CAT Questions

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The letter 'E'.

As of the above given data,

It is in the 1st place in Earth, 2nd in Heaven and in the middle of the Sea.

It appears once in a Year and twice in a Week.

It is present in February, June, September, October, November & December.

But not in the Month.

Hence, it is the letter 'E'.

Let the Original capital amount be Rs. P

Given after 2 years the amount is Rs. 10000

Rate of interest = 30%

According to the question,

$$\begin{gathered} P x 130100 x 130100 = 10000 \\ => P = 1000000169 = 5917 \end{gathered}$$

Therefore, the capital amount P = Rs. 5917

It is launched vertically with an initial speed of 110 ft/sec.

It will reach maximum height when the speed drops to 0.

Vo=110 feet per second 

ho=2 feet 

So, h(t) = -16t^2 +110t +2 

Take the derivative: h'(t) = 110 -32t 

The maximum height will be at the inflection when the derivative crosses the x-axis, then h'(t)=0. 

So, set h'(t)=0 and solve for t: 

0 = 110 -32t 

-110 = -32t 

t=3.4375 

t=3.44 seconds.

Let that something be 'x'

From the given data,

x + x/5 = 21

5x + x/5 = 21

6x = 21 x 5

=> x = 17.5

Hence, that something is x = 17.5.

The PIN code system was introduced on 15 th August 1972 by Shriram Bhikaji Velankar. Currently India has six digit post code and soon it will be converted in eight digits.

Given principal amount = Rs. 8000

Time = 3yrs

Rate = 5%

C.I for 3 yrs = $$\begin{gathered} 8000 x 105100 x 105100 x 105100 - 8000 \\ = Rs. 1261 \end{gathered}$$

Now, C.I for 2 yrs = $$\begin{gathered} 8000 x 105100 x 105100 \\ = Rs. 820 \end{gathered}$$

Hence, the required difference in C.I is 1261 - 820 = Rs. 441

= log tan10+log tan890 + log tan20+ log tan880+⋯⋯+log tan450  

= log [tan10 × tan890] + log [tan20 × tan880 ] +⋯⋯+log1  

∵ tan(90-θ)=cotθ and tan 450=1  

= log 1 + log 1 +.....+log 1 

= 0.

55 min. spaces are covered in 60 min.
60 min. spaces are covered in ((60/55) x 60) min = 720/11 min.
Loss in 65 min. = (720/11 - 65 ) = 5/11
Loss in 24 hrs = (5/11 x 1/65 x 24 x 60)min = 1440/143 min.