Question 1

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

Question 2

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

Question 3

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

Question 4

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

Question 5

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

Question 6

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

Question 7

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

Question 8

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

Question 9

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

Question 10

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

Question 11

Q11. Find the greatest number of five digits, which is exactly divisible by 7, 10, 15, 21 and 28. (1) 99840 (2) 99900 (3) 99960 (4) 99990 Solution: The least number that is exactly divisible by 7, 10, 15, 21 and 28 = L.C.M. of 7, 10, 15, 21 and 28 = 420 and all numbers satisfying the given property will be of the form 420k, where k is a natural number. Now, to find the greatest number of five digits, which is of the form 420k, we divide 100000 by 420 and subtract the remainder from 100000 and get the required number. The remainder when 100000 is divided by 420 is 40 and hence the required number is 100000 – 40 = 99960 Alternate Method: Once you know that the number is of the form 420k, and the question asks for the greatest number of five digits, start working with the options, first checking which of the numbers contain five digits and then start checking with the greatest number as to whether it is divisible by 420 or not. For example, in the given problem, all the options contain five digit numbers and so none can be eliminated using that condition. Now note that the greatest among them is 99990 but that can be eliminated just by checking that it is not divisible by 420. Since for a number to be divisible by 420, it has to be divisible by 4 and hence the last two digits of the number also has to be divisible by 4 and so 99990 is eliminated. You do not even have to perform the actual division for checking the result. Similarly you can use the divisibility test for 7 and eliminate options (a) and (b). Hence the answer is option (d). Thank You. Ravi Raja

Question 12

Q12. The number of even factors of the number N = 123^3 x 34^4 x 52^2, is (a) 144 (b) 168 (c) 24 (d) 7 Solution: N = 123^3 x 34^4 x 52^2 = {(3 x 41)^3} x {(2 x 17)^4} x [{(2)^2} x (13)]^2 or, N = (2^8) x (3^3) x (13^2) x (17^4) x (41^3) The number of divisors of the number (3^3) x (13^2) x (17^4) x (41^3) = (3 + 1)(2 + 1)(4 + 1)(3 + 1) = 4 x 3 x 5 x 4 = 240. The number of divisors of the number (2^8) x (3^3) x (13^2) x (17^4) x (41^3) = (8 + 1)(3 + 1)(2 + 1)(4 + 1)(3 + 1) = 9 x 4 x 3 x 5 x 4 = 2160. Hence the number of even factors of the number N = (2^8) x (3^3) x (13^2) x (17^4) x (41^3) = 123^3 x 34^4 x 52^2 are: 2160 – 240 = 1920 Thank You. Ravi Raja

Question 13

Q13. A person had to multiply two numbers. Instead of multiplying by 12, the person multiplied by 21, and the product went up by 270. What was the final product?

a. 252

b. 360

c. 630

d. None of these

Solution:

Let one of the numbers be x.

He was supposed to find the product 12x but by mistake he calculated the product 21x and the new result was greater than what he was required to find by 270.

That is, 21x – 12x = 270

or, 9x = 270

or, x = 270/9 = 30

Hence the final product was: 21x = (21)(30) = 630.

Thank You.

Ravi Raja

hey ravi..

hey ravi.. for Q9

answer of it will be 21 as remainder of (3^16)/100 is 21.

ashish kumar

Can u explain 10 prblm with

Can u explain 10 prblm with an example... am not able to get how (b-a)/a.b=1/b-1/a

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

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

= 1- 1/2  + 1/ 2 – 1/ 3 + . . . . + 1/10 – 1/ 11

= 1- 1/11

= 10/11

@ Nayan

Hello Nayan,

Here is the general rule:

(b - a)/a.b = (b/a.b) - (a/a.b) = 1/a - 1/b

Here it is with numbers:

(5 - 3)/3.5 = (5/3.5) - (3/3.5) = 1/3 - 1/5

I hope this is clear to you.

Thank You

Ravi Raja

@ Ashish

Hello Ashish,

Can you please show the steps as to how are you arriving at that answer that the remainder when 3^16 when divided by 100, the remainder is 21.

Here are my steps according to which I am getting a remainder of 41.

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

Diff ppl can have diff

Diff ppl can have diff approach. . the way I wld ‘ve solved it

 Reminder of 3¹⁶ divided by 100
= Reminder of 3 x 3¹⁵ divided by 100
= Reminder of 3 x 243³ divided by 100
= Reminder of 3 x 43³ divided by 100
= Reminder of 3 x (27 + 3x40x9) divided by 100
= Reminder of 3 x 27 x (1 + 40) divided by 100
= Reminder of 81 x  41 divided by 100
= Reminder of 3321 divided by 100
= 21




 

My Mistake in Question 9

I am Sorry it was my mistake. Calculation mistake. The remainder will be 21 and not 41.

Thank You

Ravi Raja

Easier Solution

Since 7 is a prime number and 100! can be written as 1*2*3*4*...98*99*100. In this multiplication digits divisible by 7 are 7 ,14,21,35,49,56,63,70,77,84,91,98 sums upto 14 and also 9 and 98 are divisible by 7 twice o number comes upto 14+2 = 16. So 7 divides 100!, 16 times perfectly and thus 7^16. Therefore to divide 100! by 7^18, 1001 needs to be multiplied by 7^2.

it should be 88

as we have 89 ones in 2^89 then after subtracting one from in v have 88 ones and one Zero

statement problem

ravi u r requested to check the statement written and  solution gvn plz correct it as soon as possible

ans....is 21its 6600-39...

ans....is 21
its 6600-39...