Quantitative Aptitude - Arithmetic Ability Questions

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.

Let the rquired number be 'P'

According to the question,

$$\begin{gathered} 100 x P100 = 20 \\ => P = 20 x 100100 = 20 \end{gathered}$$

Hence, 100 percent of the number 20 is 20.

Now, you can also try for

20 percent of what number is 100

$$\begin{gathered} 20 x P100 = 100 \\ => P = 100 x 10020 = 500 \\ => P = 500 \end{gathered}$$

Hence, 20 percent of the number 500 is 100.

17x12x19.5x12x 2660 = 2660266 = 10

 Investment ratio of A, B & C = (4000 ×6+3000×9):(5500 ×6+4000×9):4

Let the amount invested by saketh = RS. p

Now, that of sandeep = 20,000 x 6

saketh = 12 x p

Ratio of their earnings = 120000 : 12p = 6000 : (9000 - 6000)

=> $$\begin{gathered} 12000012p = 60003000 \\ => p= Rs. 5000 \end{gathered}$$

Hence, the amount investe by saketh  = Rs. p  = Rs. 5000.

This can be solved as

$$\begin{gathered} 20 1 - 4202 = 20 1 - 152 \\ = 20452 \\ = 20 1625 \\ = 645 \\ = 12.8 \end{gathered}$$

We have: l = 20 ft and lb = 680 sq. ft.

So, b = 34 ft.

Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft.

 $$\begin{gathered} = log 33 + log 23+ log 103log10×3×22 \\ \end{gathered}$$ 

=log33 12+log 23+log 10312log(10×3×22)  

=12log 33+3 log 2+12 log103log10+log3+log22  

=32log 3 + 2 log 2 + log 10log 3 + 2 log 2 + log 10 = 32