CAT Questions

---

We have : ( a + b + c) / 3 = M   or (a + b + c) = 3M.

Now. a+b+c2=3M2=9M2 .

<=>a2+b2+c2+2ab+bc+ca=9M2

a2+b2+c2=9M2 ,(ab+bc+ca=0)     

Required mean =a2+b2+c23=9M23=3M2 .

110% of S.P. = Rs. 616

S.P. = (616 x 100)/110 = Rs. 560

C.P = (110 x 560)/112 = Rs. 500

How many times can you subtract 10 from 100, for this if we go through logically, it is only 1 time. For the first time if we subtract 10 from 100, then there will no 100 to subtract from.

Hence, it is only 1 time we can subtract 10 from 100.

The quadratic equation whose roots are reciprocal of 2x2+5x+3=0 can be obtained by replacing x by 1/x.

Hence, 2(1/x)(1/x)+ 5(1/x) + 3 = 0

=> 3x2+5x+2=0

"CD-ROM is a semiconductor memory" is a FALSE statement.

Compact Disc-Read-Only Memory, a type of optical disk capable of storing large amounts of data -- up to 1GB, although the most common size is 650MB (megabytes). A single CD-ROM has the storage capacity of 700 floppy disks, enough memory to store about 300,000 textpages.

Semiconductor memory is a type of semiconductor device tasked with storing data. There are two electronic data storage mediums that we can utilize, magnetic or optical.

Magnetic storage:

* Stores data in magnetic form.

* Affected by magnetic fields.

* Has high storage capacity.

* Doesn't use a laser to read/write data.

* Magnetic storage devices are; Hard disk , Floppy disk, Magnetic tape etc.

Optical storage:

* Stores data optically, uses laser to read/write.

* Not affected by magnetic fields.

* Has less storage than a hard disk.

* Data accessing is high, compared to a floppy disc.

* Optical storage devices are; CD-ROM,CD-R, CD-RW, DVD etc.

There is also volatile memory. This is memory that loses its data once power is cut off, while non-volatile memory retains data even without power.

A red ball can be drawn in two mutually exclusive ways

 (i) Selecting bag I and then drawing a red ball from it.

(ii) Selecting bag II and then drawing a red ball from it.

Let E1, E2 and A denote the events defined as follows:

E1 = selecting bag I,

E2 = selecting bag II

A = drawing a red ball

Since one of the two bags is selected randomly, therefore 

P(E1) = 1/2  and  P(E2) = 1/2

Now, PAE1 = Probability of drawing a red ball when the first bag has been selected = 4/7

  PAE2  = Probability of drawing a red ball when the second bag has been selected = 2/6

 Using the law of total probability, we have 

 P(red ball) = P(A) = PE1×PAE1+PE2×PAE2 

                          = 12×47+12×26=1942

Number of persons between Amrita and Mukul = 50 - (10 + 25) = 15. Since Mamta lies in middle of these 15 persons, so Mamta`s position is 8th from Amrita i.e. 18th from the front.