Q. 1: Two cars travel the same distance starting at 10:00 am and 11:00 am, respectively, on the same day. They reach their common destination at the same point of time. If the first car travelled for at least 6 hours, then the highest possible value of the percentage by which the speed of the second car could exceed that of the first car is

1. 30

2. 25

3. 10

4. 20

For the highest possible value of the percentage by which the speed of the second car could exceed that of the first car the time taken by both cars should be minimum as their difference is constant (or we can say the ratio of time taken by the first car to that of 2nd car should be maximum ) which is possible only when time is taken by 1st car = 6 hr and that of 2nd car = 6-1 = 5 hr

As we know v1/v2 = t2/t1

Required percentage = (6-5)/5 *100 = 20%

1. 2/3

2. 3

3. 1

4. 1/3

given (5.55)^x = (0.555)^y = 1000 =10^3

Taking log with respect to 10 ( that is base 10)

x log 5.55 = y log (0.555) = 3 log 10 = 3

x log 5.55 = 3

x = 3/log5.55

1/x = log 5.55/3————1)

y log (0.555) = 3

y log (5.55/10) = 3

y log 5.55 – y log 10 = 3

y log 5.55 – y = 3

y = 3 /(log 5.55 -1)

1/y = (log 5.55 -1)/3————-2)

So from eq 1) & 2

1/x -1/y = 1/3

Q. 3: The product of the distinct roots of ∣x^2 − x − 6∣ = x + 2 is?

1. −16

2. -8

3. -24

4. -4

given, ∣x^2 − x − 6∣ = x + 2

Or I (x-3)*(x+2)I = (x+2)

Case 1) When x > 3

Then (x-3)*(x+2)= (x+2) or x -3 =1 or x = 4

Case 2) when – 2 <= x < 3, -(x-3)*(x+2)= (x+2)

-(x-3)*(x+2) – (x+2) =0

(x+2)(x-3+1) =0

(x-2)*(x+2) =0

x = 2 or -2

case 3 ) when x < – 2

(x-3)*(x+2) = (x+2)

x = 4 (not possible as x < -2

so required product = 4*2*(-2) = -16

Q. 4: The wheels of bicycles A and B have radii 30 cm and 40 cm, respectively. While traveling a certain distance, each wheel of A required 5000 more revolutions than each wheel of B. If bicycle B traveled this distance in 45 minutes, then its speed, in km per hour, was

1. 14π

2. 18π

3. 12π

4. 16π

Let number of revolutions of wheels of bicycle B =x

Distance traveled by bicycle A Distance traveled by bicycle B

So 2 πR(Base A) (x+5000) = 2 ΠR(Base B) *x

30*(x+5000) = 40x

150000= 10x

x = 15000

thus distance traveled by bicycle B = 2 π *40 *15000 cm = 1200000 π cm = 12 π km

time = 45 min = 45/60 hr = 3/4 hr

speed = distance /time = 12 π /(3/4) = 16 π

Q. 5: AB is a diameter of a circle of radius 5 cm. Let P and Q be two points on the circle so that the length of PB is 6 cm, and the length of AP is twice that of AQ. Then the length, in cm, of QB is nearest to

1. 7.8

2. 8.5

3. 9.1

4. 9.3

See the figure,

Given AB is diameter = 5*2 = 10 , BP =6 And AQ = AP/2

Angle P and Q will be right angled (angle in semi circle is always equals to 90 degree)

Thus in right angled triangle APB , AP = (AB^2 – BP^2)^1/2 = (10^2 – 6^2 )^1/2 = 8

So AQ = 8/2 = 4

Now in right angled triangle AQB, BQ = (AB^2 – AQ^2)^1/2 = (10^2 – 4^2 )^1/2 = (84)^1/2 = 9.1 approx

**Verbal Ability : | Q.01- Q.05 | Q.06- Q.10 | Q.11- Q.15 | Q.16- Q.20 | Q.21- Q.24 | Q.25- Q.29 | Q.30 – Q.34 |**

**Logical Reasoning : | Q.01- Q.04 | Q.05- Q.08 | Q.09- Q.12 | Q.13- Q.16 | Q.17- Q.20 | Q.21- Q.24 | Q.25 – Q.28 | Q.29 – Q.32 |**

**Quantitative Aptitude: | Q.01- Q.05 | Q.06- Q.10 | Q.11- Q.15 | Q.16- Q.20 | Q.21- Q.25 | Q.26- Q.30 | Q.31 – Q.34 |**

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