AMCAT Quantitative Ability Sample Questions and Answers (Advanced)

Soln. Let x be the total distance traveled throughout the Journey
Time taken by Andrew – Time taken by Beatrice
=15.Minutes=\frac{1}{4}Hours
\frac{x}{40}-\frac{x}{50}=\frac{1}{4}
\frac{x}{200}=\frac{1}{4}
x = 50 Km

Soln. First, we have to find out the difference between the given numbers and their respective remainders.
(20 – 14) = 6, (25 – 19) = 6, (35 – 29) = 6 and (40 – 34) = 6
Reqd. No. = (L.C.M. of 20, 25, 35, 40) - 6
∴ Required No. = 1400 – 6 = 1394

Soln. Let rotten apples,
10\times\frac{2}{5}=4
Remaining = 6 Apples

If 1 rotten apple + 2 normal apples
=^4C_{1}\times^6C_{2}=20

For 2 rotten apples + 1 normal apple
^4C_{2}\times^6C_{1}=36

For 3 rotten apples,
^4C_{3}=4

Total Outcomes = ^{10}C_{3}=120
Probability = \frac{60+36+4}{120}=\frac{100}{120}=\frac{5}{6}

Soln. Let the number be x and (184 – x). Then,
\frac{x}{3}-\frac{184-x}{7}=8
7x – 3(184 – x) = 168
10x = 720
x = 72

Soln,
Let the numbers be 4x and 5x.
Given ratio = 4: 5
Let HCF = x
We know that,
∴ HCF X LCM = Product of numbers
x X 200 = 4x X 5x
200x= 4x\times5x
200x= 20x^2
x = 10
The Numbers will be 40 and 50
And, the sum of both numbers will be 90

Explanation:
Here, Using Direct Method to find out,
Reduced Cost = \frac{A \times X}{100 \times n}
Where, A = Rate of increased quantity = 90 Rs
X = Total discount applied = 25 %
n = Increased quantity rate = 4 Kg
Reduced Cost resulting Gain = \frac{90 \times 25}{100 \times 4}
= Rs 5.62 per Kg
Now, Original Price per Kg of Kiwi = \frac{A \times X}{(100 - X) \times n}
= \frac{90 \times 25}{(100 - 25) \times 4}
= 7.5 Rs per Kg
Profit = Original Price - Reduced Code
Profit = 1.88 Rs Per Kg

Soln.
Using the rule log_a b - \log_a c = \log_a \frac{b}{c}
We will get:
log_3 49 - \log_3 7 = \log_3 \frac{49}{7} = \log_3 7

Soln. Case I: 5 males and 3 females can be selected. Then, Number of ways of selection,
^9C_{5}\times^7C_{3}=126\times35=4410

Case II: 6 males and 2 females can be selected.Then, Number of ways of selection,
^9C_{6}\times^7C_{2}=84\times21=1764

Case III: 7 males and 1 female can be selected. Then, Number of ways of selection,
^9C_{7}\times^7C_{1}=36\times7=252

Case IV: 8 males can be selected. Then, Number of ways of selection,
^9C_{8}=9

Now, Total No. of ways of selecting the members = 4410 + 1764 + 252 + 9 = 6435 ways

Soln. The average rate of percentage increase in both cases,
Weekly and Yearly will be the same.
Then, Annual % in Sales,
=\frac{21-15}{15}\times100=\frac{600}{15}=40%

Soln. SI for 2 years at rate of 15% = 15 × 2 = 30%
CI for 2 years at rate of 15%,
Applying the Net% effect for 2 years
=15+15+\frac{15\times15}{100}=32.25%
Thomas's gain % = 32.25% – 30% = 2.25%
Total gain of Thomas = 2.25% of 20000
=\frac{2.25\times20000}{100}=450