In this article certain questions have been randomly taken from different topics of the quantitative section of CAT. Each question has been selected keeping in mind certain challenges you need to manage as a test taker. These challenges may range from concept basics to application orientation and test taking strategies.

**1. If r(px+qy-rz)=q(px-qy+rz)=p(-px+qy+rz)=4, then (x+y+z)/(p+q+r)= k/pqr. Find k.**

__Topic :__ Ratios

__Approach :__Given r(px+qy-rz)=q(px-qy+rz)=p(-px+qy+rz)=4

(px+qy-rz)/1/r=(px-qy+rz)/1/q=(-px+qy+rz)/1/p=4

i.e. each= 2px/(1/r + 1/q)=2qy/(1/r + 1/p)=2rz/(1/q + 1/p)=4

Thus, 2pqrx/(r+q)=2pqry/(p+r)=2pqrz/(p+q)=4

or x/(r+q)=y/(p+r)=z/(p+q)=4/2pqr

or (x+y+z)/2(p+q+r)=4/2pqr

or (x+y+z)/(p+q+r)=4/pqr

Hence k=4

__Answer :__ Option 3

__Challenge :__ Understanding the application of rules related to ratios, componendo & dividendo

**2. A and B start running around a circular track of circumference 240m at speeds of 2m/s and 5m/s, starting from the same point at the same time in the same direction. After how much time will they meet at a point which is diametrically opposite the starting point?**

**1. 240s**

**2. 144s**

**3. 420s**

**4. They never meet**

__Topic : __Time-Speed-Distance

__Approach __:A completes one round in 120s and half the round in 60s. B completes one round in 48s and half in 24s. Let the desired point be P.

Then A will cross P after odd multiples of 60 and B will cross P after odd multiples of 24.

So, if thee meet at P, then some odd multiple of 60 should be equal to some odd multiple of 24.

Thus 60m = 24n, where m and n are odd integers

or 5m = 2n, which is not possible as 2n is always even but 5m is odd for odd m. Hence they never meet at point P.

__Answer :__ Option 4

__Challenge :__ Understanding the application of time-speed-distance to circular motion

**3. Ram opens a book of x pages exactly. He finds that one of the two consecutive pages can be expressed as (x/16) + 1 and the other as (8k+1)k where k is a natural number. If the first page of the book is numbered as 1 and is on the right side, what is the minimum number of pages in the book given that x is divisible by 32?**

**1. 544**

**2. 512**

**3. 480**

**4. Indeterminate**

__Topic : __Numbers

__Approach :__x/16 is even as x is divisible by 32.

Hence (x/16)+1 is an odd page number and on the right side

So, (8K+1)k must be the even page number on the left.

The minimum value of k=2, for which (8k+1)k= 34

Thus, (x/16)+1=35, which gives x=544.

__Note : __k cannot be 1. In that case, (8k+1)k=9 and (x/16)+1=8, which gives x=112, which is not a multiple of 32.

__Answer :__ Option 1

__Challenge :__ Understanding the application of numbers in practical situations

**4. Sonam goes for shopping. Anything from stall n costs Rs. n. She has money in the denomination of Rs. 5 and Rs. 8. What is the maximum value of k, such that she can not buy anything from k even if she has sufficient money?**

**1. 22**

**2. 27**

**3. 17**

**4. Indeterminate**

__Topic :__ Numbers

__Approach :__She can buy a thing from a stall k if k can be expressed as a an additive/multiplicative/combination function of either 5 or 8 or both 5 and 8. Thus she can buy from the following stalls:-

5 (5*1), 8 (8*1), 10 (5*2), 13 (5+8), 15 (5*3), 16 (8*2), 18(5*2 +8), 20 (5*4), 21 (8*2 +5), 23 (5*3 +8), 24 (8*3), 25 (5*3), 26(5*2 +8*2), 28 (5*4 +8), 29(8*3 +5), 30(5*6), 31(5*3 +8*2), 32(8*4), 33(28 +5), 34 (29+5), 35(30+5), 36(31+5) .....

Thus k=27

__Answer : __Option 2

__Challenge :__ Understanding application of multiples

**5. If a! + b! +c! = abc, where a, b and c are distinct decimal digits, find b.**

__Topic : __Numbers

__Approach :__Since RHS is a three digit number, none of a, b and c can take a value equal or greater than 7 because 7!=5040, which is a four digit number. Further, it is not possible to have a 6 in the system as well because 6!=720, and the possibility of 7 being in the system has been already ruled out.

The only digits left are 0,1,2,3,4 and 5. It is essential to have a 5 in the system because the maximum value of abc in the absence of 5 will be 32 (4!+3!+2!), which is a two digit number. abc can’t take a value equal to more than 150 (3!+4!+5!), so a=1, and b can’t be 5, so c=5. So the equation becomes 1!+b!+5!=1b5; trial and error with 0, 2, 3 and 4 yields 4 as the value of b. The equation is 1!+4!+5!=145

__Answer :__ Option 2

__Challenge :__ Understanding application of factorials and ability to manage apparent data insufficiency.