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 ($H$):

$$W = M_1 × D_1 × H_1 = M_2 × D_2 × H_2$$

Given data:

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

Equating the work expressions:

$$80 × 18 × 7 = M_2 × 30 × 6$$

$$10080 = 180 M_2 \implies M_2 = 10080180 = \mathbf{56}$$

Answer: (b) 56 labourers.

Method — Unitary Rate of Filling:

• Portion filled $= 35$ in $1 minute = 60 seconds$.
• Remaining portion to fill $= 1 - 35 = 25$.

Time taken to fill $35$ portion $= 60 seconds$.

$$Time to fill 1 complete cistern = 603/5 = 60 × 53 = 100 seconds$$

$$Time to fill remaining 25 = 25 × 100 = \mathbf{40 seconds}$$

Answer: (b) 40 sec.

Method — Remaining Man-Days Balance:

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

Let the remaining food last for $D$ days:

$$65 × D = 95 × 195$$

$$D = 95 × 19565 = 95 × 31 = \mathbf{285 days}$$

Answer: (b) 285 days.

Method — Inverse Variation & Algebraic Equation:

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

• Planned work $= 9 × x = 9x man-days$.
• Actual men present $= x - 6$, who took 15 days.
• Actual work $= 15 × (x - 6) man-days$.

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 $= 115$.
• Work done by B in 10 days $= 10 × 115 = 23$.
• Remaining work $= 1 - 23 = 13$.
• A's 1-day work $= 118$.

Number of days taken by A to complete $13$ of the work:

$$Time = Remaining WorkA's Rate = 1/31/18 = 183 = \mathbf{6 days}$$

Answer: (c) 6 days.

Method — Group Work Fraction & Remainder:

• Work done by B and C in 1 day $= 19 + 112 = 4 + 336 = 736$.
• Work done by B and C in 3 days $= 3 × 736 = 712$.
• Remaining work $= 1 - 712 = 512$.
• A's 1-day work $= 124$.

Time taken by A alone to finish the remaining $512$ work:

$$Time = 5/121/24 = 512 × 24 = \mathbf{10 days}$$

Answer: (c) 10 days.

Method — Speed-Distance Proportionality:

Let actual speed be $v km/hr$.

• Distance covered at actual speed in 30 hours $= 30v km$.
• Reduced speed $= v - 115v = 1415v km/hr$.
• Distance covered at reduced speed in 30 hours $= 30 × (1415v ight) = 28v km$.

Given that the difference in distance is $10 km$:

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

Answer: (b) 5 km/hr.

Method — Distance Equating Equation:

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

$$D = 8s$$

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

$$D = (s + 4) × 7.5 = 7.5s + 30$$

Equating the two expressions:

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

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

Answer: (b) 480 km.

Method — Relative Speed in Same Direction:

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

Distance gained by the policeman in $6 minutes = 660 hours = 0.1 hours$:

$$Distance Gained = 1 km/hr × 0.1 hr = 0.1 km = 100 metres$$

Initial gap between them was $200 metres$.

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

Answer: (a) 100 m.

Method — Relative Speed & Train Length Invariant:

Convert persons' speeds to $m/s$:

• Person 1: $4.5 × 518 = 1.25 m/s$.
• Person 2: $5.4 × 518 = 1.5 m/s$.

Let train speed be $u m/s$ and train length be $L metres$.

$$L = (u - 1.25) × 8.4 = 8.4u - 10.5$$

$$L = (u - 1.5) × 8.5 = 8.5u - 12.75$$

Equating the two lengths:

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

Convert back to $km/hr$:

$$Speed = 22.5 × 185 = 4.5 × 18 = \mathbf{81 km/hr}$$

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 $20 km/hr$.
• Distance covered by train A $= 20 × 1 = 20 km$.
• Remaining distance between them at 8:00 a.m. $= 110 - 20 = 90 km$.

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

$$Relative Speed = 20 + 25 = 45 km/hr$$

$$Time taken to meet = 90 km45 km/hr = 2 hours$$

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

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:

$$S_1S_2 = √T_2T_1$$

Here $T_1 = 9 hours$ and $T_2 = 16 hours$:

$$S_1S_2 = √169 = 43 = \mathbf{4 : 3}$$

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 $6055 = 1211 minutes$.

$$Time = 25 × 1211 = 30011 = \mathbf{27311 minutes past 5}$$

Answer: (a) $27311$ 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 $\implies (25 - 15) × 1211 = 10 × 1211 = 101011 min past 5$.
• Case 2 (After meeting): Minute hand is 15 spaces ahead $\implies (25 + 15) × 1211 = 40 × 1211 = 48011 = \mathbf{43711 min. past 5}$.

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

Answer: (c) $43711 min. past 5$.

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

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

$$Time = 40 × 1211 = 48011 = \mathbf{43711 minutes past 2}$$

Answer: (a) $43711$ (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): $$Spaces to gain = 20 - 4 = 16$$ $$Time = 16 × 1211 = 19211 = \mathbf{17511 min. past 4 O'clock}$$
• Case 2 (Minute hand 4 minutes ahead): $$Spaces to gain = 20 + 4 = 24 \implies 24 × 1211 = 28811 = 26211 min. past 4 O'clock$$

Case 1 matches option (a).

Answer: (a) $17511 min. past 4 O' clock$.

Method — Standard Angle Formula:

$$θ = |30H - 112M ight|$$

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

$$θ = |30(4) - 112(15) ight| = |120 - 82.5| = \mathbf{37.5^°}$$

Answer: (a) $37.5^°$.

Method — Odd Days Analysis:

• From 11 Jan 1997 to 11 Jan 1998: 1997 is an ordinary year $\implies 1 odd day$.
• From 11 Jan 1998 to 11 Jan 1999: 1998 is an ordinary year $\implies 1 odd day$.
• From 11 Jan 1999 to 11 Jan 2000: 1999 is an ordinary year $\implies 1 odd day$.
• 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 $= 1 + 1 + 1 = 3 odd days$.

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

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

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

Answer: (b) Saturday.

Method — Century & Year Odd Days Summation:

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

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.

$$Time = 15 × 1211 = 18011 = \mathbf{16411 minutes past 3}$$

Answer: (b) $16411 minutes past 3$.

Method — Minute-Space Separation:

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

• Case 1 (Minute hand 3 minutes behind): $$Spaces to gain = 25 - 3 = 22$$ $$Time = 22 × 1211 = 2 × 12 = \mathbf{24 minutes past 5}$$
• Case 2 (Minute hand 3 minutes ahead): $$Spaces to gain = 25 + 3 = 28 \implies 28 × 1211 = 33611 = 30611 min past 5$$

Case 1 exactly matches option (a).

Answer: (a) 24 minutes past 5.

Method — Angle Formula:

$$θ = |30H - 112M ight|$$

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

$$θ = |30(4) - 112(30) ight| = |120 - 165| = |-45^°| = \mathbf{45^°}$$

Answer: (c) $45^°$.

Method — Collinear Alignment Analysis:

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

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

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

Answer: (a) 44.

Step-by-Step Algebraic Derivation:

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

$$A + B = 16 \quad --- (1)$$

$$A + C = 110 \quad --- (2)$$

$$A + B + C = 14 \quad --- (3)$$

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

$$C = (A + B + C) - (A + B) = 14 - 16 = 3 - 212 = 112$$

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

$$B = (A + B + C) - (A + C) = 14 - 110 = 5 - 220 = 320$$

Combined daily work rate of B and C together:

$$B + C = 320 + 112 = 9 + 560 = 1460 = 730$$

Number of days required by B and C together:

$$Time = 1B + C = 307 = \mathbf{427 days}$$

Final Answer: B and C can complete the work together in $\mathbf{427}$ days (approx $4.29$ days).

Step-by-Step Working:

• Total distance to be cleared $= L_1 + L_2 = 150 m + 200 m = 350 metres$.
• Since both trains run in the same direction, relative speed is the difference of their speeds: $$Relative Speed = 72 - 54 = 18 km/hr$$
• Convert relative speed to $m/s$: $$18 × 518 = 5 m/s$$

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

$$Time = Total DistanceRelative Speed = 350 m5 m/s = \mathbf{70 seconds}$$

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

Step-by-Step Working:

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

Time taken until they meet:

$$Time = DistanceRelative Speed = 210 km105 km/hr = \mathbf{2 hours}$$

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

$$Distance_car = Speed_car × Time = 70 × 2 = \mathbf{140 km}$$

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

Step-by-Step Working:

Let the one-way distance between A and B be $d km$.

• Time taken to walk from A to B $= d4 hours$.
• Time taken to return from B to A $= d6 hours$.
• Total round-trip time $= 5 hours$.

Setting up the time equation:

$$d4 + d6 = 5$$

$$3d + 2d12 = 5 \implies 5d12 = 5 \implies d = \mathbf{12 km}$$

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

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 $= 8 × 120 = 820 = 25$.
• Work done by B in 8 days $= 8 × 140 = 840 = 15$.
• Combined work done by A and B $= 25 + 15 = 35$.
• Work done by C $= 1 - 35 = 25$.

Payment to C is proportional to his share of work:

$$C's Payment = 25 × Rs. 450 = 2 × 90 = \mathbf{Rs. 180}$$

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

Step-by-Step Working:

Let the total available fund reserve be $F$.

• Daily payroll expenditure for Team A $= F21$ per day.
• Daily payroll expenditure for Team B $= F28$ per day.

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

$$Combined Daily Burn Rate = F21 + F28 = F (4 + 384 ight) = 7F84 = F12$$

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

$$Number of Days = FF/12 = \mathbf{12 days}$$

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