CAT Quant Mock Test - 1 - CAT MCQ

Let the amount that spent on each variety of items by x.

Hence, total amount I spent last year = 5x.

This year I would have spent x (1.11) + x (1.22) + x (1.33) + x (1.44) + x (1.55)

Hence the total cost increased by

x x 0.11[1 + 2 + 3 + 4 + 5] = x x 0.11 x 15

∴ Percentage increase in cost

We are looking for these values of x for which

2 – 1 / y < x ≤ 5 + 1 / y, for all positive integral values ofy.

Clearly all the numbers between 2 and 5 satisfy the condition. 5 itself satisfies the condition and 2 also satisfies the condition.

But any number greater than 5 or less than 2 doesn’t satisfy the condition for all integral values ofy.

Therefore, the best description is [2, 5] or 2 ≤x ≤ 5.

Since 16 = 2⁴, in order to convert a number from base 2 to base 16, every set of four digits, starting from the left, can be directly replaced with their base 16 equivalent.

∴101110101101110)₂= [(101)(1101 )(0110)(1110)]₂

= [5DGE],₁₆.

Hence, there are two digits which are less than 9.

y > total number of Re.1 and ₹2 notes => y > 7x.

Total value of notes = 1 (4x) + 2(3x) + 10y = 10{x + y).

As x and y are integers this is divisible by 10.

As y > 7x, 10(x + y) > 80x i.e., it is greater than 780.

Going by the choices, both these conditions are satisfied by only Choice (A).

Initial volume = π x r²x h₁

After reduction, radius = 3 / 4 r

Final volume = (π x r²x h₁) / 2

(π x r²x h₁) / 2 = (π x 9 / 16 x r² x h₂)

⇒h₁ / h₂ = 9 / 16 x 2 = 9 / 8

Percentage decrease in height = 1 / 9

=

Let the sides of the triangle be a, ar and ar².

The largest side of a right angled triangle is the hypotenuse.

Hence it must be ar²

By hypotenuse theorem,

a² + (ar)² = (ar²)²

a² + a²r² = a²r⁴.

=> a²(1 + r²) = a²r⁴.

=> r⁴- r²- 1 =0.

(a) A person, say A may have shook hands with 1,3,5 etc people. This is possible.

(b)A person may have shook hands with all the others. This is possible.

(c)If each person has shaken hands with a different number of people, the number of handshakes for the 25 people has to be 0, 1, 2. ....24. But if one person has 0 handshakes, no person can have made 24 handshakes. At the most a person can have made 23 handshakes. This statement can never be true.

Height = x cm and length = breadth = 30 - 2x cm

Now to maximise this volume x(30 - 7x) (30 - 2x) must be the maximum.

If x(30 - 2x)² is maximum, then 2x(30 - 2x)² must also be maximum.

Now for any expression of the form x(K - x)n to be the maximum, x and (K - x) must be in the ratio 1 : n.

Hence for 2x (30 - 2x)² to be maximum 30 – 2x / 2x = 2 / 1

=> x = 5 => 5(30 - 2 X 5)² = 2000.

(1) (m, n) = (2, 5)

As one pile can be divided into 2, the number of piles can increase by steps of 1. It’ll be possible to get 1, 2, 3,4, 5,….piles.

(2)(m, n) = (3, 6)

In any turn, the number of piles can increase by 2 or 5.

∴It can increase by four 2’s and one 5, i.e., by 13. Hence, from 1 to 14.

Hence it is possible in this case also.

(3)(m, n) = (3, 9)

In any turn, the number of piles can increase only by 2 or 8. Hence, starting from 1 pile, we can only get an odd number (any odd number) of piles, but not 14 piles at any stage. Hence, it is not possible in this case.

(4)(m, n) = (5, 6)

In any turn, the number of piles can increase by 4 or 5.

∴ It can increase by two 4's and one 5, i.e. by 13 (say from 1 to 14)

Hence, it is possible in this case.

Given log₁₀a²b= 5 => 2log₁₀ a + log₁₀ b = 5

Also, logio a/b² = 8 => log₁₀ a - 2log₁₀ b = 8

Solving for log a, we get log₁₀ a = 18/5

=> a = (10)^{18 / 5}

=> Volume of water in B is 5a.

After the transfer of contents from A to B, the concentration of water in B becomes 66 x 2/3%: i.e., 2/3. Hence, the total volume, must have become 3/2 times the initial (i.e., 5a). Hence, the volume of milk transferred to B = 3 / 2 (5a)- 5a = 5 / 2 a = 2.5a .

∴Milk remaining in A = 4a - 2.5a = 1.5a

Now, the concentration of milk in A and B = 100% and 33 x 1/3%.

If after transferring some quantity of the contents of B to A, the concentration becomes 66 x 2/3, (i.e., mid-way between 100 and 33 x 1/3%, that means that the volume of contents from B and the volume of milk in A were mixed in equal quantities, i.e., 1.5a of milk and water solution was transferred from B to A.

∴ Required ratio = 2.5a / 1.5a = 5 : 3.

∴All the a's are odd.

s_i = a_i + a₂ +….+ a_i

If i is even, s_i is even and

If i is odd, s_i is odd

The converses are also true i.e.

If s_i is even, i is even and if s_i is odd, i is odd.

∴ (a)is false.

(b)is false. If a_i is odd (all as are odd), l_i could be odd or even.

(c)is true

(d)is false, as if s_i is odd, i is odd and a_i is always odd.

Therefore, the coordinates of B must be (-4. -5), since AB is parallel to y-axis and AB= 7.

Also, since AB is parallel to Y axis, BC must be parallel to X axis.

Now, since B is (-4, -6) and BC=7, the coordinates of C must be (3, -5).

Rate at which each worker in G₁ works = 3x units/day

Let the area of the first field be w w = 30(3x)(0.5) = 45x

Area of second field = 1,5w = 67.5x

Let the required time be t days

∴ t[30(3x)+60(x)] = 67.5x

⇒ t = 67.5 / 150 = 9 / 20

As the sum of the volumes of all the small cubes is equal to the volume of the first cube, the sum of the volumes of the spheres in these cubes is equal to the volume of the sphere is the first cube.

∴The required ratio is 1 :1.

∴300x + 400y< />

Now, if 300x rupees are spent to buy silk sarees, the number of silk sarees that could be purchased is 300x / 400 or 3 / 4x. Similarly if 400y rupees are spent to buy cotton sarees, the number of cotton sarees that could be purchased is 400y / 300 or 4 / 3 y

Given : x + y + 2 = 4 / 3y + 3 / 4x

=>4y-3x = 24. Tabulating the possible values of x and y, we have

The higher values of x and y are ignored as 300x + 400y < 10000.="" hence="" mrs.="" ahluwalia="" could="" have="" bought="" a="" maximum="" of="" 27="" />

Statement I

S₂ is a subset of Si and it has one element less than S₁.

If this element was the greatest or the least element of S₁, range of S₂< range of S₁

(Range = Max element -Min element)

If this element was neither the greatest element nor the least element of S₁, range of S₂ = range of S₁.

In any case, range of S₂ cannot be > range of S₁

Statement I is false.

Statement II

Let S₁ = {1,2, 3, 4, 5, 6, 7, 8, 9, 10} and

S₂ = {1,2, 3,4, 5,6, 7, 8,9}

Median of S₁ = 5^th element + 6^th element / 2 = 5 + 6 / 2 = 5.5

Median of S₂ = 5th element = 5

Statement II is true.

Statement III

Let S₁ = {1,2, 3, 4, 5, 6, 7, 8, 9, 10} and

S₂ = {2, 3, 4, 5, 6,7, 8, 9, 10}

MeanofS₁= 55 / 10 = 5.5 and mean of S₂ = 54 / 9 = 6

Statement III is true.

Only statements II and III are true.

(6 Ψ 13) = 78 – 19 + 1 = 60

(12 Λ 3) = 36 / 15 = 12 / 5

300 kg of fresh dates contain 240 kg water

=> 60 kg is non-water.

=> 60 kg is 80% of dried grape.

Weight of dried grapes

= 60 / 80% = 60 x 5 / 4 = 75 Kg

Now, given that

a - b = 20 => a = 20 + h and

b - c = 56 => c = b - 56

Also, a + b + c = 180 => (20 + b) + b + (b - 56) = 180

=> 3b = 216 or b = 72 => a = 20 + 72 = 92

and c = 72 - 56 = 16

=> x³ + 23 - 2(x + 2)² = 1001

=> (x + 2) [x² - 2x + 4 - 2x - 4] = 1001

=> (x + 2) (x) (x - 4) = 13 x 11 x 7 = 1001

∴ x = 11

Let the number of students in house A be x.

=> the initial number of students in house C be x - 80.

=> x / x – 80 = (x – 80) + 170 / x

=> x² = (x – 80)² + 170(x - 80)

=> x² = x² - 160x + 6400 + 170x - 13600

=> 10x = 7200

∴ x = 720

∴ The number of students in house A and house C is

x + x - 80 = 2(720) - 80 = 1360.

= HCF (3b, 3c) = HCF (3a, 3c) = 3

=> The LCM of the three numbers = 3abc

The product of the numbers = 27abc = 9(3abc) = 1620

∴ 3abc= 1620 / 9 =180

∴ The LCM of the three numbers is 180.0