Quantitative Aptitude प्रश्न और उत्तर का अभ्यास करें
8प्र: A can build a wall in 16 days while B can destroy it in 8 days. A worked for 5 days. Then B joined with A for the next 2 days. Find in how many days could A build the remaining wall ? 2032 05b5cc6cee4d2b4197774e208
5b5cc6cee4d2b4197774e208- 112 daysfalse
- 214 daysfalse
- 313 daystrue
- 416 daysfalse
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उत्तर : 3. "13 days"
व्याख्या :
Answer: C) 13 days Explanation: Now, Total work = LCM(16, 8) = 48 A's one day work = + 48/16 = + 3 B's one day work = - 48/8 = -6 Given A worked for 5 days to build the wall => 5 days work = 5 x 3 = + 15 2days B joined with A in working = 2(3 - 6) = - 6 Remaining Work of building wall = 48 - (15 - 6) = 39 Now this remaining work will be done by A in = 39/3 = 13 days.
प्र: The first 8 alphabets are written down at random. what is the probability that the letters b,c,d,e always come together ? 2032 05b5cc6d3e4d2b4197774e4d7
5b5cc6d3e4d2b4197774e4d7- 11/7false
- 28!false
- 37!false
- 41/14true
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उत्तर : 4. "1/14"
व्याख्या :
Answer: D) 1/14 Explanation: The 8 letters can be written in 8! ways.n(S) = 8!Let E be the event that the letters b,c,d,e always come together when the first 8 alphabets are written down.Now the letters a, (bcde), f, g and h can be arranged in 5! ways.The letters b,c,d and e can be arranged themselves in 4! ways.n(E) = 5! x 4!Now, the required P(E) = n(E)/n(S) = 5! x 4!/8! = 1/14Hence the answer is 1/14.
प्र: The top and bottom of a tower were seen to be at angles of depression 30° and 60° from the top of a hill of height 100 m. Find the height of the tower ? 2031 15b5cc72ce4d2b4197774f79c
5b5cc72ce4d2b4197774f79c- 142.2 mtsfalse
- 233.45 mtsfalse
- 366.6 mtstrue
- 458.78 mtsfalse
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उत्तर : 3. "66.6 mts"
व्याख्या :
Answer: C) 66.6 mts Explanation: From above diagramAC represents the hill and DE represents the tower Given that AC = 100 m angleXAD = angleADB = 30° (∵ AX || BD ) angleXAE = angleAEC = 60° (∵ AX || CE) Let DE = h Then, BC = DE = h, AB = (100-h) (∵ AC=100 and BC = h), BD = CE tan 60°=AC/CE => √3 = 100/CE =>CE = 100/√3 ----- (1) tan 30° = AB/BD => 1/√3 = 100−h/BD => BD = 100−h(√3)∵ BD = CE and Substitute the value of CE from equation 1 100/√3 = 100−h(√3) => h = 66.66 mts The height of the tower = 66.66 mts.
प्र: A man can row 6 kmph in still water. When the river is running at 1.2 kmph, it takes him 1 hour to row to a place and back. What is the total distance traveled by the man ? 2031 05b5cc6d7e4d2b4197774e728
5b5cc6d7e4d2b4197774e728- 14.58 kmsfalse
- 26.35 kmsfalse
- 35.76 kmstrue
- 45.24 kmsfalse
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उत्तर : 3. "5.76 kms"
व्याख्या :
Answer: C) 5.76 kms Explanation: Speed in still water = 6 kmph Stream speed = 1.2 kmph Down stream = 7.2 kmph Up Stream = 4.8 kmph x/7.2 + x/4.8 = 1 x = 2.88 Total Distance = 2.88 x 2 = 5.76 kms
प्र: Find out the wrong term in the given series of numbers ?6, 15, 35, 77, 165, 221 2023 05b5cc728e4d2b4197774f6dc
5b5cc728e4d2b4197774f6dc- 135false
- 2165true
- 3221false
- 477false
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उत्तर : 2. "165"
व्याख्या :
Answer: B) 165 Explanation: The given series follow the rule that : Multiplying consecutive prime numbers2,3,5,7,11,13,17,19....2x3 = 63x5 = 155x7 = 357x11 = 7711x13 = 14313x17 = 221So, the wrong term is 165.
प्र: The compound interest earned by Sunil on a certain amount at the end of two years at the rate of 8% p.a. was Rs.2828.80. Find the total amount that Sunil got back at the end of two years in the form of principal plus interest earned ? 2022 05b5cc6dee4d2b4197774ea74
5b5cc6dee4d2b4197774ea74- 1Rs. 11828.80false
- 2Rs. 19828.80true
- 3Rs. 9828.80false
- 4Rs. 19328.80false
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उत्तर : 2. "Rs. 19828.80"
व्याख्या :
Answer: B) Rs. 19828.80 Explanation: Let the sum be Rs. P P{ 1+81002- 1 } = 2828.80 It is in the form of a2-b2 = a+ba-b P(8/100)(2 + 8/100) = 2828.80 P = 2828.80 / (0.08)(2.08) = 1360/0.08 = 17000 Principal + Interest = Rs. 19828.80
प्र: The average age of A and B, 2 years ago was 26. If the age of A, 5 years hence is 40 yrs, and B is 5 years younger to C, then find the difference between the age of A and C? 2020 05b5cc6bbe4d2b4197774d86d
5b5cc6bbe4d2b4197774d86d- 111 yrsfalse
- 29 yrstrue
- 37 yrsfalse
- 413 yrsfalse
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उत्तर : 2. "9 yrs"
व्याख्या :
Answer: B) 9 yrs Explanation: Let the present ages of A and B be 'x' and 'y' respectively From the given data, [(x-2) + (y-2)]/2 = 26 => x+y = 56 But given the age of A, 5 years hence is 40 yrs => present age of A = 40 - 5 = 35 yrs => x = 35 => y = 56 - 35 = 21 => Age of B = 21 yrs Given B is 5 years younger to C, => Age of C = 21 + 5 = 26 yrs => Required Difference between ages of A and C = 35 - 26 = 9 yrs.
प्र: The sum of three consecutive odd numbers is 93. What is the middle number? 2019 05b5cc6b7e4d2b4197774d5e7
5b5cc6b7e4d2b4197774d5e7- 131true
- 233false
- 329false
- 427false
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उत्तर : 1. "31"
व्याख्या :
Answer: A) 31 Explanation: Let the three consecutive odd numbers be x, x+2, x+4 Then, x + x + 2 + x + 4 = 93 => 3x + 6 = 93 => 3x = 87 => x = 29 => 29, 31, 33 are three consecutive odd numbers. Therefore, the middle number is 31.