Chapter 2: Numbers in Day-to-Day Life

Method — Compound Proportionality / Total Person-Hours:

Total work is constant and given by the product of Men ($M$), Days ($D$), and Daily Hours ($H$):

$$W = M_1 \times D_1 \times H_1 = M_2 \times D_2 \times H_2$$

Given data:

• $M_1 = 80$, $D_1 = 18\text{ days}$, $H_1 = 7\text{ hours/day}$
• $M_2 = ?$, $D_2 = 30\text{ days}$, $H_2 = 6\text{ hours/day}$

Equating the work expressions:

$$80 \times 18 \times 7 = M_2 \times 30 \times 6$$

$$10080 = 180 M_2 \implies M_2 = \frac{10080}{180} = \mathbf{56}$$

Answer: (b) 56 labourers.

Method — Unitary Rate of Filling:

• Portion filled $= \frac{3}{5}$ in $1\text{ minute} = 60\text{ seconds}$.
• Remaining portion to fill $= 1 - \frac{3}{5} = \frac{2}{5}$.

Time taken to fill $\frac{3}{5}$ portion $= 60\text{ seconds}$.

$$\text{Time to fill } 1 \text{ complete cistern} = \frac{60}{3/5} = 60 \times \frac{5}{3} = 100\text{ seconds}$$

$$\text{Time to fill remaining } \frac{2}{5} = \frac{2}{5} \times 100 = \mathbf{40\text{ seconds}}$$

Answer: (b) 40 sec.

Method — Remaining Man-Days Balance:

• Initial provisions: $95 \text{ men} \times 200 \text{ days}$.
• After 5 days, 95 men have consumed food for 5 days.
• Remaining provisions could feed $95\text{ men}$ for $200 - 5 = 195\text{ days}$.
• Total remaining man-days $= 95 \times 195$.
• Number of men remaining after 30 leave $= 95 - 30 = 65\text{ men}$.

Let the remaining food last for $D$ days:

$$65 \times D = 95 \times 195$$

$$D = \frac{95 \times 195}{65} = \frac{95 \times 3}{1} = \mathbf{285\text{ days}}$$

Answer: (b) 285 days.

Method — Inverse Variation & Algebraic Equation:

Let the original number of men employed be $x$.

• Planned work $= 9 \times x = 9x\text{ man-days}$.
• Actual men present $= x - 6$, who took 15 days.
• Actual work $= 15 \times (x - 6)\text{ man-days}$.

Equating both expressions for total work:

$$9x = 15(x - 6)$$

$$9x = 15x - 90 \implies 6x = 90 \implies x = \mathbf{15}$$

Answer: (b) 15 men.

Method — Individual Work Fractions:

• B's 1-day work $= \frac{1}{15}$.
• Work done by B in 10 days $= 10 \times \frac{1}{15} = \frac{2}{3}$.
• Remaining work $= 1 - \frac{2}{3} = \frac{1}{3}$.
• A's 1-day work $= \frac{1}{18}$.

Number of days taken by A to complete $\frac{1}{3}$ of the work:

$$\text{Time} = \frac{\text{Remaining Work}}{\text{A's Rate}} = \frac{1/3}{1/18} = \frac{18}{3} = \mathbf{6\text{ days}}$$

Answer: (c) 6 days.

Method — Group Work Fraction & Remainder:

• Work done by B and C in 1 day $= \frac{1}{9} + \frac{1}{12} = \frac{4 + 3}{36} = \frac{7}{36}$.
• Work done by B and C in 3 days $= 3 \times \frac{7}{36} = \frac{7}{12}$.
• Remaining work $= 1 - \frac{7}{12} = \frac{5}{12}$.
• A's 1-day work $= \frac{1}{24}$.

Time taken by A alone to finish the remaining $\frac{5}{12}$ work:

$$\text{Time} = \frac{5/12}{1/24} = \frac{5}{12} \times 24 = \mathbf{10\text{ days}}$$

Answer: (c) 10 days.

Method — Speed-Distance Proportionality:

Let actual speed be $v\text{ km/hr}$.

• Distance covered at actual speed in 30 hours $= 30v\text{ km}$.
• Reduced speed $= v - \frac{1}{15}v = \frac{14}{15}v\text{ km/hr}$.
• Distance covered at reduced speed in 30 hours $= 30 \times \left(\frac{14}{15}v ight) = 28v\text{ km}$.

Given that the difference in distance is $10\text{ km}$:

$$30v - 28v = 10 \implies 2v = 10 \implies v = \mathbf{5\text{ km/hr}}$$

Answer: (b) 5 km/hr.

Method — Distance Equating Equation:

Let the uniform speed be $s\text{ km/hr}$. The distance $D$ is constant:

$$D = 8s$$

At increased speed $(s + 4)\text{ km/hr}$ and time $7.5\text{ hours}$:

$$D = (s + 4) \times 7.5 = 7.5s + 30$$

Equating the two expressions:

$$8s = 7.5s + 30 \implies 0.5s = 30 \implies s = 60\text{ km/hr}$$

$$\text{Total Distance } D = 8 \times 60 = \mathbf{480\text{ km}}$$

Answer: (b) 480 km.

Method — Relative Speed in Same Direction:

• Speed of policeman $= 11\text{ kmph}$.
• Speed of thief $= 10\text{ kmph}$.
• Relative speed of policeman with respect to thief $= 11 - 10 = 1\text{ kmph}$.

Distance gained by the policeman in $6\text{ minutes} = \frac{6}{60}\text{ hours} = 0.1\text{ hours}$:

$$\text{Distance Gained} = 1\text{ km/hr} \times 0.1\text{ hr} = 0.1\text{ km} = 100\text{ metres}$$

Initial gap between them was $200\text{ metres}$.

$$\text{Remaining distance after 6 minutes} = 200\text{ m} - 100\text{ m} = \mathbf{100\text{ metres}}$$

Answer: (a) 100 m.

Method — Relative Speed & Train Length Invariant:

Convert persons' speeds to $\text{m/s}$:

• Person 1: $4.5 \times \frac{5}{18} = 1.25\text{ m/s}$.
• Person 2: $5.4 \times \frac{5}{18} = 1.5\text{ m/s}$.

Let train speed be $u\text{ m/s}$ and train length be $L\text{ metres}$.

$$L = (u - 1.25) \times 8.4 = 8.4u - 10.5$$

$$L = (u - 1.5) \times 8.5 = 8.5u - 12.75$$

Equating the two lengths:

$$8.4u - 10.5 = 8.5u - 12.75 \implies 0.1u = 2.25 \implies u = 22.5\text{ m/s}$$

Convert back to $\text{km/hr}$:

$$\text{Speed} = 22.5 \times \frac{18}{5} = 4.5 \times 18 = \mathbf{81\text{ km/hr}}$$

Answer: (d) 81 km/hr.

Method — Distance Synchronization:

• From 7:00 a.m. to 8:00 a.m. (1 hour), train A travels alone at $20\text{ km/hr}$.
• Distance covered by train A $= 20 \times 1 = 20\text{ km}$.
• Remaining distance between them at 8:00 a.m. $= 110 - 20 = 90\text{ km}$.

From 8:00 a.m. onwards, both trains move towards each other:

$$\text{Relative Speed} = 20 + 25 = 45\text{ km/hr}$$

$$\text{Time taken to meet} = \frac{90\text{ km}}{45\text{ km/hr}} = 2\text{ hours}$$

$$\text{Meeting Time} = 8\text{:00 a.m.} + 2\text{ hours} = \mathbf{10\text{:00 a.m.}}$$

Answer: (b) 10 a.m.

Method — Inverse Square Root Formula:

If two bodies start simultaneously from points A and B towards each other and after crossing each other take $T_1$ and $T_2$ hours to complete their journeys, the ratio of their speeds is:

$$\frac{S_1}{S_2} = \sqrt{\frac{T_2}{T_1}}$$

Here $T_1 = 9\text{ hours}$ and $T_2 = 16\text{ hours}$:

$$\frac{S_1}{S_2} = \sqrt{\frac{16}{9}} = \frac{4}{3} = \mathbf{4 : 3}$$

Answer: (b) 4 : 3.

Method — Relative Minute-Space Gain:

• At 5:00, the hour hand is at 25 minute-spaces ahead of the 12 position.
• For hands to coincide, the minute hand must gain 25 minute-spaces.
• Since minute hand gains 55 min-spaces in 60 minutes, it gains 1 min-space in $\frac{60}{55} = \frac{12}{11}\text{ minutes}$.

$$\text{Time} = 25 \times \frac{12}{11} = \frac{300}{11} = \mathbf{27\frac{3}{11}\text{ minutes past 5}}$$

Answer: (a) $27\frac{3}{11}$ min past 5.

Method — 15 Minute-Spaces Separation:

At 5 o'clock, the hour hand is at 25 minute-spaces. A right angle occurs when the hands are 15 minute-spaces apart:

• Case 1 (Before meeting): Minute hand is 15 spaces behind $\implies (25 - 15) \times \frac{12}{11} = 10 \times \frac{12}{11} = 10\frac{10}{11}\text{ min past 5}$.
• Case 2 (After meeting): Minute hand is 15 spaces ahead $\implies (25 + 15) \times \frac{12}{11} = 40 \times \frac{12}{11} = \frac{480}{11} = \mathbf{43\frac{7}{11}\text{ min. past 5}}$.

Looking at the options, Case 2 is represented by option (c).

Answer: (c) $43\frac{7}{11}\text{ min. past } 5$.

Method — 30 Minute-Spaces Separation ($180^\circ$):

• At 2:00, the hour hand is at 10 minute-spaces.
• To be directly opposite ($180^\circ$), the minute hand must be 30 spaces ahead of the hour hand.
• Required minute-space position $= 10 + 30 = 40$ minute-spaces.

$$\text{Time} = 40 \times \frac{12}{11} = \frac{480}{11} = \mathbf{43\frac{7}{11}\text{ minutes past 2}}$$

Answer: (a) $43\frac{7}{11}$ (min past 2).

Method — Relative Separation:

At 4 o'clock, the hour hand is at 20 minute-spaces.

• Case 1 (Minute hand 4 minutes behind): $$\text{Spaces to gain} = 20 - 4 = 16$$ $$\text{Time} = 16 \times \frac{12}{11} = \frac{192}{11} = \mathbf{17\frac{5}{11}\text{ min. past 4 O'clock}}$$
• Case 2 (Minute hand 4 minutes ahead): $$\text{Spaces to gain} = 20 + 4 = 24 \implies 24 \times \frac{12}{11} = \frac{288}{11} = 26\frac{2}{11}\text{ min. past 4 O'clock}$$

Case 1 matches option (a).

Answer: (a) $17\frac{5}{11}\text{ min. past 4 O' clock}$.

Method — Standard Angle Formula:

$$\theta = \left|30H - \frac{11}{2}M ight|$$

Substituting $H = 4$ and $M = 15$:

$$\theta = \left|30(4) - \frac{11}{2}(15) ight| = |120 - 82.5| = \mathbf{37.5^\circ}$$

Answer: (a) $37.5^\circ$.

Method — Odd Days Analysis:

• From 11 Jan 1997 to 11 Jan 1998: 1997 is an ordinary year $\implies 1\text{ odd day}$.
• From 11 Jan 1998 to 11 Jan 1999: 1998 is an ordinary year $\implies 1\text{ odd day}$.
• From 11 Jan 1999 to 11 Jan 2000: 1999 is an ordinary year $\implies 1\text{ odd day}$.
• Note: The leap day of year 2000 occurs in February, which is after 11 January.
• Total odd days from 11 Jan 1997 to 11 Jan 2000 $= 1 + 1 + 1 = 3\text{ odd days}$.

Therefore, 11 January 2000 was $\text{Sunday} + 3 = \mathbf{\text{Wednesday}}$.

Now, 7 January 2000 is 4 days before 11 January 2000:

$$\text{Day on 7 Jan 2000} = \text{Wednesday} - 4\text{ days} = \mathbf{\text{Saturday}}$$

Answer: (b) Saturday.

Method — Century & Year Odd Days Summation:

• 1600 years: $0\text{ odd days}$.
• 300 years (1601–1900): $1\text{ odd day}$.
• 98 completed years (1901–1998): $$98 \div 4 = 24\text{ leap years}, \quad 98 - 24 = 74\text{ ordinary years}$$ $$\text{Odd days} = 24 \times 2 + 74 \times 1 = 48 + 74 = 122 \equiv 122 \pmod 7 = 3\text{ odd days}$$
• Total odd days up to 31 Dec 1998 $= 0 + 1 + 3 = 4\text{ odd days}$.
• Days in 1999 up to 5 June: $$\text{Jan (31)} \to 3, \; \text{Feb (28)} \to 0, \; \text{Mar (31)} \to 3, \; \text{Apr (30)} \to 2, \; \text{May (31)} \to 3, \; \text{June} = 5$$ $$\text{Sum} = 3 + 0 + 3 + 2 + 3 + 5 = 16 \equiv 2\text{ odd days}$$
• Total Odd Days $= 4 + 2 = 6\text{ odd days}$.

Code 6 corresponds to Saturday (0=Sun, 1=Mon, ..., 6=Sat).

Answer: (a) Saturday.

Method — Relative Minute-Space Gain:

• At 3 o'clock, the hour hand is at 15 minute-spaces.
• For the hands to coincide, the minute hand must gain 15 minute-spaces.

$$\text{Time} = 15 \times \frac{12}{11} = \frac{180}{11} = \mathbf{16\frac{4}{11}\text{ minutes past 3}}$$

Answer: (b) $16\frac{4}{11}\text{ minutes past } 3$.

Method — Minute-Space Separation:

At 5 o'clock, the hour hand is at 25 minute-spaces.

• Case 1 (Minute hand 3 minutes behind): $$\text{Spaces to gain} = 25 - 3 = 22$$ $$\text{Time} = 22 \times \frac{12}{11} = 2 \times 12 = \mathbf{24\text{ minutes past 5}}$$
• Case 2 (Minute hand 3 minutes ahead): $$\text{Spaces to gain} = 25 + 3 = 28 \implies 28 \times \frac{12}{11} = \frac{336}{11} = 30\frac{6}{11}\text{ min past 5}$$

Case 1 exactly matches option (a).

Answer: (a) 24 minutes past 5.

Method — Angle Formula:

$$\theta = \left|30H - \frac{11}{2}M ight|$$

Substituting $H = 4$ and $M = 30$:

$$\theta = \left|30(4) - \frac{11}{2}(30) ight| = |120 - 165| = |-45^\circ| = \mathbf{45^\circ}$$

Answer: (c) $45^\circ$.

Method — Collinear Alignment Analysis:

Hands of a clock are in a straight line when they are either:

• Coinciding ($0^\circ$): Occurs 11 times in 12 hours $\implies 22\text{ times in 24 hours}$.
• Directly opposite ($180^\circ$): Occurs 11 times in 12 hours $\implies 22\text{ times in 24 hours}$.

$$\text{Total Collinear Occurrences in a day} = 22 + 22 = \mathbf{44\text{ times}}$$

Answer: (a) 44.

Step-by-Step Algebraic Derivation:

Let $A, B, C$ denote the daily work rate of each person:

$$A + B = \frac{1}{6} \quad \text{--- (1)}$$

$$A + C = \frac{1}{10} \quad \text{--- (2)}$$

$$A + B + C = \frac{1}{4} \quad \text{--- (3)}$$

Subtracting equation (1) from (3) gives C's individual rate:

$$C = (A + B + C) - (A + B) = \frac{1}{4} - \frac{1}{6} = \frac{3 - 2}{12} = \frac{1}{12}$$

Subtracting equation (2) from (3) gives B's individual rate:

$$B = (A + B + C) - (A + C) = \frac{1}{4} - \frac{1}{10} = \frac{5 - 2}{20} = \frac{3}{20}$$

Combined daily work rate of B and C together:

$$B + C = \frac{3}{20} + \frac{1}{12} = \frac{9 + 5}{60} = \frac{14}{60} = \frac{7}{30}$$

Number of days required by B and C together:

$$\text{Time} = \frac{1}{B + C} = \frac{30}{7} = \mathbf{4\frac{2}{7}\text{ days}}$$

Final Answer: B and C can complete the work together in $\mathbf{4\frac{2}{7}}$ days (approx $4.29$ days).

Step-by-Step Working:

• Total distance to be cleared $= L_1 + L_2 = 150\text{ m} + 200\text{ m} = 350\text{ metres}$.
• Since both trains run in the same direction, relative speed is the difference of their speeds: $$\text{Relative Speed} = 72 - 54 = 18\text{ km/hr}$$
• Convert relative speed to $\text{m/s}$: $$18 \times \frac{5}{18} = 5\text{ m/s}$$

Time taken for the faster train to pass the slower train:

$$\text{Time} = \frac{\text{Total Distance}}{\text{Relative Speed}} = \frac{350\text{ m}}{5\text{ m/s}} = \mathbf{70\text{ seconds}}$$

Final Answer: $\mathbf{70\text{ seconds}}$ ($1\text{ minute } 10\text{ seconds}$).

Step-by-Step Working:

• Total road length $= 210\text{ km}$.
• Moving towards each other (opposite directions): $$\text{Relative Speed} = 70 + 35 = 105\text{ km/hr}$$

Time taken until they meet:

$$\text{Time} = \frac{\text{Distance}}{\text{Relative Speed}} = \frac{210\text{ km}}{105\text{ km/hr}} = \mathbf{2\text{ hours}}$$

Distance covered by the car from its starting point in 2 hours:

$$\text{Distance}_{\text{car}} = \text{Speed}_{\text{car}} \times \text{Time} = 70 \times 2 = \mathbf{140\text{ km}}$$

Final Answer: They meet after $\mathbf{2\text{ hours}}$ at a distance of $\mathbf{140\text{ km}}$ from the car's starting point.

Step-by-Step Working:

Let the one-way distance between A and B be $d\text{ km}$.

• Time taken to walk from A to B $= \frac{d}{4}\text{ hours}$.
• Time taken to return from B to A $= \frac{d}{6}\text{ hours}$.
• Total round-trip time $= 5\text{ hours}$.

Setting up the time equation:

$$\frac{d}{4} + \frac{d}{6} = 5$$

$$\frac{3d + 2d}{12} = 5 \implies \frac{5d}{12} = 5 \implies d = \mathbf{12\text{ km}}$$

Final Answer: Distance AB is $\mathbf{12\text{ km}}$.

Step-by-Step Working:

Wages are distributed in direct proportion to the quantum of work completed by each individual.

• Work done by A in 8 days $= 8 \times \frac{1}{20} = \frac{8}{20} = \frac{2}{5}$.
• Work done by B in 8 days $= 8 \times \frac{1}{40} = \frac{8}{40} = \frac{1}{5}$.
• Combined work done by A and B $= \frac{2}{5} + \frac{1}{5} = \frac{3}{5}$.
• Work done by C $= 1 - \frac{3}{5} = \frac{2}{5}$.

Payment to C is proportional to his share of work:

$$\text{C's Payment} = \frac{2}{5} \times \text{Rs. } 450 = 2 \times 90 = \mathbf{\text{Rs. } 180}$$

Final Answer: C should be paid $\mathbf{\text{Rs. } 180}$.

Step-by-Step Working:

Let the total available fund reserve be $F$.

• Daily payroll expenditure for Team A $= \frac{F}{21}$ per day.
• Daily payroll expenditure for Team B $= \frac{F}{28}$ per day.

When both teams work together simultaneously, the total daily expenditure is:

$$\text{Combined Daily Burn Rate} = \frac{F}{21} + \frac{F}{28} = F \left(\frac{4 + 3}{84} ight) = \frac{7F}{84} = \frac{F}{12}$$

Number of days the available funds $F$ will sustain both teams:

$$\text{Number of Days} = \frac{F}{F/12} = \mathbf{12\text{ days}}$$

Final Answer: Both teams can work together for $\mathbf{12\text{ days}}$.