Quantitative Aptitude - Arithmetic Ability Questions

Area = (13.86 x 10000) sq.m = 138600 sq.m

πR2=138600⇒R2=138600×722⇒R=210m 

Circumference = 2πR2=2×227×210=1320m 

Cost of fencing = Rs. (1320 x 4.40) = Rs. 5808.

Speed of Boy is B = 4.5 kmph

Let the speed of the stream is S = x kmph

Then speed in Down Stream = 4.5 + x

speed in Up Stream = 4.5 - x

As the distance is same,

=> 4.5 + x = (4.5 - x)2

=> 4.5 + x = 9 -2x

3x = 4.5

x = 1.5 kmph

perimeter of the plot = 2(90+50) = 280m
no of poles =280/5 =56m

Upstream speed = B-S

Downstream speed = B+s

B-S = 15/5 = 3 km/h

Again          B= 4S

Therefore    B-S = 3= 3S

=>             S = 1 and B= 4 km/h

Therefore    B+S = 5km/h

Therefore, Time during downstream = 15/5 = 3h

The ratio of the ages of A and B is 3 : 5.
The ratio of the ages of B and C is 3 : 5.

B's age is the common link to both these ratio. Therefore, if we make the numerical value of the ratio of B's age in both the ratios same, then we can compare the ages of all 3 in a single ratio.

The can be done by getting the value of B in both ratios to be the LCM of 3 and 5 i.e., 15.

The first ratio between A and B will therefore be 9 : 15 and
the second ratio between B and C will be 15 : 25.

Now combining the two ratios, we get A : B : C = 9 : 15 : 25.

Let their ages be 9x, 15x and 25x.
Then, the sum of their ages will be 9x + 15x + 25x = 49x

The question states that the sum of their ages is 147.
i.e., 49x = 147 or x = 3.

Therefore, B's age = 15x = 15*3 = 45

Let breadth = x metres, length = 3x metres, height = H metres. 

Area of the floor=(Total cost of carpeting)/(Rate) = (270/5) sq.m = 54 sq.m 

x×3x2=54⇒x2=54×2⇒x=6  

So, breadth = 6 m and length =362 = 9 m.

Now, papered area = (1720/10) =  172 sq.m 

Area of 1 door and 2 windows = 8 sq.m 

Total area of 4 walls = (172 + 8) sq.m = 180 sq.m 

2×9+6H=180⇒H=6