Answer: Option A
Explanation:
\({x}^{2}\) + 112x +\({y}^{2}\) = \({x}^{2}\) + 2(124)x + \({y}^{2}\)
By, \({(a + b)}^{2}\) = \({a}^{2}\) + 2ab + \({b}^{2}\)
Here, a = x and b = y = \(\frac{1}{24}\)
So, for \({x}^{2}\) + 112x +\({y}^{2}\) to be perfect square y = \(\frac{1}{24}\)
2. If x and y are natural numbers such that x + y = 2017, then what is the value of \({(-1)}^{x}\) + \({(-1)}^{y}\)?
Answer: Option C
Explanation:
Sum of two numbers can be odd only if one of them is odd and other one is even. Let x be odd and y be even
â‡’ \({(-1)}^{x}\) + \({(-1)}^{y}\) = -1 + 1 = 0
âˆ´ The value is 0
3. A root of equation a\({x}^{2}\) + bx + c = 0 (where a, b and c are rational numbers) is 5 + 3âˆš3. What is the value of \(\frac{({a}^{2}+ {b}^{2} + {c}^{2})}{(a + b + c)}\)?
Answer: Option D
Explanation:
Since one root of the equation is irrational, so another root will be 5 â€“ 3âˆš3
According to the problem statement,
â‡’ [x â€“ (5 + 3âˆš3)] [x â€“ (5 â€“ 3âˆš3)] = 0
â‡’\({x}^{2}\) â€“ 10x â€“ 2 = 0
â‡’ a = 1, b = -10 and c = -2, Put this in the required equations
âˆ´ \(\frac{({a}^{2}+ {b}^{2} + {c}^{2})}{(a + b + c)}\) = \(\frac{(1 + 100 + 4)}{ (1 â€“ 10 â€“ 2)}\) = \(\frac{-105}{11}\)
4. If the arithmetic mean of two numbers is 7 and the geometric mean of the same two number is 2âˆš10. Then find the numbers x and y respectively, such that x > y.
Answer: Option D
Explanation:
Given, Arithmetic Mean = 7
â‡’ \(\frac{(x + y)}{2}\) = 7
â‡’ x + y = 14 —-(1)
And, Geometric Mean = 2âˆš10
â‡’ âˆšxy = 2âˆš10
â‡’ xy = \({(2âˆš10)}^{2}\) = 40
Now,
â‡’ \({(x – y)}^{2}\) = \({(x + y)}^{2}\) – 4xy = 142 – (4 Ã— 40) = 196 – 160 = 36
â‡’ x – y = 6 —-(2)
Now, adding equation (1) and (2) we get
â‡’ 2x = 20
â‡’ x = 10
Now, by equation (1)
â‡’ y = 14 – x = 14 – 10 = 4
5. Cost of 8 pencils, 5 pens and 3 erasers is Rs. 111. Cost of 9 pencils, 6 pens and 5 erasers is Rs. 130. Cost of 16 pencils, 11 pens and 3 erasers is Rs. 221. What is the cost (in Rs) of 39 pencils, 26 pens and 13 erasers?
Answer: Option
Explanation:
Let the price of single pencil, pen, and eraser be x, y, and z respectively
According to question,
8x + 5y + 3z = Rs. 111 —-(1)
9x + 6y + 5z = Rs. 130 —-(2)
16x + 11y + 3z = Rs. 221 —-(3)
Subtracting equation (1) from (3)
â‡’ (16x + 11y + 3z) – (9x + 6y + 5z) = 221 – 111
â‡’ 8x + 6y = 110
â‡’ 4x + 3y = 55 —-(4)
Multiply the equation (2) by 3 and (3) by 5 and then subtracting equation (2) from (3)
â‡’ (16x + 11y + 3z) Ã— 5 – (9x + 6y + 5z) Ã— 3 = 221 Ã— 5 – 130 Ã— 3
â‡’ 80x + 55y + 15z – 27x – 18y – 15z = 1105 – 390
â‡’ 53x + 37y = 715 —-(5)
Multiply the equation (4) by 53 and (5) by 4 and then subtracting equation (4) from (5)
â‡’ 212x + 159y – 212x – 148y = 2915 – 2860
â‡’ 11y = 55
â‡’ y = 5
By putting the value of y = 5 in equation (4)
â‡’ 4x + 3 Ã— 5 = 55
â‡’ x = 10
By putting the value of y = 5 and x = 10 in equation (1)
â‡’ 8 Ã— 10 + 5 Ã— 5 + 3z = 111
â‡’ 80 + 25 + 3z = 111
â‡’ z = 2
âˆ´ Cost of 39 pencils, 26 pens and 13 erasers is 39x + 26y + 13z =39 Ã— 10 + 26 Ã— 5 + 13 Ã— 2 = Rs. 546
Answer: Option B
Explanation:
Using algebraic identities,
\({(a + b + c)}^{2}\) = \({a}^{2}\) + \({b}^{2}\) + \({c}^{2}\) + 2ab + 2bc + 2ca
By putting the respective values given in question,
â‡’ \({(9)}^{2}\) = \({a}^{2}\) + \({b}^{2}\) + \({c}^{2}\) + 2(ab + bc + ca) [âˆµ ab + bc + ca = 26]
â‡’ \({(9)}^{2}\) = \({a}^{2}\) + \({b}^{2}\) + \({c}^{2}\) + 2(26)
â‡’ \({a}^{2}\) + \({b}^{2}\) + \({c}^{2}\) = 81 – 52 = 29
Given equations,
\({a}^{3}\) + \({b}^{3}\) = 91 —-(1)
\({b}^{3}\) + \({c}^{3}\) = 72 —-(2)
\({c}^{3}\) + \({a}^{3}\) = 35 —-(3)
On adding (1), (2) and (3)
\({a}^{3}\) + \({b}^{3}\) + \({b}^{3}\) + \({c}^{2}\) + \({c}^{2}\) + \({a}^{3}\) = 91 + 72 + 35
â‡’ 2(\({a}^{3}\) + \({b}^{3}\) + \({c}^{3}\)) = 198
â‡’ \({a}^{3}\) + \({b}^{3}\) + \({c}^{3}\) = 99
Using algebraic identities,
\({a}^{3}\) + \({b}^{3}\) + \({c}^{3}\) – 3abc = (a + b + c) (\({a}^{2}\) + \({b}^{2}\) + \({c}^{2}\) – ab – bc – ca)
By putting the respective values,
â‡’ 99 – 3abc = 9 (29 – 26) [âˆµ ab + bc + ca = 26 and a + b + c = 9]
â‡’ 3abc = 99 – 27
â‡’ abc = \(\frac{72}{3}\)
âˆ´ abc = 24
2. If \(\frac{x}{a}\) + \(\frac{y}{b}\) = 3 and \(\frac{x}{b}\) â€“ \(\frac{y}{a}\) = 9, then what is the value of \(\frac{x}{y}\)?
Answer: Option
Explanation:
Here\(\frac{x}{a}\) + \(\frac{y}{b}\) = 3
â‡’ bx + ay â€“ 3ab = 0 —- (1)
Here\(\frac{x}{b}\) â€“ \(\frac{y}{a}\) = 9
â‡’ ax â€“ by â€“ 9ab = 0 —- (2)
Multiplying equation 1 by â€˜aâ€™ and equation 2 by â€˜bâ€™
â‡’ abx + \({a}^{2}\)y â€“ 3\({a}^{2}\)b = 0 —- (3)
â‡’ abx â€“ \({b}^{2}\)y â€“ 9a\({b}^{2}\) = 0 —- (4)
Subtracting equation 4 from 3,
â‡’ y =\(\frac{3ab(a â€“ 3b)}{({a}^{2} + {b}^{2})}\),
substituting value of y in equation 3,
â‡’ x =\(\frac{ 3ab(b + 3a)}{({a}^{2}+ {b}^{2})}\)
â‡’ \(\frac{x}{y}\) = \(\frac{(b + 3a)}{(a â€“ 3b)}\)
âˆ´ The value of \(\frac{x}{y}\) = \(\frac{(b + 3a)}{(a â€“ 3b)}\)
3. The value of x for which the expressions 19 – 5x and 19x + 5 become equal is ______.
Answer: Option A
Explanation:
According to the given information,
19 â€“ 5x = 19x + 5
â‡’ 19 â€“ 5 = 19x + 5x
â‡’ 14 = 24x
âˆ´ x = \(\frac{14}{24}\) = \(\frac{7}{12}\)
4. What is the sum of all the natural numbers from 1 to 35?
Answer: Option B
Explanation:
We know that Formula for Sum of natural no’s beginning from 1
= \({n}^{2}\) Ã— (1+n)
where n = last no. i.e. 35
â‡’ Sum = 352 Ã— (1 + 35)
â‡’ Sum = 35 Ã— 18 = 630
5. The sum of three consecutive odd natural numbers is 93. The smallest of three numbers is:
Answer: Option A
Explanation:
Let the three consecutive odd numbers be a, a + 2, a + 4.
Given, the sum of three consecutive odd natural numbers is 93.
âˆ´ a + a + 2 + a + 4 = 93
â‡’ 3a = 87
â‡’ a = 29
Answer: Option D
Explanation:
Let the price of chair be â€˜aâ€™.
Given, the price of 10 chairs is equal to that of 4 tables.
âˆ´ 10 Ã— a = 4 Ã— price of the table
â‡’ Price of table = 2.5a
Given, the price of 15 chairs and 3 tables together is Rs. 4500
âˆ´ 15a + 7.5a = 4500
â‡’ 22.5a = 4500
â‡’ a = Rs. 200
Price of table = 2.5a = Rs. 500
Total price of 12 chairs and 5 tables = 12 Ã— 200 + 5 Ã— 500 = Rs. 4900
2. A number is greater than twice its reciprocal by \(\frac{31}{4}\). Find the number.
Answer: Option B
Explanation:
Let the number be â€˜xâ€™.
A number is greater than twice its reciprocal by \(\frac{31}{4}\)
â‡’ x = 2(\(\frac{1}{x}\)) + \(\frac{31}{4}\)
â‡’ x – \(\frac{2}{x}\)= \(\frac{31}{4}\)
â‡’ 4\({(x -2)}^{2}\) = 31x
â‡’ 4\({x}^{2}\) – 31x – 8 = 0
â‡’ 4\({x}^{2}\) – 32x + x – 8 = 0
â‡’ 4x(x – 8) + 1(x – 8) = 0
â‡’ (x – 8)(4x + 1) = 0
x = 8 or x = – \(\frac{1}{4}\)
âˆ´ x = 8
3. If x = 2 then the value of \({x}^{3}\) + 27\({x}^{2}\) + 243x + 631 is:
Answer: Option C
Explanation:
Given equation,
f(x) = \({x}^{3}\) + 27\({x}^{2}\) + 243x + 631
â‡’ x(\({x}^{2}\) + 27x+ 243) + 631
Now, put the value of x = 2
â‡’ 2(22 + 27 Ã— 2 + 243) + 631
â‡’ 2 (4 + 54 + 243) + 631
â‡’ 2(301) + 631 = 602 + 631 = 1233.
4. If a = 2.234, b = 3.121 and c = â€“5.355, then the value of \({a}^{3}\) + \({b}^{3}\) + \({c}^{3}\) â€“ 3abc is
Answer: Option B
Explanation:
a + b + c = 2.234 + 3.121 â€“ 5.355 = 0
If a + b + c = 0, then\({a}^{3}\) + \({b}^{3}\) + \({c}^{3}\) â€“ 3abc = 0, which can be proved as under
a + b = â€“ c
Cubing both sides, we get
â‡’ \({(a + b)}^{3}\)= \({(-c)}^{3}\)
â‡’ \({a}^{3}\) + \({b}^{3}\) + 3ab(a + b) = â€“ \({c}^{3}\)
â‡’ \({a}^{3}\) + \({b}^{3}\) + 3ab(â€“ c) = â€“ \({c}^{3}\)
â‡’ \({a}^{3}\) + \({b}^{3}\) â€“ 3abc = â€“ \({c}^{3}\)
â‡’ \({a}^{3}\) + \({b}^{3}\) + \({c}^{3}\) â€“ 3abc = 0
5. If \({x}^{2}\) + \({y}^{2}\) + 1 = 2x, then the value of \({x}^{3}\) + \({y}^{5}\) is
Answer: Option D
Explanation:
\({x}^{2}\) + \({y}^{2}\) + 1 = 2x
â‡’ \({x}^{2}\) + \({y}^{2}\) + 1 â€“ 2x = 0
â‡’ \({x}^{2}\) â€“ 2x + 1 + \({y}^{2}\) = 0
â‡’ \({(x – 1)}^{2}\) + \({y}^{2}\) = 0
In the above eq. the L.H.S. can only become zero when the base of terms; (x â€“ 1) and y becomes zero because for any other value the sum of their squares will always be a positive integer.
Taking (x â€“ 1) = 0
and y = 0
Therefore, x = 1 and y = 0
âˆ´ \({x}^{3}\) + \({y}^{5}\) = 1 + 0 = 1.
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