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

• 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?

• abc, acd, ade, abe, bcf, cdf, def, bef will be the trios.
ac(b+d)+ae(b+d)+cf(b+d)+ef(b+d)=2004
{a(c+e)(b+d)+f(b+d)(c+e)}=2004
(a+f)(b+d)(c+e)=2004
2014=2^2 * 3 * 167
So split as (2,2,501) (2,6,167) (4,3,167) (2,3,334)

• Q14) The sum of the squares of the first 15 positive integers (1^2 + 2^2 + 3^2 + . . . + 15^2) is equal to 1240. What is the sum of the squares of the second 15 positive integers (16^2 + 17^2 + 18^2 + . . . + 30^2) ?
(A) 2480
(B) 3480
(C) 6785
(D) 8215
(E) 9255

• [30 * 31 * 61 - 15 * 16 * 31]/6
5 * 31 * 61 - 5 * 8 * 31
31(305 - 40)
31 * 265 = 8215

• Q15) B takes 12 more days than A to finish a piece of work. B and A starts this work and A leaves the work 12 days before the work is finished. B completes 60% of overall work.How long would have B taken to finish the work if he works alone?

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