Quantitative Aptitude प्रश्न और उत्तर का अभ्यास करें

प्र: 2/3 as a decimal 1950 0

  • 1
    0.47
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  • 2
    6.77
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  • 3
    0.67
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  • 4
    0.07
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उत्तर : 3. "0.67"
व्याख्या :

Answer: C) 0.67 Explanation: To convert a fraction into decimal, just take the number on top which we called the numerator and divide it by the number at the bottom which we called the denominator. 2/3 How it comes : 2 ÷ 3 = 0 bring down the decimal point. 20 ÷ 3 = 6 and carry the remainder 2 to make 20 20 ÷ 3 = 6 and carry the remainder 2 to make 20 20 ÷ 3 = 6 and carry the remainder 2 to make 20 etc... This gives the recurring decimal 0.6666... If we round off to two decimals , then it becomes 0.67.   Hence, decimal fraction of 2/3 = 0.67.

प्र: How many six digit odd numbers can be formed from the digits 0, 2, 3, 5, 6, 7, 8, and 9 (repetition not allowed)? 1950 0

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    8640
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  • 2
    720
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  • 3
    3620
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  • 4
    4512
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उत्तर : 1. "8640"
व्याख्या :

Answer: A) 8640 Explanation: Let the 6 digits of the required 6 digit number be abcdef Then, the number to be odd number the last digit must be odd digit i.e 3, 5, 7 or 9 The first digit cannot be ‘0’ => possible digits = 3, 5, 7, 2, 6, 8 Remaining 4 places can be of 6 x 5 x 4 x 3 ways This can be easily understood by Therefore, required number of ways = 6 x 6 x 5 x 4 x 3 x 4 = 36 x 20 x 12 = 720 x 12 = 8640 ways.

प्र: A drink vendor has 368 liters of Maaza, 80 liters of Pepsi and 144 liters of Sprite. He wants to pack them in cans, so that each can contains the same number of liters of a drink, and doesn't want to mix any two drinks in a can. What is the least number of cans required ? 1949 0

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    47
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  • 2
    46
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  • 3
    37
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  • 4
    35
    सही
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उत्तर : 3. "37"
व्याख्या :

Answer: C) 37 Explanation: The number of liters in each can = HCF of 80, 144 and 368 = 16 liters.Number of cans of Maaza = 368/16 = 23Number of cans of Pepsi = 80/16 = 5Number of cans of Sprite = 144/16 = 9The total number of cans required = 23 + 5 + 9 = 37 cans.

प्र: A six-digit number is formed by repeating a three-digit number; for example, 404404 or 415415 etc. Any number of this form is always exactly divisible by 1948 0

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    101
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  • 2
    901
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  • 3
    1001
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  • 4
    789
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उत्तर : 3. "1001"
व्याख्या :

Answer: C) 1001 Explanation: Here by trial and error method, we can obseve that 404404 = 404 x 1001; 415415 = 415 x 1001, etc. So, any number of this form is divisible by 1001.

प्र: Find the next number in the given number series? 2, 19, 0, 21, –2, ? 1948 0

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    23
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  • 2
    25
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  • 3
    27
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  • 4
    29
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उत्तर : 1. "23"
व्याख्या :

Answer: A) 23 Explanation: The given number series 2, 19, 0, 21, –2, ? follows a pattern, 2 2 + 17 = 19 19 - 19 = 0 0 + 21 = 21 21 - 23 = -2 -2 + 25 = 23   Hence, the next number in the series is 23.

प्र: A mixture contains 25% milk and rest water. What percent of this mixture must taken out and replaced with milk so that in mixture milk and water may become equal. 1946 0

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    31.8%
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  • 2
    31%
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  • 3
    33.33%
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  • 4
    29.85%
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उत्तर : 3. "33.33%"
व्याख्या :

Answer: C) 33.33% Explanation: Now, take percentage of milk and applying mixture rule 25          100        50 50            25  = 2 : 1   Hence required answer =  1/3 or 33.33%

प्र: P alone can do a piece of work in 24 days. The time taken by P to complete one-third of work is equal to time taken by Q to complete half of the work. How many days are required to complete by P and Q working together ? 1946 0

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    9 3/5 days
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  • 2
    7 days
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  • 3
    6 3/7 days
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  • 4
    8 days
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उत्तर : 1. "9 3/5 days"
व्याख्या :

Answer: A) 9 3/5 days Explanation: Let the total work be 'W' As per the given information, P can complete the work 'W' in 24 days. => one day work of P = W/24 And also given that, The time taken by P to complete one-third of work is equal to time taken by Q to complete half of the work => PW/3 = QW/2 => P's 1 day =    W/24    ?    =      W/3 =>  ?  = 24W/3W  = 8 => QW/2 = 8 days => Q alone can complete the work W in 16 days => P + Q can complete the work in 1/24 + 1/16 = 5/48 => 48/5 days = 9 3/5 days.

प्र: If A1, A2, A3, A4, ..... A10 are speakers for a meeting and A1 always speaks after, A2 then the number of ways they can speak in the meeting is 1945 0

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    9!
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  • 2
    9!/2
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  • 3
    10!
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  • 4
    10!/2
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उत्तर : 4. "10!/2"
व्याख्या :

Answer: D) 10!/2 Explanation: As A1 speaks always after A2, they can speak only in  1st  to 9th places and    A2 can speak in 2nd to 10 the places only when A1 speaks in 1st place    A2 can speak in 9 places the remaining     A3, A4, A5,...A10  has no restriction. So, they can speak in 9.8! ways. i.e   when A2 speaks in the first place, the number of ways they can speak is 9.8!.   When A2 speaks in second place, the number of ways they can speak is  8.8!.   When A2 speaks in third place, the number of ways they can speak is  7.8!. When A2 speaks in the ninth place, the number of ways they can speak is 1.8!       Therefore,Total Number of ways they can  speak = (9+8+7+6+5+4+3+2+1) 8! = 92(9+1)8! = 10!/2

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