Quantitative Aptitude - Arithmetic Ability Questions

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Milk in 1-litre mixture of A = 4/7 litre. 

Milk in 1-litre mixture of B = 2/5 litre. 

Milk in 1-litre mixture of C = 1/2 litre.

By rule of alligation we have required ratio X:Y   

   X                  :                 Y  

  4/7                                2/5  

           \                      /

              (Mean ratio)
                   (1/2) 

           /                      \

 (1/2 – 2/5)     :       (4/7 – 1/2)  

    1/10                      1/1 4 

So Required ratio = X : Y = 1/10 : 1/14 = 7:5

Let the age of A = A, B = B and C = C

From given

A = 3B

4 years ago,

C-4 = 2(A-4);

After 4 years

A + 4 = 31;

A = 27;
sub in above B, we get B = 9
C = 2(27 - 4) + 4 = 46 + 4 = 50

Hence B = 9yrs and C = 50yrs.

Let the numbers be x and y. Then,

x^2+y^2=3341......(1)

x^2-y^2=891......(2)

Adding (i) and (ii), we get : 2x^2=4232 or x^2= 2116 or x =46

Subtracting (ii) from (i), we get : 2y^2= 2450 or y^2 = 1225 or y = 35

So, the numbers are 35 and 46

Angle of triangle = 180 × 3/12

= 45 deg

Other angles of the triangle are = 60 deg, 75 deg

Angle of quadrilateral

= 360 - (45 + 60 + 75)

= 180 deg

Therefore, the four angles of the quadrilateral are 45, 60, 75 & 180 degrees.

Hence, the sum of largest and smallest angle of the quadrilateral are 180 + 45 = 225 deg.

The day of the week repeats every 7 days.

Given today is Friday. Again Friday is repeated on the 7th day, 14th,... on 7 multiple days.

Hence, Friday is on the 63rd day, as 63 is multiple of 7.

Now, the required day of the week on the 65th day is Sunday.

Let their edges be a and b. Then,

a3/b3 = 1/27 (or) (a/b)3 = (1/3)3 (or) (a/b) = (1/3).

\Ratio of their surface area = 6a2/6b2 = a2/b2 = (a/b)2 = 1/9, i.e. 1:9.

Let ABC be the isosceles triangle and AD be the altitude 

Let AB = AC = x. Then, BC = (32 - 2x). 

Since, in an isosceles triangle, the altitude bisects the base,
so BD = DC = (16 - x).
In triangle ADC,AC2=AD2+DC2 

⇒x2=82+16-x2⇒x=10 

BC = (32- 2x) = (32 - 20) cm = 12 cm. 

Hence, required area =  12*BC*AD=12*12*8=48cm2