Dena Bank PO Mock Test - 6 - Bank Exams MCQ

Let Anil takes x hrs.
⇒ Arun takes x+2

If Arun doubles the speed, he will take x-1 hrs
⇒ He needs 3 hours less.
∵ Double speed means half time.
∴ Half of the time required by Arun to cover 30 km = 3 hours

⇒ Time required by Arun to cover 30 km = 6 hour
Thus, Arun's speed = 30/6 = 5 km/h

Total time taken = (160/64) + (160/80) = 9/2 hrs
∴ Average speed = 320 x (2/9) = 71.11 km/hr.

Explanation:

a/b + b/a = (a² + b²)/ab = (a² + b² + a + b)/ab 
= [(a + b)² - 2ab]/ab
a + b = -8/1 = -8
ab = 4/1 = 4
Hence a/b + b/a = [(-8)² - 2(4)]/4 = 56/4 = 14.

Explanation:

The quadratic equation whose roots are reciprocal of 2x² + 5x + 3 = 0 can be obtained by replacing x by 1/x.
Hence, 2(1/x)² + 5(1/x) + 3 = 0
=> 3x² + 5x + 2 = 0

Explanation:

Let the price of each note book be Rs.x.
Let the number of note books which can be brought for Rs.300 each at a price of Rs.x be y.
Hence xy = 300
=> y = 300/x 
(x + 5)(y - 10) = 300 => xy + 5y - 10x - 50 = xy
=>5(300/x) - 10x - 50 = 0 => -150 + x² + 5x = 0
multiplying both sides by -1/10x
=> x² + 15x - 10x - 150 = 0
=> x(x + 15) - 10(x + 15) = 0
=> x = 10 or -15
As x>0, x = 10.

Explanation:

I.(a - 3)(a - 4) = 0
=> a = 3, 4

II. (b - 2)(b - 1) = 0
=> b = 1, 2
=> a > b

Percentage change (rise/fall) in the production of Company Y in comparison to the previous year, for different years are:

Hence, the maximum percentage rise/fall in the production of Company Y is for 1997

The percentages of production of Company Z to the production of Company Z for various years are:

Percentage increase in the production of Company Y from 1996 to 1999

Required difference = (128000 - 107000) = 21000.

From the line-graph it is clear that the productions of Company X in the years 1997, 1998, 1999, 2000, 2001 and 2002 are 119000, 99000, 141000, 78000, 120000 and 159000 and those of Company Y are 139000, 120000,100000, 128000, 107000 and 148000 respectively.

Total production of Company X from 1997 to 2002

    = 119000 + 99000 + 141000 + 78000 + 120000 + 159000

    = 716000.

and total production of Company Y from 1997 to 2002

    = 139000 + 120000 + 100000 + 128000 + 107000 + 148000

    = 742000.

Difference = (742000 - 716000) = 26000.

The difference between the productions of Companies X and Y in various years are:

For 1997 (139000 - 119000) = 20000.

For 1998 (120000 - 99000) = 21000.

For 1999 (141000 - 100000) = 41000.

For 2000 (128000 - 78000) = 50000.

For 2001 (120000 - 107000) = 13000.

For 2002 (159000 - 148000) = 11000.

Clearly, maximum difference was in 2000.

972/1972 = 243/ 493.

The required probability would be given by: All are Red OR All are white OR All are Blue = (6/18) x (5/17) x (4/16) + (4/18) x (3/17) x (2/16) + (8/18) x (7/17)x (6/16) = 480/(18 x 17 x 16) 5/51 

12.5% of 350 * 8.4% of 250 = 6.25 * 7 * 21 = 918.75

So Option D is correct

(331 + 19)*(15 – 11)*(37 + 13) = 350 * 4 * 50 = 70000

So Option B is correct

√((444)÷37 + 15 + 11 * (?)) = 7
12 + 15 + 11x = 49
11x + 27 = 49
x = 2

So Option C is correct

3/14 of 5/13 of 6/7 of 3822 = 3 * 5 * 6 * 3 = 270

So Option C is correct

69.69 – 51.54 = 18.15
18.15 + 73.65 = 91.80
91.80 – 32.42 = 59.38

So Option E is correct

3x – x = 32
x = 16
1/16 + 1/48 = 4/48
Time taken to fill the tank = 48/4 = 12 minutes

Required time = y(1-(t/x)) = 27(1-(8/24))= 18 minutes

From I : 5 or 6 or 7 or 8 or 9 or 10 days From II : 10 or 11 or 12 or 13 or 14 days From both statements, only 10 is common in them, so Bhawna took 10 days which is predicted from bot statements.

From I : B is at 12th position from right, A is third to right of B, so A is 9th from right end.
From II : H is at 14th position from right, with 8 students in between A and H – it is not sure than A is left of H or right of H, so A’s position cant be predicted.

Explanation: From I : code for ‘7’ is %
From II : code for ‘7’ is %
So it can be predicted from both alone.

From I : Suppose 1 family stays on ground floor, then the other should be on 2nd floor, and then third family on 4th floor, or there can be any other arrangement such no two families are on adjacent floors.
From II : Maximum of 4 floors means that the building can have 0 or 1 or 2 or 3 or 4 floors.
But if both statements are combined, which says that there are 3 families and they do not stay on alternate floors and maximum can be 4 floors, it is predicted that there are 4 floors in this building