Quant Boosters  Learn Quest  Set 4

N is a positive integer not more than 100. If 7^N + N^3 has a unit digit of 0, how many values can N take ?
7^n ends with 7, 9, 3, 1
7^(4k+1) > 7
7^(4k+2) > 9
7^(4k+3) > 3
7^(4k) > 1X^3 must end with 3, 1, 7 or 9
Case (i) 7^(4k+1) =>7
So we need
7 + 3 = 0
I.e x^3 ends with 3
And it is possible only when last digit is 7
here X should be of the form (4k+1)
So 17,37,57,77,97
5 possibilitiesCase (ii)
9 + 1
=> 7^(4k+2) > 9
X is even
And (even)^3 never ends with 1Case(iii)
7^(4k+3) > 3
And we need x^3 > 7
Which is possible when last digit is 3
As 3^3 => 27
So we need to take x of the form (4k+3) whose last digit is 3
So, 3,23,43,63,83
5 possibilitiesCase (iv)
7^(4k)> 1
So we need x^3 > 9
So X should end with 9
But it says X is of the form 4k
So can't be possibleSo overall 10 possibilities
In triangle PQR, PS and QT are medians to sides PR and QR respectively. If PS = 16 cm and QT = 24 cm and PS is perpendicular to QT then find the area of triangle PQR.
We know PS and QT are two Medians and they are passing through a point and we know that point is called Centroid
Now best part of centroid : It divides Triangle into 6 equal parts
Now if some how I can calculate the area of PGT, I can calculate Area of whole triangle, just multiplying it by 6
For that additional information is given to us
that is PS is perpendicular to QT
I.e PGT is a right angled Triangle
So we just need base and height
I.e length of GT and PG
For that we know the property
centroid divides the median in the ratio 2:1
So GT => 1/3 of QT => 8
And PG => 2/3 of PS => 32/3
So area of Triangle PGT =>
1/2 * (32/3) * 8
=> 128/3
And area of whole Triangle
(128/3 ) * 6 = 256A student was asked to find the sum of first n natural numbers. By mistake he missed one of the numbers in between and he got the average as 30 (7/29). Find the number he forgot to add
Avg now = 30.xx
Step 1: number of numbers will always be approx 2 * avg.
So number of numbers = 2 * 30 = near to 60
Avg= 877/29Step 2: number of numbers(after forgetting) will always be multiple of denominator.
So 29 * 2 = 58 after forgetting.
And he forgot 1, means earlier there were 58 + 1=59 numbers.Step 3:
Now sum of 59 numbers would be 1 + 2 + ... + 59 = 59 * 60/2 = 59 * 30 = 1770
Here sum for 58 numbers(after removing 1) 877 * 2= 1754Final step:
One number removed = 1770  1754 = 16In how many ways can 18 identical balls be distributed among 3 identical boxes ?
A + B + C = 18
Total solution > 20c2 => 190
Now remove cases when two of them are equal
2a + b = 18
(0,..9)
10 cases
Which include one case (6, 6 * 6)
So overall cases 9
And these cases can be arranged in 3!/2! > 3 ways
9 * 3 = 27 ways
Now (6,6,6) > 1 case
So
total number of cases
(190  27  1 )/6
=> 27 + 9 ( cases when two are equal ) + 1 ( 6,6,6)
=> 37The product of the number of diagonals of two distinct polygons with sides N1 and N2 is 40. What is the sum of N1 and N2.
Product of diagonals = 40
So possibile cases
1 * 40
2 * 20
4 * 10
5 * 8
Now case (I) can't be possible
I.e 1 diagonal can't be possible
Now we know
Number of diagonals is given by n(n3)/2
n(n3)/2 = 2
n = 4
n(n3)/2 = 20
n(n3) = 40
n = 8
Case (ii) is possible
Now case (iii)
n(n3)/2 = 4
Not possible
Case (IV)
n(n3)/2 = 5
n = 5
n(n3)/2 = 8
Not possible
So only possibility
Case (ii)
I.e polygons of side 4 and 8
So 12Ram bought a few mangoes and apples spending an amount of at most Rs.2000. If each mango costs Rs.4 and each apple costs Rs.6, and Ram bought at least one fruit of each type, how many different possible amounts could he have spent in purchasing the fruits?
4x + 6y < = 2000
2x + 3y < = 1000
atleast 1, so give 1,1 value
2x + 3y < = 1000  2  3
2x + 3y < = 995
Rhs represents amount.
Question is asking different possible values of amount. That can be anything 0 to 995 now.
x, y= 0. Amount = 0
x=1,y=0. Amount = 2
x=0,y=1. Amount = 3
x=2,y=0. Amount = 4
x=1,y=1. Amount = 5 and so on.
all amounts except amount 1 is possible
hence o to 995 => 996 values
Excluding 1, total 995 possible amount values.Beaker A and beaker B contain methanol, ethanol and phenyl in the ratio 1:3:2 and 2:1:5 respectively. Some parts of the solutions from beaker A and beaker B are thoroughly mixed and put into another beaker C.
Which of the following cannot be the ratio of methanol, phenyl and ethanol in beaker C?
(a) 10 : 23 : 15
(b) 7 : 15 : 16
(c) 6 : 13 : 13
(d) 9 : 20 : 18Methanol > (x/6 + 2y/8) => (4x +6y)/24
Phenyl > (2x/6 + 5y/8) =>
(8x +15y)/24
Ethanol > (3x/6 + y/8)
=> (12x + 3y)/24
Required ratio >
(4x +6y) : (8x +15y) : (12x +3y)
Now let's check for the possibilities
Case (I)
X = Y = 1
10: 23: 15
Option (1) possible
Similarly
Case (ii)
X = (1/4) , Y = (1/15)
Option (2) is possible
Case (iii)
X = 1/2 and y = 1/6
Option (3) is possible
So option DIf abcde is a 5 digit number divisible by 6 and a > b > c > d > e, then how many such 5 digit numbers exist ?
a + b + c + d + e = 3k and e = even and since a, b, c, d are smaller than e hence e can be 6/8
Case: e=6
We need a + b + c + d = 3k form
(1, 2, 4, 5) is a possibility.Case: e = 8
Now a + b + c + d = 3k + 1 form
we need to choose 4 numbers from (1, 2, 3, 4, 5, 6, 7) which is of the form 3k + 1
Now 1, 4, 7 = 3k + 1
2, 5 => 3k + 2
3, 6 => 3k3k + 1 can be made by two 3k + two 3k+2 => 1 way
one 3k + one 3k+2 + two 3k+1 => 2c1 * 2c1 * 3c2 = 12 waysTotal=14 ways

number of numbers will always be approx 2 * avg.
This 2 * avg is valid everywhere or only when [1,n] is given?

@gautamarya
sum of first n natural numbers = n(n+1) / 2
Average = n(n+1) / 2n = (n + 1)/2
2 * Average = (n + 1) ~ n
So this should be in [1, n]