CAT 2025 Slot 3 Question Paper

In clue 5, we are given that export volumes of P, in IC, to X and C are 600 and 1200, respectively, and P is the only country that exports to C. So, the export from X to C and ROW to C is 0.

In clue 1, we are given that normalised trade balances of P, X and C are 0%, 10%, and -20%, respectively.

For P, since the normalised trade balance is 0% the exports have to be equal to the imports according to the given formula. So, let us assume them to be 10a and 10a.

For X,

$$\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ 10$$

$$10E\ -\ 10I\ =\ E\ +\ I$$

$$11I\ =\ 9E$$

So, let us assume export to be 11b and import to be 9b.

For C,

$$\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ -20$$

$$5I\ -\ 5E\ =\ E\ +\ I$$

$$2I\ =\ 3E$$

So, let us assume export to be 2c and import to be 3c.

In clue 2, we are given that 40% of exports of X are to P and 22% of imports of P are from X.

40% of the exports of X $$=\dfrac{40}{100}\ \times\ 11b\ =\ 4.4b\ =$$ exports from X to P

22% of the imports of P $$=\dfrac{22}{100}\ \times\ 10a\ =\ 2.2a\ =$$ imports to P from X

Equating both, we get,

4.4b = 2.2a

a = 2b.

We are also given that imports for C are only from P, and we can set the rest of the imports for C to 0.

The total imports for C = 3c = 1200

c = 400

This makes the exports for C = 2c = 800

In clue 3, we are given that 90% of exports of C are to P and 4% are to ROW.

90% of the exports of C $$=\dfrac{90}{100}\ \times\ 2c\ =\ 1.8c\ =1.8\ \times\ 400\ =\ 720\ =\ $$exports from C to P

4% of the exports of C $$=\frac{4}{100}\ \times\ 2c\ =\ 0.08c\ =\ 0.08\ \times\ 400\ =\ 32\ =$$ exports from C to ROW

The exports from C to X = 800 - 720 - 32 = 48

In clue 4, we are given that 12% of exports of ROW are to X, 40% are to P.

If we assume that Imports of ROW are m and exports of ROW are n.

12% of the exports of ROW $$=\dfrac{12}{100}\ \times\ n\ =\ 0.12n\ =$$ exports from ROW to X

40% of the exports of ROW $$=\dfrac{40}{100}\ \times\ n\ =\ 0.4n\ =$$ exports from ROW to P

Placing all the values in the table, we get,

Equating the imports of P and X to the Totals, we get,

For P,

4.4b + 720 + 0.4n = 20b

15.6b = 720 + 0.4n --(1)

For X,

600 + 48 + 0.12n = 9b

9b = 648 + 0.12n --(2)

Solving (1) and (2), we get

$$\dfrac{648\ +\ 0.12n}{9}\ =\ \dfrac{720\ +\ 0.4n}{15.6}$$

$$10108.8+1.872n=6480+3.6n$$

$$3628.8=1.728n$$

$$n=2100$$

From (1),

15.6b = 720 + 0.4(2100)

15.6b = 1560

b = 100

Exports of P to ROW = 20b - 600 - 1200 = 2000 - 1800 = 200

Exports of X to ROW = 11b - 4.4b = 6.6b = 660

Substituting the values of b and n, we can figure out all the values except for Imports and Exports of ROW to ROW.

The value of exports and imports of ROW to ROW has to be equal.

We calculated the value of n to be 2100.

The exports of ROW to ROW can be calculated as,

Exports of ROW to ROW = 2100 - 840 - 252 = 1008 = Imports of ROW to ROW.

The value of m can be calculated as,

m = 200 + 660 + 32 + 1008 = 1900.

Filling up the table with all the values, we get,

Trade balance of ROW = Exports - Imports = 2100 - 1900 = 200.

Hence, the correct answer is option D.

In clue 5, we are given that export volumes of P, in IC, to X and C are 600 and 1200, respectively, and P is the only country that exports to C. So, the export from X to C and ROW to C is 0.

In clue 1, we are given that normalised trade balances of P, X and C are 0%, 10%, and -20%, respectively.

For P, since the normalised trade balance is 0% the exports have to be equal to the imports according to the given formula. So, let us assume them to be 10a and 10a.

For X,

$$\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ 10$$

$$10E\ -\ 10I\ =\ E\ +\ I$$

$$11I\ =\ 9E$$

So, let us assume export to be 11b and import to be 9b.

For C,

$$\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ -20$$

$$5I\ -\ 5E\ =\ E\ +\ I$$

$$2I\ =\ 3E$$

So, let us assume export to be 2c and import to be 3c.

In clue 2, we are given that 40% of exports of X are to P and 22% of imports of P are from X.

40% of the exports of X $$=\dfrac{40}{100}\ \times\ 11b\ =\ 4.4b\ =$$ exports from X to P

22% of the imports of P $$=\dfrac{22}{100}\ \times\ 10a\ =\ 2.2a\ =$$ imports to P from X

Equating both, we get,

4.4b = 2.2a

a = 2b.

We are also given that imports for C are only from P, and we can set the rest of the imports for C to 0.

The total imports for C = 3c = 1200

c = 400

This makes the exports for C = 2c = 800

In clue 3, we are given that 90% of exports of C are to P and 4% are to ROW.

90% of the exports of C $$=\dfrac{90}{100}\ \times\ 2c\ =\ 1.8c\ =1.8\ \times\ 400\ =\ 720\ =\ $$exports from C to P

4% of the exports of C $$=\frac{4}{100}\ \times\ 2c\ =\ 0.08c\ =\ 0.08\ \times\ 400\ =\ 32\ =$$ exports from C to ROW

The exports from C to X = 800 - 720 - 32 = 48

In clue 4, we are given that 12% of exports of ROW are to X, 40% are to P.

If we assume that Imports of ROW are m and exports of ROW are n.

12% of the exports of ROW $$=\dfrac{12}{100}\ \times\ n\ =\ 0.12n\ =$$ exports from ROW to X

40% of the exports of ROW $$=\dfrac{40}{100}\ \times\ n\ =\ 0.4n\ =$$ exports from ROW to P

Placing all the values in the table, we get,

Equating the imports of P and X to the Totals, we get,

For P,

4.4b + 720 + 0.4n = 20b

15.6b = 720 + 0.4n --(1)

For X,

600 + 48 + 0.12n = 9b

9b = 648 + 0.12n --(2)

Solving (1) and (2), we get

$$\dfrac{648\ +\ 0.12n}{9}\ =\ \dfrac{720\ +\ 0.4n}{15.6}$$

$$10108.8+1.872n=6480+3.6n$$

$$3628.8=1.728n$$

$$n=2100$$

From (1),

15.6b = 720 + 0.4(2100)

15.6b = 1560

b = 100

Exports of P to ROW = 20b - 600 - 1200 = 2000 - 1800 = 200

Exports of X to ROW = 11b - 4.4b = 6.6b = 660

Substituting the values of b and n, we can figure out all the values except for Imports and Exports of ROW to ROW.

The value of exports and imports of ROW to ROW has to be equal.

We calculated the value of n to be 2100.

The exports of ROW to ROW can be calculated as,

Exports of ROW to ROW = 2100 - 840 - 252 = 1008 = Imports of ROW to ROW.

The value of m can be calculated as,

m = 200 + 660 + 32 + 1008 = 1900.

Filling up the table with all the values, we get,

Total trade of P = 2000 + 2000 = 4000

Total trade of X = 900 + 1100 = 2000

Total trade of C = 1200 + 800 = 2000

So, both countries X and C have the least total trade.

Hence, the correct answer is option C.

This is the table given in the question with unknown values of a, b, c, d and e.

We are told that Anu and Bijay use the Xitel operator, and Chetan, Deepak, Esthan and Faruq use the Yocel operator.

The outgoing call from Anu to the Xitel operator has to be the incoming call from Xitel to Bijay because there are only two friends using Xitel, and if one of the Xitel users is calling another Xitel user, then those two have to be Anu and Bijay.

The outgoing to Xitel by Anu = 100

The incoming from Xitel for Bijay = c = The outgoing to Xitel by Anu = 100

So, the value of c in the above table is 100.

The outgoing to Xitel by Bijay = b

The incoming from Xitel for Anu = 50 = The outgoing to Xitel by Bijay = b

So, the value of b in the above table is 50.

The sum of Outgoing from Anu and Bijay to operator Yocel must be equal to the sum of Incoming minutes from Xitel for Chetan, Deepak, Esthan and Faruq.

The sum of Outgoing from Anu and Bijay to operator Yocel = a + 200 

The sum of Incoming minutes from Xitel for Chetan, Deepak, Esthan and Faruq = 250 + 275 + 100 + 100 = 725

Equating them, we get,

a + 200 = 725

a = 525

So, the value of a in the above table is 525.

The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Xitel must be equal to the sum of Incoming minutes from Yocel for Anu and Bijay.

The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Xitel = 50 + 100 + d + 0 = 150 + d

The sum of Incoming minutes from Yocel for Anu and Bijay = 225 + 125 = 350

Equating them, we get,

150 + d = 350

d = 200

So, the value of d in the above table is 200.

The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Yocel must be equal to the sum of Incoming minutes from Yocel for Chetan, Deepak, Esthan, and Faruq.

The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Yocel = 175 + 150 + 100 + e = 425 + e

The sum of Incoming minutes from Yocel for Chetan, Deepak, Esthan, and Faruq = 150 + 100 + 375 + 150 = 775

Equating them, we get,

425 + e = 775

e = 350

So, the value of e in the above table is 350.

The final table is,

The duration of calls from Bijay to Ajay is 50 minutes, which is the outgoing minutes to operator Xitel of Bijay.

Hence, the correct answer is 50.

This is the table given in the question with unknown values of a, b, c, d and e.

We are told that Anu and Bijay use the Xitel operator, and Chetan, Deepak, Esthan and Faruq use the Yocel operator.

The outgoing call from Anu to the Xitel operator has to be the incoming call from Xitel to Bijay because there are only two friends using Xitel, and if one of the Xitel users is calling another Xitel user, then those two have to be Anu and Bijay.

The outgoing to Xitel by Anu = 100

The incoming from Xitel for Bijay = c = The outgoing to Xitel by Anu = 100

So, the value of c in the above table is 100.

The outgoing to Xitel by Bijay = b

The incoming from Xitel for Anu = 50 = The outgoing to Xitel by Bijay = b

So, the value of b in the above table is 50.

The sum of Outgoing from Anu and Bijay to operator Yocel must be equal to the sum of Incoming minutes from Xitel for Chetan, Deepak, Esthan and Faruq.

The sum of Outgoing from Anu and Bijay to operator Yocel = a + 200

The sum of Incoming minutes from Xitel for Chetan, Deepak, Esthan and Faruq = 250 + 275 + 100 + 100 = 725

Equating them, we get,

a + 200 = 725

a = 525

So, the value of a in the above table is 525.

The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Xitel must be equal to the sum of Incoming minutes from Yocel for Anu and Bijay.

The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Xitel = 50 + 100 + d + 0 = 150 + d

The sum of Incoming minutes from Yocel for Anu and Bijay = 225 + 125 = 350

Equating them, we get,

150 + d = 350

d = 200

So, the value of d in the above table is 200.

The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Yocel must be equal to the sum of Incoming minutes from Yocel for Chetan, Deepak, Esthan, and Faruq.

The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Yocel = 175 + 150 + 100 + e = 425 + e

The sum of Incoming minutes from Yocel for Chetan, Deepak, Esthan, and Faruq = 150 + 100 + 375 + 150 = 775

Equating them, we get,

425 + e = 775

e = 350

So, the value of e in the above table is 350.

The final table is,

The total duration of calls made by Anu to friends having mobile numbers from Operator Yocel is Anu's outgoing minutes to Operator Yocel, which is 525 minutes.

Hence, the correct answer is 525.

This is the table given in the question with unknown values of a, b, c, d and e.

We are told that Anu and Bijay use the Xitel operator, and Chetan, Deepak, Esthan and Faruq use the Yocel operator.

The outgoing call from Anu to the Xitel operator has to be the incoming call from Xitel to Bijay because there are only two friends using Xitel, and if one of the Xitel users is calling another Xitel user, then those two have to be Anu and Bijay.

The outgoing to Xitel by Anu = 100

The incoming from Xitel for Bijay = c = The outgoing to Xitel by Anu = 100

So, the value of c in the above table is 100.

The outgoing to Xitel by Bijay = b

The incoming from Xitel for Anu = 50 = The outgoing to Xitel by Bijay = b

So, the value of b in the above table is 50.

The sum of Outgoing from Anu and Bijay to operator Yocel must be equal to the sum of Incoming minutes from Xitel for Chetan, Deepak, Esthan and Faruq.

The sum of Outgoing from Anu and Bijay to operator Yocel = a + 200

The sum of Incoming minutes from Xitel for Chetan, Deepak, Esthan and Faruq = 250 + 275 + 100 + 100 = 725

Equating them, we get,

a + 200 = 725

a = 525

So, the value of a in the above table is 525.

The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Xitel must be equal to the sum of Incoming minutes from Yocel for Anu and Bijay.

The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Xitel = 50 + 100 + d + 0 = 150 + d

The sum of Incoming minutes from Yocel for Anu and Bijay = 225 + 125 = 350

Equating them, we get,

150 + d = 350

d = 200

So, the value of d in the above table is 200.

The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Yocel must be equal to the sum of Incoming minutes from Yocel for Chetan, Deepak, Esthan, and Faruq.

The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Yocel = 175 + 150 + 100 + e = 425 + e

The sum of Incoming minutes from Yocel for Chetan, Deepak, Esthan, and Faruq = 150 + 100 + 375 + 150 = 775

Equating them, we get,

425 + e = 775

e = 350

So, the value of e in the above table is 350.

The final table is,

The total duration of calls made by Faruq to friends having mobile numbers from Operator Yocel is the outgoing minutes from Faruq to Operator Yocel, which is 350 minutes.

Hence, the correct answer is 350.

This is the table given in the question with unknown values of a, b, c, d and e.

We are told that Anu and Bijay use the Xitel operator, and Chetan, Deepak, Esthan and Faruq use the Yocel operator.

The outgoing call from Anu to the Xitel operator has to be the incoming call from Xitel to Bijay because there are only two friends using Xitel, and if one of the Xitel users is calling another Xitel user, then those two have to be Anu and Bijay.

The outgoing to Xitel by Anu = 100

The incoming from Xitel for Bijay = c = The outgoing to Xitel by Anu = 100

So, the value of c in the above table is 100.

The outgoing to Xitel by Bijay = b

The incoming from Xitel for Anu = 50 = The outgoing to Xitel by Bijay = b

So, the value of b in the above table is 50.

The sum of Outgoing from Anu and Bijay to operator Yocel must be equal to the sum of Incoming minutes from Xitel for Chetan, Deepak, Esthan and Faruq.

The sum of Outgoing from Anu and Bijay to operator Yocel = a + 200

The sum of Incoming minutes from Xitel for Chetan, Deepak, Esthan and Faruq = 250 + 275 + 100 + 100 = 725

Equating them, we get,

a + 200 = 725

a = 525

So, the value of a in the above table is 525.

The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Xitel must be equal to the sum of Incoming minutes from Yocel for Anu and Bijay.

The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Xitel = 50 + 100 + d + 0 = 150 + d

The sum of Incoming minutes from Yocel for Anu and Bijay = 225 + 125 = 350

Equating them, we get,

150 + d = 350

d = 200

So, the value of d in the above table is 200.

The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Yocel must be equal to the sum of Incoming minutes from Yocel for Chetan, Deepak, Esthan, and Faruq.

The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Yocel = 175 + 150 + 100 + e = 425 + e

The sum of Incoming minutes from Yocel for Chetan, Deepak, Esthan, and Faruq = 150 + 100 + 375 + 150 = 775

Equating them, we get,

425 + e = 775

e = 350

So, the value of e in the above table is 350.

The final table is,

Label the unknown value we need as

$$x=\text{minutes from Deepak to Chetan}$$

Use the operator groups from the table:

Xitel users: Anu, Bijay.

Yocel users: Chetan, Deepak, Eshan, Faruq.

Chetan’s incoming from Yocel = 150 minutes.

Yocel$$\to$$Chetan can come only from Deepak, Eshan, or Faruq. But the problem states Eshan $$\to$$ Chetan = 0, so

$$(\text{Deepak}\to\text{Chetan}) + (\text{Faruq}\to\text{Chetan}) = 150$$

Hence

$$x + (\text{Faruq}\to\text{Chetan}) = 150 \qquad(1)$$

Faruq’s outgoing to Yocel = 350 minutes (table). Those 350 minutes are split among Yocel recipients Chetan, Deepak and Eshan. We are told Faruq$$\to$$Eshan = 200 (given). So

$$(\text{Faruq}\to\text{Chetan}) + (\text{Faruq}\to\text{Deepak}) + 200 = 350,$$

hence

$$(\text{Faruq}\to\text{Chetan}) + (\text{Faruq}\to\text{Deepak}) = 150 \qquad(2)$$

Consider Deepak’s incoming from Yocel (table) = 100 minutes. The ones in Yocel that can call Deepak are Chetan, Eshan, and Faruq. But we are told Chetan $$\to$$ Deepak = 0 and Eshan $$\to$$ Deepak = 0, so the only Yocel caller to Deepak is Faruq. Therefore

$$(\text{Faruq}\to\text{Deepak}) = 100 \qquad(3)$$

Substitute (3) into (2):

$$(\text{Faruq}\to\text{Chetan}) + 100 = 150 \quad\Rightarrow\quad (\text{Faruq}\to\text{Chetan}) = 50.$$

Now use (1): 

$$x + (\text{Faruq}\to\text{Chetan}) = 150$$. 

With $$\text{Faruq}\to\text{Chetan} = 50$$, we get,

$$x + 50 = 150$$

$$x = 100$$

Thus Deepak $$\to$$ Chetan = 100 minutes.

Hence, the correct answer is option A.

The total number of products is $$120+135+150+165= 570$$. Therefore, the total cost must have been $$570\times 240 = 136800$$

Let the marked price for the products be $$X$$. The January, February, March, and April, selling prices would respectively be $$0.8X$$, $$0.9X$$, $$0.95X$$, and $$X$$; based on the discounts given on each of the months.

The revenue earned from selling the products in each of the months would be: $$120*0.8X + 135*0.9X + 150*0.95X + 165*X = 525X$$.

The profit earned would be: Revenue -  Cost, and therefore, 

$$525X - 136800 = 138825$$

$$525X = 275625$$

$$X = \dfrac{275625}{525} = 525$$

The correct answer is option B, Rs. 525.

Let the per hour productivity of each member from teams A, B, and C, be $$a$$, $$b$$, and $$c$$ units respectively.

Let $$W$$ be the total units of work that each of the teams does. We have:

$$W = 40*5*a = 50*8*b = 4*10*c$$

This gives, $$a= \dfrac{W}{200}$$,  $$b= \dfrac{W}{400}$$,  and $$c = \dfrac{W}{40}$$

When two members from team A, three members from team B, and one member from team C together start the job, the total per hour productivity for the first 23 hours will be;

$$\dfrac{2W}{200} + \dfrac{3W}{400} + \dfrac{W}{40} = \dfrac{17W}{400}$$.

In 23 hours, the work done will be $$\dfrac{23*17W}{400} = \dfrac{391W}{400}$$

The remaining work, after 23 hours, is $$W - \dfrac{391W}{400} = \dfrac{9}{400}$$

The current per hour efficiency of the group that comprises members from only teams A and B, is $$\dfrac{2W}{200} + \dfrac{3W}{400} = \dfrac{7W}{400}$$

In the next hour, the members from A and B will finish $$\dfrac{7W}{400}$$ units of work and $$\dfrac{9W}{400}-\dfrac{7W}{400} = \dfrac{2W}{400}$$ units of work would remain.

This work has to be finished by newly added members from team B in one hour, therefore, the number of new members required from team B would be $$\dfrac{2W}{400} \div \dfrac{W}{400} = 2$$

Option B is the correct answer.

Let the total number of students who chose science, arts, and commerce streams be $$S$$, $$A$$, and $$C$$, respectively.

We have,

$$S+A+C = 1500$$, such that, $$C = 1500-S-A$$ .....(1)

Also, $$1100S + 1000A + 800C = 15,50,000$$,  which can be simplified by dividing by $$100$$;

$$11S + 10A + 8C = 15500$$ .....(2)

Substituting the value of $$C$$ from equation (1) in equation (2), we have

$$11S + 10A + 8(1500-S-A) = 15500$$

$$\Rightarrow 3S + 2A = 3500$$

Since $$S\leq A$$, the maximum value of $$S$$ (if possible), would occur when $$S=A$$, therefore,

$$5S = 3500$$,  or $$S= 700$$.

Let the first term of the arithmetic progression be $$a$$, and the common difference be $$d$$.

We have:

$$(a+3d) + (a+6d) + (a+9d) = 99$$

$$\Rightarrow 3a + 18d = 99$$  or  $$a + 6d = 33$$ .....(1)

We are also told that the sum of the first fourteen terms is $$497$$,

$$\dfrac{14}{2}(2a + (14-1)d) = 497$$

$$\Rightarrow (2a+13d) = \dfrac{497}{7}$$

$$\Rightarrow 2a+13d = 71$$ .....(2)

Solving equations (1) and (2), we get;

$$a= 3$$ and $$d=5$$

The first five terms would therefore be: $$3, 8, 13, 18, 23$$, and their sum would be $$3+8+13+18+23 = 65$$.