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Algebra Level 3 Test 1
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Question 1 |
If x > 5 and y < –1, then which of the following statements is true?
(x + 4y) > 1 | |
x > –4y | |
–4x < 5y | |
None of these |
Question 1 Explanation:
From the question we have x > 5, y < –1
Let’s assume values: x = 6, y = –6.
For these, we see none of the statements (1, 2 and 3) are true.
Hence the correct option is (d).
Let’s assume values: x = 6, y = –6.
For these, we see none of the statements (1, 2 and 3) are true.
Hence the correct option is (d).
Question 2 |
Two men X and Y started working for a certain company at similar jobs on January 1, 1950. X asked for an initial salary of Rs. 300 with an annual increment of Rs. 30. Y asked for an initial salary of Rs. 200 with a rise of Rs. 15 every 6 months. Assume that the arrangements remained unaltered till December 31, 1959. Salary is paid on the last day of the month. What is the total amount paid to them as salary during the period?
Rs. 93,300 | |
Rs. 93,200 | |
Rs. 93,100 | |
None of these |
Question 2 Explanation:
From the given data
The amount of money given to X = 12 × 300 + 12 × 330 + ... + 12 × 570
= 12[300 + 330 + ... + 540 + 570]
= 12 ×10/2[600 + 9 x 30 ] = 52200
Amount of money given to Y is 6 × 200 + 6 × 215 + 6 × 230 + 6 × 245 + ... to 20 terms
= 6[200 + 215 + 280 ... 485]
= 6 × 20/2[400 + 19 x 15]
= 6 × 10[400 + 285]
= 60 × 685 = 41100
Therefore Total amount paid = 52200 + 41100 = Rs. 93,300.
The amount of money given to X = 12 × 300 + 12 × 330 + ... + 12 × 570
= 12[300 + 330 + ... + 540 + 570]
= 12 ×10/2[600 + 9 x 30 ] = 52200
Amount of money given to Y is 6 × 200 + 6 × 215 + 6 × 230 + 6 × 245 + ... to 20 terms
= 6[200 + 215 + 280 ... 485]
= 6 × 20/2[400 + 19 x 15]
= 6 × 10[400 + 285]
= 60 × 685 = 41100
Therefore Total amount paid = 52200 + 41100 = Rs. 93,300.
Question 3 |
x and y are real numbers satisfying the conditions 2 < x < 3 and – 8 < y < –7. Which of the following expressions will have the least value?
xy2 | |
x2y | |
5xy | |
None of these |
Question 3 Explanation:
From the given data we come to know that one value would be negative and one value would be positive.
Using the conditions 2 < x < 3 and – 8 < y < –7 , the value of y would be negative and the value of x would be positive.
Let us analyze the options:
We can see that since y has been squared in option a, its value becomes positive.
The value of xy2 would be positive and will not be the minimum.
From (b) and (c), x2y and 5xy would give negative values but we cannot find the specific negative and positive values
On comparing (a) and (c), we find that x2 < 5x in 2 < x < 3.
Since the value of y is negative therefore x2y > 5xy
Therefore 5xy would give the minimum value.
Using the conditions 2 < x < 3 and – 8 < y < –7 , the value of y would be negative and the value of x would be positive.
Let us analyze the options:
We can see that since y has been squared in option a, its value becomes positive.
The value of xy2 would be positive and will not be the minimum.
From (b) and (c), x2y and 5xy would give negative values but we cannot find the specific negative and positive values
On comparing (a) and (c), we find that x2 < 5x in 2 < x < 3.
Since the value of y is negative therefore x2y > 5xy
Therefore 5xy would give the minimum value.
Question 4 |
m is the smallest positive integer such that for any integer n ≥ m, the quantity n3 – 7n2 + 11n – 5 is positive. What is the value of m?
4 | |
5 | |
8 | |
None of these |
Question 4 Explanation:
Let the quantity y = n3 – 7n2 + 11n – 5
= (n – 1) (n2 – 6n + 5)
= (n – 1)2 (n – 5)
Now (n – 1)2 is always positive.
Expression gives a negative quantity for n < 5 the
Therefore, n will be 6.
Hence m = 6.
= (n – 1) (n2 – 6n + 5)
= (n – 1)2 (n – 5)
Now (n – 1)2 is always positive.
Expression gives a negative quantity for n < 5 the
Therefore, n will be 6.
Hence m = 6.
Question 5 |
If a, b, c and d are four positive real numbers such that abcd = 1, what is the minimum value of (1 + a)(1 + b)(1 + c)(1 + d)?
4 | |
1 | |
16 | |
18 |
Question 5 Explanation:
We have to make the product of four numbers 1.
For this we need take the value of a = b = c = d = 1.
And get the answer as the option number (c)
Try with other values, you would see why we assumed a,b,c and d as 1.
For this we need take the value of a = b = c = d = 1.
And get the answer as the option number (c)
Try with other values, you would see why we assumed a,b,c and d as 1.
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