A and B started a new company. The investment of A was thrice to that of B and the period of investment of A is twice to that of B. B got Rs. 4000 as share in annual profit.The total profit at the end of the year. was

Solution
Ratio of the equivalent capitals of A and B for 1 month
= 3x × 2t : x × t = 6 : 1
B's share = Rs. 4000
⇒^{1}⁄_{7}× Total profit = Rs. 4000
⇒ Total profit = (7× 4000)
= Rs. 28000
In a bag the 25 paise. 10 paise and 5 paise coins are in the ratio of 1 : 2 : 3. If the value of coins be Rs. 30. the number of 5 paise coins will be

Solution
Ratio of the numbers of 25 paise, 10 paise and 5 paise coins
= 1 : 2: 3
Ratio of the value = ^{1}⁄_{4}: ^{2}⁄_{10}: ^{3}⁄_{20}
= 5: 4: 3
∴ Value of the 5 paise coins
= ^{3}⁄_{12}×30 = Rs. 7.5 12
∴ Number of 5 paise coins
= 7.5 × 20 = 150
In a 20 litre mixture. the ratio of milk and water is 5 : 3. When 4 litres of mixture is substituted by 4 litres of milk. the ratio of milk and water in the new mixture will be

Solution
Quantity of milk in 16 litres of mixture = ^{5}⁄_{8} × 16 = 10 litres
Quantity of water = 6 litres
On adding 4 litres of milk,
Required ratio = 14: 6 = 7 : 3
A sum of money deposited at compound interest becomes Rs. 6690 in 3 years and Rs. 10035 in 6 years. The amount is

Solution
Let the principal be Rs. × and rate of interest
= r%per annum.
According to the question,
\(6690 = x\left ( 1+\frac{r}{100} \right )^{3}......(i)\)\(10035= x\left ( 1+\frac{r}{100} \right )^{6}.......(ii)\)
Dividing equation (ii) by (i). we have
\(\left ( 1+\frac{r}{100} \right )^{3}=\frac{10035}{6690}\)
∴ From equation (i).\(6690=x\times \frac{10035}{6690}\)
\(\Rightarrow x=\frac{6690 \times 6690}{10035}= Rs. 4460\)
A metallic cuboid of dimensions 27cm × 8cm × 1 cm is melted to form a cube. What is the difference in surface areas of two solids?

Solution
Volume of the cuboid
= 27 × 8 × 1 = 216 cm^{3}
∴ Volume of the cube = 216 cm^{3}
∴ Edge of the cube =\(\sqrt[3]{216}\)
=6cm
Total surface area of the cuboid
= 2 × (27 × 8 + 8 × 1 + 27 × 1)
= 2 (216 + 8 + 27)
= (2 × 251) cm^{2} = 502 cm^{2}
Total surface area of the cube
= 6 × 6 × 6 = 216 cm
Required difference
= (502  216) = 286 cm^{2}
One side of a parallelogram is 14 cm long.Its perpendicular distance from the opposite side is 16 cm. Area of the prallelogram is

Solution
Area of parallelogram= base × height = 14 × 16 = 224 cm^{2}
The ratio of Sonu’s age to Vijay’s age is 4 : 5 and that of Sonu’s age to Subodh’s age is 5 : 6. If the total of their ages is 69 years. how old is Vijay?

Solution
Sonu : Vijay = 4 : 5 = 20 : 25
Sonu : Subodh = 5 : 6 = 20 : 24
∴ Sonu :Vijay : Subodh
= 20: 25: 24
Sum of ratios = 20 + 25 + 24= 69
Clearly, Vijay's age = 25 years
A garrison of 200 men has provision sufficient for 24 weeks. At the end of the first week a reinforcement of 80 men arrives and the ration per day of each man is then reduced from 900 grams to 750 grams. For how many days longer can they hold out?

Solution
There is provision for 23 weeks for 200 men at the rate of 900 gm/man /per day.
We are to find, how long will it last for 280 men when each man consumes 750 gm/day.
More men, less days, Less quantity, more days
\(\Rightarrow \frac{x}{23}=\frac{200}{280}\times \frac{900}{750}\)\(\Rightarrow x=\frac{200\times 900\times 23}{280 \times 750}weeks\)
= 138 days
If 10 men or 18 boys can do a piece of work in 15 days. then 25 men and 15 boys together will do twice the work in

Solution
10 men = 18 children
⇒ 1 man = ^{18/sup>⁄10 children 25 man +15 children \(=\left ( 25\times \frac{18}{10}+15 \right )children\) = 60 children Now, more work, more days, more children, less days ∴ 1 × 60 × x = 2 × 18 × 15 \(\Rightarrow x=\frac{2\times 18\times 15}{60}=9\: days\)}