 min.

Q15. Explain why circumference ∝ radius with formula and constant value.

Ans: Circumference \(c = 2\pi r\Rightarrow c\propto r\) and constant \(k=2\pi.\)

Q16. If \(x\propto y\) and \(x=1\) when \(y=20\), find equation and value x when y=160.

Ans: \(k=1/20\). So \(x=(1/20)\times160=8.\)

Q17. If the intensity of light \(I\propto\frac{1}{d^2}\), and intensity is 100 units at 2 m, find intensity at 5 m.

Ans: \(k=I d^2 = 100\times4=400.\) At 5 m: \(I=400/25=16.\)

Q18. If \(n\) apples cost \(c\) rupees and cost is directly proportional to number, derive cost formula and compute cost if 1 apple costs ₹4 and we buy 160 apples.

Ans: \(x=4n\). For 160 → \(4\times160=₹640.\)

Q19. A job finished by 15 workers in 48 hrs. How many workers for 36 hours?

Ans: k = 15×48 = 720. Workers = 720/36 = 20 workers.

Q20. If \(x\propto y\) and when \(x=154, y=7\), find x when y=14 (complete reasoning).

Ans: \(k=x/y=154/7=22.\) Then \(x=22\times14=308.\)

PS7.1 Q1. Write statements using variation symbol:

(1) Circumference \(c\) and radius \(r\): \(c\propto r\) or \(c=kr\) (here \(k=2\pi\)).
(2) Petrol consumption \(l\) and distance \(d\): \(l\propto d\) or \(l=kd\).

PS7.1 Q2. Complete table with apples where cost per apple = ₹4 (given \(1\to4\)).

Ans: cost = \(4\times\) number of apples. So table entries:
1 → 4, 3 → 12, 8 → 32, 12 → 48, 32 → 128, 56 → 224, 160 → 640.

PS7.1 Q3. If \(m\propto n\) and when \(m=154, n=7\), find \(m\) when \(n=14\).

Ans: \(k=154/7=22\Rightarrow m=22\times14=308.\)

PS7.1 Q4. Complete the table where \(n\propto m\). (Sample shown earlier)

Ans: From example (3,5) ⇒ \(k=5/3\); compute remaining entries: if m=6 ⇒ n=10; etc.

PS7.1 Q5. If \(y \propto \sqrt{x}\) and \(x=16,y=24\), find equation.

Ans: \(\sqrt{16}=4\Rightarrow k=6 \Rightarrow y=6\sqrt{x}.\)

PS7.1 Q6. Remuneration: 4 labourers → ₹1000. Remuneration for 17 labourers?

Ans: per labourer = 250 ⇒ 17×250 = ₹4250.

PS7.2 Q1. Complete table: Number of workers vs days to finish job. Given: workers 30→days6; 20→9; 10→12; ?→36.

Ans: Work constant k = workers×days = 30×6 = 180. For 36 days: workers = 180/36 = 5 workers. (Check: 20×9=180; 10×12=120 — if book expects 120 constant then initial numbers differ; here we use 30×6=180 consistent with 20×9.)

PS7.2 Q2. Find constant and equation for:

(1) \(p\propto 1/q\) with p=15, q=4 ⇒ k=60 ⇒ \(pq=60\).
(2) \(z\propto 1/w\) with z=2.5, w=24 ⇒ k=60 ⇒ \(zw=60\).
(3) \(s\propto 1/(2t)\) with s=4 at t=5 ⇒ \(s(2t)=k\Rightarrow k=40\) ⇒ \(s=40/(2t)=20/t\).
(4) \(x\propto 1/y\) with x=15,y=9 ⇒ k=135 ⇒ \(xy=135\).

PS7.2 Q3. Boxes/apples: if 24 apples per box → 27 boxes; if 36 per box how many?

Ans: total apples = 27×24=648. Boxes at 36 each: 648/36 = 18 boxes.

PS7.2 Q4. Statements to write as variation:

(1) Wavelength \(λ\) and frequency \(f\): \(λ\propto 1/f\).
(2) Intensity \(I\) and distance \(d\): \(I\propto 1/d^2\).

PS7.2 Q5. If \(x\propto 1/y\) and x=40 when y=16. Find y when x=10.

Ans: k=40×16=640 ⇒ y = 640/10 = 64.

PS7.2 Q6. If x varies inversely as y: x=15 when y=10, find y when x=20.

Ans: k=150 ⇒ y=150/20 = 7.5.

PS7.3 Q1. Which are inverse variation? (1) workers vs time (2) number of pipes vs time (3) petrol vs cost (4) area vs radius

Ans: (1) Yes (inverse). (2) Yes (inverse). (3) No — cost ∝ petrol (direct). (4) No — area ∝ r^2 (direct with square).

PS7.3 Q2. If 15 workers build wall in 48 hours, how many workers for 30 hours?

Ans: k=15×48=720 ⇒ workers = 720/30 = 24 workers.

PS7.3 Q3. Machine fills 120 half-litre bags in 3 mins. Find time for 1800 bags.

Ans: rate = 40 bags/min ⇒ 1800/40 = 45 minutes.

PS7.3 Q4. A car at 60 km/h takes 8 h. What increase in speed for same distance if time = \(7\frac{1}{2}\) h?

Ans: distance = 480 km. Required speed = 480 / 7.5 = 64 km/h. Increase = 64 - 60 = 4 km/h.