## Wednesday, 28 December 2011

### Puzzle – 31 (XAT-2011)

On 1st March, Timon arrived in a new city and was looking for a place to stay. He met a landlady who offered to rent her apartment at a reasonable price but wanted him to pay the rent on a daily basis. Timon had a silver bar of 31 inches and an inch of the silver bar was exactly equal to a day’s rent. He agreed to pay an inch of the silver bar towards the daily rent. Timon wanted to make minimum number of pieces of silver bar but did not want to pay any advance rent. How many pieces did he make?
(1)5                 (2)8                 (3)16               (4)20               (5)31
Solution follows here:

## Tuesday, 27 December 2011

### Series-6 (CAT-2004)

Consider the sequence of numbers a1, a2, a3,... to infinity, where a1 = 81.33 and a2 = -19 and aj = aj-1 – aj-2. What is the sum of first 6002 terms of this sequence?
(1)-100.33                  (2)-30.00                    (3)62.33                     (4)119.33
Solution follows here:

## Monday, 26 December 2011

### Geometry - 19

On a semicircle with diameter AD, chord BC is parallel to the diameter. Further, each of the chords AB and CD has length 2, while AD has length 8. What is the length of BC?
(1)7.5              (2) 7                (3) 7.75                       (4) None of these

### Algebra-32 (CAT-2004)

If f(x) = x3-4x+p, and f(0) and f(1) are of opposite signs,then which of the following is necessarily true?
(1)-1 < p < 2               (2) 0 < p < 3               (3) -2 < p < 1              (4) -3 < p < 0
Solution follows here:

### Arithmetic-22 (CAT-2004)

Two boats travelling at 5 and 10 km per hour, head directly towards each other. They begin at a distance of 20 km from each other. How far apart are they (in kms) one minute before they collide?
(1)1/12                       (2)1/6             (3)1/4             (4)1/3
Solution follows here:

### Algebra - 31

A telecom service provider engages male and female operators for answering 1000 calls per day. A male operator can handle 40 calls per day whereas a female operator can handle 50 calls per day. The male and the female operators get a fixed wage of Rs.250 and Rs.300 per day respectively. In addition, a male operator gets Rs.15 per call he answers and a female operator gets Rs.10 per call she answers. To minimize the total cost, how many male operators should the service provider employ assuming he has to employ more than 7 of the 12 female operators available for the job?
(1)15   (2) 14   (3) 12   (4) 10

## Wednesday, 21 December 2011

### Algebra-30 (NCERT)

Which is larger (1.01)1000000 or 10000?
Solution follows here:
Solution:
This is a straight forward question on Binomial expansion.
(x+y)n = nC0 xn y0 + nC1 xn-1 y1 + nC2 xn-2 y2 + . + ncn x0 yn
(1.01)1000000
= (1000000)C0 + (1000000)C1 (0.01)1 + (1000000)C2 (0.01)2 +…. + (0.01) 1000000
= 1 + (1000000)(0.01) + …..
= 1+ 10000 + …
> 10000
Hence (1.01)1000000 is greater than 10000

## Tuesday, 20 December 2011

### Algebra-29

If (x2+1)/x = 5, then find the value of (x12+1)/x6?
(1)120098      (2) 5    (3) 12100       (4) 56   (5) none of these
Solution follows here:
Solution:
This is a bit changed in shape from the traditional one,
“If x + 1/x = 5 then find x6 + 1/x6
As we need to find 6th power, first we proceed for cube
x + 1/x = 5 => (x + 1/x)3 = 53
Formula to be used here : (a+b)3 = a3+b3+3ab(a+b)
=> x3 + 1/x3 + 3x(1/x)(x+1/x) = 125
=> x3 + 1/x3 + 3 (5) = 125
=> x3 + 1/x3 = 110
Now squaring on both sides,
(x3 + 1/x3)2  = 1102
Formula to be used here : (a+b)2 = a2+b2+2ab
=> x6 + 1/x6 + 2x3(1/x3) = 12100
=> x6 + 1/x6 + 2 = 12100
=> x6 + 1/x6 = 12098
Be careful at the answer options, 120098 is given to trap.

### Numbers -25

Find the remainder when 540 is divided with 11?
(1)1     (2) 0    (3) 2    (4) 10  (5) none of these
Solution follows here:

## Monday, 19 December 2011

### Algebra – 28 (CAT-2005)

Let n! = 1*2*3*...n, for integer n ≥ 1. If p = 1!(2*2!) + (3*3!) + (4*4!) +.....(10*10!), then p+2 when divided by 11! Leaves a remainder of _____
(1)10               (2)0                 (3)7                 (4)1
Solution follows here:

### P & C – 13

Let S be the set of five digit numbers formed by the digits 1,2,3,4 and 5, such that exactly two odd positions are occupied by odd digits. What is the sum of the digits in the right most position of the numbers in S?
(1)228             (2)216             (3)294             (4)192

### Numbers –24 (CAT-2005)

The right most non zero digit of the number 302720 is?
(1)1     (2)3     (3)7     (4)9

Solution follows here:

### Numbers –23 (CAT - 2005)

For a positive integer n, let Pn denotes the product of digits of n, and Sn denotes the sum of the digits of n. The number of integers between 10 and 1000 for which Pn+ Sn = n is?
(1)81               (2)16               (3)18               (4)9
Solution follows here:

## Sunday, 18 December 2011

### Numbers – 22

Find the number of integers lying between 100 and 400 (excluding the two) and divisible by any one of the numbers 2 or 3 or 5 or 7?
Solution follows here:

## Friday, 16 December 2011

### Algebra - 27 (CAT-2007)

A function f(x) satisfies f(1) = 3600, and f(1)+f(2)+........+f(n) = n2f(n), for all positive integers n>1. What is the value of f(9)?
(1)80               (2)240             (3)200             (4)100             (5)120

Solution follows here:

### Puzzle-30

Suppose you have a currency named Miso, in three denominations: 1 Miso, 10 Misos and 50 Misos. In how many ways can you pay a bill of 107 Misos?
(1)17   (2)16   (3)18   (4)15   (5)19

### Puzzle-29 (CAT-2007)

In a tournament there are n teams T1,T2,...Tn where n>5. Each team consists of k players, k>3. The following pairs of teams have one player in common:
T1 & T2, T2 & T3, T3 & T4,......... and Tn & T1
No other pairs of teams have any player in common. How many players are participating in the tournament considering all the n teams together?
(1)n(k-1)         (2)k(n-1)         (3)n(k-2)         (4)k(k-2)         (5)(n-1)(k-1)

Solution follows here:

## Thursday, 15 December 2011

### Numbers –21

John said “I have attempted 95% of the questions correctly in the examination”. What might be the minimum possible number of questions given in the examination?
(1)19   (2)10   (3)20   (4)100 (5)95
Solution follows here:

## Wednesday, 14 December 2011

### Algebra -26 (CAT-2004)

The remainder, when (1523+2323) is divided by 19 is:
(1)4                 (2)15               (3)0                 (4)18
Solution follows here:

### Numbers - 20

Using only 2,5,10,25 and 50 paise coins, What will be the minimum number of coins required to pay exactly 78 paise,69 paise and Rs 1.01 to three different persons?
(1)19               (2)20               (3)17               (4)18

## Tuesday, 13 December 2011

### Numbers-19 (XAT-2011) (Data Sufficiency)

The question is followed by two statements labelled as I and II. Decide if these statements are sufficient to conclusively answer the question. Choose the appropriate answer from the options given below:
A. Statement I alone is sufficient to answer the question.
B. Statement II alone is sufficient to answer the question.
C. Statement I and statement II together are sufficient, but neither of the two alone is sufficient to answer the question.
D. Either statement I or statement II alone is sufficient to answer the question.
E. Neither statement I nor statement II is necessary to answer the question.

Q) Given below is an equation where the letters represent digits.
(PQ).(RQ) = XXX. Determine the sum of P+Q+R+X                            (3 Marks)
I.                    X=9
II.                  The digits are unique
Solution follows here:

## Thursday, 8 December 2011

### Puzzle-28 (CAT-2004)

In the below figure, the lines represent one way roads allowing travel only northwards or only westwards. Along how many distinct routes can a car reach point B from point A?
 B ↑ North ßWest A

(1)15               (2)56               (3)120             (4)336
Solution follows here:

### Puzzle-27 (CAT-2004)

Each family in a locality has at most two adults, and no family has fewer than three children. Considering all the families together there are more adults than boys, more boys than girls and more girls than families. Then the minimum possible number of families in the locality is:
(1)4                 (2)5                 (3)2                 (4)3

### Numbers-18 (CAT-2003ii)

If a,a+2,a+4 are prime numbers, then the number of possible solutions for a is:
(1)   one                  (2) two            (3) three         (4) more than three
Solution follows here:

### Containers – Items (P and C Concepts)

There are 3 containers and 5 items. In how many ways can these items be placed in the containers?

This is a conceptual one and we can see four different cases here:

Type-I:    Similar Containers – Similar Items:

 X X X 5 0 0 4 1 0 3 1 1 3 2 0 2 2 1

5 ways

Type-II:     Different Containers – Similar Items:

Type-II(A) Each container having at least one item:

 X Y Z 3 1 1 3!/2! = 3 ways 2 2 1 3!/2! = 3 ways

## Friday, 2 December 2011

### Income Tax - 1 (XAT-2011)

Answer questions 1 and 2 based on the following information:
 Income-Tax rates for the financial year 2009-10 Individual & HUF below age of 65 Years Women below age of 65 Years Tax Rates Income upto Rs. 1,60,000/- Income upto Rs. 1,90,000/- Nil Rs. 1,60,001/- to Rs. 3,00,000/- Rs. 1,90,001/- to Rs. 3,00,000/- 10% Rs. 3,00,001/- to Rs. 5,00,000/- Rs. 3,00,001/- to Rs. 5,00,000/- 20% Above Rs. 5,00,001/- Above Rs. 5,00,001/- 30%

## Thursday, 1 December 2011

### Triangle Inequality Theorem

Length of any side of a triangle is less than the sum of the lengths of other two sides

a < b+c            b < a+c            c < a+b                                    ---(1)
From this we can easily derive the following:
a > b~c            b > a~c            c > a~b                                    ---(2)
From (1) and (2),
b~c < a < b+c             a~c < b < a+c             a~b < c < a+b

This theorem is most important for competitive exams. You can find Geometry-12 problem on this blog, which is based on this theorem.