I. The first Monday of that month fell on 4th.

If first Monday falls on 4th then the next Monday will be after 7 days that is 11th and next on 18th and next on 25thMarch. After that next Monday will fell on 1 April. So, last Monday is on 25th March.

II. The last day of that month was Saturday.If the last day of March that is 31st is Saturday then the last Monday will be on 26th March.

So, Either I or II is sufficient.

From statement I:

Lucy is half of the age of her mother who was 43 years old 5 years back.

So, now Lucy’s mother’s age is 48 years. Hence, Lucy’s age will be 24 years. But, our question cannot be answered as age of Kate is not known.

∴ Statement I alone is not sufficient to answer the question.

From statement II:

Kate is younger sister of Lucy and is one third the current age of her mother.

With this, ages of Kate and Lucy cannot be found, and hence question cannot be answered.

∴ Statement II alone is not sufficient to answer the question.

From statements I and II together:

Lucy is half of the age of her mother who was 43 years old 5 years back.

So, now Lucy’s mother’s age is 48 years. Hence, Lucy’s age will be 24 years.

Kate is younger sister of Lucy and is one third the current age of her mother.

So, Kate’s age = 48/3 = 16 years

Now, the question can be answered as ages of Kate and Lucy are known.

∴ Using both the statements together, we can answer the given question.

Let length and breadth of rectangle be L and B, respectively. Let radius of circle be R.

Area of a rectangle is twice the area of a circle.

⇒ LB = 2π R²

From statement I:

Radius of circle is equal to breadth of rectangle.

⇒ R = B

⇒ LB = 2π B²

⇒ L = 2π B

⇒ Ratio of length and breadth can be found.

∴ Statement I alone is sufficient to answer the question.

From statement II:

Perimeter of rectangle is equal to circumference of circle.

⇒ 2(L + B) = 2π R

⇒ (L + B) = π R

Also, LB = 2π R²

⇒ LB/2π = (L+B)²/π²

⇒ π LB = 2L² + 2B² + 4LB

Divide equation by B², and put L/B = T

⇒ 2T² + (4–π)T + 2 = 0

Discriminant of this equation is negative and hence no real roots exist.

∴ Statement II alone is not sufficient to answer the question.

From statement I:

All three digits of the street number are different odd numbers.

Number can be 357 or 573 or 579, and so on. It cannot be concluded if it is greater than 500 or not.

∴ Statement I alone is not sufficient to answer the question.

From statement II:

No number larger than the current street number can be formed with the three digits of the street number, by taking any other arrangement of them.

It means that if digits of number are 1, 2 and 4, then number will be 421.

Such numbers can be 421, 562, 891, 321, and so on. It cannot be concluded if it is greater than 500 or not.

∴ Statement II alone is not sufficient to answer the question.

From statements I and II together:

All three digits of the street number are different odd numbers. No number larger than the current street number can be formed with the three digits of the street number, by taking any other arrangement of them.

Smallest possible three digit number with different odd digits and satisfying this property will be 531.

⇒ Number must be greater than 500.

∴ Using both the statements together, we can answer the given question.

Let speed = v, distance = x, time = t

Time(t) = Distance/ Speed = x/v

From statement I:

If he walks 25% faster than his usual speed then he reaches 20 seconds earlier.

Then speed = v + v × 25% = v + v/4 = 5v/4

Now according to the statement,

x/(5v/4) = t – 20

⇒ 4x/5v = t – 20

⇒ x/v = (5/4) × (t – 20)

⇒ t = (5/4) × (t – 20)                     [∵ x/v = t]

⇒ 4t = 5t – 100

⇒ t = 100 seconds

Statement I alone is sufficient to answer the question.

From statement II:

If he walks half of his usual speed, he takes 100 seconds more to reach the finishing point B.

⇒x/(v/2) = t + 100

⇒ 2x/v = t + 100

⇒ 2t = t + 100                     [∵ x/v = t]

⇒ t = 100 seconds

Statement II alone is sufficient to answer the question.

From statement III:

If he walks one-fourth of his usual speed, he takes 300 seconds more to reach the finishing point B.

⇒ x/(v/4) = t + 300

⇒ 4x/v = t + 300

⇒ 4t = t + 300                      [∵ x/v = t]

⇒ 3t = 300

⇒ t = 100 seconds

Statement III alone is sufficient to answer the question.

Hence, all statements I, II and III alone is sufficient to answer the question.

From statement I:

Let number of apples with James be M and with John be N.

If John gives 7 apples to James, James will have twice the number of apples that John has.

⇒ M + 7 = 2(N – 7)

There are two variables and one equation, and hence unique values of M and N cannot be found.

∴ Statement I alone is not sufficient to answer the question.

From statement II:

Let number of apples with James be M and with John be N.

If James gives an apple to John, they will have equal number of apples.

⇒ M - 1 = N + 1

There are two variables and one equation, and hence unique values of M and N cannot be found.

∴ Statement II alone is not sufficient to answer the question.

From statements I and II together:

Let number of apples with James be M and with John be N.

If John gives 7 apples to James, James will have twice the number of apples that John has.

⇒ M + 7 = 2(N – 7)

If James gives an apple to John, they will have equal number of apples.

⇒ M - 1 = N + 1

There are two variables and two equations, which can be solved to find M and N, and subsequently M + N.

Solving, we get M = 25, N = 23.

∴ Using both the statements together, we can answer the given question.

From statement I:

After 20 minutes, the time will be 7 pm.

⇒ Time now is 6:40 pm.

∴ Statement I alone is sufficient to answer the question.

From statement II:

The next day will start after 5 hours from now.

⇒ It is five hours prior to 12 midnight, i.e., 7 pm.

∴ Statement II alone is sufficient to answer the question.

From statement I:

Largest number among them is 41.

⇒ The remaining two numbers can be at most 8 and at least 1. Possible pairs are (1, 8), (2, 7), (3, 6), (4, 5). In each of the cases the product of three numbers will be a 3 digit number.

∴ Statement I alone is sufficient to answer the question.

From statement II:

At least one of the three numbers has 2 digits.

We see that this is a statement that is always true and provides no extra information. Three natural numbers whose sum is 50 cannot be all 1 digit numbers, as sum would not exceed 27 in that case.

∴ Statement II alone is sufficient to answer the question.

From statement I:

Stacy is 9 inches shorter than Ricky. No comparison between Heath and Stacy can be drawn.

∴ Statement I alone is not sufficient to answer the question.

From statement II:

Heath is neither taller nor shorter than Ricky. So, Heath’s height is equal to Ricky. No comparison between Heath and Stacy can be drawn.

∴ Statement II alone is not sufficient to answer the question.

From statements I and II together:

Stacy is 9 inches shorter than Ricky. Heath is neither taller nor shorter than Ricky. So, Heath’s height is equal to Ricky.

Stacy is 9 inches shorter than Heath; or Heath is 9 inches taller than Stacy.

∴ Using both the statements together, we can answer the given question.

From statement I:

After 6 hours, the hour and minute hands of the clock will be perpendicular again.

⇒ Present time can be 3 o’clock or 9 o’clock. But it cannot be uniquely determined.

∴ Statement I alone is not sufficient to answer the question.

From statement II:

After 3 hours, the hands of the clock will be along a straight line.

⇒ Present time can be 3 o’clock or 9 o’clock. But it cannot be uniquely determined.

Note: Minute and Hour hands are in straight line at both 12 o’clock (in same directions) and 6 o’clock (in opposite direction).

∴ Statement II alone is not sufficient to answer the question.

From statements I and II together:

After 6 hours, the hour and minute hands of the clock will be perpendicular again. After 3 hours, the hands of the clock will be along a straight line.

⇒ Here again, present time can be 3 o’clock or 9 o’clock, even if both informations are used.

Essentially, both statement I and II provide equivalent information.

∴ Even using both the statements together, we cannot answer the given question.

From statement I:

Y leaves a remainder of 3 when divided by 8. So, Y can be 3, 11, 19, and so on. X can be any natural number.

⇒ X^ycan be 2³, whose last digit is 8, or it can be 3³, whose last digit is 7, or there can be many other cases. Last digit cannot be uniquely found.

∴ Statement I alone is not sufficient to answer the question.

From statement II:

X leaves a remainder of 7 when divided by 10. So, X can be 7, 17, 27, and so on. Y can be any natural number.

⇒ X^ycan be 7³, whose last digit is 3, or it can be 7², whose last digit is 9, or there can be many other cases. Last digit cannot be uniquely found.

∴ Statement II alone is not sufficient to answer the question.

From statements I and II together:

Y leaves a remainder of 3 when divided by 8. So, Y would be of the form 8n + 3, i.e. 4(2n) + 3, where n is a whole number.

Y leaves a remainder of 7 when divided by 10. So, X would be of the form 10m + 7, where m is a whole number.

⇒ X^y = (10m + 7)^{( 4(2n) + 3)}

We know that last digit of any natural number repeats after its every four powers.

⇒ Last digit of X^y will be same as last digit of (10m + 7)(³)

When we expand this by binomial expansion, all terms will be a multiple of 10, except 7³.

So, last digit of X^y will be last digit of 7³, i.e. 3.

∴ Using both the statements together, we can answer the given question.

From statement I:

If T is multiple of 216, then T can be 216 which is not divisible by 144 or T can be 432 which is divisible by 144. So, divisibility of T with 144 cannot be uniquely established.

∴ Statement I alone is not sufficient to answer the question.

From statement II:

T is divisible by square of cube of 2.

Square of cube of 2 = (2 × 2 × 2)² = 64

⇒ T is a multiple of 64.

So, T can be 192 which is not divisible by 144 or T can be 576 which is divisible by 144. So, divisibility of T with 144 cannot be uniquely established.

∴ Statement II alone is not sufficient to answer the question.

From statements I and II together:

T is divisible by square of cube of 2.

Square of cube of 2 = (2 × 2 × 2)² = 64

⇒ T is a multiple of 64.

Also, T is a multiple of 216.

⇒ T will be a multiple of LCM(216, 64).

⇒ T will be a multiple of 1728.

We know, 1728 = 144 x 12.

If T is divisible by 1728, it would be divisible by all its factors.

∴ T will be divisible by 144.

∴ Using both the statements together, we can answer the given question.

From statement I:

The series is of positive consecutive integers. Knowing ratio of last number and first number, we cannot determine the series uniquely. It can be numbers from 1 to 101, or from 2 to 202, or from 3 to 303, and so on. Hence, sum of series cannot be uniquely found.

∴ Statement I alone is not sufficient to answer the question.

From statement II:

There are not more than 200 terms in the series. Knowing this, we cannot determine the series uniquely. It can be numbers from 1 to 100, or from 1 to 101, or from 1 to 103, and many other options are possible. Hence, sum of series cannot be uniquely found.

∴ Statement II alone is not sufficient to answer the question.

From statements I and II together:

The series is of positive consecutive integers. Knowing that ratio of last number and first number, we can say that numbers in series would be from 1 to 101, or from 2 to 202, or from 3 to 303, and so on. Also, we know that there are not more than 200 terms in the series. So, only option possible is 1 to 101, as all other options will have more than 200 terms.

⇒ Series will be positive consecutive numbers from 1 to 101. Hence, sum of series can be uniquely found.

∴ Using both the statements together, we can answer the given question.

X² + X + C = 0. Using the formula for the roots: \(\frac{{ - b \pm \sqrt {({b^2} - 4ac} }}{{2a}}\) 

 we have

\(\frac{{ - 1 \pm \sqrt {({1^2} - 4c} }}{2}\)

 Given the equation has one root 

\(\Rightarrow \sqrt {({1^2} - 4{\rm{c}}} = 0\)

⇒ c = ¼

Hence, statement I is sufficient to answer this question.

From statement II:

The sum of the roots is -b/a. This has got nothing to do with C.

Hence, statement II is not sufficient to answer this question.

Given, log 2, log (2^x + 1) and log (2^x + 1.5) are in AP.

⇒log (2^x + 1) = \(\frac{{\log 2 + \log ({2^x} + 3)}}{2} \Rightarrow 2\log \left( {{2^x} + 1} \right) = \log 2 + {\rm{log}}\left( {{2^x} + 1.5} \right)\)

⇒ log (2^x + 1)² = log (2 × (2^x + 1.5))

Equating the terms inside the logarithms (or, simply removing the logarithms),

(2^x + 1)²= (2 × (2^x + 1.5))

⇒ (2^x)² + (1)² + 2 × 1 × 2^x = 2(2^x + 1.5)

For simplicity, let us assume 2^x = a

⇒ a² + 1 + 2a = 2(a + 1.5)

⇒a² + 1 + 2a = 2a + 3

⇒a² + 1 = 3

⇒ a² = 2

⇒ a = √2 (We do not consider -√2 because we need to equate it with 2^x and not -2^x)

Substituting the original value of a,

2^x = √2 = 2^1/2

⇒ x = ½

Hence, statement I alone is sufficient to answer this question.

From statement II:

It is only given that x is a non-negative integer. In that case, x can take infinite values.

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

Given, n = r + 2

⇒ⁿC_r= ^(r+2)C_r = \(\frac{{\left( {r + 2} \right)!}}{{\left( {r!} \right) \times \left( {2!} \right)}} = \frac{{\left( {r + 1} \right)\left( {r + 2} \right)}}{2}\)

No comprehensive answer can be derived from this expression.

Hence, statement I is insufficient to solve the question.

From statement II:

r = 25% of (30 - 40% of 25)

We use BODMAS to solve certain parts of the question to get a simplified form.

BODMAS stands for:

B – Brackets

O – Of (this simply stands for multiplication)

D – Division

M – Multiplication

A – Addition

S – Subtraction

The above is the standard order in which a given question is simplified.

On the RHS, we have

25% of (30 - 40% of 25)

As per the BODMAS, we solve the expression in the brackets first.

⇒30 - 40% of 25

Even within the bracket, we found an “OF”. Since there are no other brackets inside the bracket, we solve this part first.

40% of 25 = \(\frac{{40}}{{100}} \times 25 = 10\)

⇒ 30 – 10 = 20

Hence,

25% of 20 = \(\frac{{25}}{{100}} \times 20 = 5\;\)

Hence, r = 5.

Although the value of r has been found, the last two non-zero digits have not been found.

Hence, statement II is insufficient to solve this question.

Combining the two by substituting the value of r in \(\frac{{\left( {r + 1} \right)\left( {r + 2} \right)}}{2},\)

We have \(\frac{{\left( {r + 1} \right)\left( {r + 2} \right)}}{2} = \frac{{6 \times 7}}{2} = 21\) 

Hence, the two non-zero digits are 1 and 2.

Hence, both the statements combined can give you the solution.

From statement I:

a² + b² = 40

No values of a or b or that of the question can be arrived at using this statement.

Hence, statement I is insufficient to answer this question.

From statement II:

a × b = 12

No values of a or b or that of the question can be arrived at using this statement.

Hence, statement II is insufficient to answer this question.

Combining both,

a² + b² = 40

a × b = 12

(a - b)² = a² + b² - 2ab

⇒ (a - b)² = 40 – 2 × 12 = 40 – 24 = 16

⇒ a – b = ± 4

Since it is not known if a > b or b > a, determining the value of a – b is not possible.

Hence, both the statements combined also cannot give us the answer.

A² + A = ¾  and B² + 2B = 24

Solving both the quadratic equations to obtain the roots,

A² + A = ¾

We can solve this equation using two methods. One is to use formula \(\frac{{ - b \pm \left( {\sqrt {{b^2} - 4ac} } \right)}}{{2a}}\) 

or we can solve it by adding a certain number to both sides such that it makes the LHS a perfect square.

In this case, we add ¼ on both sides to make the LHS of the above equation a perfect square.

⇒ A² + A+ ¼ = ¾ + ¼ ⇒A² + A+ ¼ = 1

A² + A + ¼ = (A + ½)²

⇒ (A + ½)²= 1

⇒ A + ½ = ± 1

⇒ A = +1/2 or -3/2

Similarly, for B² + 2B = 24, we add 1 on both sides,

⇒ B² + 2B+ 1 = 24 + 1 = 25

⇒ (B + 1)² = 25

⇒ B + 1 = ± 5 ⇒B = 4 or – 6

The question cannot be answered by using only these two statements, since each of X and Y have two values.

Hence, statement I is insufficient to answer the question.

From statement II:

A² – 3A = -9/4 and B² – 3B = 4

We solve these equations in a method similar to the one explained previously.

We add 9/4 on both sides

⇒ A² – 3A+ (9/4) = (-9/4) + (9/4)

⇒ A² – 3A+ (9/4) = 0

⇒ (A – (3/2))² = 0

⇒ A = 3/2

Similarly, B² – 3B = 4

We add 9/4 on both sides

B² – 3B+ (9/4) = 4 + (9/4)

⇒ B² – 3B + (9/4) = 25/4

⇒ (B – (3/2))² = 25/4

⇒ B – (3/2) = ± 5/2

⇒ B = 4 or –1

Although the value of A is known, the question cannot be answered since B has 2 values, one less than A and the other greater than A.

Combining the two statements we are not able to reach to a conclusion about values of A and B, and hence the relation between them cannot be determined.

The question can not be answered even if both the statements are combined.

From Statement 1

We know that, when a number can be expressed as a₁^m1 × a₂^m2 × a₃^m3 × …. × a_n^mn,

Where, a₁, a₂, a₃… ,an are prime numbers and m₁, m₂, …, m_n are positive integers.

Total number of factors of the number = (m₁+1) × (m₂+1) × (m₃+1) × …(m_n+1)

N has a total of 16 factors.

∴ 16 = (m₁+1) × (m₂+1) × (m₃+1) × …(m_n+1)

Case 1: 16 = 1 × 16 = (0 + 1) × (15 + 1)

N can have just 1 prime factor raised to 15.

Case 2: 16 = 2 × 8 = (1 + 1) × (7 + 1)

N can have 1 prime factor raised to 7 and the other raised to 1.

Case 3: 16 = 2 × 2 × 4 = (1 + 1) × (1 + 1) × (3 + 1)

N can have two prime factors raised to 2 and 1 prime factor raised to 3

Case 4: 16 = 4 × 4 = (3 + 1)× (3 + 1)

N can have just 2 prime factors, each raised to the power of 3

Case 5: 16 = 2 × 2 × 2 × 2 = (1 + 1) × (1 + 1) × (1 + 1) × (1 + 1)

N could have 4 prime factors each raised to the power 1.

Thus, statement 1 alone is insufficient to answer the question.

From Statement 2

The prime factors of N are 2, 3, 5 and 7.

But, we do not know the total number of factors and hence we cannot determine the powers to which these prime factors are raised.

∴ We can determine only a few double-digit factors but not all of them.

Thus, statement 2 is also insufficient to answer the question.

From Statements 1 and 2

N has 16 factors in total and has prime factors 2, 3, 5and 7.

∴ Clearly, all the prime factors are raised to the power 1.

∴ the number of 2- digit factors is 9

Thus, Statement I and Statement II together are sufficient, but neither of the two alone is sufficient to answer the question.

From Statement 1

K can be 1, 2, 3, 5, 6, 10, 15, 30.

∴ K can be < or > than 5.

Thus, statement 1 alone is insufficient to answer the question.

From Statement 2

K can NOT be 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60 or 120.

As K cannot take any values between 1 and 6, (and K is positive)

∴ K > 5.

Thus, statement 2 alone is sufficient to answer the question.

Let the age of son and father be x and y respectively.

From statement 1,

5 years ago his father’s age was five times his age.

y – 5 = 5(x – 5)

⇒ y = 5x – 20

We have two variables and one equation.

∴ Statement 1 alone is not sufficient to answer the question.

From statement 2,

5 years later he would be half as old as his father’s present age.

x + 5 = y/2

2x + 10 = y

Again, we have two variables and one equation.

∴ Statement 2 alone is not sufficient to answer the question.

From statement 1 and 2:

Solving the two equations simultaneously, we get x = 10 and y = 30.

Thus both statements are required to answer the question.

In a triangle, sum of angles = 180°

From statement I,

∠QPR is 50° and ∠QRP is smaller than ∠QPR

Since ∠QPR is 50°, sum of remaining two angles = 180° – 50° = 130°

Since one of the remaining angles is 90° , the third angle must be: 130° – 90° = 40°

Since ∠QRP is smaller than ∠QPR, ∠QRP cannot be 90°.

∴ ∠PQR must be 90°.

∴ The statement I alone is sufficient to answer the question

From statement II,

Given: ∠QRP = ∠QPR

Since both angles except ∠PQR are equal therefore none of them can be 90°, because sum of all three angles must be equal to 180°

Thus, ∠PQR is 90°

∴ The statement II alone is sufficient to answer the question

Let’s assume that the age of A is x years and that of B is years.

From statement I,

y = x + 7

∴ the product of their ages = xy = x × (x + 7)

∵ we don’t know the value of either of x, y or any other relation between them, we cannot find their product with the available information.

∴ The statement I alone is not sufficient to answer the question

From statement II,

y = 14

x = y/2

⇒ x = 14/2 = 7

Product of their ages = xy

⇒ Product of their ages = 7 × 14 = 98

∴ The statement II alone is sufficient to answer the question

Let’s assume that the cost price of the calculator is Rs. C.

From statement (1):

The cost price of 5 such calculators is equal to selling price of 4 such calculators.

⇒ 5C = 4 × 2575

⇒ C = Rs. 2060

∴ Profit = Selling price – Cost price = 2575 – 2060 = Rs. 515

∴ Statement I alone is sufficient to answer the question.

From statement (2):

25% profit is earned by selling each calculator.

∵ Cost price of each calculator = Rs. C,

Profit earned by selling each calculator = 25% of Rs. C = 0.25C

Selling price = Cost price + profit

∴ Selling price = C + 0.25C = 1.25 C = 2575

⇒ C = 2575/1.25 = 2060

∴ Profit = 0.25 × 2060 = Rs. 515

∴ The statement II alone is sufficient to answer the question.

∴ Either to statement I alone or statement II alone is sufficient to answer the question.

Assume that 1 man can complete the work alone in M days.

∴ In 1 day 1 man can do 1÷M part of the work. [Assuming total work = 1 unit]

⇒In 1 day 16 man can do 16÷M part of the work.

1 woman can complete the work alone in W days.

∴ In 1 day 1 woman can do 1÷W part of the work. [Assuming total work = 1 unit]

⇒In 1 day 8 man can do 8÷W part of the work.

If in K days 16 men and 8 women together can complete the piece of work then, [Assuming total work = 1 unit]

\(K\left( {\frac{{16}}{M} + \frac{8}{W}} \right) = 1\)

From statement 1:

8 men complete the piece of work in 10 days.

\(\Rightarrow \frac{{8 \times 10}}{M} = 1\)

⇒ M = 80 days.

From statement 2:

16 women complete the piece of work in 10 days.

\(\Rightarrow \frac{{16 \times 10}}{W} = 1\)

⇒ W = 160 days.

From statement 3:

5 women complete the piece of work in 32 days.

\(\Rightarrow \frac{{32 \times 5}}{W} = 1\)

⇒ W = 160 days.

From statement 1 & 2 or 1 & 3:

M = 80 days & W = 160 days.

 We know that,

\(K\left( {\frac{{16}}{M} + \frac{8}{W}} \right) = 1\)

\(\Rightarrow K\left( {\frac{{16}}{{80}} + \frac{8}{{160}}} \right) = 1\)

⇒ K = 4

∴ In 4 days 16 men and 8 women together complete the piece of work.

∴ Only I and either II or III required to solve the question.

From statement I,

Given: He spends 13% on insurance premium, 5% on medicines and 14% on rest expenses.

Here expenditures are given but with these data we cannot reach at the salary of X.

∴ The data in Statement I alone are not sufficient to answer the question

From statement II,

Given: Even after spending of 32% of his salary, he is able to save Rs.4080

Let his salary be Rs. A.

Then, after spending 32% of his salary, remaining salary = (100 – 32)% = 68%

∴ 68% of a = 4080

\(\Rightarrow \frac{{68}}{{100}} \times a = 4080\)

\(\Rightarrow a = 4080 \times \frac{{100}}{{68}}\)

⇒ a = Rs 6000

∴ The data in Statement II alone are sufficient to answer the question

Let the digit at unit place of the two digit number be x and that at ten’s place be y.

From statement I,

⇒ x – y = 2

This relation can be true for many two digits number, like 13, 24, 35…………

∴ The statement I alone is not sufficient to answer the question

From statement II,

x + y = 8

This relation can be true for many two digits number, like 17, 28, …..

∴ The statement II alone is not sufficient to answer the question

Combining I and II,

x – y = 2

x + y = 8

On solving these two relations, we get

x = 5 and y = 3

Thus, number is 35.

∴ Both statements I and II together are needed to answer the question

We know that simple interest (SI) = principal (P) × rate of interest (r) × time

From statement I:

Given: SI = Rs. 5300

Time = 1 year

⇒ 5300 = P × r ………………. (1)

From statement II:

We know that difference between CI & SI after 2 years if rate of SI & CI are same = \({P} \times {\left( {\frac{r}{{100}}} \right)^{2\;}}\)

According to the statement: \({P} \times {\left( {\frac{r}{{100}}} \right)^{2\;}} = 1060 \ldots \ldots \ldots \ldots \ldots .\left( 2 \right)\)

[CI = compound interest, P = principal; r = rate of interest]

Statement (1) & (2) can be solved simultaneously as there are 2 variables & 2 equations.

From statement III:

Assume that the amount is Q.

The amount doubles itself in 5 years with simple interest.

∴Simple interest = amount – principal = 2Q – Q = Q

\(\Rightarrow \frac{{Q \times 5 \times R}}{{100}} = Q\)

⇒ R = 20%

[R = rate of interest]

∴The question can be answered by using only the 3rd option or using options I and II.

From statement 1:

If one diagonal of square is D units, then one side is D÷√2

⇒ Area = (D ÷√2)² = (D² ÷ 2) sq. units.

From statement 2:

 If length of one side of a square = A units, then area = A² sq. units.

From statement 3:

If perimeter of the square = P units, then length of one side of the square = P ÷4 units

⇒ Area = (P ÷4)² = (P² ÷ 16) sq. units.

∴ Any one of the three statements can be used to measure the area.

From statement I,

x > 200

Since there is no data given about y

Thus, the data in Statement I alone are not sufficient to answer the question

From statement II,

y > 190

Since there is no data given about x

Thus, the data in Statement II alone are not sufficient to answer the question

Combining I and II,

x> 200

y> 190

Even on combination of these statements, relation cannot be established because there are few value of y smaller than x and few values greater than x.

Thus, the data even in both the statements I and II together II alone not sufficient to answer the question.