# CAT 2017 QA Slot 2 | Previous Year Questions

**1. CAT 2017 QA Slot 2 | Miscellaneous**

The numbers 1,2, …, 9 are arranged in a 3 × 3 square grid in a such way that each number occurs once and the entries along each column, each row, and each of the two diagonals add up to the same value.

If the top left and the top right entries of the grid are 6 and 2, respectively, then the bottom middle entry is

Answer: 3

**Explanation** :

Sum of all digits from 1 to 9 = 45

If entries along each of the rows, columns or diagonals adds upto the same number, it implies that the sum of each row, columns or diagonal will be 45 ÷ 3 = 15.

(∵ there are 3 rows and 3 columns we divide the total of the numbers 1 to 9 i.e. 45 by 3)

In the 1^{st} row the top left cell contains the number 6 and the top right cell contains the number 2.

Therefore the middle element in the first row will be 15 – 6 – 2 = 7. Further elements of each row, columns or diagonal adds upto 15. Now this is only possible if the middle element in the entire grid i.e. the element corresponding to the 2^{nd} row and 2^{nd} column is the average of all 9 numbers i. e., 5. Now if the middle element is 5, then the bottom element in the 2^{nd} column will be 15 – 7 – 5 or 3.

Hence, 3.

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**2. CAT 2017 QA Slot 2 | Arithmetic - Time, Speed & Distance**

In a 10 km race, A, B and C, each running at uniform speed, get the gold, silver, and bronze medals, respectively. If A beats B by 1 km and B beats C by 1 km, then by how many metres does A beat C?

Answer: 1900

**Explanation** :

In a 10 km race, A beats B by 1 km. So in the time A completes 10 km, B completes

9 km. Therefore ratio of their speeds is 10 : 9.

In a similar way, ratio of speeds of B and C is also 10 : 9

A : B : C

10 : 9

10 : 9

100 : 90 : 81

To get a common ratio for speeds of A, B, C we take LCM of 9 and 10 which is equal to 90 (∵ B is common to both ratios) and equate corresponding values of A and C is also taking B as 90

So if A runs 100m, C runs 81m.

So in a 10 km race, when A runs 10000 m C will run $\frac{81}{100}\times 10000$ = 8100 m

∴ A will beat C by 10000 - 8100 or 1900 m.

Hence, 1900.

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**3. CAT 2017 QA Slot 2 | Arithmetic - Mixture, Alligation, Removal & Replacement**

Bottle 1 contains a mixture of milk and water in 7 : 2 ratio and Bottle 2 contains a mixture of milk and water in 9 : 4 ratio. In what ratio of volumes should the liquids in Bottle 1 and Bottle 2 be combined to obtain a mixture of milk and water in 3 : 1 ratio?

- A.
27 : 14

- B.
27 : 13

- C.
27 : 16

- D.
27 : 18

Answer: Option B

**Explanation** :

Bottle 1 contains 7/9 of milk

Bottle 1 contains 9/13 of milk

The final mixture contains 3/4 of milk

Now using the allegation rule, we get the ratio of quanities in bottles as:

⇒ $\frac{\mathrm{Quantity}\mathrm{in}\mathrm{bottle}1}{\mathrm{Quantity}\mathrm{in}\mathrm{bottle}2}$ = $\frac{{\displaystyle \frac{3}{4}}-{\displaystyle \frac{9}{13}}}{{\displaystyle \frac{7}{9}}-{\displaystyle \frac{3}{4}}}=\frac{{\displaystyle \frac{3}{52}}}{{\displaystyle \frac{1}{36}}}$

⇒ $\frac{\mathrm{Quantity}\mathrm{in}\mathrm{bottle}1}{\mathrm{Quantity}\mathrm{in}\mathrm{bottle}2}$ = $\frac{3}{52}+\frac{1}{36}$ = $\frac{108}{52}$ = $\frac{27}{13}$ = 27 : 13

Hence, option (b).

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**4. CAT 2017 QA Slot 2 | Arithmetic - Time, Speed & Distance**

Arun drove from home to his hostel at 60 miles per hour. While returning home he drove half way along the same route at a speed of 25 miles per hour and then took a bypass road which increased his driving distance by 5 miles, but allowed him to drive at 50 miles per hour along this bypass road. If his return journey took 30 minutes more than his onward journey, then the total distance traveled by him is

- A.
55 miles

- B.
60 miles

- C.
65 miles

- D.
70 miles

Answer: Option C

**Explanation** :

Let ‘d’ miles be the distance for the onward journey from his home to his hostel.

Now time take for onward jouney = d/60.

For the return journey, time taken to cover half distance (i.e. d/2)

⇒ $\frac{\mathrm{d}}{2}\xf725$ = $\frac{\mathrm{d}}{50}$

Now for the return journey, (after half the distance is covered) taking the bypass road will increase the distance by 5 miles i.e., the remaining distance of the journey is 'd/2 + 5' miles

Now, 'd/2 + 5' miles is covered at the speed of 50 miles per hour.

Sp time taken to cover '$\frac{\mathrm{d}}{2}$ + 5' miles is (d/2 + 5)/50 = $\frac{\mathrm{d}}{100}+\frac{1}{10}$

Total time for return journey = $\frac{\mathrm{d}}{50}$ + $\frac{\mathrm{d}}{100}+\frac{1}{10}$ = $\frac{3\mathrm{d}+10}{100}$

Now as time taken for return journey is half an hour more than onward journey.

⇒$\frac{\mathrm{d}}{50}+\frac{1}{2}$ = $\frac{3\mathrm{d}+10}{100}$

Hence, d = 30

Total distance covered (for onward and return journey) = d + d/2 + d/2 + 5

= 30 + 30/2 + 30/2 + 5 = 65 miles

Hence, option (c).

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**5. CAT 2017 QA Slot 2 | Arithmetic - Percentage**

Out of the shirts produced in a factory, 15% are defective, while 20% of the rest are sold in the domestic market. If the remaining 8840 shirts are left for export, then the number of shirts produced in the factory is

- A.
13600

- B.
13000

- C.
13400

- D.
14000

Answer: Option B

**Explanation** :

Let us assume that the number of shirts produced in the factory is ‘100x’. Now 15% of ‘100x’ or ‘15x’ shirts are defective.

So the number of remaining shirts = 100x – 15 = 85x

Now 20% of the remaining ‘85x’or ‘17x’ shirts are sold in the domestic market.

So number of shirts left for exports = 85x – 17x = 68x

As per given information, 8840 shirts are left for exports.

∴ 68x = 8840

⇒ x = 130 or 100x = 13000

So number of shirts produced in the factory is 13000.

Hence, option (b).

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**6. CAT 2017 QA Slot 2 | Arithmetic - Average**

The average height of 22 toddlers increases by 2 inches when two of them leave this group. If the average height of these two toddlers is one-third the average height of the original 22, then the average height, in inches, of the remaining 20 toddlers is

- A.
30

- B.
28

- C.
32

- D.
26

Answer: Option C

**Explanation** :

Let the average height of 22 toddlers be ‘x’ inches.

Now as the average height of the 2 toddlers who leave the group is one-third the average height of the entire group 22, their average = x/3 inches.

Further, the average height of the balance 20 toddlers in terms of x will be ‘x + 2’ inches. So the average height of all 22 toddlers can be expressed as

x = $\frac{2\left({\displaystyle \frac{x}{3}}\right)+20(x+2)}{22}$

Hence, 22x = 2x/3 + 20x + 40

⇒ x = 30

So the average height of the balance

20 toddlers will be 30 + 2 or 32 inches.

Hence, option (c).

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**7. CAT 2017 QA Slot 2 | Arithmetic - Profit & Loss**

The manufacturer of a table sells it to a wholesale dealer at a profit of 10%. The wholesale dealer sells the table to a retailer at a profit of 30%. Finally, the retailer sells it to a customer at a profit of 50%. If the customer pays Rs 4290 for the table, then its manufacturing cost (in Rs) is

- A.
1500

- B.
2000

- C.
2500

- D.
3000

Answer: Option B

**Explanation** :

Lets the manufacturers C.P. be x

As he sells it to the wholesaler at 10% profit, his S.P will be 1.1x

Now 110 is also the wholesaler’s C.P. Further, as the wholesaler sells it to a retailer at 30%, profit, his S.P will be

$\frac{130}{100}$ × 1.1x = 1.43x

Now 1.43x also the retailers C.P

As the retailer sells it to the customer at

50% profit, S.P. will be $\frac{150}{100}$ × 1.43x or 2.145x

The price paid by the customer is Rs. 4290

⇒ 2.145x = 4290

⇒ x = 2000

So Rs. 2000 is the manufacturer’s cost

Hence, option (c).

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**8. CAT 2017 QA Slot 2 | Arithmetic - Time & Work**

A tank has an inlet pipe and an outlet pipe. If the outlet pipe is closed then the inlet pipe fills the empty tank in 8 hours. If the outlet pipe is open then the inlet pipe fills the empty tank in 10 hours. If only the outlet pipe is open then in how many hours the full tank becomes half-full?

- A.
20

- B.
30

- C.
40

- D.
45

Answer: Option A

**Explanation** :

In 1 hour, the inlet pipe will fill $\frac{1}{8}$ of the tank.

Let the time taken by the outlet pipe to empty the tank be ‘x’ hours. So in 1 hour,

The outlet pipe will empty $\frac{1}{x}$ of the tank in 1 hour.

Together they fill the tank in 10 hours.

⇒ $\frac{1}{8}$ - $\frac{1}{x}$ = $\frac{1}{10}$

∴ x = 40

⇒ Outlet pipe takes 40 hours to empty the full tank.

∴ Outlet pipe will take 20 hours to empty half the tank.

Hence, option (a).

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**9. CAT 2017 QA Slot 2 | Arithmetic - Profit & Loss**

Mayank buys some candies for Rs 15 a dozen and an equal number of different candies for Rs 12 a dozen. He sells all for Rs 16.50 a dozen and makes a profit of Rs 150. How many dozens of candies did he buy altogether?

- A.
50

- B.
30

- C.
25

- D.
45

Answer: Option A

**Explanation** :

Supposing Mayank buys ‘x’ dozen candies at Rs. 15 per dozen and another ‘x’ dozen candies at Rs. 12 per dozen.

Total C.P =15(x) + 12(x) per dozen.

Now he sells these ‘x + x’ or ‘2x’ dozen candies at Rs. 16.50 per dozen.

So total S.P = 16.50 × 2x = 33x

Profit = 33x -27x = 6x

Now 6x = 150 or x = 25

As Mayank buys ‘2x’ dozen candies, number of dozens of candies bought by him

= 2 × 25 or 50

Hence, option (a).

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**10. CAT 2017 QA Slot 2 | Arithmetic - Percentage**

In a village, the production of food grains increased by 40% and the per capita production of food grains increased by 27% during a certain period. The percentage by which the population of the village increased during the same period is nearest to

- A.
16

- B.
13

- C.
10

- D.
7

Answer: Option C

**Explanation** :

Total Food Production = Total Population × Per Capita Food Production

Let the initial population be ‘p’ and the initial per capita food production be ‘c’

If f is the total food production, f = pc

This can also be rewritten as p = $\frac{f}{c}$

Now, as per given information, f increases to 1.4f and c increases to 1.27c.

So, if p_{1} is population after increase.

1.4f = p_{1}(1.27c)

$\frac{1.4f}{1.27c}$ = p_{1}

⇒ p_{1} = $\left(\frac{1.4}{1.27}\right)\left(\frac{f}{c}\right)$

or p_{1} = $\left(\frac{1.4}{1.27}\right)$(p) [$\left(\because p=\frac{f}{c}\right)$ ]

$\Rightarrow $ p_{1} ≃ 1.1p

This means that the population of the village has increased by approximately 10%.

Hence, option (c).

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**11. CAT 2017 QA Slot 2 | Arithmetic - Ratio, Proportion & Variation**

If a, b, c are three positive integers such that a and b are in the ratio 3 : 4 while b and c are in the ratio 2 : 1, then which one of the following is a possible value of (a + b + c)?

- A.
201

- B.
205

- C.
207

- D.
210

Answer: Option C

**Explanation** :

a : b = 3 : 4

b : c = 2 : 1

To get a common ratio we take LC.M (2, 4) = 4

So, we multiply each of b and c by 2

Hence, ratio of b : c = 4 : 2

∴ Ratio of a : b : c = 4 : 2

Now let a = 3x, b = c = 3 : 4 : 2

So a + b + c = 3x + 4x + 2x = 9x

As, a, b and c are all positively integers, it implies that the sum of a, b and c will be a positive integer and a multiple of 9.

Out of the given options, only 207 i.e, option (c) is a multiple of 9.

Hence, option (c).

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**12. CAT 2017 QA Slot 2 | Arithmetic - Time, Speed & Distance**

A motorbike leaves point A at 1 pm and moves towards point B at a uniform speed. A car leaves point B at 2 pm and moves towards point A at a uniform speed which is double that of the motorbike. They meet at 3:40 pm at a point which is 168 km away from A. What is the distance, in km, between A and B?

- A.
364

- B.
378

- C.
380

- D.
388

Answer: Option B

**Explanation** :

Motorbike travels 168 kms from 1 pm till 3:40 pm i.e. in 2 hours 40 mins = $\frac{8}{3}$ hours.

⇒ Speed of motorbike = $\frac{168}{8/3}$ = 63 kmph

∴ Speed of Car = 2 × Speed of Motorbike = 2 × 63 = 126 km/hr

Now as the car travels for 1 hour 40 minutes.

Distance travelled by the car = 126 × $\frac{5}{3}$ or 210 km

Distance between A and B = Distance travelled by Motorbike + Distance travelled by Car

= 168 + 210 = 378 km

Hence, option (b).

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**13. CAT 2017 QA Slot 2 | Arithmetic - Time & Work**

Amal can complete a job in 10 days and Bimal can complete it in 8 days. Amal, Bimal and Kamal together complete the job in 4 days and are paid a total amount of Rs 1000 as remuneration. If this amount is shared by them in proportion to their work, then Kamal’s share, in rupees, is

- A.
100

- B.
200

- C.
300

- D.
400

Answer: Option A

**Explanation** :

In one day, Amal, Bimal and Kamal complete $\frac{1}{4}$ of the work.

In one day, Amal and Bimal complete $\frac{1}{10}+\frac{1}{8}=\frac{4+5}{40}=\frac{9}{40}$ of the work.

So amount of work done by Kamal alone in one day is $\frac{1}{4}-\frac{9}{40}$ = $\frac{1}{40}$ of the work

∴ Fraction of work done by Kamal in 4 days = 4 × $\frac{1}{40}$ = $\frac{1}{10}$

⇒ Amount received by Kamal will be 1/10th of total amount = 1/10 × 1000 = Rs. 10

Hence, option (a).

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**14. CAT 2017 QA Slot 2 | Arithmetic - Mixture, Alligation, Removal & Replacement**

Consider three mixtures – the first having water and liquid A in the ratio 1 : 2, the second having water and liquid B in the ratio 1 : 3, and the third having water and liquid C in the ratio 1 : 4. These three mixtures of A, B and C respectively, are further mixed in the proportion 4 : 3 : 2. Then the resulting mixture has

- A.
The same amount of water and liquid B

- B.
The same amount of liquids B and C

- C.
More water than liquid B

- D.
More water than liquid A

Answer: Option C

**Explanation** :

Proportion of water in three mixtures with A, B and C is $\frac{1}{3}$, $\frac{1}{4}$ and $\frac{1}{5}$ respectively.

[Sum of these proportions is 4, 5 and 6 and LCM(4, 5, 6) = 60]

Now as three mixture with A, B and C are mixed in the ratio 4 : 3 : 2.

Let the amount of three mixtures mixed be 4 × 60, 3 × 60 and 2 × 60 = 240, 180 and 120 liters respectively.

Quantity of water in liquid A, B and C is $\frac{1}{3}$ × 240 + $\frac{1}{4}$ × 180 + $\frac{1}{5}$ × 120 = 80 + 45 + 24 = 149

Quantity of liquid A in resultant mixture = $\frac{2}{3}$ × 240 = 160

Quantity of liquid B in resultant mixture = $\frac{3}{4}$ × 180 = 135

Quantity of liquid C in resultant mixture = $\frac{4}{5}$ × 120 = 96

Now let is examine the options individually

(a) The resultant mixture has 149 liters water and 160 of liquid B.

Hence, option (a) in incorrect.

(b) The resultant mixture has 135 liters of liquid B and 96 liters of liquid C.

Hence, option (b) is incorrect.

(c) Now quantity of water in resultant mixture 149 liters and quantity of liquid B is 135 liters.

Hence, option (c) is correct.

(d) Quantity of liquid A in resultant, mixture is 160 liters and quantity of water = 149 liters.

Hence, option (d) is incorrect.

Hence, option (c).

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**15. CAT 2017 QA Slot 2 | Geometry - Quadrilaterals & Polygons**

Let ABCDEF be a regular hexagon with each side of length 1 cm. The area (in sq cm) of a square with AC as one side is

- A.
3√2

- B.
3

- C.
4

- D.
√3

Answer: Option B

**Explanation** :

ABCDEF is a hexagon with side 1 cm. Let us construct the diagram as shown below

Now if we look at triangle ABC

Further, as AB = BC, △ABC will become a 120° - 30° - 30° triangle and the ratio of it’s sides. (∵ ∠ABC is an angle of regular hexagon ABCDEF)

will be √3 : 1 : 1.

Now $\frac{AC}{AB}=\frac{\sqrt{3}}{1}$

∴ AC = √3 units

Area of square with side AC = (√3)^{2} = 3 sq.units

Hence, option (b).

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**16. CAT 2017 QA Slot 2 | Geometry - Mensuration**

The base of a vertical pillar with uniform cross section is a trapezium whose parallel sides are of lengths 10 cm and 20 cm while the other two sides are equal in length. The perpendicular distance between the parallel sides of the trapezium is 12 cm. If the height of the pillar is 20 cm, then the total area, in sq cm, of all six surfaces of the pillar is

- A.
1300

- B.
1340

- C.
1480

- D.
1520

Answer: Option C

**Explanation** :

Let us draw the base of the vertical pillar as shown below

In the above figure EF = AB = 10 cm

Also, as AD = BC, by rule of symmetry

DE = FC

Now DC = DE + EF + FC

Let DE = FC = x

20 = x + 10 + x

⇒ 2x + 10 = 20

⇒ x = 5

Now in ∆AED, AE = 12 cm and DE = 5 cm

∴ AD = $\sqrt{{5}^{2}+{12}^{2}}=13$

Also given AD = BC

∴ BC = 13 units

Now, total surface area of vertical pilar with base ABCD = Area of Rectangle with side AB and side (Height) 20 cm + Area of Rectangle side AD and side (Height) 20 cm + Area of Rectangle with side BC and side (Height) 20 cm + Area of Rectangle with side DC and side (Height) 20 cm + 2 × Area of Trapezium ABCD

⇒ 10 × 20 + 13 × 20 + 13 × 20 + 20 × 20 + 2 × $\frac{1}{2}$ × (20 + 10) × 12

⇒ 200 + 260 + 260 + 400 + 360

⇒ 1480 sq.cm.

Hence, option (c).

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**17. CAT 2017 QA Slot 2 | Geometry - Coordinate Geometry**

The points (2, 5) and (6, 3) are two end points of a diagonal of a rectangle. If the other diagonal has the equation y = 3x + c, then c is

- A.
-5

- B.
-6

- C.
-7

- D.
-8

Answer: Option D

**Explanation** :

Let ABCD be the rectangle where A = (2, 5) and C= (6, 3).

We know that AC and BD will be the diagonals of the rectangle. It can be represented using the diagram shown below

In the above figure, O is the point of intersection of the 2 diagonals and the mid point of both the diagonals.

Mid Point of AC = $\left(\frac{2+6}{2},\frac{5+3}{2}\right)\equiv $ (4, 4)

Now O also lies on diagonal BD [Diagonal of a rectangle bisect each other.]

Hence co-ordinates of O will satisfy the equation

y = 3x + c

∴ 4 = 3 × 4 + c

⇒ c = – 8

Hence, option (d).

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**18. CAT 2017 QA Slot 2 | Geometry - Quadrilaterals & Polygons**

ABCD is a quadrilateral inscribed in a circle with centre O. If ∠COD = 120 degrees and ∠BAC = 30 degrees, then the value of ∠BCD (in degrees) is

Answer: 90

**Explanation** :

Angle subtended by an arc on circle is half of the angle it makes at the center.

∴ Considering arc CD, it makes an angle of 120° at the center hence ∠CAD = 60°.

⇒ ∠BAD = 90°

Now, in a cyclic quadrilateral, sum of opposite angle = 90°.

⇒ ∠BAD + ∠BCD = 90°

∴ ∠BCD = 90°

Hence, 90.

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**19. CAT 2017 QA Slot 2 | Geometry - Mensuration | Algebra - Inequalities & Modulus**

If three sides of a rectangular park have a total length 400 ft, then the area of the park is maximum when the length (in ft) of its longer side is

Answer: 200

**Explanation** :

Let ‘x’ and ‘y’ be the dimensions of the rectangle

Let us suppose 2x + y = 400 … (1)

Area = xy.

For xy to be maximum, 2xy should also be maximum.

Now the product 2xy will be maximum when 2x = y. [AM ≥ GM]

So y + y = 400 [From (1)]

⇒ 2y = 400 or y = 200

Substituting value of y in (1) we get,

2x = 200 or x = 100

In a rectangle, length is greater than breadth, so we take y as the length.

Hence area of the park is maximum when length is 200 ft.

Hence, 200.

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**20. CAT 2017 QA Slot 2 | Geometry - Triangles**

Let P be an interior point of a right-angled isosceles triangle ABC with hypotenuse AB. If the perpendicular distance of P from each of AB, BC, and CA is 4(√2 - 1) m, then the area, in sq cm, of the triangle ABC is

Answer: 16

**Explanation** :

Now since the perpendicular distance of P from each 3 sides is the same, it is the incenter of the triangle ABC.

Let ‘x’ be one of the equal sides of the isosceles right angled triangle ABC.

So the length of the hypotenuse will be √2x.

Area of right triangle ABC = $\frac{1}{2}{\mathrm{x}}^{2}$

So the Area of triangle ABC can also be expressed as a product of the in radius ‘r’ and semi perimeter ‘s’.

s = $\frac{x+x+\sqrt{2}x}{2}$ = x + $\frac{x}{\sqrt{2}}$

∴ Area = $\left(x+\frac{x}{\sqrt{2}}\right)\left[4\right(\sqrt{2}-1\left)\right]$ = $\frac{{x}^{2}}{2}$

⇒ 4√2x - 4x + 4x - 2√2x = $\frac{{x}^{2}}{2}$

⇒ 2√2x = $\frac{{x}^{2}}{2}$

⇒ x = 4√2

Area of triangle ABC = ½ × 4√2 × 4√2 = 16 sq.units.

Hence, 16.

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**21. CAT 2017 QA Slot 2 | Algebra - Number Theory**

If the product of three consecutive positive integers is 15600 then the sum of the squares of these integers is

- A.
1777

- B.
1785

- C.
1875

- D.
1877

Answer: Option D

**Explanation** :

Since 15600 is product of 3 consecutive we can start by taking the cube of an integer closest to 15600, which is actually 15625

Now $\sqrt[3]{15625}=25$

Dividing 15600 by 25 we get 624

624 can be expressed as a product of 24 and 26. So the 3 consecutive positive integers are 24, 25 and 26

24^{2} + 25^{2 }+ 25^{2} = 576 + 625 + 676 = 1877

Hence, option (d).

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**22. CAT 2017 QA Slot 2 | Algebra - Logarithms**

If *x* is a real number such that log_{3}5 = log_{5}(2 + *x*), then which of the following is true?

- A.
0 < x < 3

- B.
23 < x < 30

- C.
x > 30

- D.
3 < x < 23

Answer: Option D

**Explanation** :

Given log_{3}5 = log_{5}(*x *+ 2)

Now log_{3}3 < log_{3}5 < log_{3}9

∴ 1 < log_{3}5 < 2

⇒ 1 < log_{5}(x + 2) < 2

⇒ log_{5}5 < log_{5}(x + 2) < 2log_{5}5

⇒ log_{5}5 < log_{5}(x + 2) < log_{5}25

⇒ 5 < 2 + *x *< 25

⇒ 3 < *x* < 23

Hence, option (d).

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**23. CAT 2017 QA Slot 2 | Algebra - Functions & Graphs**

Let f(x) = x^{2} and g(x) = 2x, for all real x. Then the value of f(f(g(x)) + g(f(x)) ) at x = 1 is

- A.
16

- B.
18

- C.
36

- D.
40

Answer: Option C

**Explanation** :

At x = 1, f[g(x)] = f[g(1)]

Now g(1) = 2^{1} = 2

f[g(1)] = f(2) = 4

At x = 1, g[f(x)] = g[f(1)]

Now f(1) = 1 and

g[f(1)] = g(1) = 2^{1} = 2

Now at x = 1

f[f(g(x)) + g(f(x))] = f(4 + 2) = f(6) = 36

Hence, option (c).

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**24. CAT 2017 QA Slot 2 | Algebra - Quadratic Equations**

The minimum possible value of the squares of the roots of the equation: x^{2} + (a + 3)x – (a + 5) = 0 is

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option C

**Explanation** :

In the given equation, let α and β be the 2 roots of the equation Now α + β = - (a + 3)

Also α × β = - (a – 5)

(α + β)^{2} = α^{2 }+ β^{2 }+ 2αβ

⇒ (a + 3)^{2} = α^{2 }+ β^{2} - 2(a + 5)

⇒ a^{2} + 6a + 9 = α^{2} + β^{2} - 2a – 10

⇒ a^{2} + 8a + 19 = α^{2} + β^{2}

⇒ (a^{2} + 8a + 16) + 3 = α^{2} + β^{2}

⇒ α^{2} + β^{2} = (a + 4)^{2} + 3

Now (a + 4)^{2} ≥ 0

So minimum value of (a + 4)^{2} will be 0 when a = -4

At a = -4, the of squares of the roots of the equation i.e., α^{2} + β^{2} will be 3.

Hence, option (c).

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**25. CAT 2017 QA Slot 2 | Algebra - Surds & Indices**

If ${9}^{x-\frac{1}{2}}$ - 2^{2x - 2} = 4^{x} - 3^{2x - 3}, then x is

- A.
$\frac{3}{2}$

- B.
$\frac{2}{5}$

- C.
$\frac{3}{4}$

- D.
$\frac{4}{9}$

Answer: Option A

**Explanation** :

${9}^{\mathrm{x}-\frac{1}{2}}$ - 2^{2x - 2} = 4^{x }- 3^{2x - 3} ...(1)

Now, ${9}^{\mathrm{x}-\frac{1}{2}}$ = ${\left({3}^{2}\right)}^{x-\frac{1}{2}}$ = 3^{2x - 1}

Also 4^{x} = (2^{2})^{x} = 2^{2x}

From (1), we get

3^{2x - 1} - 2^{2x - 2} = 2^{2x} - 3^{2x - 3}

3^{2x - 1} + 3^{2x - 3} = 2^{2x} + 2^{2x - 2}

3^{2x - 3} (3^{2} + 1) = 2^{2x - 2} (2^{2} + 1)

$\frac{{3}^{2\mathrm{x}-3}}{{2}^{2\mathrm{x}-2}}=\frac{5}{10}\mathrm{or}\frac{1}{2}$

Now this can be written as

$\frac{{3}^{2x-3}}{{2}^{2x-2}}=\frac{{3}^{0}}{{2}^{1}}$

Now since the bases of the numerator and the dominator are equal on LHS and RHS, the powers should also be equal

⇒ 3^{2x-3} = 3^{0} and 2^{2x - 2} = 2^{1}

In both cases x = 3/2

∴ x = $\frac{3}{2}$

Hence, option (a).

Workspace:

**26. CAT 2017 QA Slot 2 | Algebra - Progressions | Algebra - Logarithms**

If log (2^{a} × 3^{b} × 5^{c}) is the arithmetic mean of log (2^{2} × 3^{3} × 5), log (2^{6} × 3 × 5^{7}), and log (2 × 3^{2} × 5^{4}), then a equals

Answer: 3

**Explanation** :

Arithmetic mean of log(2^{2} × 3^{3} × 5), (log 2^{6} × 3 × 5^{7}) and (log 2 × 3^{2} × 5^{4})

= $\frac{1}{3}$[log(2^{2} × 3^{3} × 5) + log(2^{6} × 3 × 5^{7}) + log(log 2 × 3^{2} × 5^{4})]

= $\frac{1}{3}$[log(2^{2} × 3^{3} × 5 × 2^{6} × 3 × 5^{7} × 2 × 3^{2} × 5^{4})]

= $\frac{1}{3}$[log(2^{2 + 6 + 1}) × (3^{3 + 1 + 2}) × (5^{1 + 7 + 4})]

= ${\mathrm{log}}_{}\left[{2}^{9\times \frac{1}{3}}\times {3}^{6\times \frac{1}{3}}\times {5}^{12\times \frac{1}{3}}\right]$

= log [2^{3} × 3^{2} × 5^{4}]

Now log(2^{a} × 3^{b} × 5^{c}) = log(2^{3} × 3^{2} × 5^{4})

(**Note: **In the given question it should have been specified that a, b, c are integers, as there can be multiple answers for the same. Since it is a TITA question, we have to assume the same and equate power of 2, 3 and 5)

∴ Equating powers of 2, 3 and 5 we get

a = 3, b = 2 and c = 4

Hence, 3.

Workspace:

**27. CAT 2017 QA Slot 2 | Algebra - Progressions**

Let *a*_{1}, *a*_{2}, *a*_{3}, *a*_{4}, *a*_{5} be a sequence of five consecutive odd numbers. Consider a new sequence of five consecutive even numbers ending with 2*a*_{3}.

If the sum of the numbers in the new sequence is 450, then *a*_{5} is

Answer: 51

**Explanation** :

Since 2*a*_{3} is the highest number in the new sequence, the sequence of 5 even numbers starting with the lowest, when expressed (in terms of *a*_{3}) is

(2*a*_{3} - 8), (2*a*_{3} - 6), (2*a*_{3} - 4), (2*a*_{3} - 2), 2*a*_{3}

Sum of these 5 numbers = 450

⇒ 2*a*_{3 }- 8 + 2*a*_{3 }- 6 + 2*a*_{3 }- 4 + 2*a*_{3 }- 2 + 2*a*_{3} = 450

⇒ 10*a*_{3} – 20 = 450

⇒ 10*a*_{3} = 470

⇒ *a*_{3 }= 47

Now in the original sequence of 5 odd numbers, a_{3} = 47

⇒ *a*_{5} = *a*_{3 }+ 4

⇒ *a*_{5 }= 47 + 4 = 51

Hence, 51.

Workspace:

**28. CAT 2017 QA Slot 2 | Algebra - Number Theory**

How many different pairs (a, b) of positive integers are there such that a ≤ b and $\frac{1}{a}+\frac{1}{b}=\frac{1}{9}$?

Answer: 3

**Explanation** :

Given $\frac{1}{a}+\frac{1}{b}=\frac{1}{9}$

$\frac{b+a}{ab}=\frac{1}{9}$

⇒ 9b + 9a = ab

⇒ ab - 9b – 9a = 0

⇒ ab - 9b – 9a + 81 = 81 [Adding 81 both sides]

⇒ b(a – 9) – 9(a – 9) = 81

⇒ (a – 9) × (b – 9) = 81

Now 81 as a product of 2 positive integers can be written as 1 × 81, 3 × 21 or 9 × 9

If a - 9 = 1 ⇒ a = 10.

Also b - 9 = 81 ⇒ b = 90

Further, if a – 9 = 3, ⇒ a = 12

Also b – 9 = 27 ⇒ b = 36

If a – 9 = 9 ⇒ a = 18.

Also b – 9 = 9 ⇒ b = 18

∴ 3 pairs of number i.e., (10, 90), (12, 36) and (18, 18) will satisfy the given equation.

Hence, 3.

Workspace:

**29. CAT 2017 QA Slot 2 | Modern Math - Permutation & Combination**

In how many ways can 8 identical pens be distributed among Amal, Bimal, and Kamal so that Amal gets at least 1 pen, Bimal gets at least 2 pens, and Kamal gets at least 3 pens?

Answer: 6

**Explanation** :

Out of 8 pens, Amal, Bimal and Kamal get at least 1, 2 and 3 pens.

So out of 8, pens 8 – (1 + 2 + 3) = 2 identical pens are to be distributed amongst 3 people.

So the number of ways in which 2 identical pens can be distributed amongst 3 people

= ^{2+3-1}C_{3-1 }= ^{4}C_{2} or 6 ways.

Hence, 6.

Workspace:

**30. CAT 2017 QA Slot 2 | Modern Math - Permutation & Combination**

How many four digit numbers, which are divisible by 6, can be formed using the digits 0, 2, 3, 4, 6, such that no digit is used more than once and 0 does not occur in the left-most position?

Answer: 50

**Explanation** :

For the 4 digit number to be divisible by 6, the sum of the digits of the 4 digit number has to be a multiple of 3 and the unit (right most) digit has to be an even number.

The combination of 4 digits that add upto a multiple of 3 are (0, 2, 3, 4), (0, 2, 4, 6) and (2, 3, 4, 6)

**Case 1**: Combination (0, 2, 3, 4)

For this combination the required number can start with 2, 3 or 4.

First digit 2: The last digit can be 0 or 4. The 2^{nd} and 3^{rd} digit from the left can be one 0/4 or 3. So a total of 4 possibilities.

First digit 4: Here too the number of possibilities is 4 as the 2^{nd} and 3^{rd} digit from the left can be one of 0/2 or 3.

First digit 3: Digits 0, 2 and 4 can be in any order as the 2^{nd}, 3^{rd} and 4^{th} digit from the left. So a total of 3! or 6 possibilities.

So number of possibilities with combination (0, 2, 3, 4) = 4 + 4 + 6 = 14

**Case 2**: Combination (0, 2, 4, 6)

The first digit can be anyone of 2, 4 or 6 i.e., 3 possibilities. The 2nd digit can be anyone from 0/2/4/6 but apart from the one used in the left most digit i.e., 3 possibilities. The 3^{rd} digit from the left will be any one from 0/2/4/6 but apart from the 2 digits used in the leftmost or 2^{nd} leftmost digit i.e., 2 possibilities. The extreme right digit will be the last digit after 3 out of these 4 digits are used in the first 3 digits from the left.

So total number of possibilities = 3 × 3 × 2 = 18

**Case 3**: Combination (2, 3, 4, 6)

The units digit has to be one amongst 2, 4 or 6 i.e., 3 possibilities. So first 3 digits from left will be 3 and two amongst 2/4/6 (i.e., except that digit which is not chosen as the right most digit) and can be arranged in 3! or 6 ways

So number of possibilities with combination (2, 3, 4, 6) is 6 × 3 = 18

So number of 4 digits numbers that can be formed using digits 0, 2, 3, 4 and 6 is 14 + 18 + 18 = 50

Hence, 50.

Workspace:

**31. CAT 2017 QA Slot 2 | Algebra - Functions & Graphs**

If f(ab) = f(a)f(b) for all positive integers a and b, then the largest possible value of f(1) is

Answer: 1

**Explanation** :

Let f(1) = x

Suppose f(1 × 1) = f(1) × f(1)

∴ f(1) = f(1) × f(1)

⇒ x = x × x

⇒ x^{2} = x

⇒ x = 1 or x = 0

But the highest value of x = 1

Hence, 1.

Workspace:

**32. CAT 2017 QA Slot 2 | Algebra - Functions & Graphs**

Let f(x) = 2x – 5 and g(x) = 7 – 2x. Then |f(x) + g(x)| = |f(x)| + |g(x)| if and only if

- A.
$\frac{5}{2}<x<\frac{7}{2}$

- B.
$x\le \frac{5}{2}orx\ge \frac{7}{2}$

- C.
$x<\frac{5}{2}orx\ge \frac{7}{2}$

- D.
$\frac{5}{2}\le x\le \frac{7}{2}$

Answer: Option D

**Explanation** :

Now

|f(x) + g(x)| = |f(x)| + |g(x)|

This is true only if both f(x) and g(x) are both negative or both positive or both are zero

**Case 1**: Now if both f(x) and g(x) are greater than or equal to zero.

f(x) = 2x - 5 ≥ 0 or x ≥ $\frac{5}{2}$

g(x) = 7 - 2x ≥ 0 or x ≤ $\frac{7}{2}$

∴$\frac{5}{2}\le x\le \frac{7}{2}$

**Case 2**: Now if both f(x) and g(x) are less than or equal to zero.

f(x) = 2x - 5 ≤ 0 or x ≤ $\frac{5}{2}$

g(x) = 7 - 2x ≤ 0 or x ≥ $\frac{7}{2}$

This means x ≥ $\frac{7}{2}$ and x ≤ $\frac{5}{2}$.

However, this is not possible.

Hence, option (d).

Workspace:

**33. CAT 2017 QA Slot 2 | Algebra - Progressions**

An infinite geometric progression *a*_{1}, *a*_{2}, *a*_{3}, … has the property that *a _{n}* = 3(

*a*

_{n}_{+1}+

*a*

_{n}_{+2}+ …) for every

*n*≥ 1. If the sum

*a*

_{1}+

*a*

_{2}+

*a*

_{3}+ … = 32, then

*a*

_{5}is

- A.
1/32

- B.
2/32

- C.
3/32

- D.
4/32

Answer: Option C

**Explanation** :

The given series is an infinite series.

Let first term a_{1} = a and the common ratio be r.

Now, a_{n }= 3(*a*_{n+1} + *a*_{n+2} + ……..)

⇒ ar^{n-1} = 3(ar^{n} + ar^{n+1} + ...)

⇒ ar^{n-1} = 3$\left(\frac{{\mathrm{ar}}^{\mathrm{n}}}{1-\mathrm{r}}\right)$

⇒ 1 - r = 3r

⇒ r = 1/4

Now, 32 = a_{1} + a_{2} + a_{3} + ...

⇒ 32 = a + ar + ar^{2} + ar^{3} + ...

⇒ 32 = $\frac{\mathrm{a}}{1-\mathrm{r}}$ = $\frac{\mathrm{a}}{1-1/4}$ = $\frac{4\mathrm{a}}{3}$

⇒ a = 24

∴ a_{5} = ar^{4} = $24\times {\left(\frac{1}{4}\right)}^{4}$ = $\frac{24}{256}$ = $\frac{3}{32}$

Hence, option (c).

Workspace:

**34. CAT 2017 QA Slot 2 | Algebra - Progressions**

If a_{1} = $\frac{1}{2\times 5}$, a_{2 }= $\frac{1}{5\times 8}$, a_{3} = $\frac{1}{8\times 11}$, ......, then a_{1} + a_{2} + a_{3} + .... + a_{100} is

- A.
$\frac{25}{151}$

- B.
$\frac{1}{2}$

- C.
$\frac{1}{4}$

- D.
$\frac{111}{55}$

Answer: Option A

**Explanation** :

a_{1} = $\frac{1}{2\times 5}$ = $\frac{1}{3}\left(\frac{3}{2\times 5}\right)$ = $\frac{1}{3}\left(\frac{5-2}{2\times 5}\right)$ = $\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}\right)$

Similarly,

a_{2} = $\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}\right)$

a_{3} = $\frac{1}{3}\left(\frac{1}{8}-\frac{1}{11}\right)$ and so on till

a_{100} = $\frac{1}{3}\left(\frac{1}{299}-\frac{1}{302}\right)$ [First term of a_{100} = 2 + (100 - 1) × 3 = 299]

Now if we add a_{1}, a_{2}, a_{3}, ...., a_{100}

a_{1} + a_{2 }+ a_{3 }+ ... + a_{100} = $\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}\right)$ + $\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}\right)$ + $\frac{1}{3}\left(\frac{1}{8}-\frac{1}{11}\right)$ + ... + $\frac{1}{3}\left(\frac{1}{299}-\frac{1}{302}\right)$

= $\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{299}-\frac{1}{302}\right)$

Terms $\frac{1}{5}$, $\frac{1}{8}$, ... $\frac{1}{296}$, $\frac{1}{299}$ will all cancel out

⇒ a_{1} + a_{2 }+ a_{3 }+ ... + a_{100} = $\frac{1}{3}\left(\frac{1}{2}-\frac{1}{302}\right)$

⇒ a_{1} + a_{2 }+ a_{3 }+ ... + a_{100} = $\frac{1}{3}\left(\frac{1}{2}-\frac{1}{302}\right)$

So a_{1} + a_{2} + a_{3} + ... + a_{100} = $\frac{150}{906}$ = $\frac{25}{151}$

Hence, option (a).

Workspace:

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