Quantitative Aptitude - Arithmetic Ability Questions

Let the sum be Rs. p, rate be R% p.a. and time be T years.

Then,

 P×T×(R+2)100-P×T×R100=108 ⇒2PT= 10800 ...... (1)

 And, 

 P×R×(T+2)100-P×R×T100=180 ⇒2PR = 18000 ......(2)

Clearly, from (1) and (2), we cannot find the value of p
So, the data is not sufficient.

Choose 1 person for the single room & from the remaining choose 2 for the double room & from the remaining choose 4 people for the four person room, 

 Then, ⁷C₁ x ⁶C₂ x ⁴C₄ 

_{= 7 x 15 x 1 = 105}

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.

when folded along the breadth
we have 2(l/2 +b) = 34 or l+2b = 34...........(1)
when folded along the length, we have 2(l+b/2)=38 or 2l+b =38.....(2)
from 1 &2 we get l=14 and b=10
Area of the paper = 14*10 = 140 sq cm

Number of runs made by running = 120 - (3 x 4 + 8 x 6) = 120 - (60) = 60 

Required percentage = [(60/120)*100] % = 50%

Let the numbers be x and y. Then, 2x + 3y = 39...(i)and 3x + 2y = 30 ...(ii)

On solving (i) and (ii), we get :  x = 6 and y = 9.

larger number = 9

Take a square of side 4cm and a rectangle having l=6cm and b=2cm

then perimeter of square = perimeter of rectangle

area of square = 16 sq.cm

area of rectangle = 12 sq.cm

Hence a >b