Quant Boosters - Swetabh Kumar - Set 3


  • BITS Pilani | CAT 2016 - 99.93 percentile in Quant


    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

    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

    Consider 234.2 * 3 * 4 = 24.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

    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.

    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

    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

    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

    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

    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

    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 ?

    4 values
    A=(3,7), B(6,2), C(2,k)
    AB^2 = 9+25 = 34
    AC^2 = 1+(7-k)^2 = k^2-14k+50
    BC^2 = 16+(2-k)^2 = k^2-4k+20
    if AB is hyp. 2k^2-18k+70 = 34
    k^2--9k+18= 0
    2 real values here.
    if AC is hyp: k^2-4k+54 = k^2-14k+50
    10k=-4 so k=-0.4 so one value
    if BC is hyp, k^2-14k+84=k^2-4k+20 so k=6.4 so one value.
    so (2+1+1)= 4 values


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