Quant Boosters by VP  Set 2

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

Q2) 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 = 256

Q3) A 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 = 16

Q4) In 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)
=> 37

Q5) The 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 12

Q6) Ram 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.

Q7) 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 : 18

Methanol > (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 D

Q8) If 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

Q9) By selling 1000 books at 50 for a rupee a man lost 40%.How many shall he sell for a rupee to gain 20% ?

Soln: 40% loss means sp=60% of cp
Now he wants 20% profit means sp = 120% of cp
so sp should be doubled, means instead of selling 50 books for a rupee, he should sell them 50 books for 2 rupees means 25 books/rupee.

Q10) A book was sold for a certain sum and thre was a loss of 20%. Had it been sold for Rs. 12 more, there would have been a gain of 30%. What would be the profit if the book were sold for Rs 4.8 more than what it was actually sold for ?

Soln:from 20% loss to 30 % profit, difference is +30 –(20) = 50%
50% difference = 12 rupees
so 100%=cp=24 rupees
now 20% loss on cp = 20% of 24=4.8 rupees lossIf book was sold at 4.8 more this loss would be compensated and finally no loss no profit.

Q11) A BMW and a nano are travelling along a straight road. BMW overtakes nano at 9am and continues travelling till 1pm and then it returns backward and meet nano at 2pm. At what time nano will reach the point where BMW turned back.