Quantitative Aptitude - Arithmetic Ability Questions

1/15 + 1/12 + 1/R = 1/6, we got R = 60 (it means R will take 60 days to complete the work alone)

so ratio of work done by P:Q:R = 1/15 : 1/12 : 1/60 = 5 : 4 : 1

so R share = (1/10) x 1200 = 120.

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.

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

Shopkeeper sells at CP using weight of 900 gm instead of1 kg

So, 1000gm –900 gm = 100gm

⇒100/900 X 100

⇒100/9 %

Sum =Rs.(50 x 100/2x5)

= Rs. 500.

Amount

=[Rs.500x(1+5/100)²]

=Rs(500x21/20x21/20).

=Rs. 551.25

C.I

= Rs. (551.25 - 500)

= Rs. 51.25

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.