**Arithmetic Ability **Practice Question and Answer

8## Q: 1 What was the day of the week on 17th June, 1998 ? 1329 55b5cc761e4d2b4197774fd16

5b5cc761e4d2b4197774fd16- 1Mondayfalse
- 2Tuesdayfalse
- 3Wednesdaytrue
- 4Fridayfalse

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## Answer : 3. "Wednesday"

Explanation :

Answer: C) Wednesday Explanation: 17th June, 1998 = (1997 years + Period from 1.1.1998 to 17.6.1998)Odd days in 1600 years = 0Odd days in 300 years = 197 years has 24 leap years + 73 ordinary years.Number of odd days in 97 years ( 24 x 2 + 73) = 121 = 2 odd days.Jan. Feb. March. April. May. June.(31 + 28 + 31 + 30 + 31 + 17) = 168 days168 days = 24 weeks = 0 odd day.Total number of odd days = (0 + 1 + 2 + 0) = 3.Given day is Wednesday.

## Q: 2 1.12.91 is the first Sunday. Which is the fourth Tuesday of December 91 ? 682 25b5cc709e4d2b4197774f28a

5b5cc709e4d2b4197774f28a- 117.12.91false
- 224.12.91true
- 327.12.91false
- 431.12.91false

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## Answer : 2. "24.12.91"

Explanation :

## Q: 3 If the true discount on sum due 2 years hence at 14% per annum be Rs. 168, the sum due is ? 1923 25b5cc723e4d2b4197774f6a4

5b5cc723e4d2b4197774f6a4- 1Rs.948false
- 2Rs.876false
- 3Rs.768true
- 4Rs.658false

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## Answer : 3. "Rs.768"

Explanation :

Answer: C) Rs.768 Explanation: P.W. = 100×T.DR×T = 100×16814×2 = 600. Sum = (P.W. + T.D.) = Rs. (600 + 168) = Rs. 768.

## Q: 4 A milkman has 20 liters of milk. If he mixes 5 liters of water, which is freely available, in 20 liters of pure milk. If the cost of pure milk is Rs. 18 per litre, then the profit of the milkman, when he sells all the mixture at cost price, is: 905 15b5cc7d7e4d2b4197775151f

5b5cc7d7e4d2b4197775151f- 120%false
- 225%true
- 333.33%false
- 418%false

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## Answer : 2. "25%"

Explanation :

Answer: B) 25% Explanation: When the water is freely available and all the water is sold at the price of the milk, then the water gives the profit on the cost of 20 litres of milk. Since, the profit % = ProfitCostPricex100 Therefore, profit percentage = 520x100 = 25%

## Q: 5 From a tank of petrol, which contains 200 litres of petrol, the seller replaces each time with kerosene when he sells 40 litres of petrol(or its mixture). Everytime he sells out only 40 litres of petrol(pure or impure). After replacing the petrol with kerosen 4th time, the total amount of kerosene in the mixture is 7267 05b5cc7d7e4d2b4197775151a

5b5cc7d7e4d2b4197775151a- 181.92Lfalse
- 296Lfalse
- 3118.08Ltrue
- 4None of thesefalse

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## Answer : 3. "118.08L"

Explanation :

Answer: C) 118.08L Explanation: The amount of petrol left after 4 operations = 200 × 1-402004 = 200 × 454 = 200 × 256625 = 81.92 litres Hence the amount of kerosene = 200 - 81.92 = 118. 08 litres

## Q: 6 A jar was full with honey. A person used to draw out 20% of the honey from the jar and replaced it with sugar solution. He has repeated the same process 4 times and thus there was only 512 gm of honey left in the jar, the rest part of the jar was filled with the sugar solution. The initial amount of honey in the jar was: 2798 15b5cc7d7e4d2b41977751515

5b5cc7d7e4d2b41977751515- 11.25 kgtrue
- 21 kgfalse
- 31.5 kgfalse
- 4None of thesefalse

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## Answer : 1. "1.25 kg"

Explanation :

Answer: A) 1.25 kg Explanation: Let the initial amount of honey in the jar was k, then 512 = k1 - 154 or 512 = k454 => k = 512 × 625256 => k = 1250 Hence initially the honey in the jar = 1.25 kg

## Q: 7 From a container of wine, a thief has stolen 15 litres of wine and replaced it with same quantity of water. He again repeated the same process. Thus, in three attempts the ratio of wine and water became 343 : 169. The initial amount of wine in the container was: 1530 05b5cc7d7e4d2b41977751510

5b5cc7d7e4d2b41977751510- 175 litresfalse
- 2100 litresfalse
- 3150 litresfalse
- 4120 litrestrue

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## Answer : 4. "120 litres"

Explanation :

Answer: D) 120 litres Explanation: wine(left)wine(added) = 343169 It means wine(left)wine(initial amount) = 343512 (since 343 + 169 = 512) Thus, 343x = 512x1 - 15k3 343512 = 783 = 1 - 15k3 1-15k=78=1-18 Thus the initial amount of wine was 120 liters.

## Q: 8 5 men and 4 women are to be seated in a row so that the women occupy the even places . How many such arrangements are possible? 367 05b5cc7d1e4d2b4197775129f

5b5cc7d1e4d2b4197775129f- 12880true
- 21440false
- 3720false
- 42020false

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## Answer : 1. "2880"

Explanation :

Answer: A) 2880 Explanation: There are total 9 places out of which 4 are even and rest 5 places are odd. 4 women can be arranged at 4 even places in 4! ways. and 5 men can be placed in remaining 5 places in 5! ways. Hence, the required number of permutations = 4! x 5! = 24 x 120 = 2880