1. (a – b)^{2} = (a^{2} + b^{2} – 2ab)
2. (a + b)^{2} = (a^{2} + b^{2} + 2ab)
3. (a + b)(a – b) = (a^{2} – b^{2})
4. (a^{3} + b^{3}) = (a + b)(a^{2} – ab + b^{2})
5. (a^{3} – b3) = (a – b)(a2 – ab + b2)
6. (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2(ab + bc + ca)
7. (a^{3} + b^{3} + c^{3} – 3abc) = (a + b + c)(a^{2 }+ b^{2} + c^{2 }– ab – bc – ac)
Fractions in which denominators are powers of 10 are known as decimal fractions.
Thus, 1/10 = 1 tenth = .1; 1/100 = 1 hundredth = .01;
88/100 = 88 hundredths = .88; 6/1000 = 6 thousandths = .006, etc.;
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.
Annexing zeros to the extreme right of a decimal fraction does not change its value. Thus, 0.8 = 0.80 = 0.800, etc.
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.
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.
Let us 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 .
If in a decimal fraction, a figure or a set of figures is repeated continuously, then such a number is called 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 = 0.333… = 0.3; 22/7 = 3.142857142857…. = 3.142857.
Example 1: Evaluate 1082 using (a + b)2 formula.
(a) 11645
(b) 12547
(c) 11664
(d) 12745
Ans: (c)
Let’s write 108 as: 108 = 100 + 8
108^{ 2} =(100 + 8)^{ 2}
Using the formula (a + b)^{ 2} = a^{ 2 }+ 2ab + b^{ 2}
108^{ 2} = (100)^{ 2} + 2(100)(8) + (8)^{ 2}
= 10000 + 1600 + 64
= 11664
Example 2: A recipe requires 3/4 cup of sugar to make 12 cookies. How much sugar would be needed to make 36 cookies?
(a) 2.25
(b) 2.5
(c) 3.5
(d) 4.5
Ans: (a)
To find the amount of sugar needed to make 36 cookies, we can set up a proportion based on the given information:
Sugar needed / Number of cookies = 3/4 cup / 12 cookies
Let’s solve for the unknown, which is the amount of sugar needed for 36 cookies:
Sugar needed / 36 = (3/4) cup / 12
Crossmultiplying gives us:
Sugar needed * 12 = 36 * (3/4)
Sugar needed * 12 = 108/4
Simplifying the right side:
Sugar needed * 12 = 27
Dividing both sides by 12:
Sugar needed = 27 / 12
Sugar needed = 2.25 cups
Therefore, to make 36 cookies, you would need 2.25 cups of sugar.
Example 3: If 47. 2506 = 4*A + 7/B + 2*C + 5/D + 6*E, then the value of 5*A + 3*B + 6*C + D + 3*E is:
(a) 53.6003
(b) 53.603
(c) 153.6003
(d) 213.003
Ans: (c)
4*A + 7/B + 2*C + 5/D + 6*E =47.2506
=> 4 * A + 7/B + 2 *C + 5/D + 6*E = 40 + 7 + 0.2 + 0.05 + 0.0006
Comparing the terms on both sides, we get :
4*A =40, 7/B = 7, 2*C = 0.2, 5/D = 0.05, 6*E = 0.0006
A = 10,
B = 1,
C = 0.1,
D = 100,
E = 0.0001.
5*A + 3*B + 6*C + D + 3*E =(5*10) + (3*1) + (6*0.1) +100 +(3*0.0001)
=50 + 3 + 0.6 + 100 + 0.003 = 153.6003.
Example 4: If 1.5x = 0.04y, then the value of is:
(a) 73/77
(b) 7.3/77
(c) 730/77
(d) 7300/77
Ans: (a)
Example 5: Evaluate
(a) 5.2
(b) 4.8
(c) 4
(d) 5
Ans: (d)
= 3.8 + 1.2 = 5
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