# Free Number Set I - 02 Practice Test - CAT

Let N=553+173723. N is divisible by : (CAT 2000)

A. both 7 and 13
B. both 3 and 13
C. both 17 and 7
D. both 3 and 17

#### SOLUTION

Solution : D

Ans: (d).

N=553+173723=(54+1)3+(181)3723

=(51+4)3+173(68+4)3.

These two different forms of the expression are divisible by 3 and 17 both.

Elimination strategy:

Check the divisibility by 13, it is seen that it is not divisible, now check for 3, it is

divisible, hence the only option will be (d)

L Let D be a recurring decimal of the form D=0. a1a2a1a2a1a2............; where digits a1 and a2lie between 0 and 9. Further, at most one of them is zero. Which of the following numbers necessarily produces an integer, when multiplied by D? (CAT 2000)

A. 18
B. 108
C. 198
D. 288

#### SOLUTION

Solution : C

Ans: (c)

D=0. a1a2;100 D=a1a2.a1a2 (recurring); Thus, 99D=a1a2;D=(a1a299)

Required number should thus be a multiple of 99. Hence 198 is the required number.

Anita had to do a multiplication. Instead of taking 35 as one of the multipliers, she took 53. As a result, the product went up by 540. What is the new product? (CAT 2001)

A. 1050
B. 540
C. 1040
D. 1590

#### SOLUTION

Solution : D

Ans: (d)

53x - 35x = 540; 18x= 540 or x = 30; Therefore, new product=53×30=1590.

Three pieces of cakes of weights 412 lbs, 634 lbs and 715 lbs respectively are to be divided into parts of equal weights. Further, each part must be as heavy as possible. If one such part is served to each guest, then what is the maximum number of guests that could be entertained? (CAT 2001)

A. 54
B. 72
C. 20
D. None of these

#### SOLUTION

Solution : D

Ans: (d)

Total weight of three pieces: =(92+274+365)=35920=18.45 lb.

Required weight of a single piece is HCF of (92+274+365)=HCF of (9,7,36)LCMof(2,4,5)=920 lb.

Number of guests = 18.45(920)=14

Lety =12+13+12+13+....... What is the value of y?

A. -2
B. 2
C. (153)2
D. (153)2

#### SOLUTION

Solution : D

Ans: (d)

Given:y=

y =12+13+12+13+.......

Thus y = 12+13+y

y=(3+y)(6+2y+1)2y2+7y=3+y; On solving, y=(3±15)2; since the given fraction is positive, the value of y is

(153)2(CAT 2004)

Let S be a set of positive integers such that every element n of S satisfies the conditions
I. 1000n1200
II. Every digit in n is odd
Then how many elements of S are divisible by 3? (CAT 2005)

A. 9
B. 10
C. 11
D. 12

#### SOLUTION

Solution : A

Ans: (a)

The 100th and 1000th position value will be only 1.Divisibility rule for 3 is sum of digits. We already have a sum of 2, we need to make it to 6, 9 or another multiple of 3.

Units digit= a

Tens digit= b

Possible combinations of a+b= 4, 7, 10, 13, 16; where both "a and b" are odd

Thus, the possibility of unit and tens digits are (1, 3), (1, 9), (3, 1), (3, 7), (5, 5), (7, 3),(7, 9), (9, 1), (9, 7).

Let x=4+4x Then x equals : (CAT 2005)

A. 3
B. (131)2
C. (13+1)2
D. 13

#### SOLUTION

Solution : C

Ans :(c)

x=4+4x
x2=4+4x;(x24)=4x
Only option c satisfies this condition.
Shortcut : Reverse Gear
X=4+a small value ( note the minus signs inbetween )
the answer is something slightly greater than 2 .
Look for an answer option, which is slightly greater than 2
(a) 3 is much greater than 2
(b) (131)2  1.1 (this can never be the answer)
(c) (13+1)2  2.2 (this is a possible answer)
(d) 13= something greater than 3. this is also not possible, as the value is too high

Let n!=1×2×3×...........×n for integer n>1. If p=1!+(2+2!)+(3+3!)+...........+(10+10!), then p+2 when divided by 11! leaves a remainder of (CAT 2005)

A. 10
B. 0
C. 7
D. 1

#### SOLUTION

Solution : D

option (d)

If P=1!=1

Then P+2=3, when divided by 2! Remainder will be 1.

If P=1!+2×2!=5

Then, P+2=7 when divided by 3! Remainder is still 1.

Hence, P=1!+(2+2!)+(3×3!)+.....+(10×10!)

When divided by 11! Leaves remainder 1.

Alternative method :
P=1+2.2!+3.3!+....10.10!
=(21)1!+(31)2!+(41)3!+.....+(111)10!
=2!1!+3!2!+...+11!10!
=1+11!
Hence, the remainder is 1.

If R=(30652965)(30642964) , then (CAT 2005)

A. 0<R<=0.1
B. 0.1<R<=0.5
C. 0.5<R
D. R>1

#### SOLUTION

Solution : D

R=306529653064+2964
anbn=(ab)(an1+an2b+an3b2+.....+bn1)
R=(3029)[3064+(3063×29)+....+2964]3064+2964
3064+3063×29+...+2964>3064+2964
R>1
Hence, option 4.
Alternative approach: Unitary Method

R=(30652965)(3064+2964)
If you take this number to be of the form 'n' and 'n+1' instead of 30 and 29, you can solve it using simple numbers like 1(n) and 2(n+1)

i.e (2313)(22+12)=75>1. Only 1 option satisfies this. Option (d)

If x=(163+173+183+193), then x divided by 70 leaves a reminder of? (CAT 2005)

A. 0
B. 1
C. 69
D. 35

#### SOLUTION

Solution : A

Sol:

x=163+173+183+193

=(163+193)+(173+183)

=(16+19)(162+16×19+192)+(17+18)(172+17×18+182)

=35× (an odd number) +35× (another odd number)

=35× (an even number) =35×(2k).......... (k is a positive integer)

x=70k

Therefore x is divisible by 70. Remainder when x is divided by 70 = 0

Hence, option 1.

When you reverse the digits of the number 13, the number increases by 18. How many other two digit numbers increase by 18 when their digits are reversed? (CAT 2006)

A. 5
B. 6
C. 7
D. 8
E. 10

#### SOLUTION

Solution : B

Option (b):

Let the number be (10y + x)- (10x + y) = 18

9(yx)=18yx=2

So, the possible pairs of (x, y) are

(1, 3), (2, 4), (3, 5), (4, 6), (5, 7), (6, 8) and (7, 9).

But we want the number other than 13. Thus, there are 6 possible numbers, i.e., 24, 35, 46, 57, 68,79.

So, total number of possible numbers are 6.

The sum of four consecutive two digit odd numbers, when divided by 10, becomes a perfect square. Which of the following can possibly be one of these four members? (CAT 2006)

A. 21
B. 25
C. 41
D. 67
E. 73

#### SOLUTION

Solution : C

Option (c)

Sum of 4 consecutive odd numbers should result in a zero in the end, so as to be divisible by 10.

This happens in 7+9+1+3. Using options, we find that four consecutive odd numbers are 37, 39, 41 and 43, as the sum of these 4 numbers is 160, When divided by 10, we get 16 which is a perfect square.

Thus, 41 is one of the odd numbers.

A group of 630 children is arranged in rows for a group photograph session. Each row contains three fewer children than the row in front of it. What number of rows is not possible? (CAT 2006)

A. 3
B. 4
C. 5
D. 6
E. 7

#### SOLUTION

Solution : D

Let there be n rows and a students in the first row.

Number of students in the second row = + 3a Number of students in the third row = a+ 6 and so on. aThe number of students in each row forms an arithmetic progression with common difference = 3. The total number of students = The sum of all terms in the arithmetic progression.

=n[2a+3(n1)]2=630

Now consider options.

1. n=3, a=207

2. n=4,a=153

3. n=5,a=120

4. n=6,a=1952

5. n=7,a=8

Hence the only option not possible is when n=6

Let a1=p and b1=q where p and q are positive quantities. Define an=pbn1 and bn=qbn1, for even n>1
and an=pan1,bn=qan1, for odd n>1  .If  p=13 and  q=23, then what is the smallest odd n such that an+bn<0.01?            (CAT 2007)

A. 13
B. 11
C. 9
D. 15
E. 7

#### SOLUTION

Solution : C

Soln:c.
If p=13 and q=23, then p+q=1 and pq=29
a1+b1=p+q=1
a3+b3=p2q+pq2=pq(p+q)=29.
a5+b5=p3q2+p2q3=p2q2(p+q)=481.
So, in general, for odd 'n' we can write
an+bn=(29)(n12) ; we need to find least value of n such that (29)(n12)<0.01
Using the options given, if n=7, a7+b7=(29)3>0.01; if n=9, a9+b9=(29)4<0.01

How many pairs of positive integers m, n satisfy (1m)+(4n)=(112), where n is an odd integer less than 60? (CAT 2007)

A. 4
B. 7
C. 5
D. 3
E. 6

#### SOLUTION

Solution : D

Soln:d.
Given conditions :
(1m)+(4n)=(112)
M and n are positive integers
N is an odd integer less than 60.
On solving the equation, we get m=12n(n48);
Since m and n are positive integers, n>48, and from the question, 48<n<60.
Hence possible values on n=49,51,53,55,57,59.
For n=49,51 and 57 we get integral values of m. Hence the required no of integral pairs = 3