Quantitative Aptitude - Arithmetic Ability Questions

0.46 = 46/100 = 23/50.

Sum of the numerator and denominator is 23 + 50 = 73

I=prt

Here by trial and error method, we can obseve that

404404 = 404 x 1001; 415415 = 415 x 1001, etc.

So, any number of this form is divisible by 1001.

20 x 55 x 66 = L x 65 x 70
Here L = 72,600/4550 = 15.95 =~ 16
Reduction is 20-16 = 4

which is 20% reduction.

From the given data,

When the price of rice is decreased by 10%, the value of Rs. 270 would become

270 x 100/90 = 300

Now, the value 270 becomes 300

And given, the girl could buy 1 kg extra for that value

Change in value = 300 - 270 = Rs. 30

Therefore, 1 kg extra rice she can get for Rs. 30

=> The original price of rice = Rs. 30/kg.

As she gets only 9 kgs for Rs. 270 before 10% discount, now after discount she gets 10 kgs instead of 9 kgs for Rs. 270.  

A = P + Prt

Let 'M' be the maximum marks in the examination.

Therefore, Madhu got 32% of M = 32M/100 = 0.32M
And Kumar got 42% of M = 42M/100 = 0.42M.
In terms of the maximum marks Kumar got 0.42M - 0.32M = 0.1M more than Madhu. ---- (1)

The problem however, states that Kumar got 16 marks more than the cut-off mark and Madhu got 8 marks less than the cut-off mark. Therefore, Kumar has got 16 + 8 = 24 marks more than Madhu. ---- (2)

Now, Equating (1) and (2), we get
0.1M = 24 => M = 24/0.1 = 240
'M' is the maximum mark and is equal to 240 marks.
We know that Madhu got 32% of the maximum marks.
Therefore, Madhu got 32 x 240/100 = 76.8 marks.
We also know that Madhu got 8 marks less than the cut-off marks.

Therefore, the cut-off marks will be 8 marks more than what Madhu got
= 76.8 + 8 = 84.8.

Let the total applicants be 100

Then, 20% got 85 marks

i.e 20 x 85 = 1700

and 25% got 95 marks

 i.e 25 x 95 = 2375

Now, the remaining applicants are 55 and let the average marks scored by them be x.

Therefore, 2375 + 1700 + 55x  =  60×100 

6000 - 4075 = 55x

55x=1925

x=35.