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C) 0.05 miles/s

- Let the speed of the 2 dots be "a" (faster dot) and "b" (slower dot) miles/s respectively.
- When they move in opposite directions, \(\frac{1.2}{a+b}=15\)
- When they move in same direction, \(\frac{1.2}{a-b}=60\)
- Simplifying 2 equations, we get 15a + 15b = 1.2 and 60a - 60b = 1.2
- Solving 2 equations, 120a = 6 or a = 0.05 miles/s

E) 9

- Let the number of chocolates bought by one of the toddlers be "x" and the number of chocolates bought by second toddler be "y"
- Then, \(x^3 + y^3 = 189\)......... (I)
- and \(x^2 + y^2 - xy = 21\)..........(II)
- \(x^3 + y^3 = (x+y)(x^2 - xy + y^2)\)
- Thus by dividing equation (I) by equation (II) we will get the value of x + y = 9

7! onward all the terms will be divisible by 7, hence the summation of 1! till 6! should only give a remainder when divided by 7.

1! + 2! + 3! + 4! + 5! + 6! = 1 + 2 + -1 + 3 + 1 -1 = 5 is the OA

The highest score can be obtained by any of the student is +160 and minimum score can be obtained is -40, Hence the total number of scores can be obtained is 160+40 + 1 (zero) = 201

Now some of the scores are not possible such as 159, 158, 157, 154, 153 and 149, hence 201 - 6 = D) 195

It can be solved using the AP of common difference of 1, that becomes lengthy.

Let's observe a pattern here.

3 can be written in consecutive terms as 1+2 - 1 way , 5 = 2+3 - 1 way, 15 = 8+7, 1+2+3+4+5, 4+5+6 - 3 ways, we can see it is number of odd factors of the number -1

Hence 5! can only two odd factors 3 and 5, hence the number of ways should be 4-1 = 3 ways C)

5! = 120

39+40+41, 18+19+20+21+22, 1+ 2+ 3+ 4+ 5+ 6+ 7+ 8 + 9 +10 + 11+ 12+13+14+15

Also one more strategy can be applied has 120/odd number to get number of ways to find the consecutive terms

2 ways to solve the problem

Let's say A+B+C = 23
As every body should get 1 chocolate, hence assigning 1 to each, the equation becomes A+B+C = 20, and number of ways of distributing chocolates = 22C2

Now none should get more than 10, lets assign 11 to A, then the equation becomes A+B+C = 9, hence number of ways of distribution being 11C2 -> That many cases is when A gets more than 10 and similarly for all the people, the number of cases being 3*11C2

Hence number of ways being 22C2-3*11C2 D)

4

count digits from 1-2000...
1-9------9digits
10-99---90x2
100-99----900x3
1000-2000-1001x4
ans 6893

As per the equation:

X = DA + 19 if D is the divisor and A is the quotient.
5X = DB + 29 if B is the quotient.

5X = 5DA + 5*19 = 5DA + 95 but the remainder should be 29, hence 95-29 = 66 is the multiple of D. The factors of D are 1, 6,11, 66.

1, 6, 11 cannot be D as 19 and 29 are remainders, hence D should be 66

Hence 11X = 11DA + 19*11 = 11DA + 201. As D = 66, hence 11X should leave a remainder of (201/66) Remainder = 3

Correct Answer : C

The explanation is as follows :
The question doesnt state that the groups have to be equal. A group can have any number of girls between 1 and 9. So 12C3 x 9C3 x 6C3 wont be the answer. It can be solved using the following technique
- - - - - - - - - - - -
The 12 different girls can be arranged in a single line in 12! ways

To arrange them into 4 groups with atleast 1 girl in each lets consider markers which will divide the line into 4 parts. So we require 3 markers which can divide the line as follows

part1 Marker1 part2 marker2 part 3 marker3 part4

There are 11 spaces in the line of girls as shown above

-I-I-I-I-I-I-I-I-I-I-I-
The markers can be places in the above 11 places in 11C3 ways

Therefore the ways of dividing the girls into 4 groups is 11C3 X 12! i.e C

Correct Answer : D

The explanation is as follows :
The Balls are similar so the order or the arrangement will not matter. Only the ways the balls can be grouped is important. Also the balls have to be dropped into the bins so one scenario is that all the 14 balls are dropped into one bin and other 3 bins are empty
The balls can be placed as such
- - - - - - - - - - - - - -

To arrange them into 4 groups such that the no. of balls in any bin varies from 0-14.lets consider markers which will divide the line into 4 parts. So we require 3 markers which can divide the line as follows

part1 Marker1 part2 marker2 part 3 marker3 part4

So there are 14 balls and 3 markers

III and - - - - - - - - - - - - - -

In case the marker is placed as

III- - - - - - - - - - - - - -
we have all the 14 balls in the 4th bin and none in the other 3

When its like
II-I - - - - - - - - - - - - -
There are 13 balls in the 4th bin and 1 ball in the 3rd bin

So the ways of arranging these 17 items is \(\frac{17!}{(14!*3!)}\) as 14 balls and 3 markers are the same

In other words the balls can be dropped in 4 different bins in 17C3 ways which is the same as \(\frac{17!}{(14!*3!)}\)

There is one concept I want to discuss here:

Lets say A takes a days and B takes b days to complete the work.

Together they will take ab/(a+b) days to complete the work.

The difference between the days taken by A alone and days taken when A and B worked together = a - ab/(a+b) = a^2/(a+b)
Similarly difference of days in case of B = b^2/(a+b)

So if the difference is given, then the combined days can be obtained as sqrt (product of diffrence) = sqrt(a^2*b^2/(a+b)^2) = ab/(a+b)

Applying the formula, the combined number of days are sqrt(9*4) = 6 days

A number that can be expressed as a ratio of two numbers is rational. However, roots, \(\pi\), and endless non-repeating decimals are irrational numbers; they cannot be written in a ratio of two numbers.

(1) expression is not sufficient, need x
(2) expression is not sufficient, need y

(1) (2) expression reduces to \(\frac{7\pi}{22}\). \(\frac{22}{7}\) is commonly used to approximate \(\pi\), yet \(\frac{22}{7}\neq\pi\). Thus, \(\frac{x}{y}\) is not an integer.

The correct answer is C

(1) If \(t^t-1\) is not positive, then \(t\leq0\). \(t\) can be any odd negative integer or 0. Not sufficient
(2) If \(t! \neq 0\), then \(t\) can still be any positive integer. Not sufficient

(1) (2) If \(t^t -1\) is not negative and \(t! \neq 0\), then \(t=0\).
\(0^0 - 1 = 0\) and \(0! = 1\). Sufficient

Therefore, the correct answer is C.

A

X, Y & Z all Positive Integers and Z >1
Possible scenario's:- X Y Z
1 3 2
2 2 2
3 1 2
1 2 3
2 1 3
1 1 4
Favorable Scenarios = 3
Total Scenarios = 6
Thus probability = 3/6 = \(\frac{1}{2}\)

D

Max value of X can be 60.9
Max value of Y can be 65.9
Max value of X+Y can be 126.8
If we had the leeway to round off, the answer would be 127. But in this case we don't have such leeway, so the answer will be 126