# FREE IBA Quantitative Reasoning Questions and Answers

#### The average of 21 numbers is 30. The largest number is 50. If we remove the largest number, then the average of the remaining numbers will be:

Explanation:

To find the average of the remaining numbers after removing the largest one:

Find the sum of all numbers (21 numbers) = 30 * 21 = 630.

Subtract the value of the largest number (50) from the sum: 630 - 50 = 580.

Divide the adjusted sum by the new number of numbers (21 - 1 = 20): 580 / 20 = 29.

So, the average of the remaining numbers is 29.

#### A certain positive integer when divided by 7 gives a remainder of 5 and when divided by 11 gives a remainder of 8. Hence the integer can be:

Explanation:

To find a positive integer that leaves a remainder of 5 when divided by 7 and a remainder of 8 when divided by 11, we can test numbers that leave a remainder of 5 when divided by 7 until we find one that also leaves a remainder of 8 when divided by 11.

Starting with 5,12,19,26,…, we find that 19 satisfies both conditions.

#### The average grade mark of 50 students in a class is 77. There are 20 male students in the class. The average grade mark of these male students in the class is 80. The average grade marks of female students in the class is:

Explanation:

First, find the total sum of grades for the class:

Total sum of grades = Average grade × Number of students = 77 × 50 = 3850

Now, find the total sum of grades for male students:

Sum of grades for male students=80×20=1600

Subtract the sum of grades for male students from the total sum of grades to get the sum of grades for female students:

Sum of grades for female students=3850−1600=2250

Since there are 30 female students (50 total students - 20 male students), find the average grade mark for female students:

Average grade mark for female students = 2250/30 = 75

#### The ratio of boys to girls in a business class is 7: 5. If the class has a total of 60 students, how many more boys are there than girls:

Explanation:

To find out how many more boys there are than girls in the class:

1. Determine the ratio of boys to girls: 7 boys for every 5 girls.

2. Calculate the total number of parts in the ratio: 7+5=12.

3. Divide the total number of students by the total parts in the ratio to find the value of each part: 60/12 = 5.
4. Multiply the value of each part by the number of boys and girls to find their respective numbers: Boys: 7×5=35, Girls: 5×5=25.

5. Subtract the number of girls from the number of boys to find how many more boys there are: 35−25=10.

So, there are 10 more boys than girls in the class.

#### A piece of wire 80 centimeters long is bent to form a rectangle whose length is three times its width. The area of the rectangle thus formed is:

Explanation:

Length of wire = 80 cm

Length of rectangle = 3 times width

Let's denote the width of the rectangle as w cm.

Perimeter of rectangle = 80 cm

2×(length+width)=80

2×(3w+w)=80

2×4w=80

w=10

So, the width of the rectangle = 10 cm.

Length of rectangle = 3w=3×10=30 cm.

Area of rectangle = length × width = 30×10=300 square centimeters.

#### The grades of John in nine monthly tests were 58, 80, 56, 58, 64, 80, 56, 90, 80. The mode of his grades is:

Explanation:

To find the mode (the most frequently occurring value) of John’s grades:

John’s grades: 58, 80, 56, 58, 64, 80, 56, 90, 80

Count the frequency of each grade:

58 appears twice

80 appears three times

56 appears twice

64 appears once

90 appears once

The grade that appears most frequently (the mode) is 80.

#### If W < 25 then:

Explanation:

If W < 25, it means that W is less than 25. However, there is no constraint on how small W can be as long as it's less than 25. Therefore, W could also be greater than -25.

#### Suppose the mean of 30 numbers is 520. If 20 of these numbers are each increased by 12, and the remaining numbers are decreased by 6, then the mean of the new set of numbers is:

Explanation:

Mean of 30 numbers is 520.

Calculate the sum of the original 30 numbers:

Sum = Mean×Number of numbers=520×30=15600

For the new set:

20 numbers increased by 12 each, so add 20×12=240 to the sum.

10 numbers decreased by 6 each, so subtract 10×6=60 from the sum.

The net change in sum is 240−60=180.

Add this to the original sum to get the new sum: 15600+180=15780.

Now, find the new mean: New mean =
15780/30 = 526

So, the mean of the new set of numbers is 526. Therefore, the correct option is C) 5260.

#### If 4(3x - 8) = 25, what is 3x - 10?

Explanation:

First, let's solve the equation 4(3x - 8) = 25:

Expand the expression: 4 * 3x - 4 * 8 = 25

This gives us: 12x - 32 = 25

Add 32 to both sides: 12x = 57

Divide both sides by 12: x = 57/12 = 4.75

Now, let's find the value of 3x - 10 when x = 4.75:

3 * 4.75 - 10 = 14.25 - 10 = 4.25

Therefore, none of the provided options match the result.

#### Majid is now 42 years old and his brother, Zahid, is 32 years old. How many years ago was Majid three times as old as Zahid?

Explanation:

To find out how many years ago Majid was three times as old as Zahid:

Let x be the number of years ago.

At that time:

*Majid's age: 42−x
*Zahid's age: 32−x

According to the given condition, Majid was three times as old as Zahid: 42−x=3(32−x)

Let's solve for x:

42−x=96−3x

2x=54

x=27

So, 27 years ago Majid was three times as old as Zahid.

#### The perimeter of a triangle is 240. Its sides are proportional to 3:4:5. The largest side of the triangle is:

Explanation:

To find the lengths of the sides of the triangle proportional to 3, 4, and 5:

1. Sum the proportions: 3 + 4 + 5 = 12.

2. Divide the perimeter by the sum of proportions: 240 / 12 = 20.

3. Multiply each proportion by the result:

Side 1: 3 × 20 = 60

Side 2: 4 × 20 = 80

Side 3: 5 × 20 = 100

Therefore, the largest side of the triangle is 100.