Quant Boosters - Swetabh Kumar - Set 1

• speed=60, distance=120
a/v+3/4 + 3(d-a)/2v = d/v +3/2
(a+60)/v + 3/4 + 3(d-a-60)/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(1760-d)/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 30-9=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 19C11-1.
similar to ways of rearranging ABCD..should be 4! but since it says "re-arranging" 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 three-digit 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 four-digit number?
a) 40
b) 45
c) 50
d) 60

• (653-693), (703-793), (803-893), (903-993)= 30 + 5 = 35
(851-891) , (901-991) = 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^2-8n-48 = 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 ?

61

48

61

61

61

61

61