CAT 1991 Question Paper

From the below given Venn diagram, we can see that F is the conclusion of ADF

All the other options are not logically related.

In ACE : CE are reiterated, A is unrelated to CE

In AEF : All three statements are facts.

In CDF: C and F are reiterated.

Taking only 1st statement into consideration, nothing certain can be said about product Q
Similarly with the only 2nd statement, we have only 1 information about product Q and R.
But considering both statements together we can say with sufficient information that product R will be more beneficial than product Q

Using first statement alone, we won't be able to find distance as insufficient data is there.
Now using second statement alone, we can find distance d as follows:
velocity before engine failure = $$\frac{40}{2}$$ = 20 km/h and after engine failure = 5 km/hr
And we know total time exceeds 45 min.
Hence 2 + $$\frac{d}{5}$$ = $$\frac{40}{60}$$ + $$\frac{40+d}{20}$$
or $$d = \frac{40}{9}$$  km.

Considering 1st statement alone, $$x^2 + x$$ is greater than 50, a two digit number and x is a prime number, then it has to be 7.

On the other hand, we cannot uniquely determine the value of x using statement B alone. Option a) is the correct answer.

As Avg. of 3 quotations is 110, so the sum of them will be 330
Now considering 1st statement alone, if the lowest quotation is 100, then highest can not exceed 129 because then the sum of all three will exceed from 330.
So we can get the answer by considering the first statement only.

Considering first statement only, given information will not suffice to answer that how many people read both Indian express and Times of India as data regarding intersection of all and data regarding number of people reading other 2 newspapers is not given
Now if we consider only second statement then we can answer about people reading both Indian express and Times of India as follows:
Total readers = 200 = 100 (indian express) + 120 (Times of india)  + 50 (Reading none) - Reading both.
So Reading both newspapers = 270-200 = 70

Given information X=3(Y-3)
Now considering first statement alone, we will have following equation:
Y+17 = X
On solving two eq. we can find the value of X.
Hence answer will be A.

Let the side of the smallest square be x.

According to statement 1,$$10x\geqslant 60$$

$$x\geqslant 6$$

According to statement 2, $$7x*6x\leq 1512$$

$$x\leq 6$$

From both the statements x = 6 and the answer can be determined.

Considering the first statement only, we will get radius > 14. Hence, we can't find the value of radius with this information only.
Now considering the second statement only, we will get radius <= 16.​Hence, we can't find the value of radius with this information only.
Now considering both statements simultaneously, we can't find an absolute value of radius.
Hence, answer will be D
 

Considering first and second statement alone, we can't say about time difference as information will not be sufficient.
Now considering both statements together, we can say that if journey duration is of 5 hours and flight departure time is 9:00 a.m., arrival time according to new york will be 2:00 p.m. but it is given that in London, arrival time is 10:00 a.m. Hence, the time difference will be of 4 hours between NewYork and London.

Considering first statement and second statement alone, information will not suffice to find the sequence of stations,
But if we consider both statements together we can tell about the sequence of stations as between A and B, Mr. Thomas travels with Mr. Murthy but not with Mr. Rahman. That means Mr. Thomas must have departed from station C and arrived at station D.
And it is also given that between C and D there are two segments. Hence, E lies between C and D, and Mr. Rahman departed from station D.

Putting y as x and x as y in given options, only options B satisfies the function property as follows.

x = $$\frac{2y+3}{3y-2}$$

or 3xy - 2x = 2y+3

or y(3x-2) = 2x+3

or $$y = \frac{2x+3}{3x-2}$$

On solving both equations, we will get $$x = \frac{23}{2+2k}$$
now for having no solutions to system 2+2k should be 0.
Hence k=-1

For a number to be even, its unit digit should be 0,2,4,6,8
Case 1: One of the digit is 5
Hence according to question, 5 can't come in middle and at unit's place, so numbers will be 570,572,574,576,578.
Case 2: No digit is 5
Hence the hundreds place can be filled in 8 ways (except 0,5) and tens place can be filled in 9 ways (except 5).
Number of ways = 8 * 9 * 5 = 360
Hence total number of ways = 360 + 5 = 365

Suppose he appointed x persons and y of them didn't come. Hence work done by each of them increases by 32.

So $$\frac{480}{x-y} - \frac{480}{x} = 32$$

Now we can check options by putting in the above eq. 

x=10 and y=4 will be our answer

Let's say Iqbal has x cards initially and Mushtaq has y number of cards initially.

So first Mushtaq gave t cards to Iqbal, hence (x+t) = 4(y-t)

Now second time, Iqbal gave t cards to Mushtaq, hence x-t = 3(y+t)

Solving above two equations we will get x=31t and y=9t

And we know x+y<52 hence 40t<52 

because t should be a whole number it will be 1 here and x=31 and y=9

Number of people owning exactly 2 articles = 9

Number of people owning exactly 3 articles = 1

Applying AUBUC formula, we get

AUBUC = 22+15+14 - 9 -2*(1)=40

Number of people who do not own any article = 50-40 = 10

Suppose consecutive odd integers are: (a-2), a, (a+2)

Hence, 3a-6 = 2(a+2) + 3 => a=13

a+2 = 15

4 first level nodes are connected to A => 4 ways.

Each of these first level nodes are connected to 1 second level node by four connections => 4 ways

Each of these second level nodes are connected to 2 third level nodes => 2 ways.

Each of these third level nodes are connected to B by 1 route => 1 way

=> 4 * 4 * 2 * 1 = 32 ways.

Let the side of the square be 1 cm

So, area of triangle AGE = 1/2 * 1/2 * 1/2 = 1/8

Similarly, area of triangle EBF = 1/8

Area of triangle DFC = 1/2 * 1 * 1/2 = 1/4

Area of the square = 1*1 = 1

So, area of the quadrilateral = 1 - (1/8 + 1/8 + 1/4) = 1 - 1/2 = 1/2

Option a) is the correct answer.

$$2^{71} (2^2 - 2^1 - 1)$$
$$2^{71} (4-2-1)$$
$$2^{71}$$

-n+2 >= 0
or n<=2
and 2n>=4
or n>=2
So we can take only one value of n i.e. 2

let's say integers are x and y
so x+y = 10 => y = 10 - x
and $$\frac{1}{x} + \frac{1}{y} = \frac{5}{12}$$
$$\frac{1}{x} + \frac{1}{10-x} = \frac{5}{12}$$

=> (10 - x + x)*12 = 5*x(10-x)

=> $$120 = 50x - 5x^2$$

=> $$24 = 10x - x^2$$

=> x = 4, 6

=> y = 6 or 4

The bigger of the two numbers is 6.

As we know that area of the circle is directly proportional to the square of its radius.
Hence $$\frac{A_{ic}}{A_{cc}} = \frac{\frac{x^2}{4}}{\frac{x^2}{2}}$$
Where $$x$$ is side of square (say), ic is inscribed circle with radius $$\frac{x}{2}$$, cc is circumscribed circle with radius $$\frac{x}{\sqrt{2}}$$
So ratio will be 1:2

Among all options only D satisfies the given equations as follows:
$$f(y) = \frac{1-y}{1+y}$$
and for x:
$$y + xy = 1-x$$
$$x(1+y) = 1-y$$
$$x= \frac{1-y}{1+y}$$
Hence $$x=f(y)$$

The diagram for this question has been shown:

Total number of schools having either or LAB or LIB or both = a+b+x/2 - y + y + 3x = 7x/2 + a + b = 35

Here a = b = y = 0

7x/2 = 35

x = 10

Total number of schools having at least one of PG, LIB or LAB = 30+2x+x+x/2 = 30+3x+x/2 = 30+30+5 = 65

Number of schools having neither of the three = 100-65 = 35

The diagram for this question has been shown:

Total number of schools having either or LAB or LIB or both = a+b+x/2 - y + y + 3x = 7x/2 + a + b = 35

It has been given that the schools having playground don't have a Library or Laboratory. 

Hence  a = b = y = 0

7x/2 = 35

x = 10

Required ratio = 25:15 = 5:3

The expected money got by the player = 1*1/6 + 2*1/6 + 3*1/6 + 4*1/6 + 5*1/6 + 6*1/6 = 21/6 = Rs 3.5

So, the player has to pay 3.5 - 1 = Rs 2.5 to get a profit of Re 1 in the long run.

As machine B's efficiency is twice as of A's, Hence, it will complete its work in 30 hours.
And C's efficiency is putting A and B together i.e. = 20 hours $$( (\frac{1}{60} + \frac{1}{30})^{-1})$$
Now if all three work together, then it will be completed in x (say) days.

$$\frac{1}{x} = \frac{1}{20} + \frac{1}{30} + \frac{1}{60}$$

or x = 10 hours

For x<0 ; y=x+2

for 0<x<$$\frac{1}{2}$$ ; $$y=x+2$$

for $$x>\frac{1}{2}$$ ; $$y=3-x$$

Hence, $$y$$ attains its maxima at $$x=\frac{1}{2}$$  i.e. $$y$$ = 2.5

We know that x + y + z = T and x + 2y + 3z = R, where
x = number of members belonging to exactly 1 set = 70
y = number of members belonging to exactly 2 sets
z = number of members belonging to exactly 3 sets = 10
T = Total number of members
R = Repeated total of all the members = (40+50+60) = 150
Thus we have two equations and two unknowns. Solving this we get y = 25

So, 25 people belong to exactly 2 clubs.

Let the side of the square which is cut be x.

Volume of the cuboid so formed =$$(10-2x)^2*x$$

Put x = 1, 2, 3 and so on till 5

Maximum volume would be at x = 2

Volume of the cuboid so formed =$$(10-2*2)^2*2 = 72$$

The expressions in the four options can be expanded as

xyz-yz; xyz-xz; xyz-xy and xyz+xz

The closest value to xyz would be xyz-yz, as yz is the least value among yz, xz and xy.

Option a) is the correct answer.

The highest power of 5 in 80! = [80/5] + [$$80/5^2$$] = 16 + 3 = 19

So, the highest power of 5 which divides 80! exactly = 19

13*62 = 806

31*26 = 806

Hence the answer is option b

As 55 and 124 don't have any common factor, and n has to be a minimum integer, Hence, it should be 124 only. So that given equation won't have a remainder.

let's say k+4 = 7m
k = 7m-4
Now for k+2n or 7m+(2n-4) is also  multiple of 7.
or 2n-4 should be a multiple of 7 
So 2n-4 = 7p
or 2n = 7p+4
For p=2; n=9 (p cannot be 1 as n is an integer )

A calculator has two memory buttons, A and B. Value 1 is initially stored in both memory locations. The following sequence of steps is carried out five times:
add 1 to B
multiply A to B
store the result in A
What is the value stored in memory location A after this procedure?

The value in A and B before the start of the process is 1 and 1 respectively.
In 1st iteration,
After step I, A = 1 and B = 2
After steps II and III, A = 1 * 2 = 2, B = 2

In 2nd iteration,
After step I, A = 2 and B = 3
After steps II and III, A = 2 * 3 = 6, B = 3

In 3rd iteration,
After step I, A = 6 and B = 4
After steps II and III, A = 6 * 4 = 24, B = 4

In 4th iteration,
After step I, A = 24 and B = 5
After steps II and III, A = 24 * 5 = 120, B = 5

In 5th iteration,
After step I, A = 120 and B = 6
After steps II and III, A = 120 * 6 = 720, B = 6

Hence, option D is the correct answer.

Considering the centre coin which is denoted by  C1.

The surrounding coins are of equal radius as the centre coin. Since each coin is placed next to each other and considering the total number of coins = n.

The pairs formed by the coins which touch the consecutive coins is given by :

(1, 2) , (2, 3) ........... (n, 1)

The triangles formed by joining the centres of all the consecutive circles at the centre of C1 = 360 degrees.

If we join centres of 2 outer circles with the centre circle, it will make an equilateral triangle.

A single pair forms 60 degrees as it forms an equilateral triangle. Hence $$\frac{360}{n}=\ 60$$

n = 6

Let's say the amount of money Gopal has is 100x rs. With this money he can buy either 50 oranges or 40 mangoes.

Hence, the cost of 1 orange = $$\dfrac{100x}{50}$$ = Rs. 2x

Similarly, the cost of 1 mango = $$\dfrac{100x}{40}$$ = Rs. 2.5x

Taxi fare = 0.1*100x = 10x

Remaining money with Gopal = 90x
Cost incurred in buying 20 mangoes = 2.5*20 = 50x

Hence, the amount of money left with Gopal = 100x - 10x - 50x = 40x

Therefore, we can say that with this much money he can but a maximum of $$\dfrac{40x}{2x}$$ = 20 oranges. 

Since we know that Neera's husband drives at a uniform speed to and from his residence.
If he saved 10 mins overall travel time, he should have driven 5 mins less towards railway station and 5 mins less while driving towards residence.
If he saved 5 minutes in his return journey, he should have started to return 5 minutes before his actual return time.
When the husband met Neera, he should have met her 5 minutes before the actual meeting time i.e. at 5.55 PM. 
So, Neera must have walked for 55 minutes from 5PM.

Let the number of sticks assigned to each boy be N.

Let the number of boxes be M.

So, number of sticks per box = N/M

Now, if he reduces the number of sticks in each box, the equation becomes N/(M+3) = N/M - 25

So, 25 = N/M - N/(M+3)

From the options, if N = 150, then, we get 25 = 150 [ 1/M - 1/(M+3) ]

=> 1/6 = 1/M - 1/(M+3) => M = 3

So, the number of sticks assigned to each boy = 150

As we know, formulae of compound interest for 2 years  will be:
$$P(1+\frac{r}{100})^{2}$$ = 625  (Where r is rate, P is principal amount)
For 3 years:
$$P(1+\frac{r}{100})^{3}$$ = 675
Dividing above two equations we will get r=8%

Consider the following diagram: In this, the nodes E and F are connected to all the other nodes whereas the other four nodes are connected to four nodes each.

The total number of connections is 13.

If 2x+12 is perfectly divisible by x, then 12 must be divisible by x.
Hence, there are six possible values of x : (1,2,3,4,6,12)

In the diagram, A is starting point and B is ending point.

Accordingly, AB = $$\sqrt{(3^2 + 4^2)} = 5$$

Let's, say students contributed x rs.
So teachers contributed = $$\frac{3x}{2}$$
And external benefactor contributed = $$\frac{9x}{2}$$
Summation of all contribution = 4200 = $$7x$$
or $$x=600$$
Hence, teachers contributed = 900

Let the Prime number be 6n+1.

So $$(p^2 - 1)$$ = 6n(6n+2) = 12n(3n+1)

For any value of n , n(3n+1) will have a factor of 2

Hence given equation will be always be divisible by 24

For 203 :
first step = $$2\times 3^2 + 0 \times 3^1 + 3 \times 3^0$$ = 21
second step = $$2 \times 3^1 + 1 \times 3^0$$ = 7
So two steps needed to reduce it to 7

8+12 =( 20 = 2+0) = 2
7+14 =( 21 = 2+1) = 3
10+18= (28= 2+8) = 10
 

Distance between two points = $$\sqrt {(3-(-2))^2 + (8-(-7))^2}$$ = $$5\sqrt10$$

For increment in 1986-87, denominator is increased by 2 whereas in every other year it is either increased by 1 or 1.5
Since numerator has increased by 30, which is similar to that of other years, we can say lowest increment occur in the year 1986-1987.

The per capita of each of the following years is given below.

1984-85 is Rs. 3,098
1985-86 is Rs. 3,482
1986-87 is Rs. 3,786
1987-88 is Rs. 4,203
1988-89 is Rs. 4,857
1989-90 is Rs. 5,319

Hence, the percapita income is highest for the year 1989-90

The per capita of each of the following years is given below.

1984-85 is Rs. 3,098
1985-86 is Rs. 3,482 (increased by 12.42%)
1986-87 is Rs. 3,786 (increased by 8.73%)
1987-88 is Rs. 4,203 (increased by 11.00%)
1988-89 is Rs. 4,857 (increased by 15.55%)
1989-90 is Rs. 5,339 (increased by 9.92%)

The percentage increase in population over the years is fairly constant and varied between 1% and 2%. But the percentage increase in GDP per capita in the year 1988-89 is significantly higher. Hence, the correct answer is 1988-89

The population increase for the year 1985-86 from the year 1984-85 is just one crore.
In every other year, the population increase was either 1.5 crore of 2 crores.
Note that the denominators for each of the years (ie total population) is fairly same.

Hence, the correct answer is 1985-86

The per capita of each of the following years is given below.

1984-85 is Rs. 3,098
1985-86 is Rs. 3,482 (increased by Rs. 384)
1986-87 is Rs. 3,786 (increased by Rs. 304)
1987-88 is Rs. 4,203 (increased by Rs. 416)
1988-89 is Rs. 4,857 (increased by Rs. 653)
1989-90 is Rs. 5,339 (increased by Rs. 481)

Hence, the increase is highest for the year 1988-89

Let's say the principal amount is P, Rate is R.
1st withdrawal = $$PR+ \frac{P}{5}$$
Remaining money = $$\frac{4P}{5}$$

2nd withdrawal = $$(\frac{4PR}{5} + \frac{2P}{5})$$
Remaining money = $$\frac{2P}{5}$$

3rd withdrawal = $$\frac{2PR}{5} + \frac{P}{5}$$
Remaining money = $$\frac{P}{5}$$

4th withdrawal = $$\frac{P}{5} + \frac{PR}{5}$$ = 11000
Or P = 50000 (Putting R = 10%)

Let's say the principal amount is P, Rate is R (i.e. R = 10%)

1st withdrawal = $$PR+ \frac{P}{5} = \frac{15P}{50}$$

2nd withdrawal = $$\frac{4PR}{5}+ \frac{2P}{5}$$ = $$\frac{24P}{50}$$

3rd withdrawal = $$\frac{2PR}{5} + \frac{P}{5}$$ = $$\frac{12P}{50}$$

4th withdrawal = $$\frac{PR}{5} + \frac{P}{5}$$ = $$\frac{11P}{50}$$

Lowest withdrawal was after the fourth year.

Let's say the principal amount is P, Rate is R (i.e. R = 10%)

1st withdrawal = $$PR+ \frac{P}{5} = \frac{15P}{50}$$

2nd withdrawal = $$\frac{4PR}{5}+ \frac{2P}{5}$$ = $$\frac{24P}{50}$$

3rd withdrawal = $$\frac{2PR}{5} + \frac{P}{5}$$ = $$\frac{12P}{50}$$

4th withdrawal = $$\frac{PR}{5} + \frac{P}{5}$$ = $$\frac{11P}{50}$$

Maximum Interest was after the first year (i.e. PR)

Let's say the principal amount is P, Rate is R (i.e. R = 10%)

1st withdrawal = $$PR+ \frac{P}{5} = \frac{15P}{50}$$

2nd withdrawal = $$\frac{4PR}{5}+ \frac{2P}{5}$$ = $$\frac{24P}{50}$$

3rd withdrawal = $$\frac{2PR}{5} + \frac{P}{5}$$ = $$\frac{12P}{50}$$

4th withdrawal = $$\frac{PR}{5} + \frac{P}{5}$$ = $$\frac{11P}{50}$$

Hence, Maximum withdrawal was after second year.

Let's say the principal amount is P, Rate is R (i.e. R = 10%)

1st Interest = $$PR$$

2nd Interest = $$\frac{4PR}{5}$$

3rd Interest = $$\frac{2PR}{5}$$

4th Interest = $$\frac{PR}{5}$$

For having the greatest percentage change, the slope of the line should be the highest. From the graph, we can see that slope is highest for A between Feb and March. So A is the correct answer.

To answer this question, let us calculate the highest increase in market value of each share.

For share A, the highest increase in market value is from Feb to March and equals 110 - 90 = 20
For share B, the market value increases constantly over all the months and is equal to 75-70 = 5
For share C, the highest increase in market value is from Feb to March and equals 60 - 55 = 5
For share D, the highest increase in market value is from Jan to Feb and equals 50 - 40 = 10

Hence, the correct answer is March and the required share is share A

To answer this question, let us calculate the highest percentage increase in market value of each share.

For share A, the highest percentage increase in market value is from Feb to March and equals 110/90 -1 = 22.22%
For share B, the highest percentage increase in market value is from Jan to Feb and equals 75/70 -1 = 7.14%
For share C, the highest percentage increase in market value is from Feb to March and equals 60/55 -1 = 9.09%
For share D, the highest percentage increase in market value is from Jan to Feb and equals 50/40 -1 = 25%

Hence, the required answer is February and the required share is share D

Loss from the transaction from Ram's standpoint is Sum of share prices of C and D - Share price of A.

So, loss from the transaction at the end of January is (40+60) - 100 - = 0
So, loss from the transaction at the end of February is (50+55) - 90= 15
So, loss from the transaction at the end of March is (60+50)-110 = 0
So, loss from the transaction at the end of April is (65+40) - 105 = 0
So, loss from the transaction at the end of May is (60+45) -100 = 5
So, loss from the transaction at the end of June is (55+45) - 110 = 0

So, the correct answer is February.

Loss from the transaction from Ram's standpoint is Sum of share prices of C and D - Share price of A.

So, loss from the transaction at the end of January is (40+60) - 100 - = 0
So, loss from the transaction at the end of February is (50+55) - 90= 15
So, loss from the transaction at the end of March is (60+50)-110 = 0
So, loss from the transaction at the end of April is (65+40) - 105 = 0
So, loss from the transaction at the end of May is (60+45) -100 = 5
So, loss from the transaction at the end of June is (55+45) - 110 = 0

So, the correct answer is 15

We should consider three possible cases as 1. when he is using test 1
2. When he is using test 2
3. When he is using no test
Now we will choose a method where the total expenditure will be least.
So for option A, let's consider that number of defective pieces are 50.
Hence, while using test 1:
Cost of testing = $$1000 \times 2 = 2000$$
Correcting 80% of pieces = $$40 \times 25 = 1000$$
Penalty for 20% of pieces = $$10 \times 50 = 500$$
Total = 3500

For test 2:
Cost of testing = $$1000 \times 3 = 3000$$
Correcting all 50 pieces = $$ 50 \times 25 = 1250 $$
Total= 4250

For no test:
Penalty for all 50 pieces = $$ 50 \times 50 = 2500$$

Hence cost is least when he is using no test while number of defective pieces is less than 100.

 

When there are 120 widgets defective:

Using 1st test:
Cost of testing = $$1000 \times 2 = 2000$$
Cost of correcting 80% pieces = $$96 \times 25 = 2400$$
Cost of penalty on 20% pieces = $$24 \times 50 = 1200$$
Total cost = 5600

Using 2nd test:

Cost of testing = $$1000 \times 3 = 3000$$
Cost of correcting all 120 pieces = $$120 \times 25 = 3000$$
Total cost = 6000

Using no test:

Total cost = 120 x 50 = 6000

Hence, he should use 1st test only.

Let's say defective pieces are 300
Using test 1
Cost of testing= $$1000 \times 2 = 2000$$
Cost of correcting 80% pieces = $$240 \times 25 = 6000$$
Cost of penalty on 20% pieces = $$60 \times 50 = 3000$$
Total cost = 11000

Using test 2
Cost of testing = $$1000 \times 3 = 3000$$
Cost of correcting all 300 pieces = $$300 \times 25 = 7500$$
Total cost = 10500

Using no test:
Cost of penalty on all pieces = $$300 \times 50 = 15000$$

Hence, he should use 2nd test.

For 160 defective pieces:
Using 1st test:
Cost of testing = $$1000 \times 2 = 2000$$
Cost correcting 80% pieces = $$128 \times 25 = 3200$$
Cost of penalty on 20% pieces = $$32 \times 50 = 1600$$
Total = 6800

Using 2nd test:
Cost of testing = $$1000 \times 3 = 3000$$
Cost of correcting all pieces = $$160 \times 25 = 4000$$
Total = 7000

Using no test:
Cost of penalty on all pieces = $$160 \times 50 = 8000$$

Hence, he should use first test only.

When there are 200 defective pieces:
Using test 1
Cost of testing= 2000
Cost of correcting 80% pieces = 4000
Cost of penalty on 20% pieces = 2000
Total = 8000

Using test 2
Cost of testing = 3000
Cost of correcting all pieces = 5000
Total = 8000

Using no test
Cost of penalty on all pieces= 10000

Hence he can use either 1st test or second test.

In order to find the number of years for which the sum is same, we need to determine the change in production of food and fertilizer. If there is no effective change, then the sum of their productions is same.

Change in 1984 => +1 - 1.5 = -0.5

Change in 1985 => 0

Change in 1986 => -1.5 + 1.5 = 0

Change in 1987 => 0

Change in 1988 => +1.5 - 1.5 = 0

Change in 1989 => -2.5 + 0 = -2.5

Change in 1990 => 0 + 1.5 = 1.5

Change in 1991 => 0

The change is 0 in 5 years. Hence, the answer is option D.

From the graph, in the year 1988, the ratio of food production to fertilizer production is 5:3.5 i.e 10:7
Let the food production be 10X and the fertilizer production be 7X

Hence, the total is 10X+7X = 17X which equals 170
So, X = 10 and the total food production equals 10*10 = 100

Just by looking at the graphs of the previous year's it is difficult to guess how the food production in the year 1992 will be.
This is especially true as we don't have any information about the fertilizer production in the year 1992.

Hence, the correct answer is option (d)

The fertilizer and food production has been inversely proportional over the years. ie when food production increased, fertilizer production fell and vice versa. But as we can see in 1988-89, when food production remained constant, the fertilizer production fell steeply and in 89-90 when food production increased steeply, the fertilizer production stayed constant. So, 1989 is the year of the anamoly.

Let each unit in the scale be equal to X

So, fertilizer production in 1983 is 2.5X
So, fertilizer production in 1984 is 3.5X
So, fertilizer production in 1985 is 3.5X
So, fertilizer production in 1986 is 2X
So, fertilizer production in 1987 is 2X
So, fertilizer production in 1988 is 3.5X
Assuming fertilizer production in 1989 is same as that of 1988 it equals 3.5X
So, fertilizer production in 1990 is X
So, fertilizer production in 1991 is X
Hence, the total is 22.5X
But this equals 450 million tonnes. Hence, X = 20 million tonnes

Therefore, food production in 1983 = 6.5X equals 130 million tonnes

The verb is/are refers to the noun "best part". As the phrase "best part" is singular, 'is' must be used here.

Hence, option A is correct.

The usage of the phrase "as well as" is preceded by a comma.

In a sentence where "as well as" is used, the verb must agree with the noun before the "as well as" phrase. Here, the verb was/were refers to the noun 'professor'. As professor is singular, the correct verb to be used here is 'was'.

Hence, option A is correct.

Option A is wrong because two independent clauses are not joined by a conjunction.

In option C, the sentence after colon is an independent clause. But, the sentence after colon must be a dependent clause. So, option C is incorrect.

The sentence in option D must have a comma after the completion of the dependent clause. Hence it is incorrect.

Option B is the correct answer.

If "one" is used to refer a person in the dependent clause, the same word must also be used in the independent clause. Options A, C and D violate this rule.

Option B is the correct answer.

"Either ... or" and "Neither .... nor" are followed by the verb that agrees with the noun nearest to it.

In sentences A, C and D, the verbs do not agree with the nound nearest to it. Hence, option B is the correct answer.

There are three independent clauses in the sentence. Indepenent clauses can be joined by a conjunction or by using a semicolon.

Only option A satisfies this rule.

In the first two sentences, "their" must replace "his" as there are 2 persons.

"Either ... or" and "Neither .... nor" are followed by the verb that agrees with the noun nearest to it. "Ramesh" is the nearest noun to the verb and therefore, the correct verb is "has". Option D is the correct answer.

Option A contains two independent clauses without a semicolon or a conjunction.

"The fact that ..." is the correct usage. Hence, option C is wrong.

"Resulted in his having ...." is the correct usage. Option B is the correct answer.

A comma is not required after the phrase "The people of this company". Only option C satisfies this.

This sentence has only one clause. Therefore, a comma is not required. 

Option D is the correct answer.

The word who/whom refers to the "people". In this context, "who" is the correct usage.

It is essential to understand the use of pronouns here. 'Who' and 'Whom' are relative pronouns. However, they are associated differently with personal and possessive pronouns. Whenever the subject is represented via personal pronouns (like he/she/it etc.), 'who' is utilized in the sentence. If the main subject can be replaced by a possessive pronoun (like him/her/his/their etc. ), we use 'whom' in the sentence. You can ask simple questions to understand this concept. For
example:

You can ask: "Who took my pencil?"
You say: "He/She took it" (personal pronouns)
You do not say: "Him/Her took it" (possessive pronouns)

You can ask: "Whom did you meet yesterday?"
You say: "I met him/her yesterday" (possessive pronouns)
You do not say: "I met he/she yesterday" (personal pronouns)

Hence, sentences wherein 'whom' is followed by/used alongside 'he' is incorrect. Luckily, all the options, except Option A, have this usage and can be easily discarded.

The word "felicitate" in the third sentence is used as a verb. But, the word that describes a noun "contacts", must be an adjective.

The word used must be "felicitated".

"To the perceived problems" is the correct phrase to be used in the sentence since there are multiple problems. Option c) is the correct answer.

"The Indian Government's choice" is a singular noun and should have the singular verb "stems".

Hence, the sentence in option C is wrong.

The quantifier "many" cannot modify a number. The correct quantifier to be used is "most".

This is an unreal past condition. So, the verb used must be in the form of past participle.

The correct phrase to be used is "you would have definitely".

The usage of the word "requires" is wrong here. It must be replaced by "assumes".

The verb "invite" does not agree with the singular noun "choice". Hence, the correct usage of the verb is "invites".

The words "he then" are not required here. The correct sentence is "Gavaskar was a great batsman who having played more than 100 test matches, decided to call it a day".

"Sold of" must be replaced by "sold off". The remaining sentences are correct.

Option a) is the correct answer.

'Minority' should take the verb 'has been'. On the other hand, the word 'minorities' take the verb 'have been'. Option c) is the correct answer.

The use of the word "ambitious" is redundant since the word 'aspiring' has already been used. Only one of the two words has to be used in the third sentence. Option c) is the correct answer.

D introduces the paragraph. B elaborates on the idea mentioned in D. So, B follows D. This is then followed by A, which continues the idea of D-B. C is the concluding sentence. The correct order of sentences is DBAC. Option c) is the correct answer.

B is the best opening sentence - it introduces the topic of the paragraph, advertising. This is followed by C which talks about the overall effectiveness and then by D which talks about individual effectiveness. The last sentence is A. Option c) is the correct answer.

A is the best opening sentence. It introduces the topic of the paragraph - the age of pragmatism. This is followed by D, which talks about an observation that is against expectations. This is further explained in B. So, B follows D. The concluding sentence is C. So, the order of sentences is ADBC. Option c) is the correct answer.

The opening sentence is D. It introduces the topic of the paragraph - cognitive age. This is followed by B which continues the idea of cognitive age. Next is A, which talks about a past research finding. A is then followed by C, which explains the result in sentence A. The correct order of sentences is DBAC. Option b) is the correct answer.

The opening sentence of the paragraph is C - it introduces the person being talked about in the paragraph. This is followed by sentence A. It continues talking about the 'eyeballs' mentioned in C. From the options, we see that the only possibility is option c).

The best opening sentence is B. It introduces the topic of the paragraph - intelligence. This is followed by D. The word 'these' in sentence D refers to the tests in sentence B. Next is sentence A, which talks about what intelligence actually means. The concluding sentence is C. The correct order of sentences is BDAC.

C is the opening sentence of the paragraph. It introduces the topic of the paragraph - punctuation. This is followed by sentence A. It continues the idea mentioned in C and talks about how the punctuation is affected in formal speech. This is followed by D which talks about a different situation in which punctuation is affected. The concluding sentence is B. So, the correct order of sentences is CADB.

To solve such difficult and complex problems, broad and powerful thinking is needed.
Moral values cannot guarantee solution to complex problems.
Broad shoulders cannot tackle problem pertaining to human species.
Option B is negative and is not suitable.
Hence, option D is the correct answer.

As the sentence in the question talks about energy tax to raise funds, it must be for some activity which will help them come out of the energy constraints. Thus, Option D is the best fit as the concluding part.

The idiom "look before you leap means "to consider the possible consequences before committing to an action". The idiom "forewarned is forearmed" means knowing possible difficulties and dangers earlier will give one a tactical advantage. Both these idioms have a common attitude that being prepared for a possible danger is to ones advantage. Thus, option A is the right choice.

From the context we can infer that the speaker is talking to the King of Sumeria. As the King the duty to protect his subjects and ensure the economic prosperity of the kingdom will be the duty of the King. Thus, option C which reflects the same is the best choice among the options.

The sentence talks about breaking with the predominant cultural, political and social beliefs and values. In order to reconstruct something, the thing should first be broken. Thus, D logically follows the sentence.

Sentence A is incorrect as it is vague.
B is also incorrect as there is no talks of malaise in the society in the sentence.
C can also be eliminated for the same reason as B.

The given sentence talks about how entrepreneurs are never satisfied and they are always looking for something more innovative to shape the future. The quote used by the executive should be in line with this idea. C is completely unrelated to this while A distorts the meaning. Among B and D, the former one is a better choice because D talks about the business aspects which is nowhere implied in the given sentence. Hence B is the correct answer.

Sequence BDF is the most appropriate since if some nurses are qualified and if all nurses are attendants, then some attendants will also be qualified.This can be seen from the following figure:

AD will give the following figure which will have F as the conclusion.

Based on the Venn diagram shown below, option C is the correct option. ie

"All roses are plants" and "All plants need air" implies "All roses need air".

Based on the Venn diagram below, Option C is the correct option. ie "Laxman is a man" and "No man is an island" imply that "Laxman is not an Island"

Based on the Venn diagram below, option A is correct ie If "All college students are intelligent" and "Ram goes to college" are true then it holds that " Ram is an intelligent person".

Based on the Venn Diagram below, If all cigarettes are injurious to health and if Cham cham is a cigarette, then it implies that Cham cham is injurious to health.

AC will have the following figure which will have D as the conclusion.

ACF will give the following Venn diagram which will give F as the correct conclusion for the premise AC.

BDE will give the following Venn diagram which will give E as the conclusion for the premise BD.

From statements B and D we can infer E.

BEF is contradictory pair.

F is a false assumption in the pair BDF.

BD from the following Venn diagram has E as the conclusion.

BD from the following figure gives E as the conclusion.

BDF has the below given Venn diagram from which we can see that F is the conclusion for the premise BD.

The premise AB gives E as the conclusion which can be seen from the following figure:

The premise BD gives F as the conclusion which can be seen from the following Venn diagram:

From the below given Venn diagram, we can see that F is the conclusion of ADF

All the other options are not logically related.

In ACE : CE are reiterated, A is unrelated to CE

In AEF : All three statements are facts.

In CDF: C and F are reiterated.