FCI MT : Previous Year Questions Papers (Civil, Mechanical, Electrical engineering, agricultural, General, Depot, Movement, Accounts )

FCI MT : Previous Year Questions Papers (Civil, Mechanical, Electrical engineering, agricultural, General, Depot, Movement, Accounts )

FCI  management trainee papers for Civil, Mechanical, Electrical engineering, agricultural, General, Depot, Movement, Accounts etc are given below with download links. The original answer key provided by FCI has been attached at the end of the PDF.

Post Name	Subject & Post Code	Paper	Download Link
Management Trainee (General) Management Trainee (Depot) Management Trainee (Movement)	Post Code: A,B,CSubject Code: 131	Paper -I	Series A Series B Series C Series D
Management Trainee (Accounts) Management Trainee (Technical) Management Trainee (Civil Engg) Management Trainee (Mech Engg) Management Trainee (Elect Engg)	Post Code: D,E,F,G,HSubject Code: 131	Paper -I	Series A Series B Series C Series D
Management Trainee (Accounts)	Post Code: DSubject Code: 132	Paper -II	Series A Series BSeries C Series D
Management Trainee (Technical)	Post Code: ESubject Code: 133	Paper -II	Series A Series B Series C Series D
Management Trainee (Civil Engg)	Post Code: FSubject Code: 134	Paper -II	Series A Series B Series C Series D
Management Trainee (Mechanical Engg)	Post Code: GSubject Code: 135	Paper -II	Series A Series B Series C Series D
Management Trainee (Electrical Engg)	Post Code: HSubject Code: 136	Paper -II	Series A Series B Series C Series D

Mixture and Alligation

Mixture and Alligation

In mixture problems, substances with different characteristics are combined and it is necessary to determine the characteristics of the resulting mixture. To solve such problems, we can take the help of Alligation rule. This rule can also be used to find the average value of mixture when the price of two or more ingredients is mixed together.

Alligation basically means “Linking”. The rule states that when different quantities of different ingredients are mixed together to produce a mixture of a mean value, the ratio of their quantities is inversely proportional to the differences in their cost from the mean value.

Rule of Alligation: 

If 2 ingredients are mixed in a ratio, then we use the following ratio formula:

(Quantity of cheaper/ Quantity of dearer) = (Cost Price of dearer – Mean Price) / (Mean Price – Cost Price of Cheaper), Cheaper Quantity : Dearer Quantity = (d – m) : (m – c)

• m = mean price
• d = Cost price of dearer
• c = Cost Price of cheaper 

Example 1:

In what proportion must rice at Rs. 3.25 per kg be mixed with rice at Rs. 3.80 per kg, so that the mixture would be worth Rs. 3.50 per kg? 

Solution:

Here, d = Rs. 3.80, c = Rs. 3.25, m = Rs. 3.50

Quantity of cheaper / Quantity of dearer = (Cost Price of dearer – Mean Price) / (Mean Price – Cost Price of Cheaper)

Q_c / Q_d = (3.80 – 3.50) / (3.50 – 3.25) = 6/5

So, they must be mixed in the ratio 6 : 5. 

Repeated Dilution of a Mixture:

There’s a special case in mixtures in which there is repeated dilution of a mixture with one of the ingredients, by removing, say n litres of the mixture and replacing it with n litres of one of the ingredients.

Say, there are x litres of water initially. Now, n litres of the water are replaced by n litres of wine. This operation is repeated t times.

Quantity of water left in the vessel = x (1 – [n/x])^t

Here, 

• x = total quantity
• n = quantity drawn every time, and,
• t = no. of times 

Example 2:

In a vessel, there are 100 litres of water. 10 litres of water are withdrawn and replaced with milk. Again 10 litres of water are withdrawn and replaced with milk. What is the quantity of water left in the mixture?

Solution:

Here, x = total quantity = 100 litres, n = quantity withdrawn every time = 10 litres, t = no. of times = 2

Quantity of water left in the vessel = 100 (1-[10/100])² = 100 * 9 * 9/ (10*10)

Thus quantity of water left = 81 litres

Remainder Theory

Remainder Theory

Questions from Number System appear regularly in almost all competitive exams. Within number system, the questions on remainders are found to be most tricky. This article will help you learn different types of remainder questions and various approaches you can apply to solve these.

The basic remainder formula is:

• Dividend = Divisor × Quotient + Remainder

If remainder = 0, then it the number is perfectly divisible by divisor and divisor is a factor of the number e.g. when 8 divides 40, the remainder is 0, it can be said that 8 is a factor of 40.

You will understand the concept better with the help of the following examples:

Example 1:

Find the Remainder; [(12 × 13 × 14)/5]

Solution:

Remainder [(12 × 13 × 14)/5] = Remainder [2184/5] = 4. But is it the right method? Instead, find the remainder for each term when divided by 5, and replace each term with the respective remainders. Remainder (12 × 13 × 14)/5 is same as Remainder (2 × 3 × 4)/5, i.e. 24/5 = 4.

In this case,12,13 and 14 will give remainders 2, 3 and 4, respectively when divided by 5. So, replace them with the respective remainders in the expression and find the remainder.

Note: One common mistake while dealing with remainders is that, when the numbers have common factors in both dividend and divisor e.g. what is the remainder when 15 is divided by 9. Can it be solved as- 15/9 is same as 5/3.

Hence, remainder is 2. No! 15/9 will give a remainder of 6. Always remember that if you find the remainder after cancelling common term, make sure you multiply the remainder obtained with the common factors removed. In this case, you will get correct answer (6) when you multiply the remainder obtained i.e. 2 with the common factor you cancelled i.e. 3.

There are few important results relating to numbers. Those will be covered one by one in the following examples.

1. Formulas Based Concepts for Remainder:

• (aⁿ + bⁿ) is divisible by (a + b), when n is odd.
• (aⁿ - bⁿ) is divisible by (a + b), when n is even.
• (aⁿ - bⁿ) is always divisible by (a - b), for every n.

Example 2:

What is the remainder when 15³¹ + 23³¹ is divided by 19?

Solution:

15³¹ + 23³¹ is divisible by 15 + 23 = 38 (as 31 is odd). So, the remainder will be 0.

2. Concept of Negative Remainder:

By definition, remainder cannot be negative. But in certain cases, you can assume that for your convenience. But a negative remainder in real sense means that you need to add the divisor in the negative remainder to find the real remainder.

Example 2:

What is the remainder when 2¹¹ is divided by 3?

Solution:

The easiest method to solve this problem is by using the concept of negative remainders. Here, 2 when divided by 3gives a remainder of -1, which is theoretically incorrect but can be taken for the sake of convenience.

You are asked to find (-1) × (-1) ×…11 times divided by 3. Now, the product of  -1, when taken 11 times will give a product of – 1 only, which in real sense will give a remainder of – 1 + 3 = 2.

NOTE: Whenever you are getting a negative number as the remainder, make it positive by adding the divisor to the negative remainder.

3. Cyclicity in Remainders:

Cyclicity is the property of remainders, due to which they start repeating themselves after a certain point. Learn how to apply this concept with the help of following example.

Example 3:

What is the remainder when 2⁵¹ is divided by 7?

Solution:

Remainder (2¹/7) = 2; Remainder (2²/7) = 4; Remainder (2³/7) = 1; Remainder (2⁴/7) = 2.

As you can see, the remainders start repeating after the first three steps i.e. a cycle of 2, 4 and 1 is present in the remainders. So, at power 51 the remainder will be same as on power 3.

Hence, Remainder (2⁵¹/7) = 1.

4. Role of Euler’s Number in Remainders:

Euler’s Remainder theorem states that, for co-prime numbers M and N, Remainder [M^E(N) / N] = 1, i.e. number M raised to Euler number of N will leave a remainder 1 when divided by N. Always check whether the numbers are co-primes are not as Euler’s theorem is applicable only for co-prime numbers.

Example 4:

What is the remainder when 21²⁵⁶ is divided by 17?

Solution:

Remainder [21/17] = 4

Remainder [21²⁵⁶/17] = Remainder [4²⁵⁶/17]

4 and 17 are co-prime numbers and E (17) = 17 × [1 – (1/17)] = 16.

So, Euler’s theorem says Remainder [4¹⁶/17] = 1

Now, 4²⁵⁶ can be written as 4^16*16

Remainder [4²⁵⁶/17] = Remainder [4^16*16/17] = Remainder [(4¹⁶)¹⁶/17] = Remainder [(1)¹⁶/17] = 1.

With the application of Euler’s number, you can make almost 90% of the remainder questions easy and solve those in a relatively lesser time period.

Basics of Partnership

Basics of Partnership

Partnership is an association of two or more persons who put in money together in order to carry on a certain business. Partnership is of two types.

1. Simple Partnership:

When all the partners start the business at the same point of time i.e. their capitals remain in the business for the same duration of time is called simple partnership. In this kind of partnership the profit is simply distributed in the ratio of their capitals.

2. Compound Partnership:

When the capitals of the partners are invested in the business for the different time periods the partnership is known as compound. In this kind the profit sharing ratio is calculated by multiplying the capital invested with the unit of time (mostly months). Here are a few examples to understand the concept in a better manner. 

Example 1:

Ram and Shyam are partners in a business. Ram invests Rs. 300 for 12 months and Shyam invested Rs.600 for 6 months. If they gain a profit of Rs. 700 at the end of one year, what is Ram’s share?

Solution:

Ram’s total capital = 300 × 12 = 3600. 
Shyam’s total capital = 600 × 6 = 3600.
Profit sharing ratio = 3600 : 3600 ⇒ 1 : 1.
The profit is given to be Rs. 700
⇒ The share of both Ram and Shyam will be = 700 × ½ = Rs. 350.

Example 2:

Raju invested Rs. 8000 for the whole year in a business. Sonu joins after 4 months. How much he should invest so that the profits are distributed in the ratio of 2 : 1?

Solution:

Raju’s total capital = 8000 × 12 = 96000. 

Let us take the capital of Sonu = S, he invested this capital after 4 months means it remains in the business for 8 months.
Their profit sharing ratio = 2 : 1.
So the equation will be 96000/8S = 2/1
⇒ 16 S = 96000 ⇒ S = 6000.
So Sonu should invest Rs.  6000.

Example 3:

P, Q and R invest Rs. 400, 500 and 600 in a business respectively. P gets one-fourth of the profits as remuneration for managing the business. P, Q and R distribute the rest of the profits in the ratio of their investments. If in a particular year, P gets Rs. 10 less than Q and R together, what was the total profit for that year?

Solution:

After giving one-fourth of the total profit amount to P for managing the business, the rest three-fourth of total profit is divided among P, Q and R in the ratio of their investments. The share of P, Q and R in the profit will be in the ratio of 4 : 5 : 6.

Three fourth of the total profits = 4x + 5x + 6x = 15x.
That implies the total profit will be 15x × 4/3 = 20x
Total share of P = 4x + 20x / 4 = 9x....(i)
Share of Q and R= 5x + 6x = 11x. ...(ii)
The difference in (i) and (ii) above is given to be Rs. 10
( 5x + 6x ) – 9x = 10 ⇒ 2x = 10 ⇒ x = 5.
Total profit = 5 × 20 = 100.

Example 4:

M, N and L hired a ground for Rs.  12000. M used this ground for 8 cows for 3 weeks, N used it for 6 cows for 8 weeks and L used it for 18 cows for 4 weeks. What amount of rent should L pay?

Solution:

M’s total use = 8 × 3 = 24.
N’s total use = 6 × 8 = 48.
L’s total use = 18 × 4 = 72.
Their expenditure ratio = 24 : 48 : 72
⇒ 1 : 2 : 3.
⇒ L should pay 3/6 of the rent
⇒ i.e. 12000 × 3/6 = Rs.  6000.

Nobel Prizes Winners 2015

Nobel_Prizes_Winners_2015 :

1. The Nobel Prize in Physics 2015
Takaaki Kajita and Arthur B. McDonald- "for the discovery of neutrino oscillations, which shows that neutrinos have mass"
2. The Nobel Prize in Chemistry 2015
Tomas Lindahl, Paul Modrich and Aziz Sancar- "for mechanistic studies of DNA repair"
3. The Nobel Prize in Physiology or Medicine 2015
William C. Campbell and Satoshi Ōmura- "for their discoveries concerning a novel therapy against infections caused by roundworm parasites"
Youyou Tu- "for her discoveries concerning a novel therapy against Malaria"
4. The Nobel Prize in Literature 2015
Svetlana Alexievich- "for her polyphonic writings, a monument to suffering and courage in our time"
5. The Nobel Peace Prize 2015
National Dialogue Quartet- "for its decisive contribution to the building of a pluralistic democracy in Tunisia in the wake of the Jasmine Revolution of 2011"
6. The Nobel Prize in Economics 2015
British economist Angus Deaton won the 2015 economics Nobel Prize for "his analysis of consumption, poverty, and welfare,".

Quant (Concepts): DI - Tricks & Traps Bar Diagrams

Most of the time Bar Diagrams questions appear to be easy on face but  they may be  deceiving. Let us learn some typical example of bar diagram, and derive key take away points.

DIRECTIONS: The following graph gives the data about Foreign Equity Inflow (FEI) for the five countries for two years 97 and 98.FEI is taken as the ratio of foreign equity inflow to the country's GDP, which is expressed as a percentage in the following graph.

Question 1. The country with the highest percentage change in FEI in 1998 relative to its FEI in 1997, is

a. India                     b. China
c. Malaysia              d. Thailan

Solution: The basic mistake which the students do while solving this problem is just taking the difference between the figures given on the bar tops as these figures are in percentages. So, as per this the answer is China which has the maximum difference (5.96 – 4.8 = 1.16) but this is actually wrong as we need to find the percent change between the years 1997 to 1998 for the given countries.

Student can follow the book method by calculating the percent change for all the given countries and find the answer. Besides that a smarter approach can also be applied,  which is just basic visualization of the bars of two years for the countries given in the choices. We find that the bar for 1998 is less than half of 1997 in case of India, which means that the bar remaining for 1998 is less than even 50% of the bar for 1997, and no other country it is even close to this kind of a change. Hence, the answer is India in this case as the percentage decrease is more than 50%.

Question 2. Which country has maximum change in the FEI?

a. India                     b. China
c. Malaysia              d. Thailand      

Solution: In this question, you are only asked to identify the country with the maximum change and not the percent change. We can calculate this using the difference between the values given on the top of the bars. So, as per this, the answer is China which has the maximum difference i.e.(5.96 – 4.8) = 1.16.

In this question, you need to apply the basic concept of subtracting the values given. Essentially, this was the mistake which was committed while calculating the percentage change in question 1.

Question 3. Supposing in Thailand 20% of FEI in 1997 and 50% of FEI in 1998 goes to Education sector then find the ratio of the amounts allocated to Education in 1997 to 1998, assuming the GDPs of both of these years for Thailand to be same.

a.1 : 3                     b. 5 : 21
c. 6 : 11                  d. 7 : 25

Solution: Taking the data from the bar chart for Thailand it can be seen that in 1997, 5.09% of the GDP is FEI and 20% of this goes to the Education sector and in 1998, 5.82% of the GDP is FEI and its 50% goes to the Education sector. Now if you try to calculate it actually by the conventional method, it will take a lot of time and effort.

Now as you only want to find the ratio of the amount spent on Education sector in the 1997 and 1998, you can do approximation by taking 5.09% as 5% approx. for 1997 and 5.82% as 6% approx. for 1998. Now just calculate ratio as 5 × 20 × GDP : 6 × 50 × GDP = 100 : 300 = 1 : 3 approximately.

To conclude, you can notice that the questions which take long to solve by conventional method, can be solved smartly also by applying logical and calculation oriented tricks.

Quant (Concepts): Line Graphs

Quant (Concepts): Line Graphs

Problems based on line graphs appear frequently in competitive exams having data interpretation as a component. Given below are some of the popular types of questions asked from line graphs:

DIRECTIONS: Study the following graph carefully and answer the questions given below.

Example 1: In which of the following years was the percentage increase in income over the previous year the maximum?

1. 1989      2. 1990      3. 1991      4. 1992

Sol:  Percentage increase= 100 × Increase/Previous year income. Option 3 is eliminated because there’s a decrease in income in 1991.

In 1989, percentage increase = 100 × 20/30 = 66.67%.
In 1990, percentage increase = 100 × 15/50 = 30%.
In 1992, percentage increase = 100 × 17.5/57.5 = 30.43%
(All the data is in Rs ‘00 cr.)

OR

For maximum % change, numerator should be the greatest and the denominator should be smallest.The numerator for 1989 is the greatest, while the denominator for the same year is the lowest as compared to corresponding figures for 1990 and 1992. So,you need not to find the exact percent increase of all the options given.

Example 2: What was the average income of the company over the years (approximately)?

1. 4500 crores           2. 500 crores    
3. 5500 crores           4. 550 crores

Sol: Average in Rs ’00 cr. = (30 + 50 +65 + 55 + 75 + 55)/6 = 55 ⇒Average= Rs 5500 cr.

Example 3: What is the ratio of income to expenditure in the year 1990?

1. 1.2      2. 1.3      3. 1.1      4. 1.4

Sol: The income in the year 1990 is Rs. 6500 cr. and the expenditure is Rs. 5000 cr. Thus, the ratio in the year 1990 will be 6500/5000 = 1.3.

Example 4: The total expenditure in 1988 and 1989 together was equal to the income in which of the following years?

1. 1990      2. 1991      3. 1993      4. None of these

Sol: The total expenditure in 1988 and 1989 together = Rs. 6000cr, which does not match with any other year’s expenditure. So, the correct answer will be none of these.