Q11) An alloy contains zinc and copper in the ratio 7:4. If the alloy contains 14 Kg copper, then the quantity of zinc in the alloy is 1) 24.5 Kg 2) 8 Kg 3) 28 Kg 4) 17.5 Kg Solution: Zinc = 7k, copper = 4k 4k = 14; k = 3.5 Quantity of Zinc in the alloy is 7k = 7(3.5) = 24.5 Kg Hence, option 1. Q12) If 3A = 4B = 5C, then A:B:C is equal to1) 3:4:5 2) 9:8:7 3) 20:15:12 4) 1/3:1/4:1/5 Solution: A = 20, B = 15, C = 12 Hence, option 3. Q13) If , then the value of a:b is1) 2:3 2) 3:1 3) 3:2 4) 1:3 Solution: On solving, a = 3b Hence, option 2. Q14) A bag contains 1 rupee, 50 paise and 25 paise coins in the ratio 3:4:5. Total value of these coins is
Rs 24. Find the number of 50 paise coins.1) 12 2) 20 3) 16 4) 8 Solution: Let 3k be 1 rupee coins, 4k be 50 paise coins, 5k be 25 paise coins. A.T.Q. 3k(1) + 4k(0.5) + 5k(0.25) = 24 On solving, k = 4 Number of 50 paise coins = 4k = 16 Hence, option 3. ## Download: Practice Questions on Ratio ProportionQ15) If P:Q = 3:5 and Q:R = 7:8 then 5P:3Q:7R is equal to1) 6:15:25 2) 1:1:6 3) 4:7:8 4) 3:3:8 Solution: P/Q = 3/5 and Q/R = 7/8 P/Q = 21/35 and Q/R = 35/40 P = 21, Q = 35, R = 40 Hence, option 4. Q16) When 50% of a number is added to a second number, the second number increases to its four-third. What is
the ratio between the first number and the second number?1) 3:2 2) 3:4 3) 2:3 4) Data Inadequate Solution: first number = f, second number = s s + f/2 = 4s/3 f/2 = s/3 f/s = 2/3 Hence, option 3. Q17) Divide 1162 into 3 parts such that 4 times the first is equal to 5 times the second and 7 times the
third. Find the smallest part.1) 280 2) 200 3) 420 4) 490 Solution: A + B + C = 1162 4A = 5B = 7C = 4×5×7×k A = 35k, B = 28k, C = 20k 35k + 28k + 20k = 1162 k = 14 Smallest part = 20k = 20×14 = 280 Hence, option 1. Q18) Two candles of the same height are lighted at the same time. The first is consumed in 7 hours and the
second in 6 hours. Assuming that each candle burns at a constant rate, in how many hours after being lighted,
the ratio between the first and second candles becomes 3:1.1) 5 hours 36 minutes 2) 5 hours 30 minutes 3) 5 hours 4) 6 hours Solution: Height = lcm(6,7) = 42 units. As first candle is consumed in 7 hours so 6 units are consumed in an hour. As second candle is consumed in 6 hours so 7 units are consumed in an hour. 't' hours after being lighted, the ratio between the first and second candles becomes 3:1 so, (42 - 6t)/(42 - 7t) = 3/1 t = 28/5 hours = 5 hours 36 minutes Hence, option 1. Q19) From each of the two given numbers, half the smaller number is subtracted. After such subtraction,
the larger number is 4 times as large as the smaller number. What is the ratio of the numbers?1) 5:2 2) 1:4 3) 4:1 4) 4:5 Solution: Let the smaller number be x and the larger number be y. (y - x/2) = 4(x - x/2) (y - x/2) = 4x/2 y = 5x/2 y:x = 5:2 Hence, option 1. Q20) A sum of Rs 300 is divided among P, Q and R in such a way that Q gets Rs 30 more then P and R gets Rs 60
more than Q. Then, ratio of their shares is1) 2:3:5 2) 3:2:5 3) 2:5:3 4) 5:3:2 Solution: Share of P = x Share of Q = x + 30 Share of R = (x + 30) + 60 = x + 90 Sum of money with P, Q, R = 300 x + x + 30 + x + 90 = 300 3x + 120 = 300 x = 60 Required ratio = 60:(60 + 30):(60 + 90) = 60:90:150 = 2:3:5 Hence, option 3. |