Quantitative Aptitude - Arithmetic Ability Questions

Let
P's capital = p,
Q's capital = q and
R's capital = r.
Then

6p = 8q = 12r
=> 3p = 4q = 6r
=>q = 3p/4
r = 3p/6 = p/2
P : Q : R = p : 3p/4 : p/2
= 4 : 3 : 2
R's share = 4650 * (2/9) = 150*6 = Rs. 1033.33.

276+35961+50 = 10111011 = 1

Cash realised= Rs. (106.25 - 0.25)
                      = Rs. 106.

Number of questions = 5
Possibilities of choices for each question 1 to 5 respectively = 4, 4, 4, 6, 6

Reuired total number of sequences

= 4 x 4 x 4 x 6 x 6

= 2304.

= log tan10+log tan890 + log tan20+ log tan880+⋯⋯+log tan450  

= log [tan10 × tan890] + log [tan20 × tan880 ] +⋯⋯+log1  

∵ tan(90-θ)=cotθ and tan 450=1  

= log 1 + log 1 +.....+log 1 

= 0.

we can select the 5 member team out of the 8 in 8C5 ways = 56 ways.

The captain can be selected from amongst the remaining 3 players in 3 ways.

Therefore, total ways the selection of 5 players and a captain can be made = 56x3 = 168 ways.

(or)

Alternatively, A team of 6 members has to be selected from the 8 players. This can be done in 8C6 or 28 ways.

Now, the captain can be selected from these 6 players in 6 ways.

Therefore, total ways the selection can be made is 28x6 = 168 ways.

The toys are different; The boxes are identical
If none of the boxes is to remain empty, then we can pack the toys in one of the following ways
a. 2,2,1
b. 3,1,1

Case a. Number of ways of achieving the first option 2−2–1

Two toys out of the 5 can be selected in5C2 5 ways. Another 2 out of the remaining 3 can be selected in3C2 ways and the last toy can be selected in 1C1  way.

However, as the boxes are identical, the two different ways of selecting which box holds the first two toys and which one holds the second set of two toys will look the same. Hence, we need to divide the result by 2.

Therefore, total number of ways of achieving the 2−2–1 option is:

5C2*3C22=10*32=15 ways.

Case b. Number of ways of achieving the second option 3−1–1

Three toys out of the 5 can be selected in 5C3  ways. As the boxes are identical, the remaining two toys can go into the two identical looking boxes in only one way.
Therefore, total number of ways of getting the 3−1–1 option is5C3=10 ways.

Total ways in which the 5 toys can be packed in 3 identical boxes
=number of ways of achieving Case a + number of ways of achieving Case b
=15 + 10 = 25 ways

Cash price = $21 000

Deposit = 10% × $21 000 = $2100 

Loan amount = $21000 − $2100 = $18900

I=p x r x t/100 

I=11340

Total amount = 18900 + 11340 = $30240

Regular payment = total amount /number of payments