**Water is flowing at the rate of 3 km/hr through a circular pipe of 20 cm internal diameter into a circular cistern of diameter 10m and depth 2m. In how much time will the cistern be filled?**- l hour
- 1 hour 40 minutes
- 1 hour 20 minutes
- 2 hours 40 minutes

Answer: Option B.

Water flowed in 1 hour through the pipe

=22/7 * 10*10*3000/10000 meter^{3}

=660/7 meter^{3}

Volume of circular/cylindrical cistern

=660/7 * 5*5*2 = 1100/7 meter^{3}

= Required Time = 1100/7/660/7 = 5/3 hours

= 1 hour 2/3 * 60 minutes

= 1 hour 40 minutes**A bicycle wheel makes 5000 revolutions in moving 11 km. The diameter of the wheel, in cm, is**- 35
- 55
- 65
- 70

Answer: Option D.

Distance covered by wheel in one revolution = 2Πr

∴ 5000 × 2Πr = 11 × 1000

⇒ 5000 × 2 × 22/7 × r = 11000

⇒ r =11000*7 / 5000*2*22 = 0.35 metre

= 35 cm

∴ Diameter

= 2 × radius = 2 × 35 = 70 cm

**At each corner of a triangular field of sides 26 m, 28 m and 30 m, a cow is tethered by a rope of, length 7 m. The area (in m2) ungrazed by the cows is**- 336
- 259
- 154
- 77

Answer: Option C.

Area grazed by all cows 180^{°}/ 360^{°}Π r^{2}= Π r^{2}= 1/2 * 22/7 *7*7 = 77 sq.metre Semi-perimeter of triangular field(s)

= 26+28+30 / 2 = 42 metre

∴ Area of the field

√s(s-a)(s-b)(s-c)

√42(42-26)(42-28)(42-30)

√42*16*14*12

= 336 sq. metre

∴ Area ungrazed by the cows = 336 - 77 = 259 sq. metre**A shopkeeper allows 23% commission on his advertised price and still makes a profit of 10%. If he gains Rs. 56 on one item, his advertised price of the item, in Rs., is**- 820
- 780
- 790
- 800

Answer: Option D.

Let the advertised price be Rs. x. ∴ S.P. = Rs. 77x/100

∴ C.P. = Rs(77x/100 - 56)

⇒ 77x -5600/100 =(100/110 * 77x/100)

⇒77x -5600/100 = 7x/10

⇒ 7x -5600 = 70x ⇒ 7x = 5600 ⇒ x = Rs. 800**The average of 25 observations is 13. It was later found that an observation 73 was wrongly entered as 48. The new average is**- 12.6
- 14
- 15
- 13.8

Answer: Option B.

Difference of two observations = 73 - 48 = 25

∴ New average = 13 + 25/25 = 14Tata Steel Preparation Links- Sample Aptitude Questions of Tata Steel
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**If the cost price of 10 articles is equal to the selling price of 8 ar¬ticles, then gain per cent is**- 10%
- 8%
- 50%
- 25%

Answer: Option D.

Profit percent= 10-8/8 × 100 = 25%**An article is marked 40% above the cost price and a discount of 30% is allowed. What is the gain or loss percentage?**- 10% gain
- 5% gain
- 2% loss
- 12% loss

Answer: Option D.

Let the C.P. of article be Rs. 100. ∴ Marked price = Rs. 140 S.P. = 70% of 140 = Rs. 98**A man bought oranges at the rate of 8 for Rs. 34 and sold them at the rate of 12 for Rs. 57. How many oranges should be sold to earn a net profit of Rs. 45?**- 90
- 100
- 135
- 150

Answer: Option A

Let the man buy 24 (LCM of 8 and 12) oranges. ∴ C.P. of 24 oranges = 34/8 × 24 = Rs. 102 S.P. of 24 oranges = 57/12 × 24 = Rs. 114 Gain= 114- 102 = Rs. 12 ∵ Rs. 12≡ 24 oranges ∴ Rs. 45 ≡ 24/12 × 45 = 90 oranges**The number 0.121212.... in the form p/q is equal to**- 4/11
- 2/11
- 4/33
- 2/33

Answer: Option C.

0.121212 …. = 0.12 = 12/99 =4/33**if(3/5)**^{3}(3/5)^{-6}=(3/5)^{2x-1}then x is equal to- -2
- 2
- -1
- 1

Answer: Option C.

(3/5)^{3}(3/5)^{-3}(3/5)^{-3}=(3/5)^{2x-1}

⇒(3/5)^{3}(3/5)^{-3}(3/5)^{-3}=(3/5)^{2x-1}⇒(3/5)^{0}(3/5)^{-3}=(3/5)^{2x-1}⇒ 2x – 1 = - 3 ⇒ 2x = - 3 + 1 = - 2 ⇒ x = - 1