Quantitative Aptitude (QA) Test - 28

Calculation:

According to the question,

√(log_e(4x – x²)/3) to be a real number, because x is a real number 

Then, it's root should be greater than 0

⇒ √(loge(4x – x2)/3) ≥ 0

If, log_e = 2.7(Approx)

It means log_B A ≥ 0

If A ≥ 1

It's true then, B > 1

⇒ (4x – x2)/3) ≥ 1

⇒ 4x – x2 ≥ 3

⇒ x² – 4x + 3 ≤ 0

⇒ x² – 3x – x + 3 ≤ 0

⇒ (x – 1)(x – 3) ≤ 0

⇒ x belongs to (1, 3)

∴ x lies between 1 ≤ x ≤ 3.

since, the expression (x + x² + x³ + x⁴)16 has consecutive powers of x starting with x¹. and the highest power of x in the expansion is x⁴⁹.

Thus, There are 49 terms.

Concept:

n objects can be arranged in n! ways

When two or more objects are taken together they are considered as one object.

Calculation:

The total number of vowels in EQUATION is 5 (E, U, A, I, O)

So, we can consider them as single unity and they can arrange themselves in 5! ways.

Rest 3 consonants along with the vowels together can arrange in 4! ways.

So, the total number of ways the word can be arranged = 4! × 5!

let the time taken by them to cross each other be t

then time taken by A and B to reach their destinations after meeting is 12 hours and 27 hours.

⇒ t = √(12×27) = √(324) = √(18²) = 18 hours.

Calculation:

Given:

⇒ 1/a + 1/b + 1/c = 0 

Multiplying both sides by abc, we get

⇒ bc + ac + ab = 0     ---(1).

Also, (a + b + c) ² = 9²

⇒ a² + b² + c² + 2 (ab + bc + ac) = 81     ---(2)

Substituting (1) in (2), we get

⇒ a² + b² + c² = 81

We have, a³ + b³ + c³ - 3abc = (a + b + c) (a² + b² + c² - ab - ac - bc)

⇒ (9) (81 - 0) = 729

Calculation:

Abeer has 14 novels i.e Abeer = 14 novels

Abeer’s friend Sanjay takes 3 of them and gives 2 novels to Abeer, i.e Abeer = 13 novels left.

Abeer donates 7 novels but buys 4 it means Areeb has 10 novels left.

Joy takes 4 novels from Abeer and gives him 5 which means that Abeer has 11 novels left.

Abeer takes one novel from Sanjay and gives it to Joy in exchange for 3 more i.e Abeer = 14 novels.

Abeer gives those 3 novels to Sanjay and he gives Abeer a novel and a magazine, i.e Abeer = 12 novels and 1 magazine.

Rahul takes the magazine which Sanjay gave to Abeer and gives Abeer a textbook, i.e Abeer = 12 novels, and 1 textbook.

Abeer gives the textbook to Joy in exchange for a novel i.e Abeer = 13 novels.

Rahul takes the novel from Sanjay, gives it to Joy for a magazine, and gives Abeer the magazine for a novel, i.e Abeer = 1 magazine and 12 novels left.

Hence Abeer has 12 novels and 1 magazine and 0 textbooks.

Hence option 1 is correct.

for x=2/3 , x=2 , x=3 , x=3 , x=4 , x=5, the equation either becomes equal 0 or not defined. Thus, these points are the intercepts on the number line. (given below)

After plotting the number line, the region on the right of the rightmost point will be positive. and every next left region will be alternate to the preceding region.

Calculation:

The ball bounces back 90% of the previous height,

Let h be the height from which the ball is dropped.

Total distance = h + 0.9h + 0.9h + 0.9 × 0.9h + 0.9 × 0.9h + ........

⇒ h + 2 × (0.9h + 0.9 × 0.9h + ......)

⇒ h + 2 × (0.9h/(1 - 0.9)

⇒ h + 18h = 19h

⇒ 19h = 190

∴ h = 10m

Mistake Point:

Every time ball reaches a height, it will travel the same distance to reach the floor before the next bouncing.

Additional Information

The n^th term of a GP is given by ar^{n - 1}, where a is the first term and r is the common ratio of the GP.

Sum of n terms of GP with first term a and common ratio r is a ×  (rn - 1)/(r - 1).

The Sum of an infinite GP will be a/(1 - r).

Calculation:

⇒ Radius of the bucket (R) = 14 cm

⇒ Height of the bucket (H) = 15 cm

⇒ Area of the base of the cone = πr² = 5544 cm²

⇒ r = 42 cm

⇒ Volume of the bucket (V) = πr²h = 3.14 × 14² × 15 = 9231.6 cm²

⇒ Volume of the cone = 30 × volume of the bucket

⇒ 1/3 x πr²h = 30 × 9231.6

⇒ 1/3 × 3.14 × 42² h = 276948

⇒ h = 150 cm

⇒ Slant height, l² = r² + h² = 42² + 150²

⇒ l² = 1764 + 22500 = 24264

⇒ l = 155.8 cm

⇒ Curved surface area of the cone = πrl = 3.14 × 42 × 155.8 = 20542 cm²

Additional Information

Cone

⇒ Slant Height, l = √(r² + h²)

⇒ Volume(V) = ⅓ πr² h cubic units

⇒ The total surface area of the cone = πrl + πr² or Area = πr(l + r)

Cylinder

⇒ Curved Surface Area = 2πrh square units

⇒ Total surface area, A = 2πr(r + h) square units

⇒ Volume of the Cylinder, V = πr2 h cubic units

Calculation:

According to the question;

The 4 digit number having 7 before 3 can be formed 

Case I : When 7 is thousand placed, then 3 may be hundred place, tens place and unit place

From the following 73xy, 7x3y, 7xy3 are formed

Total such number formed = 3 × (8 × 7) = 168

Case II : When 7 is hundred placed, then 3 may be tens place and unit place

From the following x73y, x7y3 are formed

Total such number formed = 2 × (7 × 7) = 98

Case III : When 7 is tens place and 3 is unit place

From the following xy73 is formed

Total such number formed = 7 × 7 = 49

The total such number formed = 168 + 98 + 49 = 315

∴ The 4 digit number greater than 1000 with 7 before 3 is 315.