CAT Questions

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Let the present age's of mother be 'M' and that of her son be 'S'.

According to the given data in the question,

M - 5 = 3(S - 5)

F - 3S = -10  ...............(1)

Then given ,

M + 5 : S + 5 = 11 : 5 

=> 5M - 11S = 30 ..............(2)

From (1) & (2), we get

S = 20

Then by substituting S = 20 in (1) or (2), we get

M = 50.

Hence, the present age of mother is M = 50 years.

K's brother's wife's father-in-law is K's brother's father i.e. K's father. K's father's daughter is K's Sister.
L is daughter of K's sister => K is L's Uncle or Aunt.

$$\begin{gathered} log3437 \\ log737 \\ 13log77 \\ We know that logxx = 1 \\ => 13 \end{gathered}$$

In the given number puzzle,

Sum of Row 1 = 14

Sum of Row 2 = 23

Sum of Row 3 = 13 + ?

Sum of Column 1 = 16 

Sum of Column 2 = 23 

Sum of Column 3 = 11 + ?

Diagonal 1 sum = 17

Diagonal 2 sum = 14 + ?

From the above analysis,

Sum of Row 2 = Sum of Column 2 = 23

Sum of Row 1 = Sum of Column 3 = 14 = 11 + ?    .............(1)

Sum of Row 3 = Sum of Column 1 = 16 = 13 + ?    .............(2)

Diagonal 1 sum = Diagonal 2 sum = 17 = 14 + ?    .............(3)

Hence,

From (1), (2) & (3)

? = 3

End of first year     :  10000 x (3/5) = Rs. 6000
End of second year :  6000 x (3/5) = Rs. 3600
End of third year    :  3600 x (3/5) = Rs. 2160

Given a jeep travels a certain distance taking 6 hrs in forward journey
During the return journey, it takes 4 hrs with an increased speed 12 km/hr
Let 'x' be the distance.
Now, speed is given by distance/time.
Here the difference between both speeds = 12

 x4 - x6 = 12

 3x - 2x12 = 12

=> x = 144 km

Therefore, the distance travelled by the jeep in forward or return journey is x = 144 km.

A Clock has 3 hands i.e, an Hour hand, a Minute hand and a Seconds hand.

Let 'x' be the number of new Rs.2000 notes.

Given for one new 2000 note there are three old 500 notes.

i.e, 3x

Given 10 more new Rs.2000 notes are added to the collection
and the ratio of new Rs.2000 notes to old Rs.500 notes 

x+103x=12⇒x=20

Therefore, number of new Rs.2000 notes =  x+10 = 30.

old Rs.500 notes = 3x = 60.

Thus, the total number of notes in the collection  = 30 + 60 = 90.