Quantitative Aptitude - Arithmetic Ability Questions

Remaining work = 1 - 9/10 = 1/10

=> A & B together completes 1/10 of work in 4 days

=> 1 work can completed in ------ ? days

Let it be x days

=>  4110=x1

=> x = 40 days.

Hence, A & B together can complete the work in 40 days.

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

64/12 = 16/3 in simplest form.

(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

The given equation is in Mixed fraction.

Now to solve this type of fractions,

$$\begin{gathered} 413 + 215 + 414 =? \\ ? = 4 + 2 + 4 + 13 + 15 + 14 \\ ? = 10 + 4760 \\ ? = 104760 \\ \end{gathered}$$

From the given data,

20% of K = 32% of N

V = 4/5 N

Given Vipin's salary = 3.84 lpa = 3.84/12 = 32,000 per month

Now, N = 5/4 x 32000 = 40000

Now K = 32/20 x 40000 = 64000

Hence, karthik's monthly salary = Rs. 64000.

PW = 100×Amount100+R×T

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.