Question Bank  100 Algebra Questions From Previous CAT Papers (Solved)

Number of Questions: 100
Topic: Algebra
Solved? : Yes
Source: Previous CAT Questions

Q1) The number of solutions of the equation 2x + y = 40 where both x and y are positive integers and x ≤ y is:
(1) 7
(2) 13
(3) 14
(4) 18
(5) 20 (CAT 2006)

Solution: 2x + y = 40
x can take all the values from 1.
Let's see the point where x becomes greater than or equal to y:
2x + x = 40
x=13, y=14
So, x can take values from 113 : 13 possible solutions.

Q2) The graph of y  x against y + x is shown as below (All graphs in this question are drawn to scale and same scale has been used on each axis). which of the below options shows the graph of y against x. (CAT 2006)

Solution: Analyse the question for around 10 seconds or so, and again, there is more than one way to proceed. I will explain the easiest approach here, rather than the “theoretical” “formula oriented – slope” approach
Study the 1st graph (question, not the options carefully). These things can be observed.
When y  x = 0, y + x = 0
Take around the middle positive portion of the graph. When yx is in the middle, y+x is not that high. It has just started off. We can assume that yx is 5 and y+x will be around 1 on the same scale. ( need not be accurate values, just an approximate value)
Similarly, extending more in positive, we can see that yx = 10, y+x = 2 (again, only approximate, the only thing to bear in mind is that, yx varies more than y+x)
taking the negative portion, if we extend the graph (imaginary), y  x = 5, y + x = 1.
Similarly, extending more in the negative, y  x = 10, y + x = 2
Let us solve these simple equations, we have in hand :
y – x = 0, y + x = 0. This gives when y = 0, x = 0.
y – x = 5, y + x = 1. This gives y = 3, x = 2.
y  x = 12, y + x = 2. This gives when y = 6, x = 4
y  x = 5, y + x = 1. This gives y = 3, x = 2
y  x = 10, y + x = 2. This gives y = 6, x = 4So, when y is 0, x is 0, and when y is negative, x is positive and vice versa. So the answer is definitely option d or option e. However, in option 5, we see that the absolute value of x is large, when y is small. But, we have just proved otherwise. Absolute value of y is more, and x is less, and hence option d is the right one.
P.S: One can stop when you feel you have derived enough equations. You will never need to derive more than 5 equations. I personally recommend deriving 3 equations (one for 0, another one for positive yx and the last one for negative yx). From that we can observe the relation between y&x (negative and positive), and the absolute value variation ( y varies more than x or vice versa). In theory, this variation of yx with respect to y+x is nothing but the slope, and from this, the formulae are derived. :)

Q3) Let f (x) = max (2x + 1, 3 − 4x), where x is any real number. Then the minimum possible value of f(x) is:
(1) 1/3
(2) 1/2
(3) 2/3
(4) 4/3
(5) 5/3 (CAT 2006)

f(x) is the maximum value of 2x + 1 and 3 – 4x. Here, we see that in 1 place, x is added to something else, while in the other equation (3 – 4x), x is subtracted from something else. So, in the 1st equation, as x goes higher, f(x) will become bigger, while in the second case, when x goes bigger, the value of f(x) goes smaller, and it will be viceversa when x goes lower. So, we need to find an optimum solution, which we can obtain by equating both the equations.
2x + 1 = 3 – 4x. On solving, we get x = 1/3 and f(x) = 5/3.
If we put any value more than 1/3, 2x + 1 becomes high, and f(x) increases.
If we put any value less than 1/3, 3 – 4x becomes high, and f(x) increases.
So for x = 1/3, f(x) = 5/3 is the minimum possible value.

Q4) 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) –119
(2) –159
(3) –110
(4) 180
(5) 105 (CAT 2007)

Method 1: As f(x) is a quadratic function and f(x) attains max value of 3 at x=1, so we can express f as in this way, f(x)= k*(x 1)^2 +3.
Remember this useful property, if f is a quadratic function and f attains max/min of b at x=b, then we can write f(x)= k * (x  a)^2 + b where k is a constant.
Now, f(x)= k * (x1)^2 + 3.
Put x=0 in this eq, f(0)= k * (1)^2 + 3
or, 1= k +3
or, k = 2.
So, f(x)= 2 * (x1)^2 +3.
Now, put x=10 in this eq, f(10) = 2 * 9^2 + 3 = 162 + 3 = 159 is the ans. :)Method 2: Let the quadratic equation be ax^2 + bx + c. So, f(x) = ax^2 + bx + c
At x = 0, the value of function is 1.
x = 0, f(x) = 1
ax^2 + bx + c = a * 0 + b * 0 + c = c
c = 1.
At x = 1, f(x) = 3
x = 1, f(x) = 3
a * 1 + b * 1 + c = 3
Since c = 1, a + b = 2.
Also, we know that f(x) is maximum when x = 1. If f(x) is maximum, (dx/dt)(f(x)) = 0
Differentiating f(x), we have d/dt (ax2 + bx + c) = 2ax + b
At x = 1, 2ax + b = 0
2a + b = 0.
b = 2a
Substituting we have a + b = 2, or a + 2a = 2. a = 2. So, b = 4. So the equation is 2 x 2 + 4x + 1.
At x = 10, the value is 2 * 100 + 4 * 10 + 1 = 159

Q5) A function f (x) satisfies f (1) = 3600, and f (1) + f (2) + ... + f (n) = n² f(n), for all positive integers n >1. What is the value of f (9) ?
(1) 80
(2) 240
(3) 200
(4) 100
(5) 120 (CAT 2007)

Solution: There is more than 1 approach to solve the problem. You can either group f(n), find f(n1), gather some formula and find the solution, which might be tricky, time consuming, and “not so easy to click”. Another way is by finding each one of f(2), f(3), f(4)….till f(9), which might take upto 23 minutes depending on your speed, but is a much better choice, as you will definitely get the solution here. I approach these kinds of questions by examples initially, and then try to generalize to a solution. This method will be less time consuming than the above 2 methods, and might prove be less tedious as well.
f(1) = 3600. f(1) + f(2) = 4 f(2)
f(2) = f(1)/3. We can write this as f(2) = f(1) * 1/3.
Now f(1) + f(2) + f(3) = 9 * f(3)
f(3) = f(1) + f(2) / 8. We know f(2) = f(1)/3.
So, f(3) = f(1) + (f(1)/3) / 8.
We can write this as f(3) = f(1) * 1/3 * 2/4.
Since we already have f(1) * 1/3, we write it in such a way that the equation has f(1) * 1/3 and then what comes rest, so that we can generalize.
Calculating similarly for f(4), we get f(4) = f(1) * 1/3 * 2/4 * 3/5.
So f(9) will be f(1) * 1/3 * 2/4 * 3/5 * 4/6 * 5/7 *6/8 * 7/9 * 8/10.
f(9) = (f(1) * 1 * 2) / (9 * 10)
f(9) = 3600 * 2/ 9 * 10 = 80.

Q6) Let g (x) be a function such that g (x + 1) + g (x – l) = g (x) for every real x. Then for what value of p is the relation g (x + p) = g (x) necessarily true for every real x ?
(1) 5
(2) 3
(3) 2
(4) 6 (CAT 2005)

Solution: g (x+1) = g (x) – g (x1)
Let g (x) = p, g (x1) = q.
Then, g (x+1) = g (x) – g (x1) = p – q
g (x+2) = g (x+1) – g (x) = p – q – p = q
g (x+3) = g (x+2) – g (x+1) = qp + q = p
g (x+4) = g (x+3) – g (x+2) = q  p
g (x+5) = g (x+4) – g (x+3) = q –p +p = q = g (x1)
So, we see g (x+5) = g (x1). Thus, entry repeats after 6 times. p = 6

Q7) Let f(x) = ax^2 + bx + c, where a, b and c are certain constants and a ≠ 0. It is known that f (5) = −3f (2) and that 3 is a root of f(x) = 0.
What is the other root of f(x) = 0?
(1) −7
(2) − 4
(3) 2
(4) 6
(5) cannot be determinedWhat is the value of a+b+c?
(1) 9
(2) 14
(3) 13
(4) 37
(5) cannot be determined (CAT 2008)

Solution: First part of the question :
f(3) = 0. So 9a + 3b + c = 0
f(5) = 3f(2). So 25a + 5b + c = 3 (4a + 2b + c). Solving, we get 37a + 121b + 4c = 0.
So we have 2 equations:
9a + 3b + c = 0 > (1)
37a + 11b + 4c = 0 > (2)
Multiply (1) with 4 , we get, 36a + 12b + 4c = 0 > (3)
Subtract (3) from (2), we get, a – b = 0 or a = b. So, we got a = b.
Sum of the roots of a quadratic equation is –b/a. Here it is –a/a = 1
one of the root is 3. Sum of roots is 1. Hence the other root is 4.Second part :
We know the roots are 3 and 4.
Hence the equation is (x3) (x+4) = 0, or x^2 + x – 12 = 0.
However, even 2(x^2 + x – 12) = 0 or 100 (x^2 + x – 12) = 0 and so on will have the same roots.
Hence, we cannot find unique values of a,b,c.

Q8) Let f(x) = ax^2 – b x , where a and b are constants. Then at x = 0, f(x) is:
a. maximized whenever a > 0, b > 0
b. maximized whenever a > 0, b < 0
c. minimized whenever a > 0, b > 0
d. minimized whenever a > 0, b < 0

Method 1 :
f(x) = ax^2 – bx
x^2 and x are always non negative.
If a > 0 and b < 0, f(x) > = 0
So at x = 0, f(x) is minimum when a > 0 and b < 0
Option d.Method 2 :
f(x) = ax^2 – bx
To find max and minimum value, we have to double differentiate.
First differential f ' (x) = 2ax – b * ( x/x ). At x = 0, f ' (x) = b.
Second differential f '' (x) = 2a.
If a > 0, then f '' (x) is positive and hence the equation is minimal.
Also for the minimal value at x = 0, b is less than 0.

Q9) If f (x) = x^3 – 4x + p, and f (0) and f (1) are of opposite signs, then which of the following is necessarily true?
a. –1 < p < 2
b. 0 < p < 3
c. –2 < p < 1
d. –3 < p < 0

f(0) = p
f (1) = 1 – 4 + p = p + 3.
p and p+3 are of different signs.
Substitute for the options:
1st option, if p is a negative number, then p – 3 is also negative, and both are same sign.
For 3rd and 4th option, also it is the same.
For the 2nd option, value of p is between 0 and 3, and p – 3 is negative. Hence it agrees.
Even without substituting the answer can be found out. As p and p – 3 are different signs; p should be between 0 and 3.

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