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

Calculation

Let radius of cylinder = x cm

So, height of the cylinder = 2x cm

So, 2 × [22 /7] × r × h = 616

So, x2 = 49

So, x = 7

So, height of the cylinder = 14 cm

Cone radius = 7 × [1/2] = 3.5 cm

Quantity I: Total surface area of cylinder = 2 × [22/7] × 7 × 21 = 924 cm2

Required difference = 924 – 616 = 308 cm2

Quantity II: lateral height of cone = √196 + 12.25 = √208.25 cm

Curved surface area of the cone = [22/ 7] × 7 × √208.25 = 317.47 cm2 (approx.) So, Quantity I < Quantity II

Calculation

Speed of train P (in m/sec.) = 180 × 5 18 = 50 m/sec.

Length of train P = 50 × 4 = 200 meters

Relative speed of train P and Q = 200/10 = 20 m/sec.

So, speed of train Q = 50 – 20 = 30 m/sec.

Quantity I: Speed of train Q = 30 m/sec.

Quantity II: New speed of train P = 50 – 20 = 30 m/sec.

So, Quantity I = Quantity II

Calculation

Let the age of P after four years and the present age of Q be 5x years and 6x years respectively.

The present age of P = 5x - 4 years and the present age of R = 6x + 12 years

ATQ, (5𝑥 − 4) + 6𝑥 + (6𝑥 + 12) = 65 × 3

Or, 17𝑥 + 8 = 195

or, 17𝑥 =187

or, x = 11

Quantity I. 840

Quantity II.

Present ages of Q = 6x = 6 × 11 = 66 years

Present ages of R = 6x + 12 = 6 × 11 + 12 = 78 years

Required value = 6 × (66 + 78) = 864

So, Quantity I < Quantity II

Calculation

P2 − 10P + 25 = 0

⇒ (P−5)2 =0

⇒ P = 5

So, both roots are 5

⇒ x = 5

Now,

[x/5] = [11/ z+1]

⇒ 5/5 = 11/[z+1]

⇒ 1 = 11z + 1 ⇒ z + 1 =11

⇒ z = 10

Quantity I: 10z = 10 × 10 = 100

Quantity II: 100

Both quantities are equal.
Quantity I = Quantity II

Concept:

1) If in a right-angle triangle

Ratio of three angle = 30° : 60° : 90°  (or 1 : 2 : 3), then

The ratio of it's side opposite to this angle = 1: √3 : 2

2) AC × BD = AB × AD 

Calculation:

Statement I:

The angles of Δ are in the ratio 1 : 2 : 3

As we know in a triangle sum of all the angles is 180°

So, The angles of Δ are 30°, 60°, 90°

Therefore,

So, the sides should be in the ratio 1: √3 : 2

By using the Pythagoras' theorem

x2 + 242 = 252

⇒ x2 + 576 = 625

⇒ x2 = 625 - 576

⇒ x2 = 49

⇒ x = 7

So, the shortest side = 7 cm

But, 7, 24 & 25 are not in the ratio 1 : √3 : 2, but as per the property,

If the ratio of angle = 1 : 2 : 3 then the ratio of side = 1 : √3 : 2. 

Therefore, the information given in Statement I is not sufficient. 

Statement II:

From the Pythagoras theorem, we got x = 7 

By using the above property

BC × AD = AB × AC

7 × 24 = 25 × AD

AD = 168/25 = 6.72

Therefore, if we take the length of the shortest side equal to 7 cm, we got the length of the perpendicular drawn on the longest side of Δ from its opposite vertex is 6.72 cm.  

Statement 2 is sufficient to answer this question.

∴ Option (2) is correct.

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 

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

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 < Quantity B

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.

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 < Quantity B

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

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.