Let us assume the four numbers to be a, b, c and d in ascending order.&nbsp;Average of first two numbers is 1 more than the first number$\frac{\left(a+b\right)}{2}=a+1$$b-a=2$$b=a+2$ Average of first three numbers is 2 more than average of first two numbers$\frac{\left(a+b+c\right)}{3}=\frac{\left(a+b\right)}{2}+2$$2c=a+b+12$Substituting the value for b$2c=a+a+2+12$$2c=2a+14$$c=a+7$Average of first four numbers is 3 more than average of first three numbers.$\frac{\left(a+b+c+d\right)}{4}=\frac{\left(a+b+c\right)}{3}+3$$3d=a+b+c+36$Substituting the value of b and c$3d=a+a+2+a+7+36$$3d=3a+45$$d=a+15$d is the largest and a is the smallest and we know that d=a+15Hence the difference between the smallest and the largest values is 15.&nbsp;

We are given that average of three distinct integers is 28, that means the sum of these three integers is 28x3=84Let us write $x+y+z=84$x, y, z being the three distinct integers in ascending order.&nbsp;If the smallest number is increased by 7 and the largest number is reduced by 10$(x+7)+(y)+(z-10)=81$New arithmetic mean will be&nbsp;$\frac{81}{3}=27$And this is said to be 2 more than the middle number, meaning$27-2=y=25$$x+z=59$We are given that difference between the largest and the smallest numbers becomes 64,&nbsp;$(z-10)-(x+7)=64$$z-x=81$Adding the two equations we get,&nbsp;$2z=140$$z=70$

Let the number of total employees in the company be 100x, and the total salary of all the employees be 100y.It is given that&nbsp;20% of the employees work in the manufacturing department, and&nbsp;the total salary obtained by all the manufacturing employees is one-sixth of the total salary obtained by all the employees in the company.Hence, the total number of employees in the manufacturing department is 20x, and the total salary received by them is (100y/6)Average salary in the manufacturing department = (100y/6*20x) = 5y/6xSimilarly, the total number of employees in the nonmanufacturing department is 80x, and the total salary received by them is (500y/6)Hence, the average salary in the nonmanufacturing department = (500y/6*80x) = 25y/24xHence, the ratio is:- (5y/6x): (25y/24x)&nbsp;=&gt; 120: 150 = 4:5The correct option is A

Let us assume that A, B, C, D, and E weights are a, b, c, d, and e.1st condition$ \frac{(a+b+c)}{3} - \frac{(a+b+c+d)}{4} = x $2nd condition$ \frac{(a+b+c+e)}{4} - \frac{(a+b+c)}{3} = 2x $Adding both the equations, we get:$ \frac{(e-d)}{4} = 3x $=&gt; $ \frac{(e-d)}{4} = 3x $ =&gt; e - d = 12xGiven that 12x = 12 =&gt; x = 1.

It is given that average of three numbers is 13.Sum = $3 * 13 = 39$It is given, $\frac{39 + n}{4}$ is a odd number.Minimum value $\frac{39 + n}{4}$ can take such that n is a natural number is 11$\frac{39 + n}{4} = 11$n = 5

Let the original number of students be 'n' whose average weight is 'x'Let the number of students added be 'm' and the average weight will be x + 3We need to find the value of n : mIt is given, average weight of students in a class increased by 0.6 after new students are added.Therefore,$\frac{nx + m(x + 3)}{n + m} = x + 0.6$$nx + mx + 3m = mx + nx + 0.6n + 0.6m$$2.4m = 0.6n$$4m = n$$\frac{n}{m} = \frac{4}{1}$

a + 2b = 6From the above equation, we can say that maximum value b can take is 3 and minimum value b can take is 0.a + b + b = 6a + b = 6 - ba + b is maximum when b is minimum, i.e. b = 0Maximum value of a + b = 6 - 0 = 6a + b is minimum when b is maximum, i.e. b = 3Minimum value of a + b = 6 - 3 = 3Average =(6+3)/2 ​= 4.5The answer is option B.

Savings target in a year = 550*12 = Rs 6600Saving in first 9 months = 9(4000-3500) = Rs 4500Saving for remaining 3 months should be 6600-4500, i.e. Rs 2100Savings for each month in last 3 months = 2100/3 = Rs 700It is given, monthly expenses in last 3 months = Rs 3700This implies, his monthly earnings from 10th month should be 3700+700, i.e. Rs 4400The answer is option B

The average marks for all the students is 38.Sum = 5*38 = 190To find the minimum marks scored by Amit, we need to maximise the score of remaining students.Maximum scores sum of remaining students = 50 + 49 + 48 + 32 = 179Minimum possible score of Amit = 190 - 179 = 11It is given, Amit scored least. This implies maximum possible score of Amit is 31.Difference = 31 - 11 = 20

Let the number of students in section A and B be a and b, respectively.It is given, $a = b - 10$$\frac{32a + 60b}{a + b}$ is an integer$\frac{32a + 60(a + 10)}{a + a + 10} = k$$\frac{46a + 300}{a + 5} = k$$k = \frac{46(a + 5)}{a + 5} + \frac{70}{a + 5}$$k = 46 + \frac{70}{a + 5}$a can take values 2, 5, 9, 30, 65Difference 65 - 2 = 63

Let us assume the family spends Rs. 100 each month for the first 3 months and then spends Rs. 50 in each of the next two months.Then amount of onions bought = 10, 5, 4, 2, 1, for months 1-5 respectively.Total amount bought = 22kg.Total amount spent = 100+100+100+50+50 = 400.Average expense =&nbsp;400/22 = Rs.18.18 ≈&nbsp;18

Let Total matches played be n and in initial n-10 matches his goals be x&nbsp;so we get&nbsp;$\frac{1+x}{n}$ = 0.15we get x+1 =0.15n&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(1)From condition (2) we get :$\frac{2+x}{n}$ = 0.15we get x+2 = 0.2n&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(2)Subtracting (1) and (2)we get 1 =0.05nn =20So initially he played n-10 =10 matches

Let sum of marks of students be x&nbsp;Now therefore x = 25*50 =1250&nbsp;Now to maximize the marks of the toppers&nbsp;We will minimize the marks of 20 students&nbsp;so their scores will be (30,31,32.....49 )&nbsp;let score of toppers be yso we get 5y + $\frac{20}2$(79) = 1250we get 5y +790=12505y=460y=92So scores of toppers = 92

Given, $\frac{\text{sum of scores in n matches} + 38 + 15}{n+2} = 29$Given, $\frac{\text{sum of scores in n matches}}{n} = 30$30n + 53 = 29(n+2) $\Rightarrow n = 5$Sum of the scores in 5 matches = $29 \cdot 7 - 38 - 15 = 150$Since the batsmen scored less than 38 in each of the first 5 innings,the value of x will be minimum when the remaining four values are highest$\Rightarrow$ 37 + 37 + 37 + 37 + x = 150$\Rightarrow$ x = 2

Assume the average of 21 students other than Ramesh = aSum of the scores of 21 students other than Ramesh = 21aHence the average of 22 students = a+1Sum of the scores of all 22 students = 22(a+1)The score of Ramesh = Sum of scores of all 22 students - Sum of the scores of&nbsp;21 students other than Ramesh = 22(a+1)-21a=a+22&nbsp;= 82.5&nbsp;&nbsp;(Given)=&gt; a = 60.5Hence,&nbsp;sum of the scores of all 22 students = 22(a+1) = 22*61.5 = 1353Now the sum of the scores of students other than Gautam = 21*62 = 1302Hence the score of Gautam = 1353-1302=51

It is given that the average of the 30 integers = 5Sum of the 30 integers = 30*5=150There are exactly 20 integers whose value is less than 5.To maximise the average of the 20 integers, we have to assign minimum value to each of the remaining 10 integersSo the sum of 10 integers = 10*6=60The sum of the 20 integers = 150-60= 90Average of 20 integers =&nbsp;$\frac{90}{20}$ = 4.5

The possible average age of people&nbsp;whose ages are below 51 years will be maximum if the average age of the number of people aged&nbsp;51 years and above is minimum. Hence, we can say that that there are 30 people having same age 51 years.&nbsp; Let 'x' be the maximum average age of people whose ages are below 51.&nbsp; Then we can say that,&nbsp; $\dfrac{51*30+39*x}{30+39} = 38$ $\Rightarrow$ $1530+39x = 2622$ $\Rightarrow$ $x = 1092/39 = 28$ Hence, we can say that option D is the correct answer.&nbsp;

Let the total number of tests be 'n' and the average by 'A' Total score = n*A When 1st 10 tests are excluded, decrease in total value of scores = (nA - 20 * 10) = (nA - 200) Also, (n - 10)(A + 1) = (nA - 200) On solving, we get 10A - n = 190..........(i) When last 10 tests are excluded, decrease in total value of scores = (nA - 30 * 10) = (nA - 300) Also, (n - 10)(A - 1) = (nA - 300) On solving, we get 10A + n = 310..........(ii) From (i) and (ii), we get n = 60 Hence, 60 is the correct answer.&nbsp;

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