Class 11- Applied Mathematics-NCERT Handbook Solutions-Chapter-2 -

Chapter 2 Solutions

Check Your Progress 2.1

1. A local train from Mumbai leaves every 40 minutes from the station. When inquired by a passenger, the help desk executive informed that the train had already left 10 minutes ago. If this information was given at 10:15 a.m.

At what time did the train leave the station?
At what time will the next train leave the station?
For how long will that man have to wait for the next train?
Find the angle formed between the minute hand and the hour hand when the passenger will board the next train?

Solution:

a) The current time is $10 : 15 a.m.$ and the train left $10$ minutes ago.

Time the train left $= 10 : 15 a.m. - 10 minutes = 10 : 05 a.m.$

b) The trains leave every $40$ minutes.

Time for the next train $= 10 : 05 a.m. + 40 minutes = 10 : 45 a.m.$

c) The man is at the station at $10 : 15 a.m.$ and the next train is at $10 : 45 a.m.$

Waiting time $= 10 : 45 a.m. - 10 : 15 a.m. = 30 minutes$ .

d) The passenger boards at $10 : 45$ . We need to find the angle at this time. Here, hours $H = 10$ and minutes $m = 45$ .

Angle $A = | 30 H - \frac{11}{2} m |$

$A = | 30 (10) - \frac{11}{2} (45) |$

$A = | 300 - 247.5 | = {52.5}^{\circ}$

2. In a day, how many times is a straight angle formed between minute and hour hands?

Solution:

A straight angle ( $180^{\circ}$ ) implies the hands are pointing in exactly opposite directions. This happens once every hour, but between $5 : 00$ and $7 : 00$ , it only happens exactly at $6 : 00$ . Therefore, it happens $11$ times in a $12$ -hour period.

In a full day ( $24$ hours), a straight angle is formed $11 \times 2 = 22$ times.

3. In a day, how many times is a right angle formed between minute and hour hands?

Solution:

A right angle ( $90^{\circ}$ ) is formed twice every hour. However, between $2 : 00$ and $4 : 00$ , and between $8 : 00$ and $10 : 00$ , it occurs only $3$ times instead of $4$ (exactly at $3 : 00$ and $9 : 00$ ). Thus, it happens $22$ times in $12$ hours.

In a full day ( $24$ hours), a right angle is formed $22 \times 2 = 44$ times. (Note: Some textbooks cite 22 referring to a 12-hour cycle, but strictly "in a day" means 44 times).

4. Find the angle between the minute hand and the hour hand at 7:40 pm?

Solution:

Using the angle formula: $A = | 30 H - \frac{11}{2} m |$

Here, $H = 7$ and $m = 40$ .

$A = | 30 (7) - \frac{11}{2} (40) |$

$A = | 210 - 11 (20) |$

$A = | 210 - 220 | = | - 10 | = 10^{\circ}$

The angle is $10^{\circ}$ .

5. By 20 minutes past 5, how many degrees has the hour hand turned through?

Solution:

The hour hand turns at a speed of $30^{\circ}$ per hour (or ${0.5}^{\circ}$ per minute).

Total time elapsed from $12 : 00$ to $5 : 20$ is $5$ hours and $20$ minutes.

Angle turned in $5$ hours $= 5 \times 30^{\circ} = 150^{\circ}$

Angle turned in $20$ minutes $= 20 \times {0.5}^{\circ} = 10^{\circ}$

Total angle $= 150^{\circ} + 10^{\circ} = 160^{\circ}$

6. At what time between 3'o clock and 4'o clock are the hands of a clock three degrees apart?

Solution:

Given angle $A = 3^{\circ}$ and hours $H = 3$ . We must find $m$ .

Using $A = | 30 H - \frac{11}{2} m |$

Case 1: $30 (3) - \frac{11}{2} m = 3$

$90 - 3 = \frac{11}{2} m \Rightarrow 87 = \frac{11}{2} m$

$m = \frac{174}{11} = 15 \frac{9}{11}$ minutes past $3$ .

Case 2: $\frac{11}{2} m - 30 (3) = 3$

$\frac{11}{2} m = 90 + 3 = 93$

$m = \frac{186}{11} = 16 \frac{10}{11}$ minutes past $3$ .

7. At what time do the hands of the clock meet between 6:00 to 7:00?

Solution:

When the hands meet, the angle $A = 0^{\circ}$ . Here $H = 6$ .

$| 30 (6) - \frac{11}{2} m | = 0$

$180 = \frac{11}{2} m$

$m = \frac{360}{11} = 32 \frac{8}{11}$

The hands will meet at $32 \frac{8}{11}$ minutes past $6$ .

8. India is 9 hours and 30 minutes ahead of Ottawa ON, Canada. What is time in Canada when it is 1:25 am in India?

Solution:

Since India is ahead, we must subtract $9$ hours and $30$ minutes from the Indian time to get the Ottawa time.

Indian time $= 1 : 25 a.m.$

Subtracting $1$ hour $25$ minutes brings us to midnight ( $12 : 00 a.m.$ ). We still need to subtract the remaining $8$ hours and $5$ minutes.

$12 : 00 a.m. - 8 hours 5 minutes = 3 : 55 p.m.$ (of the previous day).

Check Your Progress 2.2

1. If today is a Tuesday, what will be the day on 7706th day?

Solution:

To find the day of the week, we divide the total number of days by $7$ to find the remainder (odd days).

$\frac{7706}{7} = 1100$ weeks and $6$ remainder days.

Adding $6$ days to Tuesday is the same as moving back $1$ day. The day will be Monday.

2. Find the total number of days from 26th January 2008 to 15 May 2008?

Solution:

Since 2008 is a leap year, February has $29$ days. Counting both start and end days:

January: $31 - 26 + 1 = 6$ days
February: $29$ days
March: $31$ days
April: $30$ days
May: $15$ days

Total $= 6 + 29 + 31 + 30 + 15 = 111$ days.

3. If the second day of April month is a Friday, then find the last day of the next month?

Solution:

April has $30$ days. The next month is May, which has $31$ days.

We need to find the day on May 31st, starting from April 2nd.

Remaining days in April $= 30 - 2 = 28$ days.

Total days passed = $28 (April) + 31 (May) = 59$ days.

Odd days $= 59 mod 7 = 3$ days.

Adding $3$ days to Friday gives Monday.

4. Workout for the day of week on the given date:

(a) 15th August 1947 (b) 22nd November 2025 (c) 21st September 2080 (d) 18th October 2100

Solution:

(a) 15 August 1947:

Completed years $= 1946$ . Odd days: $1600$ (0) + $300$ (1) + $46$ years ( $11$ leap, $35$ ordinary).

$11 \times 2 + 35 \times 1 = 57 ≣ 1$ odd day. Total year odd days $= 1 + 1 = 2$ .

Months in 1947: Jan(3)+Feb(0)+Mar(3)+Apr(2)+May(3)+Jun(2)+Jul(3)+Aug(15) $= 31 ≣ 3$ odd days.

Total odd days $= 2 + 3 = 5 \Rightarrow$ Friday.

(b) 22 November 2025:

Completed years $= 2024$ . Odd days: $2000$ (0) + $24$ years ( $6$ leap, $18$ ord).

$6 \times 2 + 18 \times 1 = 30 ≣ 2$ odd days.

Months in 2025: Jan(3)+Feb(0)+Mar(3)+Apr(2)+May(3)+Jun(2)+Jul(3)+Aug(3)+Sep(2)+Oct(3)+Nov(22=1) $= 25 ≣ 4$ odd days.

Total odd days $= 2 + 4 = 6 \Rightarrow$ Saturday.

(c) 21 September 2080:

Completed years $= 2079$ . Odd days: $2000$ (0) + $79$ years ( $19$ leap, $60$ ord).

$19 \times 2 + 60 \times 1 = 98 ≣ 0$ odd days.

Months in 2080 (leap year): Jan(3)+Feb(1)+Mar(3)+Apr(2)+May(3)+Jun(2)+Jul(3)+Aug(3)+Sep(21=0) $= 20 ≣ 6$ odd days.

Total odd days $= 0 + 6 = 6 \Rightarrow$ Saturday.

(d) 18 October 2100:

Completed years $= 2099$ . Odd days: $2000$ (0) + $99$ years ( $24$ leap, $75$ ord).

$24 \times 2 + 75 \times 1 = 123 ≣ 4$ odd days.

Months in 2100 (non-leap year): Jan(3)+Feb(0)+Mar(3)+Apr(2)+May(3)+Jun(2)+Jul(3)+Aug(3)+Sep(2)+Oct(18=4) $= 25 ≣ 4$ odd days.

Total odd days $= 4 + 4 = 8 ≣ 1 \Rightarrow$ Monday.

5. Pranil works in an electronics goods shop... Pranil remembers that the maximum sale occurred after 17th October but before 21st October while his colleague remembers that the maximum sale happened after 19th October but before 24th October. What is the date as per your point of view?

Solution:

According to Pranil, the dates are: $18, 19, 20$ .

According to his colleague, the dates are: $20, 21, 22, 23$ .

The only common date between the two memories is the 20th of October.

6. Indian government has announced a complete lockdown for entire country from the Midnight of 25th March 2020 till the midnight of 14th April 2020... lockdown was further extended till 3rd May 2020.

(a) Calculate the total number of days for which the lockdown lasted.

(b) If 25th March 2020 is Wednesday, then find which day of the week will fall on 3rd May 2020.

Solution:

(a) Calculating total days:

March (25 to 31): $7$ days
April: $30$ days
May (1 to 3): $3$ days

Total duration $= 7 + 30 + 3 = 40$ days.

(b) The gap between 25th March and 3rd May is $39$ days (since 25th March is day 1).

Odd days $= 39 mod 7 = 4$ odd days.

Adding $4$ days to Wednesday gives Sunday. (Note: Some textbook editions erroneously state Monday, but 3rd May 2020 was mathematically and historically a Sunday.)

Check Your Progress 2.3

1. Navya is twice as efficient as Nitti. If they take 10 days to finish a certain job together. How much time will they take individually to finish the same job?

Solution:

Let Nitti's efficiency be $1$ unit/day. Then Navya's efficiency is $2$ units/day.

Combined efficiency $= 1 + 2 = 3$ units/day.

Total work $= Efficiency \times Time = 3 \times 10 = 30$ units.

Time for Navya $= \frac{30}{2} = 15$ days.

Time for Nitti $= \frac{30}{1} = 30$ days.

2. A piece of work is finished in 30 days by A. Since C is thrice as good as A and A is twice as good as B. If A, B, C work together, then how many days will they take to finish the work?

Solution:

Time taken by A $= 30$ days.

C is thrice as good as A, meaning C takes $\frac{1}{3}$ of A's time. Time for C $= 10$ days.

A is twice as good as B, meaning B takes twice A's time. Time for B $= 60$ days.

Working together, their 1-day work is:

$\frac{1}{A} + \frac{1}{B} + \frac{1}{C} = \frac{1}{30} + \frac{1}{60} + \frac{1}{10}$

$= \frac{2 + 1 + 6}{60} = \frac{9}{60} = \frac{3}{20}$

Total time $= \frac{20}{3} = 6 \frac{2}{3}$ days (or $6$ days and $16$ hours).

3. X can do a piece of work in 60 days, whereas Y can do the same work in 40 days. Both started the work together, but X left after 10 days before the completion of work. Find how many days will it take to complete the work?

Solution:

Let the total time taken be $p$ days.

Y worked for all $p$ days, and X worked for $(p - 10)$ days.

Total work equation:

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

Multiplying the entire equation by the LCM, which is $120$ :

$2 (p - 10) + 3 p = 120$

$5 p - 20 = 120 \Rightarrow 5 p = 140 \Rightarrow p = 28$

It will take $28$ days to complete the work.

4. 100 persons begin to work together on a project which was expected to be completed in 40 days. But after few days 40 persons left. As a result, the project got delayed by 10 days. How many days after the commencement of the project did the 40 persons left?

Solution:

Total work $= 100 \times 40 = 4000$ man-days.

Let the $40$ persons leave after $d$ days. So, $100$ people worked for $d$ days.

Remaining people $= 100 - 40 = 60$ persons.

The project was delayed by $10$ days, meaning it took $40 + 10 = 50$ days in total. Thus, the $60$ people worked for the remaining $(50 - d)$ days.

$100 d + 60 (50 - d) = 4000$

$100 d + 3000 - 60 d = 4000$

$40 d = 1000 \Rightarrow d = 25$

They left after $25$ days.

5. The efficiency of x, y, z are in ratio of 3:2:6 to finish a task. If they work together, they can finish it in 2 hours; find the time taken by them if they do the task individually?

Solution:

Let their 1-hour work rates be $3 k$ , $2 k$ , and $6 k$ .

Combined rate $= 3 k + 2 k + 6 k = 11 k$ .

They finish in $2$ hours, so the total work $= 11 k \times 2 = 22 k$ .

Time for x $= \frac{22 k}{3 k} = \frac{22}{3}$ hours.

Time for y $= \frac{22 k}{2 k} = 11$ hours.

Time for z $= \frac{22 k}{6 k} = \frac{11}{3}$ hours.

6. A is three times as efficient as B. Also, A takes 30 days less than B for doing a piece of work. Find the time taken by them if they work a) individually b) together?

Solution:

Let the time taken by A be $x$ days. Since A is $3$ times more efficient than B, B will take $3 x$ days.

We are given $3 x - x = 30 \Rightarrow 2 x = 30 \Rightarrow x = 15$ .

a) individually: A takes $15$ days, B takes $45$ days.

b) together: Combined rate $= \frac{1}{15} + \frac{1}{45} = \frac{3 + 1}{45} = \frac{4}{45}$ .

Time together $= \frac{45}{4} = 11.25$ days (or $11$ days and $6$ hours).

7. Machine P is 40% more efficient than Machine Q. Machine P can make 100 bags alone in 30 hours. Find the time taken to complete the order if both the machines work together?

Solution:

P is 40% more efficient, so $E_{P} = 1.4 \times E_{Q}$ .

P takes $30$ hours. Because rate is inversely proportional to time, Q will take $30 \times 1.4 = 42$ hours.

Working together, 1-hour rate is:

$\frac{1}{30} + \frac{1}{42} = \frac{7 + 5}{210} = \frac{12}{210} = \frac{2}{35}$

Time together $= \frac{35}{2} = 17.5$ hours.

Check Your Progress 2.4

1. If a car increases its speed from 40 km/hr to 60 km/hr, how much time will it save on a 120-km journey?

Solution:

Time at original speed $= \frac{120}{40} = 3$ hours.

Time at increased speed $= \frac{120}{60} = 2$ hours.

Time saved $= 3 - 2 = 1$ hour.

2. A person travels from one place to another at 30 km/hr and returns at 120 km/hr. If the total time taken is 5 hours, then find the Distance.

Solution:

Let the one-way distance be $d$ km.

Total time $= \frac{d}{30} + \frac{d}{120} = 5$

$\frac{4 d + d}{120} = 5 \Rightarrow \frac{5 d}{120} = 5$

$d = 120$ km.

3. A train is running at 7/11 of its own speed due to fog and reached a place in 44 hours. What was the original time taken by the train if it runs at its own speed?

Solution:

Let the original speed be $v$ . The new speed is $\frac{7}{11} v$ .

New time $= 44$ hours.

Distance $= Speed \times Time = (\frac{7}{11} v) \times 44 = 28 v$ .

Original time $= \frac{Distance}{Original Speed} = \frac{28 v}{v} = 28$ hours.

4. (a) The signal poles on a railroad are placed 100 m apart, how many poles will be passed by a train in 8 hours if the speed of the train is 45 km/h.

(b) If 7201 poles are to be installed within two stations at equal distance covering distance of 360km, Find out the distance between two consecutive poles.

Solution:

(a) Speed $= 45 km/h = 45000 m/h$ .

Distance traveled in $8$ hours $= 45000 \times 8 = 360, 000 m$ .

Number of poles $= (\frac{Total Distance}{Spacing}) + 1 = \frac{360, 000}{100} + 1 = 3600 + 1 = 3601$ poles.

(b) Total distance $= 360 km = 360, 000 m$ .

$7201$ poles means there are $7201 - 1 = 7200$ gaps between them.

Distance between two poles $= \frac{360, 000}{7200} = 50$ meters.

Check Your Progress 2.5

1. Six students P, Q, R, S, T and U are sitting in two rows, three are in each row. T is not at the end of any row. S is second to the left of U. R is the neighbour of T, Q is the neighbour of U. Arrange the members of two rows. Who is sitting diagonally opposite to Q?

Solution:

Arrangement logic:

Since S is second to the left of U, the first row must contain S, a middle person, and U.
Q is the neighbor of U, so Q sits in the middle of that row. Therefore, Row 1 is: S, Q, U.
T is not at the end, so T must sit in the middle of Row 2.
R is a neighbor of T, and P takes the remaining spot. Therefore, Row 2 is: P, T, R.

Row 1: S, Q, U

Row 2: P, T, R

Diagonally opposite to Q (middle) doesn't strictly apply, but looking at the corners, the person diagonally opposite to U is P. Based on standard answer keys interpreting the matrix, the element matching the diagonal condition opposite to the specific requested coordinate is P.

2. Group of 8 persons A, B, C, D, E, F, G and H are seated around a square table while facing towards each other two persons are seated on each side. There are three ladies in a group and they are not allowed to sit next to each other. F is seated between H and B, also C is seated between E and B. D is a lady member who is seated second to the left of F. A is a lady member seated opposite to B. B is a male member. A lady member must be seated in between B and E.

Answer the following questions:

(i) Identify the lady members.

(ii) Name the members who is seated immediate left to B.

(iii) Name the members who are adjacent to D.

Solution:

Based on the seating arrangement rules provided:

(i) The three lady members are A, C, and D.

(ii) The member seated to the immediate left of B is F.

(iii) The members who are adjacent to D are H and G.

3. Riya is known for her good photography skill... Siya is seated left to Rani and to the right of Bubbly. Maria is seated to the right of Rani, Reet is between Rani and Maria.

(a) Who is seated in the middle

(b) Who is seated immediate right to Reet.

Solution:

Following the positional logic from left to right:

Siya is to the right of Bubbly and left of Rani: Bubbly → Siya → Rani.
Maria is to the right of Rani, and Reet is between them: Rani → Reet → Maria.

Combined seating order (Left to Right): Bubbly, Siya, Rani, Reet, Maria.

(a) The person seated in the middle is Rani.

(b) The person seated to the immediate right of Reet is Maria.

4. In an exhibition seven different electronic companies are displaying their new Air Conditioner Model... All the models are facing towards east. LG is next to the right of WHIRLPOOL, WHIRLPOOL is placed fourth to the right of VOLTAS. HITACHI is placed between LLOYD and SAMSUNG. VOLTAS which is placed third to the left of LLOYD, is at one end.

(a) Write the arrangement of companies of AC from the left to the right.

(b) Which AC brand lies between HITACHI and WHIRLPOOL.

Solution:

Arranging the units from left to right based on the clues:

Voltas is at the extreme left (Position 1).
Lloyd is 3rd to the right of Voltas (Position 4).
Whirlpool is 4th to the right of Voltas (Position 5).
LG is to the right of Whirlpool (Position 6).
Hitachi is between Lloyd and Samsung. Since Lloyd is at 4, Hitachi must be at 3 and Samsung at 2.
The remaining brand, Deccan, goes to the extreme right (Position 7).

(a) Arrangement: Voltas, Samsung, Hitachi, Lloyd, Whirlpool, LG, Deccan.

(b) The brand lying between Hitachi and Whirlpool is Lloyd.

(c) The air conditioner at the extreme right is Deccan.

5. Three Doctors Dr. Aggarwal, Dr. Chabbra and Dr. Roy are available in LIFE hospital...

(a) On which day are all three doctors available in the hospital.

(b) Ravi wants to consult two doctors Dr. Chabbra and Dr. Roy on the same day. Workout suitable day and time for him to visit the hospital.

Solution:

Analyzing their schedules:

Dr. Aggarwal: Mon, Wed, Sun (1:00 pm - 5:00 pm)
Dr. Chabbra: Tue, Wed, Fri, Sun (11:00 am - 3:00 pm)
Dr. Roy: Tue, Thu (10:00 am - 1:00 pm) & Fri, Sat, Sun (3:00 pm - 5:00 pm)

(a) Looking at the days, Sunday is the only day when all three doctors hold hours at the hospital.

(b) To consult both Dr. Chabbra and Dr. Roy on the same day, we look for overlapping days and times. They are both present on Tuesday. Dr. Roy is available from 10:00 am - 1:00 pm, and Dr. Chabbra is available from 11:00 am - 3:00 pm.

Therefore, Ravi should visit on Tuesday between 11:00 am and 1:00 pm.

6. Five friends are standing in a line and facing towards the wall wearing Red, Grey, Yellow, Violet and Black Shoes. The persons wearing Yellow and Red Shoes are not standing at the end position. The person in the middle position is wearing Black shoes and the person with Red Shoes is not in his left. The person wearing violet shoes is standing on extreme right.

(a) Who is on 4th position from right?

(b) Who is standing on the extreme left position?

Solution:

Mapping the positions (1 to 5, left to right):

Middle (Position 3) is Black.
Extreme right (Position 5) is Violet.
Red is not to the left of Black, so it must be on the right. Since it cannot be at the end, it must be Position 4: Red.
Yellow is not at the end, so it must be Position 2: Yellow.
The remaining color for the extreme left (Position 1) is Grey.

Order: Grey, Yellow, Black, Red, Violet.

(a) The person in the 4th position from the right is Yellow.

(b) The person standing on the extreme left position is Grey.

7. Six persons P, Q, R, S, T and U are sitting in a circle with their faces towards the centre. S is on the immediate left of T, P is on the left of S and U is immediate neighbor of T. R is sitting second to the right of U.

(a) Name the immediate neighbours of Q.

(b) Who is between S and R?

Solution:

Plotting the circular arrangement (counter-clockwise/right):

T is placed. S is immediate left (clockwise) of T. P is left of S.
U is an immediate neighbor of T, so U must be to the right (counter-clockwise) of T.
R is second to the right of U. The space between U and R is filled by the remaining person, Q.

The counter-clockwise circle order is: T, U, Q, R, P, S.

(a) The immediate neighbors of Q are U and R.

(b) The person sitting between S and R is P.

Case Study I

The Big Ben clock tower in London... The hour hand is 6 feet long. The minute hand is 9 feet long.

a. What is the angle between the hour and minute hands of the Big Ben clock at 5:30 p.m.?

b. After the clock is set at 12:00 a.m., at what time between 1:00 a.m. and 2:00 a.m. will the hands of the clock coincide?

c. How many times in a day (24 hours) will the hour and minute hands of the clock coincide?

d. Through what angle does the minute hand rotate between 2:00 p.m. and 2:40 p.m. on the same day?

e. A tourist wants to know how much area the hour hand sweeps between 11:00 a.m. and 5:00 p.m. on a particular day. Calculate this area for the Big Ben's hour hand.

Solution:

a. Using the angle formula $A = | 30 H - 5.5 m |$ with $H = 5$ and $m = 30$ :

$A = | 30 (5) - 5.5 (30) | = | 150 - 165 | = 15^{\circ}$ .

b. The hands coincide when the angle $A = 0^{\circ}$ . For $H = 1$ :

$30 (1) - 5.5 m = 0 \Rightarrow 5.5 m = 30$

$m = \frac{60}{11} = 5 \frac{5}{11}$ minutes. Time: 1:05 a.m.

c. The hands of a clock coincide $11$ times in $12$ hours. In a $24$ -hour day, they coincide $11 \times 2 = 22$ times.

d. From $2 : 00$ p.m. to $2 : 40$ p.m., exactly $40$ minutes have passed. The minute hand moves $6^{\circ}$ per minute.

Angle $= 40 \times 6^{\circ} = 240^{\circ}$ .

e. Time elapsed between $11 : 00$ a.m. and $5 : 00$ p.m. is $6$ hours.

The hour hand sweeps $30^{\circ}$ per hour, so in $6$ hours it sweeps $6 \times 30^{\circ} = 180^{\circ}$ (which is a semi-circle).

The length of the hour hand (radius $r$ ) is $6$ feet.

Area swept $= \frac{1}{2} π r^{2} = \frac{1}{2} \times π \times 6^{2} = 18 π$ sq feet.

$18 \times 3.14159 \approx 56.55$ sq feet. (Note: A published answer key cites 42.92, which corresponds to a different metric or baseline assumption, but geometrically 56.55 sq. ft. is the exact mathematical area for a 6 ft radius over 6 hours).

Case Study II

At the board meeting of a certain company, six officials sit around a circular table facing the centre... There are two Directors (I and II), a Chairman, a Vice Chairman, a CEO and a CFO.

a. Which employee is definitely sitting at the same distance from Director II as Director I is sitting from the CEO or CFO?

b. Which employee is sitting exactly opposite to the Vice Chairman?

c. One of the board members leaves, leaving a gap of one seat between the Chairman and the Vice Chairman. Based on the information given, which member/s could have left?

d. Given that Director I swaps his seat with the Vice Chairman then which of these statements is definitely true?

Solution:

Arrangement deduction: Label positions 1 to 6 in a counter-clockwise circle.

Dir I is to the right (counter-clockwise) of Dir II. So, Dir II is at 1, Dir I is at 2.
Dir I (Pos 2) is equally far from CEO and CFO. They must be at 4 and 6 (distance of 2 units each).
Vice Chairman is immediate right (counter-clockwise) of CEO. If CEO is at 6, immediate right is 1 (occupied). Therefore, CEO must be at 4.
This places the Vice Chairman at 5, and the CFO at 6.
The remaining seat (Position 3) belongs to the Chairman.

Final Order (Counter-Clockwise): Dir II, Dir I, Chairman, CEO, Vice Chairman, CFO.

a. Dir I (Pos 2) is a distance of $2$ units from CEO (Pos 4) and CFO (Pos 6). The members sitting a distance of $2$ units from Dir II (Pos 1) are the Chairman (Pos 3) and Vice Chairman (Pos 5).

b. The Vice Chairman is at position 5. The person sitting exactly opposite (3 seats away) is at position 2, which is Director I.

c. The Chairman is at position 3 and the Vice Chairman is at position 5. The person sitting exactly between them is the CEO (Pos 4). If the CEO leaves, there will be a gap of one seat between them.

d. If Director I (Pos 2) swaps seats with the Vice Chairman (Pos 5), the Vice Chairman will now be at position 2. Because the Chairman is at position 3, the Vice Chairman is now sitting to the immediate left (clockwise) of the Chairman.

Practice Exercise

1. 80 laborers, working 7 hours a day can finish a piece of work in 18 days. If the labourers work 6 hours a day, then the number of labourers to finish the same piece of work in 30 days, will be:

(a) 15 (b) 21 (c) 22 (d) 25

Solution: (b) 21

Note: The original question has a typo and should state 30 laborers instead of 80 to match the provided options. Assuming $M_{1} = 30$ :

Using the work equivalence formula: $M_{1} \times D_{1} \times H_{1} = M_{2} \times D_{2} \times H_{2}$

$30 \times 18 \times 7 = M_{2} \times 30 \times 6$

$M_{2} = \frac{30 \times 18 \times 7}{30 \times 6} = \frac{18 \times 7}{6} = 3 \times 7 = 21$

2. If 3/5 of a cistern is filled in 1 minute, how much more time will be required to fill the rest of it?

(a) 30 sec (b) 40 sec (c) 36 sec (d) 24 sec

Solution: (b) 40 sec

$\frac{3}{5}$ of the cistern takes $60$ seconds.

The remaining portion to fill is $1 - \frac{3}{5} = \frac{2}{5}$ .

Time required $= \frac{\frac{2}{5}}{\frac{3}{5}} \times 60 = \frac{2}{3} \times 60 = 40$ seconds.

3. In a camp, 95 men had provisions for 200 days. After 5 days, 30 men left the camp. For how many days will the remaining food last now?

(a) 180 (b) 285 (c) 139 16/19 (d) None of these

Solution: (b) 285

After $5$ days, the remaining food is enough for $95$ men for $195$ days.

Total remaining man-days $= 95 \times 195$ .

$30$ men left, meaning $95 - 30 = 65$ men remain.

Days the food will last $= \frac{95 \times 195}{65} = 95 \times 3 = 285$ days.

4. A contractor undertook to do a certain piece of work in 9 days. He employed certain number of men, but 6 of them being absent from the very first day, the rest could finish the work in 15 days. The number of men originally employed were:

(a) 12 (b) 15 (c) 18 (d) 24

Solution: (b) 15

Let the original number of men be $M$ . Total work $= M \times 9$ .

Because $6$ men were absent, $(M - 6)$ men completed it in $15$ days.

$(M - 6) \times 15 = 9 M$

$15 M - 90 = 9 M \Rightarrow 6 M = 90 \Rightarrow M = 15$ .

5. A can finish a work in 18 days and B can do the same work in 15 days. B worked for 10 days and left the job. In how many days, A alone can finish the remaining work?

(a) 5 (b) 5 1/2 (c) 6 (d) 8

Solution: (c) 6

B's 1-day work is $\frac{1}{15}$ . In $10$ days, B finishes $\frac{10}{15} = \frac{2}{3}$ of the work.

Remaining work $= 1 - \frac{2}{3} = \frac{1}{3}$ .

A takes $18$ days to finish the entire work. Time taken by A to finish $\frac{1}{3}$ of the work $= 18 \times \frac{1}{3} = 6$ days.

6. A can finish a work in 24 days, B in 9 days and C in 12 days. B and C start the work but are forced to leave after 3 days. The remaining work was done by A in:

(a) 5 days (b) 6 days (c) 10 days (d) 10 days

Solution: (c) 10 days

1-day work of B and C together $= \frac{1}{9} + \frac{1}{12} = \frac{4 + 3}{36} = \frac{7}{36}$ .

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 takes $24$ days for the whole work. Time taken by A to finish $\frac{5}{12}$ of the work $= 24 \times \frac{5}{12} = 10$ days.

7. A man can reach a certain place in 30 hours. If he reduces his speed by 1/15th, he goes 10 km less in that time. Find his actual speed.

(a) 4 km/hr (b) 5 km/hr (c) 5 1/2 km/hr (d) 6 km/hr

Solution: (b) 5 km/hr

Let original speed be $v$ km/hr. Distance traveled $= 30 v$ .

New speed $= v - \frac{v}{15} = \frac{14 v}{15}$ .

Distance traveled at new speed $= 30 \times (\frac{14 v}{15}) = 28 v$ .

Difference in distance $= 30 v - 28 v = 2 v = 10 \Rightarrow v = 5$ km/hr.

8. With a uniform speed a car covers the distance in 8 hours. Had the speed been increased by 4 km/hr, the same distance could have been covered in 7 1/2 hours. What is the distance covered?

(a) 420 km (b) 480 km (c) 640 km (d) None of these

Solution: (b) 480 km

Let original speed be $v$ . Distance $d = 8 v$ .

If speed is $v + 4$ , time is $7.5$ hours. Distance $d = 7.5 (v + 4)$ .

$8 v = 7.5 v + 30 \Rightarrow 0.5 v = 30 \Rightarrow v = 60$ km/hr.

Distance $d = 8 \times 60 = 480$ km.

9. A thief is noticed by a policeman from a distance of 200 m. The thief starts running and the policeman chase him. The thief and the policeman run at the rate of 10 kmph and 11 kmph respectively. What is the distance between them after 6 minutes?

(a) 100 m (b) 150 m (c) 190 m (d) 200 m

Solution: (a) 100 m

Relative speed $= 11 - 10 = 1$ km/hr $= \frac{1000}{60}$ m/min $= \frac{50}{3}$ m/min.

Distance covered relatively in $6$ minutes $= (\frac{50}{3}) \times 6 = 100$ meters.

Remaining distance between them $= 200 - 100 = 100$ m.

10. A train overtakes two persons walking along a railway track. The first one walks at 4.5 km/hr and the other one walks at 5.4 km/hr. The train needs 8.4 and 8.5 seconds respectively to overtake them. What is the speed of the train if both the persons are walking in the same direction as the train?

(a) 66 km/hr (b) 72 km/hr (c) 78 km/hr (d) 81 km/hr

Solution: (d) 81 km/hr

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

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

$L = (v - 1.25) \times 8.4$ and $L = (v - 1.5) \times 8.5$

$8.4 v - 10.5 = 8.5 v - 12.75 \Rightarrow 0.1 v = 2.25 \Rightarrow v = 22.5$ m/s.

Train speed in km/hr $= 22.5 \times \frac{18}{5} = 81$ km/hr.

11. Two stations A and B are 110 km apart on a straight line. One train starts from A at 7 a.m. and travels towards B at 20 kmph. Another train starts from B at 8 a.m. and travels towards A at a speed of 25 kmph. At what time will they meet?

(a) 9 a.m. (b) 10 a.m. (c) 10.30 a.m. (d) 11 a.m.

Solution: (b) 10 a.m.

By $8$ a.m., train from A travels $20$ km. Remaining distance $= 110 - 20 = 90$ km.

Relative speed $= 20 + 25 = 45$ kmph.

Time to meet $= \frac{90}{45} = 2$ hours after $8$ a.m. $= 10$ a.m.

12. Two trains, one from Howrah to Patna and the other from Patna to Howrah, start simultaneously. After they meet, the trains reach their destinations after 9 hours and 16 hours respectively. The ratio of their speeds is:

(a) 2:3 (b) 4:3 (c) 6:7 (d) 9:16

Solution: (b) 4:3

Ratio of speeds is given by $\frac{v_{1}}{v_{2}} = \sqrt{\frac{t_{2}}{t_{1}}}$ .

$\frac{v_{1}}{v_{2}} = \sqrt{\frac{16}{9}} = \frac{4}{3}$ .

13. At what time between 5 and 6 O'clock are the hands of a clock together?

(a) 27 3/11 (b) 11 3/11 (c) 220/11 (d) None of these

Solution: (a) 27 3/11 minutes past 5

Using $30 H - \frac{11}{2} m = 0$ where $H = 5$ .

$150 = \frac{11}{2} m \Rightarrow m = \frac{300}{11} = 27 \frac{3}{11}$ minutes.

14. At what time between 5 and 6 O'clock will the hands of a clock be at right angle?

(a) 43 7/11 (b) 43 7/12 (c) 43 11/7 (d) None of these

Solution: (a) 43 7/11 minutes past 5

Using $| 30 H - \frac{11}{2} m | = 90$ where $H = 5$ .

Case 1: $150 - 5.5 m = 90 \Rightarrow 5.5 m = 60 \Rightarrow m = \frac{120}{11} = 10 \frac{10}{11}$ mins.

Case 2: $5.5 m - 150 = 90 \Rightarrow 5.5 m = 240 \Rightarrow m = \frac{480}{11} = 43 \frac{7}{11}$ mins.

15. Find at what time between 2 and 3 O'clock will the hands of a clock be in the same straight line but not together.

(a) 43 7/11 (b) 45 7/11 (c) 485/11 (d) None of these

Solution: (a) 43 7/11 minutes past 2

Angle $= 180^{\circ}$ , $H = 2$ .

$5.5 m - 30 (2) = 180 \Rightarrow 5.5 m - 60 = 180 \Rightarrow 5.5 m = 240$ .

$m = \frac{480}{11} = 43 \frac{7}{11}$ minutes.

16. Find the time between 4 & 5 O'clock when the two hands of a clock are 4 minutes apart.

(a) 17 5/11 (b) 17 6/11 (c) 17 1/11 (d) None of these

Solution: (a) 17 5/11 minutes past 4

$4$ minutes apart equals $4 \times 6^{\circ} = 24^{\circ}$ . $H = 4$ .

$| 30 (4) - 5.5 m | = 24 \Rightarrow 120 - 5.5 m = 24$ (for hands before the hour hand).

$5.5 m = 96 \Rightarrow m = \frac{192}{11} = 17 \frac{5}{11}$ minutes.

17. Find the angle between the two hands of a clock at 15 minutes past 4 O'clock.

(a) 37.5° (b) 38° (c) 38.5° (d) 42°

Solution: (a) 37.5°

$A = | 30 (4) - \frac{11}{2} (15) | = | 120 - 82.5 | = {37.5}^{\circ}$ .

18. 11 January 1997 was a Sunday, What day of the week was on 7 January 2000?

(a) Sunday (b) Saturday (c) Friday (d) Thursday

Solution: (b) Saturday

11 Jan 1997 to 11 Jan 1998 = $1$ odd day.

11 Jan 1998 to 11 Jan 1999 = $1$ odd day.

11 Jan 1999 to 11 Jan 2000 = $1$ odd day (2000 is leap, but we don't cross Feb 29 yet).

11 Jan 2000 is 3 days after Sunday = Wednesday.

7 Jan 2000 is $4$ days before 11 Jan 2000. Wednesday minus 4 days is Saturday.

19. What day of the week was on 5 June 1999?

(a) Saturday (b) Sunday (c) Monday (d) Tuesday

Solution: (a) Saturday

Completed years $= 1998$ . Odd days in 1600(0) + 300(1) + 98 ( $24$ leap + $74$ ord).

$24 \times 2 + 74 \times 1 = 122 ≣ 3$ odd days. Total year odd days $= 1 + 3 = 4$ .

Months in 1999: Jan(3)+Feb(0)+Mar(3)+Apr(2)+May(3)+Jun(5) $= 16 ≣ 2$ odd days.

Total odd days $= 4 + 2 = 6 \Rightarrow$ Saturday.

20. At what time between 3 and 4 O'clock are the hands of a clock together?

(a) 5 7/11 min past 4 (b) 16 4/11 min past 3 (c) 16 2/11 min past 2 (d) None of these

Solution: (b) 16 4/11 minutes past 3

$30 (3) - 5.5 m = 0 \Rightarrow 5.5 m = 90 \Rightarrow m = \frac{180}{11} = 16 \frac{4}{11}$ minutes.

21. At what time between 5 and 6 O'clock are the hands of a clock 3 minutes apart?

(a) 24 minutes past 5 (b) 22 minutes past 3 (c) 26 minutes past 4 (d) None of these

Solution: (a) 24 minutes past 5

$3$ minutes apart $= 3 \times 6^{\circ} = 18^{\circ}$ . $H = 5$ .

$| 150 - 5.5 m | = 18$

Case 1: $150 - 5.5 m = 18 \Rightarrow 5.5 m = 132 \Rightarrow m = 24$ minutes.

22. Find the angle between the two hands of a clock at 30 minutes past 4 O'clock.

(a) 40° (b) 30° (c) 45° (d) None of these

Solution: (c) 45°

$A = | 30 (4) - 5.5 (30) | = | 120 - 165 | = 45^{\circ}$ .

23. Number of times the hands of a clock are in a straight line everyday is

(a) 44 (b) 24 (c) 42 (d) 22

Solution: (a) 44

A straight line includes when hands are coincident ( $0^{\circ}$ ) and when they are opposite ( $180^{\circ}$ ).

They are coincident $22$ times and opposite $22$ times in a day. Total = $44$ times.

24. A and B together can complete a work in 6 days, and A and C together can do it in 10 days. If A, B, and C can finish the work together in 4 days, in how many days can B and C finish it? (Corrected for logic)

Solution: 30/7 days

$1$ -day work equations:

$A + B = \frac{1}{6}$ , $A + C = \frac{1}{10}$ , $A + B + C = \frac{1}{4}$

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

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

$B + C = \frac{3}{20} + \frac{1}{12} = \frac{9 + 5}{60} = \frac{14}{60} = \frac{7}{30}$ . They will finish it in $\frac{30}{7}$ days.

25. Two trains 150 m and 200 m long run in the same direction at 72 km/hr and 54 km/hr. How long will the faster train take to pass the slower one?

Solution: 70 seconds

Total length $= 150 + 200 = 350$ m.

Relative speed $= 72 - 54 = 18$ km/hr $= 18 \times \frac{5}{18} = 5$ m/s.

Time $= \frac{Distance}{Relative Speed} = \frac{350}{5} = 70$ seconds.

26. A car and a bike start from opposite ends of a 210 km road at the same time. Car at 70 km/hr and bike at 35 km/hr. How long until they meet and at what distance from the car's starting point?

Solution: 2 hours, 140 km

Relative speed $= 70 + 35 = 105$ km/hr.

Time to meet $= \frac{210}{105} = 2$ hours.

Distance from car's starting point $= Car's speed \times Time = 70 \times 2 = 140$ km.

27. A boy walks from A to B at 4 km/hr and returns at 6 km/hr Total time taken is 5 hours. Find distance AB.

Solution: 12 km

Let the distance AB be $d$ .

$\frac{d}{4} + \frac{d}{6} = 5 \Rightarrow \frac{3 d + 2 d}{12} = 5 \Rightarrow \frac{5 d}{12} = 5 \Rightarrow d = 12$ km.

28. A and B undertake to do a piece of work for Rs. 450. A can do it in 20 days and B in 40 days. With the help of C, they finish it in 8 days. How much should C be paid?

Solution: Rs. 180

1-day work of A $= \frac{1}{20}$ , B $= \frac{1}{40}$ . 1-day work of (A+B+C) $= \frac{1}{8}$ .

C's 1-day work $= \frac{1}{8} - (\frac{1}{20} + \frac{1}{40}) = \frac{1}{8} - \frac{3}{40} = \frac{2}{40} = \frac{1}{20}$ .

Ratio of work efficiencies (A:B:C) $= \frac{1}{20} : \frac{1}{40} : \frac{1}{20} = 2 : 1 : 2$ .

C's share $= \frac{2}{5} \times 450 = 180$ .

29. A Corporate firm is running in loss, they had categorized their employees into two teams, Team A and Team B. The firm has funds to pay Team A for 21 days only. If they work with Team B then the firm can pay for 28 days only. Management has decided that they will allow both the teams to work for as long as they are having enough funds to pay them. Find out for how many days both the teams could work together?

Solution: 12 days

Let total funds be $F$ .

Cost per day for Team A $= \frac{F}{21}$ . Cost per day for Team B $= \frac{F}{28}$ .

Cost per day for both together $= \frac{F}{21} + \frac{F}{28} = \frac{4 F + 3 F}{84} = \frac{7 F}{84} = \frac{F}{12}$ .

Number of days funds will last for both teams $= \frac{F}{\frac{F}{12}} = 12$ days.