Dear CAT aspirants, if you are planning to prepare for the Number System, this article is created especially for you. Here, you will find the key properties of numbers that cover a major part of the syllabus and help you solve around 70% of the questions from this topic in the CAT Quantitative Aptitude section. In CAT 2025, you can expect 3-4 questions from Number System, and mastering these properties will give you a clear advantage. In this article, careers360 will provide you 5 most important properties of numbers for CAT that covers 70% of CAT number system questions.
This Story also Contains
Number System Topics
Property 1: Divisibility Rules
Property 2: Factors & Multiples
Property 3: Even–Odd & Prime–Composite Behaviour
Property 4: Remainder theorem (Short cut to find remainder)
Property 5: Digit Sum Properties
Why These Properties Work
Key Strategies for CAT number system preparation
Best resources & practice material for CAT Number System
CAT 2025 Preparation Resources by Careers360
CAT 2025 Number System: 5 Key Properties to Solve 70% of Questions in QA
Do you know
Importance of numbers in CAT QA?
What are the CAT number system shortcuts?
What are the CAT number system preparation strategies?
how to solve number system questions in CAT?
This article is going to help you to get your answers……
Number System Topics
The CAT Number System topics from which most of the questions are asked in exams are:
Sum the digits. If the result is divisible by 3, then the original number is divisible by 3.
4359 → 4 + 3 + 5 + 9 = 21 which is divisible by 3.
4
Examine the last two digits.
40864: 64 is divisible by 4. So, 40864 is divisible by 4.
5
The last digit is 0 or 5.
685: the last digit is 5.
4590: last digit is 0.
Both numbers are divisible by 5.
6
It is divisible by 2 and by 3.
1,458: 1 + 4 + 5 + 8 = 18, so it is divisible by 3 and the last digit is even, hence the number is divisible by 6.
7
A number is divisible by 7 if the difference between twice the unit digit and the number formed by the remaining digits is divisible by 7.
343: 34-2×3=28 which is divisible by 7. Hence, 343 is divisible by 7.
8
A number is divisible by 8 if the number formed by its last three digits is divisible by 8.
34152: Examine divisibility of just 152: 19 × 8
So, the number is divisible by 8.
9
Sum the digits. If the result is divisible by 9, then the original number is divisible by 9.
435429 → 4 + 3 + 5 + 4 + 2 + 9 = 27 which is divisible by 9.
11
A number is divisible by 11 if the difference between the sum of digits at odd places (1st, 3rd, 5th, etc.) and the sum of digits at even places (2nd, 4th, 6th, etc.) is either 0 or divisible by 11.
918,082: 9 − 1 + 8 − 0 + 8 − 2 = 22.
So, the number is divisible by 11.
Application in factor-based questions
Divisibility rules can be applied in finding the factors of a given number directly.
For example:
If a number has either 0 or 5 at unit place, it must have a factor 5.
If a number has an even digit at unit place, it must have a factor 2.
If the sum of the digits of the number is divisible by 3, it must have a factor 3.
Similarly, we can apply the divisibility rules of 6, 9, 11 etc.
If you need to find the prime factorization, use the divisibility rule of prime numbers.
Property 2: Factors & Multiples
In CAT, concept of factors and multiples are used to
Multiple: A multiple of a number is the result of multiplying that number by another whole number.
Example: Multiples of 48 = 48, 96, 144, ……
Important formula for number of factors
1. Number of factors:
Concept/ Formula: For $N=a^p×b^q ×c^r×….$ Where a, b, and c are prime factors. p, q, r are positive integers.
Number of all factors = $(p+1)(q+1)(r+1)….$
$N =1280=2×2×2×2×2×2×2×2×5=2^8×5$
Number of factors = (8 + 1) × (1 + 1) = 18
2. Sum of all factors:
Concept/ Formula: For $N=a^p×b^q ×c^r×….$ Where a, b, and c are prime factors. p, q, r are positive integers.
Sum = $(1+a^1+a^2+⋯+a^p )(1+b^1+b^2+⋯+b^q )(1+c^1+c^2+⋯.+c^r )…..$
$N =1280=2×2×2×2×2×2×2×2×5=2^8×5$
Sum of all factors = $(1+2+2^2+⋯+2^8 )(1+5) =(2^9-1)/(2-1)×6=511×6=3066$
HCF & LCM in problem solving
Highest Common Factor (HCF):
The HCF of two or more numbers is the largest positive integer that divides them without leaving a remainder. It is also known as the Greatest Common Divisor (GCD). HCF is used to solve problems related to dividing objects into smaller equal groups.
Least Common Multiple (LCM):
The LCM of two or more numbers is the smallest multiple that is evenly divisible by all of the given numbers. LCM is useful when solving problems that involve finding the least number of repetitions required for multiple events to synchronise or repeat together.
HCF and LCM of fractions:
HCF (Highest Common Factor) of fractions
The HCF of two or more fractions is the largest fraction that can exactly divide each given fraction. To find the HCF of fractions, we need to find the HCF of the numerators and the LCM of the denominators. We’ll find HCF and LCM of the numbers using:
LCM (Least Common Multiple) of fractions
The LCM of two or more fractions is the smallest fraction that is divisible by each of the given fractions. To find the LCM of fractions, we need to find the LCM of the numerators and the HCF of the denominators.
The concept of even and odd digits in CAT is useful to verify answers of number system questions.
The concept of prime and composite is useful for prime factorization, HCF and LCM questions.
Rules of operations with even & odd numbers
Classification of numbers into Even and Odd Numbers:
Non – negative integers are classified into even and odd numbers.
Odd Numbers: A whole number that is not exactly divisible by 2 is called an odd number.
For example: 3, 5, 7, 9, 11, 13, 15, are odd number.
Or a number having 1, 3, 5, 7, and 9 at its unit’s place is called an odd number.
Even Numbers: A whole number exactly divisible by 2 is called an even number.
For example: 0, 2, 4, 6, 8, 10, 12, 14, 16 .......................... are even number.
Or a number having 0, 2, 4, 6, 8 at its unit’s place is called an even number.
Important Results on Even and Odd Integers:
Prime number patterns in CAT questions
Classification of numbers into Prime and Composite Numbers:
Natural Numbers can be classified as
Prime Number: Prime numbers are those numbers which is greater than I, and have only two factors-1 and the number itself. Prime numbers are divisible only by the number 1 or itself.
Example: 2, 3, 5, 7, and 11 are the first few prime numbers.
Composite Number: Numbers having more than two factors are called composite numbers.
Example: 4, 6, 9, 10, and 15 are a few examples of composite numbers.
Note:
1. Co-Prime Numbers:
A pair of numbers that have exactly one factor in common.
Example:
16 and 25
Factors of 16: 1, 2, 4, 8, 16
Factors of 25: 1, 5, 25
Common factor: 1
2. All prime numbers are co-prime with each other.
3. 2 is the smallest prime number.
4. 2 is the only even prime number.
5. For quick answer, you should remember
Property 4: Remainder theorem (Short cut to find remainder)
The concept of remainders in CAT is used
To determine the remainder of a large number divided by another number.
To find the unit digit and last digit of a number
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Basic Remainder Theorem to find the remainder when a large number is divided by small numbers:
It is better to understand this with the help of examples:
Question: Find the remainder when $6789^{12345}$ is divided by 8.
Solution:
Divide $6789^{12345}$ by 8, remainder will be of the form $5^{12345}$ ≡ $5×5^{12344}≡5×25^{6172}$
Now, divide it by 8, remainder will be 5.
Binomial theorem to find the remainder:
The Binomial Theorem states that for any positive integer n and any real numbers a and b:
[a+bn=k=0n knCan-kbk]
Example: Find the remainder when $27^{10}$ is divided by 5.
Solution:
$27^{10}= (25+2)^{10}$, if we divide this number by 5, the remainder will be $2^{10}$.
Now divide $2^{10}$ by 5 to find the final remainder.
So, when$2^{10}$ divided by 5, remainder = 4
Use in cyclicity and last digit questions
Concept of Cyclicity to find unit digit:
Unit digit cyclicity refers to the repetition pattern of the unit digit (the digit in the one’s place) of a number. To determine the unit digit cyclicity of a number, we need to analyse the cyclicity of its unit digit (also known as the ones digit) raised to various powers.
Formula for Unit Digit Cyclicity:
The unit digit of any number raised to a power follows a specific cyclicity pattern.
To find the unit digit of an where a is the unit digit and n is any natural number.
Digit
Cyclicity
Unit digit
Remark
1
1
1
2
4
The unit digit repeats every four powers (2,4,8,6).
3
4
The unit digit repeats every four powers (3,9,7,1)
4
2
4,6
4 raised to power any even number gives 6 otherwise 4.
5
1
5
6
1
6
7
4
The unit digit repeats every four powers (7,9,3,1)
8
4
The unit digit repeats every four powers (8,4,2,6)
9
2
9,1
9 raised to power any even number gives 1 otherwise 9.
0
1
0
Example 1: $2^1=2,2^2=4,2^3=8,2^4=16,2^5=32,2^6=64$ and so on. The unit digit repeats every four powers (2,4,8,6).
Example 2: $4^1=4,4^2=16,4^3=64,4^4=256$, and so on. The unit digit alternates between 4 and 6.
Rules to determine the last two digits of the number:
Rule 1: For the numbers of the form $x^2 /(50±x)^2 / (100±x)^2…, x^2$ will be the last two digits.
Example: Find the last two digits of $96^2$.
Solution: $96^2 = (100-4)^2$
Last two digits = $4^2=16$
Rule 2: For the numbers having last digit odd except 5.
For (odd)20k, last two digits = 01, k is positive integer.
Example: Find the last two digits of 2342.
Solution: $23^42=23^40×23^2$
Last two digits = 01 × 29 = 29, [Here 29 are the last two digits of 232]
Rule 3: For the numbers having last digit 5.
For (odd5)^odd, last two digits = 75, otherwise 25.
Example: Find the last two digits of 2542 and 7575.
Solution: 2542, last two digits 25, since the digit before 5 is even in 25.
7575, last two digits 75, since the digit before 5 is odd in 75 and the power is odd.
Rule 4: For the numbers having last digit even.
For (Even)^20k, last two digits = 76, k is positive integer.
Rule 1: If the digit-sum of a number is 9, then we can eliminate the 9 straight away and the digit- sum becomes ‘zero.’ Rule 2: While solving problems if the digit sum value comes negative then just add it to 9 & result will be the actual digit sum value.
Application in divisibility and missing digit problems
A number is divisible by 9 if the sum of its digits is divisible by 9.
Example: Check whether 3987657 is divisible by 9.
Sum of the digits = 3 + 9 + 8 + 7 + 6 + 5 + 7 = 45
45 = 4 + 5 = 9 which is divisible by 9. So, the given number is divisible by 9.
Application in Missing Digit Problems:
This technique extends to solving for a missing digit by substituting it into the digit sum equation.
Example: In the number 2345x2 is divisible by 3, find possible values of x.
Digit sum = 2 + 3 + 4 + 5 + x + 2 = 16 + x
So, possible values of x are 2, 5, 8 so that sum of digits is divisible by 3 and hence number will be divisible by 3 for x = 2, 5, 8
Example: For the number 6523678pq, find the smallest possible sum of p and q so that the number is divisible by 9.
The sum of the known digits is 6 + 5 + 2 + 3 + 6 + 7 + 8 + p + q = 37 + p + q. For this number to be divisible by 9, the sum of all digits must be a multiple of 9. So, 37 + p + q must be a multiple of 9. The smallest multiple of 9 greater than 37 is 45, so p + q = 45 - 37 = 8.
Why These Properties Work
These properties work because most of the CAT numbers questions reduce to these 5 concepts.
Concept
Question Type
Divisibility & Factors
Prime factorization
HCF & LCM
Finding number of factors / sum of factors
Divisibility rules through digit sums (9, 11 tests etc.)
Remainders & Modular Arithmetic
Finding last digit / cyclicity
Remainder theorems
Divisibility-based remainder patterns
Base System & Units/Last Digits
Conversion between bases
Identifying units digit / last two digits
Place value logic
Even–Odd, Positives–Negatives & Digit Properties
Sum/product of digits properties
Solution check through even and odd digit properties
Factors and Multiples
Trailing zeros
Number of factors
Sum of factors
CAT number system questions with solutions
Q.1) A school has less than 5000 students and if the students are divided equally into teams of either 9 or 10 or 12 or 25 each, exactly 4 are always left out. However, if they are divided into teams of 11 each, no one is left out. The maximum number of teams of 12 each that can be formed out of the students in the school is:
A) 150
B) 140
C) 145
D) 145
Solution:-
Since the total number of students, when divided by either 9 or 10 or 12 or 25 each, gives a remainder of 4, the number will be in the form of LCM(9,10,12,25)k + 4 = 900k + 4. It is given that the value of (900k + 4) is less than 5000. Also, it is given that (900k + 4) is divided by 11. It is only possible when k = 2 and total students = 1804. So, the number of 12 students group = $\frac{1800}{12}$ = 150 Hence, the correct answer is option (1).
Q.2) When $10^{100}$ is divided by 7, the remainder is
A) 3
B) 4
C) 6
D) 6
Solution:-
We know that $10 \div 7$ leaves a remainder of $3$, so $10^{100}$ will leave the same remainder as $3^{100}$ when divided by $7$ Now, $3^3 = 27$ and $27 \div 7$ leaves remainder $-1$ (since $27 = 28 - 1$) So, $3^{100} = (3^3)^{33} \times 3 = 27^{33} \times 3$ This becomes $(-1)^{33} \times 3 = -3$
So, the remainder is $-3$.
Since we want a positive remainder, add $7$: $-3 + 7 = 4$ The remainder is 4.
Hence, the correct answer is option 2.
Q.3) For some natural number n, assume that (15,000)! is divisible by n!. The largest possible value of n is
A) 4
B) 7
C) 6
D) 6
Solution:-
To find the largest possible value of n, we need to find the value of n such that n! is less than 15000.
$7!=5040$
$8!=40320 > 15000$ This implies 15000! is not divisible by 40320. Therefore, the maximum value n can take is 7. Hence, the correct answer is 7.
Q.4) Minu purchased a pair of sunglasses for Rs.1000 and sold them to Kanu for 20% profit. Then, Kanu sells it back to Minu at a 20% loss. Finally, Minu sells the same pair of sunglasses to Tanu. If the total profit made by Minu from all her transactions is Rs.500, then the percentage of profit made by Minu when she sold the pair of sunglasses to Tanu is
A) 35.42%
B) 31.25%
C) 52%
D) 52%
Solution:-
Initial Cost Price for Minu = 1000 Cost Price to Kanu = 120% of 1000 = 1200 Profit made by Minu = 1200 – 1000 = 200 Again Minu purchased at 80% of 1200 = 960 The total profit made by Manu is 500. So, profit made when she sold it to Tanu = 500 – 200 = 300 Final Selling Price for Minu when sold to Tanu = 960 + 300 = 1260 Profit percent = $\frac{300}{960}$ $\times$ 100 = 31.25%
Hence, the correct answer is option (2).
Q.5) Ankita buys 4 kg cashews, 14 kg peanuts and 6 kg almonds when the cost of 7 kg cashews is the same as that of 30 kg peanuts or 9 kg almonds. She mixes all the three nuts and marks a price for the mixture in order to make a profit of Rs.1752. She sells 4 kg of the mixture at this marked price and the remaining at a 20% discount on the marked price, thus making a total profit of Rs. 744. Then the amount, in rupees, that she had spent in buying almonds is:
A) 1680
B) 1176
C) 2520
D) 2520
Solution:-
It is given,7C = 30P = 9A and Ankita bought 4C, 14P and 6A. Let 7C = 30P = 9A = 630k C = 90k, P = 21k, and A = 70k Cost price of 4C, 14P and 6A = 4(90k)+14(21k)+6(70k) = 1074k Marked up price = 1074k + 1752 $\begin{aligned} & \text { S.P }=\frac{1}{6}(1074 k+1752)+\left(\frac{4}{5}\right)\left(\frac{5}{6}\right)(1074 k+1752)=\frac{5}{6}(1074 k+1752) \\ & \text { S.P-C.P }=\text { profit } \\ & 1460-\frac{1074 k}{6}=744 \\ & \frac{1074 k}{6}=716\end{aligned}$ k = 4 Money spent on buying almonds = 420k = 420 × 4 = Rs 1680 The correct answer is Rs. 1680.
Q.6) Amal buys 110 kg of syrup and 120 kg of juice, syrup being 20% less costly than juice, per kg. He sells 10 kg of syrup at 10% profit and 20 kg of juice at 20% profit. Mixing the remaining juice and syrup, Amal sells the mixture at Rs 308.32 per kg and makes an overall profit of 64%. Then, Amal’s cost price for syrup, in rupees per kg, is:
A) 160
B) 180
C) 220
D) 220
Solution:-
Let the price of juice be Rs. $x$ per kg. Since the cost price of syrup is 20% less than the cost price of juice, the cost price of syrup is $0.8x$ per kg. Total cost price of syrup $= 110 × 0.8x = 88x$ Total cost price of juice $= 120 × x = 120x$ Total cost price $= 88x + 120x = 208x$ Since the overall profit percentage is 64%, Total profit $= 0.64(208x) = 133.12x$ Profit generated by selling 10 kg of syrup which costs Rs $0.8x$ per kg at 10% profit $= 0.1 ×10 ×0.8x = 0.8x$ Profit generated on selling 20 kg of juice which costs Rs $x$ per kg at 20% profit $= 0.2 ×20x = 4x$ The remaining profit $(133.12x - (0.8x + 4x) = 128.32x)$ is generated by selling 100 kg of syrup and 100 kg of juice at Rs 308.32 per kg. The total selling price of 100 kg of syrup and 100 kg of juice is 200 × 308.32 = 2×(30832) The cost price of 100 kg of syrup $= 0.8x ×100 = 80x$ Cost price of 100 kg of juice = $100x$ Total cost price $= 80x + 100x = 180x$ Profit = Selling price – Cost price ⇒ $128.32x = 2×(30832) - 180x$ ⇒ $308.32x = 2×(30832)$ ⇒ $x = 200$ Cost price of syrup per kg = 0.8 × (200) = Rs 160 Hence, the correct answer is option (1).
Key Strategies for CAT number system preparation
We suggest you 5 step key strategy to prepare numbers for CAT:
Step 1: Understand the weightage and question type of number system in CAT. Then break it into core concepts.
Step 2: Focus on building basic concepts like
Understanding of number properties
Quick methods to find squares, cubes, square roots, and cube roots
Hey! With an All India Rank (AIR) of 302,821 in NEET and belonging to the BCE category, it is highly unlikely to get a BDS seat in Telangana under the state quota, as the closing ranks for BCE are usually below 50,000. You may consider applying to private colleges under management quota or explore BDS seats in other states, but the chances remain very limited with this rank.
At KIMS Amalapuram, the internship stipend for MBBS students is generally reported to be around 20,000 per month, though some students have mentioned that in certain years no stipend was provided at all, which means it can vary depending on the policies in place at the time of your internship. To get the most accurate and updated information, it is always best to confirm directly with the college administration or recent interns, but on average, you can expect a stipend in the range of 18,000-20,000 per month during the compulsory rotating internship.
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The CAT 2025 exam is a national-level MBA entrance test for IIMs and top B-schools in India. It will be held on 30th November 2025 in computer-based mode across ~170 cities.
The registration is open from 1st August to 13th September 2025 on
iimcat.ac.in
.
Admit cards will be available from 5th November 2025 onward.
Graduates with at least 50 marks (45 for SC/ST/PwD) are eligible to apply.
The exam tests English, Reasoning, and Quantitative Aptitude in three timed sections.
M/s Deloitte Touche Tohmatsu Limited, one of the top four audit and accounting firms in the world with headquarters at London, UK, and with an operational presence in 153 countries, hires Management Trainees (MT) from all the premier management institutes of India thrice every year, in the months of January, May and September.
Each new group of Management Trainees (MT) have to go through a four month rigorous training schedule, after which they have to pass through a test consisting of a written assessment and a case-analysis. The top hundred ranked Management Trainees (MT) based on the performance in the test are confirmed as Management Executives (ME). The rest are given the opportunity of undergoing the training for four months one more time along with the next batch of Management Trainees (MT) and then passing through the subsequent test consisting of the written assessment and case-analysis. The Management Trainee (MT) who fails to get confirmed as a Management Executive (ME) the second time is fired.
The scatter-graph below depicts the number of Management Trainees (MT) at Deloitte taking the tests from January 2020 till May 2022, and the vis-à-vis hired Management Trainees (MT) at Deloitte who were fired :
It is also known that for the month of September 2019 at Deloitte, 96 hired Management Trainees (MT) failed to be confirmed as a Management Executive (ME) the first time, and that 36 hired Management Trainees (MT) were fired.
Question :
In which test did the minimum number of Management Trainees (MT) get confirmed as a Management Executive (ME) in the second attempt ?
Two friends Moloy and Niloy passed out from the Purulia Institute of Science and Technology with B.Tech degrees in Mechanical Engineering, but even after a year placement was hard to find. So they decided to take the challenge head-on, came down to Kolkata, rented a garage space on Park Street, and having an affinity towards making people enjoy good food, started their firm named 'B.Tech Bread-Omlette Wala'.
They started with three items on the menu. One was the French Toast which could be prepared in 3 minutes. The second was the Egg Tortillas which took 15 minutes to prepare. Any one of Moloy and Niloy could prepare any one of them at a time. The third was the Egg Bhurji with French Fries. This however was prepared on an automated fryer which could prepare 3 servings at a time and took 5 minutes irrespective of the number of servings equal to or below 3. The fryer did not need anyone to attend to it, and the time to put in the raw ingredients could be neglected. So one could tend to the preparation of other items while the Egg Bhurji with French Fries were being prepared.
They wanted to serve the orders as early as possible after the order was given. The individual items in any order were served as and when all the items were ready, and the order was then considered closed. None of the items on the menu were prepared in advance in anticipation of future orders.
On the first day, 3 groups of customers came in and ordered at 6.00 pm, 6.10 pm, and 6.13 pm. The first order was for a plate of Egg Tortillas, two plates of French Toast, and three plates of Egg Bhurji with French Fries. The second order was for a plate of French Toast and two plates of Egg Bhurji with French Fries. The third order was for a plate of Egg Tortilla and a plate of Egg Bhurji with French Fries.
On the backdrop of the above information answer the questions given :
Question:
Assuming that the next customer's order could only be attended to when the previous customer's order was closed, at what time would the first customer's order be considered closed ?
Six sticks of equal lengths were kept in the vertical position in an empty flower-vase, to be arranged at the six corners of a regular hexagon. The two ends of each of the sticks were of different colours.
The top ends of the sticks were one of each of the following colours – Red, Cyan, Pink, Brown, Black and Green. The bottom ends were one of each of the following colours – Blue, Yellow, White, Orange, Purple and Grey. Both the sets of colours mentioned were in no particular order.
It was also known that :
a) The stick with the red colour was opposite to the stick with the blue colour
b) There were exactly two sticks whose both ends had colours whose names started with the same letter
c) The stick with the grey colour was adjacent to the stick with the white colour
d) The stick with the cyan colour was adjacent to both the sticks with the brown colour and the one with the blue colour
e) The stick with the purple colour was adjacent to both the sticks with the grey colour and the one with the green colour
f) The stick with the white colour was opposite to the stick with the green colour
Question :
What was the colour of the bottom end of the stick having brown colour at the top end ?
Two friends Moloy and Niloy passed out from the Purulia Institute of Science and Technology with B.Tech degrees in Mechanical Engineering, but even after a year placement was hard to find. So they decided to take the challenge head-on, came down to Kolkata, rented a garage space on Park Street, and having an affinity towards making people enjoy good food, started their firm named 'B.Tech Bread-Omlette Wala'.
They started with three items on the menu. One was the French Toast which could be prepared in 3 minutes. The second was the Egg Tortillas which took 15 minutes to prepare. Any one of Moloy and Niloy could prepare any one of them at a time. The third was the Egg Bhurji with French Fries. This however was prepared on an automated fryer which could prepare 3 servings at a time and took 5 minutes irrespective of the number of servings equal to or below 3. The fryer did not need anyone to attend to it, and the time to put in the raw ingredients could be neglected. So one could tend to the preparation of other items while the Egg Bhurji with French Fries were being prepared.
They wanted to serve the orders as early as possible after the order was given. The individual items in any order were served as and when all the items were ready, and the order was then considered closed. None of the items on the menu were prepared in advance in anticipation of future orders.
On the first day, 3 groups of customers came in and ordered at 6.00 pm, 6.10 pm, and 6.13 pm. The first order was for a plate of Egg Tortillas, two plates of French Toast, and three plates of Egg Bhurji with French Fries. The second order was for a plate of French Toast and two plates of Egg Bhurji with French Fries. The third order was for a plate of Egg Tortilla and a plate of Egg Bhurji with French Fries.
On the backdrop of the above information answer the questions given :
Question:
Assuming that the next customer's order could only be attended to when the previous customer's order was closed, at what time would the third customer's order be considered closed ?
Two friends Moloy and Niloy passed out from the Purulia Institute of Science and Technology with B.Tech degrees in Mechanical Engineering, but even after a year placement was hard to find. So they decided to take the challenge head-on, came down to Kolkata, rented a garage space on Park Street, and having an affinity towards making people enjoy good food, started their firm named 'B.Tech Bread-Omlette Wala'.
They started with three items on the menu. One was the French Toast which could be prepared in 3 minutes. The second was the Egg Tortillas which took 15 minutes to prepare. Any one of Moloy and Niloy could prepare any one of them at a time. The third was the Egg Bhurji with French Fries. This however was prepared on an automated fryer which could prepare 3 servings at a time and took 5 minutes irrespective of the number of servings equal to or below 3. The fryer did not need anyone to attend to it, and the time to put in the raw ingredients could be neglected. So one could tend to the preparation of other items while the Egg Bhurji with French Fries were being prepared.
They wanted to serve the orders as early as possible after the order was given. The individual items in any order were served as and when all the items were ready, and the order was then considered closed. None of the items on the menu were prepared in advance in anticipation of future orders.
On the first day, 3 groups of customers came in and ordered at 6.00 pm, 6.10 pm, and 6.13 pm. The first order was for a plate of Egg Tortillas, two plates of French Toast, and three plates of Egg Bhurji with French Fries. The second order was for a plate of French Toast and two plates of Egg Bhurji with French Fries. The third order was for a plate of Egg Tortilla and a plate of Egg Bhurji with French Fries.
On the backdrop of the above information answer the questions given :
Question:
Suppose Moloy and Niloy had decided to process multiple orders at the same time, however strictly prioritising a first come first serve basis, when would the second customer's order be considered closed ?
Two friends Moloy and Niloy passed out from the Purulia Institute of Science and Technology with B.Tech degrees in Mechanical Engineering, but even after a year placement was hard to find. So they decided to take the challenge head-on, came down to Kolkata, rented a garage space on Park Street, and having an affinity towards making people enjoy good food, started their firm named 'B.Tech Bread-Omlette Wala'.
They started with three items on the menu. One was the French Toast which could be prepared in 3 minutes. The second was the Egg Tortillas which took 15 minutes to prepare. Any one of Moloy and Niloy could prepare any one of them at a time. The third was the Egg Bhurji with French Fries. This however was prepared on an automated fryer which could prepare 3 servings at a time and took 5 minutes irrespective of the number of servings equal to or below 3. The fryer did not need anyone to attend to it, and the time to put in the raw ingredients could be neglected. So one could tend to the preparation of other items while the Egg Bhurji with French Fries were being prepared.
They wanted to serve the orders as early as possible after the order was given. The individual items in any order were served as and when all the items were ready, and the order was then considered closed. None of the items on the menu were prepared in advance in anticipation of future orders.
On the first day, 3 groups of customers came in and ordered at 6.00 pm, 6.10 pm, and 6.13 pm. The first order was for a plate of Egg Tortillas, two plates of French Toast, and three plates of Egg Bhurji with French Fries. The second order was for a plate of French Toast and two plates of Egg Bhurji with French Fries. The third order was for a plate of Egg Tortilla and a plate of Egg Bhurji with French Fries.
On the backdrop of the above information answer the questions given :
Question:
Suppose Moloy and Niloy had decided to process multiple orders at the same time, however strictly prioritising a first come first serve basis, when would the third customer's order be considered closed ?
Two friends Moloy and Niloy passed out from the Purulia Institute of Science and Technology with B.Tech degrees in Mechanical Engineering, but even after a year placement was hard to find. So they decided to take the challenge head-on, came down to Kolkata, rented a garage space on Park Street, and having an affinity towards making people enjoy good food, started their firm named 'B.Tech Bread-Omlette Wala'.
They started with three items on the menu. One was the French Toast which could be prepared in 3 minutes. The second was the Egg Tortillas which took 15 minutes to prepare. Any one of Moloy and Niloy could prepare any one of them at a time. The third was the Egg Bhurji with French Fries. This however was prepared on an automated fryer which could prepare 3 servings at a time and took 5 minutes irrespective of the number of servings equal to or below 3. The fryer did not need anyone to attend to it, and the time to put in the raw ingredients could be neglected. So one could tend to the preparation of other items while the Egg Bhurji with French Fries were being prepared.
They wanted to serve the orders as early as possible after the order was given. The individual items in any order were served as and when all the items were ready, and the order was then considered closed. None of the items on the menu were prepared in advance in anticipation of future orders.
On the first day, 3 groups of customers came in and ordered at 6.00 pm, 6.10 pm, and 6.13 pm. The first order was for a plate of Egg Tortillas, two plates of French Toast, and three plates of Egg Bhurji with French Fries. The second order was for a plate of French Toast and two plates of Egg Bhurji with French Fries. The third order was for a plate of Egg Tortilla and a plate of Egg Bhurji with French Fries.
On the backdrop of the above information answer the questions given :
Question:
A fourth customer comes in and orders two plates of French Toast at 6.24 pm. Suppose Moloy and Niloy had decided to process multiple orders at the same time, however strictly prioritising a first come first serve basis. For exactly how many minutes would one of the friends be idle from 6.00 pm till serving the last customer, assuming that the four customers were the only ones to have come in within the period being discussed ?
Two friends Moloy and Niloy passed out from the Purulia Institute of Science and Technology with B.Tech degrees in Mechanical Engineering, but even after a year placement was hard to find. So they decided to take the challenge head-on, came down to Kolkata, rented a garage space on Park Street, and having an affinity towards making people enjoy good food, started their firm named 'B.Tech Bread-Omlette Wala'.
They started with three items on the menu. One was the French Toast which could be prepared in 3 minutes. The second was the Egg Tortillas which took 15 minutes to prepare. Any one of Moloy and Niloy could prepare any one of them at a time. The third was the Egg Bhurji with French Fries. This however was prepared on an automated fryer which could prepare 3 servings at a time and took 5 minutes irrespective of the number of servings equal to or below 3. The fryer did not need anyone to attend to it, and the time to put in the raw ingredients could be neglected. So one could tend to the preparation of other items while the Egg Bhurji with French Fries were being prepared.
They wanted to serve the orders as early as possible after the order was given. The individual items in any order were served as and when all the items were ready, and the order was then considered closed. None of the items on the menu were prepared in advance in anticipation of future orders.
On the first day, 3 groups of customers came in and ordered at 6.00 pm, 6.10 pm, and 6.13 pm. The first order was for a plate of Egg Tortillas, two plates of French Toast, and three plates of Egg Bhurji with French Fries. The second order was for a plate of French Toast and two plates of Egg Bhurji with French Fries. The third order was for a plate of Egg Tortilla and a plate of Egg Bhurji with French Fries.
On the backdrop of the above information answer the questions given :
Question:
Had Niloy been absent on that day, and assuming that the next customer's order could only be attended to when the previous customer's order was closed, at what time would the fourth customer's order (refer to the previous question) be considered closed ?
The bar-graph given below shows the foreign exchange reserves of Nepal (in million Rupees) from 2014 to 2021. Answer the following questions based on the graph :
Question:
What was the percentage increase (rounded to the nearest integer, if deemed necessary) in the foreign exchange reserves in 2020 over 2016 ?
The Jadavpur University’s Prince Anwar Shah Road hostel consists of two large separate buildings, one for the ladies and the other for the gents, while having a common kitchen and dining hall. It is the hostel of the CS and the EEC department of engineering students of the university.
In recognition of the growing dissatisfaction and hence complaints among the inmates of the hostel regarding the menu served for dinner, the Dean of the engineering department, Dr Aparesh Sanyal, personally decided to investigate the matter. He set about collecting information about the preference of dinner among the inmates, separately from the gents and the ladies wing of the hostel.
Dr Sanyal was able to gather the following partial information :
Hostel inmates
Menu preference for dinner
Total
Egg Meal
Fish Meal
Chicken Meal
Gents
20
Ladies
64
Total
60
The Warden of the hostel was consulted, who after investigation declared that the following facts were clear :
1. Forty percent of the hostel inmates were ladies
2. One-third of the gentlemen inmates preferred an egg meal for dinner
3. Half the hostel inmates preferred either fish meal or chicken meal
Question:
What proportion of the lady hostel inmates preferred a fish meal for dinner ?
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