Numerical Application

Given:
- Number of boys = 44, Average marks of boys = 40
- Number of girls = 36, Average marks of girls = 38

Formula used (Weighted Average):
$$\bar{x}_w = \frac{n_1 \bar{x}_1 + n_2 \bar{x}_2}{n_1 + n_2}$$

Working:
$$\text{Total marks of boys} = 44 \times 40 = 1760$$
$$\text{Total marks of girls} = 36 \times 38 = 1368$$
$$\text{Total students} = 44 + 36 = 80$$
$$\text{Average marks of class} = \frac{1760 + 1368}{80} = \frac{3128}{80} = 39.1$$

Answer: The average marks of the class = 39.1

Given:
- Average of 5 numbers = 87
- After excluding one number, average of remaining 4 numbers = 87 − 5 = 82

Working:
$$\text{Sum of 5 numbers} = 87 \times 5 = 435$$
$$\text{Sum of remaining 4 numbers} = 82 \times 4 = 328$$
$$\text{Excluded number} = 435 - 328 = 107$$

Answer: The excluded number = 107

Given:
- Nursery is closed on Sunday.
- Average plants sold Monday to Saturday (6 days) = 156
- Average plants sold Monday to Friday (5 days) = 124

Working:
$$\text{Total plants sold Mon–Sat} = 156 \times 6 = 936$$
$$\text{Total plants sold Mon–Fri} = 124 \times 5 = 620$$
$$\text{Plants sold on Saturday} = 936 - 620 = 316$$

Answer: Number of plants sold on Saturday = 316

Given:
- Average of 5 consecutive numbers = 125

Concept: For consecutive numbers, the middle (3rd) number equals the average.

Working:
Let the five consecutive numbers be $n-2,\ n-1,\ n,\ n+1,\ n+2$.
$$\text{Average} = n = 125$$
So the five numbers are: 123, 124, 125, 126, 127.

Numbers at extreme positions: 123 and 127.
$$\text{Product} = 123 \times 127 = 15621$$

Answer: Product of numbers at extreme positions = 15621

Given:
- Number of labourers = 1000
- Earlier daily wage = Rs 200
- Budget increased by Rs 15,00,000 per month
- Assume 1 month = 26 working days

Part (a): Present monthly budget
$$\text{Earlier monthly wage per labourer} = 200 \times 26 = \text{Rs } 5200$$
$$\text{Earlier total monthly budget} = 5200 \times 1000 = \text{Rs } 52,00,000$$
$$\text{Present monthly budget} = 52,00,000 + 15,00,000 = \text{Rs } 67,00,000$$

Part (b): Increase in daily income per labourer
$$\text{Increase per labourer per month} = \frac{15,00,000}{1000} = \text{Rs } 1500$$
$$\text{Increase in daily income} = \frac{1500}{26} \approx \text{Rs } 57.69$$

Part (c): New average monthly income per labourer
$$\text{New monthly income per labourer} = 5200 + 1500 = \text{Rs } 6700$$

Answers:
- (a) Present monthly budget = Rs 67,00,000
- (b) Increase in daily income ≈ Rs 57.69
- (c) New average monthly income per labourer = Rs 6700

Given:
- Number of observations = 35
- Incorrect mean = 98.6
- Misread value: 27 (incorrect) instead of 72 (correct)

Working:
$$\text{Incorrect sum} = 98.6 \times 35 = 3451$$
$$\text{Correct sum} = 3451 - 27 + 72 = 3451 + 45 = 3496$$
$$\text{Correct mean} = \frac{3496}{35} = 99.886 \approx 99.89$$

Answer: The correct mean = 99.89 (approximately)

Given:
- Average PF of defence officers = Rs 28,560
- Average PF of other officers = Rs 22,500
- Number of defence officers = 22 × (number of other officers)

Let number of other officers = $x$, so number of defence officers = $22x$.

Formula (Weighted Average):
$$\bar{x}_w = \frac{22x \times 28560 + x \times 22500}{22x + x}$$
$$= \frac{x(22 \times 28560 + 22500)}{23x}$$
$$= \frac{628320 + 22500}{23}$$
$$= \frac{650820}{23}$$
$$= \text{Rs } 28296$$

Answer: Average savings of all officers = Rs 28,296

Given:
- 5 years ago, average age of 3 children = 8 years
- A new baby is born now (age = 0 years)

Working:
$$\text{Sum of ages of 3 children 5 years ago} = 8 \times 3 = 24 \text{ years}$$
$$\text{Present sum of ages of 3 children} = 24 + 5 \times 3 = 24 + 15 = 39 \text{ years}$$
$$\text{Age of new baby} = 0 \text{ years}$$
$$\text{Total present age of all 4 children} = 39 + 0 = 39 \text{ years}$$
$$\text{Present average age} = \frac{39}{4} = 9.75 \text{ years}$$

Answer: Present average age of the 4 children = 9.75 years

Given:
- Original number of visitors = 35
- New number of visitors = 35 + 7 = 42
- Increase in total expense = Rs 42
- Decrease in average expenditure per head = Rs 1

Let original average expenditure per head = Rs $a$.

Original total expenditure = $35a$

New total expenditure = $35a + 42$

New average expenditure per head = $\frac{35a + 42}{42}$

According to the condition, new average = $a - 1$:
$$\frac{35a + 42}{42} = a - 1$$
$$35a + 42 = 42(a - 1)$$
$$35a + 42 = 42a - 42$$
$$42 + 42 = 42a - 35a$$
$$84 = 7a$$
$$a = 12$$

$$\text{Original expenditure} = 35 \times 12 = \text{Rs } 420$$

Answer: Original expenditure of the restaurant = Rs 420

Given:
- Average runs in 10 innings = 72
- Required average after 11th inning = 72 + 3 = 75

Working:
$$\text{Total runs in 10 innings} = 72 \times 10 = 720$$
$$\text{Required total runs after 11 innings} = 75 \times 11 = 825$$
$$\text{Runs needed in 11th inning} = 825 - 720 = 105$$

Answer: Virat Kohli must score 105 runs in the 11th inning.

Given:
- Radius of circular path $r = 3.5$ cm
- Time for 1 revolution $T = 3$ seconds

Working:

Circumference (total distance in one revolution):
$$d = 2\pi r = 2 \times \frac{22}{7} \times 3.5 = 22 \text{ cm}$$

Average Speed:
$$\text{Average Speed} = \frac{\text{Total distance}}{\text{Total time}} = \frac{22}{3} \approx 7.33 \text{ cm/s}$$

Average Velocity:
After one complete revolution, the ant returns to the starting point.
$$\text{Total displacement} = 0$$
$$\text{Average Velocity} = \frac{0}{3} = 0 \text{ cm/s}$$

Answer:
- Average Speed = 7.33 cm/s (approximately)
- Average Velocity = 0 cm/s

Given:
- Distance 1 = 7.2 km, Time 1 = 1.5 hours
- Distance 2 = 3.5 km, Time 2 = 2 hours

Working:
$$\text{Total distance} = 7.2 + 3.5 = 10.7 \text{ km}$$
$$\text{Total time} = 1.5 + 2 = 3.5 \text{ hours}$$
$$\text{Average Speed} = \frac{10.7}{3.5} \approx 3.06 \text{ km/h}$$

Answer: Sara's average speed = 3.06 km/h (approximately)

Given: Today is Tuesday.

Concept: To find the day after $n$ days, divide $n$ by 7 and find the remainder (odd days).

Working:
$$7706 \div 7 = 1100 \text{ weeks} + 6 \text{ remainder}$$
So 7706 days = 1100 complete weeks + 6 odd days.

Starting from Tuesday, counting 6 days forward:
Tuesday → Wednesday (1) → Thursday (2) → Friday (3) → Saturday (4) → Sunday (5) → Monday (6)

Answer: The day on the 7706th day will be Monday.

Given: Start date: 26th January 2008, End date: 15th May 2008.

Note: 2008 is a leap year (divisible by 4).

Working:
- January: 31 − 26 = 5 remaining days in January
- February (leap year): 29 days
- March: 31 days
- April: 30 days
- May: 15 days

$$\text{Total days} = 5 + 29 + 31 + 30 + 15 = 110 \text{ days}$$

Answer: Total number of days = 110 days

Given: 2nd April is Friday.

Working:
- April has 30 days.
- Last day of April = 30th April.
- Days from 2nd April to 30th April = 28 days = exactly 4 weeks.
- So 30th April is also a Friday.
- Next month is May, which has 31 days.
- 1st May = Saturday (day after Friday).
- 31st May: Days from 1st May to 31st May = 30 days.
- $30 \div 7 = 4$ weeks + 2 odd days.
- 1st May is Saturday; adding 2 days: Saturday → Sunday → Monday.

Answer: The last day of May (next month) will be Monday.

Using the standard odd-days method (reference: 1st January 0001 = Monday)

We use the formula/Zeller's approach or the standard calendar odd-days method.

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(1) 15th August 1947

Count odd days from 1st Jan 1 AD to 15th Aug 1947.

Years 1 to 1946:
- Complete centuries up to 1900: odd days in 1900 years.
- 19 centuries: $19 \times 5 = 95$ (since each non-leap century = 5 odd days, each leap century = 0)
- Century-wise: 100→5, 200→3, 300→1, 400→0 (cycle repeats every 400 years)
- 400 years = 0 odd days; 1600 years = 0 odd days.
- Remaining 300 years (1601–1900): 300→1 odd day.
- So 1900 years give 1 odd day.
- Years 1901–1946 = 46 years:
- Leap years: 1904,1908,...,1944 → $\frac{44}{4}=11$ leap years
- Ordinary years: 46 − 11 = 35
- Odd days: $35 \times 1 + 11 \times 2 = 35 + 22 = 57 = 8 \times 7 + 1 =$ 1 odd day
- Total odd days up to 31 Dec 1946 = 1 + 1 = 2 odd days.

Days in 1947 up to 15th August:
- Jan: 31, Feb: 28, Mar: 31, Apr: 30, May: 31, Jun: 30, Jul: 31, Aug 1–15: 15
- Total = 31+28+31+30+31+30+31+15 = 227 days
- $227 = 32 \times 7 + 3$ → 3 odd days

Total odd days = 2 + 3 = 5

Day code: 0=Sun, 1=Mon, 2=Tue, 3=Wed, 4=Thu, 5=Fri, 6=Sat

5 → Friday

15th August 1947 was a Friday. ✓ (historically confirmed)

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(2) 22nd November 2025

Odd days up to 31 Dec 2024:
- 2000 years = 0 odd days (every 400 years = 0)
- Years 2001–2024 = 24 years:
- Leap years: 2004, 2008, 2012, 2016, 2020, 2024 → 6 leap years
- Ordinary years: 24 − 6 = 18
- Odd days: $18 \times 1 + 6 \times 2 = 18 + 12 = 30 = 4 \times 7 + 2$ → 2 odd days
- Total up to 31 Dec 2024 = 0 + 2 = 2 odd days

Days in 2025 up to 22nd November:
- Jan:31, Feb:28, Mar:31, Apr:30, May:31, Jun:30, Jul:31, Aug:31, Sep:30, Oct:31, Nov 1–22:22
- Total = 31+28+31+30+31+30+31+31+30+31+22 = 326 days
- $326 = 46 \times 7 + 4$ → 4 odd days

Total odd days = 2 + 4 = 6 → Saturday

22nd November 2025 will be a Saturday.

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(3) 21st September 2080

Odd days up to 31 Dec 2079:
- 2000 years = 0 odd days
- Years 2001–2079 = 79 years:
- Leap years: 2004,2008,...,2076 → $\frac{76}{4}=19$ leap years
- Ordinary years: 79 − 19 = 60
- Odd days: $60 \times 1 + 19 \times 2 = 60 + 38 = 98 = 14 \times 7 + 0$ → 0 odd days
- Total up to 31 Dec 2079 = 0 odd days

Days in 2080 up to 21st September (2080 is a leap year):
- Jan:31, Feb:29, Mar:31, Apr:30, May:31, Jun:30, Jul:31, Aug:31, Sep 1–21:21
- Total = 31+29+31+30+31+30+31+31+21 = 265 days
- $265 = 37 \times 7 + 6$ → 6 odd days

Total odd days = 0 + 6 = 6 → Saturday

21st September 2080 will be a Saturday.

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(4) 18th October 2100

Odd days up to 31 Dec 2099:
- 2000 years = 0 odd days
- Years 2001–2099 = 99 years:
- Leap years: 2004,2008,...,2096 → $\frac{96}{4}=24$ leap years (Note: 2100 is NOT a leap year, but we only go to 2099)
- Ordinary years: 99 − 24 = 75
- Odd days: $75 \times 1 + 24 \times 2 = 75 + 48 = 123 = 17 \times 7 + 4$ → 4 odd days
- Total up to 31 Dec 2099 = 4 odd days

Days in 2100 up to 18th October (2100 is NOT a leap year):
- Jan:31, Feb:28, Mar:31, Apr:30, May:31, Jun:30, Jul:31, Aug:31, Sep:30, Oct 1–18:18
- Total = 31+28+31+30+31+30+31+31+30+18 = 291 days
- $291 = 41 \times 7 + 4$ → 4 odd days

Total odd days = 4 + 4 = 8 = 7 + 1 → 1 odd day → Monday

18th October 2100 will be a Monday.

Given:
- Trains leave every 40 minutes.
- At 10:15 a.m., the train had left 10 minutes ago.

Part (a): Time the train left
$$10:15 - 10 \text{ min} = \mathbf{10:05 \text{ a.m.}}$$

Part (b): Time of next train
$$10:05 + 40 \text{ min} = \mathbf{10:45 \text{ a.m.}}$$

Part (c): Waiting time
$$10:45 - 10:15 = \mathbf{30 \text{ minutes}}$$

Part (d): Angle between minute and hour hand at 10:45 a.m.

Using the formula:
$$\theta = \left|30H - \frac{11}{2}M\right|$$
where $H = 10$, $M = 45$.
$$\theta = \left|30 \times 10 - \frac{11}{2} \times 45\right| = \left|300 - 247.5\right| = 52.5°$$

Answer:
- (a) Train left at 10:05 a.m.
- (b) Next train at 10:45 a.m.
- (c) Waiting time = 30 minutes
- (d) Angle = 52.5°

Given:
- Pranil's range: after 17th October and before 21st October → 18th, 19th, 20th October
- Colleague's range: after 19th October and before 24th October → 20th, 21st, 22nd, 23rd October

Common date in both ranges:
$$\{18, 19, 20\} \cap \{20, 21, 22, 23\} = \{20\}$$

Answer: The date of maximum sale as concluded by the owner is 20th October.

Part (a): Straight angle (180°) in a day

The minute hand gains 360° over the hour hand in every 12 hours (i.e., 11 times they are opposite = straight angle in 12 hours).

In 12 hours, the hands form a straight angle 11 times.
In 24 hours: $11 \times 2 = \mathbf{22 \text{ times}}$

Part (b): Right angle (90°) in a day

In every 12 hours, the hands form a right angle 22 times (11 times at 90° and 11 times at 270°).
In 24 hours: $22 \times 2 = \mathbf{44 \text{ times}}$

Answers:
- (a) Straight angle is formed 22 times in a day.
- (b) Right angle is formed 44 times in a day.

Given: Time = 7:40 p.m., so $H = 7$, $M = 40$.

Formula:
$$\theta = \left|30H - \frac{11}{2}M\right|$$
$$\theta = \left|30 \times 7 - \frac{11}{2} \times 40\right|$$
$$= \left|210 - 220\right| = |-10| = 10°$$

Answer: The angle between the minute hand and hour hand at 7:40 p.m. = 10°

Given: Time = 5:20

Concept: The hour hand moves at $\frac{1}{2}°$ per minute.

Working:
- At 12:00, hour hand is at 0°.
- At 5:00, hour hand has moved: $5 \times 30° = 150°$
- In additional 20 minutes: $20 \times \frac{1}{2}° = 10°$
- Total degrees turned = $150° + 10° = 160°$

Answer: By 20 minutes past 5, the hour hand has turned through 160°.

Given: Time is between 3:00 and 4:00, angle between hands = 3°.

Formula:
$$\theta = \left|30H - \frac{11}{2}M\right|$$

Here $H = 3$:
$$3 = \left|90 - \frac{11}{2}M\right|$$

Case 1: $90 - \frac{11}{2}M = 3$
$$\frac{11}{2}M = 87 \Rightarrow M = \frac{174}{11} = 15\frac{9}{11} \text{ minutes}$$
Time = 3 hours $15\frac{9}{11}$ minutes

Case 2: $90 - \frac{11}{2}M = -3$
$$\frac{11}{2}M = 93 \Rightarrow M = \frac{186}{11} = 16\frac{10}{11} \text{ minutes}$$
Time = 3 hours $16\frac{10}{11}$ minutes

Answer: The hands are 3° apart at **3:$15\frac{9}{11}$ and 3:$16\frac{10}{11}$** (i.e., approximately 3:15:49 and 3:16:55).

Given:
- Lockdown start: Midnight of 25th March 2020
- Extended end: 3rd May 2020
- 25th March 2020 is Wednesday
- 2020 is a leap year.

Part (a): Total number of days
- March: 31 − 25 = 6 remaining days in March (26, 27, 28, 29, 30, 31)
- April: 30 days
- May: 3 days (1st, 2nd, 3rd)
$$\text{Total days} = 6 + 30 + 3 = 39 \text{ days}$$

Part (b): Day on 3rd May 2020
- 25th March is Wednesday.
- After 39 days: $39 = 5 \times 7 + 4$ → 4 odd days.
- Wednesday + 4 days = Wednesday → Thursday → Friday → Saturday → Sunday

Answers:
- (a) Total lockdown = 39 days
- (b) 3rd May 2020 was a Sunday

Given:
- India is 9 hours 30 minutes ahead of Ottawa, Canada.
- Time in India = 1:25 a.m.

Working:
$$\text{Time in Canada} = 1:25 \text{ a.m.} - 9 \text{ h } 30 \text{ min}$$
$$= 1:25 \text{ a.m.} - 9:30$$

Since 1:25 a.m. − 9 h 30 min goes to the previous day:
$$1:25 - 9:30 = (1:25 + 24:00) - 9:30 = 25:25 - 9:30 = 15:55$$

$$= 3:55 \text{ p.m. (previous day)}$$

Answer: When it is 1:25 a.m. in India, the time in Ottawa, Canada is 3:55 p.m. (previous day).

Given:
- Navya is twice as efficient as Nitti.
- Together they finish the job in 10 days.

Let Nitti's 1-day work = $x$, then Navya's 1-day work = $2x$.

Together:
$$x + 2x = \frac{1}{10}$$
$$3x = \frac{1}{10} \Rightarrow x = \frac{1}{30}$$

- Nitti's 1-day work = $\frac{1}{30}$ → Nitti takes 30 days
- Navya's 1-day work = $\frac{2}{30} = \frac{1}{15}$ → Navya takes 15 days

Answer: Navya takes 15 days and Nitti takes 30 days individually.

Given:
- A finishes work in 30 days → A's 1-day work = $\frac{1}{30}$
- A is twice as good as B → B's 1-day work = $\frac{1}{60}$
- C is thrice as good as A → C's 1-day work = $\frac{3}{30} = \frac{1}{10}$

Together (A + B + C):
$$\frac{1}{30} + \frac{1}{60} + \frac{1}{10}$$
$$= \frac{2}{60} + \frac{1}{60} + \frac{6}{60} = \frac{9}{60} = \frac{3}{20}$$

$$\text{Days taken} = \frac{20}{3} = 6\frac{2}{3} \text{ days}$$

Answer: A, B, and C together will finish the work in $\mathbf{6\frac{2}{3}}$ days.

Given:
- X's 1-day work = $\frac{1}{60}$
- Y's 1-day work = $\frac{1}{40}$
- X left 10 days before completion (i.e., Y worked alone for the last 10 days).

Let total days = $d$. X worked for $(d - 10)$ days, Y worked for $d$ days.

$$\frac{d-10}{60} + \frac{d}{40} = 1$$

Multiply throughout by 120:
$$2(d-10) + 3d = 120$$
$$2d - 20 + 3d = 120$$
$$5d = 140$$
$$d = 28 \text{ days}$$

Answer: The work will be completed in 28 days.

Given:
- Reduced speed = $\frac{7}{11}$ of original speed
- Time taken at reduced speed = 44 hours

Concept: Speed and time are inversely proportional (distance constant).
$$\frac{\text{Reduced speed}}{\text{Original speed}} = \frac{\text{Original time}}{\text{Reduced time}}$$
$$\frac{7}{11} = \frac{t}{44}$$
$$t = \frac{7 \times 44}{11} = 7 \times 4 = 28 \text{ hours}$$

Answer: Original time taken = 28 hours