Average

Hello Ritika, 

Find the solution.........

Let initial average =a and number of subjects = n

According to first condition,

an + 83 = (n+1) (a + 2)

an + 83 = an + 2n + a + 2

a + 2n = 81…………………1)

According to second condition,

{(a + 2) (n + 1)+ 75} = (a+2+1)(n+2)

an + a +2n + 2 + 75 = an + 2a + 3n + 6

a + n =71 …………………………..2)

From eq. 1) and 2)

n = 10 and x = 61

After paper OT and OB

Average = 61 + 2 + 1 =64

Number of subjects = 10 + 1 + 1 =12

So after getting 51 in reasoning,

Average = (64 x 12 + 51)/(13) =819/13=63

@ujjwal-dwivedi 

ANSWER 1

The concentration of sugar syrup in the mixture is 1/2. (The mixture has equal quantities of lemon juice and sugar syrup)

The concentration of sugar syrup in pure sugar syrup is 1.
The initial mixture and the pure sugar syrup are mixed in a ratio 1 : 3 to form a new mixture.
That is 2 items with sugar syrup concentrations 1/2 and 1 respectively are mixed in the ratio 1 : 3 to form a new mixture. So the concentration of sugar syrup in such a new mixture is 

{(½)*1 + 1*3} / (1+3) = 3.5/4 = 7/8

The concentration of sugar syrup in the new mixture is 7/8, that means out of 8 units of volume of the new mixture, 7 units are sugar syrup and the remaining one unit is lemon juice.
Hence the ratio of lemon juice to sugar syrup in the new mixture is 1 : 7

ANSWER 2

Let there be ‘a’ number of students originally in the class and their average weight be ‘n’ kgs.
Let ‘b’ number of students join the class having an average weight of ‘n + 3’ kgs.
This excess weight of 3 kgs brought in by each new student is shared among all the students such that each of them increases by 0.6 kgs.

3b = 0.6(a + b)
3b = 0.6 a + 0.6 b
2.4b = 0.6a
a = 4b
a : b = 4 : 1
The ratio of the number of original students to the number of new students is 4 : 1

@neraj-naiyar Sir,

It is given that, a1, a2, a3,..... Ramesh, Gautham, ...., a22 writes an examination
Given that, Average score of the 21 students other than Gautham = 62
So, Total Score - Gautham's score = 62 x 21
Ramesh scored 82.5
It is given that, when Ramesh leaves, the average score drops down by 1 mark.
Which means that Ramesh scored more than the Overall class average.
Since his departure has resulted in the decrease of the overall class average by 1, his score is 21 more than the average.
Overall Class average = 82.5 - 21 = 61.5 marks
Total Score - Gautham's score = 62 x 21
61.5 x 22 - 62 x 21 = Gautham's score
1353 - 1302 = Gautham's score
Gautham's score = 51 marks.

@neraj-naiyar

Sir,

the average of m students should be more than 40 to give total average more than 40 for maximum we will take m=5 and n=45, than 

total avg =(40*40+45*5)/40+5

=1825/45

=40.5555

hence option 2 is correct

30% of 60 = 18 . 

Let the team played n matches in a row . 

so , ( 18 + n)/(60 + n ) = 1/2 

n = 24  . 

Hence,the team won 24 matches in a row . 

after two years average will become 38 then the total age of all employees =38n

5 people of age 40 leave then total age=38n-40*5  =38n-200

after 3 more years so then total age=(38n-200) +(n-5)3            [n-5 is because five employee have already left ]

then comes 5 people of age 30 then total age become =(38n-200) +(n-5)3 +30*5

now total people are still  n and the average now is 39.7

{(38n-200) +(n-5)3 +150}/n =39.7 

41n -200 +150-15 =39.7

1.3n=65

n=50 

Hi Pulkit, 

let the average height of 22 toddlers be 3x . 

so the average height of two toddlers who left => x 

Sum of the heights of 22 toddlers = 66x

Sum of the heights of  remaining 20 toddlers = 66x - 2x = 64x 

According to question , 

64x/20 - 66x/22 = 2 

3.2x - 3x = 2 

x = 10 

hence the average height of the remaining toddlers = 3.2 x 10 = 32

answer is 50?

The average is 37  31/37 = 1400/37

we know that average of 1st n consecutive number is (n+1)/2 but average is 35 31/37 hence 
value of n should be close to 2 x 37= 74. If there are n numbers then after erasing we will have n-1 numbers then summation of the n-1 numbers = (n-1) x Avg  = 1400(n-1)/37, hence n-1 should be divisible by 37, but n is close to 74 ,  hence n = 75 so that n-1 =74 is divisible by 37. Hence n = 75.

Sum of 1st 75 natural numbers = 75 x 76/2  = 2850
Sum of 74 number after erasing one number = 1400/37 x 74 = 2800
Hence the number that was erased = 2850 - 2800 =50 .

avg of square of first n natural nos is (n+1)(2n+1)/6=130

2n^2+3n+1=780

2n^2+3n-779=0

n=[-3+√(9+6232)]/2

n=38

 sum of cube of first n natural nos. is ( n ( n +1)/2)² 

put value of n 

38*39*38*39/4

avg=sum/n

=(38*39*38*39)/4*38

get answer

Average of first n natural nos is (n+1)2=x

next n nos will be 

n , n+1, n+2, n+3.......2n 

their sum =n*n+ n(n+1)/2

average ={n*n+ n(n+1)/2}/n

average = n+x 

hence option A

4! ( ⁴C₄ + ⁵C₄ + ⁶C₄ +...+¹⁴C₄) 

Sum :  4! (¹⁵C₅) = 72072

Average : 72072/11 = 6552