Quant Boosters  Vikas Saini  Set 1

Suppose distance between X and Y is D km.
A = D+11, B = D11.
B = D+9, C = D9.
A/C = 3:2.
(AB) / (BC) = 3/2.
(D+11) (D+9) / (D11) (D9) = 3 / 2.
D^2 + 20D + 99 / D^2 – 20D + 99 = 3 / 2
2(D^2 + 20D + 99) = 3(D^2 – 20 D + 99)
D^2 – 100 D + 99 = 0.
D = 99.

Q18) A shopkeeper uses a weight of of 460 gm instead of 500 gm and sells the article at the cost price. What is profit percentage ?
a. 40%
b. 23%
c. 8 +16/23 %
d. 20%

Short approach :
Profit percentage = [ (500 – 460)/460] x 100
= (40/460) x 100

Q19) A 30% solution of alcohol is mixed with a 50% solution of alcohol to form a 10 litre solution of 45% alcohol. How much of 30% solution was used.
(a) 2 ltr
(b) 2.5 ltr
(c) 2.7 ltr
(d) 3.2 ltr
(e) 3.7 ltr

By allegation
30  50
 45 
(5045 ) : (4530)
5 : 15 = 1 : 3.
Ratio = 1 : 3
30% solution was added = 10 x 1 / (1 + 3) = 2.5 ltr.

Q20) The percentage volumes of milk in three solutions A,B and C form a geometric progression in that order. If we mix the first, second and third solutions in the ratio 2:3:4, by volume we obtain solution containing 32% milk. If we mix them in the ratio 3:2:1, by volume, we obtain a solution containing 22% milk. What is the percentage of milk in A?
a) 6%
b) 12%
c) 18%
d) 24%

Suppose volumes of milk percentage in A,B,C are a, ar, ar^2 respectively.
2a + 3ar + 4ar^2 = 32 x 9.
3a + 2ar + ar^2 = 22 X 6.
a= 12, r = 2.
The percentage of milk in A = 12%.

Q21) One fourth of a solution that was 10% sugar by weight was replaced by a second solution resulting in a solution that was 16% sugar by weight. The second solution was what percent sugar by weight ?

10  X
 16 
X16 : 16  10
1 : 1/3
X – 16 / 6 = 3 / 1
X = 34.

Q22) From a solution that has milk and water in the ratio 5 : 3, ‘x’ percent is removed and replaced with water. The concentration of milk in the resulting solution lies between 30% and 50%. Which of the following best describes the value of ‘x’?
a) 25 < x < 30
b) 20 < x < 52
c) 20 < x < 48
d) 25 < x < 60

Suppose total unit is 8. Milk = 5, water = 3.
X% of 8 is removed = 0.08x.
Where milk part is 0.05x and water part is 0.03x.
0.30 < 5 – 0.05x / 8 < 0.50
2.40 < 5 – 0.05 x < 4.0
2.60 < 0.05x < 1.
1 < 0.05x < 2.6
20 < x < 52.
Option B

Q23) The concentration of milk in 60L milk and water solution is 30%. If 6L of the solution is replaced by water then 5L solution is again replaced by water, then find the concentration of milk in the final solution.
a) 24.50 %
b) 25.00%
c) 24.75 %
d) 25.50%

The concentration in final solution = (3/10) ( 1  6/60) (1 – 5/60)
= 24.75%.

Q24) Rice of two different qualities are mixed and the mixture is sold at rs. X per kg, giving 25% profit. If higher quality rice is sold at rs X per kg, then there will be a loss of 100/11 %. If the ratio of the quantities of the lower quality rice and the higher quality rice in the mixture is 8 : 3, then what is the percentage profit when lower quality rice is sold at X per/kg ?

Cost price of low quality rice per kg is l and cost price of high quality rice per kg is h.
(8l + 3h / 11)(5/4) = h (10/11)
8l = 5h.
Cost price of lower quality rice is 5h/8.
Selling price of mixture = 10h/11.
Profit percentage = (10h/11 – 5h/8) x 100 / (5h/8) = 500/11 %.

Q25) Two containers X and Y, equal quantities of water and acid. The concentration of acid is same in both the containers X and Y. If 4 litres of solution from X is replaced by acid and the concentration of acid becomes twice what it was initially. If 8 litres of solution is from Y is replaced with pure acid then what is the ratio of final concentration of acid in solution Y to the initial concentration of acid in solution Y ?

Suppose quantity is q, concentration of acid = a, suppose quantity of water.
In container X, 4 litre acid is replaced then concentration becomes twice.
In container Y, 8 litre acid is when replaced then concentration becomes two times of twice.
Hence increases by thrice.
Ratio = 3 : 1.

Q26) A and B are two perfumes which have oil and essence in the ratio of 11:3 and 2:5 respectively.
New perfume is made by mixing A and B in the ratio of a:b such that the percentage of oil in the new perfume is 50%. Find a + b.

Oil in A = 11 / 3+11 = 11/14.
Oil in B = 2 / 2+5 = 2/7.
(11a/14) + (2b/7) = (a+b)/2
11a + 4b = 7a + 7b
a/b = 3/4
a + b = 7

Q27) The circumference of the front wheel of a cart is 40 ft long and that of the back wheel is 48 ft long. What is the distance travelled by the cart, when the front wheel has done five more revolutions than the rear wheel?
a) 950 ft
b) 1450 ft
c) 1200 ft
d) 800 ft