Quantitative Aptitude - Arithmetic Ability Questions

Let each instalment be Rs.x .

1st year =  [x + (x * 12 * 2)/100] 

2nd year = [ x + (x *12 * 1)/100] 

3rd year = x 

Then, [x + (x * 12 * 2)/100] + [ x + (x *12 * 1)/100] + x =1092 

3x + ( 24x/100 ) + ( 12x/100 )  = 1092 

336x =109200

Therefore, x = 325 

Each instalment = Rs. 325

S.I. for 1 ½ years = Rs (1164 - 1008) = Rs 156 . 

S.I. for 2 years = Rs (156 x 23 x 2)= Rs 208. 

Therefore, Principal = Rs (1008 - 208) = Rs 800. 

Now, P = 800, T= 2 and S.I. = 208.

Therefore, Rate = (100 x S.I.) / (P x T) = [ (100 x 208)/(800 x 2)]% = 13%

To be together between 9 and 10 o'clock, the minute hand has to gain 45 min. spaces.
55 min. spaces gained in 60 min.

45 min. spaces are gained in 6055×45 min  or 49111min

Let the required numbers be 33a and 33b. 

Then 33a +33b= 528   =>   a+b = 16.

Now, co-primes with sum 16 are (1,15) , (3,13) , (5,11) and (7,9).

Therefore, Required numbers are  ( 33 x 1, 33 x 15), (33 x 3, 33 x 13), (33 x 5, 33 x 11), (33 x 7, 33 x 9)

The number of such pairs is 4

Area to be plastered = 2l+b×h+l×b 

=225+12×6+25×12=744sq.m 

Cost of plastering = 744×75100=Rs.558                       

$$\begin{gathered} [(K-L)2-K+L2]4K = ab \\ K2+L2-2KL-K2+L2+2KL4K= ab \end{gathered}$$

=> - 4KL/4K = a/b

=> L = - a/b

=> bL = -a

One way to deal with fractions is to convert them all to decimals. 

In this case all you would need to do is to see which is greater than 0.5.

Otherwise to see which is greater than ½, double the numerator and see if the result is greater than the denominator. In B doubling the numerator gives us 8, which is bigger than 7. 

132 = 4 x 3 x 11, So if the number is divisible by all three numbers 4,3 and 11,then the number is divisible by 132 also.
264   => 4,3,11(/)
396   => 4,3,11(/)
462   => 11,3
792   => 4,3,11(/)
968   => 11,4
2178 => 11,3
5184 => 3,4
6336 => 4,3,11(/)
Required number of numbers=4.