At a food processing and packaging plant, identical packets of confectionary items are prepared at a constant rate by a packaging machine. Each packet contains 3 chocolates, 4 candied nuts and 2 chewing gums. The production rate of chocolates, candied nuts and chewing gums at the plant is 750, 975 and 575 units per hour respectively. All the produced items are immediately sent to a common quality checking machine that can process 3000 units of confectionary items per hour. 20 percent each of chocolates, candied nuts and chewing gums fail the quality check. The remaining items are sent to the packaging plant without any holdup or delay. Which of the following can represent the number of packets of confectionary items prepared by the packaging machine in one hour?
I. 195
II. 230
III. 280
Given:
To Find: Can the number of packets prepared by the packaging machine in 1 hour be {195, 230, 280}?
Approach:
Working out:
Operating at their respective constant rates, Photocopying machine B takes 6 minutes more than photocopying machine A to copy x pages. When machines A and B are operated simultaneously, 7x pages can be copied in 20 minutes. In how many minutes can machine A operating alone copy 2x pages?
Given:
To Find: Time taken by A alone to copy 2x pages
Approach:
1. Time taken by A alone to copy 2x pages
= 2x / A’s rate = 2x / x / a = 2a minutes
2. We’re given the combined output of A and B and the time taken for this output. Using this data, we can find the combined rate of A and B.
Then, we’ll use the equation (Combined Rate of A and B) = (Rate of A) + (Rate of B) to get an equation in terms of a.
Working out:
Looking at the answer choices, we see that the correct answer is Option B
Peter, Mark and John work in the marketing department of an elearning firm. Peter can answer 25 emails in an hour, Mark can answer 40 percent more emails in an hour than Peter and John takes 1/6 lesser time to answer the same number of emails as Peter. By what percentage should Peter increase his speed so that the three employees can together answer 1/6 more emails than they currently do?
Given:
To Find: Percentage by which Peter should increase his speed so that all 3 can answer 1 / 6th more emails than they currently do?
Approach:
1. (Percentage by which Peter should increase his speed) = Extra number of emails that he needs to answer per hour / Current number of emails answered per hour * 100
2. So, to find the percentage by which Peter should increase his speed, we should first find the extra number of emails that he needs to answer.
3. For finding the extra number of emails that he needs to answer, we need to find the total number of emails currently answered by Peter, John and Mark together in 1 hour and number of increased emails they need to answer together.
Working out:
So, Peter should increase his speed by 60%
If John produces n units of an item in t hours, what is the value of n?
(1) If John produces 4 more items per hour, he would take 1/4 lesser time to produce n units of the item.
(2) If John increases his rate of production by 25 percent, the time taken by him to produce n items would be 80 percent of the original time.
Step 1 & 2: Understand Question and Draw Inference
To Find: Value of n
Step 3 : Analyze Statement 1 independent
(1) If John produces 4 more items per hour, he would take 1 / 4 th lesser time to produce n units of the item.
n = 3t / 4 * (n / t + 4)
n = 3n / 4 + 3t
n / 4 = 3t
n = 12t
However, we do not know the value of t, hence we cannot find the value of n.
Insufficient to answer
Step 4 : Analyze Statement 2 independent
(2) If John increases his rate of production by 25 percent, the time taken by him to produce n items would be 80 percent of the original time.
n = 0.8 t * (1.22 * n / t)
n = (1.25 * 0.8) n
1 = 1
Thus, there is no new information given. Insufficient to answer
Step 5: Analyze Both Statements Together (if needed)
1. n = 12t
2. No new information given
As statement2 does not provide us any new information, the combination of the statements is insufficient to answer the question.
Transmitting data at the same constant rate, 12 identical cables can transmit x bits of digital information per minute. How many bits of information can be transmitted by 27 such cables in 25 seconds?
Given:
To Find: Number of bits (in terms of x) transmitted by 27 identical cables in 25 seconds
Approach:
Working out:
Looking at the answer choices, we see that the correct answer is Option A
The water from outlet A, flowing at a constant rate, can fill a rectangular swimming pool in 12 hours while the water from outlet B, flowing at a constant rate, can fill the same swimming pool in 18 hours. On a day when the pool was empty, outlet A was opened at 7 AM and outlet B was opened at 10 AM. Outlet A was closed at 2 PM and outlet B was closed at 4 PM. If there was no water flow from outlet A between 10:30 AM to 11:30 AM because of maintenance work, the height of the water in the pool at 4 PM was approximately what percentage of the depth of the pool?
Given:
To Find: Percentage of swimming pool that is filled?
Approach:
1. Percentage of swimming pool filled = Percentage of swimming pool filled by outlet A + Percentage of swimming pool filled by B
a. Percentage of swimming pool filled by outlet A = Rate of outlet A * Time for which outlet A was opened / Total volume of swimming pool
b. As outlet A working alone can fill the entire swimming pool in 12 hours, we can write the volume of swimming pool = Rate of outlet
A * 11
c. So, The Percentage of swimming pool filled by outlet A = Rate of outlet A * Time for which outlet A was opened / Total volume of swimming pool
d. So, The Percentage of swimming pool filled by outlet B = Rate of outlet B * Time for which outlet B was opened / Total volume of swimming pool
e. As outlet B working alone can fill the entire swimming pool in 18 hours, we can write the volume of swimming pool = Rate of outlet
A * 18
f. So, The Percentage of swimming pool filled by outlet B = Rate of outlet B * Time for which outlet B was opened / Total volume of swimming pool
2. As, we are given the time for which outlets A and B were open, we can find the percentage of swimming pool filled by outlet A and B
Working out:
a. Time at which outlet A was opened = 7 AM
b. Time for which outlet A was closed = 2 PM
c. However, outlet did not work for 1 hour between 10:30 AM  11:30 AM
d. Hence outlet A worked for 7 – 1 = 6 hours
e. So, percentage of swimming pool filled by outlet A in 6 hours = 6/12 = 50% of the swimming pool
2. Outlet B
a. Time at which outlet B was opened = 10 AM
b. Time for which outlet B was closed = 4 PM
c. Time for which outlet B worked = 4 PM – 10 AM = 6 hours
d. So, percentage of swimming pool filled by outlet B in 6 hours = 6/18 = 33% of the swimming pool
3. Therefore, outlet A and B working for 6 hours each can fill (50% +33%) = 83% of the entire swimming pool
If 30 machines, each working at the same constant rate, together can produce n/2 units of an item in 15 days, how many machines will be needed to produce 2n units of the item in 45 days?
Given:
To Find: Number of machines needed to produce 2n units of the item in 45 days
Approach:
Working out:
1. Work done by 30 machines = n / 2
2.Time taken = 15 days
3. So, we can write n / 2 = r * 15 * 30
n / r = 900
4. Substituting the value of n / 2 in equation (1), we have
X = 2 / 45 * 900 = 40
5. So, it will take 40 machines to produce 2n units of the item in 45 days
A rectangular tank has the dimensions of its base as 30 metres by 20 metres and a height of 10 metres. There are two taps attached to the tank such that each tap working alone at a constant rate can fill the tank completely in 60 hours and 90 hours respectively. One of the walls of the tank has holes along the height of the tank at a regular distance of 2 metres and the first such hole is 2 metres above the base of the tank. The rate of water outflow from each hole is 10m per hour. If both the taps are opened simultaneously in the empty tank, approximately how many hours will it take to fill the tank completely?
To Find: Time taken to fill the empty tank completely if both the taps are opened?
Approach:
Working out:
1. Total work to do done = Total volume to be filled = Volume of tank = 30 *20 *10 = 6000 m^{3}
2. Rate of Inflow
a. Time taken by tapI to fill the entire tank = 60 hours
i. Rate of water inflow from tapI = 6000 / 60 = 100 m^{3}per hour
b. Time taken by tapI to fill the entire tank = 90 hours
i. Rate of water inflow from tapI = 6000 / 90 = 200 / 3 m^{3}per hour
c. Total rate of water inflow = 100 + 200 / 3 = 500 / 3 m^{3}per hour
3. Time Taken :
The time taken to fill the tank up to 2 metres of height will be
= Vol. of water needed / rate of water filled by the taps = 30 x 20 x 2 / 500 / 3 = 36 / 5 ~7 hours
For the next 2 metres of height, the time taken would be
= 30 x 20 x 2 / 500 / 3 – 10 = 30 x 20 x 2 x 3 / 470 = 360 / 47 ~ 360 / 45 ~ 8 hours
For the next 2 metres of height, the time taken would be
= 30 x 20 x 2 / 500 / 3 – 10 = 30 x 20 x 2 x 3 / 440 = 90 / 11 ~ 8 hours
4 identical machines, which operate at the same constant rate, together started manufacturing the batch of an item at 2 PM. Did
they finish the batch before 3.30 PM?
(1) The batch size was between 200 and 210 units, inclusive, of the item
(2) One machine could manufacture between 30 and 50 units, inclusive, of the item in 1 hour
Step 1 & 2: Understand Question and Draw Inference
Given: Number of machines = 4
Let the rate of production for each machine = R units per hour
Let the batch size be B units
To Find: Did the machines finish the batch before 3:30 PM
Time from 2 PM (start time) to 3:30 PM
= 1 hour 30 minutes = 1 + 1 / 2 hours = 3 / 2 hours
So, we need to find: Is B / 4R < 3 / 2?
That is, Is B / R < 6
Step 3 : Analyze Statement 1 independent
Statement 1 says that ‘The batch size was between 200 and 210 units, inclusive
of the item’
Step 4 : Analyze Statement 2 independent
Statement 2 says that ‘One machine could manufacture between 30 and 50 units, inclusive, of the item in 1 hour’
Step 5: Analyze Both Statements Together (if needed)
B / R = Minimum possible value of B / Maximum possible value of R = 200 / 50 = 4
And, the maximum possible value of
B / R = Maximum possible value of B / Minimum possible value of R = 210 / 30 = 4
A container is completely filled with a sugar solution composed of water and sugar syrup in the ratio of 7:3. The container has 2 holes covered with filters such that from one of the holes only sugar syrup can flow out and from the other hole only water can flow out. The rates of water outflow and sugar syrup outflow from their respective holes are x cubic centimeters per hour and y cubic centimeters per half an hour respectively, such that x:y = 5:1. If the water from the solution can be drained out completely in 14 hours from the hole, how much time in hours would it take the container to be empty if both the holes are opened simultaneously?
Given:
To Find: Time taken for container to be empty, if both the holes are opened simultaneously?
Approach:
a. Time taken for water to drain out completely = Volume of water / Rate of water outflow = 14 hours
Working out:
1. Time taken for water to drain out completely = Volume of water / Rate of water outflow = 14 hours
a. 0.7S / = 14
b. S / x = 20
2. As x : y = 5 : 1,we can write x = 5y.
a. S / 5y = 20 i.e. S / y = 100
3. Hence, Time taken for sugar syrup to drain completely
= Volume of sugar syrup / Rate of sugar syrup outflow = 0.3 S / 2y = 0.3 / 2 * 100 = 15 hours
4. So, let’s compare the time taken for water and sugar syrup to drain out completely
a. Time taken for water to drain out completely = 14 hours
b. Time taken for sugar syrup to drain out completely = 15 hours
5. Hence, the time taken for the container to be empty = Max( 14, 15) = 15 hours
Therefore, the container would be completely empty in 15 hours
An automated manufacturing plant uses robots to manufacture products. A generationI robot working alone can manufacture a product in 30 hours whereas a generationII robot working alone can manufacture the same product in 20 hours. If the manager of the plant wants the product manufacturing time to be greater than equal to 5 hours with at least 1 robot each of both generations working together, how many possible combinations of the number of robots of each generation is possible?
Given:
Approach:
Working out:
GenerationI
a. Work Done = 1 product
b. Time taken = 30 hours
c. Rate of manufacturing products = products per hour
GenerationII
a. Work Done = 1 product
b. Time taken = 20 hours
c. Rate of manufacturing products = 1/20 products per hour
3. Let’s assume there are x generationI robots and y generationII robots working together to produce 1 unit of product
a. So, work done = 1 product
b. Rate of x generationI robots =
X * 1 / 30 = x / 30 products per hour
c. Rate of y generationII robots =
y * 1 / 20 = y / 20 product per hour
d. So, time taken = 1 / x / 30 + y / 30 hours
4. As the time taken ≥ 5 hours, we can write
1 / x + 30 + y / 30 ≥ 5
60 / 2x + 3y ≥ 5
2x + 3y ≤ 12
5. So, the possible combinations of {x, y} can be:
a. If y = 1, x = {1,2,3,4} – 4 options
b. If y = 2, x = {1,2,3} – 3 options
c. If x = 1, y = {1,2,3} – 3 options
d. However {x, y} = {1,1} and {1,2} is repeated.
6. So, we have a total of 10 2 = 8 possible options
a. Please note that we have not considered x, y = 0 as we are given that atleast 1 robot of each generation should be working
Answer : E
Two machines A and B worked at variable rates. The average number of items produced per day by machine A could be expressed as R_{a} (x) = ax(x + 1) and the average number of items produced per day by machine B could be expressed as R_{b} (x) = bx(x1), where x was the number of consecutive days worked and a, b were constants. If machine B produced 90 more items when it worked for 4 consecutive days than when it worked for to 3 consecutive days and machine A produced a total of 232 items when it worked in two intervals of 3 consecutive days and 4 consecutive days, how many items was produced by machines A and B working together for 10 consecutive days?
Given:
To Find: Number of items produced by machine A and B working together for 10 consecutive days
Approach:
Working out:
Answer : C
John starts reading a book from the first page onwards and reads each page only once. Every day, he reads twice the number of pages read by him the previous day. If the book has 400 pages, will John still be reading the book on the 6 day after he starts reading it?
(1) On the fifth day of reading the book, John reads more than 180 pages
(2) John reads 52 more pages on the fourth day of reading the book than he reads on the third day
Step 1 & 2: Understand Question and Draw Inference
To Find: Is 31x ≥ 400 ?
Step 3 : Analyze Statement 1 independent
Step 4 : Analyze Statement 2 independent
(2) John reads 52 more pages on the fourth day of reading the book than he reads on the third day
4x = 52, i.e. x = 13
Therefore, the answer to the question
‛Is x > 12 + 28 / 31 ?’ is YES
Sufficient to answer.
Step 5: Analyze Both Statements Together (if needed)
As we have a unique answer from step4, this step is not required.
Answer: B
Two machines A and B work at constant rates which are in the ratio of 1:2 respectively. If the two machines work together, they can complete 1/3rd of a job in 3 hours. If machine A starts working on the job and works alone for 18 hours, how many hours would machine B working alone take to complete the remaining work?
Given:
To Find: Time taken by machine B to complete the remaining work
Approach:
1. We know that Work = Rate * Time
a. So, Remaining work = Rate of machine B * Time taken
b. Therefore, Time taken = Remaining Work / Rate of machine B
c. So, to find the time taken, we either need to find the amount of remaining work and the rate of machine B or the ratio of
Remaining Work / Rate of machine B
2. For finding the remaining work, we need to find the amount of work done by machine A in 18 hours
3. We are given that machines A and B working together take 3 hours to complete W/3 units of work.
Working out:
a. W / 3 = (x + 2x) * 3, i.e. W = 27x
2. Amount of work done by machine A working alone for 18 hours = x * 18 = 18x
3. Remaining work = W – 18x = 27x – 18x = 9x
4. Amount of time taken by machine B to complete the remaining work =
9x / 2x = 4.5 hours
Answer : C
Machines M and N produce identical widgets but at different constant rates. Machine M takes 10 hours to produce 1000 widgets and Machines M and N together take 6 hours and 40 minutes to produce 2000 widgets. How much time does Machine N take to produce 3000 widgets?
Given:
To Find: Time taken by N to produce 3000 widgets
Approach:
1. Required Time = 3000 / n hours
2. We are given the Combined output of M and N in 6 hours 40 minutes. From this piece of information, we can find the combined Rate of M and N
a. Combined Rate of M and N = m + n
b. So, n = (Combined Rate of M and N) – m
c. Therefore, we need to find not only the Combined Rate of M and N, but also the value of m
3. We’re given the output of M alone in 10 hours. From this piece of information, we can find the value of m.
Working out:
= Combined output of M and N / Time taken (in hours)
= 2000 widgets / 6 hours 40 minutes
= 6 hours 40 minutes = 6 + 40 / 60 hours = 6 + 2 / 3 hours
= 20 / 3 hours
So, Combined Rate of M and N = 2000 / 20 / 3 = 300 widgets per hour
Finding m
m (in widgets per hour) = Output of Machine M / Time taken (in hours)
= 1000 widgets / 10 hours
= 100 widgets per hour
Find the Required Time
Required Time = 3000 / 200 hours
= 15 hours
Looking at the answer choices, we see that the correct answer is Option E
A rectangular reservoir is filled with water till onefifth of the height of the reservoir. If an outlet at the bottom of the reservoir is unplugged, the water in the reservoir will drain completely in 1 hour. If the outlet remains plugged and the inlet tap at the head of the reservoir is opened, the reservoir will fill to the brim in 2 hours. In how much time will the reservoir fill to the brim if the outlet is unplugged 30 minutes after the inlet tap is opened?
Given:
S, rate of draining =
Volume to be drained / Time for draining = V / 5 volume units per hour
So, rate of filling =
Volume to be filled / Time for filling = 4V / 5 / 2 = 2V / 5 volume units per hour
To Find: Time taken to fill the reservoir if outlet is unplugged 30 minutes after inlet is opened
Approach:
1. Time taken(to go from V / 5 volume of water to V volume of water in the reservoir) = 30 minutes +
Volume that remains to be filled after 30minutes of inlet / Nat Rate of filling
Working out:
= 4V / 5 – (Rate of filling) * (30 / 60 hour)
= 4V / 5 – (2V / 5) (1 / 2)
= 4V / 5 – V / 5
= 3V / 5 volume units
Finding the Net Rate of filling
Rate of filling = 2V / 5 volume units per hour
Rate of Draining = V / 5 volume units per hour
So, Net Rate of filling =
2V / 5 – V / 5 volume units per hour = V / 5 volume units per ho
Total time taken = 30 minutes + 3V / 5 / V / 5 hours
= 30 minutes + 3 hours
That is, 3 hours 30 minutes
Looking at the answer choices, we see that the correct answer is Option B
On the first day of the launch of an anticipated electronic item, a queue formed outside an exclusive electronics store that sold the item. At the time the store opened, the queue had 60 people in it and throughout the day, a new person joined the queue every 3 minutes. That day, the store served, only the people in the queue, at a constant service rate of 30 people per hour. If no person rejoined the queue upon getting served once, how many people were in the queue 4 hours after the store opened?
Given:
To Find: Number of people in the queue 4 hours after the store opened
Approach:
Working out:
Looking at the answer choices, we see that the correct answer is Option A
A is thrice as good as B in work. A is able to finish a job in 60 days less than B. They can finish the work in  days if they work together.A.
If A completes a work in 1 day, B completes the same work in 3 days
Hence, if the difference is 2 days, B can complete the work in 3 days
=> if the difference is 60 days, B can complete the work in 90 days
=> Amount of work B can do in 1 day= 1/90
Amount of work A can do in 1 day = 3 × (1/90) = 1/30
Amount of work A and B can together do in 1 day = 1/90 + 1/30 = 4/90 = 2/45
=> A and B together can do the work in 45/2 days = 22 ½ days
A can do a particular work in 6 days . B can do the same work in 8 days. A and B signed to do it for Rs. 3200. They completed the work in 3 days with the help of C. How much is to be paid to C?A.
Amount of work A can do in 1 day = 1/6
Amount of work B can do in 1 day = 1/8
Amount of work A + B can do in 1 day = 1/6 + 1/8 = 7/24
Amount of work A + B + C can do = 1/3
Amount of work C can do in 1 day = 1/3  7/24 = 1/24
work A can do in 1 day: work B can do in 1 day: work C can do in 1 day
= 1/6 : 1/8 : 1/24 = 4 : 3 : 1
Amount to be paid to C = 3200 × (1/8) = 400
6 men and 8 women can complete a work in 10 days. 26 men and 48 women can finish the same work in 2 days. 15 men and 20 women can do the same work in  days.
Let work done by 1 man in 1 day = m and work done by 1 woman in 1 day = b
Work done by 6 men and 8 women in 1 day = 1/10
=> 6m + 8b = 1/10
=> 60m + 80b = 1  (1)
Work done by 26 men and 48 women in 1 day = 1/2
=> 26m + 48b = ½
=> 52m + 96b = 1 (2)
Solving equation 1 and equation 2. We get m = 1/100 and b = 1/200
Work done by 15 men and 20 women in 1 day
= 15/100 + 20/200 =1/4
=> Time taken by 15 men and 20 women in doing the work = 4 days
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