# Free Algebra 05 Practice Test - CAT

From the set of first 10 natural numbers, three distinct prime numbers a, b and c are selected to form a quadratic equation ax2 + bx + c = 0, having real roots. Find the sum of the roots of all such possible quadratic equations that can be formed.

A. -19/2
B. 38
C. -27/5
D. - 87/10
E. -149/10

#### SOLUTION

Solution : E

The two real roots of each of the possible quadratic equations will be distinct. For real roots we must have b2 ≥ 4ac. As a, b and c distinct prime numbers from 1 to 10, only the following triplets (a, b, c) are possible: (2, 5, 3), (3, 5, 2), (2, 7, 3), (3, 7, 2), (2, 7, 5), (5, 7, 2).

Accordingly, the sum of roots of all such formed is Let F(x) = max {(x + 3), (7 – 2x)}. What is the minimum value of F(x) for 2 ≥ x ≥ 1

A. 4.50
B. 5.00
C. 4.33
D. 4.67
E. No value Satisfies the given constrins

#### SOLUTION

Solution : C

Please note that we are required to find out the minimum value of F(x), but F(x) always prefer to give the maximum of the two values (i.e., (x + 3), (7 - 2x)).

Thus there is only one possible condition to get the minimum of F(x) that is when both values i.e., (x + 3) and (7 - 2x) are equal.

(x+3) = (7-2x)

3x=4

x=4/3

This satisfies the constraint of 2 ≥ x ≥ 1. Minimum value will occur at x= 4/3 = max (13/3, 13/3) = 13/3 = 4.33

Given that a,b and c are positive real numbers, such that a+b+c= x, then which of the following is true?

A. (x-a)(x-b)(x-c)≥8abc
B. 1/a + 1/b + 1/c ≥ 9/x
C. (x-a)(x-b)(x-c)≥8/27a3
D. a,b and c
E. None of a, b and c

#### SOLUTION

Solution : D

option (d)

Assumption of values is the easiest way to solve. a=b=c=1 and x=3.

Take a=1, b=2 and c=3. x=6. all 3 still hold good. Answer option d.

Note: with answer options like “all of these” and “none of these” you need to be a bit careful. Do not mark the answer based on only one set of values. Take two cases and see which options fit in BOTH the cases.

Solve for a , | (3a-4)/(8-a)| ≤ 1

A. -2≤a≤3
B. -3≤a≤2
C. a≤-2, a≥3
D. a<6
E. a<3

#### SOLUTION

Solution : A

option (a)

Check for a=5, 11/3 1,it is not satisfied. Option (c) and option (d) are ruled out

Check for a=3, 3/3 =1, it is satisfied. Option (b) and option (e) are out

The perimeter of a rectangular sheet of paper is 84 cm. If the length of the paper is increased by 3cm and the breadth decreased by 1m, then the area of the paper is increased by 15 cm2. What is the difference in the length and breadth of the original rectangular sheet of paper?

A. 12 m
B. 15 m
C. 41 m
D. 15
E. None of these

#### SOLUTION

Solution : A

option (a)

Let the length =l and breadth =b

2(l+b)=84; l+b = 42----------(1)

Area =lb

When changes are made, (l+3)(b-1)= lb + 3b – l – 3

Given that lb + 3b  – l  –3 = lb +15

3b  – l=18--------------------(2)

by solving we get

l=27 and b=15. difference = 12

Mr. Sam has a wife and three children. Mrs. Sam is three years younger to Mr. Sam. The ages of the three children, being in AP, add up to 6/5 times that of Mr. Sam’s age. While the youngest child of Mr. Sam is a female, the sum of the ages of the three males in the family is 85 years. The age of the second son of Mr. Sam can be:

A. 22
B. 19
C. 18
D. 14
E. 21

#### SOLUTION

Solution : C

First write down the constraints given to you

Let the ages of the three children be a-d, a and a+d.

Let the age of Mr Sam be s

a-d + a + a+d = 6/5s

3a=6/5s; 5a=2s---------------(1)

2a + d + s = 85---------------(2)

Now start eliminating answer options by assuming a value for the second son whose age is “a”

‘a’ cannot be odd, as it has to be divisible by 2 (Equation 1).

option (b) and option (e) are out

consider a middle option

if a= 18 , s= 5x9 =45

from (3) 36 + d + 45 = 85 d= 85-81= 4.. the ages of the three children are 14, 18 and 22.This is the answer

a=22 will give a negative d which is not possible

a=14, d>a, which is not possible. Hence option c is correct

The complete solution set of ‘x’ for |x-1|+|x-2| + |x-3| + |x-4| > 10 is

A. (-1,2) U (3, )
B. (- ,0) U (5, )
C. (- ,-2) U (2, )
D. (- ,-1) U (4, )
E. (- ,- 1] U [5, ;)

#### SOLUTION

Solution : B

Substitute x= 0, we will get f(x) = 10. So when x = -ve we will get f(x) > 10.

Also, we will get f(x) >10 only when x > 5. hence, answer is option (b)

Or we can also go from the answer option and eliminate the wrong options.

A boy bought 3 apples, 5 mangos and 8 bananas for Rs.87. Had he bought 5 apples 7 Mangos and 6 Bananas, it would cost him Rs.121. By how much would an apple and a mango together cost more than a banana?

A. Rs. 15
B. Rs. 17
C. Rs. 19
D. Rs. 20
E. Cannot be determined

#### SOLUTION

Solution : B

3A + 5M + 8B = 87 ........................... (1)

5A + 7M + 6B = 121 ........................ (2)

Equation (2) - Equation (1)

2A + 2M -2B = 34

A+M-B=17

If x2 + y2 +z2 =1. What is the range of xy+yz+zx?

A. [-1/2, 3]
B. [-1,2]
C. [-1/2, 1]
D. [-1,1/2]
E. [-1,1/2]

#### SOLUTION

Solution : C

option (c)

x2 + y2 + z2 = 1...............................(1)

We know that

x2 + y2 + z2 + 2(xy + yzxz) = (x +y +z)2  ≥ 0

1+ 2(xy + yz +xz) ≥0

xy + yz +xz ≥ -1/2.............................(2)

since AM≥GM From above equations

x2 +y2 +z2xy + yz+ xz

This implies that xy+yz+zx<1

Therefore the interval is [-1/2, 1]

If a2b3c = 256/27 find min value of a + b + c, given a, b, c are positive real nos

A. 10
B. 11
C. 12
D. 4
E. 6

#### SOLUTION

Solution : D

Solution:option d

Take numbers as a/2, a/2, b/3, b/3, b/3,c

Now apply AM>GM on these six numbers. We have a2b3c = 256/27. Solving we get min value of a+b+c as 4. Hence correct option: d

What is the maximum number of negative real roots which can be formed for the following equation x8+9x6+18x5-9x3+4x2+2x+1=0?

A. 1
B. 2
C. 3
D. 4
E. 0

#### SOLUTION

Solution : D

To find the number of negative real roots, we need to find the number of sign changes in f(-x)

To find f(-x) replace x by –x in the equation

f(-x) = x8+9x6-18x5+9x3+4x2-2x+1

Number of sign changes = 4 Maximum number of –ve real roots = 4

Find the values of x such that x2-5x-6>0 and x2-4x+3≤0?

A. -1≤x≤2
B. x=0, x≤-4
C. 0≤x≤4
D. 1≤x≤3
E. None of these

#### SOLUTION

Solution : E

option (e)

Conventional Method:-

Factorize the equations

x2-6x+x-6>0

(x-6)(x+1)>0

x<-1 and x> 6

x2-4x+3≤0

x2-3x-x+3≤0

x(x-3)-1(x-3) ≤0

(x-1)(x-3) ≤0

1≤x≤3

there is no common solution for these equations. Hence, the answer is option (e)

**Shortcut “Reverse Gear approach”

Substitute a value which is there in two of the options and absent in the others

Substitute x=2

At x=2, equation (1) is not satisfied. Option (a) and option (d) can be eliminated

At x=0,both equations are not satisfied, hence option (b) and option (c) can also be eliminated. Answer is option (e)

Note: this is a tricky question. Whenever “none of these” is involved, you need to be a bit careful

In the equation, x6-12x5+ax4-bx3+cx2-dx+64=0. Find a+b+c+d

A. 362
B. 662
C. 562
D. 652
E. Cannot be determined

#### SOLUTION

Solution : D

Ans. (d)

Since sum of roots is given to you as 12 and product as 64. the six roots are 2,2,2,2,2,2.

Sum of roots taken two at a time (a) = 6C2x4=60

Sum of roots taken three at a time(b)= 6C3x8=160

Sum of roots taken 4 at a time (c)= 6C4x16=240

Sum of roots taken 5 at a time(d)=6C5x32=192

a+b+c+d=652

Find the constant term of a cubic equation b3+mb2 +nb + c given that it has 2 roots which are the same as that of the quadratic equation dx2-ex+f and the third root as the sum of the roots of the quadratic equation The sum of the roots of the quadratic equation = 13 and product =40

A. 365
B. 380
C. -120
D. 520
E. None of these

#### SOLUTION

Solution : E

Option (e)

Let the roots of the quadratic equation be p & q p+q=13 pxq=40

p=8 q=5.

The roots of the cubic equation are 8,5 and 13

The cubic equation can be represented as (b-8)(b-5)(b-13)=0

Constant term = (-1)3(8x5x13)= -520

X and Y are integers which lie between [-5,15] and [-5,10] respectively Which of the following represents the least value?

A. Y3X
B. X2Y2
C. XY/2
D. (X2+Y3)(-X)
E. - X

#### SOLUTION

Solution : D

Find the least value of each to find the least value amongst the four

a) Y3X = 1000 x (-5) = -5000

b) X2Y2= 0 x 0 = 0

c) XY/2= 15(-5)/2= -75/2= -37.5

d) (X2+Y3)(-X)= (225+1000)(-5)= -6125

e) –X = -15