Quantitative Aptitude - Arithmetic Ability Questions

Area to be plastered= [2(l + b) x h] + (l x b)

= [2(25 + 12) x 6] + (25 x 12)  

= (444 + 300)  

= 744 

Cost of plastering = Rs. 744 x (75/100)= Rs. 558

There are 7C2 ways of selecting two men, and 10C1 ways of selecting a woman.Since each position in the committee is different,arrange the three people in 3! ways.

So, 7C2*10C1*3! =1260

Money collected = (59.29 x 100) paise = 5929 paise.

  Number of members = 5929 = 77.

F = Rs. 498

TD = Rs. 18

PW = F - TD = 498 - 18 = Rs. 480

R = 5%

TD=PW×TR100

⇒18=480×T×5100

=> T = 3/4 years = 9 months

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.

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.  

1523+2323=19-423+19+423  

 The remainder on dividing by 19 will be -423+423= 0.