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D Divya Prakash Singh
Let the no. of winners be x. Since total no. of participants is 63, thus no. of participants who does not win = 63 - x According to the question, equation becomes:                 x(100) + (63-x)(25) = 3000 or                100x + 1575 - 25x = 3000 or                 75x + 1575 = 3000 Transposing 1575 to the RHS :                  75x = 3000 - 1575 or              75x = 1425 Dividing both...

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D Divya Prakash Singh
Let the no. of Rs.1 coin be y. And the no. of Rs.5 coin be x. Thus according to question no. of Rs.2 coin will be 3x. Also, the total no. of coins = 160 This implies : y + x + 3x = 160 or                     y + 4x = 160 Transposing 4x to the RHS                         y = 160 - 4x = No. of Rs.1 coin.   Now, it is given that total amount is Rs.300. i.e.,        1(160-4x) + 2(3x) + 5(x) =...

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D Divya Prakash Singh
Since the ratio of currency notes is 2:3:5. Therefore, Let the number of currency notes of Rs.100, Rs.50 and  Rs.10 be 2x, 3x, 5x respectively. Hence according to the question equation becomes,               1002x + 503x + 105x = 400000 or           200x + 150x + 50x  = 400000 or                   400x = 400000   Dividing both sides by 400 we get,            x = 1000 . No. of Rs.100 notes =...

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D Divya Prakash Singh
Let the rational number be x. According to the question, Transposing   to the RHS: Multiplication by 2 in both sides, we get Dividing both sides by 5, we get

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D Divya Prakash Singh
Let us assume Ravi's present age to be x years. According to the question,       x + 15 = 4x Transposing x to the RHS          15 = 3x  Now dividing both sides by 3, we get      Hence Ravi's current age is 5 years.

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D Divya Prakash Singh
Let the age of Baichung be x years. Then according to question age of Baichung's father = x + 29 years and age of Baichung's grandfather = x + 29 + 26 years The sum of their ages is given 135 years. According to that, x + x + 29 + x + 29 + 26 = 135   3x + 84 = 135 Transposing 81 on the RHS we get,             3x = 135 - 84 = 51 Dividing both sides by 3    Thus, the age of Baichung = 17...

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D Divya Prakash Singh
Let us assume the number of boys and the number of girls is 7x and 5x.  According to the given data in the question,         7x = 5x + 8 Transposing 5x to the LHS we get,      7x - 5x = 8         2x = 8 Dividing both sides by 2       x = 4 Hence number of boys = 74 = 28     and number of girls = 54 = 20 So total number of students = 28 + 20 = 48.  

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D Divya Prakash Singh
Let the current age of Rahul and Haroon be 5x and 7x repectively(Since there age ratio is given as 5:7). Four years later their ages becomes 5x + 4 and 7x + 4 repectively. According to the question,                 5x + 4 +7x + 4 = 56  or              12x + 8 = 56 Transposing 8 to the RHS we get,            12x = 48 Dividing both sides by 12:          Thus current age of  Rahul  = 54 = 20      ...

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D Divya Prakash Singh
Let the consecutive integers be x, x+1, x+2. Now according to the question equation becomes,               2x + 3(x+1) + 4(x+2) = 74   or         2x + 3x + 3 + 4x + 8 = 74  or             9x + 11 = 74 Transposing 11 to the RHS we get,          9x = 74 - 11 = 63 Divinding both sides by 9.   Therefore required consecutive integers are 7, 8, 9.  

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D Divya Prakash Singh
Let x be the multiple of 8. Then the three consecutive integers (multiple of 8) are x, x+8, x+16. Given their sum is 888, thus the equation becomes: x + x + 8 + x + 16 = 888 or          3x + 24 = 888 Transposing 24 to the RHS              3x = 888 - 24 = 864 Now dividing both sides by 3, we get       Thus the required consecutive integers are 288, 296, 304. 

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D Divya Prakash Singh
Let the consecutive integers be x, x+1, x+2. Sum of these integers is given to be 51. Thus equation becomes :  x + x + 1 + x + 2 = 51       3x + 3 = 51 Transposing 3 to the RHS,       3x = 48 Now dividing both sides by 3, we get      Hence the consecutive integers are 16, 17, 18.       

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D Divya Prakash Singh
Given that numbers are in the ratio of 5:3, so we can assume numbers to be 5x and 3x. Also, the difference between these numbers is 18, so the equation becomes        5x - 3x = 18        2x  = 18 Dividing both sides by 2, we get        x = 9 Hence the two numbers are 59 = 45 and 39 = 27  

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D Divya Prakash Singh
Assume one number to be x. Then the other number will be x + 15. Now it is given that sum of the two numbers is 95.  Thus quation becomes x + x + 15 = 95 or         2x +15 = 95      Transposing 15 to the RHS, it becomes  2x = 95 - 15 = 80 Dividng both sides by 2, we get  Thus two numbers are 40 and 55

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D Divya Prakash Singh
In isosceles triangles, we have 2 sides of equal length. Given that its perimeter is     cm. Let's assume the length of the equal side is x cm. Also,  Transposing   to the RHS side it becomes,    Now dividing both sides by 2, we get    Hence, the length of the equal sides of the isosceles triangle is     cm.  

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D Divya Prakash Singh
Let the breadth of the pool be x m. According to the question, the length of the pool = 2x + 2 m. Perimeter of recatangle = 2(l + b) = 154 m . i.e., 2(2x + 2 + x) = 154     2(3x + 2) = 154 Dividing both sides by 2, we get 3x + 2 = 77 Now transposing 2 to the RHS, we get  3x = 77 - 2 = 75  Dividing both sides by 3, we get Thus breadth of pool = 25 m     and lenght of the pool = 225 + 2 = 52 m   
Assume the number to be x. Thus according to the question,         Multiplying both sides by 2, we get    Now transposing    to the RHS, we get        Thus the number was  x = 3/4
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