Statement 1:
y - x = 20
∴ We cannot find the value of p.

Statement 2:
x + y = 1620
∴ We cannot find the value of p.

Using A and B together,
X = 800, y = 820

$$P\times\ 2\times\ R\ =\ 800$$

$$P\left(1\ +\ R\right)^2\ -\ P=\ 820$$

Solving these 2 equations, we get the value of P to be 8000.

1 and 2 together are sufficient to answer the question.

Statement 1: Total marks in 4 subjects, including English = 4×60 = 240
Statement 2: Total marks in English and Maths = 100. Total marks in Maths and Science = 160
The question can’t be answered because nothing has been said about the marks in the fourth subject.
Also, there are four unknowns, but only three equations can be formed with the given data.

Statement 1 : Length of the train = 240 m
Time taken to cross the pole = 24 seconds
Speed = $$\dfrac{240}{24}$$ = 10 m/s

Statement 2: Since the length of the platform is unknown, we cannot calculate the speed of the train using this statement.

Hence, statement 1 alone is enough to answer the question, but statement 2 alone is not enough.

Statement 1: SI in 1 year = 600

for two years = 1200

But we do not know the value of P. So, we cannot calculate the total amount just by using statement 1.

Statement 2: We are given that the value of P is 10 times the interest in 2 years.

$$P\ =\ 10\ \times\ \dfrac{P\ \times\ 2\ \times\ R}{100}$$

$$1=\ 10\ \times\ \dfrac{2\ \times\ R}{100}$$

$$R\ =\ \dfrac{100}{20}\ =\ 5\%$$

But we do not have any information regarding the value of P to calculate the total amount.

So, we cannot calculate the total amount just by using statement 2.

Using both the statements, we know the value of R is 5% and using the SI formula, we can calculate the value of P as

$$\dfrac{P\ \times\ 1\times\ 5}{100}\ =\ 600$$

$$P\ =\ 12000$$

The total amount can be can be calculated as PI + SI in 2 years = 12000 + 1200 = 13200

We can answer the question using both the statements but any statement alone is not sufficient to answer the question.

Statement 1: The number of students is more than 22 but less than 36, so it can be any number between 22 and 36.

So, just by using statement 1, we cannot answer the question.

Statement 2: If students of the class are divided into groups, each group has exactly 11 students. This means that the number of students is a multiple of 11. But it can be any number like 11, 22, 33...

So, just by using statement 1, we cannot answer the question.

Using both statements, we know that the number must lie between 22 and 36, and it also must be a multiple of 11. The only number that satisfies these conditions is 33. So, we can determine the number of students in the class to be 33 by using both statements.

Hence, both statements together are sufficient to answer the question.

Statement 1:

If the current time were 9:00, then after half an hour, the time would be 9:30.

At 9:30:

• Hour hand angle = $$9\times30+0.5\times30=270+15=285^{\circ\ }$$.

• Minute hand angle = $$30\times6=180^{\circ}$$.

• Their smaller angle = 285−180=105(or $$360-105=255^{\circ}$$, but smaller is $$105^{\circ}$$).

So at 9:30 the hands make $$105^{\circ}$$, not $$90^{\circ}$$.

Therefore, if Statement 1 is true, the current time cannot be 9:00. That lets us answer the question definitively: No, it is not 9:00.

So Statement 1 alone is sufficient.

Statement 2:

If the current time were 9:00, then 15 minutes ago it was 8:45.

At 8:45:

• Hour hand = $$8\times30+0.5\times45=240+22.5=262.5^{\circ\ }$$.

• Minute hand = $$45\times6=270^{\circ}$$

• They differ by 270−262.5=7.5, so they do not coincide.

Hence, if Statement 2 is true, the current time cannot be 9:00. So again, we can answer the question: No.

Therefore, Statement 2 alone is sufficient.

So, each statement alone is sufficient to answer the question.

Statement 1:

The interest earned in the second year at compound interest is,

$$P\left(1\ +\ \dfrac{R}{100}\right)^2\ -\ P\left(1\ +\ \dfrac{R}{100}\right)^1\ =\ P\left(1\ +\ \left(\dfrac{R}{100}\right)^2\ +\ 2\dfrac{R}{100}-\ 1\ -\ \dfrac{R}{100}\right)\ =\ P\left(\left(\dfrac{R}{100}\right)^2\ +\ \dfrac{R}{100}\right)$$

The interest earned in the second year at simple interest is,

$$\dfrac{2PR}{100}\ -\ \dfrac{PR}{100}\ =\ \dfrac{PR}{100}$$

The ratio is,

$$\dfrac{P\left(\left(\frac{R}{100}\right)^2\ +\ \frac{R}{100}\right)}{\frac{PR}{100}}=\dfrac{5}{4}$$

$$\dfrac{R}{100}+\ 1=\dfrac{5}{4}$$

$$R\ =\ 25\%$$

Hence, we can calculate the value of R just by using the statement 1.

Statement 2:

$$P\left(1\ +\ \dfrac{3R}{100}\right)\ =\ 19500$$

The value of R cannot be calculated as the value of P is unknown.

Hence, we cannot calculate the value of R just by using the statement 2.

So, statement 1 alone is enough and statement 2 alone is not enough to answer the question.

Statement 1:

Let the speed of the train be S and the length of the train be L.

When crossing a platform, the distance travelled = L + 420m.

The time taken is given to be 12 seconds. The speed S can be calculated as,

$$S\ =\ \dfrac{L\ +\ 420}{12}$$m/s

We cannot determine the value of S, as the value of L is unknown.

Statement 2:

Let the speed of the train be S and the length of the train be L.

The time taken to cross a pole is given to be 8 seconds. The speed S can be calculated as,

$$S\ =\ \dfrac{L\ }{8}$$m/s

We cannot determine the value of S, as the value of L is unknown.

Using both statements, we get,

$$\dfrac{L\ +\ 420}{12}\ =\ \dfrac{L\ }{8}$$

Solving, we get L = 840m.

The speed of the train $$S\ =\ \dfrac{L\ }{8}\ =\ \dfrac{840}{8}\ =\ 105$$m/s $$=\ 105\ \times\ \dfrac{18}{5}\ =\ 378$$kmph

Hence, the question can be answered using both statements together.

Statement 1:

85% of the students who appeared in the examination have passed with either first class, second class, or pass class.

This only tells us the total % who passed, but not how they are split among the three classes.
Alone, this is not sufficient.

Statement 2:

750 students passed in the second class. The number of students who passed in the pass class is 28% of those who passed in the second class.

Second class = 750

Pass class = $$0.28\times750=210$$

But we don’t know the total number of students, nor how many passed in first class. Alone, this is also not sufficient.

Combining Statements 1 & 2:

Let total students = N.

From Statement 1:

Total passed = 0.85N

From Statement 2:

Second class=750, Pass class=210.

First class=Total passed−(750+210)=0.85N−960

But we still don’t know the exact value of N

Even with both statements, we cannot uniquely determine the number of students who passed in the first class.

Neither statement alone nor together is sufficient to answer the question.

Statement 1: The ratio of investments in A and B is given to be 2:3. So, the percentage of the total amount that is invested in A can be calculated as $$\dfrac{2}{2\ +\ 3}\times\ 100\ =\ 40\%$$, which makes the investment in B to be 60%.

But we cannot calculate the amount invested in Scheme B just by using statement 1.

Statement 2: The percentage of total investment in Scheme A is 40% which means that investment in B is 60%.

But we cannot calculate the amount invested in Scheme B just by using statement 2.

Both statements I and II mean the same thing. ie A has 40% of the investment and B has 60% of the investment.

So, even using both statements, we cannot determine the investment in scheme B.

Statement 1:

Let A’s investment = 3x (lakhs), B’s = 5x(lakhs).
Time: A and B invested for the full 2 years = 24 months; C joined after 6 months, so invested for 24−6=18 months with Rs. 4 lakhs.

Profit-share (capital × time, in lakh-months):

• A: 3x*24 = 72x

• B: 5x*24 = 120x

• C: 4*18 = 72

So A : B : C = 72x : 120x : 72.

But we cannot calculate the profit share of B just by using statement 1.

Statement 2:

We do not have any information regarding the profit share of B.

So, we cannot calculate the profit share of B just by using statement 2.

Using both statements together:

From statement 1, profit share of B : A = 120x : 72x = 120 : 72 = 5 : 3

Given A’s share = Rs. 60,000, B’s share = $$\dfrac{5}{3}\times60,000=100,000$$.

Hence, we can answer the question using both statements.

Using statement A alone, we cannot answer the question.

Using statement B alone, we cannot answer the question. This is because the statement talks about the 'number of students', but we cannot determine the number of students in course A.

Even by using both statements, we cannot determine the number of students in course A of college X.

So, option E is the correct answer.

We cannot answer the question using either statement A or statement B alone.

However, using both statements, we can determine the answer as 83.

Hence, Option C is the correct answer.

S.P. = 100, we need to find the profit.

I : if S.P = 90, => profit % = 20%

=> $$\frac{90 - C.P.}{C.P.} * 100 = 20

=> 6 * C.P. = 450 => C.P. = 75

Therefore, Profit = 100 - 75 = Rs. 25

Thus, I alone is sufficient.

II : Let C.P. = 3x

=> Profit = 3x / 3 = x

=> 100 - 3x = x [Since, S.P. - C.P. = Profit]

=> x = 25 = Profit

Thus, II alone is sufficient.

Let digit at unit's place = y

Digit at ten's place = x

=> Number = 10x + y

Given : | x-y | = 6

I : y > x => y - x = 6

No. can be 17,28,39

Thus, I alone is not sufficient.

II : x + y = 12

case 1 : if x - y = 6

Solving the two equations, we get x = 9 & y = 3 => No. = 93

case 2 : if y - x = 6

We get x = 3 & y = 9 => No. = 39

Thus, II alone is not sufficient.

I & II : y - x = 6 & x + y = 12

Solving, we get x = 3 & y = 9

=> number = 39

Thus, both I & II together are sufficient.

I : The difference between two digits is 9 only if the digits are 0 and 9.

No. using these digits = 90 and 09.

Since, 90 is the only two digit no. Ans - 90

Thus, I alone is sufficient.

II : Let one's digit be y and ten's digit be x. => Number = 10x + y

x + y = x - y

=> y = 0 & x can be any digit.

Thus, number can be 10,20,30,40...

Thus, II alone is not sufficient.

Statement 1:

The difference between SI and CI for 2 years is Rs. 37.50.

The formula for the difference between CI and SI for 2 years at a rate R% is:

Difference = $$P\left(\dfrac{R}{100}\right)^2$$

So:

$$37.5=P\times\left(\frac{R}{100}\right)^2$$

This gives the relation between P and R.

Not sufficient alone , since we don’t know R.

Statement 2:

Simple interest for 3 years = Rs. 2250.

Formula:

$$SI=\dfrac{P\times R\times T}{100}$$

$$2250=\dfrac{P\times R\times3}{100}$$

$$P\times R=75000$$

So we get a relation between P and R.

Not sufficient alone , since P is not uniquely known.

Combining (1) and (2):

From (1):

$$37.5=P\times\dfrac{R^2}{10000}$$

$$P\times R^2=375000$$

From (2):

$$P\times R=75000$$

Divide the two equations:

$$\dfrac{P\times R^2}{P\times R}=\dfrac{375000}{75000}$$

$$R=5$$

Now substitute into (2):

$$P\times5=75000$$

$$P=15000$$

Hence, the question can be answered using both statements.

Using statement 1, since a = 1
$$\frac{a^5b^3c^2}{a^4b^6c^2} = \frac{b^3c^2}{b^6c^2}$$
Using statement 2, since b = 1
$$\frac{a^5b^3c^2}{a^4b^6c^2} =\frac{a^5c^2}{a^4c^2}$$
If we combine both the statements,
then $$\frac{a^5b^3c^2}{a^4b^6c^2} = \frac{c^2}{c^2}$$
As it is given that c is a positive integer, we can cancel $$c^2$$ from numerator and denominator. Hence, the question can be answered using both the statements.

We cannot find the value of x using any one equation.
Using both the statements:
The given equations are 2x+3y = 12, 3x+8y = 24.
Here a/d = 2/3, b/e = 3/8 and c/f = 12/24 = 1/2.
Here, a/d is not equal to b/e. Hence, the value of x can be found.
Hence, both statements together are sufficient to answer the question and each statement alone is not sufficient.

From statement 1, 4x+8y = 24
4(x+2y) = 4*6
x+2y = 6
Hence the statement can be answered using statement 1 alone.
From statement 2, 3x+8y = 9
We cannot find the value of x+2y using statement 2 alone.
Hence, the correct answer is A.

Using statement 1,
MP = 140
Discount = 10%
SP = 90% of 140 = 126.
Hence the question can be answered using statement 1 alone.
Using statement 2,
Mark up% = 40% and discount% = 10%
We can only determine the profit% but we cannot determine the SP using statement 2 alone.
Hence the question can be answered using statement 1 alone.

Based on the first statement, we can derive the equation 2x +3y = 36.
Based on the second statement, we can derive the equation $$x^2 + y^2 = 10^2$$
By solving both equations, we get two sets of values for x, y
X =6, Y = 8.

X = 5.07, Y = 8.61.

This implies X is less than Y.

Therefore, the question can be answered using both statements, but cannot be answered using each statement alone.

I : No. of students can be = 21,22,23,24,25,26

Thus, I alone is not sufficient.

II : No. of students is multiple of 5 and is between 24 and 31

=> No. of students = 25,30

Thus, II alone is not sufficient.

I & II : From above two conclusions, only common number is 25

=> No. of students = 25

Thus, I & II together are sufficient.

I : More than four but less than ten students scored more than that of B. B’s rank is an odd number

=> B's rank can be 5 or 7 or 9.

Thus, I alone is insufficient.

II : The rank of C is 16th from the last, => C's rank from top = (25+1) - 16 = 10th.

B got more marks than C. Only two students are there between B and C.

=> B's rank from top = 7th

Thus 6 students scored more marks than B.

$$\therefore$$ II alone is sufficient.

=> Ans - (B)

I : Sum of digits of the number = 8

=> Number = 17,26,35,44...

$$\because$$ There is no unique answer. Thus, I alone is insufficient.

II : Sum of the digits is 3 more than the higher digit.

E.g. Number = 36 => Sum of digits = 9 and higher digit = 6

Similarly, numbers = 36,63,35,43...

Thus, II alone is insufficient.

I & II : Combining the two cases,

The possible numbers are = 35 , 53

Still we do not have any unique answer.

Thus, even I & II together are not sufficient.

From I : It is given that $$x \times y > 85$$

Let $$x = 100 , y = 100$$

=> $$x \times y = 100 \times 100 = 10000 > 85$$

and $$x + y = 100 + 100 = 200 > 85$$

Again let us assume $$x = 10 , y = 10$$

=> $$x \times y = 10 \times 10 = 100 > 85$$

But, $$x + y = 10 + 10 = 20 < 85$$

Thus, the statement I alone is insufficient.

From II : We have no information on the second variable.

Thus, II alone is also insufficient.

From I & II : $$x \geq 84$$

and $$y \geq 2$$ because we know that y is an integer and the product of xy has to be greater than 85.

=> $$x + y \geq 86$$, which answers our question.

Thus, I & II together are sufficient.

Considering statement 1 -

Soham did not have a two-digit score, so he must have correctly attempted all 10 questions, i.e. score 100, or wrongly attempted all 10 questions, i.e. score 0 marks. But we know that he correctly attempted at least 1 question. Hence, his score must be 100.

Hence, statement 1 alone is sufficient to answer the question.

Considering statement 2 -

Suppose his present score is 70 (7 correct and 3 wrong). If he got 3 more questions wrong as compared to the present, his new score will be 40 (4 correct and 6 wrong), which is less than 60.

Suppose his present score is 80 (8 correct and 2 wrong). If he got 3 more questions wrong as compared to the present, his new score will be 50 (5 correct and 5 wrong), which is less than 60.

Hence, statement 2 alone is not sufficient to answer the question.

Hence, option A is correct.

C.I. = $$P [(1 + \frac{R}{100})^T - 1]$$

I : Interest = 100 and sum = 1000 and time = 1 year

=> $$100 = 1000 [(1 + \frac{R}{100})^1 - 1]$$

=> $$\frac{100}{1000} = [(1 + \frac{R}{100}) - 1]$$

=> $$\frac{R}{100} = \frac{1}{10}$$

=> $$R = \frac{100}{10} = 10 \%$$

$$\therefore$$ C.I. after 3 years = $$1000 [(1 + \frac{10}{100})^3 - 1]$$

= $$1000 [(\frac{11}{10})^3 - 1]$$

= $$1000 [\frac{1331}{1000} - 1] = 1000 \times \frac{331}{1000}$$

= $$331$$

Thus, I alone is sufficient.

II : Sum = 1000 , rate = $$R \%$$

S.I. after 2 years = $$\frac{P \times R \times T}{100}$$

= $$\frac{1000 \times R \times 2}{100} = 20R$$

C.I. after 1st year = $$\frac{R}{100} \times 1000 = 10R$$

C.I. after 2nd year = $$10R + \frac{R}{100} \times 10R = 10R + \frac{R^2}{10}$$

=> Required difference = 10

=> $$(10R + 10R + \frac{R^2}{10}) - (20R) = 10$$

=> $$\frac{R^2}{10} = 10$$

=> $$R^2 = 10 \times 10 = 100$$

=> $$R = \sqrt{100} = 10 \%$$

$$\therefore$$ C.I. after 3 years = $$1000 [(1 + \frac{10}{100})^3 - 1]$$

= $$1000 [(\frac{11}{10})^3 - 1]$$

= $$1000 [\frac{1331}{1000} - 1] = 1000 \times \frac{331}{1000}$$

= $$331$$

Thus, II alone is sufficient.

Thus, either statement alone is sufficient.

Clearly, we have to use both equations.

Let distance between A and B = $$d$$ km

Let speed of boat in still water = $$x$$ kmph

and speed of current = $$y$$ kmph

Using, $$time = \frac{distance}{speed}$$

=> $$\frac{d}{x + y} = 2$$ ---------Eqn(1)

and $$\frac{d}{x - y} = 4$$ ---------Eqn(2)

Dividing equation (2) by (1)

=> $$\frac{x + y}{x - y} = \frac{4}{2}$$

=> $$x + y = 2x - 2y$$

=> $$2x - x = y + 2y$$

=> $$x = 3y$$

=> $$\frac{x}{y} = \frac{3}{1}$$

$$\therefore$$ Ratio between speed of boat and current is known and not the exact value.

Thus, even both statements together are insufficient.

I : We know that ‘$’ means ‘÷’, ‘*’ means ‘x’ and ‘#’ means ‘+’

=> 144 $ 16 *7 # 9 = $$144 \div 16 \times 7 + 9$$

= $$63 + 9 = 72$$

Thus, the statement I alone is sufficient.

II : 16 $ 4 *2 # 2 =10

=> We do not have any information regarding ‘$’, ‘*’ and ‘#’  means just by using statement 2.

So, we cannot answer the question just by using statement 2.

$$\therefore$$ Statement 1 alone is sufficient.

Let number of oranges be $$o$$, apples be $$a$$ and bananas be $$b$$.

I : Total fruits = 100

Number of oranges and bananas = $$o+b=\frac{50}{100}\times100=50$$

=> $$a=100-50=50$$

Thus, I alone is sufficient.

II : We don't know the total fruits, hence II alone is insufficient.

$$\therefore$$ I alone is sufficient.

=> Ans - (A)

Brinda's rank from top = 17th

I : There are 70 students in her class.

=> Brinda's rank from last = $$(70+1)-17=54$$

Thus, I alone is sufficient.

II : Nisha's rank from top = 20th and from last = 51st

Thus, total students in class = $$(51+20)-1=70$$

=> Brinda's rank from last = $$(70+1)-17=54$$

Thus, II alone is sufficient.

$$\therefore$$ Each statement alone is sufficient.

=> Ans - (D)

I : Since, we do not know the principal amount, thus I alone is not sufficient to find the rate of interest.

II : Let principal be $$100x$$

=> Amount after 5 years = $$\frac{180}{100} * 100x = 180x$$

=> interest after 5 years = $$180x - 100x = 80x$$

$$\because$$ $$r = \frac{S.I. \times 100}{P \times T}$$

=> $$r = \frac{80x \times 100}{100x \times 5}$$

=> $$r = 16$$%

Thus, II alone is sufficient.

I : Let the number of boys and girls be $$8x$$ and $$5x$$ resp

=> $$8x - 5x = 24$$

=> $$x = \frac{24}{3} = 8$$

Hence, total number of children in the class = $$13x$$ = 13 * 8 = 104

Statement I alone is sufficient

Statement II does not give any relation between the number of boys and girls. Thus, it is not sufficient.

Let the number be $$10x + y$$

I : $$x - y = 7$$ or $$y - x = 7$$

II : $$x + y = 11$$

I & II : Solving $$x - y = 7$$ and $$x + y = 11$$

=> $$x = 9 , y = 2$$

Solving $$y - x = 7$$ and $$x + y = 11$$

=> $$x = 2 , y = 9$$

Hence, the possible numbers are 92 and 29. Since, unique answer is not determined

Ans - (E)

I : Speed = 12 kmph

= 12 * $$\frac{5}{18} = \frac{10}{3}$$ m/s

II : Circumference of ground = $$\frac{22}{7} * 700$$ = 2200 m

When both statements are combined, the required time will be

Time = $$\frac{Distance}{Speed}$$

= $$\frac{2200 * 3}{10}$$ = 660 s

From statement 1, we can just determine that the sum of all the weights that he had was equal to 15 kg.

From statement 2, we can conclude that he can weigh all the weights from 1 to 15 kg, but that does not mean that he cannot weigh the weights above 15 kg.

From statements II and I:
We can conclude that he had 1, 2, 4 and 8 kg as the weight blocks.

Hence, using both statements together, we can answer the question. Hence, C is the correct answer.

From statement 1, we can infer that 45 = 10Cx. A bit of trial and error will give 45 = 10C2. But, 10C8 = 10C2. We know that 2 students have either been selected or rejected.

From statement 2, we know that the number of people selected is less than the number of people rejected.
Combining statements 1 and 2, we can determine that 2 people must have been selected for the competition. Hence, option C is the right answer.

We cannot answer the question using I, as Mohit can be 2nd or 13th because we know that there are 5 people between Mohit and Sameer, and we also know that Sameer got 7th rank from the top.

Using only 2, we can determine Mohit's rank because we know that there are 9 people after Mohit out of 92, so his rank is 83.

Hence, statement 2 alone is enough to answer the question but statement 1 alone is not enough to answer the question.

From statement 1 alone, we can find that Mayank took 300/50= 6 days to complete the book. But we can’t tell till Wednesday how many pages he read.
From statement 2 alone, we find that Mayank reads the last 75 pages on Thursday, so till Wednesday, he must’ve read 300 - 75 = 225 pages, irrespective of which day he started reading the book. Hence, statement 2 alone is sufficient to answer the question.
Hence, option B is correct.

I : Harish missed his usual train at 10.25 a.m. and in every 5 minutes, a train comes.

Harish could have caught any train at 10.30 a.m. , 10.35 a.m. , 10.40 a.m. and so on.

There is no information about that, so it is not possible to determine which train Harish caught.

Thus, I alone is insufficient.

Even in statement II or even after combining both the statements, the same problem arises.

Thus, data in both statements even together is not sufficient to answer the question.

Let rate of interest = $$R \%$$

I : P = Rs. 25,000 and time period = 2 years

=> $$S.I. = \frac{P \times R \times T}{100}$$

= $$\frac{25000 \times R \times 2}{100} = 500 R$$

Interest obtained from compound interest = $$P [(1 + \frac{R}{100})^T - 1]$$

=> $$(500R + 250) = 25000 [(1 + \frac{R}{100})^2 - 1]$$

=> $$250 (2R + 1) = 25000 [(1 + \frac{R}{100})^2 - 1]$$

=> $$2R + 1 = 100 (1 + \frac{R^2}{100^2} + \frac{2 R}{100} - 1)$$

=> $$2R + 1 = \frac{R^2}{100} + 2R$$

=> $$R^2 = 100 \times 1 = 100$$

=> $$R = \sqrt{100} = 10 \%$$

Thus, statement I alone is sufficient.

II : Let original sum = simple interest = $$Rs. P$$ and time period = 10 years

=> $$S.I. = \frac{P \times R \times T}{100}$$

=> $$P = \frac{P \times R \times 10}{100}$$

=> $$R = \frac{100}{10} = 10 \%$$

Thus, statement II alone is also sufficient.

Thus, either statement alone is sufficient.

To find the number of plants planted, we need to find the number of columns and rows and then multiply them.

Let number of columns = $$c$$ and number of rows = $$r$$

(I): Number of columns is 4 more than rows

=> $$c - r = 4$$

But we do not know the exact value. So, statement 1 alone is not sufficient.

(II): In each column, the number of trees is even, => the number of rows is even in number

Again, we do not know the exact value. So, statement II alone is not sufficient.

Combining the above statements, we still do not know the exact number of columns and rows. So, statements I and II together are not sufficient.

=> Ans - (E)

I : Let present age of son = $$x$$ years

=> Present age of father = $$5x$$ years.

There is no other information, so statement I is not sufficient.

Similarly, II alone is not sufficient.

Combining both statements, we get :

=> $$(5x - 5) = 15 (x - 5)$$

=> $$5x - 5 = 15x - 75$$

=> $$15x - 5x = 75 - 5$$

=> $$10x = 70$$ => $$x = \frac{70}{10} = 7$$

$$\therefore$$ father's age = $$5 \times 7 = 35$$ years.

Thus, both statements together are sufficient.

If we assume the salary of Manoj to be 100x, then 20x is spent on transport, 30x is spent on food, he saves 35x, and the rest, which is 100x - 20x - 30x - 35x = 15x, is spent on entertainment.

Statement 1: If he spends 5000 on food, which we assumed to be 30x, then the amount spent on entertainment, which is 15x, can be calculated as 5000/2 = 2500. So, statement 1 alone is enough to answer the question.

Statement 2: If he saves 3500 which is 35x, then the amount spent on entertainmnet can be calculated as 15x which is 1500. So, statement 2 alone is enough to answer the question.

Hence, each statement alone is enough to answer the question.

Using the statement I alone:
Let 'a' be the HCF of x and y. Then, LCM = 20a
Hence, the product of the numbers = a*20a = 20a$$^2$$. We can see that the product depends on x. Hence, we can't find the value of x*y using statement I alone.

Using statement II alone:
We have no information regarding the LCM of x and y. Hence, using statement II alone, we can't find out the value of x*y.

Using both statements together:
From statement (I), we can see that the value of x*y = 20a$$^2$$
From statement (II), we can see that a = 30. Hence, we can find out the value of 20a$$^2$$ = 18000. Therefore, option C is the correct answer.