Quantitative Aptitude प्रश्न और उत्तर का अभ्यास करें
8प्र: A tin a mixture of two liquids A and B in the proportion 4 : 1. If 45 litres of the mixture is replaced by 45 litres of liquid B, then the ratio of the two liquids becomes 2 : 5. How much of the liquid B was there in the tin? What quantity does the tin hold? 1917 05b5cc6a7e4d2b4197774cdef
5b5cc6a7e4d2b4197774cdef- 158 lfalse
- 265 lfalse
- 350 ltrue
- 462 lfalse
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उत्तर : 3. "50 l"
व्याख्या :
Answer: C) 50 l Explanation: Let the tin contain 5x litres of liquids 5x - 45 x 455x - 45 x 15 + 45 = 25 => 5(4x - 36) = 2(x + 36) => 20x - 180 = 2x + 72 => x = 14 litres Hence, the initial quantity of mixture = 70l Quantity of liquid B = 70 - 45 x 15 + 45 = 50 litres.
प्र: The ratio of earnings of A and B is 4:5. If the earnings of A increase by 20% and the earnings of B decrease by 20%, the new ratio of their earnings becomes 6:5. What are A's earnings? 1915 05b5cc7c9e4d2b41977750ed6
5b5cc7c9e4d2b41977750ed6- 1Rs.22,000false
- 2Rs.27,500false
- 3Rs. 26,400false
- 4Cannot be determinedtrue
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उत्तर : 4. "Cannot be determined"
व्याख्या :
Answer: D) Cannot be determined Explanation:
प्र: If a student walks at the rate of 4 mts/min from his home, he is 4 minutes late for school, if he walks at the rate of 6 mts/min he reaches half an hour earlier. How far is his school from his home ? 1914 05b5cc764e4d2b4197774fd80
5b5cc764e4d2b4197774fd80- 1100 mtsfalse
- 296 mtsfalse
- 398 mtstrue
- 4120 mtsfalse
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उत्तर : 3. "98 mts"
व्याख्या :
Answer: C) 98 mts Explanation: Let the distance between home and school is 'x'.Let actual time to reach be 't'.Thus, x/4 = t + 4 ---- (1)and x/6 = t - 30 -----(2)Solving equation (1) and (2)=> x = 98 mts.
प्र: 5/9 of the part of the population in a village are females. If 30 % of the females are married. The percentage of unmarried males in the total males is ? 1913 05b5cc72ce4d2b4197774f7a1
5b5cc72ce4d2b4197774f7a1- 162.5 %true
- 2125 %false
- 384.32 %false
- 446.87 %false
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उत्तर : 1. "62.5 %"
व्याख्या :
Answer: A) 62.5 % Explanation: Let total population = p number of females = 5p/9number of males =(p-5p/9) = 4p/9married females = 30% of 5p/9 = 30x5p/100x9 = p/6married males = p/6unmarried males =(4p/9-p/6) = 5p/18Percentage of unmarried males in the total males = {(5p/18)/4p/9}x100 = (5p/18)x(9/4p) = 125/2 % = 62.5%
प्र: When the price of whear was increased by 32%, a family reduced its consumption in such a way that the expenditure on wheat was only 10% more than before. If 30 kg per month were consumed before, find the new monthly consumption of family? 1912 05b5cc61ee4d2b4197774b758
5b5cc61ee4d2b4197774b758- 127 kgsfalse
- 225 kgstrue
- 324 kgsfalse
- 421 kgsfalse
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उत्तर : 2. "25 kgs"
व्याख्या :
Answer: B) 25 kgs Explanation: Initial wheat per kg cost be Rs. 100 Then, monthly expenditure of the family = 30 x 100 = 3000 After increase in cost, Let P be new monthly consumption of family. P x 132 = 110/100 x 3000 => P = 25 kg.
प्र: The letters of the word PROMISE are to be arranged so that three vowels should not come together. Find the number of ways of arrangements? 1910 05b5cc6ade4d2b4197774d0f9
5b5cc6ade4d2b4197774d0f9- 14320true
- 24694false
- 34957false
- 44871false
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उत्तर : 1. "4320"
व्याख्या :
Answer: A) 4320 Explanation: Given Word is PROMISE. Number of letters in the word PROMISE = 7 Number of ways 7 letters can be arranged = 7! ways Number of Vowels in word PROMISE = 3 (O, I, E) Number of ways the vowels can be arranged that 3 Vowels come together = 5! x 3! ways Now, the number of ways of arrangements so that three vowels should not come together = 7! - (5! x 3!) ways = 5040 - 720 = 4320.
प्र: Find the odd man out in the following number series? 71, 88, 113, 150, 203, 277 1909 05b5cc678e4d2b4197774c158
5b5cc678e4d2b4197774c158- 1277true
- 2203false
- 3150false
- 4113false
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उत्तर : 1. "277"
व्याख्या :
Answer: A) 277 Explanation: The given number series is 71 88 113 150 203 277 +17 +25 +37 +53 +74 +8 +12 +16 +21 +4 +4 +5 The series should be 71 88 112 150 203 276 i.e, the differences of the difference should be +4. Hence, the odd man in the given series is 277.
प्र: Mr. X sold two properties P1 and P2 for Rs 1,00,000 each. He sold property P1 for 20% less than what he paid for it. What is the percentage of profit of property P2, so that he is not in gain or loss on the sale of two properties? 1905 05b5cc6aae4d2b4197774cf6f
5b5cc6aae4d2b4197774cf6f- 133.33%true
- 229.97%false
- 325%false
- 422.22%false
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उत्तर : 1. "33.33%"
व्याख्या :
Answer: A) 33.33% Explanation: Let 'A' be the cost price of property P1. Then from the given data, the selling price of P1 = Rs. 1,00,000 He got 20% loss on selling P1 ⇒A - 20A100 = 100000⇒A = 1,25,000 Therefore, the amount he lossed on selling P1 = 25,000 As ge he got no loss or gain on sale of P1 and P2, the gain on selling P2 = 25,000 But the selling price of P2 = 1,00,000 => Cost price of P2 = 75,000 Hence, the profit percentage on P2 = GainCPx 100 = 2500075000x100 = 1003= 33.33%

