Quantitative Aptitude - Arithmetic Ability Questions

 L.C.M of 5, 6, 7, 8 = 840

Therefore, Required Number is of the form 840k+3.

Least value of k for which (840k+3) is divisible by 9 is k = 2 

Therefore, Required  Number = (840 x 2+3)=1683

Given the price of a commodity is decreased by  p = 20%

And its consumption is increased by q = 10%

Now, required increase or decrease in the expenditure on the commodity can be 

p + q + pq/100 = -20 + 10 - 200/100 = -10 - 2 = -12%.

Hence, decrease in the expenditure = 12%

            => log60(3*4*5)

                              =>     log6060

                                   = 1

From the given data,

The part of honey in the first mixture = 1/4

The part of honey in the second mixture = 3/4

Let the part of honey in the third mixture = x

Then,

1/4             3/4

           x

(3/4)-x     x-(1/4) 

Given from mixtures 1 & 2 the ratio of mixture taken out is 2 : 3

=> 34-xx-14=23

=> Solving we get the part of honey in the third mixture as 11/20

=> the remaining part of the mixture is water = 9/20

Hence, the ratio of the mixture of honey and water in the third mixture is 11 : 9 .

Let the length be 'l' and breadth be 'b'.

b = l × 3/4__________(a)

2(l+b) = 1050

l+b = 525___________(b)

From equations (a) and (b),

l = 300m, b = 225 m

Area = l × b

= 300 × 225

= 67500 sq.m.

P(atleast B) = P( B or A) = P(B) + P(A) = (0.3) + (0.4) = 0.7

2nd Jan 1901 means
(1900 years and 2 day)
Now, 1600 years have 0 odd day
300 years have 1 odd day
2 days has 2 odd day
Total no. of odd days = 0 + 1 + 2 = 3 days
Hence, the day on 2nd Jan 1901 was Wednesday.

Days remaining 124 – 64 = 60 days

Remaining work = 1 - 2/3 = 1/3

Let men required men for working remaining days be 'm'

So men required = (120 x 64)/2 = (m x 60)/1 => m = 64

Men discharge = 120 – 64 = 56 men.