A, B, C are partners in a business. B’s capital is ¹⁄₆ of the total and A’s capital is equal to that of Band C together. How much does C receives out of a total profit of $2400?

Solution

Total capital is $6

A = B + C

B is ¹⁄₆ of total = 1

∴ C = 2 ∴ Profit share of C = ²⁄₆ = ¹⁄₂ = $800

The price of each article at Paul’s store is 40% greater than the price for the same article at Linda’s store. At Paul’s store, article x costs thrice as much as article y. Article x at Paul’s store costs how many times as much as article y at Linda’s store?

Solution

Method:

Let article x and y costs $300 and $100 respectively at Linda's store.

∴ The same articles cost $420 and $140 respectively at Paul's store.

∴ $\frac{420}{100}=4.2$

∴ Article x at Paul's store costs 4.2 times that of article y at Linda's store.

Method 2:

Let product x at Linda's store be 30 (any multiple of 3 ending in 0)

Harry has twice as much as Nancy. They play together and at the end of first game, Nancy wins one third of Harry’s money from Harry. What fraction of the sum that Nancy now has, must Harry win back in the second game so that they may have exactly equal money?

Solution

Let Harry have $6 and Nancy $3.

At the end of first game, Nancy has $(3 + 2) = $5 and Harry has $(6 - 2) = $4.

∴ To have equal amounts, Harry must win $0.5 from Nancy in the second game.

∴ Required fraction = $\frac{0.5}{5}=\frac{1}{10}$

Two friends undertake to drive a distance of 60 km. The first performs the journey at 6 kmph. The second starting 37 minutes later arrives 23 minutes sooner. Find the ratio of their speeds.

Solution

Time = $\frac{Distance}{Speed}$

Time required by the first person to travel 64 kms.

= $\frac{60}{6}$ = 10 hours. The second person requires 37 + 23 = 60 min. less to travel the same distance.

∴ Speed of 2^nd person = $\frac{60}{6}$ = 6.6 kmph.

∴ Ratio is 6:6.66 i.e. 60:66 = 10:11

$\sqrt{p\sqrt[3]{p}}-\sqrt[3]{p}=\sqrt{p}$ Find ‘P’ if P is a natural number.

Solution

P has to be a cube. ∴ Only possible option is 64

A hare pursued by a greyhound, is 50 of her own leaps ahead of him. While the hare takes 4 leaps, hound takes 3 leaps. In one leap, the hare goes 2 m and the greyhound 4 m. In how many leaps will the greyhound overtake the hare?

Solution

Distance covered by hare in 1 leap = 2 m

∴ Distance covered in 4 leaps = 4 × 2 = 8 m

Similarly, in the same time distance covered by hound in three leaps = 3 x 4 = 12 m

∴ In three leaps, the hound gains 12 - 8 = 4m over the hare.

The hound actually has to gain 50 leaps of hare to overtake it i.e. a distance of 50 × 2 = 100m .

∴ 4 m gained in 3 leaps.

∴ 100 m would be gained in 75 leaps of hound.

A sum of money is invested at compound interest payable annually. Interest in 5th and 6th year was $500 and $525. Find rate % per annum.

Solution

Compound interest is the interest calculated on interest. Interest in two successive years were $500 and $525 respectively.

∴ In other words, we can say that interest on $500 for 1 year = $25.

∴ Principal=$500, SJ. =$25, Number of years=1.

∴ 25 = $\frac{500\times 1\times R}{100}$ ∴ Rate = 5%

A man sells a pen at a profit of 20%. If he had bought it at 20% less and sold it for $18 less, he would have gained 25%. Find the cost price of the pen.

Solution

Let the cost price of pen = $100

∴ Selling price (SI) at 20% profit = $120

Cost price at 20% less = $80

Selling price (S₂)at 25% profit = 80 x 1.25 = $100

∴ S₁ - S₂= $20 when cost price is $100.

∴ When S₁ - S₂=$18,cost price =$\frac{18\times 100}{15}$=$90

Two clocks show the same time at 4 p.m. The first clock loses 10 minutes every 2 hours and second gains 10 minutes every hour. After how many hours will they both show the same time again?

Solution

The first clock loses 5 minutes every hour and the second clock gains 10 minutes every hour.

∴ In one hour, first clock would be behind the second dock by 15 minutes. They would both show the same time again if they are separated by 12 hours = 12 x 60 minutes.

Number of hours the first dock takes to be behind the second by 12 hrs. = $\frac{12\times 60}{15}$ = 48 hrs.

Mike, a farmer, divided 437 oranges and 342 peaches into largest possible separate baskets such that each basket has same number of fruits and only 1 type of fruit. Find the total number of baskets.

Solution

We need to find the maximum number of fruits in a basket. In other words, we need to find the HCF of 437 and 342 which is 19.

437 = 19 × 23 and 342 = 19 × 18

This implies that there are 23 baskets of oranges each consisting of 19 oranges and 18 baskets of peaches each consisting of 19 peaches

∴ Total number of baskets are 18 + 23 = 41

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