Quantitative Aptitude Practice Question and Answer
8Q: At what time between 3 and 4 o’clock will the minute hand and the hour hand are on the same straight line but facing opposite directions ? 2358 05b5cc75ee4d2b4197774fcee
5b5cc75ee4d2b4197774fcee- 13:15 2/8false
- 23:49false
- 33:49 1/11true
- 43:51false
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Answer : 3. "3:49 1/11"
Explanation :
Answer: C) 3:49 1/11 Explanation: On straight line means 180 degree angle.180 = 11/2 min – 30 hrs180 = 11/2 m – 30 × 3180 = 11/2 m – 90(180 + 90) 2 = 11 mm = 540/11 = 49 1/11 minutes.
Q: Find the missing number in the series 1, 6, ?, 15, 45, 66, 91 2352 15b5cc6afe4d2b4197774d230
5b5cc6afe4d2b4197774d230- 124false
- 228true
- 332false
- 426false
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Answer : 2. "28"
Explanation :
Answer: B) 28 Explanation: Here the given series 1, 6, ?, 15, 45, 66, 91 follows a Pattern in the series is, +5, +9, +13,...,.. +21, +25 So +4 is getting increased at every term addition. Missing Number in the series will be 15+(9+4) = 15 + 13 = 28.
Q: A container contains 50 litres of milk. From that 8 litres of milk was taken out and replaced by water. This process was repeated further two times. How much milk is now contained by the container ? 2347 05b5cc77fe4d2b419777501ce
5b5cc77fe4d2b419777501ce- 124.52 litresfalse
- 229.63 litrestrue
- 328.21 litresfalse
- 425.14 litresfalse
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Answer : 2. "29.63 litres"
Explanation :
Answer: B) 29.63 litres Explanation: Given that container has 50 litres of milk. After replacing 8 litres of milk with water for three times, milk contained in the container is: ⇒501-8503 ⇒50×4250×4250×4250 = 29.63 litres.
Q: If two fractions, each of which has a value between 0 and 1, are multiplied together, the product will be : 2331 05b5cc6cfe4d2b4197774e303
5b5cc6cfe4d2b4197774e303- 1always greater than either of the original fractionsfalse
- 2always less than either of the original fractionstrue
- 3sometimes greater and sometimes less than either of the original fractionsfalse
- 4remains the samefalse
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Answer : 2. "always less than either of the original fractions"
Explanation :
Answer: B) always less than either of the original fractions Explanation: We can easily Answer this by taking some values. The question states that the two fractions, each have values between 0 and 1.Let us say one of the fraction is 1/2 and the other fraction is 1/3 .The product of the two fractions is 1/2 x 1/3 = 1/6 . is lesser than both 1/2 and 1/3 .So, the correct answer is that the product is always less than either of the original fractions.
Q: 40% of the number is equal to three-fourth of another number. What is the ratio between first number and second number? 2330 05d1b13158a754124194a755f
5d1b13158a754124194a755f- 115:16false
- 215:8true
- 39:15false
- 48:17false
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Answer : 2. "15:8"
Q: Find the next number in the given series ? 3, 7, 6, 5, 9, 3, 12, 1, 15, . . . 2328 05b5cc71be4d2b4197774f54e
5b5cc71be4d2b4197774f54e- 12false
- 20false
- 31false
- 4-1true
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Answer : 4. "-1"
Explanation :
Answer: D) -1 Explanation: 3, 7, 6, 5, 9, 3, 12, 1, 15, ... has two different series 1) 3, 6, 9, 12, 15..... 2) 7, 5, 3, 1, ..... -1 is the next number.
Q: If a number 72k23l is divisible by 88. Then find value of k and l ? 2326 05b5cc745e4d2b4197774fa36
5b5cc745e4d2b4197774fa36- 1k=8 & l=2false
- 2k=7 & l=2true
- 3k=8 & l=3false
- 4k=7 & l=1false
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Answer : 2. "k=7 & l=2"
Explanation :
Answer: B) k=7 & l=2 Explanation: If a number to be divisile by 88, it should be divisible by both "8" and "11" Check for '8' :For a number to be divisible by "8", the last 3-digit should be divisible by "8"Here 72x23y --> last 3-digit is '23y'So y=2 [ (i.e) 232 is absolutely divisible by "8"] Chech for '11' :For a number to be divisible by "11" , sum of odd digits - sum of even digits should be divisible by "11"(7 + x + 3) - (2 + 2 + y)(7 + x + 3) - (2 + 2 + 2)(10 + x) - 6 should be divisible by "11"for x = 7=> 17 - 6 = 11 [ which is absolutely divisible by "11"] So x = 7 , y= 2.
Q: 6 men can complete a piece of work in 12 days. 8 women can complete the same piece of work in 18 days whereas 18 children can complete the piece of work in 10 days. 4 men, 12 women and 20 children work together for 2 days. If only men were to complete the remaining work in 1 day how many men would be required totally? 2325 05b5cc6ade4d2b4197774d125
5b5cc6ade4d2b4197774d125- 138false
- 272false
- 336true
- 476false
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Answer : 3. "36"
Explanation :
Answer: C) 36 Explanation: Given 4men, 12 women and 20 children work for 2 days. Workdone for 2 days by 4men, 12 women and 20 children = 4 x 26 x 12 + 12 x 28 x 18 + 20 x 218 x 10 = 12 Therefore, remaining work = 1 - 12 = 12 To complete the same work by only men in 1 day, We know that M1 x D1 = M2 x D2 Here M1 = 6 , D1 = 12 and M2 = M , D2 = 1 12 x 6 = M x 1 => M = 12 x 6 = 72 => But the remaining work = 1/2 Men required => 1/2 x 72 = 36 Only men required to Complete the remaining work in 1 day = 36.