# Arithmetic - Average - Previous Year CAT/MBA Questions

You can practice all previous year CAT questions from the topic Arithmetic - Average. This will help you understand the type of questions asked in CAT. It would be best if you clear your concepts before you practice previous year CAT questions.

**CAT 2022 QA Slot 1 | Arithmetic - Average**

The average weight of students in a class increases by 600 gm when some new students join the class. If the average weight of the new students is 3 kg more than the average weight of the original students, then the ratio of number of original students to the number of new students is

- A.
3 : 1

- B.
4 : 1

- C.
1 : 2

- D.
1 : 4

Answer: Option B

**Explanation** :

Let the number of students initially be I and the number of students joining be J.

Average of students initially be a kg, while those joining be (a + 3) kg such that final average of all students becomes (a + 0.6) kg

Initial Students Students Joining

I J

a a + 3

a + 0.6

2.4 0.6

⇒ I/J = 2.4/0.6 = 4/1

Hence, option (b).

Workspace:

**CAT 2022 QA Slot 1 | Arithmetic - Average**

The average of three integers is 13. When a natural number n is included, the average of these four integers remains an odd integer. The minimum possible value of n is:

- A.
5

- B.
1

- C.
3

- D.
4

Answer: Option A

**Explanation** :

Sum of the original 3 numbers = 3 × 13 = 39.

Now, $\frac{39+\mathrm{n}}{4}$ = odd = 2k - 1

⇒ 39 + n = 8k – 4

⇒ 43 + n = 8k

For k to be an integer least possible natural value of n = 5

Hence, option (a).

Workspace:

**CAT 2022 QA Slot 2 | Arithmetic - Average**

The average of a non-decreasing sequence of N number a_{1}, a_{2}, ..., a_{N} is 300. If a1 is replaced by 6a_{1}, the new average becomes 400. Then, the number of possible values of a_{1} is

Answer: 14

**Explanation** :

Given, $\frac{{a}_{1}+{a}_{2}+...+{a}_{N}}{N}$ = 300

⇒ a_{1} + a_{2} + ... + a_{N} = 300N ...(1)

Also, $\frac{6{a}_{1}+{a}_{2}+...+{a}_{N}}{N}$ = 400

⇒ 6a_{1} + a_{2} + ... + a_{N} = 400N ...(2)

(2) - (1)

⇒ 5a_{1} = 100N

⇒ a_{1} = 20N

Since, a_{1} is the least of the given numbers it cannot be more than the average, hence a_{1} ≤ 300.

⇒ N ≤ 15

If N = 1, a_{1} = 20 and average cannot be equal to 300.

Hence, N can take all values from 2 till 15, i.e., 14 values.

Hence, 14.

Workspace:

**CAT 2022 QA Slot 3 | Arithmetic - Average**

In an examination, the average marks of students in sections A and B are 32 and 60, respectively. The number of students in section A is 10 less than that in section B. If the average marks of all the students across both the sections combined is an integer, then the difference between the maximum and minimum possible number of students in section A is

Answer: 63

**Explanation** :

Let the number of students in section B is x and that in A is (x – 10).

⇒ $\frac{(\mathrm{x}-10)\times 32+\mathrm{x}\times 60}{2\mathrm{x}-10}$ = integer

⇒ $\frac{92\mathrm{x}-320}{2\mathrm{x}-10}$ = integer

⇒ $\frac{46\mathrm{x}-160}{\mathrm{x}-5}$ = integer

⇒ $\frac{46(\mathrm{x}-5)+70}{\mathrm{x}-5}$ = integer

⇒ (x - 5) + $\frac{70}{\mathrm{x}-5}$ = integer

⇒ $\frac{70}{\mathrm{x}-5}$ should be an integer

∴ (x – 5) should be a factor of 70 while x > 10

⇒ Highest possible value of x = 75 while lowest possible value is 12

∴ Difference between highest and lowest values of x = 75 – 12 = 63.

Hence, 63.

Workspace:

**CAT 2022 QA Slot 3 | Arithmetic - Average**

Consider six distinct natural numbers such that the average of the two smallest numbers is 14, and the average of the two largest numbers is 28. Then, the maximum possible value of the average of these six numbers is

- A.
23

- B.
22.5

- C.
24

- D.
23.5

Answer: Option B

**Explanation** :

The average of lowest two number is 14, hence their sum = 28

The average of highest two number is 28, hence their sum = 56

To maximize the average of these six numbers, we need to maximize the middle two numbers. But it is also given that all the numbers are distinct.

Since all numbers are distinct out of the two highest numbers, one must be greater than 28 (a) and the other less than 28 (b).

To maximize the middle two numbers we need to maximize ‘b’. The maximum value ‘b’ can take is 27.

∴ Maximum value of the two middle numbers can be 25 and 26.

⇒ Maximum sum of the six numbers = 28 + 25 + 26 + 56 = 135

⇒ Maximum average of the six numbers = 135/6 = 22.5

Hence, option (b).

Workspace:

**CAT 2021 QA Slot 1 | Arithmetic - Average**

Suppose hospital A admitted 21 less Covid infected patients than hospital B, and all eventually recovered. The sum of recovery days for patients in hospitals A and B were 200 and 152, respectively. If the average recovery days for patients admitted in hospital A was 3 more than the average in hospital B then the number admitted in hospital A was

Answer: 35

**Explanation** :

Let the number of patients in hospital A = x

∴ The number of patients in hospital B = x + 21

⇒ $\frac{200}{x}$ = $\frac{152}{x+21}$ + 3

⇒ 200x + 4200 = 152x + 3x^{2} + 63x

⇒ 3x^{2} + 15x - 4200 = 0

⇒ x^{2} + 5x - 1400 = 0

⇒ x = 35 or -40 (not possible)

∴ Number of patients in hospital A = 35

Hence, 35.

Workspace:

**CAT 2021 QA Slot 1 | Arithmetic - Average**

Onion is sold for 5 consecutive months at the rate of Rs 10, 20, 25, 25, and 50 per kg, respectively. A family spends a fixed amount of money on onion for each of the first three months, and then spends half that amount on onion for each of the next two months. The average expense for onion, in rupees per kg, for the family over these 5 months is closest to

- A.
20

- B.
16

- C.
18

- D.
26

Answer: Option C

**Explanation** :

Let the amounts spent by the family each month be LCM (10, 20, 25, 50) = Rs. 100 for the first 3 months and then Rs. 50 for the next two months.

Amount of onion bought during month 1 = 100/10 = 10 kgs

Amount of onion bought during month 2 = 100/20 = 5 kgs

Amount of onion bought during month 3 = 100/25 = 4 kgs

Amount of onion bought during month 4 = 50/25 = 2 kgs

Amount of onion bought during month 5 = 50/50 = 1 kgs

∴ Total amount of onion bought = 10 + 5 + 4 + 2 + 1 = 22 kgs

Total amount spend on onions = 100 + 100 + 100 + 50 + 50 = Rs. 400.

∴ Average expense for onion per kg for these 5 months = 400/22 = 18.18 ≈ Rs. 18/kg.

Hence, option (c).

Workspace:

**CAT 2021 QA Slot 2 | Arithmetic - Average**

In a football tournament, a player has played a certain number of matches and 10 more matches are to be played. If he scores a total of one goal over the next 10 matches, his overall average will be 0.15 goals per match. On the other hand, if he scores a total of two goals over the next 10 matches, his overall average will be 0.2 goals per match. The number of matches he has played is

Answer: 10

**Explanation** :

Let the number of matches played so far be ‘b’ and the goals scored in these ‘n’ matches be ‘g’.

Now if 1 goal is scored in next 10 matches

⇒ g + 1 = (n + 10) × 0.15

⇒ g = 0.15n + 1.5 …(1)

If 2 goals are scored in next 10 matches

⇒ g + 2 = (n + 10) × 0.2

⇒ g = 0.2n + 1 …(2)

Solving (1) and (2) we get,

n = 10

Hence, 10.

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**CAT 2021 QA Slot 3 | Arithmetic - Average**

The arithmetic mean of scores of 25 students in an examination is 50. Five of these students top the examination with the same score. If the scores of the other students are distinct integers with the lowest being 30, then the maximum possible score of the toppers is

Answer: 92

**Explanation** :

Let the score of 5 toppers be x each.

To maximize the score of toppers we need to minimize the scores of the remaining 20 students.

∴ Score of remaining 25 students will be 30, 31, 32, … , 49.

⇒ 25 × 50 = 30 + 31 + … + 49 + 5x

⇒ 1250 = 790 + 5x

⇒ x = 92

Hence, 92.

Workspace:

**CAT 2020 QA Slot 1 | Arithmetic - Average**

The mean of all 4-digit even natural numbers of the form ‘aabb’, where a > 0, is

- A.
5050

- B.
4864

- C.
4466

- D.
5544

Answer: Option D

**Explanation** :

‘aabb’ is an even number hence, b can be either 0 or 2 or 4 or 6 or 8.

∴ ‘aabb’ can be:

1100, 1122, 1144, 1166, 1188

2200, 2222, 2244, 2266, 2288

…

9900, 9999, 9944, 9966, 9988

Adding all these number

= 5 × (1100 + ... + 9900) + 9 × (22 + 44 + 66 + 88)

= (5 × 1100 × 45) + 9 × 22 × 10 = 5500 × 45 + 44 × 45 = 5544 × 45

∴ Average of all such numbers = (5544 × 45)/45 = 5544.

Hence, option (d).

Workspace:

**CAT 2020 QA Slot 1 | Arithmetic - Average**

Let A, B and C be three positive integers such that the sum of A and the mean of B and C is 5. In addition, the sum of B and the mean of A and C is 7. Then the sum of A and B is

- A.
4

- B.
5

- C.
6

- D.
7

Answer: Option C

**Explanation** :

Given,

$a+\frac{b+c}{2}=5$

⇒ 2a + b + c = 10 …(1)

Also, $b+\frac{a+c}{2}=7$

⇒ 2b + a + c = 14 …(2)

Solving (1) and (2) we get,

b – a = 4

⇒ b = a + 4

Substituting this in the (1)

⇒ 2a + a + 4 + c = 10

⇒ 3a + c = 6

Given all three as positive integers, only possible value for a is 1. (c cannot be 0)

So, when a = 1, c = 3 and b = 5

∴ a + b = 6.

Hence, option (c).

Workspace:

**CAT 2020 QA Slot 2 | Arithmetic - Average**

In a group of 10 students, the mean of the lowest 9 scores is 42 while the mean of the highest 9 scores is 47. For the entire group of 10 students, the maximum possible mean exceeds the minimum possible mean by

- A.
4

- B.
5

- C.
3

- D.
6

Answer: Option A

**Explanation** :

Let the highest and lowest score be h and l respectively and total score of remaining 8 students be x.

The mean of the lowest 9 scores is 42

⇒ x + l = 9 × 42 = 378 …(1)

The mean of the highest 9 scores is 47

⇒ x + h = 9 × 47 = 423 …(2)

(2) – (1)

⇒ h – l = 423 – 378 = 45

**Case 1**: Least possible average is when we minimize the highest marks. The least highest marks can be 47.

∴ Lowest score = 47 – 45 = 2

∴ Least average = $\frac{9\times 47+2}{10}$ = 42.5

**Case 2**: Highest possible average is when we maximize the lowest marks. The highest lowest marks can be 42.

∴ Highest score = 42 + 45 = 87

∴ Highest average = $\frac{9\times 42+87}{10}$ = 46.5

∴ The required difference = 46.5 – 42.5 = 4

Hence, option (a).

Workspace:

**CAT 2020 QA Slot 3 | Arithmetic - Average**

A batsman played n + 2 innings and got out on all occasions. His average score in these n + 2 innings was 29 runs and he scored 38 and 15 runs in the last two innings. The batsman scored less than 38 runs in each of the first n innings. In these n innings, his average score was 30 runs and lowest score was x runs. The smallest possible value of x is

- A.
3

- B.
4

- C.
1

- D.
2

Answer: Option D

**Explanation** :

Average score in n innings is 30 and in n + 2 innings is 29. Score in last 2 innings is 38 and 15

∴ 30n + 38 + 15 = 29(n + 2)

⇒ n = 5

∴ Total runs score in first n innings = 5 × 30 = 150

Also, he scored less than 38 runs in each of these 5 innings.

To calculate the lowest possible score in an innings we try to maximize the score of other 4 innings. Maximum runs score in each of the 4 innings can be 37.

∴ x + 4 × 37 = 150

⇒ x = 2

Hence, option (d).

Workspace:

**CAT 2020 QA Slot 3 | Arithmetic - Average**

Dick is thrice as old as Tom and Harry is twice as old as Dick. If Dick's age is 1 year less than the average age of all three, then Harry's age, in years, is

Answer: 18

**Explanation** :

Let Tom's age to be 'x' years.

∴ Dick's age = 3x.

∴ Harry's age = 6x.

Given, Dick's age is 1 year less than the average age of all three,

∴ 3x = (x + 3x + 6x)/3 – 1

⇒ 9x = 10x – 3

⇒ x = 3

∴ Harry’s age = 6x = 6 × 3 = 18.

Hence, 18.

Workspace:

**CAT 2019 QA Slot 1 | Arithmetic - Average**

A chemist mixes two liquids 1 and 2. One litre of liquid 1 weighs 1 kg and one litre of liquid 2 weighs 800 gm. If half litre of the mixture weighs 480 gm, then the percentage of liquid 1 in the mixture, in terms of volume, is

- A.
70

- B.
85

- C.
80

- D.
75

Answer: Option C

**Explanation** :

Density of liquid 1 = 1000 g/L.

Density of liquid 2 = 800 g/L.

Half litre of the mixture weighs 480 gm, so 1 L of the mixture weighs 960 gm.

So, density of the mixture = 960 g/L.

Using the alligation cross;

$\frac{Liquid1}{Liquid2}=\frac{(960-800)}{(1000-960)}$ = $\frac{4}{1}$.

Percentage of liquid 1 in the mixture = (4/5) × 100 = 80%.

Hence option 3.

Workspace:

**CAT 2019 QA Slot 1 | Arithmetic - Average**

Ramesh and Gautam are among 22 students who write an examination. Ramesh scores 82.5. The average score of the 21 students other than Gautam is 62. The average score of all the 22 students is one more than the average score of the 21 students other than Ramesh. The score of Gautam is

- A.
53

- B.
51

- C.
49

- D.
48

Answer: Option B

**Explanation** :

Let marks of Gautam be G.

∴ G + (62 × 21) = T ...(I) (where T is the total marks of all 22 students)

82.5 + (21 × x) = T ...(II) (where x is the average marks of 21 students other than Ramesh)

The average score of all the 22 students is one more than the average score of the 21 students other than Ramesh.

∴ (T/22) = 1 + x ...(III)

Solving (I), (II) and (III), we get; x = 60.5, T = 1353 and G = 51.

Hence, option 2.

Workspace:

**CAT 2019 QA Slot 2 | Arithmetic - Average**

The average of 30 integers is 5. Among these 30 integers, there are exactly 20 which do not exceed 5. What is the highest possible value of the average of these 20 integers?

- A.
4

- B.
5

- C.
4.5

- D.
3.5

Answer: Option C

**Explanation** :

Exactly 20 of the 30 integers do not exceed 5 that means 10 of the 30 integers are greater than 5.

Sum of all 30 integers = 30 × 5 = 150.

To keep the average of the 20 integers as high as possible, we need to keep the average of the 10 integers above 5 as low as possible. Since we are dealing with integers, the least value that the 10 integers above 5 can take is 6.

So, the sum of the 10 integers = 10 × 6 = 60.

Hence, the sum of the remaining 20 integers = 150 - 60 = 90

∴ The average of the remaining 20 = 90/20 = 4.5

Hence, option (3).

Workspace:

**CAT 2018 QA Slot 1 | Arithmetic - Average**

A CAT aspirant appears for a certain number of tests. His average score increases by 1 if the first 10 tests are not considered, and decreases by 1 if the last 10 tests are not considered. If his average scores for the first 10 and the last 10 tests are 20 and 30, respectively, then the total number of tests taken by him is

Answer: 60

**Explanation** :

Let there be ‘n’ tests and their average be ‘x’.

By the given conditions,

$\frac{nx-200}{n-10}$ = x + 1

$\frac{nx-300}{n-10}$ = x - 1

∴ $\frac{nx-200}{n-10}$ - 1 = $\frac{nx-300}{n-10}$ + 1

∴ $\frac{nx-200}{n-10}$ - $\frac{nx-300}{n-10}$ = 2

∴ $\frac{100}{n-10}$ = 2

∴ n = 60

Answer: 60

Workspace:

**CAT 2018 QA Slot 1 | Arithmetic - Average**

In an apartment complex, the number of people aged 51 years and above is 30 and there are at most 39 people whose ages are below 51 years. The average age of all the people in the apartment complex is 38 years. What is the largest possible average age, in years, of the people whose ages are below 51 years?

- A.
27

- B.
28

- C.
26

- D.
25

Answer: Option B

**Explanation** :

Let x is average age of 30 people whose age is 51 years and above. Clearly, x ≥ 51

Let y is average age of 39 people whose age is less than 51 years.

∴ 30x + 39y = 69 × 38

∴ 10x + 13y = 874

x ≥ 51 ⇒ 10x ≥ 510 ⇒ 10x + 13y ≥ 510 + 13y

⇒ 874 ≥ 510 + 13y ⇒ 364 ≥ 13y

⇒ 28 ≥ y

Thus, the largest possible average age of the people whose ages are below 51 years is 28 years.

Hence, option 2.

Workspace:

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

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

- A.
45

- B.
23

- C.
48

- D.
20

Answer: Option B

**Explanation** :

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

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

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

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

= 51x + 51

∴ a_{1} = x − 51

**Using options:**

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

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

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

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

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

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

Hence, option 2.

Workspace:

**CAT 2017 QA Slot 1 | Arithmetic - Average**

An elevator has a weight limit of 630 kg. It is carrying a group of people of whom the heaviest weighs 57 kg and the lightest weighs 53 kg. What is the maximum possible number of people in the group?

Answer: 11

**Explanation** :

There must be at least 1 person with weight 57 kg.

Remaining persons weigh a total of 630 - 57 = 573.

Maximum number of people in the group can be 573/53 + 1 (one who's weight is 57 kg) i.e. 10 + 1 = 11.

Hence, 11.

Workspace:

**CAT 2017 QA Slot 2 | Arithmetic - Average**

The average height of 22 toddlers increases by 2 inches when two of them leave this group. If the average height of these two toddlers is one-third the average height of the original 22, then the average height, in inches, of the remaining 20 toddlers is

- A.
30

- B.
28

- C.
32

- D.
26

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 3.

Workspace:

**CAT 2007 QA | Arithmetic - Average**

**Each question is followed by two statements A and B. Answer each question using the following instructions.**

Mark (1) if the question can be answered by using statement A alone but not by using statement B alone.

Mark (2) if the question can be answered by using statement B alone but not by using statement A alone.

Mark (3) if the question can be answered by using both the statements together but not by using either of the statements alone.

Mark (4) if the question cannot be answered on the basis of the two statements.

The average weight of a class of 100 students is 45 kg. The class consists of two sections, I and II, each with 50 students. The average weight, W_{I}, of Section I is smaller than the average weight, W_{II}, of Section II. If the heaviest student, say Deepak, of Section II is moved to Section I, and the lightest student, say Poonam, of Section I is moved to Section II, then the average weights of the two sections are switched, i.e., the average weight of Section I becomes W_{II} and that of Section II becomes W_{I}. What is the weight of Poonam?

A. W_{II} – W_{I} = 1.0

B. Moving Deepak from Section II to I (without any move from I to II) makes the average weights of the two sections equal.

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option C

**Explanation** :

Let the weights of Deepak and Poonam be d and p respectively.

(50W_{II} + 50W_{I})/100 = 45

∴ W_{II} + W_{I} = 90 ...(i)

50W_{I} + d – p = 50W_{II}

50W_{II} – d + p = 50W_{I}

∴ 50(W_{II} – W_{I}) = d – p ...(ii)

From Statement A, W_{II} – W_{I} = 1 ...(iii)

From (i), (ii) and (iii)

W_{I} and W_{II} can be found. Also, d – p = 50 ...(iv)

However this information does not give us the value of p. Statement A is insufficient to answer the question.

From Statement B,

W_{I} = W_{II} = (Sum_{I} + d)/51 = (Sum_{II} – d)/49

∴ 49(Sum_{I}) + 49d = 51(Sum_{II}) – 51d

∴ 100d = 51(50W_{II}) – 49(50W_{I})

∴ 2d = 51W_{II} – 49W_{I} ...(v)

This alone cannot help us find the value of p. Statement B is insufficient to answer the question.

Considering both statements together, we have values of W_{I} and W_{II}, which can be substituted in (v) to find d, which can be used to find p using (iv).

Hence, option (c).

Workspace:

**CAT 2007 QA | Arithmetic - Average**

Consider the set S = {2, 3, 4, ..., 2n + 1}, where n is a positive integer larger than 2007. Define X as the average of the odd integers in S and Y as the average of the even integers in S. What is the value of X – Y?

- A.
0

- B.
1

- C.
n/2

- D.
n + 1/2n

- E.
2008

Answer: Option B

**Explanation** :

Y = (2 + 4 + 6 + 8 + … + 2n)/n

X = (3 + 5 + 7 + 9 + … + (2n + 1))/n

= [(2 + 1) + (4 + 1) + (6 + 1) + (8 + 1) + … + (2n + 1)]/n

= (2 + 4 + 6 + 8 + … + 2n)/n + (1 + 1 + 1 + 1 + … n times)/n

= Y + 1

∴ X – Y = 1

Hence, option (b).

Note: The information that 'n is a positive integer larger than 2007' does not affect the answer in any way.

Workspace:

**CAT 2007 QA | Arithmetic - Average**

Ten years ago, the ages of the members of a joint family of eight people added up to 231 years. Three years later, one member died at the age of 60 years and a child was born during the same year. After another three years, one more member died, again at 60, and a child was born during the same year. The current average age of this eight member joint family is nearest to:

- A.
23 years

- B.
22 years

- C.
21 years

- D.
25 years

- E.
24 years

Answer: Option E

**Explanation** :

The sum of the ages of the members of the family ten years ago = 231

∴ The sum of the ages of the members of the family seven years ago = 231 + (3 × 8) – 60 = 195

∴ The sum of the ages of the members of the family four years ago = 195 + (3 × 8) – 60 = 159

∴ The sum of the ages of the members of the family now = 159 + (4 × 8) = 191

∴ Required average = 191/8 = 23.875 ≈ 24

Hence, option (e).

Workspace:

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