# Quant Boosters - Sagar Gupta - Set 9

• 7777777777777.........7777777(95times) mod 97

7 [ 10^95 -1 ] / 9 mod 97
10^96 mod 97 =1
10^95 * 10 mod 97 =1
10r mod 97 = 1
10r = 97b+1
r = 68
7 * 67 mod 97 = 81
81 /9 = 9

There were 2 boxes A and B. Box A contained 1000 balls, consecutively numbered from 1 to 1000 such that there was only one number printed on each ball. Box B was initially empty. All the balls on which the number printed was a multiple of 7 were transferred to the box B by Amit. After that all the balls on which the number printed was a multiple of 27 were transferred to the box B by Bineet. Finally, all the balls in the box B on which the number printed was a multiple of 37 were transferred back to the box A by Ishu. The final count of the balls in Box B was :

a : 1000
b : 0
a : 1000 - 7k
b : 7k
a : 1000 - 7k - 27m
b : 7k + 27m
a : 1000 - 7k - 27m + 37n
b : 7k + 27m - 37n
142 + 32 - 5 + 1 = 170

In a 800m race around a stadium having the circumference of 200m the top runner meets the last runner on the 5th minute of the race .If the top runner runs at twice the speed of the last runner,what is the time taken by the top runner to finish the race?
a) 20 min
b)15 min
c)10 min
d) 5 min

in 5 minute , distance covered by top runner = 5 * 2x = 10x
in 5 minute , distance covered by ast runner= 5 * x = 5x
10x -5x = 200
x = 40
speed of fast runner = 80 m/min
time taken = 800/80 = 10 min

abcd is a 4 digit number such that abcd = (ab)^2 + ( cd)^2 , ab and cd are 2 digit numbers. Find abcd

ab = x
cd = y
y + 100x = x^2 + y^2
x^2 - 100x + y^2 - y = 0
10000 - 4y^2 + 4y = Discriminant
x = 100 +/- root ( 10000 - 4y^2 + 4y ) /2
x = 50 +/- root ( 2500 - y^2+y )
2500 - y ( y-1 ) should be a perfect square
y = 33
x = 88, 12

Find the number of selections that can be made by taking 4 letters from the word 'SENTENCE' (as well as no number of arrangement.)
a.21
b. 22
c. 17
d. 10

C , EEE, NN, S, T

1. all distinct : 5C4 = 5
2. 2 same , 2 distinct : 2C1 * 4C2 = 12
3. 2 same , 2 same : 1
4. 3 Same , 1 distinct = 1 * 4 = 4

total 22

Akash, Bhanu and Sam work on a project. After they complete 1/4th of the project, Sam takes a break. For 7 days only Akash and Bhanu work on the project. After that Sam relieves both of them. He completes the project in 5 days. Bhanu works 50% faster than Akash while Sam alone can finish the entire project in 20 days. How long would Bhanu take to finish the entire project?

bhanu = 3 units/day
akash = 2 units/day
sam = x units/day
total work = 20x units
5x + 7 ( 5 ) + 5x = 20x
35 = 10x -> x=3.5
total work = 70
bhanu takes = 70/3 = 23.33 days

There are 3 solutions...
Solution 1: Consists of A and B in the volume ratio 3 : 2.
Solution 2: Consists of B and C in the volume ratio 1 : 4.
Solution 3: Consists of B and C in the volume ratio 7 : 3.
The 3 solutions are mixed in the volume ratio X : 3 : 2. If the percentage composition of A and B are equal in the resultant mixture, then what is the value of X?
a) 4
b) 5
c) 10
d) 11

solution 1 : 6ml + 4ml
solution2 : 2 ml + 8ml
solution 3 : 7ml + 3ml
Resultant : X : 3 : 2
A ( 6x ) + B ( 4x + 20 ) + C ( 30 )
6x = 4x + 20 -> x = 10

How many numbers less than 1000 have sum of digits as 12?

a + b + c = 12
total whole number solutions : 14C2 = 14 * 13/2 = 91
when a=10 : b+c =2 : 3 solutions
when a=11 : b+c =1 : 2 solutions
when a=12 : b+c = 0 : 1 solution
total 6
same when b,c = 11,12
total 6 * 3 = 18
numbers = 91- 18 = 73

A number of students appear for one of the papers of the Online Cat test series. The paper consists of 100 questions. For each correct answer the candidate is awarded 1 mark and for each wrong answer 1/5th of the marks are deducted. No marks are deducted for not attempting a question. Exactly 20% of the students scored exactly 50 marks but no two attempted the same number of questions. Find the total number of students who appeared for the test?

Total students = 5x
Students who scored exactly 50 marks = x
all these x students attempted different number of questions
Possibilities for scoring 50 marks

1. 50 correct , 0 wrong
2. 51 correct , 5 wrong
3. 52 correct , 10 wrong
4. 53 correct , 15 wrong
5. 54 correct , 20 wrong
6. 55 correct , 25 wrong
7. 56 correct , 30 wrong
8. 57 correct , 35 wrong
9. 58 correct , 40 wrong

so x=9
5x = 45

What is the remainder when 8888..88888 (written 92 times) is divided by 470

470 = 47 * 2 * 5
8888........ written 92 times mod 2 = 0 -> 2a form
8888........ written 92 times mod 5 = 3 -> 5b+3 form
8888........written 92 times mod 47
46 times 8 mod 47 =0 , so 92 times 8 mod 47 = 0
so : 2a = 5b+3 = 47c
Remainder = 188

Looks like your connection to MBAtious was lost, please wait while we try to reconnect.