Time and Work
1. A can do a half of certain work in 70 days and B one third of the same in 35 days. They together will do the whole work in.

A. 420 days B. 120 days C. 105 days D. 60 days

Answer: Option D
Explanation: A = 140 days B = 105 days [latex]\frac{1}{140}[/latex] + [latex]\frac{1}{105}[/latex] = [latex]\frac{7}{420}[/latex] = [latex]\frac{1}{60}[/latex]1/60 =&60 days
2. Anita, Indu, and Geeta can do a piece of work in 18 days, 27 days and 36 days respectively. They start working together. After working for 4 days. Anita goes away and Indu leaves 7 days before the work is finished. Only Geeta remains at work from beginning to end. In how many days was the whole work is done?

A. 16 days B. 17 days C. 18 days D. 19 days

Answer: Option A
Explanation: [latex]\frac{4}{18}[/latex] + [latex]\frac{(x -7)}{27}[/latex] + [latex]\frac{x}{36}[/latex] = 1 x = 16 days
3. A and B can do a work in 5 days and 10 days respectively. A starts the work and B joins him after 2 days. In how many days can they complete the remaining work?

A. 1 day B. 2 days C. 3 days D. 4 days

Answer: Option B
Explanation: Work done by A in 2 days = [latex]\frac{2}{5}[/latex] Remaining work = [latex]\frac{3}{5}[/latex] Work done by both A and B in one day = [latex]\frac{1}{5}[/latex] + [latex]\frac{1}{10}[/latex] = [latex]\frac{3}{10}[/latex] Remaining work = [latex]\frac{3}{5}[/latex] * [latex]\frac{10}{3}[/latex] = 2 days.
Allegation or Mixture
1. A body massage oil mix contains 60 liters of Oil A with some liters of Oil B. The rate of Oil A is Rs. 32 per liter while the rate of Oil B is 9 Rs/liter less than Oil A. A shopkeeper sells this oil mix at Rs. 28/liter. How much Oil B should be mixed in the mixture to ensure that there is no profit, no loss in the whole process?

A. 24 litres B. 36 litres C. 48 litres D. 50 litres

Answer: Option C
Explanation: Let Oil B quantity be B liters. Final mixture has '60+B' litres There is no profit no loss So, (Oil 1 quantity * rate) + (Oil 2 quantity * rate) = Mix quantity * rate Oil B is 9 Rs/- cheap. So, the rate of Oil B per litre = 23 Rs/litre ∴ 60 x 32 + B x 23 = (60+B) x 28 ∴ B = 48 litres = Oil B quantity
2. Three vessels contain a milk mixture of 30 liters each. When put in a big vessel, they result in a mixture of milk and water in the ratio 2:1. If the ratio is to be reversed to make it 1: 2, how much more water should be added to the mix?

A. 20L B. 40L C. 90L D. 100L

Answer: Option C
Explanation: 3 vessels of 30 litres each = 3 X 30 = 90 ml milk mixture Amount of milk in mixture = [latex]\frac {2}{2+1}[/latex] x 90 = 60 L Amount of water = 90-60 = 30 L To make milk to water ratio 1:2, we simply need to make water double of milk. By direct observation, we can say that we should have 60 L x 2 = 120 L water to make the required ratio We already have 30 L, we need (120-30) = 90 L more water.
3. 3 varieties of wheat were mixed at a warehouse. The rate of Type 1 wheat was Rs. 145/ Kg and rate of Type 2 wheat were Rs. 20 per kg more than Type 1. The quantities of 3 varieties of wheat were in ratio 2:1:3 respectively. The mix was finally sold at the rate of Rs. 180 per kg. Find the price of the 3rd type of wheat?

A. Rs. 196.58 B. Rs. 208.33 C. Rs. 210 D. Rs. 215.67

Answer: Option B
Explanation: Let the rate of the 3rd type of wheat be Rs. W Quantity ratio = 2 : 1 : 3 This means if we take 2 kg of Type 1 and 1 kg of Type 2, then we must take 3kg of Type 3. Also, the mix will have 2 kg + 1 kg + 3 kg = 6 kg quantity. ∴ (145 x 2) + (165 x 1) + (3 x W) = 6 x 180 ∴ W = Rs. 208.33 per kg = 3rd type price per kg

Decimal Fraction
1. How many digits will be there to the right of the decimal point in the product of 89.635 and .02218?

A. 5 B. 6 C. 7 D. 8

Answer: Option C
Explanation: Sum of decimal places = 3 + 5 = 8 The last digit in the product(digit at the extreme right) is zero (Since 5 x 8 = 40) Hence, there will be 7 significant digits to the right of the decimal point.
2. 24.39 + 562.093 + 35.96 = ?

A. 622.441 B. 622.243 C. 622.233 D. 622.443

Answer: Option D
Explanation: 24.39 + 562.093 + 35.96 = 622.443
3. What decimal of an hour is a second?

A. 0.00027 B. 0.00025 C. 0.00026 D. 0.00024

Answer: Option A
Explanation: 1 hour = 60 minutes = 3600 seconds Hence, required decimal = [latex]\frac{1}{3600}[/latex] = 0.00027
Probability
1. Ramesh throws two dices simultaneously. What is the probability that the sum of numbers on top faces in a throw is 9?

A. [latex]\frac {1}{12}[/latex] B. [latex]\frac {1}{9}[/latex] C. [latex]\frac {1}{3}[/latex] D. [latex]\frac {2}{9}[/latex]

Answer: Option B
2. India is playing 3 sets of badminton matches against Japan in Singapore. The probability of India winning the three sets is [latex]\frac{1}{7}[/latex], [latex]\frac{1}{5}[/latex], [latex]\frac{3}{4}[/latex]. What is the probability of winning at least 1 set of the game?

A. [latex]\frac {7}{24}[/latex] B. [latex]\frac {23}{24}[/latex] C. [latex]\frac {23}{35}[/latex] D. [latex]\frac {29}{35}[/latex]

Answer: Option D
3. If I roll two dices simultaneously, what is the probability of getting a sum less than 13?

A. [latex]\frac {1}{2}[/latex] B. 0.75 C. 1 D. 1.25

Answer: Option C

Average
1. Find the average price of the goods he has bought.

A. Rs. 5.5 B. Rs. 6.05 C. Rs. 8.25 D. Rs. 9

Answer: Option B
Explanation: Average =[latex]\frac {Sum of Observations}{Number of Observations}[/latex] Average = [latex]\frac {Total cost of pencils + Total cost of pens + Total cost of sharpeners}{Total pencils + Total pens + Total sharpeners}[/latex] ∴ Average price = [latex]\frac {(35 \times 4)+(30 \times 11)+(25 \times 3)}{35+30+25}[/latex] = Rs. 6.05
2. If I do not consider the earnings from selling the six-seater dining table, my average falls by Rs 2000/-. What is the cost of six-seater dining table if average earnings from selling 11 four-seater dining tables and one six-seater dining table are Rs 18000?

A. Rs. 29000 B. Rs. 30000 C. Rs. 36000 D. Rs. 40000

Answer: Option D
Explanation: Without six seater dining table, there are 11 four-seater dining tables. Average = 18000 - 2000 = 16000. Total cost of 11 small dining tables = 11 x 16000 = Rs. 176000 With big dining table, total cost of 12 tables = 12 x 18000 = Rs. 216000 Cost of six seater dining table = 216000 - 176000 = Rs. 40000
3. In an exam, the average marks for 80 students of Class V are 35. The average of marks in section A of the class is 55 while the average of marks in section B is 30. Find the number of students in Class V B.

A. 45 B. 50 C. 64 D. 70

Answer: Option C
Explanation: Total students i.e. Section A + Section B = 80 Section A = 80 - Section B = 80 - B ∴ 80 x 35 = B x 30 + (80 - B) x 55 2800 = 30B + 4400 - 55B 1600 = 25 B B = 64 ∴ B = 64 = Number of students in Class V B.
Problems on H.C.F and L.C.M.
1. Find the greatest 4-digit number exactly divisible by 3, 4 and 5?

A. 9985 B. 9960 C. 9957 D. 9975

Answer: Option B
Explanation: The greatest 4-digit number is 9999. LCM of 3, 4 and 5 is 60. Dividing 9999 with 60, we get remainder 39. Thus, the required number is 9999 - 39 = 9960.
2. Find the greatest number which, while dividing 19, 83 and 67, gives a remainder of 3 in each case?

A. 16 B. 17 C. 18 D. 19

Answer: Option A
Explanation: Subtract the remainder 3 from each of the given numbers: (19-3)=16, (67-3)=64 and (83-3)=80. Now, find the HCF of the results 16, 64 and 80, we get 16. Thus, the greatest number is 16.
3. A drink vendor has 80 liters of Maaza, 144 liters of Pepsi and 368 liters of Sprite. He wants to pack them in cans, so that each can contains the same number of liters of a drink, and doesn't want to mix any two drinks in a can. What is the least number of cans required?

A. 35 B. 37 C. 42 D. 30

Answer: Option B
Explanation: The number of liters in each can = HCF of 80, 144 and 368 = 16 liters. Number of cans of Maaza = [latex]\frac {80}{16}[/latex] = 5 Number of cans of Pepsi = [latex]\frac {144}{16}[/latex] = 9 Number of cans of Sprite = [latex]\frac {368}{16}[/latex] = 23 The total number of cans required = 5 + 9 + 23 = 37 cans.