Quantitative Aptitude - Arithmetic Ability Questions

From the given data,

containers contain mixtures of milk and water as 1365 lit, 1560 lit and 1755 lit.

Biggest measure that can measure all these different quantities exactly can be given by

HCF of (1365, 1560, 1755) 

That can be found as 

HCF of 1365, 1755 = 195

Now, HCF of 195 and 1560 = 195

Hence, biggest measure that can measure all the given different quantities exactly is 195 lit.

Let the length of the 1st train = L mts

Speed of 1st train = 48 kmph

Now the length of the 2nd train = L/2 mts

Speed of 2nd train = 42 kmph

Let the length of the bridge = D mts

Distance = L + L/2 = 3L/2

Relative speed = 48 + 42 = 90 kmph = 90 x 5/18 = 25 m/s(opposite)

Time = 12 sec

=> 3L/2x25 = 12

=> L = 200 mts

Now it covers the bridge in 45 sec

=> distance = D + 200

Time = 45 sec

Speed = 48 x5/18 = 40/3 m/s

=> D + 200/(40/3) = 45

=> D = 600 - 200 = 400 mts

Hence, the length of the bridge = 400 mts.

Let length, breadth and height of the room be 7, 3, 1 unit respectively.

Area of walls = 2(l+b)xh = 2(7+3)x1 = 20 sq. unit.

Now, length, breadth and height of room will become 3.5, 6 and 2 respectively. 

Area of walls = 2(l+b)xh = 2(3.5+6)x2 = 38 sq. unit. 

% Increase in the area of walls = (38-20)x100/20 = 90%.

X= 2/3Y 

=> Y= 3/2X

Aage of X after 6 years = 46 years

So present age of X = 46-6 = 40 years

Now, present age of Y = 3/2 * 40 = 60 years.

Let the present age of Deepa = 5p

Let the present age of Hyma = 6p

After four years their ratio = 6 : 7

=> 5p + 46p + 4 = 67

=> 35p + 28 = 36p + 24

=> p = 4

Therefore, the difference between their ages = 24 - 20 = 4 years.

 C.I. when interest compounded yearly

Rs.5000*1+4100*1+12*4100

 Rs.5000*2625*5150

=Rs.5034

C.I. when interest iscompounded half-yearly 

 Rs.5000*1+21003

= Rs. 5306.04

Difference = Rs. (5306.04 - 5304) = Rs. 2.04

From given data,

R : M = 5 : 7

M : A = 5 : 7

R : M : A = 25 : 35 : 49

25 + 35 + 49 = 109