Algebra Test - 5

(x + y)³ = x³ + y³ + 3xy ( x + y) putting value
⇒ (1)³ = x³ + y³ + 3xy (1)
⇒ x³ + y³ + 3xy = 1 Answer: (2)

Let one number be x
And second number = 17 - x

According to the given condition
(x)² - (17 - x)² = 119
x² - (289 + x² - 34x) = 119

x² - 289 - x² + 34x - 119 = 0
34x - 408 = 0

34x = 408
x = 408/34 = 12

One number = 12
∴Second number = 12 - 7 = 5. Answer: (1)

B) You might have been careless by not checking that the sum of 12 and 7 is 19.
C) You might have not known the square of 17.
D) You might have thought that end digit of square of 20 is 0 and when we subtract 1 from it, we get the last digit as 9. Answer: (1)

Let first odd number = n
Second odd number = n + 2

According to condition,
n (n + 2) = 575
n² + 2n - 575 = 0

n² + 25n - 23n - 575 = 0
n (n + 25) - 23 (n + 25) = 0

(n + 25) (n - 23) = 0
n = - 25 Rejected

1^st odd number = 23 Second odd number = 25. Sum = 23 + 25 = 48 Answer: (4)

1) You might have written the answer as 84 instead of 48 by inter changing the digits.
2) You might have written the sum of 23 and 25 as 58 wrongly.
3) You might have made some calculation mistakes while performing multiplication operations.
5) You might have read the statement incorrectly and taken the difference of numbers instead of sum of the numbers.

Let the digit at unit`s place be y and the digit at ten`s place be x
∴ Original number = 10x + y

When digits are reversed the number becomes = 10y + x

According to the conditions,
x + y = 10 …(i)
10x + y - 18 = 10y + x …(ii)

Equation (ii) reduces to
9x - 9y = 18
⇒x - y = 2 …(iii)

Adding (i) & (iii), we get
2x = 12 x = 6

Put x = 6 in equation (iii); y = 4
∴ Original number = 64. Answer: (4)
∴ Original number = 64. Answer: (4)

1) You might have guessed that the sum of 3 and 7 is 10, and 37 might be the answer, by checking first option only.
2) You might have selected 46 instead of 64 by mistake.
3) You might have tried to solve the sum by adding the digits of the number randomly.

2x² - 14x + 14 = - 10
⇒ 2x² - 14x + 24 = 0
2x² - 8x - 6x + 24 = 0
2x(x - 4) - 6(x - 4) = 0
(2x - 6) (x - 4) = 0
x = 3, 4
Answer: (3)

Let c be price of clown fish and g be that of guppy.
∴ c + 3g = 5 …….(i) and 2c + 5g = 9…………(ii)
Solving (i) and (ii), we get g = 1, c = 2 Answer: (1)

4^x + 4^{x - 1}
4^x = 4² ⇒ x = 2
x^X = 2² = 4 Answer: (2)