In the name of ALLAH, the most beneficient, the most merciful

Solved Examples Set 1 (Quantitative Ability)

  1. Rashid buys three books for $ 16 each and four books for $ 23 each, what will be the average price of books

    1. $ 18
    2. $ 20
    3. $ 22
    4. $ 24
    5. $ 16
    Price of 3 books = 3 × 16 = $ 48
    Price of 4 books = 4 × 23 = $ 92
    Total price = $ 140
    Total books = 3 + 4 = 7
    Average price of books = \(140 \over 7 \) = $ 20
  2. A and B can reap a field in 30 days, working together. After 20 days, however, B is called away and A takes 20 days more to complete the work. B alone could do the whole work in

    1. 48 days
    2. 50 days
    3. 56 days
    4. 60 days
    5. 64 days
    (A + B)'s 20 day's work = \(1 \over 30 \) × 20 = \(2 \over 3 \)
    Remaining work = 1 - \(2 \over 3 \) = \(1 \over 3 \)
    Now, \(1 \over 3 \) work is done by A in 20 days.
    Therefore, the whole work will be done by B in 20 × 3 = 60 days.
  3. A man was 32 years old when his daughter was born. He is now five times as old as his daughter. How old is his daughter now?

    1. 7 years
    2. 8 years
    3. 9 years
    4. 10 years
    5. 6 years
    Let's assume the daughter is d years old now. That means that the man is now (32 + d) years old, so that
    (32 + d) = 5d
    32 = 4d
    d = 8
  4. 8 : ? :: 1 : 4

    1. 24
    2. 16
    3. 0
    4. 32
    5. 20
    ? × 1 = 8 × 4
    ? = 32
  5. If 4a + 2 = 10, then 8a + 4 =

    1. 5
    2. 16
    3. 20
    4. 24
    5. 28
    One may answer this question by solving
    4a + 2 = 10
    4a = 8
    a= 2
    Now, plugging in 2 for a:
    8a + 4 = 8(2) + 4 = 20
    A faster way of solving this is to see the relationship between the quantity 4a + 2 (which equals 10) and 8a + 4. Since 8a + 4 is twice 4a + 2, the answer must be twice 10, or 20.
  6. A man earned an annual income of $ 245000 in 1990. He was allowed a deduction of $ 15000 relief for each of his three children and a personal relief of $ 30000. If he was charged a tax rate of 4% on first $ 50000 and 6% on his remaining income, calculate the total tax charged.

    1. $ 9200
    2. $ 8700
    3. $ 9500
    4. $ 9400
    5. $ 9000
    Total Income = $ 245000
    Total relief = 3 × 15000 + 30000 = $ 75000
    Rest income = 245000 - 75000 = 170000
    Tax on 1st 50000 = 0.04 × 50000 = $ 2000
    Tax on rest amount 120000 = 0.06 × 120000 = $ 7200
    Total tax = 200 + 7200 = $ 9200
  7. A man saves $ 500, which is 15% of his annual income. How much does he earn in one year?

    1. $ 3542.5
    2. $ 3333.33
    3. $ 3132.3
    4. $ 3075.75
    5. $ 4444.4
    Let annual income = x
    15% of x = 500
    x = \(500 \over 15\) × 100 = \(10000 \over 3\) = 3333.33
  8. A group of boys were to choose between playing hockey and badminton. The number of boys choosing hockey was three times that of those choosing badminton. Asking 12 boys who chose hockey to play badminton would make the number of players for each game equal. Find the number who chose badminton originally.

    1. 12
    2. 14
    3. 11
    4. 13
    5. 10
    Let no. of boys for badminton = x
    then no. of boys for hockey = 3x
    According to the statement,
    3x - 12 = x + 12 (12 leave hockey, 12 join badminton)
    2x = 24
    x = 12
    Hence, there were 12 boys originally choosing badminton.
  9. At a book fair, a book was reduced in price from $ 75 to $ 60. If the first price gives a 50% profit, find the percentage profit of the book sold at the reduced price.

    1. 20%
    2. 30%
    3. 40%
    4. 50%
    5. 10%
    As $ 75 (first price) gives a profit = 50%
    $ 1 gives a profit = (50/75)%
    $ 60 (reduced price) gives profit = (50/75 x 60)% = 40%
  10. A man walked for 3 hours at 4.5 km/h and cycled for some time at 15 km/h. Altogether, he traveled 21 km. Find the time taken for cycling.

    1. 1/2 hour
    2. 1 hour
    3. 1 1⁄2 hours
    4. 2 hours
    5. 2 1⁄2 hours
    The man walked the distance = 3 x 4.5 = 13.5 km. The distance cycled by the man = 21 - 13.5 = 7.5 km
    As he cyled 15 km in 1 h
    he cycled 1 km in 1/15 h
    Finally, he cycled 7.5 km in 7.5/15 = 1/2 h
  11. Which of the following expressions is equivalent to \(\frac{𝑥^2 + 3x + 1}{𝑥 + 1}\)?

    1. x + 2
    2. 𝑥 + 3
    3. 𝑥 + 2 - 1/(𝑥 + 1)
    4. 𝑥 + 3 + 1/(𝑥 + 1)
    5. 𝑥 + 4 + 5/(𝑥 + 1)
    As \(𝑥^2 + 3x + 1 = (𝑥^2 + 3x + 2) -1\)
    and
    \(\frac{𝑥^2 + 3x + 2}{x + 1} = \frac{(𝑥 + 2)(x + 1)}{x + 1} = 𝑥 + 2\)
    Therefore,
    \(\frac{𝑥^2 + 3x + 1}{x + 1} = \frac{𝑥^2 + 3x + 2}{x + 1} - \frac{1}{x + 1} = (𝑥 + 2) - \frac{1}{x + 1}\)
  12. 5.41 - 3.29 × 1.6 = ?

    1. 14.6
    2. 0.3392
    3. 0.146
    4. 3.392
    5. 1.46
    5.41 - 3.29 × 1.6 = 5.41 - 5.264 = 0.146
  13. By selling a fan for $ 475, a person loses 5%. To get a gain of 5%, he should sell the fan for:

    1. $ 500
    2. $ 525
    3. $ 535
    4. $ 575
    5. $ 505
    cost price = 100/(100 - 5) x 475 = $ 500
    sale price = (100 + 5)/100 x 500 = $ 525
  14. A basket that contains 2 apples, 3 bananas, 6 oranges, and 4 pears is in the workroom. When Ms. Hutchinson went to the workroom, other workers had already taken 1 banana, 2 oranges, and 1 pear. From the remaining fruit, Ms. Hutchinson randomly took 3 pieces of fruit separately from the basket. If each fruit is equally likely to be chosen, what is the probability that the third piece was an orange if the first two she took were also oranges?

    1. 4/165
    2. 9/11
    3. 4/11
    4. 3/11
    5. 2/9
    Ms. Hutchinson randomly takes the 3 pieces of fruit from the basket, there are 2 apples, 3 -1 = 2 bananas, 6 - 2 = 4 oranges, and 4 - 1 = 3 pears. Assuming that the first 2 pieces of fruit Ms. Hutchinson takes are oranges, there will be 2 apples, 2 bananas, 4 - 2 = 2 oranges, and 3 pears left in the basket when she selects the third piece of fruit. The probability that the third piece of fruit she selects will be an orange is \(\frac{2}{2 + 2 + 2 + 3} = \frac{2}{9}\).
  15. A boy of height 165 cm is replaced by another, which decreases the average height of the group of 34 students, by 1 cm. The height of the new student is

    1. 129 cm
    2. 130 cm
    3. 131 cm
    4. 132 cm
    5. 133 cm
    Total decreased height of 34 students = 1 × 34 = 34 cm
    Height of the replaced student = 165 - 34 = 131 cm
  16. If 3x = −9, then 3x3 − 2x + 4 =

    1. -83
    2. -71
    3. -47
    4. -17
    5. 61
    First solving 3x = −9, x = −3. Now plug into 3x3 − 2x + 4:
    3x3 − 2x + 4
    = 3(-3)3 − 2(-2) + 4
    = 3(−27) + 6 + 4
    = −81 + 6 + 4
    = −71
  17. A man is 5 times as old as his son. 2 years ago the sum of the squares of their ages was 1114. Find the present age of son.

    1. 7 years
    2. 9 years
    3. 8 years
    4. 8 1/2 years
    5. 6 years
    Let son's age = x, then
    father's age = 5x
    As before 2 years ago the sum of the squares of their ages was 1114, the equation becomes as
    \((x - 2)^2 + (5x - 2)^2 = 1114 \)
    By simplifying the equation, we have
    \(13x^2 -12x -553 = 0\)
    Now solving the equation, we have
    \(13x^2 - 12x - 553 = 0\)
    \(13x^2 - 91x + 79x -553 = 0\)
    13x(x - 7) + 79(x - 7) = 0
    (x - 7)(13x + 79) = 0
    x = 7 and x = -6.077
    As age could not be negative, hence the present age of the son is 7 years.
  18. A certain solution is to be prepared by combining chemicals X, Y and Z in the ratio 18:3:2. How many liters of the solution can be prepared by using 36 liters of X?

    1. 46 liters
    2. 47 liters
    3. 45 liters
    4. 49 liters
    5. 44 liters
    As total ratio is 18 +3 + 2 = 23
    Let total solution is x liters
    Then \(18 \over 23\) x = 36
    x = \(36 × 23 \over 18\) = 46 liters
  19. \( {𝑥 - 8 \over 24} = {3 \over 4} \)
    What is the value of 𝑥 in the equation?

    1. 10
    2. 20
    3. 26
    4. 31
    5. 40
    By cross multiplying, 4(𝑥 – 8) =3 × 24. Thus, 4𝑥 – 32 = 72, and so 4𝑥 = 104 and 𝑥 = 26.
  20. 40 arithmetic questions, each carrying equal marks, were given in a class test. A boy answered 25 questions correctly. What percentage was this? To pass a test a student must answer at least 45% of the questions correctly. Find the least number of correct answers needed to pass.

    1. 62.5%, 18
    2. 63.5%, 16
    3. 64.5%, 20
    4. 61.0%, 21
    5. 60.0%, 22
    \(x \text{% of } 40 = 25\)
    \(x \text{% } × 40 = 25\)
    \(x = {25 \over 40} × 100 \)
    x = 62.5

    \(x = 45 \text{% of } 40 \)
    \(x = 0.45 × 40 \)
    x = 18

Solved Examples Set 1
Solved Examples Set 2
Solved Examples Set 3