Quantitative Aptitude - Arithmetic Ability Questions

I = (P x r x t)/100

I = $360

A = P+I

A = 2360

I. gives, Rate = 5% p.a.

II. gives, S.I. for 1 year = Rs. 600.

III. gives, sum = 10 x (S.I. for 2 years).

Now I, and II give the sum.

For this sum, C.I. and hence amount can be obtained.

Thus, III is redundant.

Again, II gives S.I. for 2 years = Rs. (600 x 2) = Rs. 1200.

Now, from III, Sum = Rs. (10 x 1200) = Rs . 12000.

Thus, Rate =

100 x 1200

= 5% p.a.

2 x 12000

Thus, C.I. for 2 years and therefore, amount can be obtained.

Thus, I is redundant.

Hence, I or III redundant.

64/12 = 16/3 in simplest form.

Let the principal amount be Rs. x
Calculating SI:
SI for 1 year at 12% rate is Rs (12/100)*x. For 3 years it will be Rs (36/100)*x ------ (1)
Calculating CI:
For 1st year, Interest = (12/100)*x
For 2nd year, interest = (12/100)*x + (12/100)*x + (144/10000)*x
For 3rd year, interest = (12/100)*x + (12/100)*x + (12/100)*x + (144/10000)*x + (144/10000)*x + (1728/1000000)*x. ------ (2)
Subtracting equation 2 and 1 and solving further,
312*144*x/1000000 = 112.32
X = Rs 2500.
Alternative way: Using formula for Difference
between CI and SI for 3 years

we can also use the above formula to find the value of P.

Let initially milk and water in container B is 3x liter and x liter respectively

Now, 3x + (8/9) × 18 – x – (1/9) × 18 = 30

3x + 16 – x – 2 = 30

x = 8

Initial quantity is container B = 8 (3 + 1) = 32 Liter.

We know Compound Interest = C.I. = P1+r100t - 1

Here P = 2680, r = 8 and t = 2

C.I. = 26801 + 81002-1= 268027252-12= 26802725+12725-1= 2680 5225×225

= (2680 x 52 x 2)/625

= 445.95

Compound Interest = Rs. 445.95

Given speed of the bike after servicing = 55 kmph

Time taken for travelling some distance at 55 kmph = 5 hrs

Then, 

Distance = Speed x Time = 55 x 5 = 275 kms.

Now, 

Speed of the bike before servicing = 35 kmph

Distance = 275 kms

Now, Time = Distance/Speed = 275/35 = 7.85 hrs =~ 8 hrs.