**XAT Quantitative Ability & Data Interpretation Practice Questions Set-1:** This practice set has 10 questions with their answers covering topics from the areas of Quantitative Ability.

**1.** Karan picks a random integer between 1 and 999, doubles it and gives the result to Arjun. Each time Arjun gets a number from Karan, he adds 50 to the number, and gives the result back to Karan, who doubles the number again. The first person, whose result is more than 1000, loses the game. Let ‘x’ be the smallest initial number that results in a win for Karan. The sum of the digits of ’x’ is:

a) 3

b) 5

c) 7

d) 9

e) None of the above

**2.** Detergent M is produced by mixing Powder X and Powder Y in the ratio of 5:4. Powder X is prepared by mixing two raw materials, A and B, in the ratio of 1:3. Powder Y is prepared by mixing raw materials, B and C, in the ratio of 2:1. Then the final mixture is prepared by mixing 864 units of Detergent M with chemical. If the concentration of the raw material B in the final mixture is 50%, how much chemical had been added to Detergent M?

a) 328 units

b) 368 units

c) 392 units

d) 616 units

e) None of the above

**3.** In quadrilateral ABCD, AB = 5 units, BC = 17 units, CD = 5 units, and AD = 9 units. The length of the diagonal BD can be:

a) >10 and < 12

b) > 12 and < 14

c) > 14 and < 16

d) > 16 and < 18

e) Cannot be determined

**4.** In the beginning of the year 2009, a person invests some amount in a Fixed Deposit. In the beginning of 2012, the accumulated interest is Rs. 10,000 and in the beginning of 2015, the accumulated interest becomes Rs. 25,000. The interest rate is compounded annually and the annual interest rate is fixed. The principal amount is: (63 – 2015)

a) Rs. 16,000

b) Rs. 18,000

c) Rs. 20,000

d) Rs. 25,000

e) None of the above

**5.** The probability that a randomly chosen positive divisor of 10^{39} is an integer multiple of 10^{33} is x^{2}/y^{2} , then ‘y-x’ would be: (64-2014)

a) 8

b) 15

c) 21

d) 33

e) 45

**6.** Three tubes are connected to an inverted cone, with its base at the top. Two inlet tubes, X and Y, are connected to the top of the tube and can fill the empty in 8 hours and 12 hours, respectively. The outlet tube Z, connected to the bottom, can empty a filled cone in 4 hours. When the tube is completely filled with water, all three tubes are opened. Two of the three tubes remain open for 20 hours continuously and the third tube remains open for a lesser time. As a result, the height of the water inside the cone comes down to 50%. Which of the following options would be possible?

a) Tube X was open for 19 hours.

b) Tube X was open for 19 hours 30 minutes.

c) Tube Y was open for 19 hours 30 minutes.

d) Tube Z was open for 19 hours 50 minutes.

e) The situation is not possible.

**Direction (7 – 10):** Answer the questions based on the two graphs shown below.

Figure A shows the amount of work distribution, in man-hours, in a Banking Sector between office work and field job. Figure-B shows the estimated and actual work effort involved in the different office work in the same sector during the same period.

**7.** Which work requires as many man-hours as spent in sanction?

a) Office work, pre sanction and sanction

b) Office sanction

c) Post sanction

d) Office, post sanction and sanction

e) None of the above

**8.** Find the percentage of total work carried out in field job?

a) 39%

b) 19%

c) 29%

d) 9%

e) None of the above

**9.** From the given option which is nearest to the total effort in man-hours spent in field job?

a) The sum of the estimated and actual effort for office work pre sanction.

b) The estimated man-hours of office work sanction

c) The actual man-hours of office work post sanction.

d) Half of the man-hours of estimated office work sanction.

e) None of the above

**10.** Which of the following tasks will account for approximately 50 hr, if the total working hours were 100?

a) Sanction

b) Pre sanction

c) Office post sanction

d) Office post sanction plus pre sanction

e) None of the above

**Answer Key:**

**1.** (c) **2.** (b) **3.** (b) **4.** (c) **5.** (d) **6.** (c) **7.** (a) **8.** (c) **9.** (c) **10.** (a)

**Explanations:**

**Explanation 1:**

Let the smallest number be ‘X’

Steps | Karan | Arjun |
---|---|---|

Step 1 | 2X | 2X + 50 |

Step 2 | 4X + 100 | 4X + 150 |

Step 3 | 8X + 300 | 8X + 350 |

Step 4 | 16X + 700 | 16X + 750 |

Step 5 | 32X + 1500 | 32X + 1550 |

Karan has to win and X is the least possible number in the range 1 – 999.

Therefore, step 4 has to be the last step.

⇒16X + 750 > 1000

The least possible value of X = 16

Sum of the digit = 1 + 6 = 7

**Explanation 2:**

In 864 units of M,

X = 5/9 × 864 = 480 units

Y = 864 – 480 = 384 units

B in 480 units of X = 3/4 × 480 = 360 units

B in 384 units of Y = 2/3 × 384 = 256 units

Total units of B = 360 + 256 = 616

Concentration of B in the final mixture is 50%

Thus, chemical in the final mixture = (2 x 616) — 864 = 368 units.

**Explanation 3:**

Let BD = x, we get the following figure

In any triangle, the sum of any two sides must be greater than the third side. Similarly, the difference between any two sides must be smaller than the third side. Hence,

In ∆BCD,

x + 5 > 17

⇒ x > 12 ... (i)

In ∆ABD,

x < 9 + 5

⇒ x < 14 ... (ii)

Combining (i) and (ii), we get

12 < x < 14

**Explanation 4:**

Let P be the principal.

By the given conditions,

P × (1 + r/100)^{3} – P = 10000

P × (1 + r/100)^{6} – P = 25000

Let (1 + r/100)^{3} = X

Thus,

P(X - 1) = 10000 ... (i)

P (X^{2} — 1) = 25000 ... (ii)

Dividing (ii) by (i),

X + 1 = 5/2

∴ X = 3/2

Substituting value of X in (i), we get

P = Rs. 20,000

**Explanation 5:**

10 = 5×2

10^{39} = 5^{39} × 2^{39}

∴ Number of divisors = (39 + 1)(39 + 1) = 40 × 40

We need to find all the divisors of K such that 10^{39}

= K × 10^{33}

K= 10^{6} = 5^{6} × 2^{6}

∴ Number of divisors = (6 + 1)(6 + 1) = 7 x 7

The probability that a randomly chosen positive divisor of 10^{39} is an integer multiple of 10^{33}.

= 7 × 7/40 × 40 = x^{2}/y^{2}

∴ x = 7 and y = 40

∴ y – x = 40 – 7 = 33

**Explanation 6:**

Let the capacity of the tank be 24x litres.

Tubes X and Y fill 3x and 2x litres per hour while Tube Z empties 6x litres in an hour.

Let radius of the cone be r and height be h.

∴ 1/3 πr^{2}h = 24x

∴ πr^{2}h = 72x

For first 19 hours, water inside the cone

24x + 57x + 38x - 114x = 5x litres

∆XYT~∆XZS

If XZ = 2XY, ZS = 2YT

YT = r/2 and XS = h/2

After 50% reduction in the height of the water,

Volume = 1/3 π(r/2)^{2}(h/2) = (πr^{2}h)/24 = 72x/24 = 3x

**Option 1:** Tube X was open for 19 hours.

i.e., Y and Z were open for 1 more hour.

∴2x - 6x = -4x

The cone will have 5x — 4x = x litres of water.

Option 1 is eliminated.

**Option 2:** Tube X was open for 19 hours 30 minutes.

i.e., Y and Z were open for 1 more hour and X for 30 more minutes.

∴ 2x - 6x + 1.5x = -2.5x

The cone will have 5x — 2.5x = 2.5x litres of water

Option 2 is eliminated.

**Option 3:** Tube Y was open for 19 hours 30 minutes.

i.e., X and Z were open for 1 more hour and Y for 30 more minutes.

∴ 3x - 6x + x = -2x

The cone will have 5x — 2x = 3x litres of water.

Option 3 would be the possible option.

**Explanation 7:**

Man-hours spent in sanction is 400 + 100 = 500.

Now going by options, we see (a) is the only option.

**Explanation 8:**

Total work is approximately

(100 + 80) + (400 + 100) + (280 + 140) = 1100

Field job work = 80 + 100 + 140 = 320

Percentage of total work carried out in field job is 320/1100 × 100 = 29% approximately.

**Explanation 9:**

From figure the total effort in man-hours spent in field job is 320.

It is nearest to actual man-hours of office work post sanction which is 280 (approximately).

**Explanation 10:**

Total man-hours = (100 + 80) + (400 + 100) + (280 + 140) = 1100.

Total working hours = 100

Total man working = 1100/100=11

For 50 hr the total man-hours is 50 × 11 = 550, which is near to sanction (400 + 100)

Hence, (a) is the correct option.

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