Quick Math MCQ Quiz - Objective Question with Answer for Quick Math - Download Free PDF

Given:

1) m + n = 144

2) 3n + 5p = 464

3) p + m = 160

Calculations:

Substitute m = 144 - n (from Equation 1) into Equation 3:

p + (144 - n) = 160

p + 144 - n = 160

⇒ p = 16 + n

Now substitute p = 16 + n into Equation 2:

3n + 5(16 + n) = 464

⇒ 3n + 80 + 5n = 464

⇒ 8n + 80 = 464

⇒ 8n = 464 - 80

⇒ 8n = 384

⇒ n = 384 / 8 = 48

Substitute n = 48 into p = 16 + n:

⇒ p = 16 + 48 = 64

∴ The value of p is 64.

Given:

David's monthly salary = ₹13000

Monthly spending on house rent = ₹5000

Monthly spending on bills = ₹1500

Number of months in a year = 12

Savings in the birthday month = ₹0

Formula Used:

Monthly savings = Monthly salary - (Monthly spending on house rent + Monthly spending on bills)

Total savings in a year = (Number of months - 1) × Monthly savings

Calculation:

David's monthly spending = ₹5000 + ₹1500 = ₹6500

David's monthly savings (excluding birthday month) = ₹13000 - ₹6500 = ₹6500

Number of months with savings = 12 - 1 = 11 months

Total savings in one year = 11 × ₹6500 = ₹71500

David's savings in one year is ₹71500.

Given :

The number of three consecutive numbers = 87

Calculation :

Let the three consecutive natural numbers be x , (x + 1) , (x + 2)

According to the question,

x + (x + 1) + (x + 2) = 87

⇒ 3x + 3 = 87

⇒ 3x = 87 - 3

⇒ 3x = 84

⇒ x = 28

Middle term = x+1 

⇒ 28 + 1 = 29

∴ The middle number is 29.

Calculation:

Let the total number of members = x

Contribution of each member = 2x

Total collection = (Number of members) × (Contribution of each member)

3042 = x × 2x

3042 = 2x²

x² = 3042 ÷ 2

x² = 1521

x = √1521

x = 39

Therefore, the number of members present in the party was 39.

Let the greater number be x.

Then, the smaller number = (x/2) + 13 

Given:

Difference = x - [(x/2) + 13] = 30

x - (x/2) - 13 = 30

x/2 = 30 + 13

x/2 = 43

x = 86

Thus, the correct answer is 86.

Given:

10 × 20 × 30 × ... × 1000

Concept used:

Take 10 as common from each term.

Number of trailing zeroes in n! = Divide n by 5, continue this process until we get the value which is less than 5. Now, all quotients will be added and the resultant number will be the number of zeroes.

Calculations:

10 × 20 × 30 × ... × 1000

⇒ (10 × 1) × (10 × 2) × (10 × 3) × (10 × 4) ..........× (10 × 100)

⇒ 10¹⁰⁰ × (1 × 2 × 3 × ... × 100)

⇒ 10¹⁰⁰ × (100!)

Number of zeroes = 100 + {(100)/5 + (20)/5}

⇒ 100 + 20 + 4

⇒ 124

∴ The number of trailing zeroes in 10 × 20 × 30 × ... × 1000 is 124.

Given:

20% of the boys and 15% of the girls failed the exam.

Total number of students who failed = 90

Calculation:

The percentage of boys who passed = (100 - 20)% = 80%

The percentage of girls who passed = (100 - 15)% = 85%

Let, the number of appeared boys = x

The number of appeared girls = y

So, 80x/100 - 85y/100 = 70

⇒ 5 (16x - 17y) = 7000

⇒ 16x - 17y = 1400   .....(1)

Also, 20x/100 + 15y/100 = 90

⇒ 5 (4x + 3y) = 9000

⇒ 4x + 3y = 1800   .....(2)

Multiplying 4 to equation (2),

16x + 12y = 7200   .....(3)

Subtracting equation (2) from equation (3),

16x + 12y - 16x + 17y = 7200 - 1400

⇒ 29y = 5800

⇒ y = 5800/29 = 200

∴ The number of girls appeared = 200

Putting y = 200 in equation (2),

4x + 3 × 200 = 1800

⇒ 4x = 1800 - 600 = 1200

⇒ x = 1200/4 = 300

∴ The number of boys who appeared = 300

The total number of students who appeared = 300 + 200 = 500

∴ The number of students that appeared for the exam is 500

Alternate Method Calculation:-

Let the Number of boys and Girls who appeared in exam be x and y respectively,

So, according to question-

Number of boys passed in exam = [(100 - 20) × x]/100 = 0.80x

Number of girls passed in exam = [(100 - 15) × y]/100 = 0.85y

Condition (1) -  

⇒ 0.80x = 0.85y + 70

⇒ 0.80x - 0.85y = 70    ...(i)

Condition (2) -

Total failed students = 90

⇒ 0.20x + 0.15y = 90    ...(ii)

From eqⁿ (1) - [4 × eqⁿ (ii)]

⇒ (-0.85y) - 0.60y = 70 - 360 

⇒ 1.45y = 290

⇒ y = 200

Put this value in eqⁿ (i)

⇒ 0.80x = 0.85 × 200 + 70

⇒ 0.80x = 170 + 70 = 240

⇒ x = 300

So, total number of students appeared in exam = 300 + 200 = 500.

According to the question, let the following Venn diagram,

Now, 

e = 5%

b + e = 14%

⇒ b = 9%

and,

d + e = 18%

⇒ d = 13%

Therefore, 

Percentage of students who failed only in Economics = a = 41% - (b + e + d)

a = 41% - (9 + 5 + 13)%

a = 41% - 27%

a = 14%

Hence, 14% is the correct answer.

Given:

A mango kept in a basket doubles every one minute. 

The basket gets filled in 30 minutes. 

Calculation:

The basket is full (1) in 30 minutes. 

The Time required to fill the basket with mango is 30 minutes.

So, half the basket is filled in 29 minutes.

As in every minute, the basket gets doubled. So, in 29 minutes, it is half-filled and in the next minutes, it will be completely filled.

∴ Obviously, the basket will be half-filled (1/2) filled in 29 minutes. 

Alternate MethodAccording to the question,

1st min = 2¹ = 2

2nd min = 2² = 4

We have observed every min double the quantity

then in 29 min = 2²⁹ = 536870912

Last 30 min = 2³⁰ = 1073741824

We have observed the 29th min is half of the 30th min quantity.

Concept:

To find the square root of any number, factorize it.

Calculation:

4050 = 2 × 3 × 3 × 3 × 3 × 5 × 5

If we multiply by 2 in 4050

⇒ 4050 × 2 = 8100

Now, √8100 = √(2 × 2 × 3 × 3 × 3 × 3 × 5 × 5)

⇒ √8100 = 2 × 3 × 3 × 5 = 90

∴ Multiply 4050 by 2 to get 8100 of which square root is 90.

Given:

Let the number be x.

Correct value = x/2

Resultant number = 2 × x

Calculations:

According to Question,

Required percentage = Resultant number/Correct value × 100

⇒ Required percentage = [(2x)/(x/2)] × 100

⇒ Required percentage = 400%

Given:

Sum of three consecutive multiples of 5 is 285

Calculation:

Three consecutive numbers are x, x + 1 and x + 2

So, Three consecutive multiples of 5 are 5x, 5(x + 1) and 5(x + 2)

The sum of three consecutive multiple of 5 is 285

∴ 5x + 5x + 5 + 5x + 10 = 285

⇒ 15x = 270

⇒ x = 18

Now, largest number = 5(x + 2) = 5 × 20 = 100

Given:

A college hostel mess has provisions for 25 days for 350 boys.

At the end of 10 days, when some boys were shifted to another hostel, it was found that now the provisions will last for 21 more days.

Concept used:

Available provisions = Number of days × Number of boys

Calculation:

Available provisions = 25 × 350 = 8750 units

After 10 days the remaining provision = 8750 - 10 × 350

⇒ 5250 units

Let N number of boys shifted to another hostel.

According to the question,

(350 - N) × 21 = 5250

⇒ 350 - N = 5250/21

⇒ 350 - N = 250

⇒ N = 350 - 250

⇒ N = 100

∴ 100 boys were shifted to another hostel.

Given:

Total number of students = 100

Students passed in Mathematics = 50

Students passed in English = 70

Students failed in both subjects = 5

Concept used:

The number of students who passed in both subjects is found by finding the difference between the number of students required for no overlap and the given total.

Calculation:

Students who passed in both the subjects = 70 + 50 – (100 – 5)

⇒ 120 – 95 = 25

∴ The students who passed in both subjects is 25.

Alternate Method

Total number of students = 100

Number of students failed in both subject = 5

⇒ Number of students passed in any one or both subject = (100 - 5) = 95

Students passed in Mathematics = 50

⇒ Students failed in Mathematics but passed in English = (95 - 50) = 45

Students passed in English = 70

⇒ Students failed in English but passed in Mathematics = (95 - 70) = 25

Number of students passed in both subjects = (95 - 45 - 25) = 25

∴ The number of students who passed in both subjects is 25.

GIVEN:

Sum of two numbers is 90

One number exceeds the second by 16

CONCEPT:

Here we need to form an equation in one variable to get one number and the other number can be hence found.

CALCULATION:

Let the two numbers be x and x + 16.

According to the question,

x + x + 16 = 90

⇒ 2x = 74

⇒ x = 37 and x + 16 = 53