# Free Algebra 01 Practice Test - CAT

What is the value of M(M(A(M(x, y), S(y, x)), x), A(y, x)) for x = 2, y = 3? (CAT 1996)

A. 50
B. 140
C. 25
D. 70

#### SOLUTION

Solution : D

M (M (A (M (x, y), S (y, x)), A (y, x)

M (M (A (6, 1), 2), A (3, 2))

M (M (7, 2), A (3, 2))

M (14, 5) = 70

What is the value of S[M(D(A(a, b), 2), D(A(a, b), 2)), M(D(S(a, b), 2), D(S(a, b), 2))]?

A. a2 + b2
B. ab
C. a2 – b2
D. a / b

#### SOLUTION

Solution : B

Soln:(b)

S [M (D (A (a, b),2), D (A (a, b), 2)), M (D (S (a, b), 2), D (S (a, b), 2))]

S [M (D (a + b, 2), D (a + b, 2)), M (D (a – b, 2),D (a – b, 2))]

S [M(a + b)/2), (a + b) /2), M( )]

S [ ( ] = [(a +b)2 – (a – b)2 ]/ 22 Find the sum of + + ………… + (CAT 2008)

A. B. C. D. E. #### SOLUTION

Solution : A = ; = ; = . + = + = ; = ->3 - (1/3)

Thus, on similar lines, +.......... = ;

The number of common terms in the two sequences 17, 21, 25,…417 and 16, 21, 26….466 is: (CAT 2008)

A. 78
B. 19
C. 20
D. 77
E. 22

#### SOLUTION

Solution : C

Soln:

If we have 2 AP’s and are trying to find common terms, then the common terms also form a AP with a CD of LCM of the 2 common differences.

The first sequence is of the form 17 + 4a

And the second sequence is of the form 16 + 5b

By observation we see that 21 is the first term of the new sequence and the common difference is 20 ( LCM of 4 and 5)

Hence the sequence will be 21 + 20c

Now substitute values of c from the answer options and check that the last term should not cross 417.

Hence when we sub c=19 we see that the value is lesser than 417 and when we sub c=20 the value is 421 which is greater than 417.

Hence the number of terms will be 19 + first term = 20 terms or option ( c)

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?(CAT 2004)

A. -100.33
B. -30.00
C. 62.33
D. 119.33

#### SOLUTION

Solution : C

Ans: (c)

The terms of the given sequence are as follows :

a1 = 81.33       a7 = a1

a2 = -19           a8 = a2

a3 = a2 – a1       a9 = a3

a4 = - a1             a10 = a4 = -a1

a5 = -a2               a11 = a4 = - a2

a6 = - a2 + a1    a12 = a6 = -a3 and so on.

The sum of the first six terms, the next six terms and so on is 0.

The sum of the first 6002 terms can be written as the sum of first 6000 terms + 6001st term + 6002nd term. From the above explanation, the sum of the first 6000 terms is zero, 6001 term will be a1 and 6002nd term will be a2. The sum of the first 6002 terms will be a1 + a2 = 81.33 + (-19) = 62.33.

If f (0, y) = y + 1, and f(x +1, y) = f (x, f(x, y)). Then, what is the value of f(1,2) ? (CAT 2000)

A. 1
B. 2
C. 3
D. 4

#### SOLUTION

Solution : D

SOLN:(d)

f (x + 1, y) = f(x, f(x, y))

put x = 0, f (1, y) = f(0, f(0, y)) = f(0, y + 1)

= y + 1 + 1 = y + 2

Put y = 2, f (1, 2) = 4.

The function f(x) = |x - 2| + |2.5 - x| + |3.6 - x|, where x is a real number, attains minimum at

A. x = 2.3
B. x = 2.5
C. x = 2.7
D. None of these

#### SOLUTION

Solution : B

Option (b)

f(x) = | x – 2| + | 2.5 – x | + | 3.6 – x | can attain minimum value when either of the terms = 0.

Case I :

When | x – 2 | = 0 => x = 2, value of f(x) = 0.5 + 1.6 = 2.1.

Case II.

When | 2.5 – x | = 0 => x = 2.5
value of f(x)

= 0.5 + 0 + 1.1 = 1.6.

Case III.

When | 3.6 – x | = 0 => x = 3.6

f(x) = 1.6  + 1.1 + 0 = 2.7. Hence the minimum value of f(x) is 1.6 at x = 2.5.

Let u n+1 = 2un +1, (n = 0, 1, 2,…) and u0 = 0. Then u10 would be nearest to: (CAT 1993)

A. 1023
B. 2047
C. 4095
D. 8195

#### SOLUTION

Solution : A

Ans: (a)

u n+1 = 2un + 1 (n = 0, 1, 2,…..)

Put n = 0, u1 = 1

n = 1, u2 = 3

n = 2, u3 = 7

n = 3, u4 = 15

n = 4, u5 = 31

Seeing the pattern it is clear that un = 2n – 1

Hence u10 = (2)10 – 1 = 1,023.

If there are 10 positive real numbers n1 < n2 < n3 …… < n10…… How many triplets of these numbers (n1, n2, n3), (n2, n3, n4) …. can be generated such that in each triplet the first number is always less than the second number, and the second number is always less than the third number :(CAT 2002)

A. 45
B. 90
C. 120
D. 180

#### SOLUTION

Solution : C

Ans: (c)

Step 1: select the numbers

Step 2: arrange the selected numbers according to the given condition.

Three numbers can be selected and arranged out of 10 numbers in 10C3 ways = 120.

Now, only one arrangement can be possible as A>B>C, hence the solution will be 120.

Value of Ma [md (a), mn (md (b), a), mn (ab, md (ac))] where a = -2, b = -3, c = 4 is (CAT 1994)

A. 2
B. 6
C. 8
D. -2

#### SOLUTION

Solution : B

(b)

Ma [md (a), mn (md (b), a), mn (ab, md (ac))]

Ma [| - 2 |, mn ( | - 3 |, -2), mn (6, | -8 |)]

Ma [2, mn (3, -2), mn (6, 8)]

Ma[2, -2, 6] = 6

Give that a>b then the relation Ma [md (a) . mn (a, b)] = mn [a, md (Ma (a, b))] does not hold if (CAT 1994)

A. a < 0, b < 0,
B. a > 0, b > 0
C. a > 0, b < 0, | a | < | b |
D. a > 0, b < 0, | a | > | b |

#### SOLUTION

Solution : A

Soln:(a)

Ma [md (a), mn (a, b)] = mn [a, md (Ma (a, b)]

Ma [2, -3] = mn [-2, md (-2)]

2 = mn ( -2 , 2)

2 = -2

Relation does not hold for a = -2 and b = -3

Or a < 0, b < 0

fog (x) =

A. 1
B. gof (x)
C. 0
D. 1/x

#### SOLUTION

Solution : B

Option (b)

fog (x) = f{g(x)} = f {(x-3)/2} = 2 {(x-3)/2} + 3 = x

gof (x) = g{f(x)} = g (2x + 3) = (2x + 3 – 3)/2= x

therefore fog (x) = gof (x)

For what value of x; f(x) = g(x – 3)

A. -3
B. C. – 4
D. None of these

#### SOLUTION

Solution : C

Option (c)

f(x) = g(x – 3)

2x + 3 = (x – 3 –3 )/2 = (x – 6)/2

4x + 6 = x – 6

3x = -12

x = -4

What is the value of (gofofogogof) (x) (fogofog) (x)

A. x
B. x2
C. D. #### SOLUTION

Solution : B

Option (b)

{go fo fo go go f(x)} { fo go fo g (x)} from Q. 3, We have fog (x) = gof (x) = x

Therefore above expression becomes (x) . (x) = x2

What is the value of fo(fog)o(gof) (x)

A. x
B. x2
C. 2x + 3
D. #### SOLUTION

Solution : C

Option (c)

fo(fog)0(gof) (x)

We have, fog (x) = gof (x) = x

So given expression reduces to f(x), that is, 2x + 3