The Leading Maths - 7

Arithmetic 101 Types of Decimals (a) Terminating decimal number Let’s take a rational number as 3 8. Now, let’s convert the rational number into decimal number, 3 8 = 0.375, which has no more digits after 5 except 0. Therefore, it is a terminating decimal number. (b) Recurring or Repeating decimal number. Let’s take a rational number as 25 11. Now, let’s convert the rational number into decimal number, 25 11 = 2.272727..... = 2.27, Which has many more 27 after 27. Therefore, it is a repeating decimal number. (c) Non-terminating and non-recurring decimal number Let’s take a number as 15. Now, let’s evaluate the square root of 15, 15 = 3.8729......... , which has many more digits after 9 but no repeated. Therefore, it is a non-repeating and nonterminating decimal number. Conversion of Fraction and Decimal Conversion of fractions into decimals A fraction is usually converted into decimals by simply dividing the numerator by the denominator. We proceed as follows: 8 )3 0 ( 0.375 – 3 0 2 0 – 1 6 4 0 – 4 0 0 11)2 5 ( 2.2727 – 2 2 3 0 – 22 8 0 – 77 3 0 – 22 8 0 – 77 3 3 + 3 15 – 9 3.8729 68 + 8 600 – 544 767 + 7 5600 – 5369 7742 + 2 23100 – 15484 77449 + 9 761600 – 697041 77458 64559

102 The Leading Mathematics - 7 i) If there is no decimal point, we put a decimal point at the end of the last digit. ii) Add as many zeros as necessary after the last digit, iii) Start dividing as usual and iv) Put a decimal point in the quotient as soon as we put down the first figure after the decimal point in the dividend. Two cases arise. The division process may leave no remainder or a non-zero remainder at some stages. Case I. If the denominator of the fraction is 2 or 5 or both 2 and 5 (possibly repeated) as its factors, there will be no remainder (or zero as the remainder). For examples, 1 4 = 1 2 × 2 = 25 2 × 2 × 25 = 25 100 = 0.25, 3 50 = 3 × 2 50 × 2 = 6 100 = 0.06 and 1 125 = 1 125 = 1 × 8 125 × 8 = 8 10000 = 0.0008. Case II. If the denominator of the fraction is neither 2 nor 5 nor both 2 and 5 as its factors, there will be a remainder at some stages. The remainder recurs after some stages. As an example, let’s evaluate 1 7 or 1 ÷ 7. In this case, the only possible remainders at any step of the division are 0, 1, 2, 3, 4, 5, 6. If 0 is the remainder, we say that there is no remainder. In other words, the series of digits comes to an end or terminates. Otherwise, it recurs as shown alongside: If we stop after a certain number of decimal places, we shall simply get an approximation to 1 7. To show that it is recurring and we do not intend to stop at any stage, we place a dot over the recurring digit as in the case of 1 3; and if a cycle of digits recurs, we put a dot each over the first and last digits of the cycle or period. 4) 1.00 (0.25 – 0 10 – 8 20 – 20 0 50)3.00 (0.6 – 0 30 – 0 300 – 300 0 7) 1.0 (0.1428571 – 0 10 – 7 30 – 28 20 –14 60 –56 40 – 35 50 – 49 10 – 7 3

Arithmetic 103 For instances, 1 3 = 0.3 and 1 7 = 0.1428571... These types of decimals are called recurring or repeating decimal number. Conversion of decimals into fractions Conversion of decimals into fraction is rather very simple. To do so, we i. write the decimal without the decimal point as the numerator of the required fraction, ii. write 1 in the denominator followed by as many zeros as the number of decimal places in the given decimal; and iii. simplify the fraction as far as possible, i.e., write the fraction in the lowest terms. For instances, 0.1 = 1 10, 0.01 = 1 100, 1.001 = 1001 1000, 0.125 = 125 1000 = 1 8. In case, the given decimal is a recurring decimal, the procedures study in the worked out examples: CLASSWORK EXAMPLES Example 1 Convert the following fractions into decimals: (a) 3 4 (b) 5 16 (c) 22 7 (d) 5 7 9 (e) 44 7 Solution: (a) 3 4 = 3 × 25 4 × 25 = 75 100 = 0.75 (b) 5 16 = 5 × 625 16 × 625 = 3125 10000 = 0.3125 (c) 22 7 = 3.14287 14287 14287....... = 3. 14287 (d) 5 7 9 = 5.7777 = 5. 7 (e) 44 7 = 6.285714285714 = 6. 285714 3) 1.0 (0.3 –0 10 – 9 10 – 9 1

104 The Leading Mathematics - 7 Example 2 Multiply the following recurring numbers by 10, 100 and 1000 respectively. (a) 0.3 (b) 8. 249 Solution: Here, (a) 0.3 × 10 = 0.3333 ........× 10 = 3.333 ...... = 3.3 0.3 × 100 = 0.3333 ........× 100 = 33.33 ...... = 33.3 0.3 × 1000 = 0.3333 ........× 1000 = 333.3 ...... = 333.3 (b) 8.249 × 10 = 8.249249 ....... × 10 = 82.49249.... = 82.492 8.249 × 100 = 8.249249 ....... × 100 = 824.9249.... = 824.924 8.249 × 1000 = 8.249249 ....... × 1000 = 8249.249.... = 8249.249 Example 3 Convert the following decimal numbers into fractions: (a) 0.6 (b) 2.45 (c) 4.321 Solution: Here, (a) Suppose x = 0.6 = 0.666666 ................ (i) Then 10x = 6.66666 ............................ (ii) Subtracting (i) from (ii), we get 10x = 6.66666 ... (–) x = 0.666666 ...... 9x = 6.00000 ...... or, x = 6 9 = 2 3 ∴ 0.6 = 2 3 . (b) Suppose x = 2.45 = 2.454545 ................ (i) Then 100x = 245.454545 ...................... (ii) Subtracting (i) from (ii), we get 100x = 245.454545 ........ (–) x = 2.454545 ........ 99x = 243.00000 ...... or, x = 243 99 = 27 11 = 2 5 11 ∴ 2.45 = 2 5 11.

Arithmetic 105 (c) Let x = 4.321 = 4.321321321 .................... (i) Then 1000x = 4321.3231321321 ................. (ii) Subtracting (i) from (ii), we get 1000x = 4321.3231321321 ........ (–) x = 4.3213231321 ...... 999x = 4317.00000 ...... or, x = 4317 999 or, x = 1439 333 or, x = 4107 333 ∴ 4.321 = 4 107 333. EXERCISE 5.4 Your mastery depends on practice. Practice like you play. 1. Express the following fractions into decimal numbers: (a) 3 10 (b) 43 10 (c) 57 100 (d) 273 100 (e) 4572 1000 (f) 2 100 (g) 25 1000 (h) 1 2 (i) 2 5 (j) 7 4 (k) 9 25 (l) 23 50 (m) 47 8 (n) 125 16 (o) 3425 4 2. Convert each of the following fractions into decimal numbers and decide they are whether it is terminating or recurring and non-terminating decimal numbers: (a) 2 5 (b) 7 5 (c) 9 8 (d) 11 6 (e) 15 8 (f) 17 9 (g) 21 16 (h) 56 32 (i) 23 7 (j) 44 13 (k) 36 11 (l) 632 222 (m) 243 333 (n) 122 22 (o) 345 44 (p) 35 12 (q) 32 27 (r) 155 36 (s) 1 64 (t) 3 128

106 The Leading Mathematics - 7 3. Reduce the following decimal numbers into fractions: (a) 0.2 (b) 0.8 (c) 0.02 (d) 0.004 (e) 1.04 (f) 3.4.6 (g) 2.0048 (h) 4.045 (i) 10.66 (j) 12.005 (k) 0.2 (l) 0.5 (m) 2.4 (n) 4.6 (o) 5.56 (p) 4.27 (q) 5.24 (r) 0.123 (s) 4.789 (t) 10.483 ANSWERS 1. (a) 0.3 (b) 4.30 (c) 0.57 (d) 2.73 (e) 4.572 (f) 0.02 (g) 0.025 (h) 0.5 (i) 0.4 (j) 1.75 (k) 0.36 (l) 0.46 (m) 5.875 (n) 7.8125 (o) 856.25 2. (a) 0.4 (b) 1.4 (c) 1.125 (d) 1.83 (e) 1.875 (f) 1.8 (g) 1.3125 (h) 1.75 (i) 3.285714 (j) 3.384615 (k) 3.27 (l) 2.846 (m) 0.729 (n) 5.54 (o) 7.84090 (p) 2.916 (q) 1.185 (r) 4.305 (s) 0.015625 (t) 0.0234375 3. (a) 1 5 (b) 4 5 (c) 1 50 (d) 1 250 (e) 1 1 25 (f) 323 50 (g) 2 3 625 (h) 4 9 200 (i) 1033 50 (j) 12 1 200 (k) 2 9 (l) 5 9 (m) 2 4 9 (n) 4 2 3 (o) 5 56 99 (p) 4 3 11 (q) 5 8 33 (r) 123 999 (s) 4 263 333 (t) 10161 333

Arithmetic 107 5.5 Simplifications on Decimals At the end of this topic, the students will be able to: ¾ simplify the decimal numbers. Learning Objectives Four Fundamental Operations on Decimals (a) Addition and Subtraction of Decimals Decimals may first be converted into fractions and then simplified. We, however, do not do so at present. To add or find the sum of two or more decimals, we 1. convert all decimals into like decimals. 2. write the decimals (addends) in columns in such a way that the decimal points of all decimals lie in the same vertical line and the digits having the same place values lie in the same vertical column. 3. add the numbers in the same way as in the case of addition of whole numbers; and 4. put a decimal point in the sum directly under the decimal points in the vertical line. For instance, we find the sum of 1.451 , 0.509 + 10. 06 as shown below: 1.451 0.509 + 10.060 (converting 0.06 into like decimal) 12.020 = 12.01 To subtract a decimal from a larger decimal, we 1. convert the decimals into like decimals 2. write the small decimal below the larger decimal in such a way that the decimal points lie in the same vertical line, and the digits having the same place values are in the same vertical column 3. Subtract as usual as in the case of whole numbers (ignoring the decimal points)

108 The Leading Mathematics - 7 4. Put a decimal point in the result directly under the decimal points in the vertical column. For instance, we subtract 2.0352 from 5.14 as shown below: 5.1400 – 2.0352 (converting into like decimal) 3.1048 (b) Multiplication of decimals Multiplication of decimals may be broadly divided into three categories: i. Multiplication of a decimal by a whole number ii. Multiplication of a decimal by 10, 100, 1000, etc. iii. Multiplication of a decimal by a decimal (i) Multiplication of a decimal by a whole number To multiply a decimal by a whole number, we i. follow the same steps as in the case of multiplication of two whole numbers, and ii. put a decimal point in the product so that the number of the decimal places is the same as that of the given decimal. For instance, we multiply 3. 251 by 60 to find 3.251 × 60 = 195.060 = 195.06 (ii) Multiplication of a decimal by 10, 100, 100, etc. To multiply a decimal by10, 100, 1000, we simply move the decimal point in the multiplicand by one, two, three, etc. places (add zeros if necessary) to the right. For instances, we have 3.125 × 10 = 31.25; 3.125 × 100 = 312.5; 3.125 × 1000 = 3125 Remember that no decimal point is written if there is no non–zero digit at the extreme right. (iii) Multiplication of a decimal by a decimal To multiply a decimal by another decimal, 1. perform the multiplication in the same way as in the case of whole numbers, and

Arithmetic 109 2. put a decimal point in the multiplicand so as to make the number of decimal places in the product is equal to the sum of the decimal places of the multiplicand and the multipliers. For instances, we have 3.25 × 1.1 = 3.575; 3.125 × 1.1 = 3.4375; Remember that the zero after decimal point but at the extreme right can be deleted. Properties of Decimal Numbers Decimals are simply different forms of representing fractions. So, they obey the commutative, associative and distributive laws related to addition and multiplication. There exist additive as well as multiplicative identity and inverse also. For instances, 1. 1. 51 + 2.34 = 3.85 = 2.34 + 1.51; (Commutative property) 2. (1.5 + 2.3) + 1.2 = 5.0 = 1.5 + (2.3 + 1.2) (Associative property) 3. 1.555 + 0.0 = 1.5 = 0.0 + 1.555 (Additive identity : 0) 4. 1.51 + (–1.51) = 0 = 1.51 – .51 = 0 (Additive inverse of 1.5 is –1.5) 5. 1. 512 × 1.2 = 1.8144 = 1.2 × 1.512; (Commutative property) 6. (1.5 × 2.2)×1.2 = 3.30 × 1.2 = 39.60 = 1.5 × 26.4 = 1.5 × (2.2 × 1.2) (Associative property) 7. 1.543 × 1 = 1.543 = 1 × 1.543 (Multiplicative identity : 1) 8. 1.56 × 1 1.56 = 1 = 1 1.56 × 1.56 (Multiplicative inverse of 1.56 is 1 1.56 ) (c) Division of decimals Division of decimals may be broadly divided into three categories: (i) Division of a decimal by a whole number (ii) Division of a decimal by 10, 100, 1000, etc. (iii) Division of a decimal by a decimal.

110 The Leading Mathematics - 7 (i) Division of a decimal by a whole number To divide a decimal by a whole number, we i. add, if necessary, one or more zeros to the right to make the number formed by ignoring the decimal or without the decimal greater than the divisor. ii. perform division as in the case of whole numbers If there is no remainder, i. count the number of decimal places in the dividend together with the added zeros ii. start from the final digit at the extreme right of the quotient and put a decimal point so as to make the number of decimal places the same as in the dividend in step (i) (add zeros, if necessary, on the extreme left of the quotient) For instances, 1.4 ÷ 2 = 0.7 2 ) 1.4 ( 0.7 –1.4 0 0.4 ÷ 4 = 0.1 4 ) 0.4 ( 0.1 –0.4 0 0.4 ÷ 8 = 0.40 ÷ 8 = 0.05 8 ) 0.40 ( 0.05 –0.40 0 1 ÷ 2 = 1.0 ÷ 2 = 0.5 2 ) 1.0 ( 0.5 –1.0 0 1 ÷ 25 = 1.00 ÷ 25 = 0.04 25 ) 1.00 ( 0.04 – 1.00 0 If there is a remainder, (i) Repeat the Steps (1) and (2) taking the remainder as the new dividend. (ii) count the number of decimal places in the given dividend together with the total number of zeros added to the remainder.

Arithmetic 111 (iii) start from the final digit at the extreme right of the quotient and put a decimal point so as to make the number of decimal places the same as in step (ii) (add zeros, if necessary, on the extreme left of the quotient) For instances, 2) 1.3 ( 0.65 –1.2 10 – 10 0 1.3 ÷ 2 = 1.30 ÷ 2 = 0.65 ; no. of decimal place in the divided = 1 no. of 0 added to the remainder = 1 Total = 2 So, the required quotient is 0.65. (ii) Division of a decimal by 10, 100, 1000, etc. To divide a decimal by 10, 100, 1000, we simply move the decimal point in the dividend by one, two, three, etc. places (add zeros if necessary) to the left. For instances, we have 3.252 ÷ 10 = 0.3252; 3.252 × 100= 0.03252 (iii) Division of a decimal by a decimal To divide a decimal by another decimal, we 1. count the number of decimal places of the divisor. 2. shift the decimal point of the dividend to the right (add zeros if necessary) to the right by the same number of places as in the divisor. 3. perform division in the same way as in the case of whole numbers by ignoring the decimal point in the divisor. For instances, 2.4 ÷ 1.2 = 24 ÷ 12 = 2 12 ) 24 ( 2 – 24 0 2.52 ÷ 12 = 25.2 ÷ 12 = 2.1 12) 25.2 ( 2.1 –24 12 – 12 0 no. of decimal places in the divided = 1 no. of zeros added = 0 Total = 1

112 The Leading Mathematics - 7 2.562 ÷ 1.22 = 256.2 ÷ 122 = 2.1 122) 256.2 ( 2.1 – 244 122 – 122 0 no. of decimal places in the divided = 1 no. of zeros added = 0 Total = 1 33.75 ÷ 1.125 = 33750 ÷ 1125 = 30 1125) 33750 ( 30 – 3375 0 no. of decimal places in the divided = 0 no. of zeros added = 0 Total = 0 CLASSWORK EXAMPLES Example 1 Simplify: 4.205 – 2.495 + 1.289 Solution: Here, 4.205 – 2.495 + 1.289 = 1.710 + 1.289 = 2.999 Example 2 Simplify: 4.5 + 3.7 – 3.6 ÷ 1.2 Solution: Here, 4.5 + 3.7 – 3.6 ÷ 1.2 = 4.5 + 3.7 – 3.0 = 8.2 – 3.0 = 5.2 Example 3 Simplify: 7.6 – [3 + 0.5 of (3.1 – 2.3 × 1.02)] Solution: Here, 7.6 – [3 + 0.5 of (3.1 – 2.3 × 1.02)] = 7.6 – [3 + 0.5 of (3.1 – 2.346)] = 7.6 – [3 + 0.5 of 0.754], = 7.6 – [3 + 0.5 × 0.754] = 7.6 – [3 + 0.377] = 7.6 – 3.377 = 4.223

Arithmetic 113 Example 4 Simplify: 3 ÷ 16 + 1.2 × 1 4 – 1 5 + (1 – 0.8) Solution: Here, 3 ÷ 16 + 1.2 × 1 4 – 1 5 + (1 – 0.8) = 3 ÷ 16 + 1.2 × 1 4 – {0.2 + 1 – 0.8} = 3 ÷ 16 + 1.2 × 0.25 – {(1.2 – 0.8} = 3 ÷ 16 + 1.2 × 0.25 – 0.4 = 0.1875 + 1.2 × 0.25 – 0.4 = 0.1875 + 0.3 – 0.4 = 0.4875 – 0.4 = 0.0875 EXERCISE 5.5 Your mastery depends on practice. Practice like you play. 1. Add: (a) 4.205 + 2.827 (b) 7.872 + 2.4 (c) 8.809 + 6.47 + 3.007 (d) 2.097 + 0.8 + 0.47 + 3.7894 2. Subtract: (a) 2.487 – 1.527 (b) 8.924 – 8.739 (c) 8.278 – 4.379 – 1.2 (d) 4.278 – 0.27 – 0.78 – 1.278 3. Simplify the following decimals: (a) 17.709 + 9.001 – 10.02 (b) 113 + 56.28 – 74.7 (c) 72.36 – 75.07 + 21.109 (d) 53.1 – 56.028 + 4.07 4. Multiply: (a) 2.72 × 4 (b) 3.789 × 12 (c) 3.21 × 10 (d) 4.879 × 100 (e) 48.2 × 1000 (f) 4.87 × 10000 (g) 2.21 × 1.3 (h) 3.45 × 2.3 (i) 7.21 × 2.73 5. Divide: (a) 2.1 ÷ 7 (b) 7.2 ÷ 8 (c) 6.5 ÷ 0.2 (d) 0.625 ÷ 27 (e) 37.2 ÷ 10 (f) 72 ÷ 100 (g) 37.24 ÷ 100 (h) 2 ÷1000

114 The Leading Mathematics - 7 (i) 7.47 ÷ 1000 (j) 96.075 ÷ 6.3 10(k) 24.629 ÷ 1.1 (l) 14.295 ÷ 1.5 6. Simplify using BODMAS rule: (a) 18 – 4.2 ÷ 6 + 1.3 × 0.4 (b) 6.4 ÷ 1.6 of 5 + 1.3 × 3.1 – 0.07 (c) 50.3 – 5.6 ÷ 0.7 × 1.6 of 3.5 (d) 34.6 × 2.1 + 7.7 ÷ 1.1 + 4.095 7. Simplify the decimal numbers: (a) 12.8 – 0.4 of (7.2 – 3.7) + 2.4 × 3.02 (b) 23.6 – 0.6 of (9.4 – 5.6) + 0.6 × 3.06 (c) (6.4 × 6.4 – 3.3 × 3.3) ÷ (6.4 + 3.3) (d) 27.08 – [5.6 + 3 of (6.5 – 0.5 × 2.01)] (e) 19.05 – [2.06 + {2.57 + (9.91 – 5.09 – 0.76)}] (f) [(5.39)2 – (2.91)2 ] ÷ [5.39 – 2.91] (g) [(5.6)3 – (3.24)3 ] × [(5.6)2 + (5.6)(3.24) + (3.24)2 ] (h) (9.5 × 9.5 × 9.5 – 0.5 × 0.5 × 0.5) ÷ (9.5 × 9.5 + 9.5 × 0.5 + 0.5 × 0.5) (i) (8.2 × 8.2 – 2 × 8.2 × 1.4 + 1.4 × 1.4) ÷ (3.8 × 3.8 + 0.2 × 0.2 + 2 × 3.8 × 0.2) 8. (a) What must be added to 19.63 to obtain 57.87? (b) What must be added to 78.76 to get 56.981? (c) What must be subtracted from 381.6 to obtain 28.88? (d) What must be subtracted from 45.124 to get 29.98? 9. (a) What must be multiplied to 19.63 to get 33.371? (b) What must be multiplied by 78.76 to obtain 685.9996? (c) What must be divided by 11.6 to obtain 45.3? (d) What must be divided by 52.63 to get 2.8? 10. Write the following verbal problems in mathematical expressions and simplify: (a) What is the number when 5.275 is multiplied by 7 and 4.9 is added to the product? (b) Find the number when 26.275 is multiplied by 2.7 and 23.87 is subtracted from the result. (c) Find the number if 14.75 is multiplied by 348.985.

Arithmetic 115 (d) On dividing a number by 3.2, the quotient is 140.84 and the remainder 6.008. Find the number. (e) When a number is divided by 4.5, the quotient is 27.3 and the remainder is 2.75. Find the number. ANSWERS 1. (a) 7.032 (b) 10.272 (c) 18.286 (d) 7.1564 2. (a) 0.96 (b) 0.185 (c) 2.705 (d) 1.95 3. (a) 16.69 (b) 94.58 (c) 18.399 (d) 1.142 4. (a) 10.88 (b) 45.468 (c) 32.1 (d) 487.9 (e) 48200 (f) 48700 (g) 2.873 (h) 7.935 (j) 19.6833 5. (a) 0.3 (b) 0.9 (c) 32.5 (d) 0.02315 (e) 3.72 (f) 0.72 (g) 0.3724 (h) 0.002 (i) 0.00747 (j) 15.2258 (k) 22.39 (l) 9.53 6. (a) 17.82 (b) 23.96 (c) 5.5 (d) 83.755 7. (a) 18.648 (b) 23.156 (c) 3.1 (d) 4.995 (e) 10.36 (f) 8.3 (g) 2.36 (h) 9 (i) 2.89 8. (a) 38.24 (b) 21.779 (c) 352.72 (d) 15.144 9. (a) 1.7 (b) 8.71 (c) 525.48 (d) 147.364 10. (a) 41.825 (b) 47.0725 (c) 5147.52875 (d) 456.696 (e) 125.6 Project Work 5.5 Observe the money exchange rates published on the daily news paper or internet. Write the buying and selling rates of any five countries and find the differences between their exchange rates. Present it in your classroom.

116 The Leading Mathematics - 7 5.6 Verbal Problems on Decimals At the end of this topic, the students will be able to: ¾ solve the word problems related to decimal numbers. Learning Objectives To solve the word problems on fractions first express the given statement in expression of fraction then simplify using the operations of fractions. CLASSWORK EXAMPLES Example 1 The product of two numbers is 42.63. If one number is 2.1, find the other. Solution: Here, Product of two numbers = 42.63 One number = 2.1 Other number = 42.63 ÷ 2.1 = 20.3 Therefore, the other number is 20.3. Example 2 John bought 9.25 m of cloth for Rs. 425.50. Find the cost price per metre. Solution: Here, Cloth bought by John = 9.25 m Cost of 9.25 m = Rs. 425.50 Cost per metre = 425.50 ÷ 9.25 = Rs. 46 ∴ The cost of the cloth per metre is Rs. 46. Example 3 One kg Basmati rice costs Rs. 43.75. Find the cost of 17 kg of rice. Solution: Here, Cost of 1 kg rice = 43.75 Cost of 17 kg rice = 43.75 × 17 = 743.75 Therefore, 17 kg of rice cost Rs. 743.75.

Arithmetic 117 EXERCISE 5.6 Your mastery depends on practice. Practice like you play. 1. (a) Areena wanted to buy the following items: A DVD player for Rs. 49.95 A DVD holder for Rs. 19.95 Personal stereo for Rs. 21.95 Find the total amount to buy all of them. Does Areena have enough money to buy all three items if she has Rs. 90? (b) If your weekly salary is Rs. 1,015.00, how much do you have left each week after you pay for the following? Rent : Rs. 443.50 Cable TV : Rs. 23. 99 Electricity : Rs. 45.62 Groceries : Rs. 124.87 2. (a) Manisha purchased Rs. 39.46 in groceries at a store. The cashier gave her Rs. 1.46 in change from a Rs. 50 note. How much change should the cashier have given Manisha? (b) Preksha has Rs.425.82 in her checking account. How much does she have in her account after she makes a deposit of Rs. 120.75 and a withdrawal of Rs. 185.90? 3. (a) If a 10-foot piece of electrical tape has five pieces that are each 1.25 feet cut from it, what is the new length of the tape? (b) Find the total cost of two Medium Sodas, two Medium Popcorns, and two Movie tickets, whose rates are given below: Item Cost Movie Ticket Rs. 8.25 Medium Popcorn Rs. 6.00 Medium Soda Rs. 4.75 Candy Rs. 3.50 4. (a) A student earns Rs. 11.75 per hour for gardening. If she worked 21 hours this month, then how much did she earn? (b) School breakfast cost Rs. 250.50 per week. How much would pay for 15.5 weeks of breakfast cost? 5. (a) Ajay’s car gets 29.7 miles per gallon on the highway. If his fuel tank holds 10.45 gallons, then how far can he travel on one full tank? (b) A member of the school track team ran for a total of 184.5 miles in practice over 61.5 days. About how many miles did he run per day?

118 The Leading Mathematics - 7 6. (a) A store owner has 7.11 lbs. of candy. If she puts the candy into 9 jars, how much candy will each jar contain? (b) Pal will pay for his toy car in 36 monthly payments. If his car loan is for Rs.19,061, then how much will Paul pay each month? Round your answer to the nearest two decimal places. 7. (a) What is the average speed in miles per hour of a car that travels 956.4 miles in 15.9 hours? Round your answer to the nearest tenth. (b) Dilip worked 15 days for a total of 116.25 hours. How many hours did he work in average per day? 8. (a) Six cases of paper cost Rs. 159.98. How much does one case cost? Round your answer to the nearest paisa. (b) There are 2.54 centimeters in one inch. How many inches are there in 51.78 centimeters? Round your answer to the nearest thousandth. 9. (a) If a rectangular court-yard is 24.75 m long and 16.25 m wide, (i) What is the perimeter of the court-yard? (ii) What is the area of the court-yard? (b) If a rectangular field of area 3996.614 m2 is 72.35 m long then, (i) What is its breadth ? (ii) What is its perimeter? 10. (a) The circumference of the wheel of a bus is 3.25 m. (i) Find the radius of the wheel. (ii) How many revolutions are made by the wheel in covering 40.625 km? (b) The diameter of a coin is 1.075 cm. (i) Find the radius of the coin. (ii) How many coins identical to this coin must be placed in a row to cover a distance of 23.65 m? ANSWERS 1. (a) Rs. 91.85, No (b) Rs. 377.02 2. (a) Rs. 9.08 (b) Rs. 360.67 3. (a) 3.75 ft (b) Rs. 38 4. (a) Rs. 246.75 (b) Rs. 3882.75 5. (a) 310.365 miles (b) 3 miles 6. (a) 0.79 lbs (b) Rs. 529.47 7. (a) 60.15 miles (b) 7.75 hrs 8. (a) Rs. 26.66 (b) 20.39 inches 9. (a) 82 m, 402.19 m2 (b) 55.24 m, 255.19 m 10. (a) 0.517 m, 12500 (b) 0.54 cm, 22 coins

Arithmetic 119 CHAPTER 6 Ratio and Proportion Lesson Topics Pages 6.1 Ratio 120 6.2 Proportion 127 1. Compare the number of coins on the first pan and the number of apples on the second pan of the beam balance. 2. What are the fractions of yellow and blue part separately ? 3. What are the fractions of brown and red part separately ? 4. What is the meaning of fraction ? 5. What are fraction numbers? WARM-UP 4 : 8 4 12 2 : 6 2 8

120 The Leading Mathematics - 7 6.1 Ratio At the end of this topic, the students will be able to: ¾ solve the problems related to ratio. Learning Objectives Introduction Human life cannot exist in the absence of objects of one or other kind. Sometimes we have to consider part or parts of a single whole (or unit). At other times, we have to consider multiples of a certain quantity. We may also have to relate a certain quantity of one kind with some quantity of another kind. For instance, while buying something we relate some objects with money i.e., rates. Such relations are quite often expressed in terms of one hundred units, i.e., in per centum (or per cent). In particular, we do so while lending or borrowing some money, and also in calculating profits and loss during the purchase or selling of goods. Fractions or ratios do appear while dealing with such everyday activities. Ratio Division of an integer by a non-zero integer gave rise to fractions or ratios. Today, a fraction is commonly denoted by writing (i) a number above a horizontal bar (—) and another under it, e.g., 3 5 (ii) a number before a solidus “ / ” and another after it, e.g., 3/5 and (iii) a number before the colon “ : ” , e.g., 3 : 5. It is believed that the first one was introduced by the Arabs around 1200, and the second one was recommended by De Morgan in 1845. We now begin with a formal definition of the term “ratio ”. In mathematics, a ratio may be defined in the following way: A ratio of two numbers is that which tells what part or parts or multiple of a number is another number. In particular, consider pairs of numbers such as 1 and 2, 3 and 4. We can then write one number above a horizontal line and the other below it as shown below: 1 2, 3 4, 2 1 and 4 3.

Arithmetic 121 Each of them is a ratio. Here, the ratio 1 2 (1 over 2 ) or 1:2 (1 is to 2) is not the same as the ratio of 2 and 1 i.e., 1 2 or 2 : 1. Instead of the numerals, we may also use literal numbers to form a fraction or ratio. For instance, if a and b denote two numbers and b ≠ 0, we can divide a by b and write the ratio as a b or a : b. Suppose a:b a ratio that is unchanged if both of its terms are multiplied and divided by the same non-zero quantity m then, a : b = ma:mb and a:b = a m : b m . Of the two numbers in a ratio, the first one or the one above the horizontal bar is called the numerator or antecedent and the latter one or below the horizontal bar is called the denominator or consequent. A ratio is called greater in equality if the numerator or antecedent is larger. It is said to be lesser in equality if the numerator is less than the denominator or consequent. We thus have an in-equality between two numbers when the numerator and denominators are not equal. Sometimes we form a ratio from two or more given ratios just by taking the ratio of the product, the numerators to the product of the denominators. Such a ratio is called compound ratio of the given ratios. Thus, (i) Given the ratios 1 2 and 5 3 , then their compound ratio is 1 × 5 2 × 3 or 5 6 . (ii) Given the ratios a b and c d , then their compound ratio is a × b c × d . Ratio in the Lowest Term To eliminate the common factors from the antecedent and consequent of a ratio is the ratio in the simplest form. It is the ratio in the simplest form. Comparison on Ratios To compare the antecedents of the given ratios with the same consequents as we do in the case of fractions.

122 The Leading Mathematics - 7 CLASSWORK EXAMPLES Example 1 Find the compounded ratio of 4 5 and 15 8 and then reduce it to the lowest term. Solution: Here, The given ratios are 4 5 and 15 8 . So, the compound ratio = 4 × 15 5 × 8 = 3 2. Example 2 Arrange the following ratios in descending order: 2 : 3, 3 : 4, 5 : 6, 1 : 5 Solution: Here, The given ratios are 2:3, 3:4, 5:6, 1:5. The LCM of 3, 4, 6, 5 is 2 × 2 × 3 × 5 = 60 Now, 2:3 = (2 × 20) : (3 × 20) = 40:60 3:4 = (3 × 15) : (4 × 15) = 45:60 5:6 = (5 × 10) : (6 × 10) = 50:60 1:5 = (1 × 12) : (5 × 12) = 12:60 Clearly, 50:60 > 45:60 > 40:60 > 12:60 Therefore, 5:6 > 3:4 > 2:3 > 1:5 Arranging the given ratios in descending order as 5 : 6, 3 : 4, 2 : 3, 1 : 5. Example 3 If x : y = 1 : 2, find the value of (2x + 3y) : (x + 4y). Solution: Here, x : y = 1 : 2 means x y = 1 2 Now, (2x + 3y) : (x + 4y) = 2x + 3y x + 4y Dividing numerator and denominator by y, we get 2x + 3y y x + 4y y = 2. x y + 3 x y + 4 = 2. 1 2 + 3 1 2 + 4 = 1 + 3 1 + 8 2 = 4 × 2 9 = 8 9 Therefore, the value of (2x + 3y) : (x + 4y) is 8 : 9.

Arithmetic 123 Example 4 If 2a = 3b = 4C, find a : b : c. Solution: Here, Let 2a = 3b = 4c = x So, a = x 2 , b = x 3 , c = x 4 The LCM of 2, 3 and 4 is 12. Therefore, a : b : c = x 2 × 12 : x 3 × 12 : x 4 × 12 = 6x : 4x : 3x = 6 : 4 : 3 Therefore, a : b : c is 6 : 4 : 3. Example 5 Two numbers are in the ratio 3 : 4. If the sum of numbers is 63, find the numbers. Solution: Here, Sum of the terms of the ratio = 3 + 4 = 7 Sum of the numbers = 63 Therefore, the first number = 3 7 × 63 = 27 The second number = 4 7 × 63 = 36 Therefore, the two numbers are 27 and 36. Example 6 The ratio of number of boys and girls is 4 : 3. If there are 18 girls in a class, find the number of boys in the class and the total number of students in the class. Solution: Here, Number of girls in the class = 18 Ratio of boys and girls = 4 : 3 According to the question, No. of Boys No. of Girls = 4 3 or, No. of Boys 18 = 4 3 or, No. of Boys = 4 × 18 3 = 24 Therefore, total number of students is 24 + 18 = 42.

124 The Leading Mathematics - 7 Example 7 What must be added to each term of the ratio 2 : 3 so that it may become equal to 4 : 5? Solution: Let the number to be added be x, then (2 + x) : (3 + x) = 4 : 5 ⇒ 2 + x 5 + x = 4 5 or, 5(2 + x) = 4(3 + x) or, 10 + 5x = 12 + 4x or, 5x – 4x = 12 – 10 or, x = 2 Thus, the required number is 2. Example 8 Mother divided some money among Runa, Sama and Maria in the ratio 2 : 3 : 5. Maria got Rs. 150. (a) Calculate the money received by Runa and Sama. (b) Find the total amount. Solution: (a) Let the money received by Runa, Sama and Maria be 2x, 3x and 5x respectively. Given that, Maria got Rs. 150. Therefore, 5x = 150 or, x = 150 ÷ 5 = 30 So, Runa got = 2x = Rs. 2 × 30 = Rs. 60 Sama got = 3x = Rs. 3 × 60 = Rs. 90 (b) The total amount = Rs. (30 + 60 + 90) = Rs. 180. EXERCISE 6.1 Your mastery depends on practice. Practice like you play. 1. Write the ratio of the following pairs of quantities and then express in the simplest form. (a) 25 cm and 30 cm (b) 1 m and 75 cm (c) 2 kg and 1 kg 150 gm (d) 1.5 km and 850 m (e) 5 minutes and 50 second (f) 25 minutes and 10 minutes (g) 8 months and 1 year and 2 months (h) Rs. 2.5 and 80 paisa 2. Express each of the following ratios in the lowest term. (a) 24 : 30 (b) 48 : 32 (c) 3.5 : 2.8 (d) 1 4 : 5 6 (e) 4 : 6 : 8 (f) 3.2 : 2.4 : 8

Arithmetic 125 3. Find the compounded ratios of the following ratios and then reduce them to the lowest terms: (a) 3 : 5 and 25 : 9 (b) 15 : 28 and 42 : 25 4. Which ratio is greater? (a) 1: 2 and 3: 5 (b) 3 : 4 and 5 : 9 5. Arrange the following ratios in ascending order: (a) 1 : 2, 3 : 4, 1 : 6, 9 : 8 (b) 2 : 5, 3 : 10, 5 : 6, 8 : 15, 2 : 3 6. Arrange the following ratios in descending order. (a) 7 : 8, 3 : 4, 5 : 6, 11 : 12 (b) 5 : 3, 3 : 8, 7: 6, 7 : 12 , 5 : 24 7. (a) If a : b = 1 : 2, find the value of (2a + 3b) : (3a + 2b). (b) If x : y = 2 : 3, find the value of (4x – 3y) : (2x + 3y) 8. (a) If p : q = 1 : 2 and q : r = 4 : 5 find p : r (b) If x : y = 3 : 5 and y : z = 5 : 9. find x : z 9. (a) If a : b = 2 : 3 and b : c = 4 : 9, find a : b : c (b) If a : b = 5 : 4 and b : c = 5 : 6, find a : b : c 10. (a) If 2A = 3B = 6C, find A : B : C (b) If 3x = 5y = 7z, find x : y : z 11. (a) Divide 375 in the ratio 2 : 3 (b) Divide Rs.1375 in the ratio 5 : 6 12. (a) Divide 455 in the ratio 2 : 5 : 6 (b) Divide Rs.1750 in the ratio 3 : 7 : 4 13. (a) Two numbers are in the ratio 4:5. If the sum of the numbers is 225, find the numbers. (b) The ages of two brothers is 3:7. If the sum of their ages is 120 years, find their ages. 14. (a) There are 600 pupils in a school. The ratio of boys and girls in the school is 3 : 5. How many girls and boys are in the school? (b) The perimeter of a rectangle is equal to 280cm. The ratio of its length and breath is 5 : 2 . Find the area of the rectangle. 15. (a) The ratio of two numbers is 2 : 7. If the first number is 24, find the second number. (b) The ratio of the number of boys and girls in a class is 3 : 5. If the class has 45 girls, find the number of boys in the class and total number of students. 16. (a) A math class has 25 students, of which 11 are males and the rest are female. What is the ratio of males and females? (b) A group of youths has 25 boys and 23 girls. What is the ratio of girls to all the youths?

126 The Leading Mathematics - 7 17. (a) What must be added to each term of the ratio 1 : 4, so that it may becomes equal to 2 : 3 ? (b) What must be subtracted from each term of the ratio 8: 11 so that the ratio becomes 1 : 2 ? 18. (a) Two numbers are in the ratio 7 : 12. If 3 is added to each term of the numbers, the ratio becomes 2 : 3. Find the numbers. (b) Two numbers are in the ratio 5 : 11. On subtracting 2 from each number, the ratio will be 1 : 3, find the numbers. 19. (a) Dasharath divided some money among Ram, Bharat and Laxman in the ratio 5 : 4 : 2. Laxman got Rs. 2500. (i) Find the total amount distributed by Dasharath. (ii) Find the money received by Ram and Bharat. (b) A bag contains three coloured marbles of red, yellow and blue in the ratio of 7 : 5 : 8. It has 85 yellow marbles. (i) Find the total marbles in the bag. (ii) Find the number of red marbles and blue marbles. ANSWERS 1. (a) 5 : 6 (b) 4 : 3 (c) 40 : 23 (d) 30 : 17 (e) 6 :1 (f) 5 : 2 (g) 2 : 3 (h) 125 : 4 2. (a) 4 : 5 (b) 3 : 2 (c) 5 : 4 (d) 3 : 10 (e) 2 : 3 : 4 (f) 4 : 3 : 10 3. (a) 5 : 3 (b) 9 : 10 4. (a) 3 : 5 (b) 3 : 4 5. (a) 1 : 6, 1 : 2, 3 : 4, 9 : 8 (b) 3 : 10, 8 : 15, 2 : 5, 2 : 3, 5 : 6 6. (a) 11 : 12, 7 : 8, 5 : 6, 3 : 4 (b) 5 : 3, 7 : 6, 7 : 12, 3 : 8, 5 : 24 7. (a) 8 : 7 (b) – 1 : 13 8. (a) 2 : 5 (b) 1 : 3 9. (a) 8 : 12 : 27 (b) 25 : 20 : 24 10. (a) 3 : 2 : 1 (b) 35 : 21 : 15 11. (a) 150, 225 (b) Rs. 625, Rs. 750 12. (a) 17, 175, 210 (b) Rs. 375, Rs. 857, Rs. 500 13. (a) 100, 125 (b) 36 yrs, 84 yrs 14. (a) 375, 225 (b) 4000 cm2 15. (a) 84 (b) 27, 72 16. (a) 11 : 14 (b) 23 : 48 17. (a) 5 (b) 5 18. (a) 7, 12 (b) 5, 11 19. (a) Rs.13750, Rs. 6250, Rs. 5000 (b) 340, 119, 136

Arithmetic 127 6.2 Proportion At the end of this topic, the students will be able to: ¾ solve the problems related to proportion. Learning Objectives Proportion and Continued Proportion Given four numbers in succession, if the ratio of the first to the second is equal to that of the third to the fourth, the four numbers are said to be in proportion or proportional. Thus, i) 1, 2, 4, 8 are in proportion since 1 2 = 4 8 , ii) 4, 3, 8, 6 are proportional since 3 4 = 6 8 . The idea of proportion is exhibited diagrammatically by the following figures: They are also equivalent fractions. In particular, four quantities a, b, c, d are said to be proportional if a b = c d , or, a:b = c:d or, a:b :: c:d Here, a and d are called extremes and b and c are called means. The product of extremes is equal to the product of means. If, instead of four quantities, we have three quantities in succession such that the ratio of the first to the second is the same as that of the second to the third, the three numbers are said to be in continued proportion. In particular, three quantities a, b and c are said to be in continued proportion if, a b = b c. Here, a is the first proportional to b and c, b is the mean of the continued proportion, and c is the third proportional to a and b. For instance, the three numbers 1, 2, 4 are in continued proportion. Here, 1 2 = 2 4 . Direct and Indirect Proportion Proportions are said to be direct if the ratio of one kind (i.e., first pair of numbers) is equal to the ratio of another kind (i.e., second pair of numbers). 3 4 = 6 8

128 The Leading Mathematics - 7 For instance, consider the four numbers 1, 2, 4, 8. Here, the ratio of the first to the second is the same as that of the third to the fourth, i.e., 1 2 = 4 8. Proportions are said to be indirect or inverse if the ratio of one kind (i.e., first pair of numbers) is equal to the inverse of the ratio of another kind (i.e., second pair of numbers). For instance, consider the four numbers 1, 2, 8, 4. Here, the ratio of the first to the second is the same as the inverse of the ratio of the third to the fourth, i.e., or, 1 2 = 4 8 1 2 = 1 8 4 or, 1 2 = 4 8 . CLASSWORK EXAMPLES Example 1 If the four numbers 3, 4, x and 16 are proportional, find x. Solution: Here, since 3, 4, x and 16 are proportional so, 3 4 = x 16 Multiplying both sides by 16, we get 16 × 3 4 = 16 × x 16 or, 4 × 3 = x or, x = 12. Example 2 Show that the three numbers 3, 12 and 48 are in continued proportion. Solution: Here, given numbers are 3, 12 and 48. Then, 3 12 = 1 4 and 12 48 = 1 4 Thus, 3 12 = 1 4 = 12 48 Hence, 3, 12 and 48 are in continued proportion. Example 3 The first, second and third terms of the proportion are 42, 36, 35. Find the fourth term. Solution: Here, Let the fourth term be x. Thus 42, 36, 35, x are in proportion. Product of extreme terms = 42 × x Product of mean terms = 36 × 35

Arithmetic 129 Since, the numbers make up a proportion Therefore, 42 × x = 36 × 35 or, x = (36 × 35) ÷ 42 or, x = 30 Therefore, the fourth term of the proportion is 30. Example 4 Find the third proportional of 16 and 20. Solution: Let the third proportional of 16 and 20 be x. Then 16, 20, x are in proportion. This means 16 : 20 = 20 : x So, 16 × x = 20 × 20 x = 20 × 20 16 = 25 Therefore, the third proportional of 16 and 20 is 25. Example 5 If the cost of 10 kg of sugar is Rs 350, find the cost of 15 kg of sugar assuming that the changes in quantity and price are directly proportional. Solution: Here, Sugar Price 10 kg (Given) Rs 350 (Given) 15 kg (Given) x (Required) Sincerely, the changes are directly proportional, we have 10 15 = 350 x or, x = 10 15 × 350 = 15 × 35 = 525 ∴ The required cost is Rs 350. Example 6 A store has food for 200 persons for 30 days. If the number of persons and the number of days vary inversely, find how long the food will last for 150 persons. Solution: Here, Number of persons Food 120 (Given) 30 (Given) 150 (Given) x (Required)

130 The Leading Mathematics - 7 Sincerely, the changes are inversely proportional, we have 200 150 = 30 x or, 30 x = 200 150 or, x = 200 150 × 30 = 40 That is, the required number of days = 40. EXERCISE 6.2 Your mastery depends on practice. Practice like you play. 1. Show that the following triples of numbers are in continued proportion. (a) 4, 6, 8 and 12 (b) − 7, 35 and − 175 2. Find x if each of the following four numbers are proportional: (a) 1, 3, x and 9 (b) 1, x, 2 and 10 (c) x, 3, 6 and 9 (b) 2, − 15, 6 and x 3. Show that the following triples of numbers are in continued proportion. (a) 4, 16 and 64 (b) − 7, 35 and − 175 4. Find the mean proportional (i.e, x) of the following triples of numbers. (a) 3, x, 48 (b) − 7, x, − 175 5. (a) Find the third proportion to 8 and 12. (b) Find the mean proportional between 6 and 24. 6. (a) What number must be added to the first number among the numbers 1, 4, 6 and 12 so that they form in proportion? (b) What number must be subtracted from the fourth number among the numbers 4, 12, 14 and 38 so that they form in proportion? 7. Direct proportion cases: (a) 20 gm of hydrogen combines with 320 gm of oxygen to form water. Find how much hydrogen combines with 32 gm of oxygen to form water. (b) The ratio of the length to the breadth of a room is 4 : 3. If the breadth is 4 m, find the length of the room.

Arithmetic 131 (c) A gold ornament is made of 12 carat gold (i.e. ratio of gold to copper is 1 : 1). If the ornament is made of pure gold, find what carat gold it will be. (d) If father’s age is three times that of his son, find the age of the father when his son is 20 years old. (e) If the ratio of water to chemical essence in coca-cola is 99 :1 by volume, find how much water is there in 1 litre (1000 cc) of coco-cola. 8. Indirect or inverse proportional cases: (a) 1 man does a work in 10 days. In how many days 5 men will do the same work? (b) A room requires 6 m of 2 m broad carpet. Find the length of the carpet required if the breadth of a carpet is 3 m. (c) A food store has food for 100 men for 15 days. How long the same food stuff will last if there were 150 men.? (d) In a journey of 8 hours, Rijan drives at a rate of 40 km/hr. What should be his peed if he wants to complete in 6 hours? (e) If the rate of interest is 5 percent per annum, he has to pay a certain amount of interest in 3 years. If the rate of interest is 3 per cent per annum, in how many years he will have to pay the same amount of interest? ANSWERS 2. (a) 3 (b) 5 (c) 2 (d) – 45 4. (a) 12 (b) 35 5. (a) 18 (b) 12 6. (a) 1 (b) – 4 7. (a) 2 gm (b) 3 m (c) 24 (d) 60 yrs (e) 990 ml 8. (a) 2 days (b) 4 m (c) 10 days (d) 53.33 km/hr (e) 5 yrs Project Work 6.2 Take two circular papers of the same size and fold them in different parts such that they are equal in ratios. Present them in your classroom.

132 The Leading Mathematics - 7 CHAPTER 7 Profit and Loss Lesson Topics Pages 7.1 Profit and loss 133 ” What is the open market price of one umbrella? ” How much does the shopkeeper sell the umbrella to the customer for? ” Does the shopkeeper sell one umbrella for more or less price than the open market price? ” How much amount does she get if she sells 10 umbrellas? ” What is the cost of the sold cap? ” How much amount does the cap seller get if he sells 20 such caps? ” What is the price of sugar? WARM-UP Market price : 560 Selling price : 620 Market price : 270 Selling price : 250 Sugar market price : 90 Sugar selling price : 95

Arithmetic 133 7.1 Profit and loss At the end of this topic, the students will be able to: ¾ solve the problems related to profit and loss. Learning Objectives Introduction Today, man and money are almost inseparable. Man uses money. Thus, money has a very important place in society. So, man tries to get more and more money. For this, one may invest some money and get back more than what he has invested or do more income generating activities. One way of doing so is to buy things and then sell at a higher price. In other words, buying and selling could help earn more money. In general, a shopkeeper does this kind of activity. He buys things or goods from a factory, a whole-seller or from someone else by paying certain amount of money. Quite often he sells them at a higher price. Sometimes, he will have to sell at a lower price also. In the former case, he gets more than what he invested or paid in buying the things or goods; whereas in the latter case he gets less. In other words, he is said to have made profit or gain in the first case and loss in the second case. For practical purposes, we accept the following definitions: Cost Price The amount of money paid or spent or invested by a shopkeeper while buying certain things or goods from a factory or whole-seller or somebody for the purpose of selling is called the cost price for the shopkeeper. It is denoted by CP. Selling Price The amount of money at which certain articles or goods are sold by a shopkeeper is known as the selling price. It is denoted by SP. Profit or Gain If the selling price is more than the cost price, the shopkeeper is said to have made a profit or gain. The excess amount, that is, the difference between the selling price and the cost price gives the actual amount gained. In symbols, Profit = SP − CP i.e., If SP > CP, then P = SP – CP ⇒ SP = CP + P and CP = SP - P

134 The Leading Mathematics - 7 Loss If the selling price is less than the cost price, the shopkeeper is said to have suffered a loss. Then, the loss is the difference between the cost price and selling price. In symbols, Loss = CP − SP i.e., If SP < CP, then L = CP – SP ⇒ SP = CP – L and CP = SP + P Profit and loss are often expressed in percentage. Such a percentage is always calculated on the cost price of the article and never on the number of articles and selling price. We thus have, Profit percentage (P%) = Profit CP × 100% = SP – CP CP × 100% Loss percentage (L%) = Loss CP × 100% = CP – SP CP × 100% CLASSWORK EXAMPLES Example 1 Determine the unknown quantity: (a) CP = Rs. 500, Profit = Rs. 50 . SP = ? (b) Profit = Rs. 60 , SP = Rs. 560 , CP = ? (c) SP = Rs. 650, CP = Rs. 600 , Profit = ? Solution: (a) Here, CP = Rs. 500, Profit = Rs. 50 . So, SP = CP + Profit = Rs. 500 + Rs. 50 = Rs. 550. (b) Here, Profit = Rs. 60, SP = Rs. 560, So, CP = SP − Profit = Rs. 560 − Rs. 60 (c) Here, SP = Rs. 650, CP = Rs. 600, So, profit = SP − CP = Rs. 650 − Rs. 600 = Rs. 50. Example 2 Find the unknown quantity: (a) CP = Rs. 400, SP = Rs. 450, Profit percentage = ? (b) CP = Rs. 500, SP = Rs. 450, Loss percentage = ? (c) CP = Rs. 400, Profit percentage = 25%, SP = ? Solution: (a) Here, CP = Rs. 400 , SP = Rs. 450, So, profit = SP − CP = Rs. 450 − Rs. 400 = Rs. 50. So, profit percentage (P%) = Profit CP × 100 = 50 400 × 100 = 12 1 2 %

Arithmetic 135 (b) Here, CP = Rs. 500, SP = Rs. 450, So, loss = CP − SP or, loss = Rs. 500 − Rs. 450 or, loss = Rs. 50 Loss percentage (L%) = Loss CP × 100 = 50 500 × 100 = 10 % (c) Here, CP = Rs. 400, Profit percentage = 25%, Since profit percentage = Profit CP × 100 we have 25 = Profit 400 × 100 or, 25 × 4 = Profit, i.e., profit = Rs. 100. So, SP = CP + Profit = Rs. 400 + Rs. 100 = Rs. 500. Example 3 A watch is bought for Rs. 25 and is sold for Rs. 30. Find the profit and profit percent. Solution: Here, CP = Rs. 100, SP = Rs. 120. So, profit = SP − CP = Rs. 120 − Rs. 100 = Rs. 20. Also, profit percentage = Profit CP × 100 = 20 100 × 100 = 20 %. Example 4 Ronald buys a geyser for Rs. 3680 and sells it at a gain of 7¹/₂ %. For how much does he sell it? Solution: Here, CP of the geyser = Rs. 3680. Gain % = 7 1 2 % = 15 2 %. Therefore, SP of the geyser = 100 + gain% 100 × CP = 100 + 15 2 100 × 3680 = 215 200 × 3680 = Rs. 3956 Hence, Ronald sells the geyser for Rs. 3956.

136 The Leading Mathematics - 7 Example 5 The cost of 11 pencils is equal to the selling price of 10 pencils. Find the loss or profit percent, whatever may be the cost of 1 pencil. Solution: Here, the cost price of 11 pencils = SP of 10 pencils Let CP of 1 pencil be Re. 1 CP of 10 pencils = Rs. 10 SP of 10 pencils = CP of 11 pencils = Rs. 11 Profit on 10 pencils = 11 – 10 = Re. 1 Profit % = 1 10 × 100 = 10 % Example 6 A man sold an article for Rs. 280 at a loss of 16% of the cost price. Find the cost price. Solution: Here, SP = Rs. 280. Let CP = x. Loss = 16% of the cost price = 16 100 × x = 16x 100 Since CP − Loss = SP , we have x − 16x 100 = 280 or, 100x – 16x 100 = 280 , or, 84x = 28000 or, x = 24000 84 Hence, x = Rs. 333.33. EXERCISE 7.1 Your mastery depends on practice. Practice like you play. 1. Find the profit or loss: S.N. Cost Price Selling Price Profit Loss (a) Rs. 240 Rs. 247 (b) Rs. 245 Rs. 425 (c) Rs. 3421 Rs. 2341 (d) Rs. 3479 Rs. 4379 (e) Rs. 12345 Rs. 12345

Arithmetic 137 2. Determine the unknown quantity: S.N. Cost Price Selling Price Profit Loss (a) Rs. 46 ........ Rs. 12 × (b) Rs. 435 Rs. 465 ........ × (c) ........... Rs. 8765 Rs. 2145 (d) Rs. 5049 .......... × Rs. 437 (e) .......... Rs. 2345 × Rs. 1320 (f) Rs. 4509 .......... Rs. 1025 (g) .......... Rs. 2345 Rs. 569 (h) Rs. 2049 .......... × Rs. 456 (i) .......... Rs. 7345 × Rs. 1320 (j) Rs. 6320 Rs. 6230 × .......... 3. Find the profit or loss percentage in the following cases: (a) Cost Price = Rs. 350 and Selling Price = Rs. 400 (b) Cost Price = Rs. 250 and Selling Price = Rs. 235 (c) Cost Price = Rs. 300 and Profit = Rs. 75 (d) Cost Price = Rs. 720 and Loss = Rs. 63 (e) Selling Price =Rs. 1518 and Loss = Rs. 132 (f) Selling price = Rs. 540 and Gain = Rs. 60 4. Find the cost price when: (a) Selling Price = Rs. 795 and Profit % = 6% (b) Selling Price = Rs. 980 and Loss % = 12% (c) Selling Price = Rs. 66.95 and Gain % = 15.25% (d) Selling Price = Rs. 2640 and Loss = Rs. 196 5. Find the selling price when: (a) Cost Price = Rs. 875 and Profit % = 5% (b) Cost Price = Rs. 750 and Loss % = 15%

138 The Leading Mathematics - 7 (c) Cost Price = Rs. 480 and Gain % = 12 1 2% (d) Cost Price = Rs. 675 and Loss % = 92% 6. (a) Aananda purchased 120 pencils at the rate of Rs.2 per pencil. He sold 72 of them at the rate of Rs.2.5 per pencil and the remaining at the rate of Rs. 2 per pencil. Find his profit or loss per cent. (b) Mohan sold two horses for Rs. 18000 each. On one he gained 20% and on the other he lost 20%. Find his total gain or loss. 7. (a) On selling of fan for Rs. 810, Sam gains 8%. For how much did he purchase it? (b) A bed sheet was sold for Rs.483 thereby gaining 15%. Find the CP of the bed sheet. 8. (a) Amir bought an almirah for Rs.1520 and sold it at a profit of 12 1 2 %. Find the selling price of the almirah. (b) John buys a calculator for Rs. 720 and sells it at a loss of 6 1 2 %. For how much does he sell it? 9. (a) By selling a stool for Rs. 240 a carpenter loses 20%. How much percent would he/she gain or lose by selling it for Rs. 360? (b) A television is sold for Rs. 3120 at a loss of 4%. Find the gain or loss percent if the television is sold for Rs. 3445. 10. (a) On selling a bat for Rs. 371, a man gains 6%. For how much should he sell it to gain 8%? (b) By selling a camera for Rs. 2400, Ron loses 4%. At what price must he sell it to gain 12 %. 11. (a) A television set was bought for Rs. 3900. Rs. 200 was spent on transportation and Rs. 900 on repair. It was sold at a loss of 10%. Find the S.P. of television. (b) Samir buys 40 kg of wheat at Rs. 6.25 per kg and 30 kg of wheat at Rs. 7 per kg. At what rate per kg should he sell the mixture to gain 5% on the whole?  12. (a) A dealer sold a camera for Rs. 1080 gaining 1 8 of its cost price. Find: (i) the cost price of the camera, and (ii) the gain per cent earned by the dealer.

Arithmetic 139 (b) Merina sells a pen for Rs. 54 and loses 1 10 of her outlay. Find: (i) the cost price of the pen, and (ii) the loss percent. 13. (a) The selling price of 12 eggs is equal to the cost price of 15 eggs. Find the gain percent. (b) The cost price of 10 bananas is equal to the SP of 12 bananas. Find the loss percent. ANSWERS 1. (a) Rs. 7 (Profit) (b) Rs. 20 (Loss) (c) Rs. 1080 (Loss) (d) Rs. 900 (Profit) (e) Rs. 0 (neither profit nor loss) 2. (a) Rs. 52 (b) Rs. 30 (c) Rs. 6620 (d) Rs. 41612 (e) Rs. 3665 (f) Rs. 5534 (g) Rs. 1776 (h) Rs. 1593 (i) Rs. 8665 (j) Rs. 90 3. (a) 142 7 % Profit (b) 6% Loss (c) 25% (d) 83 4 % (e) 8% (f) 121 2 % 4. (a) Rs. 750 (b) Rs. 1113.64 (c) Rs. 58.09 (d) Rs. 2836 5. (a) Rs. 918.75 (b) Rs. 637.50 (c) Rs. 540 (d) Rs. 54 6. (a) 15% Profit (b) 41 6 % Loss 7. (a) Rs. 750 (b) Rs. 420 8. (a) Rs. 1710 (b) Rs. 673.20 9. (a) 20% Profit (b) 6% Profit 10. (a) Rs. 378 (b) Rs. 2800 11. (a) Rs. 4500 (b) Rs.6.90 per kg 12. (a) Rs. 960, 121 2 % (b) Rs. 60, 10% 13. (a) 25% (b) 162 3 % Project Work 7.1 Ask the price of any five same goods from two shops and write them. From which shop and which goods do you buy to get the cheapest and by how many percent ? Present them in your classroom.

140 The Leading Mathematics - 7 CHAPTER 8 Unitary Method Lesson Topics Pages 8.1 Unitary Method 141 ” What is the price of a book? ” How many books are there in a bundle? ” How much amount would pay for those books? ” How many Chocofuns are there in a packet? ” What is the cost of a packet of Chocofuns? ” What is the cost of a Chocofun? ” How many Chocofuns are there in a jar? ” What is the total cost of the 3 packets of Chocofuns? WARM-UP Price : Rs. 493 Price for 7 books = ? Rs. 60/- Total 60 pcs

Arithmetic 141 8.1 Unitary method At the end of this topic, the students will be able to: ¾ solve the simple problems of two variables of unitary method based on direct and indirect variations. Learning Objectives In our daily life, we transact about many things. When we know the cost of one article, we need to find the cost of more articles. Similarly, when we know the cost of more articles, we need to find the cost of one article. We can solve these types of problems in our dayto-day life. This is known as Unitary Method. For examples, i) If the daily wage of a staff is Rs. 1200, how much daily wages of 5 staffs is ? ii) A drive drives 63 km in an hour. How long does he/she drive in 12 hours? iii) If the cost of a dozen balls is Rs. 600, what is the cost of 1 ball? iv) If 3 girls complete a work in 5 days, find how many girls complete the same work in 1 day. v) The cost of 5 kg of apples is Rs. 1250, find the cost of 8 kg of apples. vi) If 1 man complete a work in 10 days, find how many men complete the same work in 20 days. In the examples (i) and (ii), how one unit of one quantity is related to more than one unit of another; in the examples (iii) and (iv), more than one unit of one quantity is related to one unit of another quantity, but in the examples (v) and (vi), more than one unit of one quantity is related to more units of another quantity, The unitary method is a technique in mathematics for solving a problem by finding the value of single unit, i.e., 1, (by dividing) and then finding the necessary value by multiplying the single unit value. As noted above, we meet many kinds of relations. But, we consider only those cases in which two quantities are related to each other such that; i) One quantity increases when the other increases and vice versa. This is called direction variation. ii) One quantity increases when the other decreases and vice versa. This is called indirection variation.

142 The Leading Mathematics - 7 Direct Variation or Proportion Look at the table mentioned the wage of a staff in various days as follows: No. of days 1 2 4 5 9 8 7 6 3 Income (Rs.) 1200 2400 4800 6000 10800 9600 8400 7200 3600 From the above table, if one quantity increases or decreases in a certain ratio, the other quantity also increases or decreases in the same ratio then such type of quantities are called in direct variation or proportion. Working Rule For two quantities related to each other such that one increases in the same ratio as the other: Given: 1 unit of a quantity is related to a unit of another quantity, Then, x units of the first is related to x × a units of the second. In the above table, the wage of a staff for 1 day is given, and that of days is found. The number of days increased in the ratio 1 to 2, so did the price, That is, 1 day 2 days = Rs. 1200 Rs. 2400 or, 1 2 = 1200 2400 . We may therefore take the following as an alternative way of expressing the rule described above: Variables: A B First value: a (Given) b (Given) Second value: x (Required) d (Given) Then,a x = b d . By cross multiplication, we get x = a × d b x × b = a × d Indirect Variation or Proportion Look at the table mentioned the time taken to work by certain men as follow: No. of Men 1 2 4 5 8 6 7 9 3 Days 20 10 5 4 2.5 3 2 3 2 6 7 2 2 9 6 2 3 From the above table, if one quantity increases or decreases in a certain ratio, the other quantity decreases or increases in the same ratio then such type of quantities are called in indirect variation or proportion.

Arithmetic 143 Working Rule For two quantities related to each other such that one increases in the same ratio as the other decreases: Given: 1 unit of a quantity is related to a units of another quantity, Then, x units of the first is related to a x units of the second. In the above table, the working days to work by certain men is given, and that of men is found. The number of men increased in the ratio 1 to 2, so did the price. That is, 1 man 2 man = 10 days 20 days or, 1 2 = 10 20 It is reverse or reciprocal relation between the number of men and the working days. We may therefore take the following as an alternative way of expressing the rule described above: Variables A B First value a (Given) b (Given) Second value x (Required) d (Given) Then, a x = d b . By cross multiplication, we get x = a × d b x × d = a × b CLASSWORK EXAMPLES Example 1 If the selling price of a mobile is Rs. 21000, find the selling price such one dozen mobiles. Solution: Here, The selling price of a mobile is Rs. 21000. ∴ The selling price of one dozen mobiles is Rs. 21000 × 12 = Rs. 252000. “Alternatively” Let the selling price of one dozen mobiles (12 mobiles) be Rs. x. Then, Quantity of potatoes (kg) Cost of potatoes (Rs.) 1 21000 12 x

144 The Leading Mathematics - 7 The quantity of mobiles (pcs) and the selling price of mobiles (Rs.) are in direct variation. So, 1 12 = 21000 x or, x = 21000 × 12 or, x = 252000 ∴ The selling price of one dozen mobiles is Rs. 252000. Example 2 If the cost of 10 kg of potatoes is Rs. 800, find the cost of 1 kg of potatoes. Solution: Here, The cost of 10 kg of potatoes is Rs. 800. or, The cost of 1 kg of potatoes is Rs. 800 10 = Rs. 80. “Alternatively” Let the required cost of 1 kg of potatoes be Rs. x. Then, Quantity of potatoes (kg) Cost of potatoes (Rs.) 10 800 1 x The quantity of potatoes (kg) and the cost of potatoes (Rs.) are in direct variation. So, 10 1 = 800 x or, x = 1 × 800 10 => x = 80 ∴ The cost of 1 kg of potatoes is Rs. 80. Example 3 10 workers complete a work in 12 days. (a) Find how many days 1 worker complete the same work. (b) Find how many days 40 workers complete the same work. Solution: (a) Here, 10 workers complete a work in 12 days. or, 1 worker complete the same work in 12 × 10 days = 120 days. (b) 40 workers complete the same work in 120 40 = 3 days. “Alternatively”

Arithmetic 145 Number of workers Working days 10 (Given) 12 (Given) 40 (Given) x (Required) Since, the changes are inversely proportional, we have 10 40 = x 12 or, x = 10 × 12 40 or, x = 3 That is, 40 workers complete the same work in 3 days. EXERCISE 8.1 Your mastery depends on practice. Practice like you play. 1. Identify whether each of the following is a problem of direct or indirect variation. (a) Number of pens and the cost of pens (b) Length and breadth of the rectangles of equal perimeters (c) Number of employees and working days (d) Capacity of tap to empty water and time (e) Oranges tree and number of oranges (f) People and their legs 2. (a) If 1 worker earns Rs. 125 per hour, find how much will he/she earn in 12 hours. (b) If the cost of a pencil is Rs. 15, find the cost of 15 pencils. 3. (a) The cost of 4 note copies is Rs. 180. Find the cost of 1 one note copy. (b) The cost of 1 dozen of Maths books is Rs. 6204. Find the cost of 1 Maths book. 4. (a) The cost of 1 gross of pencils is Rs. 2160. Find the cost of 1 pencil. (b) The cost of 1 quintal of steel rod is Rs. 11300. Find the cost of 1 kg of steel rod. 5. (a) A store has food-stuff for 6 men for 10 days. How long will the same food-stuff last for 1 man? (b) A food store has food for 150 men for 20 days. For how many men will the same food stuff be enough for 1 day?

146 The Leading Mathematics - 7 6. (a) A tourist travels 9 km in 3 hours. (i) How far will s/he go in 1 hour ? (ii) How far will s/he go in 7 hours ? (b) A tap can fill a 500 litres water tank in 5 minutes. (i) How many seconds will it take to fill 1 litre of water ? (ii) How long time will it take to fill a 5000 litre tank? 7. (a) A man can do a work in 15 days. (i) How many men complete the same work in 1 day? Find it. (ii) In how many days will 5 men do the same work? (b) A pipe can fill a water tank in 25 minutes. (i) How many pipes can fill the same tank in 1 minute? (ii) How many pipes can fill the same tank in 10 minutes? 8. (a) The weight of 12 sacks of rice is 60 kg. (i) What is the weight of 25 such sacks of rice ? (ii) How many do such sacks of rice weigh 15 kg? (b) Shova bought a dozen of pens for Rs. 540. (i) Find the cost of 20 such pens (ii) How many pens can be bought for Rs. 360? ANSWERS 2. (a) Rs. 150000 (b) Rs. 225 3. (a) Rs. 45 (b) Rs. 517 4. (a) Rs. 15 (b) Rs. 113 5. (a) 6 days (b) 3000 men 6. (a) 3 km, 21 km (b) 0.6 seconds, 50 minutes 7. (a) 15 men, 3 days (b) 25 pipes, 21 2 pipes 8. (a) 125 kg, 3 scales (b) Rs. 900, 8 pens Project Work 8.1 Write any 5 examples related to direct and indirect variations each used in our daily activities by asking your guardian or searching in Internet and present them in your classroom.

Arithmetic 147 Read, Understand, Think and Do 1. The numbered cards from 61 to 90 are given below. (a) List the square number and find their square root by using division method. (b) Find the cube root of 64 by using prime factorization. (c) Prove that HCF of 64 and 81 is 1. 2. A shopkeeper bought two shirts with Rs. 1680. The prices of the two shirts are given in the figure. (a) Find the HCF and LCM of 720 and 960 by factorization method. (b) Prove that the ratio of prices of two shirts = 3 4. (c) How much expensive is shirt B than shirt A in percentage? Calculate it. (d) A shopkeeper sold two shirts in Rs.2100, find his gain or loss percentage. 3. Ramila studies in grade 7 and she got 300 marks out of 600 marks but Ramesh studies in grade 10 and he got 350 marks out of 700 marks. (a) If the total marks of getting Ramila converting into 60 full marks, how much does she get? Find it. (b) Who got more percentage marks? Compare it. 4. Ramila is an officer of a company. Her salary is Rs. 42000. She spends 2 5 part of her income in food and 1 3 part of her income in education. (a) In which title does she spend more between food and education? (b) What is her monthly saving? Find it. (c) Write any two rational numbers between 2 5 and 2 6. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 Shirt A Rs. 720/- Shirt B Rs. 960/- MIXED PRACTICE–II

148 The Leading Mathematics - 7 5. The cost of 3 1 2 meters of cloth is Rs. 870 3 4 . (a) Convert 31 2 into decimals. (b) Find the cost of one meter of cloth. (c) Find the cost of 14 meters of cloths. 6. A man sold an article for Rs. 840 at a loss of 16% of the cost price. (a) Find the cost price of the article. (b) Find the cost price of such 10 articles. (c) If he sold at a profit of 16%, how much difference is between the selling prices on losing and gaining cases? Calculate it. 7. A television set was bought for Rs. 15600. Rs. 250 was spent on transportation and Rs. 150 on servicing. It was sold at a loss of 10%. (a) If SP is greater than CP, it remains profit. Is it true statement? (b) Find the selling Price of television. (c) If he had sold at in Rs. 17600, what would be the condition of profit or loss? Calculate it. (d) Find the cost of such types of 10 television sets. 8. In a company, 3 4 of a man’s income is Rs. 30000. (a) What will be 1 4 of his income? Find it. (b) Write any two rational numbers between 3 4 and 1 4. 9. In a factory, 5 men and 7 women can earn Rs. 7000 in a day. (a) How much can single man earn in a day. (b) How much can single woman earn in a day. (c) How much would 7 men and 11 women earn per day? 10. A man buys two hens for 1500 each. He sold one for 15% profit and other for 5% loss. (a) Find the selling prices of two hens in both cases. (b) Find his profit or loss percentage.

Arithmetic 149 11. Rabin bought some toys at the rate of 10 for Rs. 400 and sold them at 6 for Rs. 300. (a) Find the cost price of one toy. (b) Find the selling price of one toy. (c) Find his gain or lose percentage. 12. A man bought two bicycles for Rs. 25000 each. He sells one at a profit of 5%. (a) Write the formula for finding profit percentage. (b) Find the selling price of one bicycle with 5% profit. (c) How much should he sell the other so that he makes a profit of 20% on the whole? Calculate it. 13. A number 2 is divided by 3, the result is 0.66666666……….. (a) Define repeating decimal number. (b) Convert 0.6 into fraction. (c) Is 2 3 a rational number? Justify it. 14. Simplify (a) 6.7 – 3.5 ÷ 0.7 × 3.02 – 2.12 (b) 10.8 + 0.4(7.2 – 3.7) – 2.4 × 4.02 (c) 6 ÷ 2 of (2 + 1) (d) 12.8 – 0.5 of (7.2 – 3.7) + 2.4 × 3.02 (e) 4 7 + 2 14 – 7 8 ÷ 7 4 (f) 2 4 7 + 1 1 7 – 7 8 ÷ 2 3 4 ANSWERS 1. (a) 64, 81; 8, 9 (b) 4 2. (a) 240 (b) 33 1 3% 3. (a) 30 marks (b) same 4. (a) Food (b) Rs. 11200 5. (a) 3.5 (b) Rs. 248.79 (c) Rs. 3483 6. (a) Rs. 1000 (b) Rs. 10000 (c) Rs. 320 7. (b) Rs. 14400 (c) Rs. 1600 profit (d) Rs. 160000 8. (a) Rs. 10000 (b) 1 2, 1 8, or ... 9. (a) Rs. 1400 (b) Rs. 1000 (c) Rs. 20800 10. (a) Rs. 3150 (b) 5% profit 11. (a) Rs. 40 (b) Rs. 50 (c) 50% profit 12. (b) Rs. 26250 (c) Rs. 33750 13. (b) 2 3 14. (a) –10.52 (b) 2.552 (c) 1 (d) 18.298 (e) 153 224 (f) 523 154

150 The Leading Mathematics - 7 FM : 20 Time : 40 Min. CONFIDENCE LEVEL TEST II ARITHMETIC 1. Samjhana will try to write the counting numbers from 2 on the chess board. (a) Complete the numbers remaining to write by Samjhana. [1] (b) Write the square numbers from it. [1] (c) What are the cubic roots of 64? [1] (d) Find the and HCF and LCM of cube numbers. [2] 2. (a) Simplify: (– 3) × (– 5) + (– 14) ÷ (+ 4) [1] (b) Find two rational numbers between 2 3 and 3 4. [2] (c) Convert 0.6 into fraction. [2] 3. Observe the price of pens alongside. (a) Find the ratio of their prices of ball pen to fountain pen and fountain pen to ball pen. [1] (b) Express the above ratios in decimal forms. [1] (c) Find the cost of 5 ball pens. [1] (d) If you sold the 3 fountain pens for Rs. 50, how much did you get profit or loss? Find it. [2] 4. Rabin bought some toys at the rate of 10 for Rs. 450.50 and sold them at 6 for Rs. 310.50. (a) Find the cost price of one toy. [1] (b) Find the selling price of one toy. [1] (c) Divide the selling price of one toy by its cost price up to two decimal places. [1] (d) Find his gain or lose percentage in his traction. [2] 2 3 4 5 6 7 8 9 Rs. 24 Rs. 42 Attempt all the questions.