# Free Basic Algebra - 01 Practice Test - CAT

Find the number of terms of the sequence 32, 24, 16, 8 ... for which the sum of the terms is zero.

A. 7
B. 8
C. 6
D. 9

#### SOLUTION

Solution : D

Sn=(n2)[2a+(n1)d]
0=(n2)[64+(n1)(8)]
0=(n2)[64(n1)8]
n (36 – 4n) = 0
n = 0; n=9

As, n can not be equal to 0, so, the number of terms required is 9.

Find the sum of the first five terms of the series: 3, 12, 48...

A. 512
B. 1023
C. 198
D. 343

#### SOLUTION

Solution : B

Sn=[a(rn1)r1]

In the question a = 3, r = 4, n = 5

S5=3×(451)(41)=1023

a, b, c are the three geometric means between 96 and 6. The value of a+b+c is ___.

A. 84
B. 76
C. 86
D. 72

#### SOLUTION

Solution : A

6, _, _, _, 96

If the common ratio is r, 96=6(r)4

r4=16,  r=2

So, the geometric means are: 12, 24 and 48.

a+b+c = 84

Find the eight term of the series: 1316+112124....

A. 1192
B. 1192
C. 1384
D. 1128

#### SOLUTION

Solution : C

The series is a GP with common ratio r=(1613)=12

Eight term =ar7=13×(12)7=1384

The third term of a GP is 3. Find the product of first five terms.

A. 81
B. 243
C. 256
D. 343

#### SOLUTION

Solution : B

ar2ar, a, ar, ar2 are the first five terms.

Third term = 3

Product of first five terms = a5 = 343.

Sum of three consecutive terms in a GP is 42 and their product is 512. Find the largest of these numbers.

A. 32
B. 64
C. 128
D. none of these

#### SOLUTION

Solution : A

Let the three consecutive terms be ar,a,ar

Product = (ar)(a)(ar) =  a3 = 512,  a = 8

Sum of these numbers =a(1+r+1r)=428(1+r+1r)=42

4r217r+4=0

(4r – 1)(r – 4) = 0

r =  14or r = 4.

Hence the largest number is 32. The series is 32, 8, 2 or 2, 8, 32.

The sum of first two terms of a G.P is 53 and the sum to infinite terms is 3. What is the first term?

A.

1

B.

6

C.

9

D.

35

#### SOLUTION

Solution : A

The first two terms = a, ar

a+ar= 53

a(1r)=3

a=3(1r)

a(1+r)=53

3(1r)(1+r)=53

1r2=59

r2=49

r=23

a=3(123)=1

First term is 1.

The A.M. of two positive numbers is 15 and their GM is 12. What is the bigger number?

A. 16
B. 22
C. 24
D. 18

#### SOLUTION

Solution : C

Let the numbers be x and y.

x+y=30

xy = 144

(xy)2=(x+y)24xy

900(4×144)=324

hence (xy)=18

Solving:xy=18 & x+y=30

x = 24,       y = 6

So, the bigger number is 24.

If A.M. and G.M. of two numbers is 10 & 8 respectively, then find their H.M.

A. 6.4
B. 5.6
C. 1.56
D. 1.64

#### SOLUTION

Solution : A

G.M=(A.M×H.M)12
64=10×H.M
H.M=6.4

The sum of 3rd and 15th elements of an arithmetic progression is equal to the sum of 6th,11th,13thelements of the same progression. Then which element of the series should necessarily be equal to zero?

A. 12th
B. 11th
C. 9th
D. 8th

#### SOLUTION

Solution : A

Soln:

The 3rd  term will be x+2d

The 15th  term will bex+14d

From the information given in the question,

x+2d+x+14d=x+5d+x+10d+x+12d

Solving we get,x+11d=0, this is nothing but the 12th term of the series.

Find the roots of the following equations:
6x2x2=0

A. 12,23
B. 12,23
C. 12,23
D. 12,23

#### SOLUTION

Solution : B

6x24x+3x2=0
2x(3x2)+1(3x2)=0
(2x+1)(3x2)=0
The roots are:12,23

How many distinct real roots are possible for the below equation?
x626x327=0

A. 6
B. 2
C. 3
D. 5

#### SOLUTION

Solution : B

Let x3=y

y226y27=0

y227y+y27=0

y(y27)+1(y27)=0

(y+1)(y27)=0

The roots are y = -1, y = 27

x3=1,x=1

x3=27,x=3

How many real roots are possible for the below equation?
(x+1)(x+2)(x+3)(x+4)=24

A. 0
B. 2
C. 4
D. Cannot be determined

#### SOLUTION

Solution : B

[(x+1)(x+4)][(x+2)(x+3)]=24

(x2+5x+4)(x2+5x+6)=24

Let x2+5x=k

(k+4)(k+6)=24

k2+10k=0

k(k+10)=0

k=0,k=10

x2+5x=0,x2+5x=10

x(x+5)=0,x2+5x+10=0

x=0,x=5;D=2540=15  No real roots.

So, roots are 0, -5.

6(x2+1x2)25(x1x)+12=0

A. 12,13,2,3
B. 12,13,2,3
C. 12,13,2,3
D. 12,13,2,3

#### SOLUTION

Solution : A

Let x1x=z
(x1x)2=z2
(x2+1x22)=z2
x2+1x2=z2+2
Now substituting in the equation:
6(z2+2)25z+12=0
6z2+1225z+12=0
6z225z+24=0
6z216z9z+24=0
2z(3z8)3(3z8)=0
(2z3)(3z8)=0
z=32     ; z=83
x1x=32
x1x=83
2x2=3x2=0
3x28x3=0
2x24x+x2=0
3x29x+x3=0
2x(x2)+1(x2)=0
3x(x3)+1(x3)=0
(2x+1)(x2)=0
(3x+1)(x3)=0
x=12  ,  x=2
x=13  ,  x=3
So,the roots are:12,2,13,3

Find the roots of the following equations:
2x+3+2x=6

A.

-1, -2

B.

1, -2

C.

- 1, - 2

D.

1, 2

#### SOLUTION

Solution : C

2x.23+2x=6

Let 2x=z

8z+1z=6

8z2+1=6z

8z26z+1=0

8z24z2z+1=0

4z(2z1)1(2z1)=0

(4z1)(2z1)=0

z=14,

z=12

2x=14

x=2

2x=12

x=1

Roots are  x=2,x=1