To give CAT asirants a hands on experience on the variety of Algebra questions which frequently appear in CAT, we have listed around 60 questions practice on important topics from Algebra. All these algebra questions are with detailed explanations.Question 1: For the given pair (x, y) of positive integers, such that 4x-17y=1 and x<1000 how many integer values of y satisfy the given conditions? [1] 56 [2] 57 [3] 58 [4] 59
We first need to find out a solution for x & y. Once we get a solution, values of x would be in an AP with a common difference of 17 whereas values of y would be in an AP with a common difference of 4.
Valid Solutions:
x = 13, y = 3
x = 30, y = 7
x = 47, y = 11
.
.
x = 999, y = 235
No. of terms =\(\frac{{999 - 13}}{{17}} + 1 = \) = 58 + 1 = 59. Option D
Question 2: One year payment to the servant is Rs. 90 plus one turban. The servant leaves after 9 months and receives Rs. 65 and turban. Then find the price of the turban [1] Rs.10 [2] Rs.15 [3] Rs.7.5 [4] Cannot be determined
Payment for 12 months = 90 + t {Assuming t as the value of a turban}
Question 3: In CAT 2007 there were 75 questions. Each correct answer was rewarded by 4 marks and each wrong answer was penalized by 1 mark. In how many different combination of correct and wrong answer is a score of 50 possible? [1] 14 [2] 15 [3] 16 [4] None of these
Correct (c) + Wrong (w) + Not attempted (n) = 75
4c – w + 0n= 50
Adding the two equations we get
5c + n = 125
Values of both c & n will be whole numbers in the range [0, 50]
c (max) = 25; when n = 0
c (min) = 13; when n = 60 {Smallest value of ‘c’ which will take the marks from correct questions greater than or equal to 50}
No. of valid combinations will be for all value of ‘c’ from 13 to 25 = 13. Option D Online CAT 2021 Quant Course
Question 4: How many integer solutions exist for the equation 8x – 5y = 221 such that \(x \times y < 0\) [1] 4 [2] 5 [3] 6 [4] 8
We first need to find out a solution for x & y. Once we get a solution, values of x would be in an AP with a common difference of 5 whereas values of y would be in an AP with a common difference of 8.
Valid Solutions:
x = 32; y = 7
x = 37; y = 15
x = 42; y = 23
But we need the solutions where one variable is negative whereas the other one is positive. so, we will move in the other direction.
x = 27; y = -1
x = 22; y = -9
x = 17; y = -17
x = 12; y = -25
x = 7; y = -33
x = 2; y = -41
So, number of integer solutions where \(x \times y < 0\) is 6. Option C
Question 7: For how many positive integral values of N, less than 40 does the equation 3a – Nb = 5, have no integer solution [1] 13 [2] 14 [3] 15 [4] 12
If N is a multiple of 3, then the LHS would be divisible by 3 and RHS won’t be. Number of positive integral values less than 40 which are multiple of 3 = 13. Option A Online CAT 2021 Quant Course
Question 8: What are the number of integral solutions of the equation 7x + 3y = 123 for x,y > 0 [1] 3 [2] 5 [3] 12 [4] Infinite
Valid Solution:
x = 3; y = 34
x = 6; y = 27
.
.
x = 15; y = 6
Number of integral solutions such that x, y > 0 are 5. Option B
Question 9: The cost of 3 hamburgers, 5 milk shakes, and 1 order of fries at a certain fast food restaurant is $23.50. At the same restaurant, the cost of 5 hamburgers, 9 milk shakes, and 1 order of fries is \(\$ 39.50\). What is the cost of 2 hamburgers, 2 milk shakes, and 2 orders of fries at this restaurant? [1] 10 [2] 15 [3] 7.5 [4] Cannot be determined
Question 11: A shop stores x kg of rice. The first customer buys half this amount plus half a kg of rice. The second customer buys half the remaining amount plus half a kg of rice. Then the third customer also buys half the remaining amount plus half a kg of rice. Thereafter, no rice is left in the shop. Which of the following best describes the value of x? [1] 2 ≤ x ≤ 6 [2] 5 ≤ x ≤ 8 [3] 9 ≤ x ≤ 12 [4] 11 ≤ x ≤ 14
After first customer, amount of rice left is 0.5x – 0.5
After second customer, amount of rice left is 0.5(0.5x -0.5) – 0.5
After third customer, amount of rice left is 0.5(0.5(0.5x -0.5) – 0.5) – 0.5 = 0
Question 12: If p and Q are integers such that \(\frac{7}{10}<\frac{p}{q}<\frac{11}{15} \) , find the smallest possible value of q. [1] 13 [2] 60 [3] 30 [4] 7
The fraction lies in the range (0.7,0.733333)
We know that \(\frac{8}{{11}}\) = 0.727272.. is valid value.
The smallest value of q has to be less than or equal to 11. Only 7 fits in the range.
With a little hit and trial we get a valid value of \(\frac{p}{q}\) as \(\frac{5}{7}\)
Question 13: Given the system of equations \(\left\{ {\begin{array}{*{20}{c}}{2x + y + 2z = 4}\\{x + 2y + 3z = - 1}\\{3x + 2y + z = 9}\end{array}} \right. \), find the value of x+y+z. [1] -1 [2] 3.5 [3] 2 [4] 1
The given equations are
\(2x + y + 2z = 4 \ldots \left( 1 \right)\)
\(x + 2y + 3z = - 1 \ldots \left( 2 \right)\)
\(3x + 2y + z = 9 \ldots \left( 3 \right)\)
Take the first and the second equation :
\(\begin{array}{*{20}{c}}{2x + y + 2z = 4}\\{x + 2y + 3z = - 1}\\{}\end{array}\) multiply equation 2 by -2 , thus the 2 equations we get after multiplying are \(\begin{array}{*{20}{c}}{2x + y + 2z = 4}\\{ - 2x - 4y - 6z = 2}\\{}\end{array}\), on solving this we get \( - 3y - 4z = 6\) … (4)
Now take equation (2) and (3)
\(\begin{array}{*{20}{c}}{}\\{x + 2y + 3z = - 1}\\{3x + 2y + z = 9}\end{array}\) multiply equation (2) by -3, thus the equations will be
Question 16: A test has 20 questions, with 4 marks for a correct answer, –1 mark for a wrong answer, and no marks for an unattempted question. A group of friends took the test. If all of them scored exactly 15 marks, but each of them attempted a different number of questions, what is the maximum number of people who could be in the group? [1] 3 [2] 4 [3] 5 [4] more than 5
c + w + n = 20
4c – w = 15
Adding the two equations, we get 5c + n = 35
c(max) = 7, when n = 0 & w = 13
c(min) = 4, when n = 15 & w = 1
Maximum number of people who could be in the group = Number of possible values of ‘c’ = 4. Option B
Question 17: How many integers x with |x|< 100 can be expressed as \(x = \frac{{4 - {y^3}}}{4} \) for some positive integer y? [1] 0 [2] 3 [3] 6 [4] 4
\(x = \frac{{4 - {y^3}}}{4} = 1 - \frac{{{y^3}}}{4}\)
y3 = 4(1-x) = 4 – 4x
x = 0 or -3 or -15 or -53
No. of valid values of x = 4. Option D Online CAT 2021 Quant Course
Question 18: The number of roots common between the two equations x3+3x2+4x+5=0 and x3+2x2+7x+3=0 is: [1] 0 [2] 1 [3] 2 [4] 3
For the roots to be common to the two equation, both the equations must be equal to 0 and hence equal to each other at those values of x
x3+3x2+4x+5 = x3+2x2+7x+3
3x2+4x+5 = 2x2+7x+3
x2 – 3x + 2 = 0
x = 1 or 2
At x = 1 and at x = 2 both the equations become equal to each other
But at x = 1 or at x = 2 none of the original equations become 0.
Number of common roots = 0. Option A Online CAT 2021 Quant Course
Question 19: Let u= \({({\log _2}x)^2} - 6{\log _2}x + 12 \) where x is a real number. Then the equation xu=256, has: [1] no solution for x [2] exactly one solution for x [3] exactly two distinct solutions for x [4] exactly three distinct solutions for x
xu=256
u \(lo{g_2}x\) = 8
\(lo{g_2}x\) =\(\frac{8}{u}\)
Putting this in the first equation
u = (8/u)2 – 6×\(\frac{8}{u}\) + 12
u3 = 64 – 48u + 12u2
u3 – 12u2 + 48u – 64 =0
(u -4)3 = 0
u = 4
\(lo{g_2}x\)= \(\frac{8}{u}\) = 2
x = 2
We have exactly one solution for x. Option B Online CAT 2021 Quant Course
Question 20: Let a, b, and c be positive real numbers. Determine the largest total number of real roots that the following three polynomials may have among them: ax2 + bx + c, bx2 + cx + a, and cx2 + ax + b. [1] 4 [2] 5 [3] 6 [4] 0
For these equations to have real roots
b2 – 4ac ≥ 0
b2 ≥ 4ac
c2 – 4ab ≥ 0
c2 ≥ 4ab
a2 – 4ac ≥ 0
a2 ≥ 4ac
Multiplying the three we get
(abc)2 ≥ 64(abc)2
This is not possible for positive values of a, b & c.
So, there are no real roots for the three given polynomials. Option D
Question 22: If both a and b belong to the set (1, 2, 3, 4), then the number of equations of the form ax2+bx+1=0 having real roots is [1] 10 [2] 7 [3] 6 [4] 12
Question 23: Rakesh and Manish solve an equation. In solving Rakesh commits a mistake in constant term and finds the root 8 and 2. Manish commits a mistake in the coefficient of x and finds the roots -9 and -1. Find the correct roots. [1] 9,1 [2] -9,1 [3] -8,-2 [4] None of these
$\sqrt { 2 + \sqrt { 2 + \sqrt { 2 + \sqrt { 2 + \ldots } } } } = x$
$\sqrt { 2 + x } = x$
$2 + x = x ^ { 2 }$
$x ^ { 2 } - x - 2 = 0$
$x = 2 , - 1$
as the value of the expression will be positive, we can reject x=-1.
Hence, x=2.Online CAT 2021 Quant Course
Question 29: If a, b and c are the roots of the equation x3 – 3x2 + x + 1 = 0 find the value of \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \) [1] 1 [2] -1 [3] 1/3 [4] -1/3
Question 30: If p, q and r are the roots of the equation 2z3 + 4z2 -3z -1 =0, find the value of (1 - p) × (1 - q) × (1 - r) [1] -2 [2] 0 [3] 2 [4] None of these
If p, q and r are the roots of the equation 2z3 + 4z2 -3z -1 =0, then
f(z) = 2z3 + 4z2 -3z -1 = (z - p) × (z - q) × (z - r)
Question 31: If $\alpha, \beta$ and $\gamma$ are the roots of the equation $x^{3}-7 x+3=0$ what is the value of $\alpha^{4}+\beta^{4}+\gamma^{4}$ ? [1] 0 [2] 199 [3] 49 [4] 98
Writing the equation as
$(x-\alpha)(x-\beta)(x-\gamma)=0,$ expanding and equating coefficients we get :
Question 32: For what values of p does the equation 4x2 + 4px + 4 –3p = 0 have two distinct real roots? [1] p < -4 or p > 1 [2] -1 < p < 4 [3] p < -1 or p > 4 [4] –4 < p < 1
Question 33: If x2 + 4x + n > 13 for all real number x, then which of the following conditions is necessarily true? [1] n > 17 [2] n = 20 [3] n > -17 [4] n < 11
x2 + 4x + n > 13
x2 + 4x + 4 + n > 13 + 4 {Adding 4 to both sides}
(x + 2)2 + n > 17
Minimum value (x + 2)2 can take is 0 when x = – 2
For this to be true for all real values of x, n > 17. Option A Online CAT 2021 Quant Course
Question 34: If (x + 1)×(x – 2)×(x + 3)×(x – 4)×(x + 5)…(x – 100) = a0 + a1x + a2x2… + a100x100 then the value of a99 is equal to: [1] 50 [2] 0 [3] -50 [4] -100
a100 = 1
Sum of the roots = \(-\frac{{{a}_{99}}}{{{a}_{100}}}=-\ {{a}_{99}}\)
Question 35: If a, b, and c are the solutions of the equation x3 – 3x2 – 4x + 5 = 0, find the value of \(\frac{1}{{ab}} + \frac{1}{{bc}} + \frac{1}{{ca}} \) [1] 3/5 [2] -3/5 [3] -4/5 [4] 4/5
Question 39: A quadratic function f(x) attains a maximum of 3 at x = 1. The value of the function at x = 0 is 1. What is the value of f (x) at x = 10? [1] -105 [2] -119 [3] -159 [4] -110
If the function attains the maximum of 3 at x = 1
f(x) = p(x – 1)2 + 3
f(0) = p + 3 = 1 {It is given as 1}
p = –2
f(x) = –2(x – 1)2 + 3
f(10) = –2(10 – 1)2 + 3 = –162 + 3 = –159. Option C Online CAT 2021 Quant Course
Question 40: \(x + \frac{1}{x} = 3\) then, what is the value of \({x^5} + \frac{1}{{x{}^5}}. \) [1] 123 [2] 144 [3] 159 [4] 186
$x+\frac{1}{x}=3$
$\left.x^{2}+\frac{1}{x^{2}}+2=9 \text { \{Squaring both sides }\right\}$
Question 42: If $\alpha$ and $\beta$ are the roots of the quadratic equation $x^{2}-x-6,$ then find the value of $\alpha^{4}+\beta^{4} ?$ [1] 1 [2] 55 [3] 97 [4] none of these
$\alpha+\beta=1,$ and $\alpha \beta=-6$
Then, using the identity,
$\alpha^{4}+\beta^{4}=(\alpha+\beta)^{4}+2(\alpha \beta)^{2}-4 \alpha \beta(\alpha+\beta)^{2}$
$=(1)^{4}+2(-6)^{2}-(4 \times-6)(1)^{2}$
$=1+72+24$
$=97$Online CAT 2021 Quant Course
Question 44: If the roots of the equation x3 – ax2 + bx – c =0 are three consecutive integers, then what is the smallest possible value of b? [1] -1/√3 [2] -1 [3] 0 [4] 1/√3
b is sum of product of the roots taken 2 at a time which will be minimum when the roots are -1, 0 & 1
Question 45: Three consecutive positive integers are raised to the first, second and third powers respectively and then added. The sum so obtained is a perfect square whose square root equals the total of the three original integers. Which of the following best describes the minimum, say m, of these three integers? [1] 1 ≤ m ≤ 3 [2] 4 ≤ m ≤ 6 [3] 7 ≤ m ≤ 9 [4] 10 ≤ m ≤ 12
(a-1) + (a+1)3 = 8a2
a – 1 + a3 + 3a2 + 3a + 1 = 8a2
a3 – 5a2 + 4a = 0
a(a2 – 5a + 4) = 0
a = 0, 1, 4
0 and 1 are invalid values because a – 1 should be a positive integer
a = 4
m = a – 1 = 3. Option A Online CAT 2021 Quant Course
Question 46: The price of Darjeeling tea (in rupees per kilogram) is 100 + 0.10 n, on the nth day of 2007 (n = 1, 2, ..., 100), and then remains constant. On the other hand, the price of Ooty tea (in rupees per kilogram) is 89 + 0.15n, on the nth day of 2007 (n = 1, 2, ..., 365). On which date in 2007 will the prices of these two varieties of tea be equal? [1] May 21 [2] April 11 [3] May 20 [4] April 10
100 + 0.1n = 89 + 0.15n
0.05n = 11
n = 220
But for Darjeeling tea n cannot be more than 100.
Maximum price of Darjeeling tea = 100 + 0.1×100 = 110
Price of Ooty tea should also be 110
89 + 0.15n = 110
n = 140
On the 140th day of 2007, the prices of the Darjeeling tea and Ooty tea will be equal
Question 47: The polynomial f(x)=x2-12x+c has two real roots, one of which is the square of the other. Find the sum of all possible value of c. [1] -37 [2] -12 [3] 25 [4] 91
Question 48: Two sides of a triangle have lengths 10 and 20. How many integers can take the value of the third side length: [1] 18 [2] 19 [3] 20 [4] 21
Question 49: Which of the following is a solution to: \(6{\left( {x + \frac{1}{x}} \right)^2} - 35\left( {x + \frac{1}{x}} \right) + 50 = 0 \) [1] 1 [2] 1/3 [3] 4 [4] 6
Question 51: If $a, b, c$ are the roots of $x^{3}-x^{2}-1=0,$ what's the value of $\frac{a}{b c}+\frac{b}{c a}+\frac{c}{a b}$ ? [1] -1 [2] 1 [3] 2 [4] -2
Under the precondition you can write $(x-a)(x-b)(x-c)=0=x^{3}-x^{2}-1$ Expanding the product on the left gives $x^{3}-(a+b+c) x^{2}+(a b+a c+b c) x-a b c=x^{3}-x^{2}-1$ Now you have to compare/equate the coefficients on both sides of $\left(^{*}\right)$ and get $a+b+c=1, a b+a c+b c=0, a b c=1$ Using these and the identity $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(a b+a c+b c)$ for evaluation you get $\frac{a}{b c}+\frac{b}{a c}+\frac{c}{a b}=\frac{a^{2}+b^{2}+c^{2}}{a b c}$ $=\frac{(a+b+c)^{2}-2(a b+a c+b c)}{a b c}$ $=1$
Question 61: The equation \(x + \frac{2}{{1 - x}} = 1 + \frac{2}{{1 - x}}, \) has [1] No real root [2] One real root [3] Two equal roots [4] Infinite roots
=>\(\frac{{{x^2} - x - 2 + 1 - x + 2}}{{(1 - x)}} = 0\)
On solving the above equation : \(\frac{{{x^2} - 2x + 1}}{{1 - x}} = 0\)
The only valid solution could have been x = 1 but that is ruled out because then the denominator would become 0.
No real root exists. Option A Online CAT 2021 Quant Course
Question 63: If A.M. of the roots of a quadratic equation is 8/5 and A.M. of their reciprocals is 8/7, then the equation is [1] 5x2-16x+7=0 [2] 7x2-16x+5=0 [3] 7x2-16x+8=0 [4] 3x2-12x+7=0
Question 64: The equation x2 + ax + (b + 2) = 0 has real roots. What is the minimum value of a2 + b2? [1] 0 [2] 1 [3] 2 [4] 4
Condition for real roots is
a2 – 4(b+2) ≥ 0
a2 +b2 –(b2 +4b+8) ≥ 0
a2 +b2 ≥ (b2 +4b+8)
a2 +b2 ≥ (b+2)2 + 4
Minimum value of a2 +b2 will occur when b = -2 and it will be 4. Option D Online CAT 2021 Quant Course
