Page 1 Points to Remember : Unitary Method. The method is which a unit item helps in getting the result about any number of items is called unitary method. Rules. (i) To get more, we multiply. (ii) To get less, we divide. 750 )4000( 5 3750 250 )750( 3 750 × (HCF of 4000 and 750 = 250) = 3 16 = 16 : 3 (iv) 1.8 kg to 6 kg = 10 18 : 6 = 18 : 60 = 60 18 = 6 60 6 18 (HCF of 18, 60 = 6) = 10 3 = 3 : 10 (v) 48 minutes to 1 hour = 48 minutes : 60 minutes = 48 : 60 = 4 : 5 (vi) 2.4 km to 900 m = 2400 m : 900 m = 900 2400 = 300 900 300 2400 (HCF of 2400 and 900 = 300) ( ) EXERCISE 10 A Q.1. Find each of the following ratios in the simplest form : (i) 24 to 56 (ii) 84 paise to Rs. 3 (iii) 4 kg to 750 g (iv) 1.8 kg to 6 kg (v) 48 minutes to 1 hour (vi) 2.4 km to 900 m Sol. (i) 24 to 56 = 56 24 = 8 56 8 24 (HCF of 24, 56 = 8) = 7 3 = 3 : 7 (ii) 84 paise to Rs. 3 = 84 p : 300 p = 300 84 = 12 300 12 84 (HCF of 84 and 300 = 12) = 25 7 = 7 : 25 (iii) 4 kg to 750 g = 4000 g : 750 g = 750 4000 = 250 750 250 4000 Page 2 Points to Remember : Unitary Method. The method is which a unit item helps in getting the result about any number of items is called unitary method. Rules. (i) To get more, we multiply. (ii) To get less, we divide. 750 )4000( 5 3750 250 )750( 3 750 × (HCF of 4000 and 750 = 250) = 3 16 = 16 : 3 (iv) 1.8 kg to 6 kg = 10 18 : 6 = 18 : 60 = 60 18 = 6 60 6 18 (HCF of 18, 60 = 6) = 10 3 = 3 : 10 (v) 48 minutes to 1 hour = 48 minutes : 60 minutes = 48 : 60 = 4 : 5 (vi) 2.4 km to 900 m = 2400 m : 900 m = 900 2400 = 300 900 300 2400 (HCF of 2400 and 900 = 300) ( ) EXERCISE 10 A Q.1. Find each of the following ratios in the simplest form : (i) 24 to 56 (ii) 84 paise to Rs. 3 (iii) 4 kg to 750 g (iv) 1.8 kg to 6 kg (v) 48 minutes to 1 hour (vi) 2.4 km to 900 m Sol. (i) 24 to 56 = 56 24 = 8 56 8 24 (HCF of 24, 56 = 8) = 7 3 = 3 : 7 (ii) 84 paise to Rs. 3 = 84 p : 300 p = 300 84 = 12 300 12 84 (HCF of 84 and 300 = 12) = 25 7 = 7 : 25 (iii) 4 kg to 750 g = 4000 g : 750 g = 750 4000 = 250 750 250 4000 = 3 8 = 8 : 3 Q. 2. Express the following ratios in the simplest form : (i) 36 : 90 (ii) 324 : 144 (iii) 85 : 561 (iv) 480 : 384 (v) 186 : 403 (vi) 777 : 1147 Sol. (i) 36 : 90 = 90 36 = 18 90 18 36 (HCF of 36, 90 = 18) = 5 2 = 2 : 5 (ii) The given ratio = 324 : 144 324 144 To express it in simplest form, we divide the numerator and denominator by the H.C.F. of 324 and 144. Now, H.C.F. of 324 and 144 144 ) 324 288 ( 2 36 ) 144 144 ( 4 = 36 324 144 324 36 144 36 9 4 Required ratio = 9 : 4. (iii) The given ratio is 85 : 561 85 561 To express it in simplest form, we divided the numerator and denominator by the HCF of 85 and 561. Now, HCF of 85 and 561 = 17 85 ) 561 510 ( 6 51 ) 85 51 ( 1 34 ) 51 34 ( 1 17 ) 34 34 0 ( 2 85 561 85 17 561 17 5 33 Required ratio = 5 : 33. (iv) The given ratio = 480 : 384 480 384 To express it in simplest form, we divide the numerator and denominator by the HCF of 480 and 384. Now, HCF of 480 and 384 = 96 384 ) 480 384 ( 1 96 ) 384 384 ( 4 0 480 384 480 96 384 96 5 4 Required rato = 5 : 4. (v) The given ratio 186 403 186 403 : To express it in the simplest form, we divide its numerator and denominator by the HCF of 186 and 403. Now, HCF of 186 and 403 = 31 186 ) 403 372 ( 2 31 ) 186 186 ( 6 0 186 403 186 31 403 31 6 13 Page 3 Points to Remember : Unitary Method. The method is which a unit item helps in getting the result about any number of items is called unitary method. Rules. (i) To get more, we multiply. (ii) To get less, we divide. 750 )4000( 5 3750 250 )750( 3 750 × (HCF of 4000 and 750 = 250) = 3 16 = 16 : 3 (iv) 1.8 kg to 6 kg = 10 18 : 6 = 18 : 60 = 60 18 = 6 60 6 18 (HCF of 18, 60 = 6) = 10 3 = 3 : 10 (v) 48 minutes to 1 hour = 48 minutes : 60 minutes = 48 : 60 = 4 : 5 (vi) 2.4 km to 900 m = 2400 m : 900 m = 900 2400 = 300 900 300 2400 (HCF of 2400 and 900 = 300) ( ) EXERCISE 10 A Q.1. Find each of the following ratios in the simplest form : (i) 24 to 56 (ii) 84 paise to Rs. 3 (iii) 4 kg to 750 g (iv) 1.8 kg to 6 kg (v) 48 minutes to 1 hour (vi) 2.4 km to 900 m Sol. (i) 24 to 56 = 56 24 = 8 56 8 24 (HCF of 24, 56 = 8) = 7 3 = 3 : 7 (ii) 84 paise to Rs. 3 = 84 p : 300 p = 300 84 = 12 300 12 84 (HCF of 84 and 300 = 12) = 25 7 = 7 : 25 (iii) 4 kg to 750 g = 4000 g : 750 g = 750 4000 = 250 750 250 4000 = 3 8 = 8 : 3 Q. 2. Express the following ratios in the simplest form : (i) 36 : 90 (ii) 324 : 144 (iii) 85 : 561 (iv) 480 : 384 (v) 186 : 403 (vi) 777 : 1147 Sol. (i) 36 : 90 = 90 36 = 18 90 18 36 (HCF of 36, 90 = 18) = 5 2 = 2 : 5 (ii) The given ratio = 324 : 144 324 144 To express it in simplest form, we divide the numerator and denominator by the H.C.F. of 324 and 144. Now, H.C.F. of 324 and 144 144 ) 324 288 ( 2 36 ) 144 144 ( 4 = 36 324 144 324 36 144 36 9 4 Required ratio = 9 : 4. (iii) The given ratio is 85 : 561 85 561 To express it in simplest form, we divided the numerator and denominator by the HCF of 85 and 561. Now, HCF of 85 and 561 = 17 85 ) 561 510 ( 6 51 ) 85 51 ( 1 34 ) 51 34 ( 1 17 ) 34 34 0 ( 2 85 561 85 17 561 17 5 33 Required ratio = 5 : 33. (iv) The given ratio = 480 : 384 480 384 To express it in simplest form, we divide the numerator and denominator by the HCF of 480 and 384. Now, HCF of 480 and 384 = 96 384 ) 480 384 ( 1 96 ) 384 384 ( 4 0 480 384 480 96 384 96 5 4 Required rato = 5 : 4. (v) The given ratio 186 403 186 403 : To express it in the simplest form, we divide its numerator and denominator by the HCF of 186 and 403. Now, HCF of 186 and 403 = 31 186 ) 403 372 ( 2 31 ) 186 186 ( 6 0 186 403 186 31 403 31 6 13 Required ratio = 6 : 13 (vi) 777 : 1147 = 1147 777 To express it in simplest form, we divide its numerator and denominator by the HCF of 777 and 1147 = 37 1147 37 777 = 31 21 777 )1147( 7 777 370 )777( 2 740 37 )370( 10 370 × (HCF of 777 and 1147 is 37) = 21 : 31 Q. 3. Find the following ratios in simplest form : (i) Rs. 6.30 : Rs. 16·80 (ii) 3 weeks : 30 days (iii) 3 m 5 cm : 35 cm. (iv) 48 min : 2 hours 40 min. (v) 1 L 35 mL : 270 mL (vi) 4 kg : 2 kg 500 g. Sol. (i) The given ratio = Rs. 6·30 : Rs. 16.80 Rs. 6 . 30 Rs.16 . 80 630 1680 To express it in simplest form, we divide its numerator and denominator by the HCF of 630 and 1680. Now, HCF of 630 and 1680 = 210 630 ) 1680 1260 ( 2 420 ) 630 420 ( 1 210 ) 420 420 ( 2 0 630 1680 630 210 1680 210 3 8 Required ratio = 3 : 8 (ii) The given ratio = 3 weeks : 30 days = (3 × 7) days : 30 days = 21 days : 30 days = 21 : 30 = 7 : 10 (iii) To given ratio = 3m 5 cm : 35cm = (3 100 + 5) cm : 35 cm = 305 cm : 35 cm = 305 : 35 = 61 : 7. (iv) The given ratio = 48 min : 2 hours 40 min = 48 min : (2 60 + 40) min = 48 min : (120 + 40) min = 48 min : 160 min = 48 : 160 = 3 : 10 (v) The given ratio = 1L 35 mL : 270 mL = (1 1000 + 35) mL : 270 mL = 1035 mL : 270 mL = 1035 : 270 1035 270 To express it in simplest form, we divide its numerator and denominator by the HCF of 1035 and 270. 270 ) 1035 810 ( 3 225 ) 270 225 ( 1 45 ) 225 225 ( 5 0 Now, HCF and 1035 and 270 = 45 Page 4 Points to Remember : Unitary Method. The method is which a unit item helps in getting the result about any number of items is called unitary method. Rules. (i) To get more, we multiply. (ii) To get less, we divide. 750 )4000( 5 3750 250 )750( 3 750 × (HCF of 4000 and 750 = 250) = 3 16 = 16 : 3 (iv) 1.8 kg to 6 kg = 10 18 : 6 = 18 : 60 = 60 18 = 6 60 6 18 (HCF of 18, 60 = 6) = 10 3 = 3 : 10 (v) 48 minutes to 1 hour = 48 minutes : 60 minutes = 48 : 60 = 4 : 5 (vi) 2.4 km to 900 m = 2400 m : 900 m = 900 2400 = 300 900 300 2400 (HCF of 2400 and 900 = 300) ( ) EXERCISE 10 A Q.1. Find each of the following ratios in the simplest form : (i) 24 to 56 (ii) 84 paise to Rs. 3 (iii) 4 kg to 750 g (iv) 1.8 kg to 6 kg (v) 48 minutes to 1 hour (vi) 2.4 km to 900 m Sol. (i) 24 to 56 = 56 24 = 8 56 8 24 (HCF of 24, 56 = 8) = 7 3 = 3 : 7 (ii) 84 paise to Rs. 3 = 84 p : 300 p = 300 84 = 12 300 12 84 (HCF of 84 and 300 = 12) = 25 7 = 7 : 25 (iii) 4 kg to 750 g = 4000 g : 750 g = 750 4000 = 250 750 250 4000 = 3 8 = 8 : 3 Q. 2. Express the following ratios in the simplest form : (i) 36 : 90 (ii) 324 : 144 (iii) 85 : 561 (iv) 480 : 384 (v) 186 : 403 (vi) 777 : 1147 Sol. (i) 36 : 90 = 90 36 = 18 90 18 36 (HCF of 36, 90 = 18) = 5 2 = 2 : 5 (ii) The given ratio = 324 : 144 324 144 To express it in simplest form, we divide the numerator and denominator by the H.C.F. of 324 and 144. Now, H.C.F. of 324 and 144 144 ) 324 288 ( 2 36 ) 144 144 ( 4 = 36 324 144 324 36 144 36 9 4 Required ratio = 9 : 4. (iii) The given ratio is 85 : 561 85 561 To express it in simplest form, we divided the numerator and denominator by the HCF of 85 and 561. Now, HCF of 85 and 561 = 17 85 ) 561 510 ( 6 51 ) 85 51 ( 1 34 ) 51 34 ( 1 17 ) 34 34 0 ( 2 85 561 85 17 561 17 5 33 Required ratio = 5 : 33. (iv) The given ratio = 480 : 384 480 384 To express it in simplest form, we divide the numerator and denominator by the HCF of 480 and 384. Now, HCF of 480 and 384 = 96 384 ) 480 384 ( 1 96 ) 384 384 ( 4 0 480 384 480 96 384 96 5 4 Required rato = 5 : 4. (v) The given ratio 186 403 186 403 : To express it in the simplest form, we divide its numerator and denominator by the HCF of 186 and 403. Now, HCF of 186 and 403 = 31 186 ) 403 372 ( 2 31 ) 186 186 ( 6 0 186 403 186 31 403 31 6 13 Required ratio = 6 : 13 (vi) 777 : 1147 = 1147 777 To express it in simplest form, we divide its numerator and denominator by the HCF of 777 and 1147 = 37 1147 37 777 = 31 21 777 )1147( 7 777 370 )777( 2 740 37 )370( 10 370 × (HCF of 777 and 1147 is 37) = 21 : 31 Q. 3. Find the following ratios in simplest form : (i) Rs. 6.30 : Rs. 16·80 (ii) 3 weeks : 30 days (iii) 3 m 5 cm : 35 cm. (iv) 48 min : 2 hours 40 min. (v) 1 L 35 mL : 270 mL (vi) 4 kg : 2 kg 500 g. Sol. (i) The given ratio = Rs. 6·30 : Rs. 16.80 Rs. 6 . 30 Rs.16 . 80 630 1680 To express it in simplest form, we divide its numerator and denominator by the HCF of 630 and 1680. Now, HCF of 630 and 1680 = 210 630 ) 1680 1260 ( 2 420 ) 630 420 ( 1 210 ) 420 420 ( 2 0 630 1680 630 210 1680 210 3 8 Required ratio = 3 : 8 (ii) The given ratio = 3 weeks : 30 days = (3 × 7) days : 30 days = 21 days : 30 days = 21 : 30 = 7 : 10 (iii) To given ratio = 3m 5 cm : 35cm = (3 100 + 5) cm : 35 cm = 305 cm : 35 cm = 305 : 35 = 61 : 7. (iv) The given ratio = 48 min : 2 hours 40 min = 48 min : (2 60 + 40) min = 48 min : (120 + 40) min = 48 min : 160 min = 48 : 160 = 3 : 10 (v) The given ratio = 1L 35 mL : 270 mL = (1 1000 + 35) mL : 270 mL = 1035 mL : 270 mL = 1035 : 270 1035 270 To express it in simplest form, we divide its numerator and denominator by the HCF of 1035 and 270. 270 ) 1035 810 ( 3 225 ) 270 225 ( 1 45 ) 225 225 ( 5 0 Now, HCF and 1035 and 270 = 45 1035 270 1035 45 270 45 23 6 Required ratio = 23 : 6. (vi) The given ratio = 4 kg : 2 kg 500 g = (4 × 1000) g : (2 1000 + 500) g = 4000 g : 2500 g = 4000 : 2500 = 40 : 25 = 8 : 5 4. Mr Sahai and his wife are both school teachers and earn Rs., 16800 and Rs. 10500 per month respectively. Find the ratio of (i) Mr Sahai’s income to his wife’s income; (ii) Mrs Sahai’s income to her husband’s income; (iii) Mr Sahai’s income to the total income of the two. Sol. Earning of Sahai = Rs. 16800 and of his wife = Rs. 10500 Then total income = Rs. 16800 + 10500 = Rs. 27300 (i) Ratio in Sahai’s income and his wife = 16800 : 10500 = 10500 16800 = 2100 10500 2100 16800 (HCF of 16800 and 105000 is 2100) = 5 8 = 8 : 5 (ii) Ratio in Sahai’s wife income and her husband’s income = 10500 : Rs. 16800 = 5 : 8 (Dividing by the HCF of 10500 and 16800 = 2100) (iii) Sahai’s income and total of their income = 16800 : 27300 = 27300 16800 = 2100 27300 2100 16800 = 13 8 = 8 : 13 5. Rohit earns Rs. 15300 and saves Rs. 1224 per month. Find the ratio of (i) his income and savings; (ii) his income and expenditure; (iii) his expenditure and savings. Sol. Rohit monthly earnings = Rs. 15300 and his savings = Rs. 1224 So, his expenditure = Rs. 15300 – 1224 = Rs. 14076 Now, (i) Ratio in his income and savings = 15300 : 1224 = 1224 15300 1224 )15300( 12 1224 3060 2448 612 )1224( 2 1224 × = 612 1224 612 15300 (Dividing by HCF of 15300 and 1224) = 2 25 = 25 : 2 (ii) Ratio in his income and expenditure = 15300 : 14076 = 14076 15300 = 612 14076 612 15300 (HCF = 612) Page 5 Points to Remember : Unitary Method. The method is which a unit item helps in getting the result about any number of items is called unitary method. Rules. (i) To get more, we multiply. (ii) To get less, we divide. 750 )4000( 5 3750 250 )750( 3 750 × (HCF of 4000 and 750 = 250) = 3 16 = 16 : 3 (iv) 1.8 kg to 6 kg = 10 18 : 6 = 18 : 60 = 60 18 = 6 60 6 18 (HCF of 18, 60 = 6) = 10 3 = 3 : 10 (v) 48 minutes to 1 hour = 48 minutes : 60 minutes = 48 : 60 = 4 : 5 (vi) 2.4 km to 900 m = 2400 m : 900 m = 900 2400 = 300 900 300 2400 (HCF of 2400 and 900 = 300) ( ) EXERCISE 10 A Q.1. Find each of the following ratios in the simplest form : (i) 24 to 56 (ii) 84 paise to Rs. 3 (iii) 4 kg to 750 g (iv) 1.8 kg to 6 kg (v) 48 minutes to 1 hour (vi) 2.4 km to 900 m Sol. (i) 24 to 56 = 56 24 = 8 56 8 24 (HCF of 24, 56 = 8) = 7 3 = 3 : 7 (ii) 84 paise to Rs. 3 = 84 p : 300 p = 300 84 = 12 300 12 84 (HCF of 84 and 300 = 12) = 25 7 = 7 : 25 (iii) 4 kg to 750 g = 4000 g : 750 g = 750 4000 = 250 750 250 4000 = 3 8 = 8 : 3 Q. 2. Express the following ratios in the simplest form : (i) 36 : 90 (ii) 324 : 144 (iii) 85 : 561 (iv) 480 : 384 (v) 186 : 403 (vi) 777 : 1147 Sol. (i) 36 : 90 = 90 36 = 18 90 18 36 (HCF of 36, 90 = 18) = 5 2 = 2 : 5 (ii) The given ratio = 324 : 144 324 144 To express it in simplest form, we divide the numerator and denominator by the H.C.F. of 324 and 144. Now, H.C.F. of 324 and 144 144 ) 324 288 ( 2 36 ) 144 144 ( 4 = 36 324 144 324 36 144 36 9 4 Required ratio = 9 : 4. (iii) The given ratio is 85 : 561 85 561 To express it in simplest form, we divided the numerator and denominator by the HCF of 85 and 561. Now, HCF of 85 and 561 = 17 85 ) 561 510 ( 6 51 ) 85 51 ( 1 34 ) 51 34 ( 1 17 ) 34 34 0 ( 2 85 561 85 17 561 17 5 33 Required ratio = 5 : 33. (iv) The given ratio = 480 : 384 480 384 To express it in simplest form, we divide the numerator and denominator by the HCF of 480 and 384. Now, HCF of 480 and 384 = 96 384 ) 480 384 ( 1 96 ) 384 384 ( 4 0 480 384 480 96 384 96 5 4 Required rato = 5 : 4. (v) The given ratio 186 403 186 403 : To express it in the simplest form, we divide its numerator and denominator by the HCF of 186 and 403. Now, HCF of 186 and 403 = 31 186 ) 403 372 ( 2 31 ) 186 186 ( 6 0 186 403 186 31 403 31 6 13 Required ratio = 6 : 13 (vi) 777 : 1147 = 1147 777 To express it in simplest form, we divide its numerator and denominator by the HCF of 777 and 1147 = 37 1147 37 777 = 31 21 777 )1147( 7 777 370 )777( 2 740 37 )370( 10 370 × (HCF of 777 and 1147 is 37) = 21 : 31 Q. 3. Find the following ratios in simplest form : (i) Rs. 6.30 : Rs. 16·80 (ii) 3 weeks : 30 days (iii) 3 m 5 cm : 35 cm. (iv) 48 min : 2 hours 40 min. (v) 1 L 35 mL : 270 mL (vi) 4 kg : 2 kg 500 g. Sol. (i) The given ratio = Rs. 6·30 : Rs. 16.80 Rs. 6 . 30 Rs.16 . 80 630 1680 To express it in simplest form, we divide its numerator and denominator by the HCF of 630 and 1680. Now, HCF of 630 and 1680 = 210 630 ) 1680 1260 ( 2 420 ) 630 420 ( 1 210 ) 420 420 ( 2 0 630 1680 630 210 1680 210 3 8 Required ratio = 3 : 8 (ii) The given ratio = 3 weeks : 30 days = (3 × 7) days : 30 days = 21 days : 30 days = 21 : 30 = 7 : 10 (iii) To given ratio = 3m 5 cm : 35cm = (3 100 + 5) cm : 35 cm = 305 cm : 35 cm = 305 : 35 = 61 : 7. (iv) The given ratio = 48 min : 2 hours 40 min = 48 min : (2 60 + 40) min = 48 min : (120 + 40) min = 48 min : 160 min = 48 : 160 = 3 : 10 (v) The given ratio = 1L 35 mL : 270 mL = (1 1000 + 35) mL : 270 mL = 1035 mL : 270 mL = 1035 : 270 1035 270 To express it in simplest form, we divide its numerator and denominator by the HCF of 1035 and 270. 270 ) 1035 810 ( 3 225 ) 270 225 ( 1 45 ) 225 225 ( 5 0 Now, HCF and 1035 and 270 = 45 1035 270 1035 45 270 45 23 6 Required ratio = 23 : 6. (vi) The given ratio = 4 kg : 2 kg 500 g = (4 × 1000) g : (2 1000 + 500) g = 4000 g : 2500 g = 4000 : 2500 = 40 : 25 = 8 : 5 4. Mr Sahai and his wife are both school teachers and earn Rs., 16800 and Rs. 10500 per month respectively. Find the ratio of (i) Mr Sahai’s income to his wife’s income; (ii) Mrs Sahai’s income to her husband’s income; (iii) Mr Sahai’s income to the total income of the two. Sol. Earning of Sahai = Rs. 16800 and of his wife = Rs. 10500 Then total income = Rs. 16800 + 10500 = Rs. 27300 (i) Ratio in Sahai’s income and his wife = 16800 : 10500 = 10500 16800 = 2100 10500 2100 16800 (HCF of 16800 and 105000 is 2100) = 5 8 = 8 : 5 (ii) Ratio in Sahai’s wife income and her husband’s income = 10500 : Rs. 16800 = 5 : 8 (Dividing by the HCF of 10500 and 16800 = 2100) (iii) Sahai’s income and total of their income = 16800 : 27300 = 27300 16800 = 2100 27300 2100 16800 = 13 8 = 8 : 13 5. Rohit earns Rs. 15300 and saves Rs. 1224 per month. Find the ratio of (i) his income and savings; (ii) his income and expenditure; (iii) his expenditure and savings. Sol. Rohit monthly earnings = Rs. 15300 and his savings = Rs. 1224 So, his expenditure = Rs. 15300 – 1224 = Rs. 14076 Now, (i) Ratio in his income and savings = 15300 : 1224 = 1224 15300 1224 )15300( 12 1224 3060 2448 612 )1224( 2 1224 × = 612 1224 612 15300 (Dividing by HCF of 15300 and 1224) = 2 25 = 25 : 2 (ii) Ratio in his income and expenditure = 15300 : 14076 = 14076 15300 = 612 14076 612 15300 (HCF = 612) = 23 25 = 25 : 23 (iii) Ratio in his expenditure and savings = 14076 : 1224 = 1224 14076 = 612 1224 612 14076 = 2 23 = 23 : 2 Q. 6. The ratio of the number of male and female workers in a textile mill is 5 : 3. If there are 115 male workers, what is the number of female workers in the mill ? Sol. Let the number of male and female workers in the mill be 5x and 3x respectively. Then, 5x = 115 5 5 115 5 x (Dividing both sides by 5) x = 23 Number of female workers in the mill = 3x = 3 23 = 69. Q. 7. The boys and the girls in a school are in the ratio 9 : 5. If the total strength of the school is 448, find the number of girls. Sol. Let the number of boys and girls in the school be 9x and 5x respectively. According to the question, 9x + 5x = 448 14x = 448 14 14 448 14 x (Dividing both sides by 14) x = 32. Number of girls = 5x = 5 32 = 160. Q.8. Divide Rs. 1575 between Kamal and Madhu in the ratio 7 : 2. Sol. Total amount = Rs. 1575 Ratio in Kamal and Madhu’s share = 7 : 2 Sum of ratios = 7 + 2 = 9 Kamal’s share = Rs. 9 7 1575 = Rs. 175 × 7 = Rs. 1225 Madhu’s share = Rs. 9 2 1575 = Rs. 175 × 2 = Rs. 350 Q.9. Divide Rs. 3450 among A, B and C in the ratio 3 : 5 : 7. Sol. Total amount = Rs. 3450 Ratio in A, B and C shares = 3 : 5 : 7 Sum of share = 3 + 5 + 7 = 15 A’s share = Rs. 15 3 3450 = Rs. 230 × 3 = Rs. 690 B’s share = Rs. 15 5 3450 = Rs. 230 × 5 = Rs. 1150 C’s share = Rs. 15 7 3450 = Rs. 230 × 7 = Rs. 1610 Q. 10. Two numbers are in the ratio : 11 : 12. If their sum is 460, find the numbers. Sol. Let the numbers be 11x and 12x. Then, 11x + 12x = 460 23x = 460 23 23 460 23 x (Dividing both sides by 23) x = 20. Required numbers are (11 × 20) and 12 × 20, that is , 220 and 240. Q.11. A 35-cm line segment is divided into two parts in the ratio 4 : 3. Find theRead More

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