Quant Capsules by Shashank Prabhu  Set 4

If A is the sum of the squares of the first n natural numbers (where n < 100), then for how many values of n will A be divisible by 5?
(a) 40
(b) 60
(c) 59
(d) 39Consecutive squares starting from 1 end in: 1, 4, 9, 6, 5, 6, 9, 4, 1, 0
If you take progressive sum.... First number, first two numbers, first 3 numbers and so on, it ends in
1, 5, 4, 0, 5, 1, 0, 4, 5, 5
And the cycle will repeat
6 instances in a group of 10 numbers
But 100 is not included as n < 100
So 6×101=59Find the no. of positive integer solution for x,y for 2/x +3/y= 1/6
6(2y+3x)=xy
12y+18x=xy
12y+18xxy=0
12yx(y18)=0
As there is y18 inside the bracket, we create another y18 using the coefficient of y, 12 in this case
12(y18)x(y18)+216=0
(x12)(y18)=216
216=2^3 * 3^3 and so, has 4 * 4=16 factors and so, can be written as the product of two factors in 8 ways. As the number of ordered pairs are being asked, we get the answer to be 16.If N is a natural number how many values of N exist, such that N^2 + 24N + 21 has exactly three factors?
Let the RHS be x^2
n^2 + 24n + 21 = x^2
n^2 + 24n + 144  123 = x^2
(n+12)^12  x^2 = 123
(n+12+x)(n+12x)=123=41* 3 or 123 * 1
n=10 or n=50
So, 2 solutions.If ‘bcd’ is a three digit number whose square ends with ‘abcd’, then what is the value of a + b + c + d ?
a. 27
b. 13
c. 16
d. 20bcd = 625
It's kinda theoretical in nature. Squares of numbers ending in 005 end in 025, squares of numbers ending in 025 end in 625 and squares of numbers ending in 625 end in 625.In an island ‘Pedhauli’, people use only three symbols A, V and P to write any number.
They write 10 as PAP.
They write 15 as PVA.
They write 27 as PAAA.
What is the decimal equivalent of VPAV?
a. 65
b. 66
c. 67
d. 68They simply use base 3... 10 in base 3 is 101, 15 in base 3 is 120, 27 in base 3 is 1000.. so p=1, v=2, a=0 and vpav becomes 2102 which is equivalent to 65 in decimal.
500! + 505! + 510! + 515! is completely divisible by 5^n, where n is a natural number. How many distinct values of n are possible?
a. 120
b. 121
c. 124
d. 125500!(1 + 501 * 502 * 503 * 504 * 505 + 501* 502 *... 510 + 501 * 502 *.. * 515)
Inside bracket 5k+1 form
500! Has 124 5s
So cA contractor, intending to finish a work in 150 days, employed 75 men. They worked for 8 hours every day for 90 days and completed 2/7th part of the work. Then the contractor increased the number of men by x and thereafter all the men were made to work for 10 hours every day. If the work was completed just in time, then what is the value of x?
a. 225
b. 150
c. 75
d. None of these75 × 8 × 90 × 5/ 2
x = 150.Identical black tiles, in the shape of a square of side 3 cm, are placed along the two diagonals of a square shaped floor of side 39 cm. The rest of the floor is covered with identical white tiles of same shape and size. How many white tiles need to be replaced by the black tiles so that the black and the white tiles are in alternate positions in all the rows and columns?
Small tiles of 3 cm sides and larger floor of 39 cm sides essentially means that it is a 13 * 13 grid. So, if we see the tiles along the diagonals, we get 13 * 21=25 tiles that are black colored. Total number of black colored tiles that are required will be (13^2+1)/2=85. So, additional 60 replacements required.
Of the applicants who passed a certain test, 15 applied to both college X and Y. If 20 % of the applicants who applied to college X and 25% of the applicants who applied to college Y applied to both college X and Y, how many applicants applied only college X or college Y?
Number of people who applied to x includes those who have applied only to x and those who have applied to x and y both. The intersection has been given to be 15. 20% of people who applied to x applied to y too means that 15 is nothing but 20% of people who applied to x. Which means that 80% of people applied only to x which is 4 * 20% = 4 * 15 = 60. Similarly, 25% of people who applied to y applied to x too means that 15 is nothing but 25% of people who applied to y. Which means that 75% of people applied only to y which is 3 * 25% = 15 = 45. Total comes to 60+45=105
Sixteen persons  P1 to P16 are to be seated at a square table, which has four chairs along each side. What is the probability that P7 and P13 sit on two adjacent chairs on the same side?
first person can be selected in 4 ways as the sides of a square are the same... the rest can be arranged in 15! ways. Total of 4 * 15! ways. If two people are sitting adjacent to each other, we can place those two people together right at the start in 3*2!=6 ways. The rest can be arranged in 14! ways. Probability = 6 * 14!/(4 * 15!) = 1/10