XAT Data Sufficiency Questions

The age of Nadeem is a two-digit number. When squared yields a three-digit number whose last digit is Y. Y is a prime number.

Using statement 1:

When a number is squared :

The last digit of the number can be :

1, 2, 3, 4, 5, 6, 7, 8, 9. When squared the last digit can be :

For a number ending with 1: 1

For a number ending with 2: 4

For a number ending with 3: 9

For a number ending with 4: 6

For a number ending with 5: 5

For a number ending with 6: 6

For a number ending with 7: 9

For a number ending with 8: 4

For a number ending with 9: 1

The only possible prime number is 5.

Hence the last digit of X is 5 and Y is 5.

Using statement 2 : Y = X/3.

This alone cannot be sufficient to determine the possibilities for Y and X.

Combining both the statements :

Since Y = 5, then the value of X = 15.

The age is equal to 15.

Given, m and n are two positive integers having product less than 100

Statement 1:

mn is divisible by 6 consecutive integers. This means mn must be divisible by the LCM of these 6 integers its multiples.

Only numbers from 1 to 6 satisfy this with LCM 60. Any other set of 6 consecutive integers clearly exceeds 100 as its LCM.

Only such number satisfying the condition is 60.

60 can further be expressed as the product of two positive integers in the following ways:

1 x 60, 2 x 30, 3 x 20, 4 x 15, 5 x 12 and 6 x 10.

So, no unique values of m and n can be determined from the above.

Statement 2:

m + n is a perfect square.

From this statement alone too no unique set of solutions can be determined.

From Statements 1 and 2:

Only the pair 6 x 10 satisfies the condition.

Hence, both statements taken together are sufficient to answer the question, but neither statement alone is sufficient.

From statement 1:

Red + Blue + Green = 17

Since, this ensures that there are at least 3 green marbles, so, remaining marbles must be red and blue. 

$$\therefore $$ Red + Blue = 17 - 3 = 14

Blue = 14 - Red = 14 - 8 = 6

But, this doesn't give us the total number of green marbles. So, statement 1 alone is not sufficient.

From statement 2:

Red + Blue + Green = 19

Since, this ensures that there are at least 2 red marbles, so, remaining marbles must be blue and green 

$$\therefore $$ Blue + Green = 19 - 2 = 17

Again, Green = 17 - Blue = 17 - 8 = 9.

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

Sum of the ages of the 6 friends is the square of a prime number.

Statement I states that all the members were aged between 50 and 85 years. There are 6 friends in total. Therefore, the sum of the ages can vary from 6*50 to 6*85.
Sum of the ages of the 6 persons should lie between 300 and 510. 
361 is the only square of a prime number (19) in the given range. Therefore, statement I alone is sufficient to answer the question. 

Statement II states that the standard deviation of their ages is 4.6. SD gives us how much the values deviate from the mean. We cannot obtain the mean using the SD. Therefore, statement II alone is insufficient to answer the question without knowing the values in the data set. 

Therefore, option A is the right answer. 

Harry and Sunny have cards numbered from 1 to 10.
Sum of the number of cards picked by them = 1+2+3+...+10 = 55.

Sum of the numbers on the cards picked by Harry is 5 more than that of Sunny.
Let x be the sum of the numbers picked by Sunny. 
x+x+5 = 55
x = 25
25 can be written as 10+9+1+2+3 or 9+7+5+1+3 etc. (There are many such combinations). We cannot determine the person who has card number 2. 
Statement I alone is insufficient. 

Statement II states that one of the 2 persons has exactly 4 even-numbered cards.
The person with 4 even-numbered cards might or might not contain card number 2. Therefore, statement II alone is insufficient. 

Combining both the statements, we know that the sum of the numbers on the cards with Harry is 5 more than the sum of the number on the cards with Sunny. Therefore, the sum of the number of cards with Sunny should be 25 and Harry should be 30.

Sum of the 5 even numbers = 2+4+6+8+10 = 30.
Sum of the 5 odd numbers = 1+3+5+7+9 = 25.
By replacing one of the odd numbers with even numbers, we have to make the sums 30 and 25. We cannot replace an odd number with an even number and still get the sum as 25. Therefore, the set with 4 odd numbers must add up to 30 and the set with 4 even numbers must add up to 25. We can interchange 1 with 6 or 3 with 8 or 5 with 10. The card with number 2 will always remain with the person with 4 even cards. The card with number 25 will always remain with Sunny. Therefore, we can determine the answer using both the statements together and hence, option D is the right answer.

Since only the specialist bowlers complete their quota of 4 overs, the maximum over a non-specialist bowler can bowl is three.

Total number of overs bowled by non-specialist bowlers = 8

If one of the three bowlers bowls 1 over, the other two will bowl 3 and 4 overs. Since the maximum over a non-specialist bowler can bowl is three, this is an invalid case.

If one of the non-specialist bowlers bowls 2 overs, the other two will bowl 3 overs each.

Thus, the minimum a non-specialist bowler can bowl is two overs.

Since the above calculation did not need either of S1 and S1,

The correct answer is E.

As per  Statement given  paragraph the economy rates are 6,6,7,9 which exclude best bowler and worst bowler

And  Economy rates of specialist bowlers are lower than that of nonspecialist bowlers.

Let’s take economy rates of specialist bowlers as 5, 6, 6.

Hence their cumulative runs would be (5 × 4) + (6 × 4) + (6 × 4) = 68
As per Statement 2, Total runs for non specialist bowlers =69. And their
economy rates are 7, 9. Let’s take economy rate of worst bowler be x
 (7 × 3) + (9 × 2) + (x × 3) = 69        

=>   x =10
For the lowest values of the specialist bowler we take 4. Then value of the worst bowler will be less than 9 which is incorrect.

Hence only a single case is possible.
So we can find the economy rate of worst bowler using both the statements.
Hence answer is option D. 

Let A,B,C,D,E,F represents Anita, Biplove, Cheryl, Danish, Emily and Feroze respectively.

Statement I : It is not sufficient as nothing is mentioned about D and F.

Statement II : E = F, thus E and F cannot score the lowest.

Also, C scored higher than at least 2 and B scored more than D while A scored the highest.

The only person left with the lowest marks can be D.

Thus, statement II alone is sufficient.

If the object does not change direction at point C then it will be at D above the person. But since it is given that object changed direction and continued flying upwards, thus object would reach point E. 

=> CG is perpendicular bisector to AB => BG = 300 m

In $$\triangle$$ BCG

=> $$cos 30 = \frac{BG}{BC}$$

=> $$\frac{\sqrt{3}}{2} = \frac{300}{BC}$$

=> $$BC = \frac{600}{\sqrt{3}} = 200 \sqrt{3}$$

Now, according to statement I, time can increase only when the angle with ground level will increase.

But, according to statement II, if CD = $$200 \sqrt{3}$$ (= BC), then at constant speed, it will not take any additional time and thus there should not be any increase in angle.

But in first statement direction has been changed whereas in second statement direction has not been changed. 

So both the statements are inconsistent with each other. 

Median of 11 integers is 15, => In ascending order 6th integer = 15

=> Numbers = 6,8,12,13,14,15,20,22

Statement I : Average of four smallest = 6 + 8 + 12 + 13

= $$\frac{39}{4} = 9.75$$

It is given that, avg of 4 largest - avg of 4 smallest = 13.25

=> Average of 4 largest = 13.25 + 9.75 = 23

=> Sum of 4 largest numbers = 23 * 4 = 92

So, we can easily allocate other three numbers different minimum values but more than 15 and maximize the remaining one value

Thus, statement I is sufficient.

Statement II : Sum of 11 integers = 11 * 16 = 176

Sum of given 8 integers = 6+8+12+13+14+15+20+22 = 110

Sum of remaining numbers = 176 - 110 = 66

So, we can easily allocate other three numbers different minimum values but more than 15 and maximize the remaining one value

Thus, statement II is sufficient.

$$\therefore$$ Either statement I or II is sufficient.

Statement iii and iv are redundant. Thus, we can cancel out first three options.

From (i) , (ii) , (v), we have : 

Let A,B,C be the ages of Jose's sons in ascending order

and $$A + B + C = 13$$

and $$A + B = 4$$

=> $$C = 13 - 4 = 9$$

Now, from (vi) we get $$A = B$$

=> $$A = B = \frac{4}{2} = 2$$

$$\therefore$$ i, ii, v and vi are required to answer the question.

Using the property that, tangents from same point to a circle are equal in lengths.

In above quadrilateral, PA = PM + MN, => $$d = a + MN$$ ----------Eqn(I)

RC = RN + NM

=> $$PS = d + e$$

$$SR = e + c$$

$$QR = b + c + MN$$

$$PQ = a + b$$

From statement I : 

We can conclude that, $$w + x + y + z = w + x + y + z$$

=> $$(w + z) + (x + y) = (w + x) + (y + z)$$

=> $$PQ + SR = PS + QR$$

Substituting values from above equation, we get :

$$\therefore a + b + e + c = d + e + b + c + MN$$

Using eqn(I),

=> $$a = a + MN + MN$$

=> $$MN = 0$$

Thus, statement I alone is sufficient.

Statement II alone is not sufficient, for we can have more than one value of MN possible.

Given relationship is $$(PQ)(RQ) = XXX$$

Since, X can take nine values from 1 to 9, thus we have 9 possibilities.

$$111 = 3 \times 37$$                    $$666 = 18 \times 37$$

$$222 = 6 \times 37$$                    $$777 = 21 \times 37$$

$$333 = 9 \times 37$$                    $$888 = 24 \times 37$$

$$444 = 12 \times 37$$                  $$999 = 27 \times 37$$

$$555 = 15 \times 37$$

But, out of these 9 cases, only in 999, we get the unit's digit of the two numbers same. Since it is a unique value,

Thus, we need neither statement I nor II to answer the question.

In the figure, $$\triangle AQR \sim \triangle APS$$

=> $$\frac{AQ}{AP} = \frac{QR}{PS} = \frac{AR}{AS} = k$$ --------Eqn(I)

Statement I : PQ = 3 cm , RS = 4 cm , $$\angle$$ QPS = 60°

In right $$\triangle$$ PQM

=> $$sin 60^{\circ} = \frac{QM}{QP}$$

=> $$\frac{\sqrt{3}}{2} = \frac{QM}{3}$$

=> $$QM = \frac{3 \sqrt{3}}{2} = RN$$

Similarly, $$sin (\angle RSN) = \frac{3 \sqrt{3}}{8}$$

=> $$\angle RSN = sin^{-1} (\frac{3 \sqrt{3}}{8})$$

$$\therefore$$ In $$\triangle$$ APS

=> $$\angle PAS = 180^{\circ} - \angle APS - \angle PSA$$

=> $$\angle PAS = 120^{\circ} - sin^{-1} (\frac{3 \sqrt{3}}{8})$$

Thus, statement I alone is sufficient. 

Statement II : PS = 10, QR = 5

From eqn(I), $$k = \frac{1}{2}$$

But, we do not know anything regarding the measures of the remaining sides or any of the angles.

So, statement II is not sufficient.

Expression : $$A_{n+1}=A_{n}^{2}+1$$ ---------Eqn(I)

Statement I : $$A_{0} = 1$$

Putting n = 0 in Eqn (I), => $$A_1 = A_0^2 + 1 = 1 + 1 = 2$$

Similarly, $$A_2 = A_1^2 + 1 = 4 + 1 = 5$$ and so on

We can find the values of $$A_{900}$$ and $$A_{1000}$$ and also their greatest common divisor.

Thus, statement I alone is sufficient.

Statement II : We have $$A_{1} = 2$$

In the above manner, we can determine $$A_{900}$$ and $$A_{1000}$$ and also their greatest common divisor.

Thus, statement II alone is sufficient.

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