Q2) A dishonest seller uses a weight of 700 gm in place of 1 kg and adds 15% impurities in pulse. What would be his profit percentage if he claims to be selling at cost price?
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RE: Question Bank  Arithmetic  Hemant Malhotra

Number Of Solutions For Equations Involving Difference/Sum Of Perfect Squares  Hemant Malhotra
Number of ways in which a natural number can be expressed as difference of two perfect square
let number = N = x^2  y^2
N = (x  y) (x + y)Case 1 :
let N = even number * even number ( product of 2 even number let 4 = 2 * 2)
so (x  y) (x + y) = even * even
so x  y = even, x + y = even
2x = even + even
so 2x = even (because sum of even + even = even))
so x = even/2 (which will be an integer)
so in this case we will get a integral value of x and yCase 2 :
N = even * odd
x^2  y^2 = even * odd
(x  y) (x + y) = even * odd
x  y = even
x + y = odd
2x = even + odd
2x = odd (because even + odd = odd)
so x = odd/2 ( odd number /2 will not be an integer so in this case we will not an integer value of x and y)Case 3 :
N = odd * odd
(x  y)(x + y) = odd * odd
x  y = odd
x + y = odd
2x = odd + odd
2x = even (because odd + odd = even as 3 + 3 = 6)
so x = even/2
which will give integral valueso This is just basic thing that when number is expressed as even * even or odd * odd then only we could find integral solutions and number could be expressed as difference of two perfect square number
Type 1 : Number which is a multiple of 4
Find number of ways in which 20 could be expressed as difference of two perfect square numbers
Case 1 :
x^2  y^2 = 20
(x  y)(x + y) = 20 = 1 * 20 = 2 * 10 = 4 * 5
case1  (x  y) (x + y) = 2 * 10 (even * even)
so we will find integral value here
x  y = 2
x + y = 10
so x = 6
y = 4Case 2 :
(x  y) (x + y) = 4 * 5 ( even * odd not useful for us ))
so total number of positive values of x and y is 1 (6, 4)
total solution = 4 * positive integral solutions = 4 * 1 = 4
so total number of solutions = 4Why we multiplied by 4 here ???
x = 6 and y = 4
so x^2 = 36 and y^2 = 16
so 36  16 = 20
but if x = 6 and y = 4 then also x^2 = 36 and y^2 = 16
if x = 6 and y = 4 (same case)
if x = 6 and y = 4 (same case)
so there will be 4 possible integral values.Direct Formula ( number which is multiple of 4)
Find the number of positive integer solutions of equation x^2 – y^2 = 20
solution: (No of factors after dividing by 4)/2.
Here number = 20 and N/4 = 20/4 = 5 and number of factors of 5 = (1 + 1) = 2
so ans = 2/2 = 1 ( so positive integer values=1 ) and TOTAL = 4 * 1 = 4Type 2 : Number which is a perfect square
x^2  y^2 = 16
(x  y) (x + y) = 16 = 1 * 16 = 2 * 8 = 4 * 4Case 1 :
(x  y) (x + y) = 1 * 16 ( odd * even case rejected)Case 2 :
(x  y)(x + y) = 2 * 8 (even * even case)
x  y = 2
x + y = 8
so x = 5 and y = 3
(5,3)Case 3 :
(x  y) (x + y) = 4 * 4
x  y = 4
x + y = 4
so x = 4 and y = 0
(4,0)so total positive solutions here = 1 only because (4,0) is not a positive solution
Now the other values (5,3) (5,3)(5,3)(5,3) (4,0)(4,0)
so total solutions = 6
Look here total number of solutions are not 4 * positive numbers but 4 * positive + 2Direct approach : {(No of factors on the right hand side of given after dividing by 4) – 1 }/2.
Example:
x^2  y^2 = 16
Divide 16 by 4 = 4
Number of factors of 4 ( 2^2 ) = 2 + 1 = 3
NOW ( 3  1) / 2 = 1
Total positive solutions = 1
Total Integral solutions = 4 * positive + 2 (in case of perfect square)
so 4 * 1 + 2 = 6 total integral solutionsPractice question : x^2  y^2 = 36 find number of total integral solution [ Answer is 6 ]
Type 3 : N = 4k+2 form
let number =18 = 4 * 4 + 2 form
x^2  y^2 = 18
(x  y)(x + y) = 1 * 18 = 2 * 9 = 3 * 6
so all cases are even * odd so there will be no solution for thisNOTE  Any number of 4k + 2 form can't be expressed as difference of two perfect square
So x^2  y^2 = 26 can't be expressed in this form for any integral values of x and yType 4 : When number is Odd
x^2  y^2 = 85
so (x  y)(x + y) = 1 * 85 = 5 * 17
so odd * oddCase 1 :
(x  y)(x + y) = 1 * 85
x  y = 1
x + y = 85
so x = 43
y = 42Case 2 :
(x  y)(x + y) = 5 * 17
x  y = 5
x + y = 17
x=11, y=6
so two positive solutions
and 2 * 4 = 8 total integral solutionsDirect Formula : (No of factors) /2
x^2  y^2 = 85
85 = 5 * 17 so factors = 2 * 2 = 4
so 4/2 = 2 (number of positive integral values =2)
and total = 2 * 4 = 8Practice Questions  Set 1
 x^2  y^2 = 124 ( find total number of integral solutions)
 Find number of ways in which 256 could be expressed as difference of two perfect square numbers .
 x^2  y^2 = 21 ( find number of positive integral solutions)
 x^2  y^2 = 122 ( find number of integral solutions)
 x^2  y^2 = 100 ( find the no of non negative integral solution)
Number of ways in which a natural number can be expressed as Sum of two perfect squares
Case 1  when N is not a perfect square and prime factors in the form 4k + 1
Example : x^2 + y^2 = 5^2 × 13^3
Here 5 and 13 are both are both of form 4k + 1
Number of positive integral solutions = Number of Factors of 5^2 × 13^3 = 12
Total integral solutions = 4 × 12 = 48Case 2 : when N is a perfect square
Example : x^2+y^2= 5^2 ×13^2
Here number is perfect square and 5 and 13 both are in 4n + 1 form so
Positive integral solutions = Number of Factors  1= 9  1 = 8
Total integral solutions= 4 × 8 + 4 = 36 (we will consider only 4k + 1 form of prime factors)Extra addition of 4 is for x = +/ 5 * 13 and y = +/ 5 * 13
Case 3 when x^2 + y^2 = 3^2 × 7^3
here 3 and 7 are 4k + 3 form and no 4k + 1 form
So number of integral solutions = zeroCase 4  x^2 + y^2 = 5^2 × 3^4
here 3 is 4k + 3 form and 5 is 4k+1 form
now 2 cases arise when 4k + 3 form has odd power then number of integral solutions =0
 when 4k + 3 form has even power then ignore that and find number of factors of 4k + 1 form
Example x^2 + y^2= 5^3 ×7^2
Here 7 is 4k + 3 form but power is even so ignore that now find factors of 5^3 which is 4.
Number of positive integral solutions = 4
and total = 4 × 4 = 16Case 5 : When number is of perfect square form and no 4k + 1 form also then
Number of integral solutions will be 4 onlyExample= x^2 + y^2 = 81 = 3^4
Here number of integral solutions is 4Practice Questions  Set 2
 x^2 + y^2 = 80
 x^2 + y^2 = 80
 x^2 + y^2 = 121
 x^2 + y^2 = 126
 x^2 + y^2 = 120

RE: Question Bank  Arithmetic  Hemant Malhotra
Q1) There is a clock that has a special way of telling the time. It does not have any hands or numbers on it, but it has a chimer. If the time is 1 o'clock, it chimes once. If the time is 2 o'clock, it chimes twice, and so forth. The time gap between any two chimes is 4 seconds. How many seconds would it take you to know the time, after the first chime is heard, if it is 11 o'clock?

RE: Question Bank (1)  Logical Reasoning  Hemant Malhotra
Q29) Four students  Kali, Ranu, Dia and Dev  were asked the names of the heads of four labs  Alpha, Beta, Theta and Gama  in a college. The heads of these four labs are Mr. Hari, Mr. Karan, Mr. Ram and Mr. Arjun, in no particular order. Ranu named Mr. Ram as Alpha's head and Mr. Arjun as Theta's head. Both Kali and Dia named Mr. Hari as Gama's head. Kali didn't name Mr. Karan as Theta's head. No two students named the same person as the head in case of lab Beta. Dia and Dev named the same person as the head in case of exactly two of the four labs. No other pair of students, apart from Dia and Dev, named the same person as the head in case of these two particular labs. The person, who was named as the head of a different lab by each student, was identified correctly by Dev and it was Dev's only correct answer. Ranu gave exactly two correct answers
Who is the head of lab Alpha?
a Mr. Ram
b Mr. Arjun
c Mr. Hari
d Mr. KaranFor which lab did none of the students give the name of the correct head?
a Alpha
b Beta
c Theta
d GamaWhich student did not give any correct answer?
a Dia
b Kali
c Either (a) or (b)
d Both (a) and (b)

RE: Question Bank  Geometry  Hemant Malhotra
Q1) Centroid of triangle is at (1,1) while its orthocenter is at (5,3) then circumcentre of triangle could be
(1) (1,3)
(2) (8/3 , 0)
(3) (0,8/3)
(4) (7/3,1/3) 
RE: Question Bank  Number Theory  Hemant Malhotra
Q1) If a! + (a+1)! has 19 zeroes at the end , what is the value of a
a) 77
b) 78
c) 79
d) 80
e) CBD 
Cauchy Schwarz Inequality  Hemant Malhotra
Cauchy Schwarz Inequality :
If A_{1},A_{2},A_{3}......A_{n} are n positive real number and B_{1},B_{2},B_{3}......B_{n} are n positive real numbers then
(A_{1}B_{1 } + A_{2}B_{2 + ... + }A_{n}B_{n})^{2 }≤ (A_{1}^{2} + A_{2}^{2} + A _{3}^{2}......A_{n}^{2} ) x ( B_{1}^{2} + B_{2}^{2 }+ B_{3}^{2} ... + B_{n}^{2})
If a^{2} + b^{2} + c^{2} = p^{2}+ q^{2} + r^{2} = 101, where a, b, c, p, q and r are all distinct real numbers, then which of the following inequalities is true?
1) ap + bq + cr < 99
2) ap + bq + cr < 101
3) ap + bq + cr < 202
4) ap + bq + cr < 200
5) None of theseApply CauchySchwarz Inequality
(a^{2} + b^{2} + c^{2} ) (p^{2}+ q^{2} + r^{2}) > = (ap + bq + cr)^{2}
so (101) × (101) > = (ap + bq + cr)^{2}
Now distinct
(101) ×(101) > (ap+bq+cr)^{2 } so (ap+bq+cr) < 101
15x + 20y = 375. what is the minimum value of (x^{2}+y^{2})^{1/2}
A) 17
B ) 14
C) 16
D) 15Method1 Apply Cauchy
(15 * x + 20 * y)^{2 } < = (15^{2 }+ 20^{2}) (x^{2 }+ y^{2})
375^2 < = (625) (x^{2} + y^{2})
(x^{2} + y^{2}) > = (375/25)^2
so (x^{2} + y^{2})^{1/2} > = 15
Method 2 Alternate 15x + 20y = 375 so 3x+4y = 75
we need min value of (x^{2}+y^{2})^{1/2 }which is nothing but distance of (x,y) from (0,0)
so basically we want to find perpendicular length to 3x + 4y = 75 from (0,0)
75/(3^{2}+4^{2})^{1/2 } = 15
If x^{2 }+ y^{2} =14x + 6y + 6, what is the largest and minimum possible value that 3x + 4y can have
Method 1 Cauchy theorem
x^{2 }+ y^{2}  14x 6y  6 = 0
(x7)^{2} + (y3)^{2} = 64
Now apply Cauchy
we want 3x + 4y
and we have value of (x7)^{2} and (y3)^{2}
so break 3x+4y in x7 and y3 form to apply Cauchy
( 3 * (x 7 ) + 4 * (y3))^2 < = ( 3^{2} + 4^{2}) * ( (x  7)^{2} + (y  3)^{2} )
(3x+4y33)^{2} < = (25) * (64)
40 < = (3x  4y  33) < = 5 * 8
40 < = 3x  4y 33 < = 40
7 < = 3x  4y < = 73
So minimum value is 7 and maximum is 73
Method 2 2nd approach to solve this kind of questions ( question involving equation of circle))
x^{2} + y^{2 } 14x  6y  6 = 0(x7)^{2} + (y3)^{2 }64 = 0
center = (7,3) and radius =8
now let 3x + 4y = k now this line will touch circle then only it will be maximum or minimum
perpendicular from center to line will be equal to radius
The length of perpendicular from a point (x1, y1) to a line ax + by + c = 0 is (ax1 + by1 + c)/ root(a^{2}+b^{2})
=  (3 * 7 + 4 * 3  k) / root(3^{2}+4^{2}) = 8
so 33k = 40
so 33k = + 40
so k = 7 or 73
Method 3 
x^{2} + y^{2 } 14x  6y  6 = 0 , let 3x+4y=k so x= (k4y)/3
put this value in equation  > (k4y/3)^{2}+y^{2}14((k4y)/36y6=0
now form quadratic in y and then D > =0 and u will find range of k
Method 4  x^{2} + y^{2 } 14x  6y  6 = 0
circle with centre 7,3 and radius 8
7 + 8 * cos x = X
3 + 8 * sin x = Y
3 (7 + 8 cos x) + 4 (3 + 8 sin x)
33 + 24 cos x + 32 sin x
33 +/ sqrt (24^{2 }+ 32^{2})
33 +/ 40 so 7 minimum and 73 maximum
Given that the real numbers a,b,c,d and e satisfy simultaneously the relations a+b+c+d+e = 8 and a^{2}+b^{2}+c^{2}+d^{2}+e^{2} = 16, determine the maximum and minimum value of a.
Method we want value of a so apply cauchy in other terms and form inequality in terms of a
so (b*1+c*1+d*1+e*1)^2 < = (b^{2}+c^{2}+d^{2}+e^{2})*( 1^{2}+1^{2}+1^{2}+1^{2})
(8a)^{2} < = (16a^{2})*4
so 5a^{2}16a < =0
a (5a16) < =0
0 < =a < =16/5

RE: Quant Boosters  Hemant Malhotra  Set 12
Method 1 :
y=x^2+x+3/(x^2+x+1))
yx^2+yx+y=x^2+x+3
x^2(y1)+x(y1)+y3=0
x real so D > =0
(y1)^24*(y1)(y2) > =0
y1)(y14y+12) > =0
(y1)(3y+11) > =0
(y1)(3y11)
so (1,11/3]Method 2 :
y=1+((2/(x^2+x+1))
now x^2+x+1=x^2+x+1/41/4+1
(x+1/2)^2+3/4
so min value is 3/4 and maax infinity
so y=1+2/inf
y=1
and y=1+2/(3/4))
y=1+8/3=11/3 
RE: Question Bank  Arithmetic  Hemant Malhotra
Q43) Ravi and Akash run a 2 km long race. Ravi beats Akash by 2 minutes. If Akash increases his speed by 2 km/hr and Ravi decreases his speed by 2 km/hr, Akash will beat Ravi in the same race by 2 minutes. What are the speeds of Ravi and Akash?
a) 15 km/hr, 12 km/hr
b) 10 km/hr, 8 km/hr
c) 8 km/hr, 6 km/hr
d) 12 km/hr, 10 km/hr 
RE: Question Bank (2)  Logical Reasoning  Hemant Malhotra
Q16) Four cities W, X, Y and Z together have 11 hotels.Five of which viz; P, Q, R, S and T, are fivestar hotels and the rest viz; A, B, C, D, E and F are threestar hotels.
(i) P and Q are situated in different cities and they are the only fivestar hotels in the cities in which they are situated.
(ii) E and F are the only hotels in city Y.
(iii)P is not situated in the city which has largest number of hotels.
(iv) Fivestar hotels are situated in more number of cities than threestar hotels.
(v) City Z has more number of fivestar hotels than W, which does not have any threestar hotels.
(vi) Each city has a different number of hotels.How many hotels are there in city Z?
How many cities have threestar hotels?
Which of the following conditions are sufficient to determine the number of hotels in city W?
A.Only (i) and (v)
B. Only (ii), (v) and (vi)
C.Only (i),(ii) and (v)
D.Only (i), (v) and (vi)

RE: Question Bank  Algebra  Hemant Malhotra
Q26) The quadratic polynomial f(x) = ax^2 + bx + c has integer coefficients such that f(1), f(2), f(3), and f(4) are all perfect squares of integers but f(5) is not. What is the value of a, b and c?

RE: Question Bank  Algebra  Hemant Malhotra
Q 3) Find minimum value of x + y + z when xyz = 45, where x, y, z are different positive integers

RE: Question Bank  Modern Math  Hemant Malhotra
Q3) In an examination, there are 100 questions. A student is awarded 12 marks for each correct answer.He loses 3 marks for each wrong answer and loses 2 marks for each unattempted question. The net score of Hari in that test is 625. The number of ways in which Hari could have attempted the exam is ?

RE: Question Bank  Arithmetic  Hemant Malhotra
Q25) Farook marks up the price of an article by 50% and then offers a discount of 20% to Shahrukh. Shahrukh sells it for Rs.20 more than the price at which he purchased it. If Shahrukh’s selling price is 30% more than the original cost price of the article, then Shahrukh’s profit percentage is?

Question Bank  Geometry  Hemant Malhotra
Number of Questions: 100
Topic: Geometry
Answer key available?: No
Source: Elite's Grid forum 
RE: Question Bank  Geometry  Hemant Malhotra
Q19) In a cyclic quadrilateral ABCD, if AB = 2, BC = 3, CD = 4 and DA = 5, what is the ratio of the lengths of the diagonals?
(1) 7 : 11
(2) 11 : 13
(3) 10 : 11
(4) 13 : 15 
Quant Boosters  Hemant Malhotra  Set 11
Number of Questions  30
Topic  Quant Mixed Bag
Solved ?  Yes
Source  Elite's Grid Prep Forum 
RE: Quant Boosters  Hemant Malhotra  Set 11
Q1) Find maximum number of real roots (+ve , ve) for the equations x^3 + 6x^2 + 11x  6 = 0

RE: Question Bank  Arithmetic  Hemant Malhotra
Q66) A dishonest trader marks up his goods by 80% and gives a discount of 25%. Besides he gets 20% more amount per kg from wholesaler and sells 10%less per kg to customer. What is the over all profit %

RE: Question Bank  Modern Math  Hemant Malhotra
Q24) There are ten 50 paise coins placed on a table. Six of these show tails and four show heads. A coin is chosen at random and flipped over (not tossed). This operation is performed seven times. One of these coins is then covered. Of the remaining nine coins, five show tails and four show heads. The covered coin shows:
a) A head
b) A tail
c) More likely a head
d) More likely a tail