# PE 3 - Mathematical Reasoning | LR - Mathematical Reasoning

**Answer the next 3 questions based on the information given below.**

The currency of the country Papaya is Laugh (Lh). There are exactly five denominations in which the currency is available – Lh 2, Lh 4, Lh 5, Lh 7 and Lh 11.

Each of five persons, A through E, had notes of exactly one denomination with him and no two persons among the five had notes of the same denomination. On a particular day, the five persons visited a shop and each of them purchased a different product. The amounts that A through E paid the shopkeeper were Lh 280, Lh 278, Lh 336, Lh 300 and Lh 660, in that order. Further, the number of notes that each person gave the shopkeeper was distinct. However, the amount that each person paid the shopkeeper was more than the price of the product that he purchased. Therefore, the shopkeeper returned the excess amounts to each of A through E in the form of 3 notes, 1 note, 2 notes, 4 notes and 2 notes respectively, but in denomination(s) different from that in which the person had paid the shopkeeper.

**PE 3 - Mathematical Reasoning | LR - Mathematical Reasoning**

What is the total number of notes that the five persons paid the shopkeeper?

Answer: 378

**Explanation** :

Given:

Denominations available: 2, 4, 5, 7 and 11.

No two persons gave same number of notes.

278 is divisible by only 2, hence B paid Lh. 278 in denominations of 2 (i.e., 139 notes).

Of all the amounts given, only 660 is divisible by 11, hence E paid Lh.r 660 in denominations of 11 (i.e., 60 notes).

Now, 300 is divisible by 4 or 5. But if D pays 300 in denomination of 5, he would give 60 notes which is not possible since E gave 60 notes.

∴ D paid Lh. 300 in denominations of 4.

Now out of 280 and 336, only 280 is divisible by 5.

∴ A paid Lh. 280 in denominations of 5 (i.e., 56 notes) and C paid Lh. 336 in denominations of 7 (i.e., 48 notes).

∴ Total number of notes given by them = 56 + 139 + 48 + 75 + 60 = 378.

Hence, 378.

Workspace:

**PE 3 - Mathematical Reasoning | LR - Mathematical Reasoning**

Which of the following cannot be the price (in Lh) of the product that E purchased?

- (a)
651

- (b)
648

- (c)
653

- (d)
644

Answer: Option D

**Explanation** :

E gave Lh. 660 and was returned 2 notes.

The notes returned should not be of same denomination as that given by the E.

Option (a): If the price is 651, E will get (660 – 651 =) Lh. 9 back.

He can get a note of 4 and 5 each.

∴ Price of Lh. 651 is possible.

Option (b): If the price is 648, E will get (660 – 648 =) Lh. 12 back.

He can get a note of 5 and 7 each.

∴ Price of Lh. 648 is possible.

Option (a): If the price is 653, E will get (660 – 653 =) Lh. 7 back.

He can get a note of 2 and 5 each.

∴ Price of Lh. 653 is possible.

Option (a): If the price is 644, E will get (660 – 644 =) Lh. 16 back.

He will get a note of 5 and 11 each (which is not possible).

∴ Price of Lh. 644 is not possible.

Hence, option (d).

Workspace:

**PE 3 - Mathematical Reasoning | LR - Mathematical Reasoning**

How many of the following statements are definitely true?

I. The price of the product that A purchased is an even number.

II. The price of the product that B purchased is an even number.

III. The price of the product that C purchased is an odd number.

IV. The price of the product that D purchased is an even number.

- (a)
0

- (b)
1

- (c)
2

- (d)
3

Answer: Option A

**Explanation** :

I. A paid Lh. 280 in denominations of 5 and gets 2 notes back. These notes can be of any denomination out of 2, 4, 7 and 11.

∴ The amount he gets back may be even or odd.

⇒ Statement I is not necessarily true.

Applying the same logic for all the other statements we can figure out that none of the given 4 statements is definitely true.

Hence, option (a).

Workspace:

**Answer the next 4 questions based on the information given below.**

Four employees P, Q, R and S are evaluated on four different parameters – Coding, Technical Skills, Networking and Temperament. Each employee was given a score, which was an integer between 1 and 7 (both inclusive) on every parameter. No employee scored equal on any two of the four parameters. Similarly, no two employees scored equal on any of the four parameters.

Further it is known that:

- The average scores of the four employees in the parameters Coding, Technical Skills, Networking and Temperament were 3, 5, 4 and 5 respectively.
- The sequence of the employees, when arranged in descending order of the scores received in Coding was R, P, S and Q.
- The sequence of the employees, when arranged in descending order of the scores received in Technical Skills was Q, S, R and P.
- The sequence of the employees, when arranged in descending order of the scores received in Networking was P, Q, R and S.
- The sequence of the employees, when arranged in descending order of the scores received in Temperament was S, Q, P and R.
- P did not get 6 in any of the parameters while R got 6 in exactly one parameter.

**PE 3 - Mathematical Reasoning | LR - Mathematical Reasoning**

Which of the following employees did not score 7 out of 7 in any of the given parameters?

- (a)
Only R

- (b)
Only S

- (c)
Both S and R

- (d)
Neither S nor R

Answer: Option A

**Explanation** :

**Coding:**

R got 6 in one parameter, this can only be Coding. [In none of the other parameters, R can get 6 since he is ranked 3rd or 4th in other parameters]

Total score in Coding = 4 × 3 = 12

∴ P, Q and S score a total of (12 – 6 =) 6 in Coding.

Also, in Coding, R > P > S > Q

∴ R = 6, P = 3, S = 2 and Q = 1

**Technical Skills:**

Total score in Technical Skills = 4 × 5 = 20.

Q > S > R > P

Sum of 20 is possible in two ways (7, 6, 5, 2) or (7, 6, 4, 3)

P scores the least in Technical skills, hence, he get a score of 3 since he already scored 3 in Coding.

∴ Score of (Q, R, S, P) is (7, 6, 5, 2)

**Networking:**

Total score in networking = 4 × 4 = 16

P > Q > R > S

P is first in networking but he did not score 6 in networking, hence he would have scored either 7, 5 or 4 in networking.

If P scores 4, others will have to score 16 – 4 = 12 which is not possible (max sum possible is 1 + 2 + 3 = 6).

If P scores 5, others will have to score 16 – 5 = 11 which is not possible (max sum possible is 2 + 3 + 4 = 9).

If P scores 7, others will have to score 16 – 7 = 9 which is possible when (Q, S, R) is (5, 3, 1) or (4, 3, 2)

Since S scored 2 in coding he cannot score 2 in Networking, hence

(P, Q, S, R) = (7, 5, 3, 1)

**Temperament:**

Total score in Temperament = 4 × 5 = 20.

S > Q > P > R

Sum of 20 is possible in two ways (7, 6, 5, 2) or (7, 6, 4, 3)

Since R scores 3 in Networking, he cannot score 3 in Temperament.

∴ Score of (S, Q, P, R) is (7, 6, 5, 2)

We get the final score table:

R did not score 7 on 7 in any of the parameters.

Hence, option (a).

Workspace:

**PE 3 - Mathematical Reasoning | LR - Mathematical Reasoning**

How many employees received a score of 4 out of 7 in any of the given parameters?

- (a)
0

- (b)
1

- (c)
2

- (d)
3

- (e)
Cannot be determined

Answer: Option A

**Explanation** :

Consider the solution to first question of this set.

No employee scored 4 out of 7 in any of the parameters.

Hence, option (a).

Workspace:

**PE 3 - Mathematical Reasoning | LR - Mathematical Reasoning**

Score of Q in all four parameters put together is:

(Type 0 if your answer is ‘Cannot be determined’).

Answer: 19

**Explanation** :

Consider the solution to first question of this set.

Total score for Q = 1 + 7 + 5 + 6 = 19.

Hence, 19.

Workspace:

**PE 3 - Mathematical Reasoning | LR - Mathematical Reasoning**

In all the parameters which score was obtained maximum number of times.

- (a)
4

- (b)
5

- (c)
2

- (d)
More than one of the above

Answer: Option C

**Explanation** :

Consider the solution to first question of this set.

Score of 7, 6, 5 and 3 appeared maximum (3) times.

Hence, option (c).

Workspace:

**Answer the next 2 questions based on the information given below.**

Garvita has 10 cards labeled 2, 4, 6, 8, 10, 12, 14, 16, 18 and 20 with her. She randomly picks 4 of these 10 cards and notes down the sum of the numbers on them. After 6 rounds of picking the sum in each round was 46, 24, 42, 64, 34, 48.

She also noted something peculiar. Although she picked cards randomly, 2 of these 10 cards were never picked and each of the other card was picked thrice.

**PE 3 - Mathematical Reasoning | LR - Mathematical Reasoning**

Which card was never picked by Garvita?

- (a)
2

- (b)
8

- (c)
16

- (d)
Both 8 and 16

Answer: Option D

**Explanation** :

Sum of all the cards she picked in total 6 rounds = 46 + 24 + 38 + 64 + 38 + 48 = 258.

Sum if all 10 cards were picked thrice = 3 × (2 + 4 + 6 + … + 20) = 330.

∴ The sum of the 2 cards that were never picked = (330-258)/3 = 24

Now, these two cards can be (20 & 4) or (18 & 6) or (16 & 8) or (14 & 10) … (1)

The least sum of 4 cards picked is 24. This is possible when she picks

Case 1: 2 + 4 + 6 + 12

Case 2: 2 + 4 + 8 + 10

∴ Cards 1 and 2 should definitely be picked.

From (1): The two cards never picked cannot be (20 & 4) …(2)

The highest sum of 4 cards picked is 64. This is possible when she picks

Case 1: 10 + 16 + 18 + 20

Case 2: 12 + 14 + 18 + 20

∴ Cards 18 and 20 should definitely be picked.

From (1): The two cards never picked cannot be (18 & 6) …(3)

From (1), (2) and (3)

The two cards never picked can be (16 & 8) or (14 & 10)

For the sum of cards to be 64 in round 4, at least one of 10 or 14 must be picked.

∴ The two cards never picked can only be (16 & 8)

Hence, option (d).

Workspace:

**PE 3 - Mathematical Reasoning | LR - Mathematical Reasoning**

Which card was not picked by Garvita in round 4?

- (a)
12

- (b)
14

- (c)
16

- (d)
20

Answer: Option C

**Explanation** :

Consider the solution to first question of this set.

Garvita picked cards 12, 14, 18 and 20 in 4^{th} round.

Hence, option (c).

Workspace:

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