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

• 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

• 18 ft of submarine's height + accounting for the 2 ft underwater torpedo means that the submarine has to sink by 20 ft. So, it will have to lose 20000 gallons. So, it will need 20000/3000 ie 20/3 minutes to go under. Distance to be travelled is 150 km. So, speed should be 150/(20/3) = 45/2 km/min or 1350 kmph.

• @shashank_prabhu can we use partition method here?

• @Naman-Jain-0
Partition method is applicable for cases like dividing n identical objects among r distinguishable parts right ? How would you apply it here ?

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