Free Mixed Bag 1 Practice Test - CAT 

Euler’s number of  $92=92(1-\frac{1}{23})(1-\frac{1}{2})=88$

$\frac{{19}^{88}}{92}|r\times \frac{{19}^4}{92}|r=1\times \frac{49}{92}|r=49$  (from frequency)

OPTION E

To find the common root, equate the two equations

2x²+gx+1=8x²+2x+g

-6x²+x(g-2)+(1-g)=0

Now find the discriminant of this equation and equate that to 0(D=0 means that the roots are equal and there exists a single root, acording to the question also, there is a single root which is common, hence D=0)

we will get it as : g² - 28g +28 = 0

Check with answer options at this stage

None of the values of x is being satisfied, hence option e.

Going from answer options is the best choice.

**Observe that only options (a) and (c) are in an AP.

By Assumption technique; taking a=2, b=3 and c=4

b-a=1, c-b=1 and a=2, which does not form a GP. Hence this option can never be true

Taking a=1, b=2 and c=3.

b-a=1, c-b=1 and a=1 is in a GP. Hence, answer option (c) is the right one

Option e

The number of AP’s = (Number of factors of 997 - 1), which is prime.
Hence only one AP’s of 998 terms can be formed.
Answer is option a

Two perpendicular lines intersecting at the center of the square will divide the square in four equal areas. Area of the square = 100. So, area of shaded region = $\frac{100}{4}$ = 25.

Option (a)

Digit Sum = Remainder  when  a  number is  divided by 9

$\frac{25!}{9}Rem=0\Rightarrow$ Digit Sum=9

Join AD. As AB is the diameter $\angle$ ADB = 90. And hence $\Delta$ADB is a right angle triangle.

Triangles AEB and DEC are similar hence if the ratio of area is 4: 1, ratio of sides will be 2:1.

So, AE: DE = 2: 1. In triangle ADE, hypotenuse AE = 2, DE = 1 AD${}^2$ = 4 – 1 = 3. AD = $\sqrt{3}$

So, $\angle$ AED = 60 ${}^{\circ }$ .

Option c

Using CRT

5A+3=8B+5

5A=8B+2, B=1 and A=2 .the AP is of the form 40K +13

40K+13=7C+4

7C=40K+9

7c=5K+2  $\Rightarrow$ K=1 and C=7. The AP is of the form 53+280N

Three digit numbers possible in this AP are 53+280=333 , 53+2(280)=613  and  53+3(280)=893.

option (a)

Take a=0. the inequality satisfies for this value as(2<3). Hence option (c) and option (e) are ruled out.

Take a= 3 => 1<0, Which is not true. Hence, option (b) and option (d) are ruled out. Answer is option (a)

Option a and b can be eliminated directly, as from the statement sum of its first 6 terms = 5 times the sum of its next six terms, we can deduce that the Common Difference is negative (decreasing AP).

Go from answer options :

Take option c

Sum to $1^{st}$ 6 terms is given by the formula $\frac{n}{2(2a+[n-1\left]d\right)}$

= 3(200-50)= 450

Sum to $2^{nd}$ 6 terms = Sum to 12 terms – sum to 6 terms

= 6( 200-110) -450

= 540 – 450 = 90

5 $\times$ Sum to $2^{nd}$ 6 terms = 450

Hence option c

$(148{)}^{1084}=(37{)}^{1084}\times (4{)}^{1084}$

$=({37}^4{)}^{271}\times (2{)}^{2168}$

$=({37}^2\times {37}^2{)}^{271}\times ....56$

$=(..69\times ..69{)}^{271}\times .....56$

$=(61{)}^{271}\times ....56=....61\times ....56=......16$
 

Assume a=1, b=2, c=3

X=-3Y=-5Z=2

XY+YZ+XZ = 15-10-6 = -1

Assume a=2,b=3 and c=4

X= -5Y= -7Z= 3

XY+YZ+XZ= 35-21-15= -1

Option(c)
If one of the angles of a cyclic quadrilateral is 90${}^{\circ }$ , the diagonal will be as long as the diameter = 25 cm (12.5 x 2)

Option (c)

Let the first term = 1, cd = 1

Take x = 2, 2x = 4, the progression is 1, 2, 3, 4

New progression = 1, 3

Sum to x terms of progression 1 = 1+2+3+4 = 10

Sum to new progression = 4

$Ratio=\frac{10}{4}=\frac{5}{2}$.
Only option c satisfies this condition.