Quant Boosters by Anurag Singh Chauhan  Set 1

Q.1 find a natural number n for which 3^9+3^12+3^15+3^n is a perfect cube.
Solution: First of all we will try to find out cube
Now, 3^9+ 3^12 +3^15 +3^n = 3^9( 1 + 3^3 + 3^6 +3^n9 )
= (3^3)^3 [ 1+ 3.3^2 + 3.(3^2)^2 +(3^2)^3 + 3^(n9) – 3(3^2)^2]
= (3^3)^3[(1+3^2)^3]
Since, 3^(n9) – 3^5 = 0
3^(n9) = 3^5
n9 = 5
so, n = 14
hence given number is a pefect cube for n = 14
Q.2 if a, b, c, d, e are positive real numbers such that a+b+c+d+e = 8 and a^2 + b^2 + c^2 + d^2 + e^2 = 16 find range of e.
Solution: Concept  We know that these of type of questions can be solved by inequality.
If a, b, c, d are positive real numbers then [( a+ b+ c + d)/4]^2 ≤ (a^2 +b^2 + c^2 + d^2)/4 ... eqn(1) { by Techebycheff’s Inequality}
Now, a + b + c + d + e = 8
a + b + c + d = 8  e
a^2 + b^2 + c^2 + d^2 = 16 – e^2
putting above value in equation (1)
(8e)^2/ 16 ≤ ( 16 – e^2)/ 4
On solving this , e(5e – 16) ≤ 0
Using number line rule .
0 ≤ e ≤ 16/5
Q.3 find the period of f(x) = Sin^4 x + cos^4 x
Solution. Since Sin x and Cos x both has a period Pi, but f(x) is even function and Sin x and Cos x are complementary.
So f(x) has period = 1/2 {LCM of (Pi, Pi) } = Pi/2
Q. 4. How many values of x in 19 < x < 98 satisfy Cos^2 x + 2Sin^2 x = 1
Sol. This type of questions are frequently asked in mocks and MBA exams. Like CAT, MAT, GMAT.
Now we have, Cos^2 x + 2Sin^2 x = 1
We know , Cos^2 x + Sin^2 x = 1
On subtracting we get ,
Sin^2 x = 0
Sin x = 0 then x = nPi {general solution}
So, 19 < n Pi < 98
Dividing by Pi we get, 6.04 < n < 31.19
So, total number solutions equal to n = 38
Q.5 The cost of 3 Pizzas, 5Samosas, 1 order of milk ata a restaurant is 23.5 rupees . In the same restaurant the cost of 5 pizzas, 9 samosas, and 1 order of milk is 39.50 rupees. What is the cost of 2 pizzas, 2 samosas, 2 milk at restaurant.
Solution. Also these of questions often seen in management exams, Now we will discuss two approaches for this questions
Let, Pizza cost = P, Samosa cost = S, Milk cost = M
Now, Making equation
3P + 5S + M = 23.5 …… (A)
5P + 9S + M = 39.5 …… (B )
Multiplying equation (A) by 2 & subtracting equation (B )
P + S + M = 7.5
So required answer = 2* 7.5 = 15
Second approach  Solve equation (A) & (B ) by elimination method and get values of P &S . It is bit lengthy .
So this method will give you exact value of P , S & M . try yourself .
Q. 6 Find all Positive integers x &y satisfying 1/root(y) + 1/root(x) = 1/root(50)
Solution.
1/root(y) =  1/root(x) + 1/root(50) = (root(x)  root(50))/ root(50x)
Squaring both sides, 1/y = {x + 50 2*5 * root(2)x}/ 50x
Now L.H.S is rational so also R.H.S must be rational So root(2x) must rational now x e N
So 2x = (2a)^2
x= 2a^2
Also y = 2b^2
Now equation reduces ,
1/aroot(2) + 1/broot(2) = 1/5root(2)
1/a + 1/b = 1/5
5( a + b) = ab
5a + 5b – ab = 0
5a + 5b – ab + 25 – 25 = 0
a(5 – b) + 5( b – 5) = 25
(b5)(a5) = 25
a = 10 , b = 10 or (10, 10) and (30,6) and (6, 30)
Q.7 if a, b, c, d are the sides of a quadrilateral find the minimum value of (a^2 +b^2+c^2)/d^2 .
Solution these types of questions can be easily derived by help of inequality
We know that
(ab)^2 + (bc)^2 + (ca)^2 ≥ 0
2(a^2 +b^2 +c^2) ≥ 2 (ab+bc+ca)
Adding a^2+b^2+c^2 to both sides
3(a^2+b^2+c^2) ≥ (a^2+ b^2+c^2) + 2(ab+bc+ca)
3(a^2+b^2+c^2) ≥ (a+b+c)^2
Since sum of any three sides of an quadrilateral is greater than fourth sides
3(a^2+b^2+c^2) ≥ (a+b+c)^2 > d^2
(a^2+b^2+c^2)/d^2 > 1/3 so minimum value is 1/3
Q.8– Find x^5 + y^5 + z^5 if it is given that x+y+z=1 , x^2+y^2+z^2=2, x^3+y^3+z^3= 3 .
Solution these types of question are frequently asked in many competitive exam
Let us we solve it by cubic equation
Now we know if x,y,z are the roots of the equation
at^3+bt^2+ct+d = 0 , (a≠0)
we are given
x+y+z = 1 = (S1)
x^2+y^2+z^2 = S2
x^3+y^3+z^3 = S3
we know , S1= b/a
aS1+b = 0
a*1 + b = 0
b = a
(x+y+z)^2  2(xy+yz+zx) = (x^2+y^2+z^2)
1^2 – 2*c/a = S2
12c/a = 2
c= a/2
similarly by using identity
d = a/6
now putting values b,c,d in cubic equation
at^3 – at^2 – at/2 – a/6 = 0
dividing by a
t^3 t^2 t/21/6 = 0 equation………. (A)
6t^3 – 6t^2 – 3t – 1= 0
Multiplying both sides by t
6t^4 – 6t^3 – 3t^2 – t = 0
Putting t = x, y, z in succession and adding
6 S4 – 6 S3 – 3 S2 S1 = 0
So ,S4 = 25/6
Multiplying by t^2
6t^5 – 6t^4 – 3t^3 t^2 = 0
Putting t= x,y,z in succession and adding
6 S5 – 6 S4 – 3 S3 – S2 = 0
6 S5 = 36
S5 = 6
Q.9 Find all pairs of (a, b) of positive integers such that 2a + 7b divides 7a + 2b .
Solution: These type of questions can be solved by hit and trial method but this method gives you approximation value not exact values. So we will try to find by proper method
Now, let H be the H.C.F of (a, b)
So, let a = hx ,b =hy
Now, 2a+7b divides 7a+2b
Also 2x+7y divides 7x+2y
2x+7y divides 49x+14y
And 2x+7y divides 2(2x+7y)
Now 2x+7y divides 49x+14y – 4x14y i.e 45x
Similarly 2x+7y divides 45y
Hence 2x+7y divides h.c.f of 45x, 45y
So 2x+7y divides 45
2x+7y should be 9,15,45
Possible pair of x,y (1,1) (4,1) (19,1) (12,3) (5,5) so all solutions are (h,h) (4h,h) (19h,h) (12h,3h) (5h,5h)
Q 10 A leaf is torn from back of a book. The sum of remaining pages is 15000 . what are the page numbers on the torn leaf.
Solution – this question often asked in mocks and aptitude exams like CAT, MAT, GMAT etc..
Now, let the numbers of pages be n
Since , number of pages after a leaf is torn is 15000, then the sum of the numbers on the pages will be greater than 15000
Now, n(n+1)/2 > 15000
n (n+1) > 30000
(n+1)^2 > n (n+1) > 30000 > 173^2
So , 172 < n < 176
n could be 173, 174, 175
n = 173 then n(n+1)/2 = 173*174/2 = 15051
Sum of the numbers on torn pages.
= 15051 – 15000 = 51
And this would be y + ( y +1)
= 2y + 1 = 51
So page numbers are 25, 26
If n = 174 the n(n+1)/2 = 15225
Sum on torn pages
15225 – 15000 = 225
Page numbers on torn pages are 112, 113
When = 175, n(n+1)/2 = 15400
Sum on the torn pages = 400 which is even . So not possible .
Now , we have to know that the smaller number on the torn pages should be odd. So, the numbers on the torn pages are 25 and 26 and the number of pages is 173.
Q.11 find all positive integers for which n^2 + 96 is perfect square
Solution these types of question can be asked in GMAT , CAT, NMAT etc.
We solve it by two approaches
Now suppose k is a positive integer such that
n^2+ 96 = k^2
k^2 –n^2 = 96
(kn) (k+n) = 96
Since (kn) , (k+n) must be both odd and even
But odd case is not possible
Only possible cases are..
kn = 2, k+n = 48
kn = 4, k+n = 24
kn = 6, k+n = 16
kn = 8 , k+n = 12
on solving these equations we get
pairs of k,n are… (25,23) (14,10) (11,5) (10,2)
second approach
positive solutions = (no of factors of 96 after dividing by 4) /2
= 2^3*3=total factors are 8
So positive solutions = 8/2=4
Q.12 Find all integral solutions of a^4 + b^4 + c^4 = 1991 + d^ 4.
Solution. In this type of question we would apply divisibility on L.H.S and R.H.S in most cases also these questions can be solved orally
Now, we will learn new concept.
Let n be an integer, When it is odd.
Let n^4 – 1 = ( n^2 1)(n^2 +1)
= (n1)(n+1)(n^2+1)
It must be divisible by 16 for any odd n.
Also (n1) , (n+1) are consecutive even integers and one of them is divisible by 4.
If n is even n^4 must be divisible by 16.
Thus n^4 gives remainder 0 and 1 .
Possible remainders of a^4 + b^4 + c^4 – d^4 when divided by 16 are { 1, 0, 1, 2, 3}
But 1991 gives remainder 7 when divided by 16 so no integral solution exist because L.H.S & R.H.S have different remainders