Que: According to the table given below, a state has a total of 23,000 numbers of companies from seven regions. By what percent of the total number of companies in the region, the number of companies of Region V is more than the number of companies of Region T, approximately?

Region-wise distribution of companies in the state
Region P ------------- 1,100
Region Q ------------- 2,525
Region R ------------- 1,800
Region S ------------- 2,100
Region T ------------- 2,750
Region U ------------- 150
Region V ------------- 3,500

(A) 25.1%
(B) 44%
(C) 19.7%
(D) 5.4%
(E) 27.27%
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Solution: Number of companies in region V = 3,500 and the number of companies in region T = 2,750

Thus, Percent change: After – Before/ Before * 100(%) [After:3,500 ; Before: 2,750, since ~than Before, so Region T becomes Before]

=> \(\frac{(3,500 – 2,750) }{ 2,750}\) * 100(%)

=> \(\frac{750}{2,750}\) * 100(%) = 27.27%

Therefore, E is the correct answer.

Answer E
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Que: With the increase of 20% in the price of milk, a housewife can buy 5 liters less quantity for $60 than she was buying before the increase. What was the initial price per liter of milk?

(A) $2.00
(B) $2.50
(C) $2.75
(D) $3.00
(E) $3.50
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Solution: The price of milk is increased by 20%.

Let the original price of milk per liter be $10.

Increased price: \(\frac{120}{100} * $10 = $12\)

Number of liters of milk: \(\frac{Total Price }{ Price per liter }\)

Number of liters of milk before increase: \(\frac{$60}{$10}\)\(\)=6

Number of liters of milk after increase: \(\frac{$60}{$12}\)=5

The difference in the quantity of milk obtained: 6 - 5 = 1

Thus, for the $10 initial price, the difference is 1 liter.

For 5 liters difference, \(\frac{10}{5}\) = $2

Therefore, A is the correct answer.

Answer A
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Que: According to the data obtained for the last year, $a can buy a ‘b’ number of items. If the average cost of each item increased by 30 percent and also 'b' items can be bought this year, then the number of items can be bought with $8a equals

(A) 4.8b
(B) 2.4b
(C) 6.15b
(D) 8b
(E) 5.50b
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Solution: Cost of b items = $a

Since the cost increases by 30%, the new cost of b items = \(\frac{130}{100} * $a = $\frac{13a}{10}\)

Thus, with a budget of $\(\frac{13a}{10}\), b items can be bought this year. Thus, with the budget of $8q, the numbers of items that can be bought

=> \(\frac{8ab}{\frac{13a}{10}}\) (Since $\(\frac{13a}{10}\):b = $8a:x, we get x = \(\frac{8ab}{\frac{13a}{10}}\) = 6.15b

Therefore, C is the correct answer.

Answer C
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Que: If p is the product of the reciprocals of integers from 150 to 250, inclusive, and q is the product of the reciprocals of integers from 150 to 251, inclusive, what is the value of (p^{−1} + q^{−1} ) in terms of p?

(A) \(\frac{p }{ (251)^2}\)

(B) 251 × 252 × p

(C) 252p

(D) \(\frac{252}{ p}\)

(E) 251 × 252 × \(p^2\)
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Solution: \(p = \frac{1}{150} * \frac{1}{151} * ….. * \frac{1}{250}\)

\(q = (\frac{1}{150} * \frac{1}{151} * ….. * \frac{1}{250}) * \frac{1}{251} = p* \frac{1}{251} = \frac{p}{251}\).

Thus, we have \(p^{−1} + q^{−1} = \frac{1}{p} + \frac{1}{q}\)

=> \(\frac{1}{ p} + \frac{251 }{ p} = \frac{252}{p}\)

Therefore, D is the correct answer.

Answer D
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Que: A set is such that if m is in the set, \(m^2 + 3\) is also in the set. If −1 is in the set, which of the following is also in the set?

I. −2
II. 4
III. 19

(A) Only I
(B) Only II
(C) Only I and II
(D) Only II and III
(E) I, II, and III
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Solution: According to the problem: If m is in the set, \((m^2 + 3)\) is also in the set.

However, it does NOT imply that if \((m^2 + 3)\) is in the set, then m must be in the set.

What it does imply is that: If \((m^2 + 3)\) is NOT in the set, m is NOT in the set.

Thus, if we have m = −1 as a member of the set, \((m^2 + 3)\) = \((-1)^2 + 3 = 4\) is also a member of the set. Thus, statement II is correct.

Proceeding in the same way: Since m = 4 is a member of the set, then \((m^2 + 3)\) = \((4^2 + 3)\) = 19 is a member of the set. Thus, statement III is also correct.

Therefore, D is the correct answer.

Answer D
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Que: A machine can be repaired for $800 and will last for one year, while the new machine would cost $3,600 and will last for three years. The average cost per year of the new machine is what percent greater than the cost of repairing the current machine?

(A) 15%
(B) 25%
(C) 116.67%
(D) 50%
(E) 66.67%
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Solution: The cost of repairing the current machine = $800. The cost of new machine = $3,600

Since the new machine lasts for three years, the average cost per year = \(\frac{$3,600}{ 3} = $1,200\).

Thus, the required percentage = \(\frac{(1,200 – 800)}{ 800}\) × 100(%) = 50%.

Therefore, D is the correct answer.

Answer D
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Que: In a class, 65 percent of the boys and 78 percent of the girls play basketball. If 72 percent of all the students play basketball, what is the ratio of the number of girls to the number of boys?

(A) \(\frac{4}{3}\)

(B) \(\frac{7}{6}\)

(C) \(\frac{8}{7}\)

(D) \(\frac{9}{8}\)

(E) \(\frac{13}{11}\)
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Solution: Let the number of the boys be 100b and the number of girls is 100g, then we get the total number of students to be 100(b+g).

As we are dealing with percent, according to the IVY approach take ‘100’ as total, and b and g are the initials of the words ‘boys’ and “girls” respectively.

Also, this question is applied by the IVY Approach, Each Each Together

=> 65 percent of the boys: \(\frac{65}{100} * 100b = 65b (Each)\)

=> 78 percent of the girls: \(\frac{78}{100} * 100g = 78g (Each)\)

=> 72 percent of all the students: \(\frac{72}{100} * 100(b+g) = 72(b+g) (Together)\)

Then, we get 65b + 78g = 72(b+g) = 72b + 72g.

Rearranging gives us 78g - 72g = 72b - 65b, 6g = 7b or g = \(\frac{7b}{6}\).

Ratio of number of girls : Ratio of number of boys:

=> 100g : 100b = g : b = \(\frac{7b}{6}\) : b = \(\frac{7}{6}\) : 1

=> 7 : 6

Thus, the ratio of number of girls to the number of boys: 7:6

Therefore, B is the correct answer.

Answer B
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Que: 44, 52, 56, 65, 73, 75, 77, 95, 96, 97

The list above shows the scores of 10 students obtained on a scheduled test. If the standard deviation of the 10 scores is 20.50, how many of the scores are greater than one standard deviation above the mean of the 10 scores?

(A) None
(B) One
(C) Two
(D) Three
(E) Four

Solution: Question asks the number of scores > (Mean + S.D.)

=> Mean = (44 + 52 + 56 + 65 + 73 + 75 + 77 + 95 + 96 + 97) / 10 = 730 / 10 = 73

=> S.D. = 20.50. Thus,

=> Mean + S.D. = 73 + 20.50 = 93.50

It is clear that three scores (95, 96, and 97) are greater than 93.50.

Therefore, D is the correct answer.

Answer D
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Que: The number of cars this year decreased by 10% compared to last year. The number of mobiles this year has increased by 12% compared to last year. If the total number of cars and mobiles this year increased by 5% compared to last year, what was the ratio of the number of cars to the number of mobiles last year?

A. 1:2
B. 15:7
C. 9:11
D. 11:9
E. 7:15
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Que: Approximately how many seconds will it take for a car with a constant rate of 55 miles per hour to travel a distance of 750 feet? (1 mile = 5,280 feet)

A. 9.3 sec
B. 10 sec
C. 11.2 sec
D. 16 sec
E. 21 sec
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Que: Approximately how many seconds will it take for a car with a constant rate of 55 miles per hour to travel a distance of 750 feet? (1 mile = 5,280 feet)

A. 9.3 sec
B. 10 sec
C. 11.2 sec
D. 16 sec
E. 21 sec
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Solution: Now let's solve this PS question using IVY Approach.

IVY Approach -> IVY Approach 3-2: 100c = Cars; 100m= Mobiles

=> Percentage question ->100 -> c = cars ; m = mobiles

=> This year cars = 90% of 100c = 90c (∵ 10% decrease)

=> This year mobiles = 112% of 100m = 112m (∵ 12% increase)

=> Total Vehicles this year: 105% of(100c + 100m) = 105c + 105m (∵ 2% increase)

=> 90c + 112m = 105c + 105m

=> 112m – 105m = 105c – 90c

=>7m = 15c

=> \(c = \frac{7}{15} m\)

=> \(∴ 100c : 100m =c :m=\frac{7}{15} m: m = 7 :15 \)

Therefore, E is the correct answer.

Answer E
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