Theory of Equations Test

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Theory of Equations Test - 5

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Question 1
1 / -0

Karan and Nikhil are basketball players. Their ratio of shooting the ball to the basket is 8 : 3, where Karan shoots 8 balls when Nikhil shoots 3. But, Karan converts one basket in four shots, while Nikhil converts one basket in two shots. If Nikhil has missed 39 shots, how many shots has Karan converted?

A
52

B
53

C
54

D
55

Solution

Karan Nikhil
Total shots 8k 3k
If Nikhil missed 39, then he would have made 39 also ( 1 out of 2 are made)
Total shots Nikhil shot = 78
Total shots Karan shot = 78 × = 208
Total shots converted by Karan = (1 out of 4) = 52
Thus, answer option 1 is correct
Question 2
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The taxi charges in a city comprise of a fixed charge with the charge for the distance covered. For a journey of 10 km, the charge paid is Rs. 75 and for a journey of 15 km, the charge paid is Rs. 110. What will a person have to pay for travelling a distance of 25 km?

A
Rs. 180

B
Rs. 100

C
Rs. 90

D
Rs. 50

Solution

Let fixed charges = Rs. x
And charges per distance = Rs. y/km
According to I^st condition,
x + 10y = 75 … (i)
According to II^nd condition,
x + 15y = 110 … (ii)

y =
Putting the value of y in equation (i),
x + 70 = 75 ⇒ x = Rs. 5
For travelling 25 km, he has to pay = 5 + 25 × 7 = 5 + 175 = Rs. 180.
Question 3
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When you reverse the number 13, the number increases by 18. How many other two-digit numbers increase by 18 when their digits are reversed?

A
5

B
6

C
7

D
8

Solution

Let the number be 10x + y.
After reversing the digits, the number formed = 10y + x.
A.T.Q.
10y + x = (10x + y) + 18
∴ (y - x) = 2

Total 7 pairs including 13 and 6 excluding 13
Question 4
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How many pairs of positive integers (m, n) satisfy , where n is an odd integer less than 60?

A
6

B
4

C
5

D
3

Solution

(n is odd < 60)
m = ...(1) (m and n are +ve integers.)
So, n – 48 ≥ 1
So, the values of n for which (1) is satisfied are 49, 51 and 57.
Question 5
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In a shop, the cost of 3 burgers, 7 milkshakes and French fries is Rs. 120. In the same shop, the cost of 4 burgers, 10 milkshakes and French fries is Rs. 164.5. Find the cost of 1 burger, 1 milkshake and French fries in the shop.

A
Rs. 41

B
Rs. 21

C
Rs. 31

D
Rs. 51

Solution

3B + 7M + F = 120 .……..(1)
4B + 10M + F = 164.5 …….. (2)
(2) – (1)
⇒ B + 3M = 44.5 ……... (3)
2B + 6M = 89
Putting (3) in (1), we get
⇒ (2B + 6M) + B + M + F = 120
⇒ B + M + F = 31
Question 6
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A two digit number becomes five-sixth of itself if the digits are reversed. The two digits differ by one. The number is

A
56

B
54

C
45

D
65

Solution

(10a + b) = 10b + a ……….. (1)
And a - b = 1 ………. (2)
From (1),
50a + 5b = 60b + 6a
⇒ 4a = 5b
or we have
b = 4
a = 5
Question 7
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How many positive integer pairs (a, b) satisfy the equation ab = a + b + 20?

A
0

B
2

C
3

D
4

Solution

ab = a + b + 20
ab - a = b - 1 + 21
a(b - 1) - 1 (b - 1) = 21
(a - 1) (b - 1) = 21
= 1 × 21, 3 × 7, 7 × 3, 21 × 1
∴ a = 2, b = 22
a = 4, b = 8
a = 8, b = 4
a = 22, b = 2
Total of 4 pairs satisfy the given equation.
Question 8
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The number of solutions of the equation 7x + 3y = 123, where x, y > 0, is

A
3

B
4

C
5

D
6

Solution

7x + 3y = 123
When, y = 6, x = 15
y = 13, x = 12
y = 20, x = 9
y = 27, x = 6
y = 34, x = 3
Here, above solutions are ordered pairs.
So, we have used the concepts of ordered pair.
The only possible pairs when x, y > 0 is 5.
Question 9
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A piece of string is 40 cm long. It is cut into three pieces. The longest piece is three times as long as the middle-sized and the shortest piece is 23 cm shorter than the longest piece. Find the length of the shortest piece.

A
27 cm

B
5 cm

C
4 cm

D
9 cm

Solution

Let the largest piece of string be 3x cm, then middle-sized and the shortest piece would be x cm and
(3x - 23) cm,
respectively.
Or 3x + x + (3x - 23) = 40
⇒ x = 9
Therefore, the shortest piece = (3 × 9 - 23) = 4
Question 10
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A businessman knows that the price of a commodity will increase by Rs. 5 per packet. He bought some packets of the commodity for Rs. 4,500. If he bought them on new price, then he got 10 packets less. Find the number of packets bought by him.

A
90

B
100

C
50

D
125

Solution

Let the price of commodity be x per packet.
So, he can buy packets.
When the price of commodity is increased by Rs. 5, he would buy packets
According to question,
- = 10
Or = 10
45 × 50 = x (x + 5)
So x = 5
So, the number of packets bought by him is 50. Hence, option (3) is the answer
Question 11
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The number of ordered pairs of natural numbers (x, y) satisfying the equation 2x + 3y = 100 is

A
13

B
14

C
15

D
16

Solution

2x + 3y = 100
The first pair, (x, y), satisfying the above equation is (2, 32).
The other ordered pairs can be derived from the above pair by following a pattern (Adding 3 to every value of x and subtracting 2 from every value of y).
Thus, the pairs are (5, 30); (8, 28); (11, 26); (14, 24); (17, 22); (20, 20); (23, 18); (26, 16); (29, 14); (32, 12); (35, 10); (38, 8); (41, 6); (44, 4); (47, 2). The total pairs are 16.
Question 12
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Examine the nature of roots of
(i) x² + 9x + 27 = 0
(ii) 6x² - 13x - 5 = 0

A
(i) Real (ii) Real

B
(i) Real (ii) Imaginary

C
(i) Imaginary (ii) Real

D
(i) Imaginary (ii) Imaginary

Solution

(i) D = b² - 4ac = (+9)² - 4 Â´ 1 Â´ 27 = 81 - 108 = - 27, which is
negative. Hence, the roots will be imaginary and conjugate.
(ii) D = b² - 4ac = (- 13)² - 4 Â´ 6 Â´ (- 5) = 169 + 120 = 289,
which is a perfect square. Hence, the roots will be real, rational
and unequal.
Question 13
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How many integral roots does the equation x – = 3 – have?

A
0

B
1

C
2

D
3

Solution

x – = 3 –
x = 3 – + = 3
When x = 3, is not defined (As denominator is zero).
Hence, this equation has no integral roots.
Question 14
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Which one of the following pairs cannot be the roots of the quadratic equation?

A
4, - 4

B
5 + , 5 -

C
3,

D
2, - 3

Solution

Since, irrational roots occur in pairs,
So, 3 and is not possible.
Question 15
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The square of one-eighth of a number when added to 12 results in the number itself. Find the number.

A
16

B
48

C
16 or 48

D
64

Solution

+ 12 = x
+ 12 = x
x² - 64x + 768 = 0
(x - 16) (x - 48)
So, the required answer is (16, 48)
Question 16
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If 9^z - 10 × 3^z+ 9 = 0, then z is equal to

A
2 or 0

B
1 or 3

C
1 or 9

D
1 or - 2

Solution

9^z - 10 ´ 3^z + 9 = 0 3^2z - 10 ´ 3^z + 9 = 0 ………….. (1)
Putting 3^z = y in the equation (1), we get
∴ y² - 10 y + 9 = 0 (y - 9) (y - 1) = 0 y = 1 or 9
Either 3^z = 1 3^z = 3⁰ z = 0
Or 3^z = 9 3^z = 3² z = 2
Question 17
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Two farmers Ravi and Ram have 40 goats in total. They sell their goats at different prices, but each of them receives the same amount of sum. If Ravi had sold his goats at Ram's price, he would have received Rs. 5400. If Ram had sold his goats at Ravi's price, he would have received Rs. 2400. How many goats does Ravi have in total?

A
24

B
26

C
28

D
22

Solution

Assume that Ravi and Ram have n and 40 – n goats, respectively.
Ravi's selling price = Rs. x and Ram's selling price = Rs. y
So, xn = y(40 – n) ……….. (1) (As given)
If prices are interchanged, then
ny = 5400 …….. (2)
And x(40 – n) = 2400……… (3)

Substitute the value of x and y from equation 2 and 3 into equation 1, we get

Take square root on both side, we get

±

5n = 120 or,
n = 24 n = 120 (not possible)
Ravi have 24 goats in total.
Question 18
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If the roots of the equation ax² + bx + c = 0 are in the ratio of m : n, where b = c, then the value of + + is

A
0

B
1

C
– 1

D
b/a

Solution

ax² + bx + c = 0 [Let a, b be the roots]
b = c and : = m : n, =
+ + = + +
⁼ = 0
Question 19
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If α and β are the roots of the equation |x|² + |x| - 6 = 0, then

A
α + β = 0

B
α =

C
α + β = 1

D
α² = β

Solution

Put |x| = y.
Given equation is y² + y - 6 = 0.
By solving, we get y = 2 or - 3, i.e. |x| = 2 or - 3
Modulus cannot be negative, therefore |x| = 2.
Roots (α and b) of the given equation are 2 and 2.
Hence, α + β = 0
Question 20
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If both x and y are real, and x² + y² + 2x – 10y = –26, then find the value of x + y.

A
1

B
2

C
3

D
4

Solution

x² + y² + 2x – 10y + 26 = 0
⇒ (x + 1)² + (y – 5)² = 0
Since both x and y are real, so (x + 1)² = 0.
⇒ x = –1
And (y – 5)² = 0
⇒ y = 5
∴ x + y = 4