# Quant Boosters - Shashank Prabhu, CAT 100 Percentiler - Set 6

• When A is at Finish Point, gap between B and C is 80m. When B covers the remaining 100m, the gap increases by 10m. Thus to make the gap of 10m, B has to cover 100m.
Hence, to make the gap of 90m, B must cover 900m

• Q4) A bag contains 9 red cards numbered 1, 2, 3 …, 9 and 9 black cards numbered 1, 2, 3 …, 9. In how many ways can we choose 9 out of the 18 cards so that there are exactly 3 duos, where a duos means a red card and a black card with the same number?
a. 80640
b. 13440
c. 18480
d. 1680

• The 3 numbered cards can be chosen in 9c3 = 84 ways. Post that, we need to select cards such that the numbers don't repeat. So, the first card can be chosen from the remaining 12 cards in 12 ways, the second card in 10 ways and the third card in 8 ways. However, as the order is not important, we will have to divide it by 3!. So, 84 * 12 * 10 * 8/3! = 84 * 16 * 10 = 13440

• Q5) The angles of a convex pentagon are in an arithmetic progression. Which of the following can never be the value of any of its angles?
(1) 36°
(2) 35°
(3) 34°
(4) All of these

• 4 is perfect. If the angles are a-2d, a-d, a, a+d and a+2d, a=108. The largest angle will be less than 180 as it is a convex pentagon. So, 108+2d greater than 36

• Q6) The number of distinct points at which the curve y^3 - 5y^2 + x^2 + 6y - 5x = 0 intersects either the x-axis or the y-axis is

• If x=0, y^3-5y^2+6y=0, y=0 or 2 or 3
If y=0, x^2-5x=0, x=0 or 5
(0,0)(0,2)(0,3)(5,0) are the only points (0,0) was common to both the cases.

• Q7) Number of ways in which 270 can be written as 2 co-primes?

• 270 = 2 * 3^3 * 5
2^(3 - 1) = 4
1 * 270
2 * 135
27 * 10
5 * 54

[There's a direct formula - 2^(no of prime factors - 1)]

• Q8) Gautam decided to go to a temple only on first and the last day of a year. He continues to go to the temple in this fashion till the time he finds that he has visited the temple atleast once on each of the different days of a week. The minimum number of days required to achieve this is

• Starting from the last day of a non leap year you get x and x+1, if the next year is a leap year, you get x+2 as the last day. Then x+3, x+3, x+4, x+4, x+5, x+5, as the first and the last days and finally x+6 as the first day of the new year. 4 years and 2 days... 1463 days.

• Q9) What is the smallest possible positive integer such that the product of all its digits equals 9! ?

• 9! = 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9
2^7 * 3^4 * 5 * 7
5 and 7 cannot be increased further so they have to be present as it is. We can reduce the numbers by joining various powers of 2 and 3. The highest power of 2 that is a single digit is 8 and the highest power of 3 that is a single digit is 9. So, we are left with 2 * 8 * 8 * 9 * 9 * 5 * 7. We cannot tweak these numbers any further and so, it will be 2578899.

• Q10) Three elements a, b and c are selected from the set A = {2, 3, 5, 6, 7} to form a three-digit number ‘abc’, where a < b < c. Similarly, two elements p and q are selected from the set B = {0, 1, 8, 9} to form a two-digit number ‘pq’, where p > q. Let, M be the total number of all the possible values of ‘abc’ and N be the total number of all the possible values of ‘pq’. What is the value of (M-N)?

• From set A, the number of three digit numbers that have unequal digits is 5c3×3!=60. But, out of every 6 cases, only one would be in the form of a>b>c. So, number of cases satisfying the criteria is 60/6=10. For set B, the number if two digit numbers with distinct digits that can be formed is 4c1×3c1=12... half of these will have p>q and so, 6 cases in total. So, 10-6=4.

Understand that i have not considered the first digit to be non 0 in the second case because there won't be a number starting with a 0 that will satisfy p>q and the fact that it reduces our work.

• Q11) Find the sum of all numbers between 200 and 1500 that cannot be written as a sum of two or more consecutive numbers?

• Any odd number can be expressed as a sum of two consecutive integers
Any even number that is in the form of 2^a * x^b * y^c... (in short, an even number not in the form of 2^n) can be expressed as a sum of three or more consecutive integers
The only type that remains is numbers in the form of 2^n. These cannot be expressed as a sum of at least 2 consecutive integers
So power of 2 lying between 200-1500 => 256+512+1024 =>1792

• Q12) There are 10 students out of which three are boys and seven are girls. In how many different ways can the students be paired such that no pair consists of two boys?

• 3 boys can be treated as 3 different units. Now, the girl corresponding to the first boy can be chosen in 7 ways, the girl corresponding to the second boy can be chosen in 6 ways and the girl corresponding to the third boy can be chosen in 5 ways. The remaining 4 girls can be teamed in 4c2/2!=3 ways.
Total 7 * 6 * 5 * 3 = 630 ways.

• Q13) A natural number is written on each face of a cube so that the sum of the numbers on all the faces is S. A small triangular portion is sliced off from each corner of the cube. The product of the numbers on the faces that meet at a particular corner is written on the portion sliced off at that corner. The sum of the numbers written on all the sliced off portions is 2004. How many different values can S take?

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