Quantitative Aptitude - Arithmetic Ability Questions

Speed in downstream = (14 + 4) km/hr = 18 km/hr;

Speed in upstream = (14 – 4) km/hr = 10 km/hr.

Let the distance between A and B be x km. Then,

x/18 + (x/2)/10 = 19 ⇔ x/18 + x/20 = 19 ⇒ x = 180 km.

Required Number = (L.C.M  of 12, 16, 18,21,28)+7

= 1008 + 7

= 1015

Assume the weight of alloy a is 100 kg 

Therefore, The weight of alloy B is 400kg

             Gold          silver         copper

    A       40kg          60kg          0kg

    B      140kg         160kg        100kg

  total    180kg          220kg        100kg

Therefore, Ratio of gold and silver in new alloy = 180500:200500= 36% :44%

Let cp= 100,
35 % increase in sp=135
10 % discount in 135((135*10)/100)=13.5
so 1st sp=(135-13.5)=121.5, again 15 % discount in 1st sp((121.5*15)/100)=18.225
2nd sp=(121.5-18.225)=103.275,

so finally cp=100,sp=103.275 ,gain by 3.27%

From 5 consonants, 3 consonants can be selected in 5C3 ways.

From 4 vowels, 2 vowels can be selected in 4C2 ways.

Now with every selection, number of ways of arranging 5 letters is 5P5ways.

Total number of words = 5C3*4C2*5P5

                                = 10x 6 x 5 x 4 x 3 x 2 x 1= 7200

Let the present age of the man be 'P' and son be 'Q',
Given, P + Q = 64 or Q = (64 - P)
Now the man says "I am five times as old as you were, when I was as old as you are",
So, P = 5[B - (P - Q)]
We get 6P = 10Q,
Substitute value for Q,
6P = 10(64 - P),
Therefore P = 40, Q = 24.