Aptitude - Simplification
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- Simplification - Formulas
- Simplification - General Questions
Let number of notes of each denomination be x.
Then x + 5x + 10x = 480
16x = 480
x = 30.
Hence, total number of notes = 3x = 90.
Let the number of students in rooms A and B be x and y respectively.
Then, x - 10 = y + 10 x - y = 20 .... (i)
and x + 20 = 2(y - 20) x - 2y = -60 .... (ii)
Solving (i) and (ii) we get: x = 100 , y = 80.
The required answer A = 100.
Let the cost of a chair and that of a table be Rs. x and Rs. y respectively.
Then, 10x = 4y or y = | 5 | x. |
2 |
15x + 2y = 4000
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5 | x = 4000 |
2 |
20x = 4000
x = 200.
So, y = | ![]() |
5 | x 200 | ![]() |
= 500. |
2 |
Hence, the cost of 12 chairs and 3 tables = 12x + 3y
= Rs. (2400 + 1500)
= Rs. 3900.
2ab = (a2 + b2) - (a - b)2
= 29 - 9 = 20
ab = 10.
Let the price of a saree and a shirt be Rs. x and Rs. y respectively.
Then, 2x + 4y = 1600 .... (i)
and x + 6y = 1600 .... (ii)
Divide equation (i) by 2, we get the below equation. => x + 2y = 800. --- (iii) Now subtract (iii) from (ii) x + 6y = 1600 (-) x + 2y = 800 ---------------- 4y = 800 ---------------- Therefore, y = 200. Now apply value of y in (iii) => x + 2 x 200 = 800 => x + 400 = 800 Therefore x = 400
Solving (i) and (ii) we get x = 400, y = 200.
Cost of 12 shirts = Rs. (12 x 200) = Rs. 2400.