Official Quant thread for CAT 2013

MBA CAT 2024
9821
@gnehagarg said:
a*x^2+b*x+c=0a*x^2+b*x+210=0sum of two roots =-b/aProduct of two roots=210/aa is number of factors of 210210=2*3*5*7Number of factors =2*2*2*2=16a can be negativeTotal number is 32
I haven't solved the problem, but I see a problem with your solution.
Say, take a = 35, Product of the roots then would be 6.
and Integer roots could be (1,6),(-1,-6) or (2,3) or (-2,-3), resulting in a different value of b,
for each case, whereas you have counted it as only one case.
To make my point, more clear, you have clearly ignored "b", as b would change the equation would change, as explained above.
9822

1. Doctors have advised Renu, a chocolate freak, not to take more than 20 chocolates in one day. When she went to the market to buy her daily quota, she found that if she buys chocolate from the market complex she would have to pay Rs. 3 more for the same number of chocolates than she would have spent had she bought them from her uncle Scrooge €™s shop, getting two sweets less per rupee. She finally decided to get them from uncle Scrooge €™s sho paying only in one-rupee coins. How many chocolates did she buy?

A. 12

B. 9

C. 18

D. 15

E. Data Insufficient

2. how much would she have spent at the market complex?

a. 6 b 12 c 9 d 5.

9823
@IIM-A2013 said:
1. Doctors have advised Renu, a chocolate freak, not to take more than 20 chocolates in one day. When she went to the market to buy her daily quota, she found that if she buys chocolate from the market complex she would have to pay Rs. 3 more for the same number of chocolates than she would have spent had she bought them from her uncle Scrooge €™s shop, getting two sweets less per rupee. She finally decided to get them from uncle Scrooge €™s sho paying only in one-rupee coins. How many chocolates did she buy?A. 12B. 9C. 18D. 15E. Data Insufficient2. how much would she have spent at the market complex?a. 6 b 12 c 9 d 5.
1. A. 12?

Let n be the number of chocolates she bought.
Price at market = 1/(x-2)

Price at Scrooge's shop = 1/(x)

n/(x-2) = n/(x) + 3
2n = 3x(x-2)
n = 3x(x-2)/2

She buys from Scrooge's shop
n/x is an integer
3(x-2)/2 is an integer
x can only be 4 ( maximum value of n = 20)
n = 12

2. a. 6?

At the market complex
n/(x-2) = 12/2 = 6
9824
If 'a' is the number of positive integral solutions of equation 5x + 4y + 3z = 3012 and 'b' is the number of positive integral solution of equation 5x + 4y + 3z = 3000, then find the value of a - b.
9825
@chillfactor said:
If 'a' is the number of positive integral solutions of equation 5x + 4y + 3z = 3012 and 'b' is the number of positive integral solution of equation 5x + 4y + 3z = 3000, then find the value of a - b.
EDITED :
600????
9826
@chillfactor said:
If 'a' is the number of positive integral solutions of equation 5x + 4y + 3z = 3012 and 'b' is the number of positive integral solution of equation 5x + 4y + 3z = 3000, then find the value of a - b.
602? (trying to do orally, may be wildly off)

regards
scrabbler
9827
@chillfactor said:
If 'a' is the number of positive integral solutions of equation 5x + 4y + 3z = 3012 and 'b' is the number of positive integral solution of equation 5x + 4y + 3z = 3000, then find the value of a - b.
Sir, Kaise karna hai ise??
9828
If a, b, c, d are integers, then find number of all quadruplets (a, b, c, d) such that a*b*c*d = 1050.= 7*3*5*5*2
9829
Archimedes planned to count all of the prime numbers between 2 and 1000 using the Sieve of Eratosthenes as follows:
(a) List the integers from 2 to 1000.
(b) Circle the smallest number in the list and call this p.
(c) Cross out all multiples of p in the list except for p itself.
(d) Let p be the smallest number remaining that is neither circled nor crossed out. Circle p.
(e) Repeat steps (c) and (d) until each number is either circled or crossed out.


At the end of this process, the circled numbers are prime and the crossed out numbers are composite. Unfortunately, while crossing off the multiples of 2, Archimedes accidentally crossed out two odd primes in addition to crossing out all the even numbers (besides 2). Otherwise, he executed the algorithm correctly. If the number of circled numbers remaining when Archimedes finished equals the number of primes from 2 to 1000 (including 2), then what is the largest possible prime that Archimedes accidentally crossed out?

I thought it should be 997 as it is largest prime and if we cut it while crossing off it doesn't make any difference and number of primes from 2 to 1000 remains same.
9830
@sparklingaubade said:
If a, b, c, d are integers, then find number of all quadruplets (a, b, c, d) such that a*b*c*d = 1050.= 7*3*5*5*2


4^5 ?


9831
@ScareCrow28 said:
Sir, Kaise karna hai ise??
my approach.

5x+4y+3z=3012 --------> 1
5x+4y+3z=3000---------> 2

for eqn 1

at z=1 eqns becomes
5x+4y=3009
here solutions are x=601,597..........1 i.e 151

for z= 2
5x+4y = 3006
solutions are x=598,594........2 i.e 150

for z= 3
5x+4y=3003
soln x= 599,595..... 3 i.e 150

for z= 4
5x+4y = 3000
soln are x= 596,592..... 4 i.e 149


for all solutions after that both eqns will have equal no. of solutions. as only for
eqn 1 z=5 and for
eqn 2 z=1 (only values of z will change)


hence total solutions are 151+150+150+149 = 600


(ignore calc mistakes)
9832
@sparklingaubade said:
If a, b, c, d are integers, then find number of all quadruplets (a, b, c, d) such that a*b*c*d = 1050.= 7*3*5*5*2
a*b*c*d = (2)(3)(5^2)(7)

a = (2^x1)(3^y1)(5^z1)(7^w1)
b = (2^x2)(3^y2)(5^z2)(7^w2)
c = (2^x3)(3^y3)(5^z3)(7^w3)
d = (2^x4)(3^y4)(5^z4)(7^w4)

x1 + x2 + x3 + x4 = 1, so 4 ways
y1 + y2 + y3 + y4 = 1, so 4 ways
z1 + z2 + z3 + z4 = 2, so C(5, 3) = 10 ways
w1 + w2 + w3 + w4 = 1, so 4 ways

Total = 4*4*10*4 = 640 positive integral ordered quadruplets

a, b, c, d can be negative also, so {1 + C(4, 2) + C(4, 4)}*640 = 5120 integral ordered quadruplets
@bullseyes said:
EDITED : 600????
@scrabbler said:
602? (trying to do orally, may be wildly off)regardsscrabbler
600 is correct.

If (p, q, r) is a solution of equation 3x + 4y + 5z = 3012, then (p - 1, q - 1, r - 1) will be a solution for the equation 3x + 4y + 5z = 3000 (as 3 + 4 + 5 = 12)

So, to get a - b, we just have to find all those solutions when atleast one of x, y, z are 1 in 3x + 4y + 5z = 3012 (as then x, y, z will be 0 for equation 3x + 4y + 5z = 3000 and we need only positive integral solutions)

When x = 1, y = 5k + 1, where k varies from 0 to 150, so 151 cases
When y = 1, x = 5k + 1, where k varies from 0 to 200, so 201 cases
When z = 1, x = 4k + 1, where k varies from 0 to 250, so 251 cases

But x = y = 1, y = z = 1, z = x = 1 are counted twice

So, (151 + 201 + 251) - 3 = 600 cases

Hence, a - b = 600
9833
Two bags contain some numbers, and the total number of numbers is prime. When we transfer the number 170 from bag 1 to bag 2, the average in both bags increases by one. If the total sum of all numbers is 2004, find the number of numbers.
9834
Start with n = 2. Two players A and B move alternately by adding a proper divisor of n to the current n. The player in whose move 1990 or a greater number is achieved will win. Both the players are intelligent and good at maths. Who will win and how??

Note: Except the number itself all other factors are proper divisors.
9835
@sparklingaubade said:
Two bags contain some numbers, and the total number of numbers is prime. When we transfer the number 170 from bag 1 to bag 2, the average in both bags increases by one. If the total sum of all numbers is 2004, find the number of numbers.
167 ?
9836
@sparklingaubade said:
Archimedes planned to count all of the prime numbers between 2 and 1000 using the Sieve of Eratosthenes as follows: (a) List the integers from 2 to 1000. (b) Circle the smallest number in the list and call this p. (c) Cross out all multiples of p in the list except for p itself. (d) Let p be the smallest number remaining that is neither circled nor crossed out. Circle p. (e) Repeat steps (c) and (d) until each number is either circled or crossed out.At the end of this process, the circled numbers are prime and the crossed out numbers are composite. Unfortunately, while crossing off the multiples of 2, Archimedes accidentally crossed out two odd primes in addition to crossing out all the even numbers (besides 2). Otherwise, he executed the algorithm correctly. If the number of circled numbers remaining when Archimedes finished equals the number of primes from 2 to 1000 (including 2), then what is the largest possible prime that Archimedes accidentally crossed out?I thought it should be 997 as it is largest prime and if we cut it while crossing off it doesn't make any difference and number of primes from 2 to 1000 remains same.
I think you didn't understand the question.

Suppose while crossing out multiples of 3, he accidentally crossed out odd prime numbers 'a' and 'b', then the numbers that are composite but will become prime are a^2, ab and b^2

But question says that number of circled numbers equals the number of primes from 2 to 1000. This means only two of a^2, ab and b^2 are less than 1000.

I think now its easy to proceed.
9837
@sparklingaubade said:
Two bags contain some numbers, and the total number of numbers is prime. When we transfer the number 170 from bag 1 to bag 2, the average in both bags increases by one. If the total sum of all numbers is 2004, find the number of numbers.
Bag 1 has 'a' numbers and bag 2 has 'b' numbers and average be 'x' and 'y'

ax - 170 = (a - 1)(x + 1)
a - x = -169

by + 170 = (b + 1)(y + 1)
b + y = 169

Also, ax + by = 2004

a(a + 169) + b(169 - b) = 2004
(a + b)(a - b + 169) = 2004

It is given that the total number of numbers is prime, so (a + b) is prime

As 2004 = 4*3*167

So, only possible values of a + b are 2, 3 and 167 (2 and 3 are not possible, as then a - b > a + b which is not possible)

So, a + b = 167
@sparklingaubade said:
Start with n = 2. Two players A and B move alternately by adding a proper divisor of n to the current n. The player in whose move 1990 or a greater number is achieved will win. Both the players are intelligent and good at maths. Who will win and how??Note: Except the number itself all other factors are proper divisors.
Note that if a player get an odd number n from his opponent, then he can increase the number by atmost n/3, but if he gets an even number n, then he can increase it by at most n/2

Play around using this, I'm sure you will get the logic.

S = {1, 2, 3, 4, .., n} is a set of first n natural numbers. For each of its non-zero subset, 'strange sum' is defined as :-
Arrange all the elements of the subsets in decreasing order and then add and subtract successive terms alternately such that largest term is always added. For eg:- 'strange sum' for (1, 2, 6, 7} is 7 - 6 + 2 - 1 = 2 and for {5}, its just 5

Find the sum of all such alternating sums for:-
a) n = 7
b) Deduce a general form for n.
9838
@chillfactor said:
S = {1, 2, 3, 4, .., n} is a set of first n natural numbers. For each of its non-zero subset, alternating sum is defined as :-Arrange the elements in subsets in decreasing order and then alternately add and subtract successive terms such that largest term is always added. For eg:- alternating sum for (1, 3, 5, 7} is 7 - 5 + 3 - 1 = 4 and for {3}, its just 3Find the sum of all such alternating sums for:-a) n = 7b) Deduce a general form for n.
a) 448
b) 2^(n-1) * n ?

Take n = 2
subsets : {1} {2} {1,2}
sums = 1 + 2 + 2 - 1 = 2*2
The number of subsets that contain 2 are 2^(2-1)

Take n = 3
subsets : {1} {2} {3} {1,2} {2,3} {3,1} {1,2,3}
sums = 1 + 2 + 3 + (2 - 1) + (3 - 2) + (3 - 1) + (3 - 2 + 1) = 4*3
The number of subsets that contain 3 are 2^(3-1)

So general formula = 2^(n-1) * n
9839
@chillfactor said:
S = {1, 2, 3, 4, .., n} is a set of first n natural numbers. For each of its non-zero subset, 'strange sum' is defined as :-Arrange all the elements of the subsets in decreasing order and then add and subtract successive terms alternately such that largest term is always added. For eg:- 'strange sum' for (1, 2, 6, 7} is 7 - 6 + 2 - 1 = 2 and for {5}, its just 5Find the sum of all such alternating sums for:-a) n = 7b) Deduce a general form for n.
I tried for n = 1, 2, 3 and 4...(trial and error)

I found that the alternating sums come:
n=1: 1 sum =1
n=2: 1, 1, 2 sum = 4
n=3: 1, 1, 1, 2, 2, 2, 3 sum = 12
n=4: 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 4 sum = 32

From this I deduced that there is a binomial coefficient/pascal's triangle relationship; for example the number of times each number r from 1 to n exists is nCr. For 4 we get 4 1s, 6 2s, 4 3s, and 1 4. So for 5 we should get 5 1s, 10 2s, 10 3s, 5 4s and 1 5, totaling 80.

The second thing to note is that the totals also follow a pattern:

n = 1: sum =1 = 1 x 2^0
n = 2: sum = 4 = 2 x 2^1
n = 3: sum = 12 = 3 x 2^2
n = 4: sum = 32 = 4 x 2^3
n = 5: sum = 80 = 5 x 2^4

From which I conclude that
(a) sum for n = 7 will be 7 x 2^6 = 448 and
(b) sum for n in general will be n x 2^(n-1)

regards
scrabbler
9840

Yup, 448 and n*2⁽ⁿ⁻¹⁾ is correct for last question.


Sum of the areas of all the triangles having vertices as vertices of cube having dimension 1 x 1 x 1 is (m + √n + √p). Find (m, n, p).