Quantitative Aptitude - Arithmetic Ability Questions

Quantity1-
Rakhi’s marks= 115
Meenakshi’s marks= 115 - 20 = 95
Puneeta’s marks= 95 + 65= 160
Ankita’s marks=160 - 35= 125
Simpy’s marks= 125+ 67= 192
Total maximum marks= 192 + 108= 300
Required percentage marks of Simpy
= 192300×100=64%
Quantity2-
Let length and breadth be 100.

After increase in length it become 160, then
reduction in breadth be ‘x’
Now, 160*x= 100*100
Hence, x = 10000160=62.5

100-62.5=37.5%

Let the sum be Rs. x. Then,
C.I. = x ( 1 + ( 10 /100 ))^2 - x = 21x / 100 ,
S.I. = (( x * 10 * 2) / 100) = x / 5
(C.I) - (S.I) = ((21x / 100 ) - (x / 5 )) = x / 100
( x / 100 ) = 632 * x = 63100.
Hence, the sum is Rs.63,100.

Each of the 10 letters can be posted in any of the 5 boxes.

So, the first letter has 5 options, so does the second letter and so on and so forth for all of the 10 letters.

i.e. 5*5*5*….*5 (upto 10 times) = 5 ^ 10.

Let C.P. = Rs. C.
Then, 832 - C = C - 448
2C = 1280 => C = 640
Required S.P. = 150% of Rs. 640 = 150/100 x 640 = Rs. 960.

Cost price of 25 kg = Rs. (15 x 14.50 + 10 x 13) = Rs. 347.50.
Sell price of 25 kg = Rs. (25 x 15) = Rs. 375.
profit = Rs. (375 — 347.50) = Rs. 27.50.

                                              A     +      B        =      C     +     D 

                                              |              |                 |             | 

    Ratio of efficiency         10x   +    5x               9x     +   6x 

                                             |________|                 |_________|   

                                                   15x                           15x 

Therefore , ratio of efficiency of A:C  =10:9 

Therefore,  ratio of days taken by A:C = 9:10 

Therefore, number of days taken by A = 18 days                 

Other number = 11×7700275= 308