CAT Questions

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Let the original price of rice be Rs. x

Let rice a man can buy for Rs. 500 at rs. x/kg be = R kgs

From given data, for Rs. 500

x ---- R kgs

(x + 25x/100) ---- (R - 5) kgs

=> Rx = (R - 5)(x + 25x/100)

=> Rx = (R - 5)(125x/100)

=> 100Rx = 125Rx - 625x

=> R = 25

So at rate of Rs. x/kg, man get 25 kgs for Rs. 500

=> x = 500/25 = Rs. 20

Given 2000 ---- 54 days

The provisions for 2000 men for 39 days can be completed by 'm' men for only 20 days.

i.e, 2000 ----- 39 days == m ---- 20 days

=> m x 20 = 2000 x 39

m = 3900

So total men for 20 days is 3900

=> 2000 old and 1900 new reinforcement.

Hence, reinforcement = 1900.

By mistake = x/10 

Actual value = x * 10 

% Change = 10x-x1010x×100 = 99100×100  = 99% (negative)

Since actual value is greater than the wrong value.

Passing marks are 0.6x 

so 0.3x + 30 = 0.6x   => x = 100

Q's mother is sister of P &  daughter of M.[sex of P and M are not clear]

S is daughter of P and sister of T.  [sex of P and T is not clear]

M's children : Q's mother and P

P's children  :S[female] and T

M is either Grandmother or Grandfather of T

Given 4men, 12 women and 20 children work for  2 days.

Workdone for 2 days by 4men, 12 women and 20 children = 4 x 26 x 12 + 12 x 28 x 18 + 20 x 218 x 10 = 12

Therefore, remaining work = 1 - 12 = 12

To complete the same work by only men in 1 day,

We know that M1 x D1 = M2 x D2

Here M1 = 6 , D1 = 12 and M2 = M , D2 = 1

12 x 6 = M x 1

=> M = 12 x 6 = 72

=> But the remaining work = 1/2

Men required => 1/2 x 72 = 36

Only men required to Complete the remaining work in 1 day = 36.

Let the side of the square plot be 'a' ft.
Given area of the plot (a x a) = 289 => a = 17
Length of the fence = Perimeter of the plot = 4a = 68 ft.
Cost of building the fence = 68 x 58 = Rs. 3944.