1. Find the ratio of milk and water in the mixture obtained when one vessel containing milk and water in ratio of 3:1 is mixed with another vessel containing milk and water in the ratio 5:2. Both the vessel are of equal capacity.

Ans:2
Let vessels be of unit capacity.
Milk in first vessel = 3/4
Water in first vessel = 1/4
Milk in second vessel = 5/7
Water in second vessel = 2/7
Required ratio = (3/4 + 5/7)/(1/4 + 2/7)
= 41/15

2. Some small cubes are placed inside a large box which is in the shape of cuboid having dimension 420 cm * 300 cm * 240 cm. The length of each small cube is 6 cm. Find the maximum number of cubes that can be placed inside the box?

Ans:3
Length of each cube = 6 cm
Length of box = 420 cm, width = 300 cm and height = 240 cm
Number of cubes that can be kept inside the box = (420 *300*240)/(6*6*6)=70*50*40 = 140000

3. Ramesh distributed 10% of his income to his daughter, 15% to wife and 12% to his son. After this distribution, he contributes 5% from the remaining to Cancer institute and left with Rs 47880. Find the total income of Ramesh?

Ans:2
Ramesh distributes 37% of his salary to his family.
From remaining 63%, he contributes 5% to Cancer Institute i.e. 3.15%
So, Ramesh is left with 59.85% of his salary which is equal to Rs 47880
Therefore, Salary of Ramesh = 47880 * 100/59.85 = Rs 80,000

4. A sum of prize money is to be distributed among four contestants in the ratio of 20:15:12:8 according to their rank in the competition. If winner got Rs.40000 more than the 3rd rank holder then find the average of prize money of all contestants.

Ans:2
Let share of each = 20x, 15x, 12x and 8x.
Amount that winner got more than 3rd rank holder = 20x – 12x = 8x = 40000 (Given)
=> x = 5000
Average of prize money of all contestant = (20x + 15x + 12x + 8x)/4 = 13.75x = 13.75 * 5000 = Rs.68750

5. A shopkeeper sells its item at 10% loss from cost price, but he uses false weight instead of 1 kg weight and gets an overall profit of 20%. By what false weight he replaced 1 kg weight?

Ans:3
Let CP = p
SP = 0.9p
Let he solds ‘y’ grams instead of ‘1000’ grams
Then CP = py
SP = 0.9p*1000 = 900p
(900p – py)/py = 0.2
900 – y = 0.2y
900 = 1.2y
y = 900/1.2 = 750 grams

6. The probability that it rains on any given day next week is 2/9. Find the probability that it rains only on Wednesday and Saturday.

Ans:3
Probability that it rains on any of the given day of next week = 2/9
Probability that it does not rain on any given day of next week = (1- 2/9) = 7/9
So, required probability = (7/9) (7/9) (2/9) (7/9) (7/9) (2/9) (7/9)
= (7^5 * 2^2)/(9^7)

7. If the difference of 75% of Y and 33(1/3)% of X is equal to the sum of 50% of X and 12.5% of Y, then fin the ratio of X and Y.

Ans:4
According to question-
(75% of Y – 33(1/3)% of X) = (50% of X + 12.5% of Y)
75Y/100 – 100X/300 = 50X/100 + 12.5Y/100
75Y – 100X/3 = 50X + 12.5Y
62.5Y = 250X/3
X:Y = 3:4

8. Rakhi deposited Rs 18000 in SBI and got Rs 24000 after 4 years when bank is offering simple interest. If interest rate is decreased by 1.33 Percentage points then in how much time Rs 12000 will become Rs 18000?

Ans:4
Interest = 24000-18000= Rs 6000
Original interest rate offered by the bank = (6000 X 100)/(18000 X 4)=8.33%
Decreased interest rate in this case = (8.33 – 1.33)% = 7%
Total % increase in the amount from Rs 12000 to Rs 18000 = 6000/12000 X 100% = 50%
Time taken by this = 50/7=7 1/7 years

9. In what ratio should the profit be divided among three partners after 14 months who invested initially in ratio 3: 4: 7 and the one with minimum investment withdraws 30% after every 4 months and the one with maximum investment withdraws 20% after 8 months.

Ans:1
Let the investments be 3a, 4a and 7aRatio of profit share after 14 months is given as,=> (3a*4 + 3a*0.7*4 + 3a*0.72*4 + 3a*0.73*2): (4a*14): (7a*8 + 7a*0.8*6)=> 28338: 56000: 89600

=> 14169: 28000: 44800

10. A swimmer can swim the distance downstream which is thrice the distance which he swims upstream in the same time. Calculate the speed of swimmer in still water if speed of current is 2 km/h?

Ans:5
Let distance upstream = x and speed of swimmer = y
Downstream distance = 3x
3x/(y+2) = x/(y-2) => 3y – 6 = y + 2
y = 4

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