Quantitative Aptitude - Arithmetic Ability Questions

(4a+7b)(4c-7d) = (4a-7b)(4c+7d)

(4a+7b)/(4a-7b) = (4c+7d)/(4c-7d)

Using componendo and dividendo

(4a+7b)+(4a-7b) / (4a+7b)-(4a-7b)

= (4c+7d)+(4c-7d) / (4c+7d)-(4c-7d)

Or 8a/14b = 8c/14d

Or a/b = c/d

I = prt = [15000 × 0.07 × (214/365) ]=615.52

Future value, S = P + I = $15 000 + $615.52 = $15 615.52

Let the greater and smaller number be p and q respectively.

4q = p + 5 ------ (I)

p = 3q+2 ------- (II)

From equation (I) and (II)

q = 7

p = 23

Let the total number of documents to be printed be 12.

 The number of documents printed by P in 1 day = 4.

 The number of documents printed by Q in 1 day = 3.

 The number of documents printed by R in 1 day = 2.

 
Thus, the total number of documents that can be printed by all the machines working simultaneously in a single day = 9.

Therefore, the number of days taken to complete the whole work = 12/9 = 4/3 days.

Let S be the sample space and E be the event of selecting 1 girl and 2 boys.

Then, n(S) = Number ways of selecting 3 students out of 25

                = 25C3  = 2300.

         n(E)= 10C1*15C2 = 1050. 

P(E) = n(E)/n(s) = 1050/2300 = 21/46

Let the breadth of the rectangle = b mts

Then Length of the rectangle = b + 6 mts

Given perimeter = 84 mts

2(L + B) = 84 mts

2(b+6 + b) = 84

2(2b + 6) = 84

4b + 12 = 84

4b = 84 - 12

4b = 72

b = 18 mts

=> Length = b + 6 = 18 + 6 = 24 mts

Now, required Area of the rectangle = L x B = 24 x 18 = 432 sq. mts

Let the number of students = 100

Number of students who failed in one or more subjects = 14 + 22 + 18 = 54%

The percentage of students who passed in both the subjects = 100 - 54 = 46%.

We have P = 15 000, r = 0.07 and since the actual date the loan was taken out
is not given, we use t =7/12

I = Prt=$15 000* 0.07* 7/12 = $612.50

Amount repaid = Future or accumulated value,

S = P + I = $15 000 + $612.50 = $15 612.50