## Wednesday, 8 June 2016

### IMPORTANT FACTS AND FORMULAE OF SIMPLE INTEREST

1.. Principal: The money borrowed or lent out for a certain period is called the

principal or the sum.

2. Interest: Extra money paid for using other's money is called interest.

3. Simple Interest (S.I.) : If the interest on a sum borrowed for a certain period is reckoned uniformly, then it is called simple interest.

Let Principal = P, Rate = R% per annum (p.a.) and Time = T years. Then,

(i)                 S.I. =  (P*R*T )/100

(ii)        P=(100*S.I)/(R*T) ;R=(100*S.I)/(P*T) and T=(100*S.I)/(P*R)

### IMPORTANT FACTS AND FORMULAE OF PIPES AND CISTERNS

1. Inlet: A pipe connected with a tank or a cistern or a reservoir, that fills it, is known as an inlet.

Outlet: A pipe connected with a tank or a cistern or a reservoir, emptying it, is

known as an outlet.

2. (i) If a pipe can fill a tank in x hours, then : part filled in 1 hour = 1/x

(ii) If a pipe can empty a full tank in y hours, then : part emptied in 1 hour = 1/y

(iii) If a pipe can .fill a tank in x hours and another pipe can empty the full tank in y hours                                              (where y> x), then on opening both the pipes, the net part filled in 1 hour = (1/x)-(1/y)

(iv) If a pipe can fill a tank in x hours and another pipe can empty the full tank in y hours (where x > y), then on opening both the pipes, the net part emptied in 1 hour = (1/y)-(1/x)

### IMPORTANT FACTS AND FORMULAE OF RATIO AND PROPORTION

I. RATIO: The ratio of two quantities a and b in the same units, is the fraction a/b and we write it as a:b.

In the ratio a:b, we call a as the first term or antecedent and b, the second  term or consequent.

Ex. The ratio 5: 9 represents 5/9 with antecedent = 5, consequent = 9.

Rule: The multiplication or division of each term of a ratio by the same non-zero number does not affect the ratio.

Ex.  4: 5 = 8: 10 = 12: 15 etc. Also, 4: 6 = 2: 3.

2. PROPORTION: The equality of two ratios is called proportion.

If a: b = c: d, we write, a: b:: c : d and we say that a, b, c, d are in proportion . Here a and d are called extremes, while b and c are called mean terms.

Product of means = Product of extremes.

Thus, a: b:: c : d <=> (b x c) = (a x d).

3. (i) Fourth Proportional: If a : b = c: d, then d is called the fourth proportional

to a, b, c.

(ii) Third Proportional: If a: b = b: c, then c is called the third proportional to

a and b.

(iii) Mean Proportional: Mean proportional between a and b is square root of ab

4. (i) COMPARISON OF RATIOS:

We say that (a: b) > (c: d) <=>  (a/b)>(c /d).

(ii) COMPOUNDED RATIO:

The compounded ratio of the ratios (a: b), (c: d), (e : f) is (ace: bdf)

5.   (i) Duplicate ratio of (a : b) is (a2 : b2).

(ii) Sub-duplicate ratio of (a : b) is (a : b).

(iii)Triplicate ratio of (a : b) is (a3 : b3).

(iv) Sub-triplicate ratio of (a : b) is (a : b ).

(v) If (a/b)=(c/d), then  ((a+b)/(a-b))=((c+d)/(c-d))    (Componendo and dividendo)

6. VARIATION:

(i) We say that x is directly proportional to y, if x = ky  for some constant k and

we write, x µ y.

(ii) We say that x is inversely proportional to y, if xy = k for some constant k and

we write, x∞(1/y)

### IMPORTANT FACTS AND FORMULAE OF PERCENTAGE

1.      Concept of Percentage : By a certain percent ,we mean that many hundredths. Thus x percent means x hundredths, written as x%.

To express x% as a fraction : We have , x% = x/100.

Thus, 20% =20/100 =1/5; 48% =48/100 =12/25, etc.

To express a/b as a percent : We have, a/b =((a/b)*100)%.

Thus, ¼ =[(1/4)*100] = 25%; 0.6 =6/10 =3/5 =[(3/5)*100]% =60%.

2.      If the price of a commodity increases by R%, then the reduction in consumption so asnot to increase the expenditure is

[R/(100+R))*100]%.

If the price of the commodity decreases by R%,then the increase in consumption so as to decrease the      expenditure is

[(R/(100-R)*100]%.

3.    Results on Population : Let the population of the town be P now and suppose it increases at the rate of

R% per annum, then :

1.      Population after nyeras = P [1+(R/100)]^n.

2.      Population n years ago = P /[1+(R/100)]^n.

4.    Results on Depreciation :  Let the present value of a machine be P. Suppose  it depreciates  at the rate

R% per annum. Then,

1.      Value of the machine after n years = P[1-(R/100)]n.

2.      Value of the machine n years ago = P/[1-(R/100)]n.

5.      If A is R% more than B, then B is less than A by

[(R/(100+R))*100]%.

If  A is R% less than B , then B is more than A by

[(R/(100-R))*100]%.

### IMPORTANT FACTS AND FORMULAE OF SURDS AND INDICES

1. LAWS OF INDICES:

(i)                 am x an = am + n

(ii)    am­ / an = am-n

(iii)             (am)n = amn

(iv)             (ab)n = anbn

(v)     ( a/ b )n = ( an / bn )

(vi)             a0 = 1

2. SURDS: Let a be a rational number and n be a positive integer such that a1/n = nsqrt(a)

is irrational. Then nsqrt(a)  is called a surd of order n.

3. LAWS OF SURDS:

(i)  n√a = a1/2

(ii) n √ab = n √a * n √b

(iii) n √a/b = n √a  /  n √b

(iv) (n √a)n = a

(v) m√(n√(a)) = mn√(a)

(vi) (n√a)m = n√am

### NUMS ON AVERAGE

Ex.1:Find the average of all prime numbers between 30 and 50?

Sol: there are five prime numbers between 30 and 50.

They are 31,37,41,43 and 47.

Therefore the required average=(31+37+41+43+47)/5  ó199/5  ó 39.8.

Ex.2. find the average of first 40 natural numbers?

Sol:    sum  of first n natural numbers=n(n+1)/2;

So,sum of 40 natural numbers=(40*41)/2  ó820.

Therefore  the required average=(820/40) ó20.5.

Ex.3. find the average of first 20 multiples of 7?

Sol:    Required average =7(1+2+3+…….+20)/20   ó(7*20*21)/(20*2) ó(147/2)=73.5.

Ex.4. the average of four consecutive even numbers is 27. find the largest of these numbers?

Sol:    let the numbers be x,x+2,x+4 andx+6. then,

(x+(x+2)+(x+4)+(x+6))/4) = 27

ó(4x+12)/4 = 27

óx+3=27      ó x=24.

Therefore the largest number=(x+6)=24+6=30.

Ex.5. there are two sections A and B of a class consisting of 36 and 44 students respectively. If the average weight of section A is 40kg and that of section B is 35kg, find the average weight of the whole class?

Sol:   total weight of(36+44) students=(36*40+44*35)kg  =2980kg.

Therefore weight of the total class=(2980/80)kg  =37.25kg.

Ex:6.nine persons went to a hotel for taking their meals 8 of them spent Rs.12 each on their meals and the ninth spent Rs.8 more than the average expenditure of all the nine.What was the total money spent by them?

Sol: Let the average expenditure of all nine be Rs.x

Then 12*8+(x+8)=9x or 8x=104 or x=13.

Total money spent = 9x=Rs.(9*13)=Rs.117.

Ex.7: Of the three numbers, second is twice the first and is also thrice the third.If the average of the three numbers is 44.Find the largest number.

Sol: Let the third number be x.

Then second number = 3x.

First number=3x/2.

Therefore x+3x+(3x/2)=(44*3) or x=24

So largest number= 2nd number=3x=72.

Ex.8:The average of  25 result is 18.The average of 1st 12 of them is 14 & that of last 12 is 17.Find the 13th result.

Sol: Clearly 13th result=(sum of 25 results)-(sum of 24 results)

=(18*25)-(14*12)+(17*12)

=450-(168+204)

=450-372

=78.

Ex.9:The Average of 11 results is 16, if the average of the 1st 6 results is 58 & that of the last 63. Find the 6th result.

Sol: 6th result = (58*6+63*6-60*11)=66

Ex.10:The average waight of A,B,C is 45 Kg. The avg wgt of A & B be 40Kg & that of B,C be 43Kg. Find the wgt of B.

Sol. Let A,B,c represent their individual wgts.

Then,

A+B+C=(45*3)Kg=135Kg

A+B=(40*2)Kg=80Kg & B+C=(43*2)Kg=86Kg

B=(A+B)+(B+C)-(A+B+C)

=(80+86-135)Kg

=31Kg.

Ex. 11. The average age of a class of 39 students is 15 years. If the age of the teacher be included, then the average increases by3 months. Find the age of the teacher.

Sol. Total age of 39 persons = (39 x 15) years

= 585 years.

Average age of 40 persons= 15 yrs 3 months

= 61/4 years.

Total age of 40 persons = (_(61/4 )x 40) years= 610 years.

:. Age of the teacher = (610 - 585) years=25 years.

Ex. 12. The average weight of 10 oarsmen in a boat is increased by 1.8 kg when one of the crew, who weighs 53 kg is replaced by a new man. Find the weight of the new

man.

Sol. Total weight increased =(1.8 x 10) kg =18 kg.

:. Weight of the new man =(53 + 18) kg =71 kg.

### IMPORTANT FACTS AND FORMULAE OF DECIMAL FRACTIONS

I.    Decimal Fractions : Fractions in which denominators are powers of 10 are known as decimal  fractions.

Thus ,1/10=1 tenth=.1;1/100=1 hundredth =.01;

99/100=99 hundreths=.99;7/1000=7 thousandths=.007,etc

II.  Conversion of a Decimal Into Vulgar Fraction : Put 1 in the denominator under the decimal point and annex with it as many zeros as is the number of digits after the decimal point. Now, remove the decimal point and reduce the fraction to its lowest terms.

Thus, 0.25=25/100=1/4;2.008=2008/1000=251/125.

III.  1.  Annexing zeros to the extreme right of a decimal fraction does not change its value
Thus, 0.8 = 0.80 = 0.800, etc.

2.  If numerator and denominator of a fraction contain the same number of decimal
places, then we remove the decimal sign.

Thus,   1.84/2.99 = 184/299 = 8/13;    0.365/0.584 = 365/584=5

IV.  Operations on Decimal Fractions :

1.  Addition and Subtraction of Decimal Fractions : The given numbers are so
placed under each other that the decimal points lie in one column. The numbers
so arranged can now be added or subtracted in the usual way.

2.  Multiplication of a Decimal Fraction By a Power of 10 : Shift the decimal
point to the right by as many places as is the power of 10.

Thus, 5.9632 x 100 = 596,32; 0.073 x 10000 = 0.0730 x 10000 = 730.

3.Multiplication of Decimal Fractions : Multiply the given numbers considering
them without the decimal point. Now, in the product, the decimal point is marked
off to obtain as many places of decimal as is the sum of the number of decimal
places in the given numbers.

Suppose we have to find the product (.2 x .02 x .002). Now, 2x2x2 = 8. Sum of decimal places = (1 + 2 + 3) = 6. .2 x .02 x .002 = .000008.

4.Dividing a Decimal Fraction By a Counting Number : Divide the given
number without considering the decimal point, by the given counting number.
Now, in the quotient, put the decimal point to give as many places of decimal as
there are in the dividend.

Suppose we have to find the quotient (0.0204 + 17). Now, 204 ^ 17 = 12. Dividend contains 4 places of decimal. So, 0.0204 + 17 = 0.0012.

5. Dividing a Decimal Fraction By a Decimal Fraction : Multiply both the dividend and the divisor by a suitable power of 10 to make divisor a whole number. Now, proceed as above.

Thus, 0.00066/0.11 = (0.00066*100)/(0.11*100) = (0.066/11) = 0.006V

V.  Comparison of Fractions : Suppose some fractions are to be arranged in ascending or descending order of magnitude. Then, convert each one of the given fractions in the decimal form, and arrange them accordingly.

Suppose, we have to arrange the fractions  3/5, 6/7 and 7/9  in descending order.

now, 3/5=0.6,6/7 = 0.857,7/9 = 0.777....

since  0.857>0.777...>0.6, so 6/7>7/9>3/5

VI. Recurring Decimal : If in a decimal fraction, a figure or a set of figures is repeated continuously, then such a number is called a recurring decimal.

In a recurring decimal, if a single figure is repeated, then it is expressed by putting a dot on it. If a set of figures is repeated, it is expressed by putting a bar on the set

Thus 1/3 = 0.3333….= 0.3;  22 /7 = 3.142857142857.....= 3.142857