Verify the following:
(i) 60 : 105 : : 84 : 147
(ii) 91 : 104 : : 119 : 136
(iii) 108 : 72 : : 129 : 86
(iv) 39 : 65 : : 141 : 235

Open in App

Solution

(i) 60:105::84:147
60 = 60 ÷ 15 = 4 (H.C.F. of 60 and 105 is 15.)
105 105 ÷ 15 7
84 = 84 ÷ 21 = 4 (H.C.F. of 84 and 147 is 21.)
147 147 ÷ 21 7
Hence, 60:105::84:147 are in proportion.
(ii) 91:104::119:136
91 = 91 ÷ 13 = 7 (H.C.F. of 91 and 104 is 13.)
104 104 ÷ 13 8
119 = 119 ÷ 17 = 7 (H.C.F. of 11 and 136 is 17.)
136 136 ÷ 17 8
Hence, 91:104::119:136 are in proportion.
(iii) 108:72::129:86
108 = 108 ÷ 36 = 3 (H.C.F. of 108 and 72 is 36.)
72 72 ÷ 36 2
129 = 129 ÷ 43 = 3 (H.C.F. of 129 and 86 is 43.)
86 86 ÷ 43 2
Hence, 108:72::129:86 are in proportion.
(iv) 39:65::141:235
39 = 39 ÷ 13 = 3 (H.C.F. of 39 and 65 is 13.)
65 65 ÷ 13 5
141 = 141 ÷ 47 = 3 (H.C.F. of 141 and 235 is 47.)
235 235 ÷ 47 5
Hence, 39:65::141:235 are in proportion.

Suggest Corrections

Similar questions

Q. Evaluate

17 {39+(38-12) (104-91)}

Evaluate:
(i) $(38)^{2} - (37)^{2}$
(ii) $(75)^{2} - (74)^{2}$
(iii) $(92)^{2} - (91)^{2}$
(iv) $(105)^{2} - (104)^{2}$
(v) $(141)^{2} - (140)^{2}$
(vi) $(218)^{2} - (217)^{2}$

Q. Find the odd man out in the series.

64, 71, 80, 91, 104, 119, 135, 155

Q. For a reaction

E_{a}

= and

K = 3.2 \times 10^{4} S^{- 1}

at 300 k.

Q. If 35% of a number added to 39 is the number itself, the number is

(a) 60
(b) 65
(c) 75
(d) 105

Join BYJU'S Learning Program

Verify the following: (i) 60 : 105 : : 84 : 147 (ii) 91 : 104 : : 119 : 136 (iii) 108 : 72 : : 129 : 86 (iv) 39 : 65 : : 141 : 235

Verify the following:
(i) 60 : 105 : : 84 : 147
(ii) 91 : 104 : : 119 : 136
(iii) 108 : 72 : : 129 : 86
(iv) 39 : 65 : : 141 : 235