Quantitative Aptitude - Arithmetic Ability Questions

Step (i): Let x be the number of boys and y be the number of girls.
Given total number of boys and girls = 100
x+y=100 -------------- (i)

Step (ii): A boy gets Rs. 3.60 and a girl gets Rs. 2.40
The amount given to 100 boys and girls = Rs. 312
3.60x + 2.40y = 312 -------------- (ii)

Step (iii):
Solving (i) and (ii)
3.60x + 3.60y = 360 --------- Multiply (i) by 3.60
3.60x + 2.40y = 312 --------- (ii)
1.20y = 48
y = 48 / 1.20
= 40

 Number of girls = 40

Let B's present age = x years. Then, A's present age = (x + 9) years.
(x + 9) + 10 = 2(x - 10)

 => x + 19 = 2x - 20

 => x =39.

Total distance travelled in 12 hours    =(35+37+39+.....upto 12 terms)
This is an A.P with first term, a=35, number of terms,
n= 12,d=2.
Required distance    = 12/2[2 x 35+{12-1) x 2]
=6(70+23)
= 552 kms.

Let the quantity of alcohol and water be 4x litres and 3x litres respectively

4x/(3x+5) = 4/5 

20x = 4(3x+5)

8x = 20

x = 2.5

Quantity of alcohol = (4 x 2.5) litres = 10 litres.

By the rule of alligation:

Profit of first part                         Profit of second part

So, ratio of 1st and 2nd parts = 4 : 6 = 2 : 3. 

Therefore, Quantity of 2nd kind = (3/5 x 1000)kg = 600 kg.

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

4 o'clock, the hands of the watch are 20 min. spaces apart.
To be in opposite directions, they must be 30 min. spaces apart.
Minute hand will have to gain 50 min. spaces.
55 min. spaces are gained in 60 min 

50 min. spaces are gained in 6055×50 min. or 54611 

Required time = 54611 min. past 4.

1 man's 1 day work =1/96 ; 1 woman's 1 day work = 1/192

Work done in 6 days=6896+8192=6×18 =34 

Remaining work = 1/4

(8 men +8 women)'s 1 day work = 1896+8192 =1/8

Remaining work =1/4 -  1/8 = 1/8

 1/96 work is done in 1 day by 1 man

Therefore, 1/8 work will be done in 1 day by 96 x (1/8) =12 men