# Free Mixed Bag 9 Practice Test - CAT

There is a square of side “a” which has a circle inscribed in it. This circle has another square inscribed in it. Another circle is inscribed in this square. A third square is inscribed in this circle and a third circle is inscribed in this square. Find the ratio of the areas of the outermost square and the innermost circle

A. a:2
B. a:4
C. a2: π
D. a: π
E. π

#### SOLUTION

Solution : C

If a=4, the square-circle-square….. ratio will be 4:π:2:π2: 1:π4

(m+1)2 = 24n+ 2m +2. m is less than 100 and n is any whole number. Find the number of possible values of m ?

A. 10
B. 32
C. 18
D. none of these

#### SOLUTION

Solution : D

Option (d)

m2 -1 =24n. This means that when m2 is divided by 24 it leaves a remainder -1.

m is of the form (6p+1)2 or (6p-1)2.. This is equal 12p(3p+/-1)+1..

We can see that the term 12p(3p+/-1) Is always divisible by 24..,Therefore we have to find the possible values of (6p+1) and ( 6p-1) less than 100.

The receipts on railway travel vary as the excess of speed of the train over 30 kmph. The expenses vary as the square of that excess. What is the speed at which the profits will be greatest if at 60 kmph is the expenses are just covered?

A. 40
B. 35
C. 45
D. None of these

#### SOLUTION

Solution : C

Option(c)
Let the excess of speed over 30
kmph=S

Receipts= R and Expenses=E

Then

R=K1 S

And E= K2S2

Also, at 60kmph, R=E

Thus, K1=30 K2

We need to maximize R-E= K1S- K2S2

=SK2(30-S)

S+30-S, the sum is constant, the product will be maximum when they are equal

Thus S=30-S S=15. Speed = 30+15= 45

Let x1,x2,x3,...,xn be a sequence of integers such that:
i) -1  xi  2 for i = 1, 2 ..., n
ii) x1+x2+x3+...+xn = 19
iii) x21+x22+x23+...+x2n = 99
Determine the minimum and maximum possible values of x31+x32+x33+...+x3n

A. 20,99
B. 19,133
C. 0,391
D. 4, 99

#### SOLUTION

Solution : B

Option (b)

Let a, b and c denote the number of -1s, 1s and 2s in the sequence respectively. So,

-a + b + 2c = 19 and a + b + 4c = 99 (neglecting the zeroes)

So, a = 40 – c and b = 59 – 3c where 0 < c < 19 (as b > 0)

So, x31+x32+x33+...+x3n = -a + b + 8c = 19 + 6c

When c = 0 (a = 40, b = 59), the minimum value is 19; when c = 19 (a = 21, b =2), maximum value is 133.

There is a number which is formed from 6 natural numbered tiles arranged in a row. The value of the first tile (leftmost) is 1. If this tile is moved from the leftmost place to the rightmost place, then the number obtained is three times of the original number. Then the sum of the digits of this special number is?

A. 21
B. 20
C. 19
D. 27

#### SOLUTION

Solution : D

Option(d)

3 * (1ABCDE)  = ABCDE1

The only possibility for the last digit for the number is 7, which when multiplied by 3 gives 1 with a carry over 2

Proceeding this way,the number can be determined as

E = 7, D = 5, C = 8, B = 2, A = 4. The sum of the digits = 27

The remainder obtained when 100*(9910) is divided by (100*99) + 1 is

A. 0
B. 1
C. 100
D. 9900
E. none of these

#### SOLUTION

Solution : D

Option (d)

Go from unitary method, try for smaller numbers. Consider 10x910 = (10.9)99

(10x9)+1

Consider the remainder when 9991. go by frequency method.

9191R=9

9191R=-10

9391R= -90/91|R=+1

Therefore 9991R=1

Remainder will be only the first part of the numerator=10x9, in the case we have considered = 90

And in the question = 100x99= 9900

Digit sum of a number is defined as the sum of the digit till a single digit number is obtained.N = 2! +3! +4! + 5! +……………….10 !. Find the digit sum of N ?

A. 0
B. 9
C. 1
D. 5

#### SOLUTION

Solution : A

Option (a)

Digit sum is nothing but the remainder obtained when the number is divided by 9. Thus from 6! To 10! the number is completely divisible by 9. Hence, digit sum is 0 1!+2!+3!+4!+5! = 1+2+6+24+120 = 153, which is completely divisible by 9.
Hence, digit sum is 0

Four distinct numbers are selected from first 100 natural numbers. How many numbers are there which are divisible by both 3 and 5?

A. 6
B. 7
C. 4
D. None of these

#### SOLUTION

Solution : D

Option(d)

A number divisible by both 3 as well as 5 has to be divisible by 15. In the first 100 natural numbers, there are 6 such numbers-> 15, 30, 45, 60, 75,
90

Favourable number of ways = 6C4

Triangle ABC is a right angled triangle at B. A point O is lying on the side BC. A semicircle with centre O is inscribed inside the triangle such that it touches AC at D. Find the area of quadrilateral ABOD if length AB = 30 units and   BCA = 30.

A. 519.6
B. 432
C. 449
D. 259.8
E. 459

#### SOLUTION

Solution : A BCA = 30

BAC = 60

ODC = 90 (radius is perpendicular to the tangent)

COD = 60

BOD = 120

Now the sides OB = OD (Radii of same circle)

AB = AD (tangents from the same external point)

so, triangle ABO is congruent to triangle ADO

BOA = DOA = 60

Now AB = 30, so OB = 303, so area ( AOB) = area (AOD) = 12 * 30 * 30 3 = 150 3

So area (ABOD) = 300 3 = 519.6.

How many 4 digit numbers can be written as difference of squares in at least one way ?___

#### SOLUTION

Solution :

Option (d)

All numbers of the form 4k+2 cannot be expressed as difference of squares.

Thus, those 4 digit numbers which can be expressed as difference of squares in atleast one way= number of four digit numbers - 4 digit numbers of the form 4k+2

= 9000 -2250= 6750.