Data Sufficiency MCQ Quiz - Objective Question with Answer for Data Sufficiency - Download Free PDF

Let the four consecutive even integers be:

x, (x + 2), (x + 4), (x + 6)

The average of these numbers =

⇒ [x + (x + 2) + (x + 4) + (x + 6)] ÷ 4 = (4x + 12) ÷ 4 = x + 3

Statement I: The largest number is 6 more than the smallest number. 

⇒ (x + 6) = x + 6 ⇒ Always true for consecutive even numbers.

This confirms the numbers are in the pattern we assumed, but does not give the value of x.

Thus, Not sufficient to find the average.

Statement II: The second smallest number is 20% less than the third number.

Second smallest = (x + 2)

Third number = (x + 4)

⇒ (x + 2) = (x + 4) – 20% of (x + 4)

⇒ x + 2 = 0.8(x + 4)

⇒ x + 2 = 0.8x + 3.2

⇒ x - 0.8x = 3.2 - 2 ⇒ 0.2x = 1.2 ⇒ x = 6

Now we can find all the numbers:

⇒ x = 6 ⇒ Numbers: 6, 8, 10, 12

⇒ Average = (6 + 8 + 10 + 12) ÷ 4 = 36 ÷ 4 = 9

Thus, Sufficient to find the average.

Therefore, the data in statement II alone is sufficient to answer the question, while the data in statement I alone is not sufficient.

Let: x = amount invested in Scheme X 10000 – x = amount invested in Scheme Y

From Statement I alone: We only know x + (10000 – x) = 10000, and Scheme Y’s rate and period, but no interest amounts ⇒ cannot find x.

From Statement II alone: We know Scheme X rate (8% compounded annually for 2 years) and Scheme Y rate (12% simple for 3 years), and total interest = ₹2390, but we don’t know the total principal (10000) ⇒ cannot determine x.

Using I and II together:

Interest from X = x × [(1.08)² – 1] = 0.1664x

Interest from Y = (10000 – x) × (12% × 3) = 0.36 (10000 – x) = 3600 – 0.36x

Total interest = 0.1664x + (3600 – 0.36x) = 3600 – 0.1936x = 2390 ⇒ 3600 – 2390 = 0.1936x ⇒ 1210 = 0.1936x ⇒ x = 6250

⇒ Amount in Y = 10000 – 6250 = 3750

⇒ Difference = 6250 – 3750 = ₹2500

Thus, the data given in both Statements I and II together are necessary to answer the question.

Let that total number be T.

Statement I:

Male : Female in X = 4 : 3 → So in X, male = 4x and female = 3x ⇒ total in X = 7x

Male : Female in Y = 5 : 2 → So in Y, male = 5y and female = 2y ⇒ total in Y = 7y

Now, we are given that total employees in X = total employees in Y
⇒ 7x = 7y ⇒ x = y

Now:

Female in X = 3x

Female in Y = 2x (since y = x)

Total females = 3x + 2x = 5x

But x is unknown.

Even the ratio of junior to senior male employees in X (3:5) is irrelevant here unless we know a value.

So, Statement I alone is NOT sufficient.

Statement II:

Ratio of junior to senior female in Y = 1:2

Let junior female = a, senior female = 2a ⇒ total female in Y = a + 2a = 3a

Senior male in X + senior female in Y = 690

We don’t know total male in X or how many are senior → cannot extract female number in X.

Also, we cannot connect this info to total employees (T) or get specific values.

So, Statement II alone is NOT sufficient.

Now combine both statements:

From Statement I:

Female in X = 3x

Female in Y = 2x

Total females = 5x

From Statement II:

Senior male in X + senior female in Y = 690
→ From Statement I, male in X = 4x
→ Ratio of junior to senior male = 3:5
⇒ So, senior male in X = 5 parts out of (3+5) = 5/8 of 4x
⇒ Senior male in X = (5/8) × 4x = (20x)/8 = 2.5x

Female in Y = 2x
→ From Statement II, ratio of junior:senior = 1:2 ⇒ senior = 2 parts out of 3 ⇒ 2/3 of 2x = (4x)/3

Now:

2.5x + (4x)/3 = 690

(7.5x + 4x)/3 = 690

(11.5x)/3 = 690

11.5x = 2070

x = 180

Now we can compute:

Female in X = 3x = 540

Female in Y = 2x = 360

Total females = 900

Thus, Both statements together are necessary.

Calculation

Let, Number of female managers = x

Then,

Male managers = 1.2x (20% more than female managers)

Male non-managers = 25% more than male managers

= 1.25 × 1.2x = 1.5x

Female non-managers = 25% more than female managers

= 1.25 × x = 1.25x

Total males and total females

Total males = male managers + male non-managers

= 1.2x + 1.5x = 2.7x

Total females = female managers + female non-managers

= x + 1.25x = 2.25x

Difference between total males and females = 90

Or, 2.7x − 2.25x = 0.45x = 90

⇒ x=90 / 0. 45 = 200

Female managers = x = 200

Male managers = 1.2x = 240

Male non-managers = 1.5x = 300

Female non-managers = 1.25x = 250

Male clerks = 50% of male non-managers = 50% of 300 = 150

Female clerks = male clerks + 25 = 150 + 25 = 175

Quantity I: Female clerks as % of male managers

[175/240] × 100 ≈ 72.92% ≈ 73%

Quantity II: Male clerks as % of female managers

[150/200] × 100 = 75%

⇒75%

So, Quantity I

Calculation

Let A = 4x

Given:

D = 125% of A

⇒ D = [5/4] × A = 5x

B = D

⇒ B = 5x

C = B + 5 = 5x + 5

Average of A, B, and C = 39
⇒ A + B + C = 3 × 39 = 117

So, A + B + C = 4x + 5x + (5x + 5) = 14x + 5 = 117

⇒ 14x = 112

⇒ x = 112/14=8

A = 4x =32

B = D = 5x = 40

C = 5x + 5 = 45

Quantity I: A + C = 32 + 45 = 77

Quantity II: B + D = 40 + 40 = 80

Quantity I

Statement I:

⇒ x + 2y = 6   

Here, we cannot find the value of x with the help of only one equation

Hence, Statement I alone is insufficient

Statement II:

⇒ 3x + 6y = 18

Here, we cannot find the value of x with the help of only one equation

Hence, Statement II alone is insufficient

From Statement I and II:

⇒ x + 2y = 6      ----(1)

⇒ 3x + 6y = 18      ----(2)

Multiplying equation (1) by 3 we get

⇒ 3(x + 2y) = 6 × 3

⇒ 3x + 6y = 18      ----(3) 

Here, both equations (2) and (3) is same so we can not find the value of x

∴ Statements I and II together are not sufficient

Confusion Points

The second equation is only the multiple of first, so we cannot find the values of x and y 

Quantity A:

⇒ y = (100 – 62.5)% of 840

⇒ y = 37.5% of 840

⇒ y = 3/8 × 840 = 315

Now,

⇒ x = (100 + 20)% of y

⇒ x = 1.2 × 315 = 378

⇒ Quantity A = 378

Quantity B: 420

∴ Quantity A

From statement 2,

Earning of Z = Rs. 500

Earning of X and Y = Rs. 500 + 40 = Rs. 540.

⇒ Required average of daily wages = (X + Y + Z)/3 = (540 + 500)/3 = Rs. 1040/3

∴ 2 alone is sufficient while 1 alone is insufficient.

From Statement 1: X - 2Y = 5

cannot find the value of X and Y.

From statement 2: X2 – 25 = 4XY - 4Y2

X2 – 25 = 4XY - 4Y2  -------(1)

X2 - 4XY + 4Y2 = 25

(X - 2Y)² = 25

X - 2Y = 5

cannot find the value of X and Y.

So, same equation in both the statements.

Hence, option (3) is the correct answer.

Confusion PointsHere, after calculation, we got only 1 equation, hence we cannot conclude the exact values of X and Y. 

Statement 1:

X – 15 = integer

⇒ X is also an integer

Statement 2:

X – 10 = odd integer

⇒ X is an odd integer.

⇒ (X – 5) is even.

∴ Statement 2 alone is sufficient while statement 1 alone is insufficient.

Quantity A -

Let the volume of tank = LCM of (15, 20, 12) = 60 units.

X's capacity = 60 / 15 = 4 units.

Y's capacity = 60 / 20 = 3 units.

Hole's emptying capacity = 60 / 12 = 5 units.

Time taken to fill (3 / 4)th tank = 45 / (4 + 3) = 6.42 units

Time taken for remaining (1 / 4)th tank = 15 / (4 + 3 - 5) = 7.5 units

Total time = 6.42 + 7.5 = 13.92 hours

Quantity B - 14 hours.

Hence, Quantity A

Confusion Points The statement that a hole at the bottom would have caused the tank to empty in 12 hours was made to give readers a sense of the pipe's flow rate, not to imply that the hole is at the bottom.

Statement 1∶

y = mx + 2

We cannot find anything with statement 1.

Statement 2∶

Line passes thgough (2, 1)

We cannot find anything with statement 2.

Combining statement 1 and 2∶

∵ Line passes through (2, 1), it will satisfy the equation of line y = mx + 2

∴ Putting x = 2 and y = 1 in the equation of line

⇒ 1 = 2m + 2

⇒ m = -1/2

∴ Both statements 1 and 2 are sufficient.

Calculation:

Since angles by a chord on two different points on the same segment of a circle are equal.

∵ ∠D = 60°

So, ∠ACB = ∠D = 60°

Hence, Both 1 and 2 are sufficient (Option 2 is correct)

Statement A:

⇒ One-third of each boxes’ weight is 2 kg

⇒ Weight of each box = 6 kg

⇒ So, the total weight of 6 boxes = 36 kg

Statement B:

The total weight of four boxes is 12 kg more than the total weight of two boxes

Let the weight of 1 box be x.

⇒ Given, 4x - 12 = 2x

⇒ x = 6 kg

⇒ So, the total weight of 6 boxes = 36 kg

∴ Both Statements 1 and 2 alone are sufficient

From I

We know that a + b is positive ⇏ a is positive as b can be a large positive value when a is negative

For example,

Let the number of b be 2,

Then the number of a be -3

so, according to the statement

a + b = 2 + (-3) = -1 is negative

From II

We know that a - b is positive ⇏ a is positive as b can be a large negative value when a is negative

For example,

Let the number of b be 2,

Then the number of a be -3

so, according to the statement

a - b = 2 - (-3) = 5 is positive

Now, adding both i.e. I + II

(a + b) + (a - b) = positive

a is positive

Both the statements prove a is positive.