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.

Suppose the man works overtime for x hours.

Now, working hours in 4 weeks = (5 x 8 x 4) = 160.

Therefore,  160 * 2.40 + x * 3.20 = 432

 => 3.20x = 432 - 384 = 48

 => x = 15.

 Hence, total hours of work = (160 + 15) = 175.

Dog : Hare = (3*3) leaps of hare : 5 leaps of hare = 9 : 5.

required area = (area of an equilateral triangle of side 7 cm)- (3 * area of sector with à = 60 degrees and r = 3.5cm)
34*7*7-3*227*3.5*3.5*60360 sq cm

=34*49-11*0.5*3.5 sq cm

= 1.967 sq cm

Since it is an isosceles trapezium

So angle C = angle D = x

let A = 180 –D = 180 –x (since AB is parallel to CD)

B = 180-x

A+C = 180 –x + x = 180 degrees (Property of cyclic quadrilateral)

ABCD is cyclic parallelogram with AB // CD and AD // BC

Considering angles A = C = y (Property of parallelogram) and B=D = x Also since it is cyclic A+C = B+D = 180 degrees

So x=y=90 degrees And also opposite sides are equal being a parallelogram

Thus ABCD is a rectangle

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%