Quant Boosters  Swetabh Kumar  Set 1

speed=60, distance=120
a/v+3/4 + 3(da)/2v = d/v +3/2
(a+60)/v + 3/4 + 3(da60)/2v = d/v+1
subtract.
60/v  90/v = 1/2
30/v = 1/2 so v=60.
so 1/2 +3d/2v = 2d/2v+3/2
d/2v = 1 so d=2v = 2 * 60=120

Q21) A and B run a 1760 m race ending in a dead heat. At first A runs 20% faster than B. B then quickens his pace , and for remaining distance runs 20% faster than A. When B quickens his speed, how much distance has A run?

B is v and 36v/25 A is V.
d/v + 25(1760d)/36v = 5 * 1760/6v
36d + 25(1760)25d = 30 * 1760
11d = 5 * 1760 so d=800 so A has gone: 6/5 * 800 = 960

Q22) Three cities are connected such that each city has at least one direct route to each of the other. If one wants to go from A to B, either a direct route (AB) or an indirect one (ACB) can be used. Totally, there are 33 routes from A to B and 23 routes from B to C. How many indirect routes are there from A to C?

A to B direct: x
B to C: y
A to C: z
x+yz=33 and y+xz=23 so y=5, x=3, z=6
so A to C indirect= xy = 3*5=15

Q23) How many 3 digit numbers are there such that their product is 24?
a) 30
b) 21
c) 24
d) 17
e) 15

abc= 2^3 * 3
5C2 * 3C2 = 30 ordered.
(12,2,1): 6 ways and (24,1,1): 3 ways not allowed. so 309=21

Q24) Priyanka has 11 different toys and supriya has 8 different toys. Find the number of ways in which they can exchange their toys so that each keeps her initial number of toys?

Total ways of distributing 19 as 11 and 8 between the girls would be 19C11 * 8C8 = 19C11
but since they have to exchange, so original distribution would not count. so 19C111.
similar to ways of rearranging ABCD..should be 4! but since it says "rearranging" so original would not count. hence 4!1 = 23. likewise, here, the word "exchange" is used.

Q25) For how many ordered triplet (x,y,z) of positive integers less than 10 is the product xyz divisible by 20?

20=5 * 2^2 so one 5 should be there and rest we have to distribute
(5,2,2)....(5,2,8) so 3+6 * 3 = 21
(5,4,4)...(5,4,8) so 3+2 * 6 = 15
(5,6,6,), (5,6,8) = 3+6=9
(5,8,8)= 3 so till here= 48.
ab (5,4,1/3/5/7/9) = 3+6 * 4 = 27
and (5,8,1/3/5/7/9) = 3+6 * 4 = 27
so 54 extra.
hence 48+54=102

Q26) How many threedigit odd numbers ending with 1 or 3 are possible such that the sum of the number and the number obtained by reversing the digits is a fourdigit number?
a) 40
b) 45
c) 50
d) 60

(653693), (703793), (803893), (903993)= 30 + 5 = 35
(851891) , (901991) = 10 + 5 = 15 so 35 + 15 = 50

Q27) Find the number of zeroes in : 100^1 * 99^2 * 98^3 * 97^4 * ... * 1^100

(96+91+86...6+1)+(76+51+26+1)
= (5 * 19 + 1 + 5 * 18 + 1 ... 5 * 0 + 1) + 154
= 20 + 5 * 190 + 154
= 1124

Q28) In an arithmetic sequence, a1 = 4. If a17, a37 and a77 form a geometric sequence in that order, find a500.

503 or 4
(4+36d)^2 = (4+16d)(4+76d)
16+1296d^2 + 288d = 16+368d + 1216d^2
80d = 80
d=1 or d=0
a+499d= 4+499=503 or 4+0 = 4

Q29) The numbers 6, m, n form an arithmetic progression and the numbers m, n, 16 form a geometric
progression. Find the maximum value of m + n.

2m=6+n or 16m=48+8n
and n^2 = 16m
n^28n48 = 0
n=12 and m = 9 so 21

Q30) The points (3,7), (6,2) and (2,k) are the vertices of a triangle. For how many values of k is the triangle a right triangle ?