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Test: Linear Equations in One Variable - 1 - ACT MCQ


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10 Questions MCQ Test Mathematics for ACT - Test: Linear Equations in One Variable - 1

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Test: Linear Equations in One Variable - 1 - Question 1

If 50 is subtracted from two-third of a number, the result is equal to sum of 40 and one-fourth of that number. What is the number?

Detailed Solution for Test: Linear Equations in One Variable - 1 - Question 1

Let the number be x.

According to Question,

2x/3 - 50 = x/4 + 40

⇒ 5x/12 = 90

⇒ x = (90 x 12)/5

⇒ x = 216

∴ The number is 216.

Test: Linear Equations in One Variable - 1 - Question 2

A man has equal number of five, ten and twenty rupee notes amounting to Rs. 385. Find the total number of notes?

Detailed Solution for Test: Linear Equations in One Variable - 1 - Question 2

Given:

Number of Five rupee note = Number of Ten rupee note = Number of Twenty rupee note

Calculation:

Let the equal number of five, ten and twenty rupee notes be x

⇒ 5x + 10x + 20x = 385

⇒ 35x = 385

⇒ x = 385/35 = 11

Total number of notes = 11 notes of Rs 5 + 11 notes of Rs 10 + 11 Rs of 20 =  33 notes

Check:

11 notes of Rs 5 + 11 notes of Rs 10 + 11 Rs of 20 = 11 × 5 + 11 × 10 + 11 × 20 = 55 + 110 + 220 = Rs. 385

Total number of notes = 11 × 3 = 33 

Therefore the correct answer is 33.
Alternate Method:

Since the number of Rs 5, Rs 10, and Rs 20 notes are equal, the ratio of the number of notes is = 1 ∶ 1 ∶ 1

Or, we can say, the total number of notes must be multiple of 3.

 Only two options are multiple of 3, others can be eliminated.

option 2 : 33. That means the number of notes of each dimension is 33/3 = 11.

5 × 11 + 11 × 10 + 11 × 20 = 385

∴ The total number of notes is 33.

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Test: Linear Equations in One Variable - 1 - Question 3

If the sum and product of two numbers is 34 and 288 respectively, find the sum of their cubes.

Detailed Solution for Test: Linear Equations in One Variable - 1 - Question 3

Let the numbers be ‘x’ and ‘y’

Given, sum = x + y = 34

Product = xy = 288

we need to find   x3 + y3

As we know, x3 + y3 = (x + y)3 – 3xy(x + y) = (x + y)[(x + y)2 – 3xy]

put the values of x + y and xy in to the equation

∴ Sum of their cubes = 34 × [(34)2 – 3 × 288] = 34 × (1156 – 864) = 34 × 292 = 9928

Test: Linear Equations in One Variable - 1 - Question 4

If x6 + x5 + x4 + x3 + x2 + x + 1 = 0, then find the value of x5054 + x6055 - 7

Detailed Solution for Test: Linear Equations in One Variable - 1 - Question 4

Given:

x6 + x5 + x4 + x3 + x2 + x + 1 = 0

Calculation:

Considering the given equation

x6 + x5 + x4 + x3 + x2 + x + 1 = 0      -----(1)

In equation (1) multiplying by x

x7 + x6 + x5 + x4 + x3 + x2 + x = 0      -----(2)

Equation (2) – (1)

⇒ x7 - 1 = 0

⇒ x7 = 1

x5054 + x6055 - 7

⇒ (x7)722 + (x7)865 – 7

⇒ (1)722 + (1)865 – 7

⇒ 1 + 1 – 7

⇒ - 5

∴ Required value is – 5

Test: Linear Equations in One Variable - 1 - Question 5

Lata makes 20 kg paneer at home and sells it every morning. From one litre of milk, she makes 200 gms of paneer. The cost of a litre of milk is Rs. 40. How much money does she spend in procuring milk every day?

Detailed Solution for Test: Linear Equations in One Variable - 1 - Question 5

⇒ Milk used to make 200 gms of panner = 1 litre

⇒ Milk used to make 20 kgs paneer = (20 × 1000/200) × 1 = 100 litre

⇒ Cost of 1 litre milk = Rs. 40

⇒ Cost of 100 litre Milk = 100 × 40 = Rs. 4000

Test: Linear Equations in One Variable - 1 - Question 6

In a group of 1300 students, every student reads 5 subjects and every subject is read by 65 students. The number of subjects are:

Detailed Solution for Test: Linear Equations in One Variable - 1 - Question 6

1 student reads = 5 subjects

1300 students read = 5 × 1300 = 6500 subjects

1 subject is read by = 65 students

So, total different subjects read = 6500/65 = 100

Test: Linear Equations in One Variable - 1 - Question 7

When you reverse the digits of the number 13, the number increases by 18. How many other two‐digit numbers increase by 18 when their digits are reversed?

Detailed Solution for Test: Linear Equations in One Variable - 1 - Question 7

Calculation:

Let the unit digit be x and tens digit be y

According to the question,

⇒ 10y + x - (10x + y) = 18

⇒ 9y - 9x = 18

⇒ y - x = 2

It means that difference of digits of two-digit numbers is 2.

∴ Six cases other than (13, 31) are possible

(24, 42) (35, 53) (46, 64) (57, 75) (68, 86) (79, 97)
Mistake Points:
Here we cannot take (20, 2)
Because 2 is not a two-digit number. Normally we do not write numbers from 1 - 9  as 01, 02, 03, 04, 05, 06, 07, 08, and 09.

Test: Linear Equations in One Variable - 1 - Question 8

The solution of equation 10x + 26 = 0 is a / an________.

Detailed Solution for Test: Linear Equations in One Variable - 1 - Question 8

Given:

The given equation is 10x + 26 = 0

Concept:

Rational numbers are the numbers that can be written in the form of p/q, where q is not equal to zero. 

Calculation:

10x + 26 = 0

⇒ 10x = -26

⇒ x = (-26)/10

⇒ x = -13/5

∴ 10x + 26 = 0 is a rational number.

Test: Linear Equations in One Variable - 1 - Question 9

Difference of two numbers is 50% of the smaller number. If greater number is 120, then find sum of both numbers is:

Detailed Solution for Test: Linear Equations in One Variable - 1 - Question 9

Given:

Greater number = 120

Calculation:

Let smaller number be x

According to the question

120 – x = x × (50/100)

⇒ 120 – x = x/2

⇒ 240 – 2x = x

⇒ x + 2x = 240

⇒ 3x = 240

⇒ x = 240/3

⇒ x = 80

∴ Sum of both numbers = 120 + 80 = 200

Test: Linear Equations in One Variable - 1 - Question 10

If x2 - 7x + 1 = 0 then find the value of (x + 1/x).

Detailed Solution for Test: Linear Equations in One Variable - 1 - Question 10

Given:

x2 - 7x + 1 = 0

Calculation:

x2 - 7x + 1 = 0

Dividing by x:

⇒ x - 7 + 1/x = 0

⇒ x + 1/x = 7

∴ Value of x + 1/x = 7 

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