Quantitative Aptitude - Arithmetic Ability Questions

Length of largest tile = H.C.F. of 1517 cm and 902 cm = 41 cm. 

Area of each tile = 41×41cm2  

Required number of tiles  = 1517×90241×41= 814

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

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.

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

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

9M + 12B ----- 12 days ...........(1)
12M + 12B ------- 10 days........(2)
10M + 10B -------?
108M + 144B = 120M +120B
24B = 12M => 1M = 2B............(3)

From (1) & (3)

18B + 12B = 30B ---- 12 days

20B + 10B = 30B -----? => 12 days.