Quantitative Aptitude - Arithmetic Ability Questions

(A+B+C) do 1 work in 10 days.

So (A+B+C)'s 1 day work=1/10 and as they work together for 4 days so workdone by them in 4 days=4/10=2/5

Remaining work=1-2/5=3/5

(B+C) take 10 more days to complete 3/5 work. So( B+C)'s 1 day work=3/50

Now A'S 1 day work=(A+B+C)'s 1 day work - (B+C)'s 1 day work=1/10-3/50=1/25

A does 1/25 work in in 1 day

Therefore 1 work in 25 days.

We know that 

$$\begin{gathered} 40% = 40100 = 25 \\ ⇒25 x 244 x ? = 48 \\ ? = 48 x 5 x 42 x 24 = 20 \end{gathered}$$

Originally, let the number of seats for Mathematics, Physics and Biology be 5x, 7x and 8x respectively.
Number of increased seats are (140% of 5x), (150% of 7x) and (175% of 8x).
⇒ [(140/100) × 5x],[(150/100) × 7x] and [(175/100) × 8x]
⇒ 7x, 21x/2 and 14x.

⇒ The required ratio =7x : 21x/2 : 14x
⇒ 14x : 21x : 28x
⇒ 2 : 3 : 4

We know that,

Odd days --> days more than complete weeks

Number of odd days in 400/800/1200/1600/2000 years are 0.

Hence, the number of odd days in first 1600 years are 0.

Number of odd days in 300 years = 1

Number of odd days in 49 years = (12 x 2 + 37 x 1) = 61 days = 5 odd days

Total number of odd days in 1949 years = 1 + 5 = 6 odd days

Now look at the year 1950

Jan 26 = 26 days = 3 weeks + 5 days = 5 odd days

Total number of odd days = 6 + 5 = 11 => 4 odd days

Odd days :-

0 = sunday ;

1 = monday ;

2 = tuesday ;

3 = wednesday ;

4 = thursday ;

5 = friday ;

6 = saturday

Therefore, Jan 26th 1950 was Thursday.

Given a x a = 12

asked to find a x a x a x a =  (a x a) x (a x a)

a⁴ = a² x a² 

= 12 x 12

= 144.

More :: Mathematical operations questions and answers.

Let C.P. be Rs. x.

Then,= >1920-xx*100=x-1280x*100

=>  1920 - x = x - 1280

=>  2x = 3200

=>  x = 1600

 Required S.P. = 125% of Rs. 1600 =Rs(125/100*1600) = Rs2000