Quantitative Aptitude प्रश्न और उत्तर का अभ्यास करें
8प्र: How many iron rods, each of length 7 mts and diameter 2 cms can be made out of 0.88 cubic metre of iron ? 1457 05b5cc77fe4d2b419777501c4
5b5cc77fe4d2b419777501c4- 1500false
- 2600false
- 3400true
- 4300false
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उत्तर : 3. "400"
व्याख्या :
Answer: C) 400 Explanation: Given length of iron rod =7 mts and and diameter of rod = 2 cms => r = 1/100 mts Therefore, Volume of one rod = πr2h cu. m = 227×11002×7cu. m = 115000 cu. m Given volume of iron = 0.88 cu. m Therefore, Number of rods = 0.88×500011 = 400.
प्र: 5 sin(x)+ 12 cos(x) + r is always greater than or equal to zero. What is the smallest value satisfied by 'r' ? 1399 05b5cc77fe4d2b419777501bf
5b5cc77fe4d2b419777501bf- 1-13false
- 211false
- 313true
- 4-11false
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उत्तर : 3. "13"
व्याख्या :
Answer: C) 13 Explanation: Given, 5sinx + 12cosx ≥ -r ⇒13513sin x + 1213cos x ≥ -rcosx) ≥ -r 513= cosA => sinA =1213 => 13(sinx cosA + sinA cosx) ≥ -r => 13(sin(x + A)) ≥ -r we know that , -1 ≤ sin (angle) ≤ 1 => 13sin (x + A) ≥ -13 ⇒rmin =13
प्र: How many numbers with distinct digits are possible product of whose digits is 18? 1983 05b5cc77fe4d2b419777501ba
5b5cc77fe4d2b419777501ba- 112false
- 28true
- 36false
- 44false
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उत्तर : 2. "8"
व्याख्या :
Answer: B) 8 Explanation: Two digit numbers: The two digits can be 2 and 9: Two possibilities 29 and 92.Three-digit numbers: The three digits can be 1, 2 and 9 => 3! Or 6 possibilities.We cannot have three digits as (3, 3, 2) as the digits have to be distinct.We cannot have numbers with 4 digits or more without repeating the digits.So, there are totally 8 numbers.
प्र: The difference between the squares of two consecutive odd integers is always divisible by ? 1139 05b5cc77fe4d2b419777501b5
5b5cc77fe4d2b419777501b5- 18true
- 22false
- 36false
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उत्तर : 1. "8"
व्याख्या :
Answer: A) 8 Explanation: Let the two consecutive odd integers be (2x + 1) and (2x + 3)Then, (2x + 3)2 - (2x + 1)2 = (2x + 3 + 2x + 1) (2x + 3 - 2x - 1) = (4x + 4)(2)= 8 (x + 1), which is always divisible by 8
प्र: What should be the maximum value of B in the following equation?8A9 – 6B2 + 4C6 = 723 1598 05b5cc77fe4d2b419777501b0
5b5cc77fe4d2b419777501b0- 14false
- 21false
- 35false
- 47true
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उत्तर : 4. "7"
व्याख्या :
Answer: D) 7 Explanation: 1 1 8 A 9 + 4 C 6 - 6 B 2 7 2 3 We may represent the given sum, as shown below:1 + A + C - B = 12 A + C - B = 11 Now,giving the maximum values to A and C, i.e.A = 9 and C = 9, we get B = 7.
प्र: How many prime numbers exist in 65×359×113 ? 2557 05b5cc77fe4d2b419777501ab
5b5cc77fe4d2b419777501ab- 129false
- 231true
- 333false
- 427false
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उत्तर : 2. "31"
व्याख्या :
Answer: B) 31 Explanation: Given 65×359×113 ⇒2×35×5×79×113 ⇒25×35×59×79×113 Thus, there are (5 + 5 + 9 + 9 + 3)= 31 prime numbers.
प्र: The ratio of Karthik’s and Kalyan’s ages is 4: 5. If the difference between the present age of Kalyan and the age of Karthik 5 years hence is 3 years, then what is the total of present ages of Karthik and Kalyan? 2807 05b5cc77fe4d2b419777501a1
5b5cc77fe4d2b419777501a1- 156 yearsfalse
- 262 yearsfalse
- 369 yearsfalse
- 472 yearstrue
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उत्तर : 4. "72 years"
व्याख्या :
Answer: D) 72 years Explanation: Let us consider Karthik as K1 and Kalyan as K2. Given k18+k1 and K2 - (K1 + 5) = 3 K2 - K1 = 8 K2 = 8 + K1 Now, k18+k1= 45 => K1 = 32 years Therefore K2 = 40 years. K1 + K2 = 72 years.
प्र: Ten years ago, Kumar was thrice as old as Sailesh was but 10 years hence, he will be only twice as old. Find Kumar’s present age ? 3298 05b5cc77fe4d2b419777501a6
5b5cc77fe4d2b419777501a6- 150 yearsfalse
- 270 yearstrue
- 360 yearsfalse
- 440 yearsfalse
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उत्तर : 2. "70 years"
व्याख्या :
Answer: B) 70 years Explanation: Let Kumar’s present age be "x" years and Sailesh’s present age be "y" years.Then, according to the first condition, x - 10 = 3(y - 10) => x - 3y = - 20 ........(1) Now, Kumar’s age after 10 years = (x + 10) years. Sailesh’s age after 10 years = (y + 10) years. (x + 10) = 2 (y + 10) => x - 2y = 10 ......(2) Solving (1) and (2), we get x = 70 and y = 30 Kumar’s present age = 70 years and Sailesh’s present age = 30 years.