Quantitative Aptitude - Arithmetic Ability Questions

B runs 45/2mts in 6 sec

B covers 300m in (6*2/45*300)sec =80 sec

sum=Rs.x

C.I=[x(1+4/100)^2-x]=(676/625x-x)=51/625

S.I=(x*4*2)/100=2x/25

x=625

Let the two consecutive odd integers be (2x + 1) and (2x + 3)

Then,

(2x + 3)² - (2x + 1)²

= (2x + 3 + 2x + 1) (2x + 3 - 2x - 1)

= (4x + 4)(2)
= 8 (x + 1), which is always divisible by 8

Given the three mixtures ratio as (1:2),(2:3),(3:4)

(1+2),(2+3),(3+4)

Total content = 3,5,7

Given equal quantities of the three mixtures are mixed, then LCM of 3, 5, 7 = 105

105/3 = 35 , 105/5 = 21 , 105/7 = 15

Now, the individual equal quantity ratios are (35x1, 35x2), (21x2, 21x3), (15x3, 15x4)

 i.e (35,70), (42,63), (45,60)

So overall mixture ratio of milk and water is

35+42+45 : 70+63+60

122:193

But in the question asked the ratio of water to milk = 193 : 122

Cost Price of 5 kg = Rs.(14*2 + x*3) = (28 + 3x).

Sell price of 5 kg = Rs. (22x5) = Rs. 110.

[{110 - (28 + 3x)}/(28 + 3x) ]* 100 =10

[82-3x/28 + 3x]= 1 / 10

820 - 30x = 28 +3x 

33x = 792 

x = 24

The largest 4 digit number is = 9999
After dividing 9999 with 88, we get 55 as remainder
so largest 4 digit number divisible by 88 = 9999-55 = 9944.

The first letter is E and the last one is R.

Therefore, one has to find two more letters from the remaining 11 letters.

Of the 11 letters, there are 2 Ns, 2Es and 2As and one each of the remaining 5 letters.

The second and third positions can either have two different letters or have both the letters to be the same.

Case 1: When the two letters are different. One has to choose two different letters from the 8 available different choices. This can be done in 8 * 7 = 56 ways.

Case 2: When the two letters are same. There are 3 options - the three can be either Ns or Es or As. Therefore, 3 ways.

Total number of possibilities = 56 + 3 = 59