Quant Boosters  Anubhav Sehgal, NMIMS Mumbai  Set 3

This is a question that tests your practical acumen as well along with mathematical skills.
When we removed page numbers 311 to 458, we should have lost (458 – 311 + 1) = 148 pages.
But since it says that 350 pages were left. It means that 150 pages were lost. What does this mean?
This means that the page numbered as 311(odd) was on the left side, making us lose 310 as well and similarly page numbered as 458 (even) made us lose 459 as well.
Now if we remove page 216 to page 242, we will lose (242 – 216 + 1) = 27 pages. Don’t make the same mistake twice. Since page 216 is even numbered, 217 is also lost but that has already been accounted for. But, tearing off page 242 would make us lose 243 as well. Hence in total we lose 28 more pages and hence are left with 350 – 28 = 322 pages

Q26) Two different twodigit natural numbers are written beside each other such that the larger number is written on the left. When the absolute difference of the two numbers is subtracted from the fourdigit number so formed, the number obtained is 5481. What is the sum of the two twodigit numbers?

abcd  (ab cd ) = 5481
1000a +100b + 10c + d  10a + 10cb+d = 5481
990a + 99b + 20c + 2d = 5481
Clearly a=5 as b, c and d all are digits and a cd).
And sum = 55 + 18 = 73

Q27) A natural number x leaves a remainder 1 when divided by p. The resultant quotient, when divided by q, leaves a remainder 2. The resultant quotient, when divided by r, leaves a remainder 3 and the quotient, thus obtained, is exactly divisible by 5. If p, q and r are all natural numbers, which of the following is the least possible value of x?

It is a case of successive division. In such a case the number is written as:
x= (p(q(r(5k) + 3) + 2) + 1.
As the number needs to be smallest, we must take as small values as possible.
Now, remainder with p is 1, so p >1. So, p = 2.
Remainder with q = 2, so, q > 2, so q = 3.
Remainder with r = 3, so, r > 3, so, r = 4.
And, k = 0. x = 2(3(4(0) + 4) + 2) + 1 = 2(3(4) + 2) + 1 = 2(14) + 1 = 29.

Q28) One of the smaller sides of a right angled triangle is (2^2) * (3^3) * (4^4) * (5^5) * (6^6) * (7^7). It is known that other two sides are integers. How many triangles of this type are possible?

b^2 = (2^4) * (3^6) * (4^8 ) * (5^10) * (6^12) * (7^14)
= (2^32) * (3^18 ) * (5^10) * (7^14)
=> (a + c)(a  c) = (2^32) * (3^18 ) * (5^10) * (7^14)
Since both (a + c) and (a  c) are even, say 2k and 2n, then
k * n = (2^30) * (3^18 ) * (5^10) * (7^14)Now, k is greater than n, so we just have to write (2^30) * (3^18 ) * (5^10) * (7^14) as product of two numbers where k is the greater one and n is smaller one, then we can get values for a and c
So, answer will be (31 * 19 * 11 * 15  1)/2 = 48592

Q29) In a test consisting of 15 questions, 3 marks are awarded for a correct answer, 1 mark is deducted for an incorrect answer and no mark is awarded for an unattempted question. If a student attempts at least one question in the paper, what is the number of distinct scores that he can get?

Basically if no of marks are +n for correct and 1 for incorrect, then
Number of score which are not possible is given by (n  1) + (n  2) + ... + 1 = n * (n  1)/2 = C (n, 2)
Assume a +6 and 1 marking scheme.
When one gets 30 correct; he gets 180.
And with 29 correct, he gets 174.
So, for a score of more than174, one needs 30 correct.
And with 30 correct i.e. all correct, you can have 0 wrong.
So, 30 correct, 1 wrong or 2 wrong or 3 wrong or 4 wrong or 5 wrong is not possible.
With 29 correct, 1 wrong is possible. But 2, 3 , 4 or 5 is not.
With 28 correct, 3, 4, 5 wrong is not possible.
So, basically, when we have +6 and 1: Number of invalid scores are: 1 + 2 + 3 + 4 + 5 = 15When we have +5 and 1:
Number of invalid scores are: 1 + 2 + 3 + 4 = 10.When we have +4 and 1:
Number of invalid scores are: 1+2+3 = 6.Here, 15 questions are there with +3 and 1 marking scheme.
Highest marks possible = 15 * 3 = 45
Lowest possible marks = 15 * 1 = 15
Total marks = 45 – (15) + 1 = 61 possible scores.
Out of which, C (3, 2) = 3 are not possible.
So, number of valid scores = 61 – C (3, 2) = 58.How many integers ‘n’ are there between 0 and 10^99, such that the unit digit of n^3 is 1?
Each number from 1 to 10 results in distinct unit digit for cube.
So, each digit in cube appears 1/10 times.
So, from 1 to 10^99 it will appear 10^98 times

Q30) How many subsets of the set {1, 2, 3, 4 ... 30} have the property that the sum of the elements of a subset is less than or equal to 232?

Sum of all elements of set S = 465.
For any subset having sum of elements x, there will be one complimentary set which is having sum of elements equal to (465 – x).Example let there be a set {1, 2, 3, 4}.
Then subset which is having sum =3 will be {1, 2} {3}: (2)
and subset having sum (1+2+3+4) 3 =7 will also be 2: {3,4} {1,2,4}So, number of subsets having sum of elements x = Number of subsets having sum of elements (465  x)
=> No of subsets having sum 1 = No of subsets having sum 464
Also, No of subsets having sum 2 = No of subsets having sum 463
….
No of subsets having sum 232 = No of subsets having sum 233
Adding them we will get,
No of subsets having sum less than or equal to 232
= No of subsets having sum more than 233
= 2^30/2
= 2^29