### Miscellaneous quiz for SBI PO

Que. 1

Three cooks have to make 80 idlis. They are known to make 20 pieces every minute working together. The first cook began working alone and made 20 pieces having worked for some time more than three minutes. The remaining part of the work was done by the second and the third cooks working together. It took a total of 8 minutes to complete the 80 idlis. How many minutes would it take the first cook alone to cook 160 idlis for a marriage party the next day?

1.16 minutes

2.24 minutes

3.

**32 minutes**

4.40 minutes

5.None of these

Solution

Answer:

We have, Time × rate = work

We have, Time × rate = work

Let, 1st cook completed 20 idlis in x minutes.

Total work of 80 idliswas done in 8 minutes.

So, remaining 60 idlis were made by 2nd and 3rd cook together in (8 – x) minutes.

Rate of 1ST cook = r1 = 20/x

Total rate of 2nd and 3rd cook = r2 + r3 = 60/(8 – x)

When they all work together, they work at a rate of (r1 + r2 + r3).

Their rate working together = 20 pieces/minute.

∴(r1 + r2 + r3) = 20

⇒20x+60(8−x)=20⇒8−x+3xx(8−x)=1 20x608x208x3xx8x1

⇒ 8 + 2x = 8x – x2

⇒ x2 – 6x + 8 = 0

⇒ x2 – 4x – 2x + 8 = 0

⇒ x(x – 4) – 2(x – 4) = 0

⇒ (x – 2)(x – 4) = 0

⇒ x = 2 or x = 4

But, given that, x is greater than 3 minutes.

So, x = 4

Thus r1 = 20/4 = 5 pieces/minute.

So, for 160 pieces, he needs 160/5 = 32 minutes.....

Que. 2

A rope makes 70 rounds of a circumference of a cylinder whose radius of the base is 14 cm. How many times can it go around a cylinder with radius 20 cm?

A rope makes 70 rounds of a circumference of a cylinder whose radius of the base is 14 cm. How many times can it go around a cylinder with radius 20 cm?

1.72

2.7

3.

4.51

5.None of these

Solution

2.7

3.

**49**4.51

5.None of these

Solution

Answer:

We know that, formulae:

We know that, formulae:

Circumference of a circle = 2πr

Where, r = radius of the circle

Let the required number of rounds be x.

More radius, less rounds (indirect proportion)

∴ 20 : 14 ∷ 70 : x

⇒ (20 × x) = (14 × 70)

⇒ x=14×7020⇒x=49x147020x49.

Hence, the required number of rounds = 49......

Que. 3

A rectangular plate is of 6 inch breadth and 12 inch length. Two apertures of 2 inch diameter each and one aperture of 1 inch diameter have been made with the help of a gas cutter, what is the area of the remaining portion of the plate?

A rectangular plate is of 6 inch breadth and 12 inch length. Two apertures of 2 inch diameter each and one aperture of 1 inch diameter have been made with the help of a gas cutter, what is the area of the remaining portion of the plate?

1.62.5 sq. inch

2.68.5 sq. inch

3.64.5 sq. inch

4.66.5 sq. inch

5.

Solution

2.68.5 sq. inch

3.64.5 sq. inch

4.66.5 sq. inch

5.

**None****of****these**Solution

Answer:

Given that the length of the rectangular plot = 12 inches

Given that the length of the rectangular plot = 12 inches

Breadth of the rectangular plot = 6 inches

∴ Area of the rectangular plot = Length × Breadth = 12 × 6 = 72 sq. inch

Given, diameter of 2 aperture = 2 inch

∴ Radius of aperture = 1 inch

⇒ Area of two apertures with 2 inches diameter each = 2 × π × r2 = 2 × π × (1)2 = 2π Sq .inch

Given, diameter of 1 aperture = 1 inch

∴ Radius of aperture = 1/2 inch

⇒ Area of one aperture with 1 inch diameter = π × r2 = π × (1/2)2 = π /4 sq .inch

Area of remaining portion after removing the apertures,

⇒ 72 – 2π - π/4 = 72 – ((9/4) × π) = 64.93 sq. inch

∴The area of the remaining portion is 64.93 sq. inch........

Que. 4

Two workers A and B working together completed a job in 5 days. If A worked twice as efficiently as he actually did and B worked 1/3 as efficiently as he actually did, the work would have been completed in 3 days. To complete the job alone, A would require :

Two workers A and B working together completed a job in 5 days. If A worked twice as efficiently as he actually did and B worked 1/3 as efficiently as he actually did, the work would have been completed in 3 days. To complete the job alone, A would require :

1.5¹/5days

2.

3.7½days

4.8¾days

5.None of these

Solution

2.

**6¼days**3.7½days

4.8¾days

5.None of these

Solution

Answer:

Let A completes the work alone in x days

Let A completes the work alone in x days

∴ A’s 1 day work = 1/x

Let B completes the work alone in y days

∴ B’s 1 day work = 1/y

Since both completes the work in 5 days.

∴ (A+B)’s 1 day work = 1/5

⇒ (1/x) + (1/y) = 1/5 ………………………………….(1)

If A works twice as efficiently, he will complete the work in x/2 days.

So, Now A’s 1 day work = 1/(x/2) = 2/x

If B works with 1/3 efficiency, he will complete the work in 3y days

So, Now B’s 1 day work = 1/3y

Since, now both complete the work in 3 days.

∴ (A+B)’s 1 day work = 1/3

⇒ (2/x) + (1/3y) = 1/3 …………………………….(2)

We now have 2 equations in 2 variables that can be solved as any other system of equations. To make solving the equations appear more natural, we can write 1/x as ‘a’ and 1/y as ‘b’. These are dummy variables.

a + b = 1/5 ………………..(3)

and 2a + b/3 = 1/3 ………………..(4)

Solving these 2 equations, we get

a = 4/25

⇒ x = 1/a = 25/4 = 6.25 = 6 1/4 days

So, to complete the work alone, A would require 6 1/4 days......