Quant Boosters  Vikas Saini  Set 1

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

Let total distance travelled by cart = D.
D/40 – D/48 = 5.
D = 1200 ft.

Q28) A and B go by Bus from X to Z which is on the way to Y. C goes from Y to Z by Auto. A and B's Bus goes at 75km/hr and C's auto moves at 15 km/hr. All the three start at 6:00 am.B and A go ahead, meet C and pick him up. They return immediately to Z at the same time. The distance between X and Y is 600 km and XZ is 400 km.
What is the total distance travelled by A ?
What is time of entire journey ?

Speed of A & B = 75 kmph.
Speed of C = 15 kmph.
Relative speed = 75+15 = 90kmph.
Time they will meet = 600/90 = 6.66 = 6 hr 40 min.
But after meeting with C, they pick up him and return to Z.
At meeting point distance travelled by A & B = 75 x 6.66 = 500 km.
Now Z is 100 km farther than this.
Total distance travelled by A = 500+100 = 600 km. (Answer)
Time of entire journey = 6.66 + 100/75 = 8 hr (Answer)

Q29) When an object is dropped, the number of feet N that it falls is given by the formula N=(1/2)gt^2 where t is the time in seconds from the time it was dropped and g is 32.2. If it takes 5 seconds for the object to reach the ground, how many feet does it fall during the last 2 seconds?

In 5 seconds travelled by object = (1/2) x 32.2 x (5)^2.
In first 3 seconds it travelled = (1/2) x 32.2 x (3)^2
In last 2 seconds travelled = (1/2) x 32.2 x [5^2 – 3^2]
= 16.1 x 16
= 257.6

Q30) Two men are walking towards each other alongside a railway track. A freight train overtakes one of them in 20 seconds and exactly 10 minutes later meets the other man coming from the opposite direction. The train passes this man is 18 seconds. Assume the velocities are constant throughout. How long after the train has passed the second man will the two men meet?

Let assume length of the train is ‘L’ and speed of train = a.
Speed of man 1 = b.
Speed of man 2 = c.
20 = L / a – b.
18 = L / a + c.
18a + 18c = 20a – 20b.
a = 10b + 9c.
Distance of two men = 600 x (a + c).
Time = 600(a+c) – 600(b+c) / (b+c)
= 600 ( a – b) / (b + c)
= 600 (10b + 10c – b) / (b + c)
= 5400 seconds
= 90 min (ans)