Quantitative Aptitude - Arithmetic Ability Questions

Interest received by A from B = 10% of half of Rs.50000 = 10% of Rs. 25000 = Rs.2500.

Amount received by A per annum for being a working partner = 1500 x 12 = Rs.18000

Let 'P' be the part of the remaining profit that A receives as his share.

So,total income of A = (Rs.2500 + Rs.18000 + Rs. P )

Total income of B = only his share from the remaining profit = 'P', as  A and B share the remaining profit equally.

We know that income of A = Twice the income of B 

So, (2500 + 18000 + P ) = 2(P)

P = 20500

Thus, the total profit = 2P + Rs.18000

= 2(20500) + 18000 = Rs.59000.

(2 x 14) men +(7 x 14) boys = (3 x 11) men + (8 x 11) boys

 =>5 men= 10 boys    => 1man= 2 boys

 Therefore, (2 men+ 7 boys) = (2 x 2 +7) boys = 11 boys

( 8 men + 6 boys) = (8 x 2 +6) boys = 22 boys.

 Let the required  number of days be x.

 More boys , Less days     (Indirect proportion)

 More work , More days    (Direct proportion)

 Boys22:11Work1 : 3⋮⋮ 14:x

 Therefore, (22 * 1 * x) = (11 * 3 * 14)

=> x = 21 

Hence, the required number of days = 21

let ABC be the isosceles triangle, the AD be the altitude 

Let AB = AC = x then BC= 32-2x       [because parameter = 2 (side) + Base] 

since in an isoceles triange the altitude bisects the base so 

BD = DC = 16-x 

In a triangle ADC, AC2=AD2+DC2 

x2=82+16-x2 ⇒x=10 

BC = 32-2x = 32-20 = 12 cm 

Hence, required area = 12*BC*AD= 12*12*10 = 60 sq cm

987 = 3 * 7 * 47.
So, the required number must be divisible by each one of 3, 7, 47
553681 => (Sum of digits = 28, not divisible by 3)
555181 => (Sum of digits = 25, not divisible by 3)
555681 is divisible by each one of 3, 7, 47.

Cost Price of 2 dozen oranges Rs. (5 + 4) = Rs. 9.
Sell price of 2 dozen oranges = Rs. 11.
If profit is Rs 2, oranges bought = 2 dozen.
If profit is Rs. 50, oranges bought = (2/2) * 50 dozens = 50 dozens.

Angle traced by hour hand in214 21 hrs = 36012×2140 =157102 

Angle traced by min. hand in 15 min = 36012×150 = 900 

 Required Angle = 157120-900 = 67102

let the sum lent at 5% be Rs.x and that lent at 8% be Rs.(1550-x). then, 

Interest on x at 5% for 3 years + interest on (1550-x) at 8% for 3 years = 300

 x*5*3100+1500-x*8*3100=300 

 x=800

Required ratio =  x : (1550-x) = 800 : (1550-800) = 800 : 750 = 16 : 15

Milk = 3/5 x 20 = 12 liters, water = 8 liters

If 10 liters of mixture are removed, amount of milk removed = 6 liters and amount of water removed = 4 liters.

Remaining milk = 12 - 6 = 6 liters

Remaining water = 8 - 4 = 4 liters

10 liters of pure milk are added, therefore total milk = (6 + 10) = 16 liters.

The ratio of milk and water in the new mixture = 16:4 = 4:1

If the process is repeated one more time and 10 liters of the mixture are removed,
then amount of milk removed = 4/5 x 10 = 8 liters.

Amount of water removed = 2 liters.

Remaining milk = (16 - 8) = 8 liters.

Remaining water = (4 -2) = 2 liters.

Now 10 lts milk is added => total milk = 18 lts

The required ratio of milk and water in the final mixture obtained

= (8 + 10):2 = 18:2 = 9:1.