Quantitative Aptitude - Arithmetic Ability Questions

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.

It won't requires that much solving.

Here given the increase in weekly collection by Rs. 1,05,000

That is this increase denotes for all 7 days.

=> Gross Collection increase per day = Rs. 1,05,0007 = Rs. 15,000

Hence, the gross collection increase per day = Rs. 15,000.

Let, Distance between A and B = d

Distance travelled by P while it meets Q = d + 11

Distance travelled by Q while it meets P = d – 11

Distance travelled by Q while it meets R = d + 9

Distance travelled by R while it meets Q = d – 9

Here the ratio of speeds of P & Q => SP : SQ = d + 11 : d – 11

The ratio of speeds of Q & R => SQ : SR = d + 9 : d – 9

But given Ratio of speeds of P & R => P : R = 3 : 2

 SPSR = SPSQxSQSR = d+11d+9d-11d-9

=> d+11d+9d-11d-9 = 3/2

=>  d = 1, 99

=> d = 99 satisfies.

Therefore, Distance between A and B = 99

99  = 1 x 3 x 3 x 11

101= 1 x 101

176= 1 x 2x 2 x 2 x 2 x 11

182= 1 x 2 x 7 x 13

So, divisors of 99 are 1, 3, 9, 11, 33, 99

      divisors of 101 are 1,101

      divisors of 176 are 1, 2, 4, 8, 16, 22, 44, 88, 176

      divisors of 182 are 1, 2, 7, 13, 14, 26, 91, 182

Hence , 176 hasthe most number of divisors. 

Suppose they meet after 'h' hours

Then

3h + 4h = 17.5

7h = 17.5

h = 2.5 hours

So they meet at => 4 + 2.5 = 6:30 pm

Speed of lorry = 36012 = 30 kmph

Speed of van =250100 x 30 = 75 kmph

Speed of bike = 35 x 75 = 45 kmph

Therefore, now required average speed of bike and van  = 75 + 452= 60 kmph.

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.)

Let the actual distance travelled be x km.
Then x/8=(x+20)/12
=> 12x = 8x + 160
=> 4x = 160
=> x = 40 km.