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This is the first post in the CAT 2016 Sprint Preparation Series – Algebra. We have posted 20 questions from previous year CAT papers, forums, mock tests, and other entrances that are on par with the level of difficulty you can expect in CAT 2016. We will be posting the solutions and traps/things to look at while solving similar questions so that you are avoid making silly mistakes during the test.

You can go through the entire series by clicking on this link: CAT 2016 Sprint Preparation Series by Learningroots.

# CAT 2016 Sprint Preparation Series – Algebra

1. The number of non-negative real roots of 2^x – x – 1 = 0 equals

2. When the curves y = log10x and y = x–1 are drawn in the x-y plane, how many times do they intersect for values x ≥ 1?

3. A test has 50 questions. A student scores 1 mark for a correct answer, –1/3 for a wrong answer, and –1/6 for not attempting a question. If the net score of a student is 32, the number of questions answered wrongly by that student cannot be less than

a. 6

b. 12

c. 3

d. 9

4. Let a, b, c, d be four integers such that a+b+c+d = 4m+1 where m is a positive integer. Given m, which one of the following is necessarily true?

a. The minimum possible value of a2 + b2 + c2 + d2 is 4m2–2m+1

b. The minimum possible value of a2 + b2 + c2 + d2 is 4m2+2m+1

c. The maximum possible value of a2 + b2 + c2 + d2 is 4m2–2m+1

d. The maximum possible value of a2 + b2 + c2 + d2 is 4m2+2m+1

5. How many pairs of positive integers m, n satisfy 1/m + 4/n = 1/12, where, ‘n’ is an odd integer less than 60?

a. 6

b. 4

c. 7

d. 5

e. 3

6. A quadratic function ƒ(x) attains a maximum of 3 at x = 1. The value of the function at x = 0 is 1. What is the value ƒ(x) at x = 10?

a. –119

b. –159

c. –110

d. –180

e. –105

Q 7-8.

7. Which of the following best describes an + bn for even ‘n’?

a. q(pq)n/2 – 1(p + q)

b. qpn/2 – 1 (p + q)

c. qn/2 (p + q)

d. qn/2 (p + q)n/2

e. q(pq)n/2 – 1(p + q)n/2

8. If p = 1/3 and q = 2/3, then what is the smallest odd ‘n’ such that an + bn < 0.01?

a. 7

b. 13

c. 11

d. 9

e. 15

9. If the roots of the equation x^3 – ax^2 + bx – c = 0 are three consecutive integers, then what is the smallest possible value of b?

a. – 1/√3

b. –1

c. 0

d. 1

e. 1/√3

Q 10-11. 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.

10. What is the other root of f(x) = 0?

a. –7

b. – 4

c. 2

d. 6

e. Cannot be determined

11. What is the value of a + b + c?

a. 9

b. 14

c. 13

d. 37

e. Cannot be determined

12. Let g(x) be a function such that g(x + 1) + g(x – 1) = 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?

a. 5

b. 3

c. 2

d. 6

13. 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

14.

a. ½

b. – 1

c. ½ or – 1

d. – ½ or – 1

15. Consider the sequence of numbers a1, a2, a3, … to infinity where a1 = 81.33 and a2 = –19 and

aj = aj–1 – aj–2 for j ≥ 3. What is the sum of the first 6002 terms of this sequence?

a. –100.33

b. –30.00

c. 62.33

d. 119.33

16. If x^2 * y^3 = 8, where x, y > 0 then what is the minimum value of 4x + 3y?

17. If a, b are integers then how many ordered pairs (a, b) satisfy the equation a^2 + ab + b^2 = 1?

18. If the sum of the first ‘n’ terms of an Arithmetic Progression is 100 and the sum of the next ‘n’ terms of the Arithmetic Progression is 300, then what is the ratio of the first term and the common difference?

19. If a, b are the roots of x^2 + x + 1 = 0 and c, d are the roots of x^2 + 3x + 1 = 0, then what is the value of (a – c)(b + d)(a + d)(b – c)?

20. There were 4 parcels all of whose weights were integers (in kg). The weights of all the possible pairs of parcels were noted down and amongst these the distinct values observed were 94 kg, 97 kg, 101 kg and 104 kg. Which of the following can be the weight of one of the parcels?

a. 40 kg

b. 45 kg

c. 48 kg

d. 53 kg

Solutions

1. Probably the best way to do this is by observing how the equation behaves by putting in values:

If x = 0 or 1, it is balanced. For x less than 0, or greater than 1, it will be unbalanced. So, 2 solutions in total.

2. ylog10 = (logx) and y = 1/x

So, x^x = 10. As x is positive, there will be only one value for which, x^x will be equal to 10.

3. Let the number of correct attempts be a, incorrect ones be b and the unattempted ones be c.

a – b/3 – c/6 = 32

6a – 2b – c = 192

a + b + c = 50

We have to find the minimum possible value of b.

7a – b = 242

For b to be minimum, a has to be minimum as well. The minimum value of a is 35 for which, b will be 3 and so, that is our answer.

4. For any expression of the form a + b + c + d, the minimum value of a^2 + b^2 + c^2 + d^2 will be when a = b = c = d. Or in other words, the four numbers have to be as close to the average as much as is possible. As the sum is in the form of 4m + 1, we can say that the four numbers would give the minimum value of the above expression if they are m, m, m and m+1.

So, we get the sum of their squares to be:

4m^2 + 2m + 1 and so, option b is correct.

5. This is a textbook CAT question.

1/m + 4/n = 1/12

48m + 12n = mn

48m + 12n – mn = 0

m(48 – n) – 12(48 – n) = -576

(m – 12)(n – 12) = 576

If n is an odd integer, the second part should be odd. So, the first part has to be even.

Now, we know that 576 = 2^6 * 3^2 and so, total of 21 factors. We also know that there are 3 odd factors (1, 3, 9). As n can take only 3 values, that will be the number of the solution sets to the equation.

6. Let f(x) = ax^2 + bx + c

f(1) = a + b + c = 3 which is also the maximum value.

f(0) = c = 1.

f(10) will be 100a + 10b + c

Now, when a quadratic expression has the maximum value, we can say that the first order derivative will be 0.

So, 2ax + b = 0

As x = 1, we get that b = -2a

Substituting in the above equations, we get

a + b + 1 = 3

a = -2. So, b = 4 and c as we already know is 1.

f(10) = -200 + 40 + 1 = -159.

7. This looks extremely complicated and like something that should be on Sheldon Cooper’s board and not in CAT. But like many things in life, ‘All that is gold does not glitter’.

Basically,

a2 = p*b1 and b2 = q*b1

a3 = p*a2 and b3 = q*a2

a4 = p*b3 and b4 = q*b3

a5 = p*a4 and b5 = q*a4

So, let n = 2, a2 + b2 = p*b1 + q*b1 = q(p + q)

For n = 4, a4 + b4 = b3 * (p + q) = pq^2 * (p + q)

Option a.

8. an + bn for odd n will be given by (p + q)*an-1

So, according to the inequality,

an-1 < 0.01

So aeven < 0.01

aeven is nothing but pq, (pq)^2 and so on.

(2/9)^(n/2) should be less than 0.01

We get n/2 = 4 for which this happens for the first time. So, n = 8. The next odd number will be 9 and so, that is our answer.

9. b is nothing but sum of products of roots taken 2 at a time. If the roots are (n – 1), n and (n + 1), we get

n(n – 1) + (n – 1)(n + 1) + n(n + 1) = b

n^2 – n + n^2 – 1 + n^2 + n = b

3n^2 – 1 = b.

As 3n^2 will always be non negative, the least value of the expression is -1.

10. 3 is a root of f(x) = 0

9a + 3b + c = 0

f(5) = 25a + 5b + c

f(2) = 4a + 2b + c

So, according to the given information

25a + 5b + c = -12a – 6b – 3c

37a + 11b + 4c = 0

From the first equation, we get

36a + 12b + 4c = 0

On subtracting, we get

a – b = 0

Or a = b.

Also, we get c = -12a if we substitute it back.

So, the original f(x) becomes

ax^2 + ax – 12a = 0

a(x^2 + x – 12) = 0

a(x + 4)(x – 3) = 0

So, the other root is -4.

11. a + b + c is a + a – 12a = -10a and so, it cannot be determined.

12. g(1) = g(0) + g(2)

g(2) = g(1) + g(3)

g(3) = g(2) + g(4)

and so on…

So, g(0) = -g(3), g(1) = -g(4) and so on…

As we can see the terms are separated by 3 but their signs are inverted. So, the signs will be restored after a gap of 6. So, p should be equal to 6.

13. f(0) = p, f(1) = p – 3

Looking at the options, we can see that only option b fits.

14. This is another brilliant CAT question. A serious aspirant will definitely know that if a/b = c/d = e/f = r, then (a + c + e)/(b + d + f) = r.

Now, what happens when add all those fractions?

(a + b + c)/2(a + b + c) = r

If we cancel out a + b + c, we get r = ½

But is that our answer? Nope.

When we cancelled out (a + b + c), we assumed that a + b + c will not be equal to 0. What if it is equal to 0?

Then we get a = -(b + c), b = -(a + c) and c = -(a + b)

So, we get r = -1 and the second part of our answer.

15. a1 = 81.33

a2 = -19

a3 = a2 – a1 = -100.33

a4 = a3 – a2 = -81.33

a5 = a4 – a3 = 19

a6 = a5 – a4 = 100.33

a7 = a6 – a5 = 81.33 and the cycle will continue…

As we can see, the sum of the first 6 terms is 0, the sum of the next 6 terms will be 0 and so on. So, we are concerned about only the 6001st and the 6002nd term which will be 81.33 and -19. The sum will be 62.33 which will be our answer.

16. This is the reverse of what we normally do for AM >= GM questions.

Let the numbers be 2x, 2x, y, y and y. So, using the AM >= GM inequality, we get

(2x + 2x + y + y + y)/5 >= {(2x)(2x)(y)(y)(y)}^1/5

(4x + 3y)/5 >= (4x^2*y^3)^1/5

4x + 3y >= 10.

So, our answer will be 10.

17. a^2 + ab + b^2 = 1

So, (a + b)^2 = ab + 1

Now, as (a + b)^2 is always positive and greater than 1 and a and b are integers, we can say that either ab = 0 or ab = 1 for it to be satisfied.

The solution set that we get is (1, 0) (0, 1) (-1, 0) (0, -1) (1, -1) (-1, 1). 6 sets.

18. Sum of first n terms is 100. So,

Sn = {2a + (n – 1)d}n/2 = 100

S2n = {2a + (2n – 1)d}n = 400

On division, we get

½ = {2a + (n – 1)d}/{2a + (2n – 1)d}

4a + 2nd – 2d = 2a + 2nd – d

2a = d OR a/d = ½

19. We get that

a + b = -1 and c + d = -3

Also, ab = 1 and cd = 1

(a – c)(b + d)(a + d)(b – c)

(ab + ad – bc – cd)(ab – ac + bd – cd)

(ab)^2 – ab*ac + ab*bd – abcd + ad*ab – ad*ac + ad*bd – ad*cd – ab*bc + bc*ac – bc*bd + bc*cd – abcd + cd*ac – cd*bd + (cd)^2

1 – ac + bd – 1 + ad – ad*ac + ad*bd – ad – bc + bc*ac – bc*bd + bc – 1 + ac – bd + 1

-a^2 + d^2 + c^2 – b^2

We know that (a + b)^2 = a^2 + b^2 + 2 and (c + d)^2 = c^2 + d^2 + 2

So, we get the above expression to be

1 + 7 = 8.

20. Let the weights be a, b, c and d. Now, there are 4c2 = 6 possible cases with two weights each. As this is equal to 4 different values, it is obvious that two items weight the same. Let the weights be a, a, b and c.

So, some combination will be 2a. The only even numbers are 94 and 104 and so, a could be equal to either 47 or 52.

Case i. a = 47. Then b = 50 and c = 54

Case ii. a = 52. Then b = 45 and c = 49

Option b is the only weight that matches.

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