Quant Boosters - Shashank Prabhu, CAT 100 Percentiler - Set 3

• For two overlapping sets, we always use the form n(A) + n(B) - n(A & B ) as the intersection is counted twice in the n(A) + n(B) part.

In case of three overlapping sets, we use n(A) + n(B) + n(C) - n(A & B ) - n(B & C) - n(A & C) + n(A & B & C).
For those who are unclear about this, draw up a random Venn diagram and check out the total number of elements and calculate it again using the above expression. In this case, we simply need to figure out all the numbers that are not divisible by one or more of 5, 7 and 8 in the first 780 minutes (13 hours). So, going by the above expression,

n(5) = 780/5 = 156
n(7) = 780/7 = 111
n(8) = 780/8 = 97
n(5 & 7) = 780/35 = 22
n(7 & 8 ) = 780/56 = 13
n(5 & 8 ) = 780/40 = 19
n( 5 & 7 & 8 ) = 780/280 = 2

So, 156+111+97-22-13-19+2=312

The remaining ones will be 780-312=468.

PS: Mind that for overlapping cases, we take the LCMs and not the direct products. So, say the pepperoni topping happens every 2 minutes and the olive topping happens every 6 minutes, the overlap would happen every 6 minutes and not every 12 minutes.

• Q21) For how many ordered triplets (x, y, z) of positive integers less than 10 is the product xyz divisible by 20?
a) 72
b) 120
c) 90
d) 102
e) None of these

• 5, 4, odd - 3! * 4 + 3 * 1 = 27
5,4,even (except 8 ) - 3! * 2 + 3 * 1 = 15
5, 8, odd - 3! * 4 + 3 * 1 = 27
5, 8, even (except 4) - 3! * 2 + 3 * 1 = 15
5, 4, 8 - 3! = 6
5,6,6 = 3
5,6,2 = 6
5,2,2 = 3
Total 102 cases

Alternatively
5, 4x,(odd number)
odd number is 5 -> 2 * 3 = 6 cases
odd number is not 5 -> 3! * 8 = 48 cases

5,(even number),(even number)
same even numbers -> 4 * 3 = 12 cases
different even numbers -> 6 * 3! = 36 cases

Total = 6 + 48 + 12 + 36 = 102 cases

• Q22) A certain number 'C' when divided by N1 it leaves a remainder of 13 and when it is divided by N1 it leaves a remainder of 1,where N1 and N2 are +ve integers .Then the value of N1+N2 is ,if N1/N2 =5/4:
a.36
b.27
c.54
d. can't be determined uniquely

• C = N1 * x + 13
C = N2 * y + 1
N1 * x+13=N2 * y+1
N2 * y - N1 * x = 12
N2(y-5x/4)=12
N2(4y-5x)=48
N2=48/(4y-5x)
N1=60/(4y-5x)
N1+N2=108/(4y-5x)
(x,y)=(3,4)
(x,y)=(6,8)
(x,y)=(9,12) and so on all satisfy. So, multiple solutions are possible.

• Q23) Two horses start trotting towards each other, one from A to B, other from B to A. They cross each other after one hour and the first horse reaches B , 5/6 hour before the second horse reaches A. If the distance between A and B is 50 km, what is the speed of the slower horse?

• Let speeds be p and q respectively for the two horses.
50/q-50/p=5/6
50/(p+q)=1
p+q=50
60p-60q=pq
60(50-q)-60q=(50-q)q
3000-120q=50q-q^2
q^2-170q+3000
q=150 or q=20
q cannot be 150 as then p would be negative. So, q=20.

• Q24) Rudolph bikes at a constant rate and stops for a five-minute break at the end of every mile. Jennifer bikes at a constant rate which is three-quarters the rate that Rudolph bikes, but Jennifer takes a five-minute break at the end of every two miles. Jennifer and Rudolph begin biking at the same time and arrive at the 50-mile mark at exactly the same time. How many minutes has it taken them?

• Let speed of Rudolph be 4r
Speed of Jennifer will be 3r
50/4r + 245 = 50/3r + 120
50/12r = 125
So, 50/3r + 120 = 500+120 = 620 minutes.

• Q25) What is the remainder when 21^3 + 22^3 + 23^3 + 24^3 is divided by 90?

• (a^n + b^n + c^n ...)/(a + b + c ...) where n is odd and a, b, c .. are in an AP will give remainder 0.

Alternatively
21^3 + 24^3 + 22^3 + 23^3
45 × (odd number) + 45 × (odd number)
45 × even number
Divisible by 90.

• Q26) N = 202 x 20002 x 200000002 x 200000...2 (15 zeros) x 200000...2 (31 zeros)
Find the sum of digits of N ?

• 202 * 20002 = 4040404 this is a 7 digit number and so, when multiplied by 200000002, which has 7 zeros in between will be 808080808080808 which is a 15 digit number and so on... So, sum in the first case is 4 * 4=16, second case is 8 * 8=64, then it will be 16 sixteens and so, 7 * 16=128 and finally, 32 thirty twos and so, 32 * 5=160. Of course pure digital sum might be the key if the options are set like that... the option which adds up to 4^5=1024=7 will be the one.

• Q27) In a survey conducted, it was found that, of the 100 persons surveyed, 20 persons do not read any of the three magazines - B.W , B.I , B.T. 45 people read B.W, 40 people read B.I., and 45 people read B.T.

What is the maximum number of people who read exactly one magazine?
a) 45
b) 50
c) 55
d) 60
e) 65

What is the maximum possible number of people who read only B.W.?
a) 5
b) 25
c) 15
d) 20
e) 35

If no person reads only B.T, what is the maximum possible number of people reading all the three magazines?
a) 5
b) 35
c) 10
d) 15
e) 7

• a+2b+3c=130
a+b+c=80
b+2c=50
I) For a to be max, c has to be max. Max c=25. So, a=55
II) For only BW to be max, as much as possible of BI and BT should overlap. For BI and BT to have max overlap, 40 (minimum between 40 and 45) should be shared between the two sets. The remaining 5 from BT will go to the only BT sector. So, we are left with 80-40-5=35 people who will read only BW

• Q28) How many integer pair of a and b satisfy the equation 5|a| + 6|b| = 70

• Consider 5a+6b=70 where a and b are non negative integers This gives us (a,b) as (14,0)(8,5)(2,10). As either of the integers or both of them could be positive or negative, we get 2 cases for each value, except 0. Essentially (8,5)(-8,5)(8,-5) and (-8,-5). Similarly for (2,10) we get 4 cases. But (14,0) will give us 2 cases only (14,0) and (-14,0). So total of 10 cases.

• Q29) If a * b * c * d * e = 1050, where a,b,c,d,e are natural numbers, then how many solutions are possible for this equation?

• Best way would probably be to take 5 different numbers and add the powers of 2, 3, 5 and 7 to 1, 1, 2 and 1 respectively.
a=(2^p1) * (3^q1) * (5^r1) * (7^s1)
b=(2^p2) * (3^q2) * (5^r2) * (7^s2)
c=(2^p3) * (3^q3) * (5^r3) * (7^s3)
d=(2^p4) * (3^q4) * (5^r4) * (7^s4)
e=(2^p5) * (3^q5) * (5^r5) * (7^s5)

p1+p2+p3+p4+p5 = 1 (5 solutions)
q1+q2+q3+q4+q5 = 1 (5 solutions)
r1+r2+r3+r4+r5 = 2 (15 solutions)
s1+s2+s3+s4+s5 = 1 (5 solutions)

5 * 5 * 15 * 5 = 1875

• Q30) A submarine needs to take in 1000 gallons of water if it wants to submerge 1 ft in water. The height of the submarine is 20 ft and it is currently submerged 2 ft in water. A battleship 150 km away sees it and fires a torpedo 2 ft below sea-level towards the submarine. If the submarine can take a maximum 3000 gallons of water in one minute, what should be the minimum constant speed of the torpedo for it to hit the submarine, if the submarine starts submerging just as the torpedo is fired?
a) 1080 km/hr
b) 1500 km/hr
c) 1350 km/hr
d) 1250 km/hr
e) 1099 km/hr

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