Quantitative Aptitude - Arithmetic Ability Questions

Profit ratio X & Y = (700 × 3) + (700 × 5/7× 3) + (700× 5/7 + 200 × 3/5 )× 6 : 600 ×12
X:Y= 7320 : 7200= 183:180
∴ X’s share from profit = 183 × 726/(183+180) = Rs. 366.

If the queen of spades and the four of diamonds must be in hand,we have 50 cards remaining out of which we are choosing 3.

So, 50C3

Here 85 of 352 => 85 is what percent in 352

Let p% of 352 is 85

Percentage(%) is generally calculated in the total of 100.

=> p x 352/100  = 85

=> p = 8500/352  = 24.14%

Hence, 85 is 24.14% of 352.

Let the initial speed of Akhil be '4p' kmph

Then speed after decrease in speed = '3p' kmph

We know that, 

Change in speed == Change in time

According to the given data,

$$\begin{gathered} 243p - 244p = 35 - 2560 \\ => p = 12 \end{gathered}$$

Hence, the initial speed of Akhil = 4p = 4 x 12 = 48 kmph.

Let the original Speed be "s" kmph
And the usual time be "t" hrs
Given that if the bus is running at 9s/10 kmph the time is 22 hrs
=> [9s/10] x 22 = t x s
=> 99/5 = t
=> t = 19.8 hrs

Hence, if bus runs at its own speed, the time saved = 22 - 19.8 = 2.2 hrs.

Manju borrows Rs. 5000 for 2 years at 4% p.a. simple interest 

She also lends it at 6 1⁄4% p.a for 2 years 

=> Total Gain = 6 1/4% − 4% = 2 1/4%

So her gain in the transaction for 1 year

= The simple interest she gets for Rs.5000 for 1 year at 2 1⁄4% per annum

= PTR100=5000×94×1100 = Rs. 112.5/ year.

Difference in amount = 514 - 415 = 99

99 = (415 x 4 x R)/100

R = 5.96%

Given L1 = 140 m

L2 = 160 m

S1 = 60 km/hr

S2 = 80 km/hr

From the question we get,

S1 + S2 = (L1 + L2) / T

=> (60 + 80) 5/18 m/s = 140 + 160/T

=> T = 54/7 = 7.71 sec