Quantitative Aptitude - Arithmetic Ability Questions

Let 1 woman's 1 day work = x.

Then, 1 man's 1 day work = x/2 and 1 child's 1 day work  x/4.

So, (3x/2 + 4x + + 6x/4) = 1/7

28x/4 = 1/7 => x = 1/49

1 woman alone can complete the work in 49 days.

So, to complete the work in 7 days, number of women required = 49/7 = 7.

 As per the BODMAS rule, the priority in which the operations should be done is:

Note: Addition and subtraction can be treated on same priority (from left to right) when they are in consecutive order.

(115/5) + 12 × 6 = ? + (64/4) – 35 23 + 72 = ? + 16 – 35 95 = ? - 19 95+19 = ? ? = 114

Here the given number series is 270, 261, 279, 252, 289, 243

270 - 9 = 261

261 + 18 = 279

279 - 27 = 252

252 + 36 = 288 (not equals to 289)

288 - 45 = 243

Hence, the odd number in the given number series is 289.

Product of numbers = 11 x 385 = 4235

Let the numbers be 11a and 11b . Then , 11a x 11b = 4235  =>  ab = 35

Now, co-primes with product  35 are (1,35) and (5,7)

So, the numbers are ( 11 x 1, 11 x 35)  and (11 x 5, 11 x 7)

Since one number lies 75 and 125, the suitable pair is  (55,77)

Hence , required number = 77

5 : 4 = 5/4 = ( (5/4) x 100 )% = 125%.

Pass percentage  =[(252/270)*100] % =280/3 % =9313 9 %

Lets say shyam sells at 100,

Since Ram sells 25% cheaper than Shyam,

Therefor Ram sells at less than 25% of100 or 75 rs.

Ram sells 25% dearer than Bram or 125% of Ram =100% of Bram or

125% of x (say x price of bram )=75rs.
or 100% of x =60rs.
hence price of bram is 60rs.

now Bram's good is cheaper than Shyam's as (100-60)x100/100% or 40%.

Hence Brams Good is 40% cheaper than that of Shyam's good.

It is given that last three digits are randomly dialled. then each of the digit can be selected out of 10 digits in 10 ways.
Hence required probability =1103 = 1/1000