Quantitative Aptitude - Arithmetic Ability Questions

You can find the integers which when divided by 5 have a remainder 2 by adding 2 to all multiples of 5. So we have n = 7 , 12, 17, 22 etc.

From this series we can see that n does not have to be odd.

Also n + 1 can be a prime because, for example, 12 + 1 = 13

And (n + 2) / 7 has a remainder 2 in some cases but not all.

Remember the question asks us for what MUST be true, and we see that none of the statements are true in all cases. However, adding 3 to any of the values of n will always give a multiple of 5.

Let us calculate both the length and width of the room in centimeters.

Length = 6 meters and 24 centimeters = 624 cm

width = 4 meters and 32 centimeters = 432 cm

As we want the least number of square tiles required, it means the length of each square tile should be as large as possible.Further,the length of each square tile should be a factor of both the length and width of the room.

Hence, the length of each square tile will be equal to the HCF of the length and width of the room = HCF of 624 and 432 = 48

Thus, the number of square tiles required = (624 x 432 ) / (48 x 48) = 13 x 9 = 117

let inner radius be r meters. 

$$\begin{gathered} 2πr=440⇒r=440×744=70m \\ radius of outer circle =70+14m \end{gathered}$$  

Therefore, Radius of outer circle = 70 + 14 = 84m

Given rate of wheat at cheap = Rs. 2.90/kg

Rate of wheat at cost = Rs. 3.20/kg

Mixture rate = Rs. 3/kg

Ratio of mixture = 

          2.90                              3.20

                              3

(3.20 - 3 = 0.20)            (3 - 2.90 = 0.10)

0.20 : 0.10 = 2:1

Hence, wheat at Rs. 3.20/kg be mixed with wheat at Rs. 2.90/kg in the ratio of 2:1, so that the mixture be worth Rs. 3/kg.

Since  Area of Rectangle = side1 x side2
Therefore, error% in area
= x + y + xy100 %
= [20 - 10 + ( -10 x 20)/100]% or 8%
i.e. 8% excess.

Average = (11 + 22 + 33 + 44 + 55 + 66 + 77 + 88 + 99) / 9

             =( (11 + 99) + (22 + 88) + (33 + 77) + (44 + 66) + 55) / 9

             = (4 * 110 + 55)/9 =  495 / 9 = 55. 

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.