Quantitative Aptitude - Arithmetic Ability Questions

Amount = Rs.8000*1+51002=Rs.8820

surface area = [2(10*4 + 4*3 + 10*3)]sq.cm = 164 sq.cm

Given 2000 ---- 54 days

The provisions for 2000 men for 39 days can be completed by 'm' men for only 20 days.

i.e, 2000 ----- 39 days == m ---- 20 days

=> m x 20 = 2000 x 39

m = 3900

So total men for 20 days is 3900

=> 2000 old and 1900 new reinforcement.

Hence, reinforcement = 1900.

There are seven positions to be filled.

The first position can be filled using any of the 7 letters contained in PROBLEM.

The second position can be filled by the remaining 6 letters as the letters should not repeat.

The third position can be filled by the remaining 5 letters only and so on.

Therefore, the total number of ways of rearranging the 7 letter word = 7*6*5*4*3*2*1 = 7! ways.

Let the original price of rice be Rs. x

Let rice a man can buy for Rs. 500 at rs. x/kg be = R kgs

From given data, for Rs. 500

x ---- R kgs

(x + 25x/100) ---- (R - 5) kgs

=> Rx = (R - 5)(x + 25x/100)

=> Rx = (R - 5)(125x/100)

=> 100Rx = 125Rx - 625x

=> R = 25

So at rate of Rs. x/kg, man get 25 kgs for Rs. 500

=> x = 500/25 = Rs. 20

πr2h2πr2h = 924264  

=> r = (24/364) x 2 = 7m 

2πrh = 264 ⇒ h = (264 × 722×12 ×17) = 6m   

Required Rato =2r/h  =14/6 = 7:3

Given 4men, 12 women and 20 children work for  2 days.

Workdone for 2 days by 4men, 12 women and 20 children = 4 x 26 x 12 + 12 x 28 x 18 + 20 x 218 x 10 = 12

Therefore, remaining work = 1 - 12 = 12

To complete the same work by only men in 1 day,

We know that M1 x D1 = M2 x D2

Here M1 = 6 , D1 = 12 and M2 = M , D2 = 1

12 x 6 = M x 1

=> M = 12 x 6 = 72

=> But the remaining work = 1/2

Men required => 1/2 x 72 = 36

Only men required to Complete the remaining work in 1 day = 36.

d=2*s=>d=2*69696