Test: CAT Quant 2018 (Shift - 2) - CAT MCQ

In the first case, 10 type A and 45 type B pipes fill the tank in 30 mins.

So, the capacity of the tank = 1/2 (10a + 45b)…..(i)

In the second case, 8 type A and 18 type B pipes fill the tank in 1 hour.

So, the capacity of the tank = (8a + 18b)…. (ii)

Equating (i) and (ii), we get

10a + 45b = 16a + 36b

=>6a=9b

From (ii), capacity of the tank = (8a + 18b) = (8a + 12a) = 20a

In the third case, 7 type A arid 27 type B pipes fill the tank.

Net efficiency = (7a + 27b) = (7a + 18a) = 25a

Time taken = 20a/25a hour = 48 minutes.

Hence, 48 is the correct answer.

So, 5x = 52 - 2x²

or, 2x² + 5x - 52 = 0

on solving, we get x = 4 or 13/-2

But, it is given that x is a positive number.

So, x = 4

and the minimum value = 5*4 = 20

Hence, 20 is the correct answer.

Therefore, at the time of their meeting car 1 must have travelled 200 km and car 3 must have travelled 100 km.

As the time is same, ratio of speed of, car 1 to speed of car 3 = 2:1.

Car 3 meets car 2 at P, which is 100 km from A.

Therefore, at the time of their meeting car 2 must have travelled 100 km and car 3 must have travelled 200 km.

As the time is same, ratio of speed of car 2 to speed of car 3 = 1 : 2.

Speed of car 1: speed of car 3 = 2:1

And speed of car 2: speed of car 3 = 1: 2

So, speed of car 1: speed of car 2: speed of car 3 = 4:1: 2

Hence, option D is the correct answer.

We can further factorize n²-9n+14 as (n-2) (n-7).

(n³ - 11n² + 32n - 28 = (n - 2)²(n - 7)

=>n³- 11n² + 32n - 28 > 0

=> (n - 2)²(n - 7) > 0

Therefore, we can say that n-7>0

Hence, n_min = 8

Let the scores be increased by x

So, after increment, AmaTs score = 11 k + x and Bimal's score = 14k + x

According to the question,

11k – x = 47

14k + x = 56

On solving, we get x = 42/9 k

Ratio of Bimal's new score to his original score

= (14k +x) /14k

= 168k/14*9k

= 4/3

Hence, option A is the correct answer.

We will get number 'ba' after interchanging its digit.

It is given that 10a+b > 3*(10b + a)

7a > 29b

If b = 1, then a = {5,6,7,8,9}

If b = 2, then a = {9}

If b = 3, then no value of 'a' is possible. Hence, we can say that thereare a total of 6 such numbers.

PΔQ ={1,4, 5, 6}

R = {1,3,7,8,9} and S = {2,4,9, 10}

RΔS = {1,2, 3,4,7, 8,10}

(PΔQ) Δ(RΔS) = {2, 3, 5, 6,7, 8,10}

Thus, there are 7 elements in (PΔQ)A(RΔS).

hence, 7 is the correct answer.

we can see that area of parallelogram ABCD = 2*Area of triangle ACD

48 = 2*Area of triangle ACD

Area of triangle ACD = 24

(1/2) * CD * DA * sin ADC = 24

AD* sin ADC 6

we know that sinϴ ≤ 1, Hence, we can say that AD ≥ 6

=> s ≥ 6

When they are mixed, quantity of ethanol in the mixture

= (20 + 300S)

Let this solution be mixed with equal volume i.e. 400 of third solution in which the strength of ethanol is 20%.

So, the quantity of ethanol in the final solution

= (20 + 300S + 80) = (300S + 100)

It is given that, 31.25% of 800 = (300S + 100)

or, 300S + 100 = 250

or S = 1/2 = 50%

Hence, 50 is the correct answer.

Let 'n' be the number of girls on junior level. It is given that the number of girl versus girl matches in junior level is 1530.

=>nC₂ = 153

=> n (n-1)/2 = 153

=>n (n-1) = 306

=> n² – n-306 = 0

=> (n+17) (n-18) =0

=>n=18 (rejecting n=-17)

Therefore, number of boys on junior level = 43 -18 = 25.

Let 'm' be the number of boys on senior level. It is given that the number of boy versus boy matches in senior level is 276

=> mC₂ = 276

m = 24

Therefore, number of girls on senior level = 51 - 24 = 27.

Hence, the number of matches a boy plays against a girl = 18*25+24*27 = 1098

We are given that AB = 5 cm and

Let us draw OM such that OM ⊥AB.

In right angle triangle AMO

sin30° = AM/AO

=> AO = 2*AM = 2*2.5 = 5 cm.

Therefore, we can say that the radius of the circle = 5 cm.

In right angle triangle PNO,

Sin60◦ = PN/PO

=>

Therefore, PQ = 2*PN = 5√3 cm

a₁ + a₂ + a₃ + ... + a₅₂ = 52x ... (1)

Therefore, average of a₂, a₃, ....a₅₂ = x+1

a₂ + a₃ + a₄ + ... a₅₂ = 51 (x+1)... (2)

From equation (1) and (2), we can say that

a₁ + 51(x + 1) = 52x

a₁ = x - 51.

We have to find out the largest possible value of a_1.a₁ will be maximum when 'x' is maximum.

(x+1) is the average of terms a₂, a₃ ...a₅₂. We know that a₂< a₃< ... < a₅₂ and a₅₂ = 100.

Therefore, (x+1) will be maximum when each term is maximum possible. If a₅₂ = 100, then a₅₂ = 99, a₅₀ = 98 ends so on.

a₂ = 100 + (51 -1)*(-1) = 50.

Hence, a₂ + a₃ + a₄ + ... + a₅₂ = 50+51+...+99+100 = 51 (x+1)

51 * (50 + 100)/2 = 51(x + 1)

x = 74

Therefore, the largest possible value of a₁ = x - 51 =74- 51 = 23.

Nth term of the series can be written as Tn = (4n + 3) * (4n + 7)

Last term, (4n+3) = 95 i.e. n = 23

(4n + 3)*(4n + 7)

16n² + 40n + 21

16* (23*24*47)/6 + 40 * (23 * 24)/2 + 21* 23

= 80707

It is given that N^N = 2¹⁶⁰.

We can rewrite the equation as N^N = (2⁵)^160/5 = 32³²

=> N = 32

N² + 2^N = 32² + 2³² = 2¹⁰ + 2³² = 2¹⁰ * (1 + 2²²)

Hence, we can say that N² + 2^N can be divided by 2¹⁰

Therefore, X_max = 10

On Monday, An alone completed filling the tank at 8 pm. Therefore, we can say that pump A worked for (8 -1) hours. Hence, the volume of the tank = a*(8 -1) liters.

Similarly, on Tuesday, B alone completed filling the tank at 6 pm. Therefore, we can say that pump B worked for (6 -1) hours. Hence, the volume of the tank = b*(6 -1) liters.

On Wednesday, An alone worked till 5 pm, and then B worked alone from 5 pm to 7 pm, to fill the tank. Therefore, we can say that pump A worked for (5 -t) hours and pump B worked for 2 hours. Hence, the volume of the tank = a*(5 - t) +2b liters.

We can say that a*(8-t) = b*(6-t) = a*(5- t) + 2b

a*(8 -1) = a*(5 -1) + 2b

3a = 2b ... (1)

a*(8 -1) = b*(6-1)

Using equation (1), we can say that

a * (8 - t) = 3a/2 * (6 -t)

t = 2

Therefore, we can say that the tank gets emptied at 2 pm daily. We can see that A takes 6 hours and pump B takes 4 hours alone.

Hence, working together both can fill the tank in = frac{6*4}{6+4} = 2.4 hours or 2 hours and 24 minutes.

The pumps started filling the tank at 2:00 pm. Hence, the tank will be filled by 4:24 pm.

Time taken to cover second 50 km at 50 km/hr = 1 hr.

Time taken to cover last 50 km at 25 km/hr = 2 hr.

When car 2 starts, car 1 has already covered 20 km.

So, time taken by car 1 to reach B after car 1 starts = total time - time required to travel first 20 km

= 3 hr 30 min -12 min = 3 hr 18 min

Distance travelled by car 2 = (50 + 50 + 45) = 145 km

Distance from B = (150 -145) km = 5 km

Hence, 5 is the correct answer.

Let 'a' and 'b' are those two numbers.

=> a² + b² = 97

=>a² + b² = 2ab = 97 — 2 ab

=> (a – b) ² = 97 - 2ab

we know that (a² – b²) ≥ 0

=>97-2ab ≥ 0

=>ab ≤ 48.5

Hence, ab ≠ 64.

Therefore, option D is the correct answer.

Area of the rectangle = a*b

Perimeter of the rectangle = 2*(a+b)

=>a*b/(2*(a + b))² = 1/25

=> 25ab = 4(a + b)²

=> 4a² - 17ab + 4b² = 0

=> (4a - b)(a - 4b) = 0

=> a = 4b or b/4

We initially assumed that a > b, therefore a ≠ 4.

Hence, a = 4b

=> b: a = 1: 4

We can see that triangle BPC and BQC are inscribed inside a semicircle. Hence, we can say that

Therefore, we can say that BQ ⊥ AC and CP ⊥ AB.

In triangle ABC,

Area of triangle = (1/2)*Base*Height = (1/2)*AB*CP = (1/2)*AC*BQ

=> BQ = AB *CP/AC = 30 * 20/25 = 24 cm.

Therefore, the height of the perpendicular drawn from point A to BC = 2*32/8 = 8 units.

Let us draw a possible diagram of the given triangle.

We can see that if A coincide with (-4, 0) then the distance between A and (0, 0) = 4 units.

If we move the triangle up or down keeping the base BC on x = 4, then point A will move away from origin as vertical distance will come into factor whereas horizontal distance will remain as 4 units.

Hence, we can say that minimum distance between A and origin (0, 0) = 4 units.

Positive Mark: 3

Negative Mark: 0

=>1/2 * π *(AB²/ 4) = 72 * π

=> AB = 24cm

Given that area of the rectangle ABCD = 768 sq.cm

=>AB*BC = 768

=>BC = 32 cm

We can see that the perimeter of the remaining shape = AD + DC + BC + Arc(AB)

=> 32+24+32+π * 24/2

=> 88 + 12π

For n = 1, the value is 0, for n = 2, the value is 1225 which is the 35th multiple of 3.

If we observe set B, it consists of all the multiples of 35 including 0.

So, we can say that every member of set A will be in B while every member of set B will not necessarilybe in set A.

Hence, option A is the correct answer.

Therefore, final quantity of water in the mixture = 875 - 567 = 308 ml

Hence, we can say that the percentage of water in the mixture = 308/875 x 100 = 35.2%

Let f(x) = 2x² - ax + 2. We can see that f(x) is a quadratic function.

For, f(x) > 0, Discriminant (D) < />

=> (-a)² – 4*2*2< />

=> (a-4)(a+4) < />

=> a ε (-4,4)

Therefore, integer values that 'a' can take = {-3, -2, -1,0,1,2,3}

Let g(x) = x²- bx + 8. We can see that g(x) is also a quadratic function.

For, g(x)≥0, Discriminant (D) ≤ 0

=> (-b)²– 4*8* 1 < />

=>(b - √32 ) (b + √32 ) < />

=> b ε(-√32 ,√32 )

Therefore, integer values that 'b' can take = {-5, -4, -3, -2, -1,0,1,2,-3,4,5}

We have to find out the largest possible value of 2a - 6b.

The largest possible value will occur when 'a' is maximum and 'b' is minimum.

a_max = 3, b_min = -5

Therefore, the largest possible value of 2a - 6b – 2*3 – 6*(-5) = 36.

Therefore, we can say that 1/log₂100 = log₁₀2/ log₁₀100

=>1/Iog₂100-1/Iog₄100 – 1/Iog₅100 - log₁₀100 + 1/Iog₂₀100 -1/Iog₂₅100 + 1/Iog₅₀100

=> log₁₀2/log₁₀100 - log₁₀4/log₁₀100 + log₁₀5/log₁₀100 - log₁₀10/log₁₀100 + log₁₀20/log₁₀100 - log₁₀25/log₁₀100 + log₁₀50/log₁₀100

We know that log₁₀100 = 2

=> 1/2 * [log₁₀2 -log₁₀4+ log₁₀5-log₁₀10+ log₁₀20-log₁₀25+ log₁₀50]

=>1/2 *[log₁₀ (2*5*20*50)/(4 *10 *25)]

=> 1/2 * [log₁₀10]

=>½

Given that, p³ = q⁴ = r⁵ = s⁶

p³=s⁶

=> log_s(pqr)

By substituting value of p, q, and r from equation (1), (2) and (3)

 A is the correct answer.

Let us assume that in drum 2, A and B are in the ratio x: 1.

It is given that drums 1 and 2 are mixed in the ratio 3:4 and in this final mixture, A and B are in the ratio 13:7.

By equating concentration of A

3 * 18/ (18 + 7 + 4) * x/(x + 1)

=>3 + 4=13/ (13 + 7)

=>54/25 + 4x/x+1 = 91/20

=> 4x/(x+1)= 239/100

=>x= 239/161

Therefore, we can say that in drum, 2, A and B are in the ratio 239/161 :1 or 239:161.

It is given that they can together complete a work in 16 days. Hence, total amount of work = 16(R+G)... (1)

For first 7 days both of them worked together. From 8th day, Ramesh worked at 70% of his original efficiency whereas Ganesh worked at his original efficiency. It took them 17 days to finish the same work, i.e., Ramesh worked at 70% of his original efficiency for 10 days.

=> 16(R+G) = 7(R+G)+10(0.7R+G)

=> 16(R+G) = 14R+17G

=> R = 0.5G ... (2)

Total amount of work left when Ramesh got sick = 16(R+G) - 7(R+G) = 9(R+G) = 9(0.5+G) = 13.5G

Therefore, time taken by Ganesh to complete the remaining work = 13.5/G =13.5 days.