# CRE 1 - Basics | Arithmetic - Average

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**CRE 1 - Basics | Arithmetic - Average**

The current average age of a family of five members is 37 years. Find the average age (in years) of the family after 3 years.

- (a)
42

- (b)
40

- (c)
39

- (d)
None of these

Answer: Option B

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**Explanation** :

Since, age of every member of the family will increase by 3 years, the average of the whole family will also increase by 3 years.

Hence, option (b).

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**CRE 1 - Basics | Arithmetic - Average**

The rainfall recorded in a city for the ten year period 1981 – 1990 is as follows (in cms):

89, 95, 70, 102, 29, 79, 63, 85, 72, 50

The mean rainfall is

- (a)
bigger than 100

- (b)
less than 100

- (c)
between 95 and 100

- (d)
less than 40

Answer: Option B

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**Explanation** :

Mean rainfall = $\frac{\mathrm{89+95+70+102+29+79+63+85}}{10}$ = 73.4 cm

Hence, option (b).

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**CRE 1 - Basics | Arithmetic - Average**

The population of six villages is 621, 679, 697, 603, 649 & 651. Find the average population of the villages.

- (a)
649

- (b)
650

- (c)
672

- (d)
691

Answer: Option B

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**Explanation** :

Average = $\frac{621+679+697+603+649+651}{6}$ = 650

**Alternately,**

All these numbers are around 650.

Assume the base to be 650 and add the average deviation.

Average = Assumed average + average deviation

Average = 650 + $\frac{\left(621-650\right)+(679-650)+(697-650)+(603-650)+(649-650)+(651-650)}{6}$

Average = 650 + $\frac{-29+29+47-47-1+1}{6}$

Average = 650 + 0 = 650

Hence, option (b).

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**CRE 1 - Basics | Arithmetic - Average**

The average temperature from Monday to Thursday was 48° and from Tuesday to Friday in the same week was 47°. If the temperature on Monday was 42°, find the temperature on Friday.

- (a)
40°

- (b)
38°

- (c)
36°

- (d)
34°

Answer: Option B

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**Explanation** :

Given, Mon to Thur, (4 days) avg. Temp = 48°C

Thus, cumulative temp = 48 × 4 = 192°C.

Since temp, on Mon was 42°C

∴ Tue to Thur, (3 days), cumulative temp = 192 – 42 = 150°

Let temp. on Friday be x° C.

Then from the problem, (150 + x)/4 = 47

⇒ x = 38°C

Hence, option (b).

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**CRE 1 - Basics | Arithmetic - Average**

The average of $32\frac{1}{3},32\frac{1}{4},31\frac{2}{3},33\frac{3}{4}$ is?

- (a)
$32\frac{1}{2}$

- (b)
32

- (c)
$32\frac{1}{3}$

- (d)
$32\frac{3}{4}$

- (e)
$32\frac{2}{3}$

Answer: Option A

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**Explanation** :

Sum of the numbers = $32\frac{1}{3}+32\frac{1}{4}+31\frac{2}{3}+33\frac{3}{4}$

= 32 + 32 + 31 + 33 + $\left[\frac{1}{3}+\frac{1}{4}+\frac{2}{3}+\frac{3}{4}\right]$

= 128 + $\left[\frac{1}{4}+\frac{3}{4}+\frac{1}{3}+\frac{2}{3}\right]$

= 128 + 2 = 130.

Average = $\frac{130}{4}$ = $32\frac{1}{2}$.

Hence, option (a).

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**CRE 1 - Basics | Arithmetic - Average**

The average of 8 numbers is 23. If each of the numbers is multiplied by 8, find the average of the new set of numbers.

- (a)
8

- (b)
21

- (c)
29

- (d)
184

Answer: Option D

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**Explanation** :

Cumulative score of 8 numbers = 8 × 23

New cumulative score = 8 × (8 × 23)

Avg. score of new set = (8 × 8 × 23)/8 = 184.

**Alternately,**

Since each number is multiplied with 8, the average also will become 8 times.

∴ New average = old average × 8 = 23 × 8 = 184.

Hence, option (d).

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**CRE 1 - Basics | Arithmetic - Average**

Aakash was supposed to find the total weight of 20 packets of sweets, not all of which weighted the same. In the process he calculated average of all the possible sets of three packets at a time and found that the average reading that he obtained was 72 kg. Find the sum of the weights of the 20 packets.

- (a)
720 kgs

- (b)
2880 kgs

- (c)
1440 kgs

- (d)
Can't be determined

Answer: Option C

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**Explanation** :

We know that average of average of all possible sets of 3 (or any other number) would always be equal to the average of initial numbers.

Hence, over here the average of 20 weights is 72 kg. Hence, their sum = 72 × 20 = 1440 kg.

Hence, option (c).

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**CRE 1 - Basics | Arithmetic - Average**

Mr. Sharma’s family consists of 6 people. Mr Sharma, Mrs. Sharma and four Son’s. Average age of the family immediately after the birth of 1^{st}, 2^{nd}, 3^{rd} and 4^{th} son was 16, 15, 16 and 15 years respectively. Find the age of eldest son if the present average age of the entire family is 16 years.

- (a)
14

- (b)
12

- (c)
15

- (d)
11

Answer: Option B

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**Explanation** :

Sum of the ages immediately after the birth of 1^{st} child = 16 × 3 = 48

Sum of the ages immediately after the birth of 2^{nd} child = 15 × 4 = 60

The difference between the sum of the ages = 60 – 48 = 12 is due to 3 persons growing by same number of years. Hence, the gap between birth of 1^{st} child and 2^{nd} child = 12/3 = 4 years.

Similarly, the difference between the birth of 2^{nd} and 3^{rd} child = (16 × 5 – 15 × 4)/4 = 5 years

Similarly, the difference between the birth of 3^{rd} child and 4^{th} child = (15 × 6 – 16 × 5)/5 = 2 years

Hence, when the 4^{th} child was born, the 1^{st} child would be = 4 + 5 + 2 = 11 years old.

When 4^{th} child was born the average age of the family was 15 years and current average age is 16 which implies that each of the family member grows 1 year. Hence, the current age of the 1^{st} child = 11 + 1 = 12 years

Hence, option (b).

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**CRE 1 - Basics | Arithmetic - Average**

Average of 8 consecutive odd natural numbers is 36. Find the highest of these 8 numbers.

Answer: 43

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**Explanation** :

Let the first odd number be x.

∴ The numbers are: x, x + 2, x + 4, x + 6. x + 8, x + 10, x + 12 and x + 14.

Since these numbers are in Arithmetic Progression the average of these 8 numbers will be same as average of first and the last number.

∴ (x + x + 14)/2 = 36

⇒ 2x + 14 = 72

⇒ x = 29

∴ The highest number = x + 14 = 43.

**Alternately**,

Since consecutive odd numbers will be in Arithmetic Progression and there avereage is 36 which will also be the average of 4^{th} and 5^{th} numbers.

∴ The 4^{th} term will be 35 and 5^{th} will be 37. Now, the 6^{th} number will be 39, 7^{th} will be 41 and 8^{th} will be 43.

Hence, 43.

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