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

# Solved Examples Set 1 (Quantitative Ability)

1. $${𝑥 - 8 \over 24} = {3 \over 4}$$
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.
2. A train takes 50 minutes for a journey if it runs at 48 km/hr. The rate at which the train must run to reduce the time to 40 minutes will be

1. 50 km/hr
2. 55 km/hr
3. 60 km/hr
4. 57 km/hr
5. 65 km/hr
$$50 × 48 \over 40$$ = 60 $$km \over hr$$
3. A single discount equivalent to a discount series of 20%, 10% and 25% is

1. 55%
2. 54%
3. 46%
4. 42%
5. 50%
If 3 succesive discounts are a%, b% and c%
then single discount = a + b + c – ($$ab \over 100$$ + $$bc \over 100$$ + $$ca \over 100$$ – $$abc \over 10000$$)
a = 20, b = 10, c = 25, solving we get, 46%.
4. 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
5. A third-grade class is composed of 16 girls and 12 boys. There are 2 teacher-aides in the class. The ratio of girls to boys to teacher-aides is

1. 16:12:1
2. 8:6:2
3. 8:6:1
4. 8:3:1
5. 4:3:1
Girls to boys to teacher-aides are in proportion 16 to 12 to 2. Reduced to lowest terms, 16:12:2 equals 8:6:1.
6. On a trip to visit friends, a family drives 65 miles per hour for 208 miles of the trip. If the entire trip was 348 miles and took 6 hours, what was the average speed, in miles per hour, for the rest of the trip?

1. 44
2. 50
3. 51
4. 58
5. 60
As the first part of the trip took $$\frac{208 \text{ miles}}{65 \text{ } \frac{miles}{hour}} = 3.2 \text{ hours},$$ so the remaining 140 miles (348 - 208) took 2.8 hours (6 - 3.2). The average speed for the rest of the trip was $$\frac {140 \text{ miles}}{2.8 \text{ hours}} = 50$$ miles per hour.
7. By selling 60 chairs, a man gains an amount equal to selling price of 10 chairs. The profit percentage in the transaction is

1. 10%
2. 15%
3. 16.67%
4. 20%
5. 22%
selling price of 60 chairs = selling price of 10 chairs
profit of 60 chairs = profit of 10 chairs
profit of 6 chairs = profit of 1 chair
profit of 1 chair = profit of 1/6 chair
profit %age = 1/6 x 100 = 16.67%
8. 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
9. 1.02 - 0.20 + ? = 0.842

1. 0.222
2. 232
3. 2
4. 0.022
5. 0.012
1.02 - 0.20 + ? = 0.842
0.82 + ? = 0.842
? = 0.842 - 0.82 = 0.022
10. A primary school had an enrollment of 850 pupils in January 1970. In January 1980 the enrollment was 1,120. What was the percentage increase for the enrollment?

1. 31.76%
2. 33.50%
3. 30.65%
4. 34.76%
5. 30.55%
Percentage increase for the enrollment = $$1120 - 850 \over 850$$ × 100 = 31.76
11. $${5.76 \over 1.6} - 2.4 = ?$$

1. 1.2
2. 2.4
3. 7.2
4. 0.12
5. 0.012
$${5.76 \over 1.6} - 2.4 =$$ 3.6 - 2.4 =1.2
12. If n! = n ⋅ (n − 1) ⋅ (n − 2) ⋅ (n − 3) . . . 2 ⋅ 1, what is the value of $$\frac{(6!)(4!)}{(5!)(3!)}$$

1. 5/4
2. 8/5
3. 10
4. 24
5. 1152
$$\frac{(6!)(4!)}{(5!)(3!)}$$ = $$\frac{(6 . 5 . 4 . 3 . 2 . 1)(4 . 3 . 2. 1)}{(5 . 4 . 3 . 2 . 1)(3 . 2 . 1)}$$ = $$\frac{6 . 4}{1}$$ = 24
13. Which set of ordered pairs represents a function?

1. {(−5,5),(4,8),(−5,−6)}
2. {(−1,−1),(−1,6),(−1,−10)}
3. {(−3,7),(2,5),(−7,7)}
4. {(2,3),(−2,4),(−2,−5)}
5. {(2,3),(3,2),(2,5)}
For a set of ordered pairs to be a function, no single 𝑥-coordinate can be mapped to two distinct 𝑦-coordinates. This is not the case for option A, where 𝑥=−5 is mapped to both 𝑦=5 and 𝑦=−6. Similarly, in options B (𝑥=−1), D (𝑥=−2), and E (𝑥=2), an 𝑥 value is mapped to two different 𝑦 values.
14. Which expression is equivalent to $$\frac{6𝑥^2 + 4𝑥}{2𝑥}$$?

1. 7x
2. 5x2
3. 3x + 2
4. 6x2 + 2
5. 3x2 + 2x
As $$\frac{6𝑥^2}{2𝑥} = 3𝑥,$$ and $$\frac{4𝑥}{2𝑥} = 2,$$ so then $$\frac{6𝑥^2 + 4𝑥}{2𝑥} = 3𝑥 + 2$$
15. In the series 8, 9, 12, 17, 24 . . . the next number would be

1. 29
2. 30
3. 33
4. 35
5. 41
In the series, 8, 9, 12, 17, 24 . . .
9 − 8 = 1
12 − 9 = 3
17 − 12 = 5
24 − 17 = 7
Hence, the difference between the next term and 24 must be 9 or
x − 24 = 9, and
x = 33
Hence, the next term in the series must be 33
16. How much would I have to pay for a book which cost $72 to product, if the printing company sold it to a bookseller at 20% profit and in return the bookseller sold it to me at a profit of 25%? 1.$ 104
2. $106 3.$ 108
4. $110 5.$ 109
Cost of the book product = $72 Profit of printing company = 20% of 72 = 0.2 x 72 = 14.4 Now the cost of the book = 72 +14.4 =$ 86.4
Profit of the bookseller = 25% of 86.4 = 21.4
Finally, the cost of the book = 86.4 + 21.4 = $108 17. A fruit-seller has 120 oranges. Given that he has 20% more apples than oranges and 40% less oranges than pears, find the number of apples and the number of pears the fruit seller has. 1. 144, 200 2. 148, 380 3. 149, 220 4. 140, 190 5. 142, 190 No. of apples = 120 + 20% of 120 = 120 + 0.2 × 120 = 144 Let x = No. of pears x - 40% of x = 120 x - 0.4x = 120 0.6x = 120 x = $$120 \over 0.6$$ = 200 Hence, no. of pears = 200 18. $${396 \over 11}$$ + 19 = ? 1. 19.8 2. 36 3. 55 4. 33 5. 50 $${396 \over 11}$$ + 19 = 36 + 19 = 55 19. A shop owner blends three types of coffees, A, B and C, in the ratio 3:5:7. Given that type A coffee costs$ 70 per kg, type B coffee costs $100 per kg and type C coffee costs$ 130 per kg, calculate the cost per kg of the blended mixture.

1. $106 2.$ 108
3. $109 4.$ 110
5. $105 Cost per kg = 70 x 1/5 + 100 x 1/3 + 130 x 7/15 =$ 108 per kg
20. 10 men can complete a job in 14 days. How long will it take 4 men to finish the same job if they work at the same rate?

1. 33 days
2. 35 days
3. 37 days
4. 39 days
5. 31 days
$$14 × 10 \over 4$$ = 35 days