Quantitative Aptitude - Arithmetic Ability Questions

By trial and error method from the given options,

as 54 - 45 =  (1/11) x (45+54)  = 99/11  = 9

Hence this satisfies the given conditions.

Then the woman's husband age is 54.

The cow can graze an area of 2826 sq.m i.e it is able to cover area of circle of 2826 sq.m

Therefore, πr2 =2826
r2 =(2826×7/22) = 899.18 = 900
r = 30 m.

Therefore, the radius of the circle implies the length of the rope = 30 mts.

Let the total distance be 'x' km.

Time taken to cover remaining 40% of x distance is   t1=40×x100×48

But given time taken to cover first 60% of x distance is   t2=t1+2060hrs

 $$\begin{gathered} t2=60×x100×48 \\ \end{gathered}$$

∴60×x100×48= 40×x100×48+2060

 x=80 km.

p=d×1002R2 

Where d is difference, r is rate of interest. 

=> (48 x 100 x 100) / 8 x 8 = Rs.7500

Speed = 78 x 5/18 = 65/3 m/sec.
Time = 1 min = 60 sec.
Let the length of the train be x meters.
Then, (900 + x)/60 = 65/3
x = 400 m.

x^z = y²        10^(0.48z) = 10^{(2 x 0.70)} = 10^1.40

 0.48z = 1.40

 z = 140/18 = 35/12 = 2.9 (approx.)

Maximum earning will be only when he will won on the maximum yielding table.

     A ----> 10:1

     B ----> 20:1

     C ----> 30:1

 i.e, he won on B and C but lost on A

 20 x 200 + 30 x 200 -1 x 200 = 9800

 minimum earning will be when he won on  table A and B and lose on that table 3.

Therefore, 10 x 200 + 20 x 200 - 1 x 200  =  6000-200 = 5800

Therefore, Difference= 9800 - 5800 = 4000

Total work is given by L.C.M of 72, 48, 36

Total work = 144 units

Efficieny of A = 144/72 = 2 units/day

Efficieny of B = 144/48 = 3 units/day

Efficieny of C = 144/36 = 4 units/day

According to the given data,

2 x p/2 + 3 x p/2 + 2 x (p+6)/3 + 3 x (p+6)/3 + 4 x (p+6)/3 = 144 x (100 - 125/3) x 1/100

3p + 4.5p + 2p + 3p + 4p = 84 x 3 - 54

p = 198/16.5

p = 12 days.

Now, efficency of D = (144 x 125/3 x 1/100)/10 = 6 unit/day

(C+D) in p days = (4 + 6) x 12 = 120 unit

Remained part of work = (144-120)/144 = 1/6.