# Quant Boosters - Shashank Prabhu, CAT 100 Percentiler - Set 5

• We can form 7 digit numbers only.
There are 2 cases now: either the number starts with a 0 or not. So we take the total number of cases irrespective of whether it starts with a 0 and then subtract the cases that start with a 0.
0 _ _ _ _ _ _ in 6!/(3! * 2!)=60 ways
_ _ _ _ _ _ _ in 7!/(3! * 2!)=420 ways
Numbers greater than 10^6=420-60=360

• Q5) All three-digit numbers, in which the ten’s digit is a natural number and is a perfect square, are formed using the digits 1 to 9. The sum of all such numbers is

• _ 1 _
_ 4 _
_ 9 _
3 * 9 * 45 * 100 + 3 * 9 * 45 * 1 + 81 * 10 + 81 * 40 + 81 * 90
121500+1215+11340
134055

• Q6) The X-mansion has a rectangular plot of area 900 sq.m and has to be fenced using non-magnetic materials to ward off Magneto. Logan comes up with an idea to fortify two adjoining sides with bricks and the remaining two with a wooden fence. One metre of the wooden fence costs 10 Dollar and one metre of the brick fence costs 25 Dollar. Ted Mosby, the architect, has been given 2,000 Dollar to complete the task. Should Ted take up the contract?
a. Yes
b. No
c. Doesn't matter
d. Data insufficient

• Ted would not take up the contract if even in the best case, where the material used will be minimum, he would end up making a loss. Best case is when all the sides are equal and so, the costs will be optimized. So 30 * 2 * 25 + 30 * 2 * 10 = 2100 which is still less than 2000. So, Ted won't take it up.

• Q7) In how many ways can 1000 be written as a product of 3 factors?

• abc=1000
a=2^a1 * 5^a2
b=2^b1 * 5^b2
c=2^c1 * 5^c2
a1+b1+c1=3... 10 solutions
a2+b2+c2=3... 10 solutions
Total solutions=100. But these will include ordered solution sets. So we need to eliminate these
1 case (10,10,10)
3 cases (1,1,1000)(2,2,250)(5,5,40) with two repeated so 9 cases in the original 100
Remaining 90 cases divided by 3! so 15 ways.
Total of 19 ways

• Q8) In a mile race, Akshay can be given a start of 128 m by Bhairav. If Bhairav can give Chinmay a start of 4 m in a 100 m dash, then who out of Akshay and Chinmay will win a race of one and half miles, and what will be the final lead given by the winner to the loser? (One mile is 1,600 m.)
a. Akshay, 1/12 miles
b. Chinmay, 1/32 miles
c. Akshay, 1/24 miles
d. Chinmay, 1/16 miles

• D is right.
A/B=1472/1600
B/C=100/96=1600/1536
A/C=1472/1536
So, C will win. If C covers 2400 m, A will cover 1472/1536 * 2400=25 * 92=2300 m. So, C wins by 100 m or 1/16th of a mile.

• Q9) What is the value of the following expression?
[(1/ (2^2 – 1) + (1/ (4^2 – 1) + 1/ (6^2 – 1) + … 1/ (20^2 – 1)]

• It will be a series in the form of
1/(1 * 3)+1/(3 * 5)...1/(19 * 21)
1/2{2/(1 * 3)+2/(3 * 5)...2/(19 * 21)}
1/2{(3-1)/(3 * 1)+(5-3)/(5 * 3)...(21-19)/(21 * 19)}
1/2(1-1/3+1/3-1/5...1/19-1/21)
1/2(1-1/21)
10/21

• Q10) Basanti is walking beside a railway track between Pune and Baramati at a constant speed towards Baramati. Local trains ply between the two cities at equal intervals in both directions. She encounters a train going from Pune to Baramati after every 8 minutes and a train going from Baramati to Pune after every 6 minutes. What is the time interval between the two consecutive trains going from Pune to Baramati? (Trains between Pune and Baramati in both the directions run at the same speed)

• Distance between two trains be d. Speed of Basanti be s, speed of trains be t.
In same direction, time between two trains = d/(t-s) = 8.
In the opposite direction, time taken to move from one train to the next one = d/(t+s) = 6
8t-8s = 6t+6s
t = 7s
d = 48s
d/t = 48/7

• Q11) Consider a four digit number for which the first two digits are equal and the last two are also equal. How many such numbers are perfect squares.

• The last two digits of any square will show a cyclicity around multiples of 50 for eg: 23^2, 27^2, 73^2, 77^2, 123^2, 127^2... will all end in 29. So, if you look at the squares of the first 25 numbers, you can see that there are only 2 cases where the last two digits repeat (00 and 44). Now 00 is not possible because you cannot have a square in the form of aa. So, we are left with 44 only. Need to try for 38^2, 62^2 and 88^2. 38^2 will be between 1225 and 1600 so, 1144 is not possible, 62^2 will be between 3600 and 4225 so again 3344 or 4444 are not possible. 88^2 will be between 7225 and 8100 and so, 7744 is possible. You need to check for only one value this way. And if you know a few numbers and their properties, it would be of some help. The likes of 1729, 145, 1001, 10001, 40585, 7744 and so on.

• Q12) There are four machines in a factory. At exactly 8 pm, when the mechanic is about to leave the factory, he is informed that two of the four machines are not working properly. The mechanic is in a hurry, and decides that he will identify the two faulty machines before going home, and repair them next morning. It takes him twenty minutes to walk to the bus stop. The last bus leaves at 8:32 pm. If it takes six minutes to identify whether a machine is defective or not, and if he decides to check the machines at random, what is the probability that the mechanic will be able to catch the last bus?
a. 0
b. 1/6
c. 1/4
d. 1/3
e. 1

• There are 2 defective machines and 2 good machines. He can check two machines before he leaves for home. So, there are 2 cases that are in his favour. If the first two machines are good, he can identify the defective ones as the last two and leave. Also, if he gets the first two machines as defective, he would have known what he needed to know. The problem lies if he gets only one defective out of the first two and so, will have to check the third one. The four machines can be arranged in 6 ways (arrangements of ddgg). So, 2/6=1/3.

• Q13) In a convex hexagon, two diagonals are drawn at random. What is the probability that the diagonals intersect inside the hexagon?
(1) 5/12
(2) 1/2
(3) 7/12
(4) 2/5

• Number of diagonals is n(n-3)/2 ie. 9
Two diagonals can be chosen in 9c2=36 ways
For two diagonals to intersect inside the hexagon, the vertices should form a quadrilateral. Possible ways are 6c4=15
Probability is 15/36 = 5/12

• Q14) The product of 3 natural numbers(N1, N2, N3) is 12 times of their HCF. Find the number of ordered triplets (N1, N2, N3).

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