Algebra plays a crucial role in the CAT, as an average of 35–40% of the 'Quantitative Aptitude' section's problems are algebra-related, and the questions posed from this section are not particularly calculative and can be rapidly solved using options. According to question papers from previous years, despite being algebraic, the queries in this section are extremely logical.
For this, you must first master the fundamentals, then solve a multitude of problems, so that you are familiar with the variety and diversity of problems.
Remainder Theorem
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If a polynomial (i.e., a rational integral function) f(x) is divided by (x – a), then the remainder is f(a), which is obtained by substituting a for x in f(x).
- If a polynomial f(x ) is divided by (x + a ), then remainder is f(−a)
- If a polynomial f(x ) is divided by (x − a ), then remainder is f(a)
- If a polynomial f(x) is divided by (ax + b), then the remainder is f(\(-\frac{b}{a}\))
Factor Theorem
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If f(x) be a polynomial (i.e., rational integral function) in x
and f(a) = 0, then (x − a) is a factor of f (x).
By remainder theorem
if f(x) be divided by (x − a) the
remainder is f(a). But f(a) = 0, i.e., there is no remainder.
∴ f(x) is exactly divisible by (x − a)
Hence, (x − a) is a factor of f(x).
- If f(x)be a polynomial and f(−a) = 0,then (x+a) is a factor of f(x).
- If f(x) be a polynomial and f(\(-\frac{b}{a}\)) = 0,then(ax + b)is a factor of f(x).
HCF and LCM of Polynomials
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- A polynomial D(x) is a divisor of the polynomial P(x) if it is a factor of P(x). Where Q(x) is another polynomial such that P(x) = D(x)⋅Q(x)
- HCF/GCD (Greatest Common Divisor): A polynomial h(x) is called the HCF or GCD of two or more given polynomials, if h(x) is a polynomial of highest degree dividing each one of the given polynomials.
- LCM (Least Common Multiple): A polynomial P(x) is called the LCM of two or more given polynomials, if it is a polynomial of smallest degree which is divided by each one of the given polynomials.
For any two polynomials P(x) and Q(x).We have :
P(x).Q(x) = [HCF of P(x) and Q(x)] × [LCM of P(x) and Q(x)]
Important Formulae
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- (a + b)2 = a2 + b2 + 2ab
- (a − b)2 = a2 + b2 − 2ab
- (a + b)2 = (a − b)2 + 4ab
- (a – b)2 = (a + b)2 − 4ab
- (a – b)2 = \(\frac{1}{2}\)[ (a + b)2 + (a – b)2 ]
- a2 – b2 = (a + b)(a − b)
- (a + b)3 = a3 + b3 + 3ab(a + b)
- (a − b)3 = a3 − b3 − 3ab(a − b)
- a3 + b3 = (a + b)(a2 + b2 − ab)
- a3 − b3 = (a − b) (a2 +b2 + ab)
- (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ac)
- a3 + b3 + c3 − 3abc = (a + b + c)(a2 + b2 + c2 − ab − bc − ac)
- if a + b + c = 0, then a3 + b3 + c3 = 3abc
- a4 + b4 + a2b2 = (a2 + b2 + ab)(a2 + b2 − ab)
- a4 − b4 = (a − b)(a + b)(a2 + b2)
- a8 − b8 = (a − b)(a + b)(a2 + b2)(a4 + b4)
Basic facts to remember
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To factorize cyclic expressions, one must:
(a) Arrange the words in descending or ascending order by the power of one of the characters (or literals).
(b) combining two or three concepts, you can identify a common factor.
(c) Rewrite the elements of the other factor with the next letter's decreasing or increasing power.
(d) Repeat steps until all variables are identified.
Sigma (Σ) notation
Σx = x + y + z, for letters x, y, z
Σxy = xy + yz + zx
Σx2(y−z) = x2(y−z) + y2(z−x) + z2(x− y)
Some important results
- Σ(x−y) = 0
- Σ(x2 − y2) = 0
- Σ(x3 − y3) = 0
- Σx(y − z) = 0
- Σx2(y2 − z2) = 0
- Σx2(y − z) = − (x − y)( y − z)(z − x)
- Σx(y2 − z2 ) = (x−y)(y−z)(z−x)
- Σ(x−y)3 = 3(x−y)(y−z)(z−x)
Linear Equations
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An equation of the form ax + by = c, where a, b and c are real numbers is called a linear equation in two variables x and y.
The graph of a linear equation ax + by = c is a straight line.
The value of the variables that satisfy the equation is called the solution (or solution set) of the equation.
Key Features of Linear Equation with Two Variables:
(i) The graph of an equation of the form x = k (where k is a constant) is a straight line parallel to the y-axis k units away from the y-axis.
(ii) The graph of a y = k equation (where k is a constant) is a straight line parallel to the x-axis k units away from the x-axis.
(iii) The solution to the two equations is found at the site of intersection of the two lines.
(iv) the number of solutions to a single linear equation with two variables is infinite.
Simultaneous Linear Equations
Two or more linear equations in two variables form a system of linear simultaneous equations.
e.g., a1 x + b1 y = c1 and a2 x + b2 y = c1
Consistent System: A system is said to be consistent if it has at least one answer and is made up of two simultaneous linear equations.
Inconsistent System: A system with two simultaneous linear equations that does not have a solution is said to be inconsistent.
If two equations, a1 x + b1 y = c1 and a2 x + b2 y = c2, are true at the same time, then
\(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\) \(\implies\) Unique Solution (Independent Equation)
\(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\) \(\implies\) Infinite solution (Dependent Equation)
\(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\) \(\implies\) No Solution (Inconsistent Equation)
Properties of graphs of a1 x + b1 y = c1 and a2 x + b2 y = c2
- intersecting if the system has a unique solution.
- coincident if the system has infinite number of solution
- parallel if the system has no solution.
Algebraic Methods of Solving Simultaneous Equations in Two Variables
Ways to solve simultaneous equations with two variables:-
- Elimination Method
- Substitution Method
- Comparison Method
- Elimination Method
- Cramer’s Rule (Cross multiplication method)
Previous Year CAT Questions
Ques 1: When you reverse the digits of the number 13, the number increases by 18. How many other two digit numbers increase by 18 when their digits are reversed? (CAT 2006)
- 5
- 6
- 7
- 8
- 10
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Solution: (B)
Let the two digit number be 10x + y, x,y>0 and x,y<10
Reversing the digits the number becomes = 10y + x
|10y + x - 10x - y| = 18
:. 9 |y- x| = 18
:. ly - x| = 2
Thus y and x can be (1, 3), (2, 4), (3, 5), (4, 6), (5, 7), (6, 8) and (7, 9)
Other than 13, there are 6 such numbers.
Ques 2: A confused bank teller transposed the rupees and paise when he cashed a cheque for Shailaja, giving her rupees instead of paise and paise instead of rupees. After buying a toffee for 50 paise, Shailaja noticed that she was left with exactly three times as much as the amount on the cheque. Which of the following is a valid statement about the cheque amount? (CAT 2007)
- Over Rupees 13 but less than Rupees 14
- Over Rupees 7 but less than Rupees 8
- Over Rupees 22 but less than Rupees 23
- Over Rupees 18 but less than Rupees 19
- Over Rupees 4 but less than Rupees 5
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Solution: (D)
Let the amount on Shailaja's cheque be Rs. x and paise y = (100x + y) paise
The teller gives her (100y + x) paise
Now, 100y + x - 50 = 3(100x + y)
97y - 299x = 50
y = (50 + 299x)/97 = [50 + 8x + 291x]/97 = [(50 + 8x)/97] + 3
(50 + 8x) has to be a multiple of 97, y 99
50 + 8x = 97k
x = 12k - 6 + [(k - 2)/8]
At K = 2,10, 18..
x = 18, 115, 212…..
x = 18 is the only possible value.
So at x= 18, y = 5
The amount on Shailaja's cheque is over Rs. 18 but less than Rs. 19.
Ques 3: In an examination, there were 75 questions. 3 marks were awarded for each correct answer, 1 mark was deducted for each wrong answer and 1 mark was awarded for each unattempted question. Rayan scored a total of 97 marks in the examination. If the number of unattempted questions was higher than the number of attempted questions, then the maximum number or correct answers that Rayan could have given in the examination is: (CAT 2022 Slot 2)
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Solution: Let the number of correct, wrong and un-attempted questions be c, wand u respectively.
Since u > c + w
=u > 75 - u
= u > 37.5
.:. Least possible value of u is 38
=c+w+u = 75 ...(1)
=3c -w+ u = 97 ... (2)
(2) + (1)
=> 4c + 2u = 172
2c + u = 86
2c = 86 - u
c will be greatest when u is least i.e., u = 38
2c = 86 - 38 = 48
c= 24
Maximum value c can take is 24.
Ques 4: The donation box can receive notes of only Rs. 100, Rs. 250 and Rs. 500. One good day, the donation box was found to contain exactly 100 notes amounting to a total of Rs.15250. Then the maximum possible number of notes of Rs.500 that the donation box may have contained is: (CAT 2022 Slot 3)
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Solution: Let the number of 100, 250 and 500 notes be a, b and c respectively.
a+ b+ c = 100 ……… (i)
100a + 250b + 500c = 15250
2a + 5b + 10c = 305 ………... (ii)
Multiplying 2 with Equation (i) and subtracting from Equation (ii)
3b + 8c = 105
c = \(\frac{105-3b}{8} = \frac{104+ 1- 3b}{8}= 13 + \frac{1-3b}{8}\)
when b is minimum, c is maximum. The least value b can take is 3 for c to be an integer.
c = \(13 + \frac{1-9}{8} = 12\)
Ques 5: A basket of 2 apples, 4 oranges and 6 mangoes costs the same as a basket of 1 apple, 4 oranges and 8 mangoes, or a basket of 8 oranges and 7 mangoes. Then the number of mangoes in a basket of mangoes that has the same cost as the other baskets is __ (CAT 2021 Slot 1)
- 13
- 12
- 11
- 10
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Solution: (A) Let the cost of each apple, orange and mango be ‘a’, ‘o’ and ‘m’ respectively
According to the question
:: 2a + 4o + 6m = a +40 + 8m = 8o + 7m
2a + 4o + 6m = 80 + 7m
4o + m = 2a ...(1)
As, a + 40 + 8m = 80 + 7m
40 - m = a ….(2)
Solving (1) and (2)
80 = 3a and m = a/2
Value of each basked in terms of mangoes
2a + 4o + 6m
= 4m + 3a/2 + 6m
= 10m + 3/2 x 2m
= 10m + 3m
= 13m
Value of each basket is same as value of 13 mangoes.
Ques 6: How many three-digit numbers are greater than 100 and increase by 198 when the three digits are arranged in the reverse order?(CAT 2021 Slot 1)
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Solution: Let the three-digit number be 'abc'.
Based on the question:-
'abc' + 198 = 'cba'
100a + 10b + c + 198 = 100c + 10b + a
99c – 99a = 198
c – a = 2
:. Value of (c, a) can be (9, 7) or (8, 6) or (7, 5) or (6, 4) or (5, 3) or (4, 2) or (3, 1) i
For each of these 7 values of c and a, b can taken from 0 to 9.
: possible numbers will be = 7x 10 = 70.
Ques 7: A box has 450 balls, each either white or black, there being as many metallic white balls as metallic black balls. If 40% of the white balls and 50% of the black balls are metallic. then the number of non-metallic balls in the dox Is: (CAT 2021 Slot 2)
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Solution: 250
Let total number of black balls and white balls be 100b and 100w respectively.
So,100w + 100b = 450
w+ b = 4.5 ...(1)
As given in the question:-
Number of metallic balls for white and black balls are equal
White metallic = 40% of 100w = 40w
Black metallic = 50% of 100b = 50b
Now, 40w = 50b
w= 1.25b ...(2)
Putting Eqaution (2) in (1) we get
1.25b + b = 4.5
b= 2 and w = 2.5
:: non-metallic white balls = 60% of 100w = 150
non-metallic black balls = 50% of 100b = 100
: Total non-metallic balls = 150 + 100 = 250
Ques 8: A gentleman decided to treat a few children in the following manner. He gives half of his total stock of toffees and one extra to the first child, and then the half of the remaining stock along with one extra to the second and continues giving away in this fashion. His total stock exhausts after he takes care of 5 children. How many toffees were there in his stock initially? (CAT 2020 Slot 1)
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Solution: 62
Let there be x toffees after 4th child.
As the man gives half of x to 5th child,he will be left with x/2 toffees.
After giving 1 chocolate to 5th child, he is left with Zero toffees
x/2 - 1 = 0
Similarly, he will be left with 6 toffees after 3rd, 14 toffees after 2nd and 30 toffees after 1st and 62 toffees at the beginning.
Ques 9: Aron bought some pencils and sharpeners. Spending the same amount of money as Aron, Aditya bought twice as many pencils and 10 less sharpeners. If the cost of one sharpener is & 2 more than the cost of a pencil, then the minimum possible number of pencils bought by Aron and Aditya together is: (CAT 2020 Slot 2)
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Solution: Let the price of a pencil = x
Price of a sharpener = x + 2.
Aron bought ‘p’ no of pencils and ‘s’ no of sharpeners.
Aditya buys ‘2p’ pencils and ‘s – 10’ sharpeners.
:: px + s(x + 2) = 2px + (s - 10) (x + 2)
px + sx + 2s = 2px + sx + 2s - 10x - 20
20 = px - 10x
x(p - 10) = 20
the least value of p can be 11 to keep the value of x positive.
As Aron and Aditya bought 3p pencils together
So 3p = 3 x 11 = 33.
Ques 10: Let k be a constant. The equations kx + y = 3 and 4x + ky = 4 have a unique solution if and only if__ . (CAT 2020 Slot 3)
- |k| = 2
- k = 2
- I k| \(\neq\) 2
- k \(\neq\) 2
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Solution: (C)
To have a unique solution
\(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)
so, \(\frac{k}{4} = \frac{1}{k}\) ( From the given equations)
k2 = 4
K = \(\pm\) 2 or |k| \(\neq\) 2
Ques 11: If x and y are integers, then the equation 5x + 19y = 64 has (CAT 2014)
- No solution for x < 300 and y < 0
- No solution for x > 250 and y > −100
- A solution for 250 < x ≤ 300
- A solution for −59 < y < −56
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Solution: (C)
if y = 1, x=9
if y changes (increases or decreases) by 5, x will change (decrease or increase) by 19.
if x = 256, we get y = 64
Using these values, we see options (a), (b) and (d) are eliminated and also that there exists a solution for 250 < x ≤ 300.
Ques 12: For which of the following values of k will 3x + (k + 3)y = 1 and kx + 6y = 4 have a unique solution? (CAT 2014)
- 3
- −6
- 6
- Any value except 3 and −6
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Solution:
As the system has a solution, the coefficients are not proportional.
\(\frac{3}{k} \neq \frac{k+3}{6}\)
k(k+3) \(\neq 18\)
If k(k + 3) = 3 (6) or (−6)(−3), i.e. if k = 3 or −6, the system does not have a unique solution.
Hence, any value other than k = 3 and k = − 6 will result in a unique solution.
Ques 13: Which one of the following conditions must p, q and r satisfy so that the following system of linear simultaneous equations has at least one solution, such that p + q + r ≠ 0 (CAT 2003)
x+2y−3z = p 2x + 6y − 11z = q
x − 2y + 7z = r
- 5p−2q−r = 0
- 5p+ 2q+ r = 0
- 5p+ 2q−r = 0
- 5p-2q+r = 0
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Solution: (D)
Going by the choices,
5 p − 2q − r
= (5 x + 10y − 15z ) − (4 x + 12 y − 22z ) − ( x − 2y + 7z ) = 0
Only option (A) satisfies the condition.
Ques 14: If x and y are integers, then the equation 5x + 19y = 64 has (CAT 2003)
- no solution for x < 300 and y < 0
- no solution for x > 250 and y > −100
- a solution for 250 < x < 300
- a solution for −59 < y < −56
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Solution: (C)
5x + 19 y = 64
At x = 256, we get y = −64
So at y=-64 and x= 256 solution can be found for the above equation.
Ques 15: If x, y and z are real numbers, such that x + y + z = 5 and xy + yz + zx = 3. What is the largest value that x can have? (CAT 2002)
- \(\frac{5}{3}\)
- \(\sqrt 19\)
- \(\frac{13}{3}\)
- None of these
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Solution: (B)
(x+ y+z)2 =x2+ y2+z2 +2(xy+ yz+zx)
x2+y2+z2=19
At y = z = 0, the value of y is maximum.
As xy + yz + zx ≠ 0 both the values cannot be zero.
So, x2<19
X can be (!3/3) as x2 = (169/9)=18.8
How to approach Basic Basic Algebra and Linear Equation questions on CAT
- Go through the remainder and factor theorem and Important formulae before attempting questions.
- Students should be thorough with solving simultaneous equations.
- Logical approach to all the questions is necessary in all the exams.
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