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We’ll look at a few more problems. Hope you have gone through the earlier Quick Quant posts. We are solving questions with shortcuts and non-conventional methods. See if you can crack this set in five minutes.

Q.1 A milkman mixes 20 L of water with 80 L of milk. After selling one-fourth of this mixture, he adds water to replenish the quantity that he has sold. What is the current proportion of water to milk? (CAT 2004)

(1) 2:3 (2) 1:2 (3) 1:3 (4) 3:4

Q.2 Two boats, traveling at 5 and 10 kms per hour, head directly towards each other. They begin at a distance of 20 kms from each other. How far are they (in kms) one minute before they collide? (CAT 2004)

(1) 1/12 (2) 1/6 (3) 1/4 (4) 1/3

Q.3 In a tournament, there are n teams T1, T2… Tn with n>5. Each team consists of k players, k>3. The following pairs of teams have one player in common: T1 and T2, T2 and T3, T3 and T4… Tn-1 and Tn, Tn and T1. No other pair of teams has any player in common. How many players are participating in the tournament, considering all the n teams together? (CAT 2007)

(1) n(k-1) (2) k(n-1) (3) n(k-2) (4) k(k-2) (5) (n-1)(k-1)

Q.4 Len n! = 1 * 2 * 3 * 4 *… * n for integer n ≥ 1. If p = 1! + 2*2! + 3*3! + … 10*10! Then p + 2 when divided by 11! Leaves a remainder of (CAT 2005)

(1) 10 (2) 0 (3) 7 (4) 1

Q. 5 Consider a sequence of seven consecutive integers. The average of first five integers is n. The average of all seven integers is: (CAT 2000)

(1)n (2) n+1 (3) K*n, where K is a function of n (4) n+(2/7)

Solution:

Q.1 A milkman mixes 20 L of water with 80 L of milk. After selling one-fourth of this mixture, he adds water to replenish the quantity that he has sold. What is the current proportion of water to milk? (CAT 2004)

Solution:

Instead of solving this question using equations, we can use simple logic to get this done. Ratio is always expressed in the simplest form and hence sum of water and milk should divide 100 completely. This rules our option 2 {1+2=3} and option 4 {3+4=7}

Option 1 = 2:3 = 40:60
Option 3 = 1:3 = 25:75

Since he is replacing only with water, water quantity will increase by a large margin and milk will fall by a large margin too. This rules our 25:75. Answer is 40:60

Another way of solving this question is if milk was 80 L and 1/4th was removed, 3/4th was the remaining quantity = 80*3/4 = 60. Remaining will be water. Hence, 40:60

Q.2 Two boats, traveling at 5 and 10 kms per hour, head directly towards each other. They begin at a distance of 20 kms from each other. How far are they (in kms) one minute before they collide? (CAT 2004)

Solution:

As they are moving in opposite direction, relative speed will be sum of the two speeds = 5 + 10 = 15 kmph.

This means that they cover 15 kms in one hour or in 60 minutes together. Hence, in 1 minute, they cover 15/60 = ¼ km together.

Using this logic, a minute before colliding, the distance between them should be ¼ km. Answer = ¼

Q.3 In a tournament, there are n teams T1, T2… Tn with n>5. Each team consists of k players, k>3. The following pairs of teams have one player in common: T1 and T2, T2 and T3, T3 and T4… Tn-1 and Tn, Tn and T1. No other pair of teams has any player in common. How many players are participating in the tournament, considering all the n teams together? (CAT 2007)

Solution:

This can be written in simple alphabetic form. Considering our limits, let’s start with the minimum number. Let n = 6, k = 4
Meaning, every team has 4 members, there are 6 teams

Team 1 – A B C D
Team 2 – D E F G
Team 3 – G H I J
Team 4 – J K L M
Team 5 – M N O P
Team 6 – P Q R A

18 players, when n = 6 and k = 4. The only answer option giving 18 as answer is n(k-1)

Q.4 Len n! = 1 * 2 * 3 * 4 *… * n for integer n ≥ 1. If p = 1! + 2*2! + 3*3! + … 10*10! Then p + 2 when divided by 11! Leaves a remainder of (CAT 2005)

Solution:

As we are dividing the sum by 11 which is the next natural number after 10, we can generalize this as:

(1*1! + 2*2! + 3*3! + … + n*n!  + 2) / (n+1)!

if n is 1, we get 1 + 2 = 3 mod 2 = 1
if n is 2, we get 5 + 2 = 7 mod 6 = 1
if n is 3, we get 23 + 2 = 25 mod 24 = 1

In every case, we get 1 as the answer. Hence, 1! + 2*2! + 3*3! + … + 10*10! + 2 divided by 11! will give 1 as the remainder.

Q. 5 Consider a sequence of seven consecutive integers. The average of first five integers is n. The average of all seven integers is: (CAT 2000)

Solution:

Let the series be a-3, a-2 ,a-1, a, a+1, a+2, a+3.

Then the average of the first 5 numbers will be 5a-5/5 = a-1 = n

Average of the 7 numbers is a = n + 1

Option 2 🙂

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