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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
Pure milk costs Rs. 40 per litre. A milkman adds water to 35 litres of pure milk and sells the mixture at Rs. 35 per litre. How many litres of water does he add?1). 2 litres2). 5 litres3). 7 litres4). 11 litres |
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Answer» Given that, Pure milk costs Rs. 40 per litre. A milkman adds water to 35 litres of pure milk and SELLS the mixture at Rs. 35 per litre. According to the question, By applying the RULE of the allegation, The ratio of milk and water = (35 – 0) : (40 – 5) = 35 : 5 So, the ratio of milk and water is 35 : 5 i.e. 7 : 1. Hence, In 7 litres milk, added water = 1 litre So, in 35 litres milk, added water $(\Rightarrow \frac{{35}}{7} = 5\;litres)$ ∴ 5 litres of water milkman will add in the milk. |
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| 52. |
A receptacle is filled with a mixture of two liquids L1 and L2 in the ratio of 7 : 5. Due to poor handling, 9 litres of mixture fell from the receptacle and the volume was made up with L2, thus the ratio of the both liquids in the mixture becomes 7 : 9. Find the initial quantity of L1 in the mixture.1). 24 litres2). 21 litres3). 26 litres4). 28 litres |
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Answer» Let the receptacle initially contains 7x and 5x of mixtures L1 and L2 respectively. Remaining quantity of L1, = (7x – 7 × 9/12) = (7x - 21/4) litres Remaining quantity of L2, = (5x – 5 × 9/12) = (5x - 15/4) litres According to the QUESTION, [(7x - 21/4)/ (5x - 15/4) + 9] = 7/9 ⇒ (28x - 21)/ (20X + 21) = 7/9 ⇒ 252x - 189 = 140x + 147 ⇒ 112x = 336 ⇒ x = 3 ∴ Initial quantity of L1 in the mixture = 7 × 3 = 21 litres |
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| 53. |
1). 2 : 52). 1 : 33). 5 : 34). 4 : 3 |
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Answer» First of all, we use given quantities and PLACE them in the formula, to know the required quantities. Cost of 1 litre milk of 1ST quality = Cost price of dearer = d = Rs. 13.70 Cost of 1 litre milk of 2nd quality = Cost price of cheaper = C = Rs. 11.90 Desired cost of 1 kg of the mixture = Mean price = m = Rs. 12.50 $(\begin{array}{l} {\rm{Required\;RATE}} = \frac{{{\rm{Quantity\;of\;Cheaper}}}}{{{\rm{Quantity\;of\;Dearer}}}} = \frac{{{\rm{d}} - {\rm{m}}}}{{{\rm{m}} - {\rm{c}}}}\\ \Rightarrow \frac{{{\rm{Quantity\;of\;Cheaper}}}}{{{\rm{Quantity\;of\;Dearer}}}} = \frac{{13.70 - 12.50}}{{12.50 - 11.90}} = \frac{{1.20}}{{0.60}} = \frac{2}{1} \end{array})$ ∴ The ratio of the mixture should be 1 : 2 |
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| 54. |
1). 2.07 kg sugar2). 2.07 kg water3). 5 kg water4). 5 kg sugar |
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Answer» Ten kg of solution prepared by Adya will initially contain : Amount of sugar = $(\frac{{29}}{{64}} \times 10 = 4.53125\;kg)$ Amount of water = 10 − 4.53125 = 5.46875 kg She has been instructed to PREPARE the solution by using sugar and water in the RATIO 35 ? 29 but she prepared in the ratio 29 ? 35 which clearly MEANS, she used less quantity of sugar. Keeping the quantity of water same, let she adds X kg of sugar to it. $(\begin{array}{l}\THEREFORE {\rm{\;}}\frac{{Amount\;of\;sugar}}{{Amount\;of\;water}} = \frac{{35}}{{29}}\\ \Rightarrow \frac{{4.53125 + x}}{{5.46875}} = \frac{{35}}{{29}}\end{array})$ ⇒ 131.40625 + 29x = 191.40625 ⇒ 29x = 60 ∴ x = 2.07 kg ∴ Adya will have to add 2.07 kg of sugar more to prepare the syrup as per instructions. |
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| 55. |
In what ratio must two kinds of tea worth Rs. 18 and Rs. 31 per kg be mixed so as to gain 25 per cent by selling the mixture at Rs. 36 per kg?1). 7 : 122). 11 : 543). 5 : 544). 3 : 24 |
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