Simple Equations Test

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Simple Equations Test - 9

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Weekly Quiz Competition

Question 1
1 / -0

You want to distribute 20 sweets among your classmates on your birthday. You are required to distribute equal number of sweets in each of the 5 rows of the class. Which of the following equations would you use to find the number of sweets (s) to be distributed in each row?

A
5s = 20

B
5 + s = 20

C
20 + s = 5

D
= 20

Solution

Total number of sweets = Number of sweets distributed in each row × Number of rows
20 = s × 5
5s = 20
Question 2
1 / -0

Solve the following equation for q:

25 - 2q = 15

A
q = 4

B
q = 5

C
q = 6

D
q = 7

Solution

25 - 2q = 15
25 - 15 = 2q
2q = 10
q = 5
Hence, option (2) is correct.
Question 3
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Subtracting 10 from 15 less than thrice a number gives 20 as the result. The number is

A
20

B
25

C
15

D
30

Solution

Let the number be x.
According to the question:
3x - 15 - 10 = 20
3x = 45
x = 15
Hence, option (3) is correct.
Question 4
1 / -0

Which of the following equations does not give 7 as the solution for s?

A
36 + 4s = 64

B

C
71 - s = 64

D
12s = 84

Solution

(A):
36 + 4s = 64
4s = 64 - 36
4s = 28
s = 7

Statement (A) has 7 as its solution.

(B): Given,

s = 10.28

Statement (B) does not have 7 as its solution.

(C): 71 - s = 64
s = 71 - 64
s = 7

Statement (C) has 7 as its solution.

(D):
12s = 84
s = 7
Hence, option (2) is correct.
Question 5
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184 toy cars are to be distributed between two classes in such a way that of the number of toy cars for one class is more than of the number of toy cars for the other class by 8. Find the greater number of toy cars between the two.

A
122

B
120

C
130

D
152

Solution

Let one class get x toy cars.
Then, the second class will get 184 - x.

According to the question,

= 54

i.e. One class will get 54 toy cars and the other will get 184 - 54 = 130.

The greater number of toy cars between the two is 130.
Question 6
1 / -0

^If ^{, then find the value of x.}

A
1

B
2

C
3

D
5

Solution

5x = 6 + 3x
2x = 6
x = 3
Hence, option (3) is correct.
Question 7
1 / -0

Find the value of 'a' for the following equation to be true:

18 - (20 - 16 ÷ 4 × 4) + a = 60

A
46

B
56

C
66

D
76

Solution

The given equation is:
18 - (20 - 16 ÷ 4 × 4) + a = 60
18 - (20 - 4 × 4) + a = 60
18 - (20 - 16) + a = 60
18 - 4 + a = 60
14 + a = 60
a = 60 - 14
a = 46
Hence, option (1) is correct.
Question 8
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When of a number, of the same number and of the same number are added together, the sum obtained is 45. Find the number.

A

B

C

D

Solution

Let the number be x.

x +x +x = 45

LCM of 3, 5 and 7 = 105

So,

x =

x =

Hence, option (3) is correct.
Question 9
1 / -0

One of the two supplementary angles is 35°. Find out the other angle.

A
95°

B
105°

C
135°

D
145°

Solution

Sum of two supplementary angles = 180°
Missing angle = 180° - 35°
= 145°
Question 10
1 / -0

Find out the value of 'x' in the following equation:

A

B

C

D

Solution

Hence, option (2) is correct.
Question 11
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If , then what is the value of x?

A
6

B
10

C
4

D
None of the above

Solution

6x - 9 = 27

6x = 27 + 9

6x = 36

x = 6
Question 12
1 / -0

If of a number is 150, then what is of the same number?

A
10

B
25

C
30

D
35

Solution

Let the number be a.

So, x a = 150

Or, a = = 90
Now, of 90 = x 90 = 10

Hence, option (1) is correct.
Question 13
1 / -0

Find the value of 'x' in the following expression:

A
4

B
-4

C
-5

D
None of the above

Solution

Hence, option (2) is correct.
Question 14
1 / -0

Which of the given options is not one of the steps in solving the following equation?

4x + 99 = -1

A
4x = 100

B

C
4x = -1 - 99

D
None of the above

Solution

The solution of the equation is as follows:
4x + 99 = -1
4x = -1 - 99
4x = -100

Therefore, expression in option 1 is not one of the steps in solving the given equation.
Hence, option (1) is correct.
Question 15
1 / -0

Among the following equations of the form, y = mx + c, where y, m and c are constants, which one gives x = as the solution?

A
5 = 12x - 3

B
3 = 8x - 5

C
5 = 16x - 3

D
None of the above

Solution

(1) 5 = 12x - 3, so x = =

(2) 3 = 8x - 5
8x = 8 or x = 1

(3) 5 = 16x - 3
16x = 8
x =
Hence, option (3) is correct.
Question 16
1 / -0

A's age is half the age of his brother. 5 years from now, the difference between the ages of the two brothers will be only 3 years. Find the current age of A's brother.

A
5

B
6

C
7

D
None of the above

Solution

Let the age of A be x years.
So, present age of A's brother = 2x years
Age of A after 5 years = (x + 5) years
Age of A's brother after 5 years = (2x + 5) years
As per question statement:
(2x + 5) - (x + 5) = 3
x = 3
So, the present age of A's brother = 2x = 2 × 3 = 6 years.
Hence, option (2) is correct.
Question 17
1 / -0

Represent the following scenario in a mathematical equation:

There are two types of chocolates: Big and Small. The length of one big chocolate is 5 cm more than the combined length of 5 small chocolates. If the length of one big chocolate is 0.30 m, then form an equation to determine the length of a small chocolate.

A
5 = 5x + 30

B
30 = 5x + 5

C
5 = 30x + 5

D
None of the above

Solution

Length of one big chocolate = 0.3 m = 30 cm
Let the length of one small chocolate be x cm.
So, according to the question:
30 = 5x + 5
Hence, option (2) is correct.
Question 18
1 / -0

At a railway station, the number of women is 200 more than thrice of number of men. Find the number of men at the station if the number of women is 500.

A
250

B
200

C
100

D
None of the above

Solution

Let the number of men at station be M.
Now, according to the question:
3M + 200 = W ...(1)
Number of women = 500
So, 3M + 200 = 500
M =
Hence, option (3) is correct.
Question 19
1 / -0

In a company, it is announced that the maximum incentive to be given to an employee is Rs. 600 more than 5 times the minimum incentive. If the amount of maximum incentive is Rs. 3000, then find the amount (in Rs.) of minimum incentive.

A
360

B
480

C
500

D
None of the above

Solution

Let the minimum incentive be L, and the maximum incentive be M.
So, according to the question:
M = 5L + 600 ... (1)
Now, maximum incentive = Rs. 3000
So, 3000 = 5L + 600

Hence, option (2) is correct.
Question 20
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Some children are sitting in a park. They have a certain number of chocolates. If each child eats 2 chocolates, then 2 extra chocolates will be left. If eat child eats 3 chocolates, then they will be short by 2 chocolates. Find the number of children and the number of chocolates, respectively.

A
3, 10

B
4, 10

C
5, 12

D
None of the above

Solution

Let number of children = x
and number of chocolate = y
So, y = 2x + 2 ... (1)
Also, y = 3x - 2 ... (2)
From (1) and (2), we get
2x + 2 = 3x - 2
x = 4
Thus, y = 3(4) - 2 = 12 - 2 = 10
So, there are 4 children and they have 10 chocolates.
Hence, option (2) is correct.
Question 21
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Sum of two digits of a number is 7. When 27 is subtracted from this number, its digits get interchanged. Which of the following are correct steps to find the number?

I. Let tens digit be m.

II. Then, ones digit = (7 - m)
Number = 10 × m + (7 - m) = 9m + 7

III. Subtracting 27 from the number, we get 9m - 20.

IV. Number with its digits interchanged = 10 × (7 - m) + m

V. 9m – 20 = 70 – 9m

VI. Hence, the number = 52

A
Steps II and VI

B
Steps I and III

C
Steps IV and V

D
All of the given steps

Solution

Lets take all the steps.
I. Let tens digit be m. - Correct.

II. Then, ones digit = (7 - m)
Number = 10 × m + (7 - m) = 9m + 7 - Correct.

III. Subtracting 27 from the number, we get 9m - 20.
i.e 9m + 7 - 27 = 9m - 20 - Correct.

IV. Number with its digits interchanged = 10 × (7 - m) + m - Correct.

V. 9m – 20 = 70 – 9m
i.e. 18m = 90, m = 5 - Correct.

VI. Hence, the number = 52
i.e m = 5 (tens digit) and 7 - 5 = 2 (ones digit)
Hence, the number is 52 - Correct.
Question 22
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Select the incorrect statement:

a. In an equation, whatever operation happens on the right side of equals sign also happens on the left side.
b. If a number is subtracted from one side of the equation, the same number is subtracted from the other side as well.
c. If one side of the equation is multiplied by a number, then the other side of the equation is multiplied by any other number.
d. If a number is added to one side of the equation, then the same number is added to the other side as well.

A
(a)

B
(b)

C
(c)

D
(d)

Solution

a) In an equation, whatever operation happens on the right side of equals sign also happens on the left side.
This means if one side is halved, then the other side will be halved as well, which is true.

b) If a number is subtracted from one side of the equation, the same number is subtracted from the other side as well.
This means that if 3 is subtracted from one side, it is also subtracted from the other side, which is true.

c) If one side of the equation is multiplied by a number, then the other side of the equation is multiplied by any other number.
This means that if one side is multiplied by 2, then the other side can be multiplied by any number other than 2, which is incorrect.

d) If a number is added to one side of the equation, then the same number is added to the other side as well.
This means that if 5 is added to one side, then 5 is added to the other side too, which is true.

So, statement (c) is incorrect.

Hence, option (3) is the answer.

Question 23

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Directions: Read the following statements and choose the correct option with the help of the table given below.

Sum of two numbers is 92. One number is thrice the other number.
I. If the bigger number is d, then find the other number.
II. Make an equation representing the given statement.
III. Find the smaller number.

Options	I	II	III
(A)	3d	3d + d = 92	21
(B)		d + = 92	23
(C)	2d	2d + d = 92	28
(D)		d + = 92	28

(A)

(B)

(C)

(D)

Solution

I. If the bigger number is d, then find the other number.
Now, we are given that the smaller number is 1/3 of the bigger number. Therefore, if the bigger number is d, then the smaller number is .

II. Make an equation representing the given statement.
Now, if the bigger number is d and the smaller number is ,then

III. Find the smaller number.
After solving the equation, we get d = 69.
So, smaller number = 69 ÷ 3 = 23

Question 24
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A shoe store has 48 pairs of shoes in black and brown colours only. Each black pair costs Rs. 1500 and each brown pair costs Rs. 1900. If the total earnings from the sale of all the pairs are Rs. 79,200, then

I. form an equation representing the case
II. find the number of pairs of black shoes
III. find the number of pairs of brown shoes

A
I. 1500 × m + 1900 × (48 - m) = 79,200
II. 30
III. 18

B
I. 15 × m + 19 × m = 48
II. 30
III. 18

C
I. 1900 × m + 1500 × (48 - m) = 79,200
II. 38
III. 10

D
None of these.

Solution

Let the number of black shoe pairs be m and the number of brown shoe pairs be 48 - m.

1. According to question statement:
1500 × m + 1900 × (48 - m) = 79,200

2. Number of black shoe pairs:
1500 × m + 1900 × (48 - m) = 79,200
1500m + 91200 - 1900m = 79,200
- 400m = - 12000
400m = 12000
m = 30

3. Number of brown shoe pairs = 48 - m = 48 - 30 = 18

Hence, option (1) is correct.

Question 25

1 / -0

Match the following:

Column - 1	Column - 2
A) The sum of ages of 5 children born at regular intervals of 3 years is 50 years. What is the age of the youngest child of them?	a) 10
B) A is 2 years older than B, who is twice as old as C. If the total age of A, B and C is 27 years, then how old is B?	b) 14
C) Sachin is 4 years younger than Rahul. If their ages are in the respective ratio of 7 : 9, then how old is Sachin?	c) 4

A) - a, B) - c, C) - b

A) - a, B) - b, C) - c

A) - c, B) - a, C) - b

A) - c, B) - b, C) - a

Solution

(A) Let the ages of children be x, (x+ 3), (x+ 6), (x+ 9) and (x+ 12) years.
Then, x+ (x+ 3) + (x+ 6) + (x+ 9) + (x+ 12) = 50
5x= 20
x= 4.
Therefore, age of the youngest child = x= 4 years, i.e. (c).

B) Let C's age be xyears. Then, B's age = 2xyears. A's age = (2x+ 2) years.
Hence, (2x+ 2) + 2x+ x= 27
5x= 25
x= 5.
Hence, B's age = 2x= 10 years, i.e. (a).

C) Let Rahul's age be xyears.
Then, Sachin's age = (x- 4) years

Hence,

9x- 36 = 7x
2x= 36
x= 18
Hence, Sachin's age = (x- 4) = 14 years, i.e. (b).

So, option (3) is correct.

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Re-Attempt Weekly Quiz Competition

Simple Equations Test - 9

Result

Simple Equations Test - 9

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SHARING IS CARING

Question 1

5s = 20

5 + s = 20

20 + s = 5

= 20

Solution

Question 2

q = 4

q = 5

q = 6

q = 7

Solution

Question 3

20

25

15

30

Solution

Question 4

36 + 4s = 64

71 - s = 64

12s = 84

Solution

Question 5

122

120

130

152

Solution

Question 6

1

2

3

5

Solution

Question 7

46

56

66

76

Solution

Question 8

Solution

Question 9

95°

105°

135°

145°

Solution

Question 10

Solution

Question 11

6

10

4

None of the above

Solution

Question 12

10

25

30

35

Solution

Question 13

4

-4

-5

None of the above

Solution

Question 14

4x = 100

4x = -1 - 99

I. 1500 × m + 1900 × (48 - m) = 79,200
II. 30
III. 18

I. 15 × m + 19 × m = 48
II. 30
III. 18

I. 1900 × m + 1500 × (48 - m) = 79,200
II. 38
III. 10