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The average of 5 consecutive numbers is n. If the next two numbers are also included, the average will :


AIncreases by 1

BRemain the same

CIncrease by 1.4

DIncrease by 2

Answer: Option A

Explanation:

let x is first no.
So, x,x+1,x+2,x+3,x+4 are consecutive no.
avg. of this 5 no.
=> (x+x+1+x+2+x+3+x+4)/5=n
x+2=n; n=x+2
As per the question:
Add next 2 no.
so new avg = (x+x+1+x+2+x+3+x+4+x+5+x+6)/7=m
x+3=m ;m=x+3
New avg-Old avg=x+3-x-2=1
Increased by 1

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The average of four consecutive even numbers is 27. The largest of these numbers is :


A36

B32

C30

D28

Answer: Option C

Explanation:

(x+(x+2)+(x+4)+(x+6))/4=27
=>x=24
so consecutive even numbers are 24,26,28,30

i.e 30 is largest

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A candidate scores an aggregate of 60% marks, scoring an average of 56% in 4 of the papers and 68% in the others. How many papers were there totally?


A6

B7

C10

D9

ENone of these

Answer: Option A

Explanation:

Lets assume no. of subject scoring avg. 68% = y
so, (56*4 + 6x * y)/(4+y) = 60
=> y=2
Therefore total subject = (2+4) = 6

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Of the three numbers, second is twice the first and is also thrice the third. If the average of the three numbers is 44, the largest number is :


A24

B36

C72

D108

Answer: Option C

Explanation:

Lets assume numbers as x, y, and z.
Given, 2nd is twice the 1st and three times the 3rd (i.e so y is the largest).
x = (1/2)*y
z = (1/3)*y
Average = y[1/2 + 1 + 1/3]/3
=> (11/18)y = 44. => y = 18*44/11 = 72.

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The average age of 30 students of a class is 12 years. The average age of a group of 5 of the students is 10 years and that of another group of 5 of them is 14 years. What is the average age of the remaining students ?


A8 years

B10 years

C12 years

D14 years

Answer: Option C

Explanation:

Given average of 30 students = 12 years.
So Total age of 30 students = 30*12 = 360 years.
First group total age = 5*10 = 50 years.
Second group total age = 5*14 = 60 years.
Remaining = 360-(50+70)=240
So, Average = 240/20 = 12 years.

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The average set of numbers is 46, if 4 numbers whose average is 52 are subtracted from this set,the average becomes 44.5.find the original number in the set?


A22

B20

C18

D16

Answer: Option B

Explanation:

Let us consider total no's in a set = n.
Sum of set= 46 * n.
=> 46*n-4*52=44.5*(n-4)
=> 46n-208=44.5n-178
=> 1.5n=30 =>n=20.
Hence, total no. in a set = 20.

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The average of 7 numbers is 50. The average of three of them is 40, while the average of the last three is 60. What must be the remaining number


A40

B60

C50

D55

Answer: Option C

Explanation:

Sum of last 3 numbers=180.
Sum of other 3 numbers=120.
Sum of all seven numbers=350.
=> 180+120+x=350
=> x=50.

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Gavaskar's average in first 50 innings was 50. After the 51st innings his average was 51. How many runs he made in the 51st innings?


A89 runs

B101 runs

C100 runs

DNone of these

Answer: Option B

Explanation:

Lets assume he made runs in 51st innings = x.
Given, Gavaskar got 50 in 50 innings
so, total score = 50*50 = 2500 runs.
Now at 51st inining, average = (2500+x)/51 = 51
=> x = (2601-2500) = 101 runs.

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In seven given numbers,the average of first four numbers is 4 and that of last four numbers is also 4. if the average of these seven numbers is 3, the fourth number is?


A3

B4

C7

D11

Answer: Option D

Explanation:

Lets assume 7 numbers as a,b,c,d,e,f,g.
Given Average of first 4 nums are 4.
=> (a+b+c+d)/4=4.
=> a+b+c+d=16 ------------ (1)
Given, Average of last 4 nums are also 4.
=> (d+e+f+g)/4=4.
=> d+e+f+g=16 -------- (2)
Given, Average of 7 nums are 3.
(a+b+c+d+e+f+g)/7=3
=> a+b+c+d+e+f+g=21 ---------- (3)
By eq. (1), (2) and (3)
=> 16+16-d=21 => d=32-21 => d=11.

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A train travels at an average speed of 90 km/hr without any stoppages. However, its average speed decrease to 60km/hr on account of stoppages. On an average, how many minutes per hour does the train stop?


A12 minutes

B18 minutes

C24 minutes

D20 minutes

Answer: Option D

Explanation:

Speed Difference = (90-60) = 30km
(30/90)*60min = 20min

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