Olympiad Test: Elementary Mensuration-II - 2 - Class 7 MCQ

To determine the capacity of the cylindrical tank, we need to find the volume of the cylinder, which is given by the formula:

Volume=πr²h

Where:
- r is the radius of the base of the cylinder
- h is the height of the cylinder

Step 1: Analyze the Statements

1. Statement I: The area of the base is 61600sq. cm.
- The area of the base of a cylinder is given by the formula:
Area=πr²
- From this statement, we can find the radius r as follows:

- Thus, Statement I is necessary to find the radius.

2. Statement II: The height of the tank is 1.5 times the radius.
- This gives us a direct relationship between height and radius:
h=1.5r
- Therefore, Statement II is also necessary to find the height.

3. Statement III: The circumference of the base is 880cm.
- The circumference of the base is given by:
Circumference=2πr
- From this statement, we can find the radius r as follows:

- Thus, Statement III is also necessary to find the radius.

Step 2: Determine Necessary Statements

To calculate the volume of the cylinder, we need both the radius r and the height h.

- From Statement I, we can find the radius.
- From Statement II, we can find the height based on the radius.
- From Statement III, we can also find the radius.

Conclusion

To determine the capacity of the cylindrical tank, we need:
- Statement I (Area of the base) or Statement III (Circumference of the base) to find the radius.
- Statement II (Height in terms of radius) is necessary to find the height.

Thus, the necessary statements to answer the question are:
- Statement II is essential, and either Statement I or Statement III is also required.

Final Answer

The necessary statements are:
- Statement II (Height is 1.5 times the radius)
- Either Statement I (Area of the base) or Statement III (Circumference of the base).

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Part filled by (A + B + C) in 3 minutes = 3 (1/30 + 1/20 + 1/10) = (3 × 11/60) = 11/20
Part filled by C in 3 minutes = 3/10
∴ Required ratio = (3/10 × 20/11) = 6/11

Net part filled in 1 hour (1/5 + 1/6 – 1/12)
= 17/60
∴ The tank will be full in 60/17 hours i.e.,

Work done by the leak in 1 hour
= (1/2 – 3/7) =1/14
∴ Leak will empty the tank in 14 hrs.

Let B be turned off after x minutes. Then,
Part filled by (A + B) in x min. + Part filled by
A in (30 – x) min. = 1.
∴ x( 2/75 + 1/45) + (30 – x)2/75 = 1
⇒ 11x/225 + (60 – 2x)/75 = 1
⇒ 11x + 180 – 6x = 225
⇒ x = 9

Suppose, first pipe alone takes x hours to fill the tank .
Then, second and third pipes will take (x – 5) and (x – 9) hours respectively to fill the tank.
1/x + 1/(x – 5) = 1/(x – 9)
(x – 5 + x)/x(x – 5) =1/(x – 9)
⇒ (2x – 5)(x – 9) = x(x – 5)
⇒ x² – 18x + 45 = 0
⇒ (x – 15)(x – 3) = 0
⇒ x = 15 [Neglecting x = 3]

Work done by the waste pipe in 1 minute
=1/15 – (1/20 + 1/24)
= (1/15 – 11/20)
= – 1/40 [–ve sign means emptying]
∴ Volume of 1/40 part = 3 gallons
Volume of whole = (3 × 40) gallons
= 120 gallons

Suppose pipe A alone takes x hours to fill the tank.
Then, pipes B and C will take x/2 and x/4 hours respectively to fill the tank.
1/x + 2/x + 4/x = 1/5
7/x = 1/5
⇒ x = 35 hrs

Let the cistern be filled by pipe A alone in x hours.
Then, pipe B will fill it in (x + 6) hours.
1/x + 1(x + 6) = 1/4
(x + 6 + x)/x (x + 6) = 1/4
⇒ x² – 2x – 24 = 0
⇒ (x – 6)(x + 4) = 0
⇒ x = 6
[Neglecting the negative value of x]

Part filled by A in 1 min =1/20
Part filled by B in 1 min =1/30
Part filled by (A + B) in 1 min
= (1/20 + 1/30) =1/12
∴ Both pipes can fill the tank in 12 minutes. 

Part filled in 4 minutes = 4(1/15 + 1/20)
= 7/15
Remaining part = (1 – 7/15) = 8/15
Part filled by B in 1 minute = 1/20
∴ 1/20 : 8/15 : : 1: x
x = (8/15 × 1 × 20) = min
= 10 min. 40 sec.
The tank will be full in (4 min. + 10 min. + 40 sec.) = 14 min. 40 sec.

Let the slower pipe alone fill the tank in x minutes.
Then, faster pipe will fill it in x/3 minutes.
1/x + 3/x = 1/36
⇒ 4/x = 1/36
⇒ x = 144 min.

Part filled by (A + B) in 1 minute
= ( 1/60 + 1/40) =1/24
Suppose the tank is filled in x minutes.
Then, x/2( 1/24 + 1/40) = 1
x/2 × x/15 =1
x = 30 min.

Time taken by one tap to fill half of the tank = 3 hrs.
Part filled by the four taps in 1 hour
= (4 × 1/6) = 2/3
Remaining part = (1 – 1/2) = 1/2
∴ 2/3 : ½ :: 1 : x
⇒ x = (1/2 × 1 × 3/2) = 3/4 hours = 45 minutes
So, total time taken = 3 hrs. 45 minutes

(A + B)’s 1 hour’s work = (1/12 + 1/15)
= 3/20
(A + C)’s hour’s work = (1/12 + 1/20)
= 2/15
Part filled in 2 hrs = (3/20 + 2/15) = 17/60
Part filled in 6 hrs = (3 × 17/60) = 17/20
Remaining part = (1 – 17/20) = 3/20
Now, it is the turn of A and B and 3/20 part is filled by A and B in 1 hour.
∴ Total time taken to fill the tank
= (6 + 1) hrs = 7 hrs.

Part filled in 2 hours = 2/6 = 1/3
Remaining part = (1 – 1/3) = 2/3
∴ (A + B)’s 7 hour’s work = 2/3
(A + B)’s 1 hour’s work = 2/21
∴ C’s 1 hour’s work = {(A + B + C)’s 1 hour’s work} – {(A + B)’s 1 hour’s work}
= (1/6 – 2/21) = 1/14
∴ C alone can fill the tank in 14 hours.

I. Time taken to fill the cistern without leak = 9 hours.
Part of cistern filled without leak in 1 hour = 1/9
II. Time taken to fill the cistern in presence of leak = 10 hours.
Net filling in 1 hour =1/10
Work done by leak in 1 hour = (1/9 – 1/10) =1/90
∴ Leak will empty the full cistern in 90 hours.
Clearly, both I and II are necessary to answer the question.
∴ Correct answer is (d).

I. A’s 1 minute’s filling work = 1/16
II. B’s 1 minute’s filling work =1/8
(A + B)’s 1 minute’s emptying work
= (1/8 – 1/16) =1/16
∴ Tank will be emptied in 16 minutes.
Thus, both I and II are necessary to answer the question.
∴ Correct answer is (d).

II. Part of the tank filled by A in 1 hour = 1/4
III. Part of the tank filled by B in 1 hour = 1/6
(A + B)’s 1 hour’s work = (1/4 + 1/6) = 5/12
∴ A and B will fill the tank in 12/5hrs
= 2 hrs 24 min.
So, II and III are needed.
∴ Correct answer is (b).

Number of balls = Volume of big sphere /
Volume of one small sphere
= (4 π / 3 × 8 × 8 × 8)/
(4 π / 3 × 0.5 × 0.5 × 0.5)
= 4096