Category Archives: IITJEE Foundation mathematics

Algebra for IITJEE Mains: harder factors: Mathematics Hothouse

IIT JEE Foundation video lecture: Plane Geometry: From Mathematics Hothouse

How to solve equations: Dr. Vicky Neale: useful for Pre-RMO or even RMO training

Dr. Neale simply beautifully nudges, gently encourages mathematics olympiad students to learn to think further on their own…

A nice dose of practice problems for IITJEE Foundation math and PreRMO

It is said that “practice makes man perfect”.

Problem 1:

Six boxes are numbered 1 through 6. How many ways are there to put 20 identical balls into  these boxes so that none of them is empty?

Problem 2:

How many ways are there to distribute n identical balls in m numbered boxes so that none of the boxes is empty?

Problem 3:

Six boxes are numbered 1 through 6. How many ways are there to distribute 20 identical balls between the boxes (this time some of the boxes can be empty)?

Finish this triad of problems now!

Nalin Pithwa.

IITJEE Foundation Math and PRMO (preRMO) practice: another random collection of questions

Problem 1: Find the value of \frac{x+2a}{2b--x} + \frac{x-2a}{2a+x} + \frac{4ab}{x^{2}-4b^{2}} when x=\frac{ab}{a+b}

Problem 2: Reduce the following fraction to its lowest terms:

(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}) \div (\frac{x+y+z}{x^{2}+y^{2}+z^{2}-xy-yz-zx} - \frac{1}{x+y+z})+1

Problem 3: Simplify: \sqrt[4]{97-56\sqrt{3}}

Problem 4: If a+b+c+d=2s, prove that 4(ab+cd)^{2}-(a^{2}+b^{2}-c^{2}-d^{2})^{2}=16(s-a)(s-b)(s-c)(s-d)

Problem 5: If a, b, c are in HP, show that (\frac{3}{a} + \frac{3}{b} - \frac{2}{c})(\frac{3}{c} + \frac{3}{b} - \frac{2}{a})+\frac{9}{b^{2}}=\frac{25}{ac}.

May u discover the joy of Math! 🙂 🙂 🙂

Nalin Pithwa.

A cute illustrative problem on basics of inequalities: IITJEE Foundation Maths


Graph the functions f(x)=\frac{3}{x-1} and \frac{2}{x+1} together to identify the values of x for which we can get \frac{3}{x-1} < \frac{2}{x+1}

Also, confirm your findings in the above graph algebraically/analytically.

Also, then find the points of intersection of these graphs using a TI graphing calculator and also algebraically/analytically.


Just a suggestion, it helps to have TI graphing calculator, a bit pricey, but …it might help a lot…If you wish, you can grab one from the following:

Of course, there might be excellent freeware software packages that plot graphs, but I am strongly biased: I like the TI graphing calculator toy very much!! I go to the extent of claiming that it is better for a kid to have a nice graphing calculator like a TI than a smartphone! 🙂 🙂 🙂

Later on, I hope to show how a detailed study of graphs helps right from high school level. In fact, an immortal Russian mathematician, I. M. Gel’fand had written a book titled “Functions and Graphs” for high-school children. You can check if it is available in Amazon India or Flipkart or Infibeam, etc.

Oh, one more thing: I love this topic of inequalities because of several reasons. Just now, one of my students, Darpan Gajra asked me certain questions about inequalities when he tried to solve the above problem —- I think his questions were good, and I hope this little explanation that I gave him helps many of other students also.


Consider the given inequality: \frac{3}{x-1} < \frac{2}{x+1}.

You might be tempted to solve this very fast in the following way:

Just cross-multiply: So, we get 3(x+1)<2(x-1), which in turn means 3x+3<2x-2, that is, x<-5. This is the answer by “fluke”!!!

In fact, the problem requires a detailed solution as follows:

Firstly, let us see what is wrong with the above fast solution:

Consider any inequality a<b. Now, we know that only if x>0, then ax<bx. But, if x<0, then ax>bx. Now, in the present question, if you simply cross-multiply, you are not clearing assuming whether (x-1)>0 or (x-1)<0; (x+1)>0 or (x+1)<0.

Also, one simple yet, I call it a golden rule is: even though it is not mentioned explicitly in the question, when you solve the question, immediately write the big restrictions: x \neq 1, x \neq -1 as these values make the denominators of the two fractions zero. Develop this good habit.

So, now, coming to the solution:

\frac{3}{x-1} < \frac{2}{x+1}, where x \neq \pm 1

\frac{3}{x-1} - \frac{2}{x+1}<0



Case I: (x^{2}-1)>0: Multiplying both sides of the above fraction inequality by (x^{2}-1) gives us:

x+5<0, that is, x<-5, but also (x^{2}-1)>0 is a restriction.  So, it means that x<-5 AND (x-1)(x+1)>0.

Subcase Ia: (x+1)>0 and (x-1)>0, which gives, x>-1 and x>1, which together imply that x>1, but we need x<-5 also as a restriction/assumption. So, in this subcase Ia, solution set is empty.

Subcase Ib: x+1<0 and x-1<0, which in turn imply, x<-1 and x<1, that is, x<-1, but we need x<-5 also. Hence, in subcase Ib: x<-5 and x<-1, we get x<-5 as the solution set.

Case II: (x^{2}-1)<0: Multiplying both sides of \frac{x+5}{x^{2}-1}<0 by (x^{2}-1), we get:

(x+5)>0, hence, x>-5. But, we also need the restriction/assumption: (x^{2}-1)<0, which implies that (x+1)(x-1)<0.

Subcase IIa: (x+1)>0 and (x-1)<0, that is, -1<x<1

So, in subcase IIa, we have (x>-5) \bigcap (-1<x<1), that is, (-1<x<1).

Subcase IIb: (x+1)<0, and (x-1)>0, that is, x<-1 and x>1. But, this itself is an empty set. Hence, in subcase IIb, the solution set is empty.

Now, the final solution is \{ case I\} OR \{ case II\}, that is, \{ case I\} \bigcup \{ case II\}, that is, \{ x<-5\}\bigcup {-1<x<1}


a) Graph the function f(x)=\frac{x}{2} and g(x)=1+\frac{4}{x} together to identify the values of x for which \frac{x}{2}>1+\frac{4}{x}

(b) Confirm your findings in (a) algebraically.

More later,

Nalin Pithwa.

Are kids no longer learning multiplication tables in school?

This topic is very close to my heart and head. I had written a remark about this in an earlier blog article. 

Today, I suddenly found some “echoes” or “resemblances” to my views. Kindly let me re-blog, or share the opinions about this from today’s EETimes, authored by Max Maxfield, Designline Editor. The URL is pasted below:;

Thanks Max and EETimes !

From Nalin Pithwa.

System of Equations — IITJEE Foundation Maths

Some problems are presented below as warm-up exercises:

Question 1:

Solve : x+y = 15 and xy=36

Question 2:

Solve: x-y=12 and xy=85

Question 3:

Solve: x^{2}+y^{2}=74 and xy=35

Question 4:

Solve : x^{2}+y^{2}=185 and x+y=17

Some other standard set of systems of equations and techniques for solving them are presented below:

Any pair of equations of the form

x^{2} \pm pxy + y^{2}=a^{2}….Equation I

x \pm y = b…Equation II

where p is any numerical quantity, can be reduced to one of the cases in the warm-up exercises; for, by squaring II and combining with I, an equation to find xy is obtained, the solution can then be completed by the aid of equation II.

Example 1:

Solve x^{3}-y^{3}=999…Equation 1

and x-y=3…Equation 2.

By division, x^{2}+xy+y^{2}=333…Equation 3.

From Equation 2, x^{2} - 2xy + y^{2} = 9;

by subtraction, 3xy=324 and xy=108…Equation 4.

From Equation (2) and (4), x=12, y=9 or x=9, y = -12

Example 2:

Solve: x^{4}+x^{2}y^{2}+y^{4}=2613…Equation I

and x^{2}+xy+y^{2}=67…Equation 2

Dividing (1) by (2), x^{2}-xy+y^{2}=39…Equation 3.

From Equation (2) and (3), by addition, x^{2}+y^{2}=33; by subtraction xy=14, hence, we get the following: x = \pm 7, x = \pm 2, or y = \pm 2, y = \pm 7

Example 3:

Solve: \frac{1}{x} - \frac{1}{y} = \frac{1}{3}…Equation (1)

and \frac{1}{x^{2}} + \frac{1}{y^{2}} = \frac{5}{9}…Equation (2)

From (1), by squaring, \frac{1}{x^{2}} - \frac{2}{xy} + \frac{1}{y^{2}} = \frac{1}{9},

and by subtraction, \frac{2}{xy} = \frac{4}{9}

adding to (2), \frac{1}{x^{2}} + \frac{2}{xy} + \frac{1}{y^{2}} = 1

and hence, \frac{1}{x} + \frac{1}{y} = \pm 1.

Combining with (1), \frac{1}{x} = \frac{2}{3}, or -\frac{1}{3}; \frac{1}{y} = - \frac{1}{3}, or -\frac{2}{3}; so, the solutions are: x = \frac{3}{2}, -3 and y =3, -\frac{3}{2}.

The following method of solution may always be used when the equations are of the same degree and homogeneous.


Solve x^{2}+ xy + 2y^{2} = 74 …Equation 1

and 2x^{2} + 2xy + y^{2} = 73…Equation 2.

Put y = mx, and substitute in both equations. Thus,

x^{2}(1+m+2m^{2}) =74…Equation 3.

x^{2}(2 + 2m + m^{2}) =73…Equation 4.

By division, \frac{1+m+2m^{2}}{2+2m+m^{2}} = \frac{74}{73}

73 + 73m + 146m^{2} = 148 + 148m + 74m^{2}

Hence, 72m^{2} -75m  - 75=0; or 24m^{2} - 25m - 25 = 0; hence, (8m + 5)(3m - 5) = 0; so m = -\frac{5}{8} or \frac{5}{3}.

Case (i):

Take m= -\frac{5}{8} and substitute in either (3) or (4).

From (3): x^{2}(1- \frac{5}{8} + \frac{50}{64}) = 74

x^{2} = 64, so x = \pm 8; so y = mx = - \frac{5}{8}x = \mp 5.

Case (ii):

Take m = \frac{5}{3}, then from (3) x^{2}(1 + \frac{5}{3} + \frac{50}{9}) = 74, so we get x^{2} = 9, so x = \pm 3; and hence, y = mx = \frac{5}{3}x = \pm 5.

When one of the equations is of first degree and the other of a higher degree, we may from the simple equation, find the value of one of the unknowns in terms of the other, and substitute in the second equation.


Solve 3x - 4y = 5…Equation (1)

and 3x^{2} - xy - 3y^{2} = 21…Equation (2)

From (1), we have x = \frac{5+4y}{3}

and substituting in (2), 3\frac{(5+4y)^{2}}{9} - \frac{y(5+4y)}{3} - 3y^{2}= 21

75 + 120y +48y^{2} - 15y - 12y^{2} - 27y^{2} = 189

9y^{2} + 105y -114 = 0

3y^{2} + 35y - 38 = 0

(y-1)(3y+31) = 0

Hence, y = 1, -\frac{38}{3}, and x = 3, -\frac{137}{9}.

The examples we have presented above will be sufficient as a general explanation of the methods to be employed; but, in some cases, special artifices are necessary.


Solve: x^{2} + 4xy + 3x = 40 - 6y -4y^{2}…Equation (1)

and 2xy - x^{2} = 3…Equation (2)

From (1) and (2), we have x^{2} + 4xy + 4y^{2} + 3x + 6y = 40;

that is, (x+2y)^{2} + 3(x+2y) - 40 = 0;

or, ((x+2y)+8)((x+2y)-5) = 0, hence, x+2y = -8, 5


Combining x+2y =5 with (2), we obtain 2x^{2} - 5x + 3=0, so x = 1, \frac{3}{2}; and by substituting in x + 2y = 5, we get y = 2, -\frac{7}{4}.


Combining  x+2y = -8 with (2), we obtain 2x^{2} + 8x + 3 = 0; hence, x = \frac{-4 \pm \sqrt{10}}{2}, and y = \frac{-12 \mp \sqrt{10}}{4}


Solve: x^{2}y^{2} - 6x = 34 - 3y…Equation (1)

and 3xy + y = 2((9+x)…Equation (2)

From (1), we have x^{2}y^{2} -6x + 3y = 34;

From (2), we have 9xy - 6x + 3y = 54;

By subtraction, x^{2}y^{2} - 9xy + 20 = 0; that is, (xy - 5)(xy -4) = 0; so xy = 5, 4.


Substituting xy = 5 in (2), gives y -2x = 3. From these equations, we obtain x =1 or -\frac{5}{2}; y = 5, or -2.


Substituting xy =4 in (2), gives y - 2x = 6. From these equations, we obtain x = \frac{-3 \pm \sqrt{17}}{2}, and y = 3 \pm \sqrt{17}.

Hope you enjoyed it,

Nalin Pithwa.










Some problems for IITJEE Foundation mathematics

Problem 1.

A cistern can be filled by two pipes in 33\frac{1}{3}; if the larger pipe takes 15 minutes less than the smaller to fill the cistern, find in what time it will be filled by each pipe singly.

Problem 2.

By rowing half the distance and walking the other half, a man can travel 24 km. on a river in 5 hours with the stream, and in 7 hours against the stream. If there were no current, the journey would take 5\frac{2}{3} hours; find the rate of his walking, and rowing and the rate of the stream.

Problem 3.

Factorize: 2a^{2}x^{2}-2(3b-4c)(b-c)y^{2}+abxy

Problem 4:

Find the fourth root of 81x^{4}-216x^{3}y+216x^{2}y^{2}-96x^{3}y+16y^{4}

Problem 5:

Find the sixth root of


More interesting stuff on IITJEE foundation maths later,

Nalin Pithwa




A little note : IITJEE foundation maths


Solve 2(x^{2}-6)=3(x-4)


We have 2x^{2}-12=3x-12

that is, 2x^{2}=3x call this as Equation (1)

Transposing, 2x^{2}-3x=0


Hence, x=0, or 2x-3=0

Thus, the roots are 0 and \frac{3}{2}


In equation (1), above, we might have divided both sides by x and obtained the simple equation 2x=3, hence, x=\frac{3}{2}, which is one of the solutions of the given equation. But, the student must be particularly careful to notice that whenever an x, or a factor containing x, is removed by division from every term of an equation, it must not be neglected, since the equation is satisfied by x=0, which is therefore, one of the roots.

More on IITJEE foundation maths later,

Nalin Pithwa