Quant Boosters - Nitin Gupta - Set 1
The eight digit number satisfying the given condition is 99984999 (9998 is twice as large as 4999).
Now, the biggest hint for finding factors of this number is in the question itself.
∴ 9998 = 4999 × 2
Multiplying both the sides by 10000, we get,
99980000 = 4999 × 20000
Adding 4999 to both the sides, we get,
99984999 = 4999 × 20000 + 4999
∴ 99984999 = 4999 × 20001
It is easy to see that 20001 is divisible by 3, so we divide it by 3 to find that 20001 = 6667 × 3
Now, instead of trying to determine rigorously whether 4999 and 6667 are prime or not, we take a
hint from option 2 and option 3, and see that 59 × 113 = 6667
∴ 6667 is not prime.
Hence, option 5.
Q20) Dr. Numerez is reading the newspaper in his living room when the doorbell rings. He opens it to find a salesman trying to sell him a new product. Dr. Numerez is not interested, but when the salesman insists, he tells him, "You can go to my brother. He might be interested in buying this. His house is in this very block, and its number is composite, but relatively prime to the number of my house. You can ask me just one more question, which I should be able to answer with a 'yes' or 'no', and then leave." The salesman knows that Dr. Numerez's house is number 10 in the block, which has houses numbered 1 to 26. Assuming that Dr. Numerez will answer him correctly, which of the following questions should the salesman ask to uniquely determine the number of his brother's house?
a) Is the number a perfect square?
b) Does the number have exactly 4 distinct divisors?
c) Does the number have exactly 3 distinct divisors?
d) Is the number lesser than the number of your house?
e) Any of these
Dr. Numerez has already told him that the number is composite, but relatively prime to his house
Looking for such numbers from 1 to 26, we find only two: 9 and 21.(These two numbers can be found out faster by using the knowledge that none of the even or prime numbers would satisfy the conditions stated by Dr. Numerez.)
9 is a perfect square while 21 is not.
9 has 3 distinct divisors while 21 has 4.
9 is lesser than 10 while 21 is more.
∴ Any of the given questions will help the salesman uniquely determine the answer.
Hence, option 5.
Q21) For any two natural numbers x and y, if x^2 + y^2 is divisible by 21, what are the remainders given by x and y respectively on division by 21?
a) 3, 4
b) 2, 5
c) 0, 0
d) 1, 6
e) None of these
Note that if x gives a remainder of r on division by any number, then x^2 will give a remainder of r^2. In this case, if x^2 + y^2 is divisible by 21, that means it is divisible by 3 as well as by 7.
Now, all natural numbers, on division by 3, give a remainder of 0, 1 or 2. Therefore, their squares,give a remainder of 0 or 1 (since a remainder of 4 is further divisible by 3, giving a remainder of 1).
Therefore, the sum of two squares will give a remainder of 0 (0 + 0), 1 (0 + 1 or 1 + 0) or 2 (1 + 1). In other words, when the sum of two squares gives a remainder of 0 on division by 3, the only way this can happen is if both the squares themselves are divisible by 3, and therefore, both the original numbers are divisible by 3.
Similarly, all natural numbers, on division by 7, give a remainder of 0, 1, 2, 3, 4, 5 or 6. Therefore,their squares give a remainder of 0, 1, 2 or 4 (prove this). Now, note that the only way the sum of two squares can give a remainder of 0 on division by 7 is if both squares are themselves divisible by
7, and so, both the original numbers are divisible by 7.
Thus, if x^2 + y^2 is divisible by 21, both x and y will be divisible by 21.
Hence, option 3.
Note: Elimination of options may be tried, but it will not help in choosing between options 3 and 5
Q22) If there are 30 factors between 1 and √N, find the total number of factors of N. It is given that N is a natural number but √N is not a natural number.
The question is based on the logic of relative placement of factors. If factor at 12th position multiplied with factor at 25th position results in N, then factor at 11th position multiplied with 26th position will also result in N. That should be sufficient to find the answer. There would be 11 factors on the left hand side of 12th factors, and since factors exist in pairs and factors equidistant from centre always result in number. There would be 11 factors on the right of factor at 25th position, so N has total 36 factors
Q23) 20 teams of five archers each compete in an archery competetion. An archer finishing in kth place contributes k points to his team, and there are no ties. The team that wins will be the one that has the least score. Given that, the 1st position team's score is not the same as any other team, the number of winning scores that are possible is
(e) none of the foregoing
The teams’ scores must sum to 1 + 2 + . . . + ..+99+100 =5050.
The winning score must be no larger than 1/20 *5050 = 252.5 and is at least 1 + 2 + 3 + 4 + 5 = 15.
However, not all scores between 15 and 252 inclusively are possible because all teams must have integer scores and no team can tie the winning team. If the winning score is s, the sum of all teams’ scores is at least s + 19(s + 1) = 20s +19
so solving gives s < = 251. Hence, 251 − 15 + 1 = 237 winning scores are possible.
Hence, choice (b) is the right answer
Q24) A function f(x) is defined as f(x) = f(x – 2) + x(x – 2) for all the integer values of ‘x’. Given that f(1) + f(6) = 0. What is the value of f(1) + f(2) + f(3) + f(4) + f(5) + f(6)?
OA - Option D
Q25) Three boys A, B and C start running at constant speeds from the same point P along the circumference of a circular track. The speeds of A, B and C are in the ratio 5:1:1. A and B run clockwise while C runs in the anticlockwise direction. Each time A meets B or C on the track he gives them a card. What is the difference in the number of cards received by B and C if A distributes 33 cards in all?
Let the speed of A, B and C be 5v, v and v respectively.
Let the circumference of the track be 12vt.
Instance when A meets:
B – 3t, 6t, 9t, 12t, 15t, 18t ....
C – 2t, 4t, 6t, 8t, 10t, 12t ....
In every 6t units of time A meets B twice while A meets
C thrice. So in 36t A would give:
B – 2 × 6 = 12 cards
C – 3 × 6 = 18 cards. (total 30)
Now, at 38t A gives C a card (his 31st) at 39t A gives
B a card (his 32nd) and at 40t A gives C a card
So the required difference = (18 + 2) – (12 + 1) = 7.
Q26) A car travels downhill at 72 kmph (kilometers per hour), on the level at 63 kmph, and uphill at only 56 kmph The car takes 4 hours to travel from town A to town B. The return trip takes 40 minutes more. What is the distance between the two towns in kilometers?
(1) can not be determined
(5) none of these
a/72 + b/63 + c/56 = 4
a/56 + b/63 + c/72 = 14/3
Add them to get:-
2a/63 + 2b/63 + 2c/63 = 26/3
=> a + b + c = 21 * 13 = 273
Q27) India and Brazil play the Soccer World Cup final in which India defeats Brazil 4 – 2. In how many different ways could the goals have been scored given that Brazil never had a lead over India during the match?
(d) None of these
If Brazil never gets a lead over India, the first goal of the match should be scored by India. At this stage India leads by 1 – 0, and the only way in which Brazil can take a lead is by scoring the next two goals i.e. a scoring pattern like IBBIII. In all other cases Brazil would never be able to lead India. Total cases where India scores the first goal
5!/ 3!2! = 10
Different scoring patterns possible = 10 – 1 = 9.
Q28) N girls and 2N boys played a chess tournament. Every player played every other player exactly once. The boys won 7/5 times as many matches as the girls (and there were no draws). Then which among the following is definitely false? (Assume 1 point for a win and 0 for a loss)
(a) Boys pocketed prime number of points against girls
(b) Girls always won twice or more matches than boys won against them
(c) The sum of the scores of top 3 individual players was not between 25 and 33
(d) The sum of the scores of top 3 individual players was 69
(e) none of the foregoing
Total number of matches among boys were 2nC2, among girls were nC2 and between boys and girls were n*2n. Please note that 2nC2 + nC2 + 2n^2 = 3nC2. Assume 1 point for a win and 0 for a loss.
=> Girls pocketed nC2 points amongst themselves and boys pocketed 2nC2 points among themselves. Let boys take k points from their matches against girls => girls take 2n^2 - k from their matches gainst boys.
=> 2nC2 + k = 7/5*(nC2 + 2n^2 - k), solving we get 8k = n(5n+1).
For n = 3, k = 6.
For n = 8, k = 41
For n = 11, k = 77.
(a) can be true as for n = 8, k = 41.
(b) can be true as can be seen for for n = 3, 8, 11, ...
(c) is true as for n = 3, top 3 can score 8+7+6 = 21 points and for n = 11, when 33 matches are played top 3 will always score more than 16+15+14 = 45.
(d) is true, For n = 11, we can have the top 3 score as 23+23+23 = 69.
Hence, choice (e) is the right answer
Q29) Find the number of positive integral solutions for a + b + c = 20 if a ≤ b ≤ c