Math Tricks For Competitive Exams

Sunday, May 27, 2018

Tricks Development

9:03 AM

Q. Radius of a circle is increased by 10 %, find the percentage increase in area.

Ans. You can get the answer by using the formula below:

Percentage increase in the area = $\small 2m + \frac{m^{2}}{100}$

Percentage increase in the area = $\small 2\times 10 + \frac{100}{100}$

Percentage increase in the area = 21

The answer is 21.

But how do I get the formula?

Let me explain below.

Area of a circle = A= $\pi r^{2}$

Suppose the radius is increased m%

New Area = $\pi \times \left ( r + \frac{rm}{100} \right )^{2}$

New Area =   $\pi \times \left ( \frac{100r+rm}{100} \right )^{2}$

New Area =   $\pi \times \left ( \frac{10000r^{2}+200r^{2}m+r^2{m^{2}}}{10000} \right )$

New Area =    $\frac{10000\pi r^{2}+200\pi r^{2}m+\pi r^{2}m^{2}}{10000}$

Increase in the area = New Area – Original Area

Increase in the area = $\frac{10000\pi r^{2}+200\pi r^{2}m+\pi r^{2}m^{2}}{10000}$ – $\pi r^{2}$

Increase in the area = $\frac{200\pi r^{2}m+\pi r^{2}m^{2}}{10000}$

Percentage increase in the area = Increase X 100
Orignal

Percentage increase in the area = $\frac{\frac{200\pi r^{2}m+\pi r^{2}m^{2}}{10000}}{\pi r^{2}} \times 100$

Percentage increase in the area = $2m + \frac{m^{2}}{100}$

So now you understand where it came from.

Friday, May 11, 2018

Profit and Loss

12:22 PM

Q. A man sales an article at 10% gain. Had he sold at for Rs. 60 more he could have gained 20%. What is the C.P of the article.

Ans. The Formula is as under.

C.P = more gain X 100 = 60 X 100 = 600 Rs
Diff in profit 10

Q. A merchant who marked his goods up by 50% subsequently offered a discount of 20% .What is the percentage profit that the merchant make after offering the discount.

Ans. The Formula is as under.

Profit Percent = $X + Y + \frac{XY}{100}$ = $50 - 20 - \frac{50\times 20}{100}$ = 20%

The negative values(-20) are because there is a discount.

Q. A shopkeeper marks his goods at 20% above the C.P and allow of 10% discount.
Find the profit percent.

Ans. The Formula is as under.

Profit Percent = $X + Y + \frac{XY}{100}$ = $20 - 10 - \frac{20\times 10}{100}$ = 8%

Positive answer shows Profit

Negative answer shows Loss

Q. If the cost of petrol increases by 40%.By what percent the person should reduce the consumption considering expenditure on petrol remains same.

Ans. The Formula is as under.

Decrease in consumption = $\frac{X}{100+X} \times 100$ =   $\frac{40}{100+40} \times 100$ =   $28 \frac{4}{7}$ percent

Q. If the cost of petrol decreases by 10%.By what percent the person should increase the consumption considering expenditure on petrol remains same.

Ans. The Formula is as under.

Increase in consumption =    $\frac{X}{100-X} \times 100$ = $\frac{10}{100-10} \times 100$ =   $11\frac{1}{9}$   percent

Q. A man sold two cows at Rs. 1995 each. One he lost 10% and on the other he gained 10%. What is his gain or loss percent.

Ans. The Formula is as under.

Loss Percent =   $\frac{X^{2}}{100}$ = $\frac{10^{2}}{100}$ = 1 %

There is always loss if the loss and gain percentages are same.

Q. A shopkeeper marks the price of his goods 20% higher than the original price. Then he gives 20% discount on goods.Find his profit or loss percent.

Ans. The Formula is as under.

Loss Percent =   $\frac{X^{2}}{100}$ = $\frac{20^{2}}{100}$ = 4%

Q. What price should a shopkeeper mark on an article costing him Rs. 153 to get
20% profit after allowing discount of 15%.

Ans. The Formula is as under.

M.P = $\frac{100+P}{100-D} \times C.P$ =    $\frac{100+20}{100-15} \times 153$ = 216

Q. 'A' sells a good to 'B' at a profit of 15%. 'B' Sells it to 'C' at a profit of 20%.
What is the resultant profit.

Ans. The Formula is as under.

Resultant Profit = $X + Y + \frac{XY}{100}$ = $15+20+\frac{15\times 20}{100}$ = 38%

Q. A merchant professes to sell goods at 20% profit ,but he uses weight of 900 gm in place of a Kilogram. What is his profit percentage.

Ans. The Formula is as under.

Profit Percent = $(100+20)\times \frac{1000}{900}-100$ = $33\frac{1}{3}$

Trains

10:47 AM

Q. What is the time required by a 110 m long train to cross a tunnel 220 m
long at rate of 22 m/s

Ans. The formula for this is as below.

Time = Distance = 220+110 = 15 s
Speed 22

Q. A train running at the speed of 60 Km/hr crosses a pole in 9 sec.
What is the length of the train in meters.

Ans. The formula for this is as below.

Length of the train = Speed X Time = 60 X 9 X 5   = 150 m
18

Km = 1000 = 5   meter
Hr 3600 18 Sec

Q. How many seconds will a train , 60 m in length, travelling at the rate of 42 Km an hour take to pass another train 84 m long proceeding in the same direction at the rate of 30 Km an hour.

Ans. Time = Total Length   = (60 + 84) X 18 = 43.2 sec
Relative Speed (42-30) X 5

Q. Two Trains are running at 40 Km/hr and 20 Km/hr respectively in the same direction. Fast train completely passes a man sitting in the slower train in 5 sec.
What is the length of fast train.

Ans. Length of the fast train = Relative Speed X Time

= (40-20) X  5 X 5 = 22  7  m
18 9

Km = 1000 = 5   meter
Hr 3600 18 Sec

Q. A train 50 m long passes a platform 100 m long in 10 sec.
Find the speed of the train in Km/hr.

Ans.
Speed = Distance = 50+100 = 15 m/s X 18  = 54 Km/hr
Time 10 5

Q. A 180 m long train crosses a man standing on the platform in 6 sec.
What is the speed of the train.

Ans.

Speed = Distance = 180  X 18  = 108 Km/hr
Time 6 5

Q. Two trains whose lengths are 180 m and 220 m respectively are running in direction opposite to one another with respective speeds of 40 Km/hr and 50 Km/hr.
Find the time taken by them to cross each other.

Ans.

Time taken = Total Length    = 220+180 X 18 = 16 s
Relative Speed 50+40 5

Thursday, May 3, 2018

Compound interest

10:40 AM

Q. The difference between the compound interest and simple interest on a certain sum at 10% p.a for 2 years is Rs. 631.Find the sum.

Ans. The formula is as under.

Sum = Diff X 100 X 100
r X r

Sum = 631 X 100 X100
   10 X 10

Sum = 63100

The above formula can be used only for 2 years period.

The formula for 3 years will be

Sum =  Diff X 100 X 100 X100
r X r X(300+r)

Q. The height of an equilateral triangle is   $8\sqrt{3}$   cm.Find it's area.

Ans. The formula is as under.

A = $\frac{h^{2}\times \sqrt{3}}{3}$ h is the height

A =    $\frac{64\times \sqrt{3}\times \sqrt{3}\times \sqrt{3}}{3}$

A = 64 $\sqrt{3}$

Q. A circle is inscribed in an equilateral triangle whose radius is   $4\sqrt{3}$ cm.
What is the perimeter of the equilateral triangle?

Ans. The formula is as under.

The perimeter of an equilateral triangle = 3 X side
= 3 X 2r X $\sqrt{3}$
= 3 X 2 X 4 X $\sqrt{3}\times \sqrt{3}$

The perimeter of an equilateral triangle = 72

The side of an equilateral triangle = radius of circle X   $2\sqrt{3}$
when the circle is inscribed in it

Wednesday, May 2, 2018

Tax

10:56 AM

Q.The tax on a commodity is diminished by 20% and its consumption
increases by 15%. Find the effect on revenue.

Ans. X = -20 since the tax is diminished
Y = 15 since consumption is increased

Effect on revenue = X + Y + XY = -20+15- 20x15 = -8
100 100

Since the answer is negative the revenue is decreased by 8%

If the answer is positive the revenue will increase.

If the tax diminished and the consumption increased is by the same percentage
then the formula will be as below

Revenue decreased = $\frac{X^{2}}{100}$

Q. If the compound interest on a certain sum for 2 years at 10%p.a is Rs. 2100.

Find the simple interest on it at the same rate of interest for 2 years.

Ans. The formula for finding the simple interest for the same rate of

interest is as below.

S.I = C.I X 200

200+r

S.I = 2100 X 200

210

S.I = 2000 Rs.

Q. The compound interest on a sum of money for 2 years is Rs. 52 and simple interest for 2 years at the same rate is Rs. 50. Find the rate of interest.

Ans. The formula for finding the rate of interest when the compound and

simple interest is given is as below.

R = 200 X Diff Diff = 52-50 = 2

S.I

R = 200 x 2 = 8

By using the above formulae you can solve the questions quickly

by avoiding lots of steps.

Tuesday, May 1, 2018

Geometry

11:16 PM

Q. Circumference of a circle is 44 cm.Find it's area.

Ans. The formula for finding the area of the circle when the circumference is given is as below.

A = $\large \frac{c^{2}}{4\pi }$

A = $\large \frac{44\times 44\times 7}{4\times 22}$

A = $\large 154 cm^{2}$

If you find the radius from the circumference and then calculate area from radius you have to do 6 steps,3 steps for calculating the radius and 3 steps for calculating the area.
With the above formula, you avoid 4 steps and get the answer in just 2 steps.

Q. Perimeter of a semicircle is 72 cm.Find it's radius.

Ans. The formula for finding the radius of the semicircle when the perimeter is given is as below.

$r = \frac{P\times 7}{36}$

'P' is the perimeter of the semicircle and 'r ' is the radius.

$r = \frac{72\times 7}{36}$

r = 14 cm

If you find the radius from the perimeter of the semicircle you have to do it in 4 steps.
But with the above formula, you get the answer in just one step.

Q. The breadth of a rectangle is increased by 20%. At what percent the length be decreased to maintain the same area.

Ans. The formula for finding the percentage decrease in length is as below.

The percentage decrease in length = $\frac{100X}{100+X}$

Where 'X' is the percentage increase in breadth.

The percentage decrease in length = $\frac{100\times 20}{120}$

The percentage decrease in length = 50/3 %

If you find the percentage decrease in length by doing the calculation you have to do it in 6steps. But with the above formula, you get the answer in just two steps.

Math Tricks For Competitive Exams

Sunday, May 27, 2018

Simple Interest

Tricks Development

Friday, May 11, 2018

Profit and Loss

Trains

Thursday, May 3, 2018

Compound interest

Wednesday, May 2, 2018

Tax

Tuesday, May 1, 2018

Geometry

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