![]() ![]() If the difference between square of a negative number and itself is equal to 20, find the negative number. Therefore, the required positive number is 25. Given : Difference between y and √y is equal to 20. If the difference between a positive number and its positive square root is equal to 15, find the positive number. The present ages of A and B are 15 years and 10 years respectively. Given : A's age 5 years ago is equal to ⁵⁄₄ of B's age 2 years ago. ![]() Let a and b be the present ages of A and B. Find the present the present ages of A and B. A's age 5 years ago is equal to ⁵⁄₄ of B's age 2 years ago. Therefore, the cost of 100 pencils is $275.Ī is older than B by 5 years. Subsitute x = 100 to estimate the cost of 100 pencils. Substitute A = 1.75 and B = 100 in (1) to get the cost function. Given : The total cost of 128 pencils is $324. Given : The total cost of 84 pencils is $147. (Here y = total cost, x = number of units) Since the cost function is linear, we can write it as given below. If the cost function be linear, find the equation of the cost function and then use it to estimate the cost of 100 pencils. Therefore, the required three digit number is 807.Ī pencil manufaturing company produces 84 pencils at a cost $247 and 128 pencils at a cost of $324. Given : The number formed by switching the digits at hundreds place and tens place is less than the original number by 99.ġ00(y) + 10(0) + 1(x) = 100(x) + 10(0) + 1(y) - 99 Given : Sum of the digits at hundreds place and tens place 15. The number formed by switching the digits at hundreds place and tens place is less than the original number by 99. In a three digit number, the middle digit is zero and sum of the other two digits is 15. Therefore, the number of adults tickets sold was 350 and kids tickets was 250. Given : Cost of each adult ticket was $12 and kid ticket was $7 and tickets were sold for a total of $5950. Given : A total of 600 tickets were sold. Let x be the number of adult tickets and y be the number of kids tickets sold. If a total of 600 tickets were sold for a total of $5950, find the number of adults tickets and kids tickets were sold. Given : If the numerator is increased by 2 and the denominator by 1, the fraction becomes 1/2įor a concert, it was charged $12 for an adult and $7 for a kid. Given : Sum of the numerator and denominator is 6. If the numerator is increased by 2 and the denominator by 1, the fraction becomes 1/2. In a fraction, sum of the numerator and denominator is 6. In triangle ABC, it is given that the average of m ∠A and m∠B is 75 °. In triangle ABC, the average of m ∠A and m ∠B is 75 °. Given : The width of the rectangle is two-fifth of its length. Given : Perimeter of the rectangle is 28 cm. Let x and y be the length and width of the rectangle respectively. If the width is two-fifth of the length, find the area of the rectangle. Multiply both sides of the equation above by 15 to get rid of the denominators 3 and 5. Least common multiple of the two denominators (3, 5) is 15. Given : One-third of the number exceeds its sixth part by 4. If one third of a number exceeds its fifth part by 8, find the number. The equilibrium price is the price in which the quantity of goods demanded is equal to the quantity of goods supplied. The supply equation giving the supply s in lbs. The demand equation for a certain item is ' d = 15( p - 10)', where d is demand in lbs and p is price in dollars. Given : Switching the digits results a number which is less than the original number by 45. Let x and y be the digits at tens place and ones place respectively. Switching the digits results a number which is less than the original number by 45. Sum of the digits in a two digit number is 11. It is given that adding 1 to both numerator and denominator makes the fraction ⅖. If 1 be added to both numerator and denominator, the fraction becomes ⅗. The denominator of a fraction is 1 less than twice the numerartor. Therefore, the present age of Liam is 24 years. If four times of Liam's age 9 years ago be subtracted from thrice of his age 4 years hence, the result would be equal to his present age. It is given that decreasing 23 from 5 times of a number results 57. When 23 is decreased from 5 times a number.
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