CAT Quant based DI Questions & Solutions

A sample of real CAT Quant based DI past-year questions, each with a worked solution and the correct answer highlighted. AthenaPrep has 59 Quant based DI questions in all — sign up free to practise them timed.

Passage
Three countries — Pumpland (P), Xiland (X) and Cheeseland (C) — trade among themselves and with the (other countries in) Rest of World (ROW). All trade volumes are given in IC (international currency). The following terminology is used:
• Trade balance = Exports - Imports
• Total trade = Exports + Imports
• Normalized trade balance = Trade balance / Total trade, expressed in percentage terms

The following information is known.
1. The normalized trade balances of P, X and C are 0%, 10%, and -20%, respectively.
2. 40% of exports of X are to P. 22% of imports of P are from X.
3. 90% of exports of C are to P; 4% are to ROW.
4. 12% of exports of ROW are to X, 40% are to P.
5. The export volumes of P, in IC, to X and C are 600 and 1200, respectively. P is the only country that exports to C.
Q1.CAT 2025
How much is exported from C to X, in IC?
Answer: 48
Show solution
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 $=\dfrac{4}{100}\ \times\ 2c\ =\ 0.08c\ =\ 0.08\ \times\ 400\ =\ 32\ =$ 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,

Exports from C to X are 48 IC.

Hence, the correct answer is 48.
Q2.CAT 2025
Which among the countries P, X, and C has/have the least total trade?
- Only X
- Only C
- Both X and C
- Only P
Show solution
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.
Passage
Two students, Amiya and Ramya are the only candidates in an election for the position of class representative. Students will vote based on the intensity level of Amiya’s and Ramya’s campaigns and the type of campaigns they run. Each campaign is said to have a level of 1 if it is a staid campaign and a level of 2 if it is a vigorous campaign. Campaigns can be of two types, they can either focus on issues, or on attacking the other candidate.

If Amiya and Ramya both run campaigns focusing on issues, then
• The percentage of students voting in the election will be 20 times the sum of the levels of campaigning of the two students. For example, if Amiya and Ramya both run vigorous campaigns, then 20 × (2+2)%, that is, 80% of the students will vote in the election.
• Among voting students, the percentage of votes for each candidate will be proportional to the levels of their campaigns. For example, if Amiya runs a staid (i.e., level 1) campaign while Ramya runs a vigorous (i.e., level 2) campaign, then Amiya will receive 1/3 of the votes cast, and Ramya will receive the other 2/3. The above-mentioned percentages change as follows if at least one of them runs a campaign attacking their opponent.
• If Amiya runs a campaign attacking Ramya and Ramya runs a campaign focusing on issues, then 10% of the students who would have otherwise voted for Amiya will vote for Ramya, and another 10% who would have otherwise voted for Amiya, will not vote at all.
• If Ramya runs a campaign attacking Amiya and Amiya runs a campaign focusing on issues, then 20% of the students who would have otherwise voted for Ramya will vote for Amiya, and another 5% who would have otherwise voted for Ramya, will not vote at all.
• If both run campaigns attacking each other, then 10% of the students who would have otherwise voted for them had they run campaigns focusing on issues, will not vote at all.
Q3.CAT 2024
What is the maximum possible voting margin with which one of the candidates can win?
- 20%
- 29%
- 28%
- 26%
Show solution
We are looking for the minimum possible number of votes that Ramya can get and maximise the number of votes that Amiya can get.

We can borrow the scenario from the previous question where Ramya runs an attacking campaign, and we minimised the number of votes she can get.

To minimise the number of votes, we can have Ramya run a staid campaign to minimise the votes, so minimum intensity, which will get her 20% of the votes if she ran with issues. Now that she is running with attacking, she will loose 20% of the votes to Amiya and 5% of the votes will not vote anymore.

That is a total 25% loss. Remaining votes she will get is 75% of the 20% which will leave her with 15% of the votes.

And to maximise the number of votes Amiya can get, we will have her run an vigorous issues campaign, which will give her 2x20% of the votes, that is 40% of the votes. And since Ramya has been running an attacking campaign, 20% of her votes are transferred to Amiya. 20% of the 20% of the votes which is 4% that were going to Ramya will now go to Amiya. That will bring up Amiya's tally up to 44% leaving Ramya's tally at 15%.

The difference in the votes will be 44-15=29%.

This is the maximum possible vote difference between the two candidates that is possible.
Passage
The following table represents addition of two six-digit numbers given in the first and the second rows, while the sum is given in the third row. In the representation, each of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 has been coded with one letter among A, B, C, D, E, F, G, H, J, K, with distinct letters representing distinct digits.
Q4.CAT 2019
Which digit does the letter A represent?
Answer: 1
Show solution

The value of F can only be 0 as F+F=F can only hold if F=0.

Also, A can only be 1(in the second column) because to get a carry of more than 1, B has to be a double-digit number which is not possible. (A carry is a digit that is transferred from one column of digits to another column of more significant digits.)

So the data can be tabulated as follows:

Since the last row in the third column is 0, the carry to the second column must have been 1, Hence B+1+1=11 => B=9

In the 4th column, H+H = 10 since a carry 1 has gone to the 3rd column. Hence H=5.

G+K must be 11 and the carry 1 goes to the next column, so C=1+1=2.

Now, G,K can take values (3,8), (4,7) and (5,6) in any order.

From 5th column G=J+1 => J=G-1

Case: G=3 and K=8, here J =2 which is not possible as C =2

Case: G=8 and K=3, J=7, a possible case.

Case: G=4 and K=7, J=3 possible

Case: G=7 and K=4, J=6 possible

Case: G=5 and K=6, J=4 not possible as H =5.

Case: G=6 and K=5, J=5 both J and K are same, not possible.

Hence the cases can be tabulated as follows:

The letter A represents 1.
Passage

Three pouches (each represented by a filled circle) are kept in each of the nine slots in a 3 × 3 grid, as shown in the figure. Every pouch has a certain number of one-rupee coins. The minimum and maximum amounts of money (in rupees) among the three pouches in each of the nine slots are given in the table. For example, we know that among the three pouches kept in the second column of the first row, the minimum amount in a pouch is Rs. 6 and the maximum amount is Rs. 8.

There are nine pouches in any of the three columns, as well as in any of the three rows. It is known that the average amount of money (in rupees) kept in the nine pouches in any column or in any row is an integer. It is also known that the total amount of money kept in the three pouches in the first column of the third row is Rs. 4.
Q5.CAT 2019
How many pouches contain exactly one coin?
Answer: 8
Show solution
We can make the following table from "the total amount of money kept in the three pouches in the first column of the third row is Rs. 4."

If the minimum and maximum value are 1, then the sum of the three pouches in the middle will be Rs 3.

If we calculate the maximum and minimum value possible for each slot in column 1. For the slot, column 1 and row 1, the maximum value possible is 10{2,4,4} while the minimum value possible is 8{2,2,4}.

Similarly, for the slot, column 1 and row 2, the maximum value possible is 13{3,5,5} while the minimum value possible is 11{3,3,5}.

It is known that the average amount of money (in rupees) kept in the nine pouches in any column or in any row is an integer. Thus the sum of coins in a row or column must be a multiple of 9.

So, we can iterate that 10,13,4 ...{27} is the only sum possible for the slots of column 1.

We now know two elements of row 2, thus we can iterate from the maximum and the minimum value possible for the slot {cloumn 3, row 2} that 38 is the only value possible for the slot.

We can make the following table:

Similarly, we can find the amount for Column 2.

For the slot, column 2 and row 1, the maximum value possible is 22{6,8,8} while the minimum value possible is 20{6,6,8}.

For the slot, column 2 and row 3, the maximum value possible is 5{1,2,3} while the minimum value possible is 4{1,1,2}.

Thus {20,3,4} is the only solution possible.

We can similarly make the following table for the last column.

Answer 8
Passage
There are only four brands of entry level smartphones called Azra, Bysi, Cxqi, and Dipq in a country.
Details about their market share, unit selling price, and profitability (defined as the profit as a percentage of the revenue) for the year 2016 are given in the table below:

In 2017, sales volume of entry level smartphones grew by 40% as compared to that in 2016. Cxqi offered a 40% discount on its unit selling price in 2017, which resulted in a 15% increase in its market share. Each of the other three brands lost 5% market share. However, the profitability of Cxqi came down to half of its value in 2016. The unit selling prices of the other three brands and their profitability values remained the same in 2017 as they were in 2016.
Q6.CAT 2018
The brand that had the highest profit in 2017 is:
- Bysi
- Azra
- Cxqi
- Dipq
Show solution
Let '100x' be the number of smartphones sold in year 2016. Then the number of smartphones sold in 2017 = 1.4*100x = 140x

It is given that Cxqi offered a 40% discount on its unit selling price in 2017 i.e. selling price in 2017 = 0.6*30000 = Rs. 18000

Also Cxqi's merket share increased by 15% whereas the other three brands lost 5% market share.

Amount of profit generated by Azra = $\dfrac{10}{100}$ *15000*49x = 73500x

Amount of profit generated by Bysi = $\dfrac{30}{100}$ *20000*28x = 168000x

Amount of profit generated by Cxqi = $\dfrac{20}{100}$ *18000*42x = 151200x

Amount of profit generated by Dipq = $\dfrac{30}{100}$ *25000*21x = 157500x

We can see that brand Bysi generated maximum profit in year 2017. Hence, option A is the correct answer.
Passage
The base exchange rate of a currency X with respect to a currency Y is the number of units of currency Y which is equivalent in value to one unit of currency X. Currency exchange outlets buy currency at buying exchange rates that are lower than base exchange rates, and sell currency at selling exchange rates that are higher than base exchange rates.

A currency exchange outlet uses the local currency L to buy and sell three international currencies A, B, and C, but does not exchange one international currency directly with another. The base exchange rates of A, B and C with respect to L are in the ratio 100:120:1. The buying exchange rates of each of A, B, and C with respect to L are 5% below the corresponding base exchange rates, and their selling exchange rates are 10% above their corresponding base exchange rates. The following facts are known about the outlet on a particular day:

1. The amount of L used by the outlet to buy C equals the amount of L it received by selling C.
2. The amounts of L used by the outlet to buy A and B are in the ratio 5:3.
3. The amounts of L the outlet received from the sales of A and B are in the ratio 5:9.
4. The outlet received 88000 units of L by selling A during the day.
5. The outlet started the day with some amount of L, 2500 units of A, 4800 units of B, and 48000 units of C.
6. The outlet ended the day with some amount of L, 3300 units of A, 4800 units of B, and 51000 units of C.
Q7.CAT 2018
What was the buying exchange rate of currency C with respect to currency L on that day?
- 1.10
- 0.95
- 2.20
- 1.90
Show solution
It is given that the base exchange rates of A, B and C with respect to L are in the ratio 100:120:1. Let us assume that base exchange rates are '100a', '120a' and 'a' in that order.

It is given that the buying exchange rates of each of A, B, and C with respect to L are 5% below the corresponding base exchange rates. Therefore, we can say that the buying exchange rates are 95a, 114a, 0.95a.

It is given that the selling exchange rates of each of A, B, and C with respect to L are 10% above their corresponding base exchange rates. Therefore, we can say that the selling exchange rates are 110a, 132a, 1.1a.

We know about the opening and closing units in stock for each currency. Let us draw the table accordingly.

Let 'p', 'q' and 'r' be the number of units of currency A, B and C bought by the outlet on that day.

Then, we can say that the outlet sold 'p - 800', 'q' and 'r-3000' units of currency A, B and C respectively.

It is given that the amount of L used by the outlet to buy C equals the amount of L it received by selling C.

$\Rightarrow$ 0.95a*r = 1.1a*(r - 3000)

$\Rightarrow$ 0.15r = 3300

$\Rightarrow$ r = 22000

It is also given that the amounts of L used by the outlet to buy A and B are in the ratio 5:3.

$\Rightarrow$ $\dfrac{p*95a}{q*114a} = \dfrac{5}{3}$

$\Rightarrow$ p = 2q

Also, the amounts of L the outlet received from the sales of A and B are in the ratio 5:9.

$\Rightarrow$ $\dfrac{(p-800)*110a}{q*132a} = \dfrac{5}{9}$

$\Rightarrow$ $\dfrac{(2q-800)*110a}{q*132a} = \dfrac{5}{9}$

$\Rightarrow$ q = 600

Therefore, p = 2q = 2*600 = 1200.

It is given that the outlet received 88000 units of L by selling A during the day.

$\Rightarrow$ (p-800)*110a = 88000

$\Rightarrow$ (1200-800)*110a = 88000

$\Rightarrow$ 44000a = 88000

$\Rightarrow$ a = 2

We can fill the entire table and answer all the questions.

From the table we can see that the buying exchange rate of currency C with respect to currency L was 1.9. Hence, we can say that option D is the correct answer.
Passage
A study to look at the early learning of rural kids was carried out in a number of villages spanning three states, chosen from the North East (NE), the West (W) and the South (S). 50 four-year old kids each were sampled from each of the 150 villages from NE, 250 villages from W and 200 villages from S. It was found that of the 30000 surveyed kids 55% studied in primary schools run by government (G), 37% in private schools (P) while the remaining 8% did not go to school (O).

The kids surveyed were further divided into two groups based on whether their mothers dropped out of school before completing primary education or not.. The table below gives the number of kids in different types of schools for mothers who dropped out of school before completing primary education:

It is also known that:
1. In S, 60% of the surveyed kids were in G. Moreover, In S, all surveyed kids whose mothers had completed primary education were in school.
2. In NE, among the O kids, 50% had mothers who had dropped out before completing primary education.
3. The number of kids in G in NE was the same as the number of kids in G in W.
Q8.CAT 2017
What percentage of kids from S were studying in P?
- 37%
- 6%
- 79%
- 56%
Show solution
Let us make note of the information given in the set.
The set states that rural kids were surveyed from 3 regions NE, West and South.
50 students each were surveyed from 150 villages in the NE, 250 villages from the West and 200 villages from the South.

Total number of kids from the NE = 150*50 = 7500
Total number of kids from the West = 250*50 = 12500
Total number of kids from the South = 200*50 = 10,000.

The table, given in the question, gives the number of students whose mothers dropped out before completing primary education. Therefore, the mothers of the remaining students should have completed primary education.

There are 7500 students from the NE in total. The mothers of 5000 of those students dropped out before completing primary education. Therefore, the remaining 2500 kids should have mothers who have completed primary education.

Let us make 2 tables - one representing the number of students and the other representing the number of students whose mother has completed primary education.

55% of the surveyed kids studied in schools run by the Government.
Number of kids studying in Govt. School should be 0.55*30000 = 16,500

From the given table, we know that the mothers of 13,500 kids who went to Govt. Schools dropped out of primary school.
Therefore, the remaining 16,500 - 13,500 = 3000 kids should have mother who have completed primary school.

37% of the surveyed kids study in private schools.
Number of kids studying in private schools should be 0.37*30000 = 11,100.
Number of kids in private schools whose mothers have completed primary school = 11,100 - 2700 = 8400.

8% of the surveyed kids did not go to school.
Number of kids who did not go to school = 0.08* 30000 = 2400
Number of kids not going to schools whose mothers have completed primary school = 2400 - 1800 = 600.

In S, 60% of the surveyed kids were in G.
Therefore, 6000 kids in S must be from G.

In NE, among the O kids, 50% had mothers who had dropped out before completing primary education.
There are 300 O kids whose mothers dropped out before completing primary education. These kids represent 50% of the total number of O-kids. Therefore, there must be 600 O-kids from NE - 300 kids should have mothers who dropped out before completing primary education and 300 kids should have mothers who have completed primary education.

The number of kids in G in NE was the same as the number of kids in G in W. Therefore, the number of kids in G in NE and the number of kids in G in W should be equal to 5250. Total number of kids:

Number of kids whose mothers have completed primary education:

We have been given that in S, all surveyed kids whose mothers had completed primary education were in school. Therefore, the number of kids not going to school whose mothers have completed primary education in S should be 0. Filling the tables, we get,

Total number of kids:

Number of kids whose mothers have completed primary education:

As we can see from the table, 3700 kids out of the 10,000 kids from S are studying in P. 37% of the total number of students from S were studying in P. Therefore, option A is the right answer.
Passage
In a square layout of site 5m ~ 5m 25 equal-sized square platforms of different heights are built. The heights (in metre) of individual platforms are as shown below:

Individuals (all of same height) are seated on these platforms. We say an individual A can reach individual B, if all the three following conditions are met;

(i) A and B are In the same row or column
(ii) A is at a lower height than B
(iii) If there is/are any individuals (s) between A and B, such individual(s) must be at a height lower than that of A.

Thus in the table given above, consider the Individual seated at height 8 on 3rd row and 2nd column. He can be reached by four individuals. He can be reached by the individual on his left at height 7, by the two individuals on his right at heights of 4 and 6 and by the individual above at height 5.

Rows in the layout are numbered from top to bottom and columns are numbered from left to right.
Q9.CAT 2017
Which of the following is true for any individual at a platform of height 1 m in this layout?
- They can be reached by all the individuals in their own row and column.
- They can be reached by at least 4 individuals.
- They can be reached by at least one individual.
- They cannot be reached by anyone.
Show solution
Since, we have been given that a person can be reached only by those who are smaller than him. Hence, 1 cannot be reached by anyone. Thus, option D is the correct answer.
Passage
An old woman had the following assets:
(a) Rs. 70 lakh in bank deposits
(b) 1 house worth Rs. 50 lakh
(c) 3 flats, each worth Rs. 30 lakh
(d) Certain number of gold coins, each worth Rs. 1 lakh
She wanted to distribute her assets among her three children; Neeta, Seeta and Geeta.
The house, any of the flats or any of the coins were not to be split. That is, the house went entirely to one child; a flat went to one child and similarly, a gold coin went to one child.
Q10.CAT 2017
The value of the assets distributed among Neeta, Seeta and Geeta was in the ratio of 1:2:3, while the gold coins were distributed among them in the ratio of 2:3:4. One child got all three flats and she did not get the house. One child, other than Geeta, got Rs. 30 lakh in bank deposits.
How many gold coins did the old woman have?
- 72
- 90
- 180
- 216
Show solution
Let the total number of gold coins with the old woman be '9n'.
Total value of the assets with the old woman = 50 + 3*30 + 70 + 9n = 210+9n.
We know that the assets have been distributed in the ratio 1:2:3.
Therefore, Neeta must have received 35+1.5n (by value), Seeta must have received 70+3n and Geeta must have received 105+4.5n.
Further, it has been given that the gold coins distributed were in the ratio 2:3:4.
Therefore, the number of gold coins with Neeta must be '2n', Seeta must be '3n' and Geeta must be '4n'.
Seeta has '3n' gold coins. Therefore, the total value of the assets with her must be 70. Seeta could not have inherited all the flats. Therefore, Seeta must have received the house ( worth 50 lakh) and 20 lakh from bank deposits.

We know that Geeta did not receive Rs. 30 lakh from the bank deposits. Therefore, Neeta must have received Rs. 30 lakh.
The remaining 5 lakh must be contributed by the gold coins (Since there is no other asset worth 5 lakh).
=> 5 + 1.5n = 2n
=> 0.5n = 5
=> n = 10
The old-woman must have had 10*9 = 90 gold coins. Therefore, option B is the right answer.
Passage
Directions for the following five questions: Answer the following questions based on the information given below:

Abdul, Bikram and Chetan are three professional traders who trade in shares of a company XYZ Ltd. Abdul follows the strategy of buying at the opening of the day at 10 am and selling the whole lot at the close of the day at 3 pm. Bikram follows the strategy of buying at hourly intervals: 10 am, 11am, 12 noon, 1 pm and 2 pm, and selling the whole lot at the close of the day, Further, he buys an equal number of shares in each purchase. Chetan follows a similar pattern as Bikram but his strategy is somewhat different. Chetan’s total investment amount is divided equally among his purchases. The profit or loss made by each investor is the difference between the sale value at the close of the day less the investment in purchase. The “return” for each investor is defined as the ratio of the profit or loss to the investment amount expressed as a percentage.
Q11.CAT 2008
One day, two other traders. Dane and Emily joined Abdul, Bikram and Chetan for trading in the shares of XYZ Ltd. Dane followed a strategy of buying equal numbers of shares at 10 am. 11 am and 12 noon, and selling the same numbers at 1 pm, 2 pm and 3 pm Emily, on the other hand, followed the strategy of buying shares using all her money at 10 am and selling all of them at 12 noon and again buying the shares for all the money at 1 pm and again selling all of them at the close of the day at 3 pm. At the close of the day the following was observed.

i. Abdul lost money in the transactions.

ii. Both Dane and Emily made profits.

iii. There was an increase in share price during the closing hour compared to the price at 2 pm.

iv. Share price at 12 noon was lower than the opening price

Share price was at its highest at
- 10 am
- 11 am
- 12 noon
- 1 pm
- cannot be determined
Show solution
Let the share prices at 10,11,12,1,2,3 be p,q,r,s,t,u
Since Abdul lost all money, p>u
From Dane it can be observed, s+t+u>p+q+r
p>r and u>t
From Emily it can be observed, r+u>p+s
From these inequalities it can be inferred that the p is the highest.
Passage
Directions for the following four questions:

A health-drink company’s R&D department is trying to make various diet formulations, which can be used for certain specific purposes. It is considering a choice of 5 alternative ingredients (O, P, Q, R, and S), which can be used in different proportions in the formulations.

The table below gives the composition of these ingredients. The cost per unit of each of these ingredients is O: 150, P: 50, Q: 200, R: 500, S: 100.

Q12.CAT 2007
The company is planning to launch a balanced diet required for growth needs of adolescent children. This diet must contain at least 30% each of carbohydrate and protein, no more than 25% fat and at least 5% minerals. Which one of the following combinations of equally mixed ingredients is feasible?
- O and P
- R and S
- P and S
- Q and R
- O and S
Show solution
Two ingredients are mixed in equal proportion. So, the minimum amount of protein and carbohydrate should be 60 each, the minimum amount of minerals is 10 and the maximum amount of fat is 50. The combination of O and S satisfies this requirement.

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CAT Quant based DI Questions & Solutions

Passage

Three countries — Pumpland (P), Xiland (X) and Cheeseland (C) — trade among themselves and with the (other countries in) Rest of World (ROW). All trade volumes are given in IC (international currency). The following terminology is used:
• Trade balance = Exports - Imports
• Total trade = Exports + Imports
• Normalized trade balance = Trade balance / Total trade, expressed in percentage terms

The following information is known.
1. The normalized trade balances of P, X and C are 0%, 10%, and -20%, respectively.
2. 40% of exports of X are to P. 22% of imports of P are from X.
3. 90% of exports of C are to P; 4% are to ROW.
4. 12% of exports of ROW are to X, 40% are to P.
5. The export volumes of P, in IC, to X and C are 600 and 1200, respectively. P is the only country that exports to C.

Q1.CAT 2025

How much is exported from C to X, in IC?

Answer: 48

Show solution

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 $=\dfrac{4}{100}\ \times\ 2c\ =\ 0.08c\ =\ 0.08\ \times\ 400\ =\ 32\ =$ 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,

Exports from C to X are 48 IC.

Hence, the correct answer is 48.

Q2.CAT 2025

Which among the countries P, X, and C has/have the least total trade?

Only X
Only C
Both X and C
Only P

Show solution

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.

Passage

Two students, Amiya and Ramya are the only candidates in an election for the position of class representative. Students will vote based on the intensity level of Amiya’s and Ramya’s campaigns and the type of campaigns they run. Each campaign is said to have a level of 1 if it is a staid campaign and a level of 2 if it is a vigorous campaign. Campaigns can be of two types, they can either focus on issues, or on attacking the other candidate.

If Amiya and Ramya both run campaigns focusing on issues, then
• The percentage of students voting in the election will be 20 times the sum of the levels of campaigning of the two students. For example, if Amiya and Ramya both run vigorous campaigns, then 20 × (2+2)%, that is, 80% of the students will vote in the election.
• Among voting students, the percentage of votes for each candidate will be proportional to the levels of their campaigns. For example, if Amiya runs a staid (i.e., level 1) campaign while Ramya runs a vigorous (i.e., level 2) campaign, then Amiya will receive 1/3 of the votes cast, and Ramya will receive the other 2/3. The above-mentioned percentages change as follows if at least one of them runs a campaign attacking their opponent.
• If Amiya runs a campaign attacking Ramya and Ramya runs a campaign focusing on issues, then 10% of the students who would have otherwise voted for Amiya will vote for Ramya, and another 10% who would have otherwise voted for Amiya, will not vote at all.
• If Ramya runs a campaign attacking Amiya and Amiya runs a campaign focusing on issues, then 20% of the students who would have otherwise voted for Ramya will vote for Amiya, and another 5% who would have otherwise voted for Ramya, will not vote at all.
• If both run campaigns attacking each other, then 10% of the students who would have otherwise voted for them had they run campaigns focusing on issues, will not vote at all.

Q3.CAT 2024

What is the maximum possible voting margin with which one of the candidates can win?

Show solution

We are looking for the minimum possible number of votes that Ramya can get and maximise the number of votes that Amiya can get.

We can borrow the scenario from the previous question where Ramya runs an attacking campaign, and we minimised the number of votes she can get.

To minimise the number of votes, we can have Ramya run a staid campaign to minimise the votes, so minimum intensity, which will get her 20% of the votes if she ran with issues. Now that she is running with attacking, she will loose 20% of the votes to Amiya and 5% of the votes will not vote anymore.

That is a total 25% loss. Remaining votes she will get is 75% of the 20% which will leave her with 15% of the votes.

And to maximise the number of votes Amiya can get, we will have her run an vigorous issues campaign, which will give her 2x20% of the votes, that is 40% of the votes. And since Ramya has been running an attacking campaign, 20% of her votes are transferred to Amiya. 20% of the 20% of the votes which is 4% that were going to Ramya will now go to Amiya. That will bring up Amiya's tally up to 44% leaving Ramya's tally at 15%.

The difference in the votes will be 44-15=29%.

This is the maximum possible vote difference between the two candidates that is possible.

Passage

The following table represents addition of two six-digit numbers given in the first and the second rows, while the sum is given in the third row. In the representation, each of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 has been coded with one letter among A, B, C, D, E, F, G, H, J, K, with distinct letters representing distinct digits.

Q4.CAT 2019

Which digit does the letter A represent?

Answer: 1

Show solution

The value of F can only be 0 as F+F=F can only hold if F=0.

Also, A can only be 1(in the second column) because to get a carry of more than 1, B has to be a double-digit number which is not possible. (A carry is a digit that is transferred from one column of digits to another column of more significant digits.)

So the data can be tabulated as follows:

Since the last row in the third column is 0, the carry to the second column must have been 1, Hence B+1+1=11 => B=9

In the 4th column, H+H = 10 since a carry 1 has gone to the 3rd column. Hence H=5.

G+K must be 11 and the carry 1 goes to the next column, so C=1+1=2.

Now, G,K can take values (3,8), (4,7) and (5,6) in any order.

From 5th column G=J+1 => J=G-1

Case: G=3 and K=8, here J =2 which is not possible as C =2

Case: G=8 and K=3, J=7, a possible case.

Case: G=4 and K=7, J=3 possible

Case: G=7 and K=4, J=6 possible

Case: G=5 and K=6, J=4 not possible as H =5.

Case: G=6 and K=5, J=5 both J and K are same, not possible.

Hence the cases can be tabulated as follows:

The letter A represents 1.

Passage

Three pouches (each represented by a filled circle) are kept in each of the nine slots in a 3 × 3 grid, as shown in the figure. Every pouch has a certain number of one-rupee coins. The minimum and maximum amounts of money (in rupees) among the three pouches in each of the nine slots are given in the table. For example, we know that among the three pouches kept in the second column of the first row, the minimum amount in a pouch is Rs. 6 and the maximum amount is Rs. 8.

There are nine pouches in any of the three columns, as well as in any of the three rows. It is known that the average amount of money (in rupees) kept in the nine pouches in any column or in any row is an integer. It is also known that the total amount of money kept in the three pouches in the first column of the third row is Rs. 4.

Q5.CAT 2019

How many pouches contain exactly one coin?

Answer: 8

Show solution

We can make the following table from "the total amount of money kept in the three pouches in the first column of the third row is Rs. 4."

If the minimum and maximum value are 1, then the sum of the three pouches in the middle will be Rs 3.

If we calculate the maximum and minimum value possible for each slot in column 1. For the slot, column 1 and row 1, the maximum value possible is 10{2,4,4} while the minimum value possible is 8{2,2,4}.

Similarly, for the slot, column 1 and row 2, the maximum value possible is 13{3,5,5} while the minimum value possible is 11{3,3,5}.

It is known that the average amount of money (in rupees) kept in the nine pouches in any column or in any row is an integer. Thus the sum of coins in a row or column must be a multiple of 9.

So, we can iterate that 10,13,4 ...{27} is the only sum possible for the slots of column 1.

We now know two elements of row 2, thus we can iterate from the maximum and the minimum value possible for the slot {cloumn 3, row 2} that 38 is the only value possible for the slot.

We can make the following table:

Similarly, we can find the amount for Column 2.

For the slot, column 2 and row 1, the maximum value possible is 22{6,8,8} while the minimum value possible is 20{6,6,8}.

For the slot, column 2 and row 3, the maximum value possible is 5{1,2,3} while the minimum value possible is 4{1,1,2}.

Thus {20,3,4} is the only solution possible.

We can similarly make the following table for the last column.

Answer 8

Passage

There are only four brands of entry level smartphones called Azra, Bysi, Cxqi, and Dipq in a country.
Details about their market share, unit selling price, and profitability (defined as the profit as a percentage of the revenue) for the year 2016 are given in the table below:

In 2017, sales volume of entry level smartphones grew by 40% as compared to that in 2016. Cxqi offered a 40% discount on its unit selling price in 2017, which resulted in a 15% increase in its market share. Each of the other three brands lost 5% market share. However, the profitability of Cxqi came down to half of its value in 2016. The unit selling prices of the other three brands and their profitability values remained the same in 2017 as they were in 2016.

Q6.CAT 2018

The brand that had the highest profit in 2017 is:

Bysi
Azra
Cxqi
Dipq

Show solution

Let '100x' be the number of smartphones sold in year 2016. Then the number of smartphones sold in 2017 = 1.4*100x = 140x

It is given that Cxqi offered a 40% discount on its unit selling price in 2017 i.e. selling price in 2017 = 0.6*30000 = Rs. 18000

Also Cxqi's merket share increased by 15% whereas the other three brands lost 5% market share.

Amount of profit generated by Azra = $\dfrac{10}{100}$ *15000*49x = 73500x

Amount of profit generated by Bysi = $\dfrac{30}{100}$ *20000*28x = 168000x

Amount of profit generated by Cxqi = $\dfrac{20}{100}$ *18000*42x = 151200x

Amount of profit generated by Dipq = $\dfrac{30}{100}$ *25000*21x = 157500x

We can see that brand Bysi generated maximum profit in year 2017. Hence, option A is the correct answer.

Passage

The base exchange rate of a currency X with respect to a currency Y is the number of units of currency Y which is equivalent in value to one unit of currency X. Currency exchange outlets buy currency at buying exchange rates that are lower than base exchange rates, and sell currency at selling exchange rates that are higher than base exchange rates.

A currency exchange outlet uses the local currency L to buy and sell three international currencies A, B, and C, but does not exchange one international currency directly with another. The base exchange rates of A, B and C with respect to L are in the ratio 100:120:1. The buying exchange rates of each of A, B, and C with respect to L are 5% below the corresponding base exchange rates, and their selling exchange rates are 10% above their corresponding base exchange rates. The following facts are known about the outlet on a particular day:

1. The amount of L used by the outlet to buy C equals the amount of L it received by selling C.
2. The amounts of L used by the outlet to buy A and B are in the ratio 5:3.
3. The amounts of L the outlet received from the sales of A and B are in the ratio 5:9.
4. The outlet received 88000 units of L by selling A during the day.
5. The outlet started the day with some amount of L, 2500 units of A, 4800 units of B, and 48000 units of C.
6. The outlet ended the day with some amount of L, 3300 units of A, 4800 units of B, and 51000 units of C.

Q7.CAT 2018

What was the buying exchange rate of currency C with respect to currency L on that day?

1.10
0.95
2.20
1.90

Show solution

It is given that the base exchange rates of A, B and C with respect to L are in the ratio 100:120:1. Let us assume that base exchange rates are '100a', '120a' and 'a' in that order.

It is given that the buying exchange rates of each of A, B, and C with respect to L are 5% below the corresponding base exchange rates. Therefore, we can say that the buying exchange rates are 95a, 114a, 0.95a.

It is given that the selling exchange rates of each of A, B, and C with respect to L are 10% above their corresponding base exchange rates. Therefore, we can say that the selling exchange rates are 110a, 132a, 1.1a.

We know about the opening and closing units in stock for each currency. Let us draw the table accordingly.

Let 'p', 'q' and 'r' be the number of units of currency A, B and C bought by the outlet on that day.

Then, we can say that the outlet sold 'p - 800', 'q' and 'r-3000' units of currency A, B and C respectively.

It is given that the amount of L used by the outlet to buy C equals the amount of L it received by selling C.

$\Rightarrow$ 0.95a*r = 1.1a*(r - 3000)

$\Rightarrow$ 0.15r = 3300

$\Rightarrow$ r = 22000

It is also given that the amounts of L used by the outlet to buy A and B are in the ratio 5:3.

$\Rightarrow$ $\dfrac{p*95a}{q*114a} = \dfrac{5}{3}$

$\Rightarrow$ p = 2q

Also, the amounts of L the outlet received from the sales of A and B are in the ratio 5:9.

$\Rightarrow$ $\dfrac{(p-800)*110a}{q*132a} = \dfrac{5}{9}$

$\Rightarrow$ $\dfrac{(2q-800)*110a}{q*132a} = \dfrac{5}{9}$

$\Rightarrow$ q = 600

Therefore, p = 2q = 2*600 = 1200.

It is given that the outlet received 88000 units of L by selling A during the day.

$\Rightarrow$ (p-800)*110a = 88000

$\Rightarrow$ (1200-800)*110a = 88000

$\Rightarrow$ 44000a = 88000

$\Rightarrow$ a = 2

We can fill the entire table and answer all the questions.

From the table we can see that the buying exchange rate of currency C with respect to currency L was 1.9. Hence, we can say that option D is the correct answer.

Passage

A study to look at the early learning of rural kids was carried out in a number of villages spanning three states, chosen from the North East (NE), the West (W) and the South (S). 50 four-year old kids each were sampled from each of the 150 villages from NE, 250 villages from W and 200 villages from S. It was found that of the 30000 surveyed kids 55% studied in primary schools run by government (G), 37% in private schools (P) while the remaining 8% did not go to school (O).

The kids surveyed were further divided into two groups based on whether their mothers dropped out of school before completing primary education or not.. The table below gives the number of kids in different types of schools for mothers who dropped out of school before completing primary education:

It is also known that:
1. In S, 60% of the surveyed kids were in G. Moreover, In S, all surveyed kids whose mothers had completed primary education were in school.
2. In NE, among the O kids, 50% had mothers who had dropped out before completing primary education.
3. The number of kids in G in NE was the same as the number of kids in G in W.

Q8.CAT 2017

What percentage of kids from S were studying in P?

Show solution

Let us make note of the information given in the set.
The set states that rural kids were surveyed from 3 regions NE, West and South.
50 students each were surveyed from 150 villages in the NE, 250 villages from the West and 200 villages from the South.

Total number of kids from the NE = 150*50 = 7500
Total number of kids from the West = 250*50 = 12500
Total number of kids from the South = 200*50 = 10,000.

The table, given in the question, gives the number of students whose mothers dropped out before completing primary education. Therefore, the mothers of the remaining students should have completed primary education.

There are 7500 students from the NE in total. The mothers of 5000 of those students dropped out before completing primary education. Therefore, the remaining 2500 kids should have mothers who have completed primary education.

Let us make 2 tables - one representing the number of students and the other representing the number of students whose mother has completed primary education.

55% of the surveyed kids studied in schools run by the Government.
Number of kids studying in Govt. School should be 0.55*30000 = 16,500

From the given table, we know that the mothers of 13,500 kids who went to Govt. Schools dropped out of primary school.
Therefore, the remaining 16,500 - 13,500 = 3000 kids should have mother who have completed primary school.

37% of the surveyed kids study in private schools.
Number of kids studying in private schools should be 0.37*30000 = 11,100.
Number of kids in private schools whose mothers have completed primary school = 11,100 - 2700 = 8400.

8% of the surveyed kids did not go to school.
Number of kids who did not go to school = 0.08* 30000 = 2400
Number of kids not going to schools whose mothers have completed primary school = 2400 - 1800 = 600.

In S, 60% of the surveyed kids were in G.
Therefore, 6000 kids in S must be from G.

In NE, among the O kids, 50% had mothers who had dropped out before completing primary education.
There are 300 O kids whose mothers dropped out before completing primary education. These kids represent 50% of the total number of O-kids. Therefore, there must be 600 O-kids from NE - 300 kids should have mothers who dropped out before completing primary education and 300 kids should have mothers who have completed primary education.

The number of kids in G in NE was the same as the number of kids in G in W. Therefore, the number of kids in G in NE and the number of kids in G in W should be equal to 5250. Total number of kids:

Number of kids whose mothers have completed primary education:

We have been given that in S, all surveyed kids whose mothers had completed primary education were in school. Therefore, the number of kids not going to school whose mothers have completed primary education in S should be 0. Filling the tables, we get,

Total number of kids:

Number of kids whose mothers have completed primary education:

As we can see from the table, 3700 kids out of the 10,000 kids from S are studying in P. 37% of the total number of students from S were studying in P. Therefore, option A is the right answer.

Passage

In a square layout of site 5m ~ 5m 25 equal-sized square platforms of different heights are built. The heights (in metre) of individual platforms are as shown below:

Individuals (all of same height) are seated on these platforms. We say an individual A can reach individual B, if all the three following conditions are met;

(i) A and B are In the same row or column
(ii) A is at a lower height than B
(iii) If there is/are any individuals (s) between A and B, such individual(s) must be at a height lower than that of A.

Thus in the table given above, consider the Individual seated at height 8 on 3rd row and 2nd column. He can be reached by four individuals. He can be reached by the individual on his left at height 7, by the two individuals on his right at heights of 4 and 6 and by the individual above at height 5.

Rows in the layout are numbered from top to bottom and columns are numbered from left to right.

Q9.CAT 2017

Which of the following is true for any individual at a platform of height 1 m in this layout?

They can be reached by all the individuals in their own row and column.
They can be reached by at least 4 individuals.
They can be reached by at least one individual.
They cannot be reached by anyone.

Show solution

Since, we have been given that a person can be reached only by those who are smaller than him. Hence, 1 cannot be reached by anyone. Thus, option D is the correct answer.

Passage

An old woman had the following assets:
(a) Rs. 70 lakh in bank deposits
(b) 1 house worth Rs. 50 lakh
(c) 3 flats, each worth Rs. 30 lakh
(d) Certain number of gold coins, each worth Rs. 1 lakh
She wanted to distribute her assets among her three children; Neeta, Seeta and Geeta.
The house, any of the flats or any of the coins were not to be split. That is, the house went entirely to one child; a flat went to one child and similarly, a gold coin went to one child.

Q10.CAT 2017

The value of the assets distributed among Neeta, Seeta and Geeta was in the ratio of 1:2:3, while the gold coins were distributed among them in the ratio of 2:3:4. One child got all three flats and she did not get the house. One child, other than Geeta, got Rs. 30 lakh in bank deposits.
How many gold coins did the old woman have?

Show solution

Let the total number of gold coins with the old woman be '9n'.
Total value of the assets with the old woman = 50 + 3*30 + 70 + 9n = 210+9n.
We know that the assets have been distributed in the ratio 1:2:3.
Therefore, Neeta must have received 35+1.5n (by value), Seeta must have received 70+3n and Geeta must have received 105+4.5n.
Further, it has been given that the gold coins distributed were in the ratio 2:3:4.
Therefore, the number of gold coins with Neeta must be '2n', Seeta must be '3n' and Geeta must be '4n'.
Seeta has '3n' gold coins. Therefore, the total value of the assets with her must be 70. Seeta could not have inherited all the flats. Therefore, Seeta must have received the house ( worth 50 lakh) and 20 lakh from bank deposits.

We know that Geeta did not receive Rs. 30 lakh from the bank deposits. Therefore, Neeta must have received Rs. 30 lakh.
The remaining 5 lakh must be contributed by the gold coins (Since there is no other asset worth 5 lakh).
=> 5 + 1.5n = 2n
=> 0.5n = 5
=> n = 10
The old-woman must have had 10*9 = 90 gold coins. Therefore, option B is the right answer.

Passage

Directions for the following five questions: Answer the following questions based on the information given below:

Abdul, Bikram and Chetan are three professional traders who trade in shares of a company XYZ Ltd. Abdul follows the strategy of buying at the opening of the day at 10 am and selling the whole lot at the close of the day at 3 pm. Bikram follows the strategy of buying at hourly intervals: 10 am, 11am, 12 noon, 1 pm and 2 pm, and selling the whole lot at the close of the day, Further, he buys an equal number of shares in each purchase. Chetan follows a similar pattern as Bikram but his strategy is somewhat different. Chetan’s total investment amount is divided equally among his purchases. The profit or loss made by each investor is the difference between the sale value at the close of the day less the investment in purchase. The “return” for each investor is defined as the ratio of the profit or loss to the investment amount expressed as a percentage.

Q11.CAT 2008

One day, two other traders. Dane and Emily joined Abdul, Bikram and Chetan for trading in the shares of XYZ Ltd. Dane followed a strategy of buying equal numbers of shares at 10 am. 11 am and 12 noon, and selling the same numbers at 1 pm, 2 pm and 3 pm Emily, on the other hand, followed the strategy of buying shares using all her money at 10 am and selling all of them at 12 noon and again buying the shares for all the money at 1 pm and again selling all of them at the close of the day at 3 pm. At the close of the day the following was observed.

i. Abdul lost money in the transactions.

ii. Both Dane and Emily made profits.

iii. There was an increase in share price during the closing hour compared to the price at 2 pm.

iv. Share price at 12 noon was lower than the opening price

Share price was at its highest at

10 am
11 am
12 noon
1 pm
cannot be determined

Show solution

Let the share prices at 10,11,12,1,2,3 be p,q,r,s,t,u
Since Abdul lost all money, p>u
From Dane it can be observed, s+t+u>p+q+r
p>r and u>t
From Emily it can be observed, r+u>p+s
From these inequalities it can be inferred that the p is the highest.

Passage

Directions for the following four questions:

A health-drink company’s R&D department is trying to make various diet formulations, which can be used for certain specific purposes. It is considering a choice of 5 alternative ingredients (O, P, Q, R, and S), which can be used in different proportions in the formulations.

The table below gives the composition of these ingredients. The cost per unit of each of these ingredients is O: 150, P: 50, Q: 200, R: 500, S: 100.

Q12.CAT 2007

The company is planning to launch a balanced diet required for growth needs of adolescent children. This diet must contain at least 30% each of carbohydrate and protein, no more than 25% fat and at least 5% minerals. Which one of the following combinations of equally mixed ingredients is feasible?

O and P
R and S
P and S
Q and R
O and S

Show solution

Two ingredients are mixed in equal proportion. So, the minimum amount of protein and carbohydrate should be 60 each, the minimum amount of minerals is 10 and the maximum amount of fat is 50. The combination of O and S satisfies this requirement.

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