CAT4MBA

Q1. The number FIVE as written using block capitals contains exactly 10 strokes or segments of a straight line. Find a number which when written out in words (using no tricks) contains as many strokes as the number says.

Solution:

TWENTY – NINE is the number which is written with as many strokes as the value of the number it states.

2 + 4 + 4 + 3 + 2 + 3 + 3 + 1 + 3 + 4 = 29

Thank You.

Ravi Raja

Q2. The number of 1's in the binary notation of (2^89) – 1 is

(a) 89

(b) 88

(c) 90

(d) 1

Solution:

(2^1) – 1 when written in the binary notation is written as 1 and thus contains one 1.

(2^2) – 1 when written in the binary notation is written as 11 and thus contains two 1s.

(2^3) – 1 when written in the binary notation is written as 111 and thus contains three 1s.

(2^4) – 1 when written in the binary notation is written as 1111 and thus contains four 1s.

……………………

……………………

……………………

……………………

and so on.

Now, make a note of the pattern that is being followed and we can conclude that (2^89)- 1 when written in the binary notation, will contain eighty – nine 1s.

Thank You.

Ravi Raja

Q3. The highest power of 2 in 10! + 11! + 12! + 13! + ...+ 1000! is
(a) 8
(b) 9
(c) 10
(d) 11

Solution:

The highest power of 2 contained in the sum of a set of factorials of different numbers is equal to the highest power of 2 contained in the least of those numbers because the highest power of 2 contained in the rest of them will always be greater than or equal to the highest power of 2 contained in the least of them.

Hence in the given problem, the highest power of 2 will be equal to the highest power of 2 contained in 10! and that is equal to 8.

Thank You.

Ravi Raja

Q4. Assume that all bricklayers work at the same rate of speed. If it takes nine bricklayers (all working at the same time) fourteen days to do a job, how long would it take for the job to be done by

a) Seven bricklayers

b) Three bricklayers

Solution:

Here we will be using the result (M1 x D1)/(W1) = (M2 x D2)/(W2)

Where M denotes the number of men involved in doing the work, D denotes the number of days required to complete the work and W denotes the quantity of work done or to be done (and is generally taken as 1, unless the exact work is specified)

Solution (a):

M1 = 9, D1 = 14, W1 = 1, M2 = 7, W2 = 1 (as the same work is to be done)

(M1 x D1)/(W1) = (M2 x D2)/(W2)

or, (9 x 14)/(1) = (7 X D2)/(1)

or. D2 = (9 x 14)/(7) = 9 x 2 = 18 days, which is the required answer.

Solution (b):

M1 = 9, D1 = 14, W1 = 1, M2 = 3, W2 = 1 (as the same work is to be done)

(M1 x D1)/(W1) = (M2 x D2)/(W2)

or, (9 x 14)/(1) = (3 X D2)/(1)

or. D2 = (9 x 14)/(3) = 3 x 14 = 42 days, which is the required answer.

Thank You.

Ravi Raja

Q5. If both 11^2 and 3^3 are factors of the number (a) * (4^3) * (6^2) * (13^11), then what is the smallest possible value of a?

1. 121

2. 3267

3. 363

4. 33

Solution:

The expression (a) * (4^3) * (6^2) * (13^11) can also be written as

(a) * {(2^2)^3} * {(3 * 2)^2} * (13^11)

= (a) * (2^6) * (3^2) * (2^2) * (13^11)

= (a) * (2^8) * (3^2) * (13^11)

Now, it is given that both 11^2 and 3^3 are factors of the given number. That is, (11^2) * (3^3) is a factor of the given number.

Now in the simplified form of the given number, we can see that it already contains the factor 3. Hence ‘a’ has to contain at least 11^2 and 3 as its factors and so, the minimum value of ‘a’ has to be (11^2) * (3) = 121 * 3 = 363.

Thank You.

Ravi Raja

Q6. Let (x, y) be co-prime numbers. Then
(a) x + y and x – y have no common factor other than 1 and 3 for all values of x and y.
(b) x + y and x – y have no common factor other than 1 and 2 for all values of x and y.
(c) x + y and x – y have no common factor other than 1 for all values of x and y.
(d) None of the above

Solution:

Now, note that 1 is a common factor of all pairs of integers (a, b). Hence we just have to check whether 2 or 3 is a common factor of x + y and x – y.

Suppose that 3 is a common factor of x + y and x – y.

Then both x + y and x – y can be written in the forms 3p and 3q, where p and q are co – prime integers.

That is,
x + y = 3p, for some integer p and
x – y = 3q, for some integer q, where p and q are co – prime integers.

Now, by adding and subtracting both the equations, we get:
2x = 3(p + q) ------------- (1)
2y = 3(p – q) ------------- (2)

Now, x, y, p and q all being integers, it is clear from both (1) and (2) that both x and y are divisible by 3.

But that is not possible since it is given that x and y are co – prime numbers.

Hence option (a) is ruled out.

Now, similarly, suppose that 2 is a common factor of x + y and x – y.

Then both x + y and x – y can be written in the forms 3p and 3q, where p and q are co – prime integers.

That is,
x + y = 2p, for some integer p and
x – y = 2q, for some integer q, where p and q are co – prime integers.

Now, by adding and subtracting both the equations, we get:
2x = 2(p + q)
2y = 2(p – q)

x = (p + q) ------------- (1)
y = (p – q) ------------- (2)

Now, p and q being co – prime numbers, we can just check by taking examples and see that both p + q and p – q can have a common factor 2 and that implies that both x and y can have a common factor 2, which again contradicts the given condition of the problem that x and y are co – prime numbers.

Hence option (b) is also ruled out.

Now, consider the general form that, suppose that k is a common factor of x + y and x – y.

Then both x + y and x – y can be written in the forms kp and kq, where p and q are co – prime integers.

That is,
x + y = kp, for some integer p and
x – y = kq, for some integer q, where p and q are co – prime integers.

Now, by adding and subtracting both the equations, we get:
2x = k(p + q) ------------- (1)
2y = k(p – q) ------------- (2)

Now, x, y, p and q all being integers, it is clear from both (1) and (2) that both x and y are divisible by k, which again is not possible since it is given that x and y are co – prime numbers.

Hence k cannot take any value other than 1 and this implies that x + y and x – y have no common factor other than 1 for all values of x and y.

Thank You.

Ravi Raja

Q7. What is the least number that should be multiplied to 100! to make it perfectly divisible by 7^18?
a. 1
b. 7
c. 21
d. 49

Solution:

Check that the highest power of 7 contained in 100! Is:
[ 100/7 ] + [ 100/49 ] + [ 100/343 ] + [ 100/2401 ] + ……
= 14 + 2 + 0 + 0 + ……
= 16

(Note: Here [ x ] denotes the Greatest Integer Function)

Hence for 100! to be perfectly divisible by 7^18, it should contain two more sevens in it and that can be obtained by multiplying it by 7^2 = 49 and that is the least such number to be multiplied to 100! to make it perfectly divisible by 7^18.

Thank You.

Ravi Raja

Q8. A car has traveled 24,000 km and, in that distance, has worn out 6 tyres. Each tyre travelled the same distance. How far did each separate tyre travel?

Solution:

The number of ways in which 4 tyres can be selected out of 6 are 6C4 = 15 ways and thus each tyre has covered 6/15 of the total distance covered and that is equal to (6 x 24,000)/(15) = 6 * 1600 = 9,600 km.

Thank You.

Ravi Raja

Q9. The remainder when 3^256 is divided by 100 is

(a) 61

(b) 21

(c) 41

(d) 81

Solution:

The number of integers less than 100 and prime to 100 are 40.

Using Euler – Fermat Theorem, we see that 3^40 when divided by 100, the remainder will be 1 and therefore, (3^40)^6 = 3^240 when divided by 100, the remainder will be 1.

So basically we have to find the remainder when 3^16 is divided by 100 and then multiply the result by 1 and get that as my required answer.

3^16 = (3^4)^4 = 81^4 = (81^2)^2 = (6561)^2 = (6600 – 29)^2

So now it is clear that the remainder when 3^16 is divided by 100 will be the same as the remainder when (6600 – 29)^2 is divided by 100. That is, the remainder when 29^2 = 841 is divided by 100 and that is 41, which is the required answer.

Thank You.

Ravi Raja

Q10. Find the sum 1/(1.2) + 1/(2.3) + . . . .+ 1/(10.11)
a. 1
b. 2
c. 10/11
d. 11/12
e. 2531/2520

Solution:

1/(1.2) + 1/(2.3) + . . . .+ 1/(10.11)

= (2 – 1)/(1.2) + (3 – 2)/(2.3) + …… + (11 – 10)/(10.11)

= [{2/(1.2)} – {1/(1.2)}] + [{3/(2.3)} – {2/(2.3)}] + …… + [{11/(10.11)} – {10/(10.11)}]

(Since (a – b)/(a.b) = {a/(a.b)} – {b/(a.b)} = (1/b) – (1/a)

= (1/1) – (1/2) + (1/2) – (1/3) + (1/3) – (1/4) + … … + (1/9) – (1/10) + (1/10) – (1/11)

Now, note that all the terms between (1/1) and – (1/11) get cancelled.

Hence the required sum is: (1/1) – (1/11) = (11 – 1)/(1.11) = 10/11.

Thank You.

Ravi Raja