CAT 2017 QA Slot 2 | Previous Year CAT Paper

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.

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.

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:

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

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

Hence, option (b).

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)

⇒ $\frac{\mathrm{d}}{2}÷25$ = $\frac{\mathrm{d}}{50}$

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).

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).

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).

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).

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).

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).

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 p₁ is population after increase.

1.4f = p₁(1.27c) 

$\frac{1.4f}{1.27c}$ = p₁ 

⇒ p₁ = $\left(\frac{1.4}{1.27}\right)\left(\frac{f}{c}\right)$

or p₁ = $\left(\frac{1.4}{1.27}\right)$(p) [$\left(\because \mathrm{ p}=\frac{f}{c}\right)$ ]

$\Rightarrow$ p₁ ≃ 1.1p 

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

Hence, option (c).

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).

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).

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. 100

Hence, option (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).

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).

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).

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).

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.

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.

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.

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

24² + 25² + 25² = 576 + 625 + 676 = 1877

Hence, option (d).

Answer: Option D

Explanation :

Given log₃5 = log₅(x + 2)

Now log₃3 < log₃5 < log₃9

∴ 1 < log₃5 < 2
⇒ 1 < log₅(x + 2) < 2
⇒ log₅5 < log₅(x + 2) < 2log₅5
⇒ log₅5 < log₅(x + 2) < log₅25
⇒ 5 < 2 + x < 25
⇒ 3 < x < 23

Hence, option (d).

Answer: Option C

Explanation :

We have to calculate f [f(g(x)) + g(f(x))]

First we need to calculate f(g(x)) and g(f(x))

At x = 1, f[g(x)] = f[g(1)]
Now g(1) = 2¹ = 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) = 2¹ = 2

Now at x = 1
f [f(g(x)) + g(f(x))] = f(4 + 2) = f(6) = 36

Hence, option (c).

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 = α² + β² + 2αβ

⇒ (a + 3)² = α² + β² - 2(a + 5)
​​​​​​​⇒ a² + 6a + 9 = α² + β² - 2a – 10
⇒ a² + 8a + 19 = α² + β²
⇒ (a² + 8a + 16) + 3 = α² + β²
⇒ α² + β² = (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., α² + β² will be 3.

Hence, option (c).

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 = (2²)^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} (3² + 1) = 2^{2x - 2} (2² + 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 = 3⁰ and 2^{2x - 2} = 2¹

In both cases x = 3/2

∴ x = $\frac{3}{2}$

Hence, option (a).

Answer: 3

Explanation :

Arithmetic mean of log(2² × 3³ × 5), (log 2⁶ × 3 × 5⁷) and (log 2 × 3² × 5⁴)

= $\frac{1}{3}$[log(2² × 3³ × 5) + log(2⁶ × 3 × 5⁷) + log(log 2 × 3² × 5⁴)]

= ​​​​​​​$\frac{1}{3}$[log(2² × 3³ × 5 × 2⁶ × 3 × 5⁷ × 2 × 3² × 5⁴)]

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

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

= log [2³ × 3² × 5⁴]

Now log(2^a × 3^b × 5^c) = log(2³ × 3² × 5⁴)

(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.

Answer: 51

Explanation :

Since 2a₃ is the highest number in the new sequence, the sequence of 5 even numbers starting with the lowest, when expressed (in terms of a₃) is
(2a₃ - 8), (2a₃ - 6), (2a₃ - 4), (2a₃ - 2), 2a₃ 

Sum of these 5 numbers = 450

⇒ 2a₃ - 8 + 2a₃ - 6 + 2a₃ - 4 + 2a₃ - 2 + 2a₃ = 450
⇒ 10a₃ – 20 = 450
⇒ 10a₃ = 470
⇒ a₃ = 47

Now in the original sequence of 5 odd numbers, a₃ = 47 
⇒ a₅ = a₃ + 4
⇒ a₅ = 47 + 4 = 51

Hence, 51.

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.

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+3-1C_3-1 = ⁴C₂ or 6 ways.

Hence, 6.

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.

Answer: 1

Explanation :

Let f(1) = x

Suppose f(1 × 1) = f(1) × f(1)
∴ f(1) = f(1) × f(1)
⇒ x = x × x
⇒ x² = x
⇒ x = 1 or x = 0

But the highest value of x = 1

Hence, 1.

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).

Answer: Option C

Explanation :

The given series is an infinite series.
Let first term a₁ = a and the common ratio be r.

Now, a_n = 3(a_n+1 + a_n+2 + ……..)
⇒ ar^n-1 = 3(arⁿ + arⁿ⁺¹ + ...)
⇒ ar^n-1 = 3$\left(\frac{{\mathrm{ar}}^{\mathrm{n}}}{1-\mathrm{r}}\right)$
⇒ 1 - r = 3r
⇒ r = 1/4

Now, 32 = a₁ + a₂ + a₃ + ...
⇒ 32 = a + ar + ar² + ar³ + ...
⇒ 32 = $\frac{\mathrm{a}}{1-\mathrm{r}}$ = $\frac{\mathrm{a}}{1-1/4}$ = $\frac{4\mathrm{a}}{3}$
⇒ a = 24

∴ a₅ = ar⁴ = $24\times {\left(\frac{1}{4}\right)}^4$ =  $\frac{24}{256}$ = $\frac{3}{32}$

Hence, option (c).

Answer: Option A

Explanation :

a₁ = $\frac{1}{2\times 5}$ = $\frac{1}{3}\left(\frac{3}{2\times 5}\right)$ = $\frac{1}{3}\left(\frac{5-2}{2\times 5}\right)$ = $\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}\right)$

Similarly, 
a₂ = $\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}\right)$

a₃ = $\frac{1}{3}\left(\frac{1}{8}-\frac{1}{11}\right)$ and so on till

a₁₀₀ = $\frac{1}{3}\left(\frac{1}{299}-\frac{1}{302}\right)$ [First term of a₁₀₀ = 2 + (100 - 1) × 3 = 299]

Now if we add a₁, a₂, a₃, ...., a₁₀₀

a₁ + a₂ + a₃ + ... + a₁₀₀ = $\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}\right)$ + $\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}\right)$ + $\frac{1}{3}\left(\frac{1}{8}-\frac{1}{11}\right)$ + ... + $\frac{1}{3}\left(\frac{1}{299}-\frac{1}{302}\right)$

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

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

⇒ a₁ + a₂ + a₃ + ... + a₁₀₀ = $\frac{1}{3}\left(\frac{1}{2}-\frac{1}{302}\right)$

⇒ a₁ + a₂ + a₃ + ... + a₁₀₀ = $\frac{1}{3}\left(\frac{1}{2}-\frac{1}{302}\right)$

So a₁ + a₂ + a₃ + ... + a₁₀₀ = $\frac{150}{906}$ = $\frac{25}{151}$

Hence, option (a).