Test: CAT Quantitative Aptitude- 7 (October 28) - CAT MCQ

Consider the figure,

The radius of hemispherical bowl = r Radius of cylinder below the surface of water = Radius of cylinder PQRS = r/3
Height of cylinder = r

Volume of the water displaced = Volume of the cylinder immersed.

Maximum height of the cylinder that can be immersed = AB - AC = BC

Using Pythagoras theorem in ΔPOT, we get,
∴ BC = 2√2r / 3
∴ Height of cylinder below the surface of water 
= 2√2π / 3 
∴ Volume of Liquid displaced = Volume of the cylinder immersed
= π x r² / 9 x 2√2π / 3 = 2√2πr³ / 27
Hence, option 2.

Murugan takes two hours to complete the journey in which he travels from home to school and back.

∴ Car takes 1 hour to travel from home to school and vice versa.
When son starts walking towards home, 20 minutes saved in whole journey i.e., 10 minutes journey from home to school and vice versa.

∴ Distance travelled by son = Distance travelled by car in 10 minutes.

As car takes 1 hour to reach from home to school of which 10 minutes distance is covered by son, car travels for 50 minutes from home to school.

∴ Son walks for 50 minutes.

Distance from school to car = 5 km

Speed of the boy = 6 km/hr

Hence, option 2.

Given the problem, Krunal played 11 matches, and the average runs scored by him in these 11 matches was 16. The problem states that the runs scored in the nnnth match was 2 less than that of the (n−1)th match.

Step 1: Establish the relationship between matches

Let the runs scored in the first match be x.

Then, according to the problem:

• Runs in the second match = x−2
• Runs in the third match = x−4
• ...
• Runs in the nnnth match = x−2(n−1)

Step 2: Calculate the sum of runs

The sum of runs scored over 11 matches is:

S=x+(x−2)+(x−4)+⋯+[x−2(11−1)]

This is an arithmetic sequence with the first term xxx and the common difference −2. The sum of an arithmetic sequence can be calculated using the formula:

Let the speed of stream be x km/hr.

∴ Length of the river = (5 + x) x 5 = (5x + 25) km

Also, [5 - (x + 1)] x 40 = 5x + 25

Solving this we get, x = 3 km/hr

∴ Length of the river = (5 x 3 + 25) = 40 km

Answer: 40

Hence Option 2

Let the free baggage allowed on board = f kg
It is seen that Mr. P was charged double the amount of Mr. Q.
Thus, as the rate per kg is the same, the excess baggage of Mr. P must be 2 times that of Mr. Q.
Let the excess baggage of Mr. Q be q kg.
Thus, the excess baggage of Mr. P = 2q kg
Let the rate charged for the excess baggage be Rs. r/kg
So, the quantity of baggage carried by Mr. P = (f + 2q) kg and that carried by Mr. Q = (f + q) kg
Now, as the total baggage carried is 50 kg, we have f + 2q + f + q = 50
Or 2f + 3q = 50 .... (i)
As Mr. Q was charged Rs. 1,400, we have
qr = 1,400 .... (ii)
Now, when 1 of them has all the baggage, the total baggage becomes (2f + 3q) kg.
As the free baggage allowed is halved, the charge becomes Rs. (1.5r + 3qr) = Rs. 6,300
Or 1.5r + 3qr = 6,300 ....(iii)
Plugging in the value of qr from equation (ii) into equation (iii), we have
1.5r + 4,200 = 6,300 .... (iv)
Or fr = 1,400 .... (v)
Now, from equations (ii) and (v), we get that
f = q .... (vi)
Plugging in the value of f as q from equation (vi) into equation (i) and solving for q, we get
q = 10
Thus, f = 10 as well.
Now, total weight of Mr. P's baggage = (f + 2q) kg
= (10 + 2 × 10) kg
= (10 + 20) kg
= 30 kg

Let the negative marking per mistake for the first 20 mistakes be n₁ and for all the subsequent mistakes, let it be n₂.
Jack attempted 160 questions and got only 80 correct and Jill attempted 150 questions and got only 100 correct.
So, 80(1) - 20(n₁) - 60(n₂) = 55
100(1) - 20(n₁) - 30(n₂) = 85
20 + 30n₂ = 30
n₂ = 1/3 = 0.33

If a point (p, q) lies inside the circle x² + y² = 11, then p² + q² < 11.
If p = 0, q can be –3, –2, –1, 0, 1, 2 or 3.
If p = 1 or –1, q can be –3, –2, –1, 0, 1, 2 or 3.
If p = 2 or –2, q can be –2, –1, 0, 1 or 2.
If p = 3 or –3, q can be –1, 0 or 1.
So, total = 7 + 14 + 10 + 6 = 37

Given:                           

Probability of leap year having 53 Sundays is to be determined.

Concepts used:

Number of days in leap year = 366

Calculation:

A week has 7 days and total days are 366.

⇒ Number of Sundays in a leap year = 366/7 = 52 Sundays + 2 days

⇒ Total outcomes with 2 days = (Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday) = 7

⇒ Number of outcomes without Sundays = 5

⇒ Probability of leap year with 53 Sundays = 2/7

∴ Probability of leap year with 53 Sundays is 2/7.

Let the first term and common ratio be a and r respectively.
Forth term = ar³ = 378

Similarly, Fifth term = ar⁴ =1134 Now,

r = ar⁴lar³ = 1134/378 = 3

∴ a =1134/81 = 14

Answer: 14

Hence Option 2

Let x = 4.
So we get,

f(4) = -f(5)
Now we will try the options.
Option 1: f(x) = 1 - x ⇒ f(4) = 1 - 4 = - 3 and - f(5) = -[1 - 5] = 4 As, f(4) ≠ - f(5)
Hence, option 1 is incorrect.
Option 2: f(x) = 1 - x² ⇒ f(4) = 1 - (4)² = 1 - 16 = -15 and -f(5) = -[1 - (5)²] = 24 As, f (4)  -f(5)
Hence, option 2 is incorrect.
Option 3: f(x) =x²(1 -x)² ⇒
f(4) = 1 - (4)² = 1 - 4)² = 16 x 9 and -f(5) = -(5)² x [1 - (5)]² = 25 x 16
∴ f(4) ≠ -f(5)
Hence, option 3 is incorrect.
Hence, option 4.

The angles are 180°and 180°.
So, 58(r – 2)s = 59r(s – 2)
Or r = 116s/(118 – s)
But r is positive; so s  117.
s = 117 (Maximum)

Let the cost price for cookies A and B be 2x and 5x respectively and the profits for cookies A and B be 4y and 5y respectively.

Thus, for cookie B, SP = 5x + 5y = 100

∴ x + y = 20

For cookie A, SP = 2x + 4y But we do not know the value of x andy hence the selling price of cookie A cannot be determined.
Hence, option 4.

The question cannot be answered using either of the statements alone. We need not combine the statements as statement II implies statement I.
Answer: 5

Hence Option 5

Let A₁, A₂, A₃, A₄, …, A_m be 'm' A.M.s between 1 and 31.
Therefore, 1, A₁, A₂, …, A_m, 31 are in A.P.
Let d be the common difference of the A.P.
Here, the total number of terms is m + 2 and T_{m + 2} = 31.
1 + (m + 2 - 1)d = 31
(m + 1)d = 30
d = 
A₇ = T₈ = a + 7d = 1 + 7  =  = 
…………………
…………………
A_{m - 1} = T_m = 1 + (m - 1)d
= 1 + (m - 1)  = 
= 
 =  =  =  [Given]

 = 

9m + 1899 = 155m - 145
146m = 2044
m =  = 14
Hence, m = 14

Numbers between 1 and 100 divisible by 3 are 3, 6, 9, …, 99.
Sum = 3 + 6 + 9 + 12 + ... + 99
Sum = 3(1 + 2 + 3 + 4 + ... + 33)
Now, using the formula,
Sum of first n natural numbers = 

1 + 2 + 3 + 4 + ... + 33 =  = 561

Sum = 3(561) = 1683

Let the marked price of the wholesaler be Rs. x.
After 20% discount, shopkeeper buys the goods at Rs. 0.8x To fetch 50% profit,
Shopkeeper marks the goods at 0.8x + 0.4x = Rs. 1.2x
To sell the goods at the marked price of wholesaler, shopkeeper should offer
∴ discount of 0.2x Discount percentage = (0.2x/1.2x) x 100 = 16.67%
Hence, option 2.

Let x be the height of the cone.
∴ Diameter = 3x/2

∴ Radius = 3x/4

So we get,
25² = x² + (3x/4)²

∴ x = 20 cm

∴ Radius of the cone = (3 x 20)/4 = 15 cm

Volume of the cone = (1/3) x π x 15² x 20 = 1500π cm³ 
Hence, option 1.

Statement I alone is not sufficient as xy may or may not be less than 5.
Statement II restricts x to 1/2 < x < 2/3 and y to the interval - 4 < y< 4 .

Thus, largest possible value of xy would be less than 8/3, which is less than 5.

Thus, Statement II is sufficient to answer.
Hence, option 2.

Please refer to the Venn diagram. Here P stands for Physics, C stands for Chemistry, M stands for Maths and B stands for Biology.

We can write the given statements in form of ratios as following:
P C M B
2x 4x
1x 2x 3x 1x

∴The ratio of students taking up P:C:M:B will be 1:2:3:1.

Since we are supposed to maximize the number of students taking up all subjects, place lx the minimum of them in the intersection area of all four subjects. After placing the remaining students we get the values for different areas as given in the Venn diagram.

∴The total number of students will be 3x = 120

∴ x = 40
Hence, option 2.

DE = BF = 10 km
tan 60° = √3 = y / EF
∴ EF = y / √3 = DB
Let CF be y km.
tan 60° = √3 = y / EF
∴ EF = y / √3 = DB
AB = 10√3 + y / √3
BC = 10 +y
AC = 10√13 km
AB²+BC² = AC²  ...(Pythagoras theorem)
(10√3 + y / √3)² + (10 + y)² = (10√13)²
Solving, y = 15 km

∴ Height = 10 + 15 = 25 km

Hence, option 3.

The bigger triangle is a right triangle.
So, area = [1/2] × 120 × 22 = 1320 m² [1]
For the triangle having the lengths of sides as 22 m, 24 m and 26 m,
s =  [1/2]
Area of the triangle =  [1/2]
= 
= 
= 245.92 m² [1]
Therefore, area of the shaded region
= (1320 – 245.92) m²
≈ 1074 m² (Approx.) [1]
1074 is the nearest integer answer.