# CAT 2018 QA Slot 2

**1. CAT 2018 QA Slot 2 | Arithmetic - Time, Speed & Distance**

On a long stretch of east-west road, A and B are two points such that B is 350 km west of A. One car starts from A and another from B at the same time. If they move towards each other, then they meet after 1 hour. If they both move towards east, then they meet in 7 hrs. The difference between their speeds, in km per hour, is

Answer: 50

**Explanation** :

If both cars travel towards east, they meet in 7 hours. Therefore the faster car starting from A covers the distance of separation of 350 km in 7 hours. Therefore,

$a-b=\frac{350}{7}=50$

Workspace:

**2. CAT 2018 QA Slot 2 | Arithmetic - Time, Speed & Distance**

Points A and B are 150 km apart. Cars 1 and 2 travel from A to B, but car 2 starts from A when car 1 is already 20 km away from A. Each car travels at a speed of 100 kmph for the first 50 km, at 50 kmph for the next 50 km, and at 25 kmph for the last 50 km. The distance, in km, between car 2 and B when car 1 reaches B is

Answer: 5

**Explanation** :

The speeds of both the cars in the three legs of the journey are equal. Therefore the second car will reach point B as many minutes after the first car as it started after the first car.

For the first 50 km of the journey, the speed of the cars is 100 kmph. Therefore, the first car reaches a distance of 20 km

from point A after = $\frac{20}{100}=0.2$ hours

= 0.2 × 60 = 12 minutes.

When the first car reaches point B, the second car is 12 minutes away from point B. The speed of the car in the last 50 km distance is 25 kmph. Therefore the distance of the second car from point

$B=25\times \frac{12}{60}=5km$

Hence, 5.

Workspace:

**3. CAT 2018 QA Slot 2 | Algebra - Progressions**

The arithmetic mean of x, y and z is 80, and that of x, y, z, u and v is 75, where u = (x + y)/2 and v = (y + z)/2. If x ≥ z, then the minimum possible value of x is

Answer: 105

**Explanation** :

As the AM of x, y and z is 80,

x + y + z = 80 × 3 = 240 … (I)

As the AM of x, y, z, u and v is 75, x + y + z + u + v = 75 × 5 = 375 … (II)

Subtracting equation (I) from (II), we get

u + v = 135

$\therefore \frac{x+y}{2}+\frac{y+z}{2}=135$

∴ x + 2y + z = 270 … (III)

Solving equations (I) and (III), we get y=30.

Therefore x + z = 210.

Now, since x ≥ z, the minimum value of

x = 105.

Workspace:

**4. CAT 2018 QA Slot 2 | Arithmetic - Time & Work**

A water tank has inlets of two types A and B. All inlets of type A when open, bring in water at the same rate. All inlets of type B, when open, bring in water at the same rate. The empty tank is completely filled in 30 minutes if 10 inlets of type A and 45 inlets of type B are open, and in 1 hour if 8 inlets of type A and 18 inlets of type B are open. In how many minutes will the empty tank get completely filled if 7 inlets of type A and 27 inlets of type B are open?

Answer: 48

**Explanation** :

Suppose the inlet pipes of type A fill in water at the rate ‘a’ units per minute and the inlet pipes of type B fill in water at the rate ‘b’ units per minute.

Therefore we have the following

30(10a + 45b) = 60(8a + 18b)

∴ 300a + 1350b = 480a + 1080b

∴ 180a = 270b

∴ a = 1.5b

Total capacity of the tank

= 300a + 1350b = 300(1.5b) + 1350b

= 1800b

If 7 inlet pipes of type A and 27 inlet pipes of type B are opened, the volume of water filled in every minute

= 7a + 27b = 7(1.5b) + 27b = 37.5b

Therefore the number of minutes taken to fill the tank

$=\frac{1800}{37.5}=48$

Hence, 48.

Workspace:

**5. CAT 2018 QA Slot 2 | Arithmetic - Simple & Compound Interest**

Gopal borrows Rs. X from Ankit at 8% annual interest. He then adds Rs. Y of his own money and lends Rs. X+Y to Ishan at 10% annual interest. At the end of the year, after returning Ankit’s dues, the net interest retained by Gopal is the same as that accrued to Ankit. On the other hand, had Gopal lent Rs. X+2Y to Ishan at 10%, then the net interest retained by him would have increased by Rs. 150. If all interests are compounded annually, then find the value of X + Y.

Answer: 4000

**Explanation** :

The only difference in the two situations is that in the second situation, Gopal lent Rs. Y more to Ishan than in the first situation. Therefore the additional interest retained by Gopal = 0.1Y.

Therefore 0.1.Y = 150 or Y = 1500.

Now, Ankit lent Gopal Rs. X at 8% interest. Therefore the interest retained by Ankit = Rs. 0.08X.

Gopal lent Ishan Rs. (X + 1500) at 10% interest.

Therefore the interest retained by Gopal = 0.1X + 150 - 0.08X = 0.02X + 150

Therefore, 0.08X = 0.02X + 1500 or X = 2500.

Therefore the value of X + Y = 4000.

Workspace:

**6. CAT 2018 QA Slot 2 | Arithmetic - Time & Work**

Ramesh and Ganesh can together complete a work in 16 days. After seven days of working together, Ramesh got sick and his efficiency fell by 30%. As a result, they completed the work in 17 days instead of 16 days. If Ganesh had worked alone after Ramesh got sick, in how many days would he have completed the remaining work?

- A.
12

- B.
11

- C.
13.5

- D.
14.5

Answer: Option C

**Explanation** :

Suppose ‘R’ and ‘G’ are the number of units of work completed by Ramesh and Ganesh everyday. Therefore the total quantum of work to be completed = 16(R + G) = 16R + 16G

For the first 7 days, both work at 100% efficiency and for the remaining 10 days, only Ramesh works at 70% efficiency. Therefore the total quantum of work to be completed = 7(R + G) + 10(0.7R + G) = 14R + 17G

∴ 16R + 16G = 14R + 17G or G = 2R

When Ramesh fell sick, the total quantum of work could have been done in 9 more days with both working at 100% efficiency. Since Ganesh is twice as efficient as Ramesh is, if Ganesh had worked alone, he would have been able to finish the work in

$9\times \frac{3}{2}=13.5$ days

Hence, option 3.

Workspace:

**7. CAT 2018 QA Slot 2 | Arithmetic - Time, Speed & Distance**

Points A, P, Q and B lie on the same line such that P, Q and B are, respectively, 100 km, 200 km and 300 km away from A. Cars 1 and 2 leave A at the same time and move towards B. Simultaneously, car 3 leaves B and moves towards A. Car 3 meets car 1 at Q, and car 2 at P. If each car is moving in uniform speed then the ratio of the speed of car 2 to that of car 1 is

- A.
1 : 2

- B.
2 : 7

- C.
1 : 4

- D.
2 : 9

Answer: Option C

**Explanation** :

We have the following:

The cars 3 and 1 meet at point Q. Therefore, the car 3 covers 100 km in the time the car 1 covers 200 km. Therefore the ratio of the speeds of cars 3 and 1 is 1 : 2.

The cars 2 and 3 meet at point P. Therefore, the car 2 covers 100 km in the time the car 3 covers 200 km. Therefore the ratio of the speeds of cars 3 and car 2 is 2 : 1.

Combining the two statements, the ratio of the speeds of cars 2 and 1 is 1 : 4.

Hence, option 3.

Workspace:

**8. CAT 2018 QA Slot 2 | Arithmetic - Mixture, Alligation, Removal & Replacement**

A jar contains a mixture of 175 ml water and 700 ml alcohol. Gopal takes out 10% of the mixture and substitutes it by water of the same amount. The process is repeated once again. The percentage of water in the mixture is now

- A.
25.4

- B.
20.5

- C.
35.2

- D.
30.3

Answer: Option C

**Explanation** :

Originally, the volume of alcohol = 4 times the volume of water. Therefore, original percentage of alcohol = 80% and original volume of water = 20%.

When 10% of the mixture is removed and replaced with water, the percentage of alcohol remained in the mixture

$=80\%\times \left(\frac{9}{10}\right)=64.8\%$

Therefore the percentage of water in the mixture = (100 - 64.8)% = 35.2%.

Hence, option 3.

Workspace:

**9. CAT 2018 QA Slot 2 | Arithmetic - Time & Work**

A tank is emptied everyday at a fixed time point. Immediately thereafter, either pump A or pump B or both start working until the tank is full. On Monday, A alone completed filling the tank at 8 pm. On Tuesday, B alone completed filling the tank at 6 pm. On Wednesday, A alone worked till 5 pm, and then B worked alone from 5 pm to 7 pm, to fill the tank. At what time was the tank filled on Thursday if both pumps were used simultaneously all along?

- A.
4:36 pm

- B.
4:12 pm

- C.
4:48 pm

- D.
4:24 pm

Answer: Option D

**Explanation** :

The time taken by pump A alone to fill the tank completely = t hours.

The time taken by pump B alone to fill the tank completely = ‘t - 2’ hours.

On Wednesday, pump A was used for ‘t-3’ hours and pump B was used for 2 hours to completely fill the tank.

Therefore we have, $\frac{t-3}{t}+\frac{2}{t-2}=1$

$\therefore \; (t\; -\; 3)\; (t\; -\; 2)\; +\; 2t\; =\; t(t\; -\; 2)$

∴ t^{2} - 5t + 6 + 2t = t^{2} - 2t

Solving for t, t = 6.

Therefore pump A takes 6 hours and pump B takes 4 hours to completely fill the tank. Both the tanks are started at 2 pm.

Suppose the capacity of the tank is 12 litres, in one hour A fills 2 litres and B fills 3 litres. Therefore if both pumps A and B are used, they will together fill the tank in

$\frac{12}{2+3}=2.4hours$

= 2 hours and 24 minutes

Therefore on Thursday, the tank will be completely full at 4.24 pm.

Hence, option 4.

Workspace:

**10. CAT 2018 QA Slot 2 | Arithmetic - Mixture, Alligation, Removal & Replacement**

The strength of a salt solution is p% if 100 ml of the solution contains p grams of salt. If three salt solutions A, B, C are mixed in the proportion 1 : 2 : 3, then the resulting solution has strength 20%. If instead the proportion is 3 : 2 : 1, then the resulting solution has strength 30%. A fourth solution, D, is produced by mixing B and C in the ratio 2 : 7. The ratio of the strength of D to that of A is?

- A.
1 : 3

- B.
1 : 4

- C.
2 : 5

- D.
3 : 10

Answer: Option A

**Explanation** :

Suppose the concentrations of the solutions A, B and C are ‘a’, ‘b’ and ‘c’ respectively.

If the three solutions are mixed in the ratio 1:2:3, the resultant solution has 20% concentration. Therefore, we have

$\frac{a+2b+3c}{1+2+3}=20ora\; +\; 2b\; +\; 3c\; =\; 120\; ...(I)$

If the three solutions are mixed in the ratio 3 : 2 : 1, the resultant solution has 30% concentration. Therefore, we have

$\frac{3a+2b+c}{3+2+1}=30or3a+2b+c$ = 180 … (II)

Multiplying equation (I) with 3 and (II) with 2 and equating both, we get

3a + 6b + 9c = 6a + 4b + 2c

⇒ 2b + 7c = 3a

Now, concentration of solution D = $\frac{2b+7c}{9}$ = $\frac{3a}{9}$ = $\frac{a}{3}$

Therefore the ratio of D's concentration to A's concentration $=\frac{a/3}{a}=\frac{1}{3}$

Hence, option 1.

Workspace:

**11. CAT 2018 QA Slot 2 | Arithmetic - Average**

Let a_{1}, a_{2}, ... , a_{52} be positive integers such that a_{1} < a_{2} < ... < a_{52}. Suppose, their arithmetic mean is one less than the arithmetic mean of a_{2}, a_{3}, ..., a_{52}. If a_{52} = 100, then the largest possible value of a_{1} is

- A.
45

- B.
23

- C.
48

- D.
20

Answer: Option B

**Explanation** :

Given numbers are a_{1},a_{2},a_{3},...,a_{51},100. If the arithmetic mean of these 52 numbers is ‘x’, we have

a_{1} + a_{2} + a_{3} + ... + a_{51} + 100 = 52x

Given: the arithmetic mean of a_{2}, a_{3}, a_{4},...,a_{51}, 100 is ‘x + 1’

a_{2} + a_{3} + ⋯ + a_{51} + 100 = 51(x+1)

= 51x + 51

∴ a_{1} = x − 51

**Using options:**

If a_{1} = 48 or x = 99. Therefore, a_{2} + a_{3} + ⋯ + a_{51} = 52 × 99 − 48 − 100 = 5000.

We cannot find 50 unique numbers greater than 48 but less than 100 such that they add up to 5000.

If a_{1} = 45 or x = 96. Therefore, a_{2} + a_{3} + ⋯ + a_{51} = 52 × 96 − 45 − 100 = 4847.

We cannot find 50 unique numbers greater than 45 but less than 100 such that they add up to 4847.

If a_{1} = 23 or x = 74. Therefore, a_{2} + a_{3} + ⋯ + a_{51} = 52 × 74 − 23 − 100 = 3725.

We can find 50 unique numbers greater than 23 but less than 100 such that they add up to 3725.

Hence, option 2.

Workspace:

**12. CAT 2018 QA Slot 2 | Arithmetic - Ratio, Proportion & Variation**

The scores of Amal and Bimal in an examination are in the ratio 11 : 14. After an appeal, their scores increase by the same amount and their new scores are in the ratio 47 : 56. The ratio of Bimal’s new score to that of his original score is

- A.
4 : 3

- B.
3 : 2

- C.
8 : 5

- D.
5 : 4

Answer: Option A

**Explanation** :

Suppose the original scores of Amal and Bimal are 11x and 14x respectively.

Suppose ‘y’ is the increase in their marks.

Therefore the new marks of Amal and Bimal are ‘11x + y’ and ‘14x + y’ respectively.

Therefore, we have $\frac{11x+y}{14x+y}=\frac{47}{56}$

∴ 616x + 56y = 658x + 47y

Solving for x and y, we get 14x = 3y

Therefore Bimal’s old marks = 14x = 3y and his new marks = 14x + y = 4y. Therefore the required ratio = 4:3.

Hence, option 1.

Workspace:

**13. CAT 2018 QA Slot 2 | Arithmetic - Mixture, Alligation, Removal & Replacement**

There are two drums, each containing a mixture of paints A and B. In drum 1, A and B are in the ratio 18 : 7. The mixtures from drums 1 and 2 are mixed in the ratio 3 : 4 and in this final mixture, A and B are in the ratio 13 : 7. In drum 2, then A and B were in the ratio

- A.
220 : 149

- B.
251 : 163

- C.
229 : 141

- D.
239 : 161

Answer: Option D

**Explanation** :

Suppose the mixture from drum 1 = 300x litres and the mixture from drum 2 = 400x litres. Therefore, we have the following

Therefore the volume of Paint A in Drum B = 455x – 216x = 239x and the volume of Paint B in Drum B = 245x − 84x = 161x.

Therefore the required ratio is 239:161.

Hence, option 4.

Workspace:

**14. CAT 2018 QA Slot 2 | Arithmetic - Mixture, Alligation, Removal & Replacement**

A 20% ethanol solution is mixed with another ethanol solution, say, S of unknown concentration in the proportion 1 : 3 by volume. This mixture is then mixed with an equal volume of 20% ethanol solution. If the resultant mixture is a 31.25% ethanol solution, then the unknown concentration of S is

- A.
52%

- B.
50%

- C.
48%

- D.
55%

Answer: Option B

**Explanation** :

When a 20% solution is mixed with x% solution in the ratio 1 : 3, the resultant concentration (in %) of the solution

$=\frac{20\left(1\right)+x\left(3\right)}{1+3}=\frac{3x+20}{4}=0.75x+5$

When this solution is mixed with 20% solution having equal volume, the resultant concentration (in %) of the solution

$=\frac{0.75x+5+20}{2}=\frac{3}{8}x+12.5\phantom{\rule{0ex}{0ex}}\therefore \frac{3}{8}x+12.5=31.25$

Solving for x, x = 50.

Hence, option 2.

Workspace:

**15. CAT 2018 QA Slot 2 | Geometry - Coordinate Geometry**

A triangle ABC has area 32 sq units and its side BC, of length 8 units, lies on the line x = 4. Then the shortest possible distance between A and the point (0,0) is

- A.
8 units

- B.
4 units

- C.
$2\sqrt{2}$ units

- D.
$2\sqrt{4}$ units

Answer: Option B

**Explanation** :

The area of the triangle is 32 sq units.

Therefore, 32 = ½ × BC × height =

Givne, BC = 8

⇒ Height = $\frac{2\times 32}{8}$ = 8 cm

Therefore Height of the triangle = 8 cm.

Side BC lies on the line x = 4 or it is parallel to y-axis. Therefore the height of the triangle will be parallel to x axis. In order to minimize the distance of point A from the origin, point B should be (4, 4) and point C should be (4, -4), as shown.

Therefore point A is (-4,0) and the distance of point A from the origin = 4 units.

Hence, option 2.

Workspace:

**16. CAT 2018 QA Slot 2 | Geometry - Circles**

A chord of length 5 cm subtends an angle of 60° at the centre of a circle. The length, in cm, of a chord that subtends an angle of 120° at the centre of the same circle is

- A.
$5\sqrt{3}$

- B.
$6\sqrt{2}$

- C.
2π

- D.
8

Answer: Option A

**Explanation** :

The triangle formed by joining the endpoints of the chord of a circle with the center of the circle is an isosceles triangle because the two sides of the triangle (radii of the circle) are congruent. If the angle subtended by the chord at the center of the circle is 60 degrees, the triangle is an equilateral triangle. Therefore the side of the triangle and the radius of the circle is equal to the side of the triangle = 5 cm.

Now we have the following

OP is perpendicular to the chord AB. Therefore P is the midpoint of chord AB. Further, OA and OP are the radii of the circle. Therefore triangles OAP and OBP are congruent. Therefore triangle OAP and OBP are 30-60-90 triangles.

We have, $\mathrm{cos}30=\frac{\sqrt{3}}{2}=\frac{AP}{OA}=\frac{AP}{5}$

Therefore, $AP=\frac{5}{2}\sqrt{3}$

Therefore, $\mathcal{l}\left(AB\right)=2\times \mathcal{l}\left(AP\right)=5\sqrt{3}$

Hence, option 1.

Workspace:

**17. CAT 2018 QA Slot 2 | Geometry - Quadrilaterals & Polygons**

A parallelogram ABCD has area 48 sqcm. If the length of CD is 8 cm and that of AD is s cm, then which one of the following is necessarily true?

- A.
s ≠ 6

- B.
s ≥ 6

- C.
5 ≤ s ≤ 7

- D.
s ≤ 6

Answer: Option B

**Explanation** :

Area of parallelogram = Base × Height

∴ 48 = 8 × Height

∴ Height = 6 cm

If the parallelogram is a rectangle, AD is its height. In that case, s = 6.

Otherwise, using Pythagoras theorem,

AD > 6 or s > 6.

Therefore, s ≥ 6.

Hence, option 2.

Workspace:

**18. CAT 2018 QA Slot 2 | Geometry - Quadrilaterals & Polygons**

The area of a rectangle and the square of its perimeter are in the ratio 1 ∶ 25. Then the lengths of the shorter and longer sides of the rectangle are in the ratio ?

- A.
1 : 3

- B.
2 : 9

- C.
1 : 4

- D.
3 : 8

Answer: Option C

**Explanation** :

Suppose l = length of the rectangle and b = breadth of the rectangle.

Therefore, the area of the rectangle = lb and the perimeter of the rectangle = 2(l + b)

Therefore,

$\frac{\mathcal{l}b}{{\left[2\right(l+b\left)\right]}^{2}}=\frac{1}{25}i.e.\frac{\iota b}{{(l+b)}^{2}}=\frac{4}{25}$

Substituting the options, we can see that only option [3] i.e. 1:4 matches.

Hence, option 3.

Workspace:

**19. CAT 2018 QA Slot 2 | Geometry - Mensuration**

From a rectangle ABCD of area 768 sq cm, a semicircular part with diameter AB and area 72π sq cm is removed. The perimeter of the leftover portion, in cm, is

- A.
88 + 12π

- B.
82 + 24π

- C.
80 + 16π

- D.
86 + 8π

Answer: Option A

**Explanation** :

The area of the semicircle = 72π sq. cm. If ‘r’ is the radius of the semi-circular portion removed,

We get $\pi \times \frac{{r}^{2}}{2}=72m$

Therefore r = 12 cm. Therefore the diameter of the semi-circle = AB = 24 cm.

Area of rectangle ABCD = 768 sq. cm.

Therefore the length ofside AC $=\frac{768}{24}=32cm.$

The perimeter of the leftover portion

= Perimeter of the semicircle + l(AD) + l(DC) + l(CB)

= 12π + 32 + 24 + 32 = 88 + 12π

Hence, option 1.

Workspace:

**20. CAT 2018 QA Slot 2 | Geometry - Triangles | Geometry - Circles**

On a triangle ABC, a circle with diameter BC is drawn, intersecting AB and AC at points P and Q, respectively. If the lengths of AB, AC, and CP are 30 cm, 25 cm, and 20 cm respectively, then the length of BQ, in cm, is

Answer: 24

**Explanation** :

Since points P and Q lie on the circumference of the circle drawn with BC as diameter, angles BPC and BQC are right angled triangles.

Now, area of triangle ABC = $\frac{1}{2}\times AB\times PC$ = $\frac{1}{2}\times AC\times BQ$

Therefore, BQ = $\frac{\mathrm{AB}\times \mathrm{PC}}{\mathrm{AC}}$ = $\frac{30\times 20}{25}$ = 24

Answer: 24

Workspace:

**21. CAT 2018 QA Slot 2 | Algebra - Logarithms**

If p^{3} = q^{4} = r^{5} = s^{6}, then the value of log_{s}(pqr) is equal to

- A.
47/10

- B.
16/5

- C.
24/5

- D.
1

Answer: Option A

**Explanation** :

${\mathrm{log}}_{\mathrm{s}}\left(\mathrm{pqr}\right)={\mathrm{log}}_{\mathrm{s}}\mathrm{p}+{\mathrm{log}}_{\mathrm{s}}\mathrm{q}+{\mathrm{log}}_{\mathrm{s}}\mathrm{r}\phantom{\rule{0ex}{0ex}}\mathrm{Given}:{\mathrm{p}}^{3}={\mathrm{s}}^{6}\mathrm{or}\mathrm{p}={\mathrm{s}}^{2}$

Therefore, ${\mathrm{log}}_{s}p=2$

On similar lines, ${\mathrm{log}}_{\mathrm{s}}\mathrm{q}=\frac{3}{2}\mathrm{and}{\mathrm{log}}_{\mathrm{s}}\mathrm{r}=\frac{6}{5}$

Therefore, ${\mathrm{log}}_{s}\left(pqr\right)$

$={\mathrm{log}}_{s}p+{\mathrm{log}}_{s}q+{\mathrm{log}}_{s}r=\frac{47}{10}$

Hence, option 1.

Workspace:

**22. CAT 2018 QA Slot 2 | Algebra - Logarithms | Algebra - Inequalities & Modulus**

The smallest integer n for which 4n > 17^{19} holds, is closest to

- A.
33

- B.
37

- C.
39

- D.
35

Answer: Option C

**Explanation** :

We have, 4^{n} > 17^{19}

Taking log on both sides to the base 4,

n log_{4} 4 > 19 log_{4} 17

Now, 17 > 4^{2.}

∴ log_{4} 17 > 2.

Also, log_{4} 4 = 1

Therefore, n > 38.

Hence, option 3.

Workspace:

**23. CAT 2018 QA Slot 2 | Algebra - Logarithms**

$\frac{1}{l{\mathrm{og}}_{2}100}-\frac{1}{{\mathrm{log}}_{4}100}+\frac{1}{{\mathrm{log}}_{5}100}-\frac{1}{{\mathrm{log}}_{10}100}+\frac{1}{{\mathrm{log}}_{20}100}-\frac{1}{{\mathrm{log}}_{25}100}+\frac{1}{{\mathrm{log}}_{50}100}?$

- A.
-4

- B.
10

- C.
0

- D.
1/2

Answer: Option D

**Explanation** :

$\frac{1}{l{\mathrm{og}}_{2}100}-\frac{1}{{\mathrm{log}}_{4}100}+\frac{1}{{\mathrm{log}}_{5}100}-\frac{1}{{\mathrm{log}}_{10}100}+\frac{1}{{\mathrm{log}}_{20}100}-\frac{1}{{\mathrm{log}}_{25}100}+\frac{1}{{\mathrm{log}}_{50}100}?$

$={\mathrm{log}}_{100}2-{\mathrm{log}}_{100}4+{\mathrm{log}}_{100}5-{\mathrm{log}}_{100}10+{\mathrm{log}}_{100}20-{\mathrm{log}}_{100}25+{\mathrm{log}}_{100}50\phantom{\rule{0ex}{0ex}}={\mathrm{log}}_{100}\left(\frac{2\times 5\times 20\times 50}{4\times 10\times 25}\right)={\mathrm{log}}_{100}\left(10\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}$

Hence, option 4.

Workspace:

**24. CAT 2018 QA Slot 2 | Algebra - Inequalities & Modulus**

If the sum of squares of two numbers is 97, then which one of the following cannot be their product?

- A.
16

- B.
48

- C.
–32

- D.
64

Answer: Option D

**Explanation** :

Given: a^{2} + b^{2} = 97.

If a^{2} = b^{2} = 48.5, 6.5 < |a| = |b| < 7.

Therefore any or neither or a and b can be negative.

Therefore the product of a and b must follow –49 < ab < 49.

Out of the given options, 64 definitely lies outside this range.

Hence, option 4.

Workspace:

**25. CAT 2018 QA Slot 2 | Algebra - Number Theory**

How many two-digit numbers, with a non-zero digit in the units place, are there which are more than thrice the number formed by interchanging the positions of its digits?

- A.
6

- B.
8

- C.
7

- D.
5

Answer: Option A

**Explanation** :

If ‘x’ is the digit in the ten’s place and ‘y’ is the digit in the unit’s place of the two digit number, we have the following:

$10x+y>3(10y+x)\phantom{\rule{0ex}{0ex}}\therefore x>\frac{29}{7}y$

When y = 1, x = 5, 6, 7, 8 or 9 (Total 5 values)

When y = 2, x = 9 (Total 1 value)

Therefore the required answer is 5 + 1 = 6.

Hence option 1.

Workspace:

**26. CAT 2018 QA Slot 2 | Algebra - Surds & Indices**

If N and x are positive integers such that N^{N} = 2^{160} and N^{2} + 2N is an integral multiple of 2^{x}, then the largest possible x is

Answer: 10

**Explanation** :

From observation, N^{N} = 2^{160} = 32^{32} or N = 32.

Now, N^{2} + 2^{N} = 32^{2} + 2^{32} = (2^{5 })^{2} + 2^{32}

= 2^{10} + 2^{32} = 2^{10} (1 + 2^{22} ).

Here, (1 + 2^{22} ) is an odd number, hence the higest power of 2 is 10.

Therefore the largest value of x = 10.

Hence, 10.

Workspace:

**27. CAT 2018 QA Slot 2 | Algebra - Quadratic Equations | Algebra - Inequalities & Modulus**

If a and b are integers such that 2x^{2} − ax + 2 > 0 and x^{2} − bx + 8 ≥ 0 for all real numbers x, then the largest possible value of 2a − 6b is

Answer: 36

**Explanation** :

We have to maximize the value of 2a - 6b. Therefore let us look for the largest possible value of a and the smallest possible value of b.

If 2x^{2} – ax + 2 > 0 for all value of x, the graph of 2x^{2} – ax + 2 is above the x-axis or the roots of the quadratic equations

2x^{2} – ax + 2 = 0 are imaginary or its discriminant is less than 0. Therefore, we have a^{2} – 16 < 0 or a^{2} < 16 or –4 < a < 4. Therefore the largest possible value of a is 3.

If x^{2} – bx + 8 ≥ 0, the discriminant of the equation is less than or equal to zero.

Therefore, ${b}^{2}-32\le 0or-4\sqrt{2}\le b\le 4\sqrt{2}or-5.64\le b\le 5.64.$

Therefore the smallest possible value of b is –5.

Therefore the maximum possible value of 2a - 6b = 2(3) – 6(–5)=36.

Answer: 36

Workspace:

**28. CAT 2018 QA Slot 2 | Algebra - Quadratic Equations**

The smallest integer n such that n^{3} – 11n^{2 }+ 32n – 28 > 0 is

Answer: 8

**Explanation** :

The sum of the coefficients of n^{3} -11n^{2} + 32n - 28 is not zero.

Similarly the sum of the odd coefficients of n^{3} -11n^{2} + 32n - 28 is not equal to its even coefficients. Therefore it is not divisible by (n - 1) and (n + 1).

If we put n = 2, n^{3} – 11n^{2} + 32n – 28 = 8 – 44 + 64 – 28 = 0.

Therefore (n – 2) is a factor of the polynomial n^{3} - 11n^{2} + 32n – 28.

We have, (n - 2) is a factor of n^{3} – 11n^{2} + 32n – 28

∴ (n^{2} – 9n + 14) = (n – 2)(n – 2)(n - 7) = (n – 2)^{2} (n – 7).

Therefore n = 2 and n = 7 are the two roots of the equation n^{3 }- 11n^{2} + 32n - 28 = 0.

Also, at n = 0, n^{3} – 11n^{2} + 32n - 28 = –28 < 0. Therefore the curve touches the x-axis at 2 and intersects the x-axis at 7.

Therefore the smallest integer for which n^{3 }- 11n^{2} + 32n - 28 > 0 is 8.

Answer: 8

Workspace:

**29. CAT 2018 QA Slot 2 | Modern Math - Permutation & Combination**

In a tournament, there are 43 junior level and 51 senior level participants. Each pair of juniors play one match. Each pair of seniors play one match. There is no junior versus senior match. The number of girl versus girl matches in junior level is 153, while the number of boy versus boy matches in senior level is 276. The number of matches a boy plays against a girl is

Answer: 1098

**Explanation** :

The number of matches played in a group of ‘n’ people

$=\frac{n(n-1)}{2}$

The number of girl vs girl matches in junior level = 153.

Therefore, $\frac{n(n-1)}{2}=153orn=18$

Therefore there are 18 girls at junior level.

The number of boy vs boy matches in senior level = 276.

Therefore, $\frac{\mathrm{n}(\mathrm{n}-1)}{2}=276\mathrm{or}\mathrm{n}=24$

Therefore there are 24 boys at senior level.

Therefore, we have

Therefore, the number of boys vs girls matches = 25 × 18 + 24 × 27 = 1098

Answer: 1098

Workspace:

**30. CAT 2018 QA Slot 2 | Algebra - Progressions**

Let t_{1}, t_{2},… be real numbers such that t_{1} + t_{2} +…+ t_{n} = 2n^{2} + 9n + 13, for every positive integer n ≥ 2. If t_{k} = 103, then k equals

Answer: 24

**Explanation** :

For n = 2, t_{1} + t_{2} = 39

For n = 3, t_{1} + t_{2} + t_{3} = 58

For n = 4, t_{1} + t_{2} + t_{3} + t_{4} = 81

For n = 5, t_{1} + t_{2} + t_{3} + t_{4} + t_{5} = 108

For n = 6, t_{1} + t_{2} + t_{3} + t_{4} + t_{5} + t_{6} = 139.

Therefore, t_{3} = 19, t_{4} = 23, t_{5} = 27, t_{6} = 31 and so on. Thus t_{k} = 7 + 4k

If t_{k} = 103, 7 + 4k = 103 or k = 24.

Answer: 24

Workspace:

**31. CAT 2018 QA Slot 2 | Algebra - Functions & Graphs**

Let f(x) = max{5x, 52 –2x^{2}}, where x is any positive real number. Then the minimum possible value of f(x) is

Answer: 20

**Explanation** :

he graph of the function 52 – 2x^{2} will be of ‘inverted U’ shape, while the graph of the function y = 5x will be a straight line with a positive slope. Therefore, the minimum value of the required function will be obtained at a point of intersection of y = 52 – 2x^{2} and y = 5x.

Therefore, 52 – 2x^{2} = 5x or 2x^{2 }+ 5x – 52 = 0.

$\therefore (x-4)(2x+13)=0\phantom{\rule{0ex}{0ex}}\therefore x=4orx=-\frac{13}{2}$

Since x is a positive real number, x = 4.

At x = 4, 5x = 52 – 2x^{2} = 20

Answer: 20

Workspace:

**32. CAT 2018 QA Slot 2 | Algebra - Progressions**

The value of the sum 7 × 11 + 11 × 15 + 15 × 19 + ...+ 95 × 99 is

- A.
80730

- B.
80773

- C.
80751

- D.
80707

Answer: Option D

**Explanation** :

The first number in the terms are 7, 11, 15, …, 95, which form an AP with a = 7 and d = 4.

The nth term for this AP is Tn = 3 + 4n and the number of terms is

$n=\frac{95-7}{4}+1=23$

The second number in the terms are 11,15,19,...,99, which form an AP with a = 11 and d = 4. The nth term for this AP is Tn= 7 + 4n and the number of terms is

$n=\frac{99-11}{4}+1=23$

Therefore the required sum

$=\sum _{n=1}^{23}(3+4n)(7+4n)\phantom{\rule{0ex}{0ex}}=\sum _{n=1}^{23}(16{n}^{2}+40n+21)\phantom{\rule{0ex}{0ex}}=16\sum _{n=1}^{23}{n}^{2}+40\sum _{n=1}^{23}n+21\sum _{n=1}^{23}1\phantom{\rule{0ex}{0ex}}=16\times \frac{23\times 24\times 47}{6}+40\times \frac{23\times 24}{2}+21\times 23$

= 69184 + 11040 + 483

= 80707

Hence, option 4.

**Alternatively,**

The given series is: 7 × 11 + 11 × 15 + 15 × 19 + 19 × 23 + 23 × 27 + 27 × 31 + 31 × 35...

The unit’s place of the term 7 × 11 is 7.

The unit’s place of the term 11 × 15 is 5.

The unit’s place of the term 15 × 19 is 5.

The unit’s place of the term 19 × 23 is 7.

The unit’s place of the term 23 × 27 is 1.

The unit’s place of the sum of these 5 terms is 5.

After 23 × 27, the same unit’s place repeat in that order upto 83 × 87.

Thus up-to 83 × 87, the series of five terms each appears four times. Therefore the sum of all the terms up-to 83 × 87 is 4 × 5 = 20 or 0.

Now, we are left with three terms 87 × 91 + 91 × 95 + 95 × 99. The unit’s place in the sum of these three terms is 7 + 5 + 5 = 17 (or 7).

There is only one option 80707 that ends in 7.

Hence, option 4.

Workspace:

**33. CAT 2018 QA Slot 2 | Algebra - Number Theory | Modern Math - Sets**

If A = {6^{2n} -35n -1: n = 1,2,3,...} and B = {35(n - 1) : n = 1,2,3,...} then which of the following is true?

- A.
Every member of A is in B and at least one member of B is not in A

- B.
At least one member of A is not in B

- C.
Every member of B is in A.

- D.
Neither every member of A is in B nor every member of B is in A

Answer: Option A

**Explanation** :

For n = 1, A_{1} = 0

For n = 2, A_{2} = 1225 = 35 × 35

Also, 6^{2n}- 35n - 1 = (36n-1) - 35n.

The term in the bracket 36^{n} - 1 is always divisible by (36 - 1)or by 35. Similarly 35n is also always divisible by 35. Therefore each term in the set A is divisible by 35. However, not all positive multiples of 35 are present in set A.

For n = 1, B_{1} = 35(1 - 1) = 0

For n = 2, B_{2} = 35(2 - 1) = 35 and so on.

Thus all whole number multiples of 35 are present in set B.

Thus every member of set A is present in every member of set B but at least one member of set B is not present in set A.

Hence, option 1.

Workspace:

**34. CAT 2018 QA Slot 2 | Modern Math - Sets**

For two sets A and B, let AΔB denote the set of elements which belong to A or B but not both. If P = {1,2,3,4}, Q = {2,3,5,6,}, R = {1,3,7,8,9}, S = {2,4,9,10}, then the numberof elements in (PΔQ)Δ(RΔS) is

- A.
6

- B.
8

- C.
9

- D.
7

Answer: Option D

**Explanation** :

The set (PΔQ) = {1,4,5,6}.

The set (RΔS) = {1,2,3,4,7,8,10}

Therefore the set (PΔQ) Δ (RΔS)

= {2,3,5,6,7,8,10}. Thus there are 7 elements in the set.

Hence, option 4.

Workspace:

## Feedback

**Help us build a Free and Comprehensive CAT/MBA Preparation portal by providing
us your valuable feedback about Apti4All and how it can be improved.**