# B tech standard sem2 quantitative ability solution

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QUANTITATIVE ABILITY – SOLUTION HSEM2BTECHSTANDARDQA1119 3x 100  300x RATIO AND PROPORTION 1. 4x 50  200x Ans: [c] 12x 25  300x Let the two numbers be 4x and 13x. 300x + 200x + 300x = 60,000 Given LCM = 312 800x = 60,000 Here HCF = x x = 75 52x = 312 Number of 25p coins = 12(75) = 900 x=6 So largest number is 13(6) = 78 2. 6. 79.20 is divided among 7 men, 11 women, 5 boys Ans: [c] 7M + 11W + 5B = 79.20 Let the digits of two digit number be x and y. Given W = 3B So number is 10x + y M=W+B By interchanging digits we get 10y + x M = 4B Difference = 36 7(4B) + 11(3B) + 5B = 79.20 10x + y – (10y + ) = 36 66B = 79.20 9x – 9y = 36 B = 1.2 x–y=4 M = 4B = 4(1.2) Given x and y are in the ratio 2:1 Share of men = 4.80 x = 2a and y = a 2a – a = 4 3. 7. Let A’s income be 4x. Numbers are 8 and 4 Income = Savings + Expenditure  (8 + 4) – (8 – 4) A’s expenditure = 4x – 25  12 – 4 = 8 Let B’s income be 5x Ans: [c] Given Ratio of expenses = 5:6 B’s expenditure = 5x – 50 24x – 150 = 25x – 250 Ratio of decrease of labourers = 15:11 x = 100 Total ratio of wages of labourers = 2215 : 1511 = 333:275 A’s income is 400 and B’s 500. 8. Ans: [c] Given present bill = 5000 Given A;B and B:C = 3:2 i.e for 275 Ratio wages = 5000 A:B = 3:2 5000 for ‘1’  275 B:C = 3:2 5000 for ‘330’   330 = 6000 275 B:C = 6:4 Ans: [a] Runs made by A  Let x be the num ...
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