Quantitative Aptitude Practice Question and Answer
8Q: How many parallelograms will be formed if 7 parallel horizontal lines intersect 6 parallel vertical lines? 2719 05b5cc754e4d2b4197774fbb4
5b5cc754e4d2b4197774fbb4- 1215false
- 2315true
- 3415false
- 4115false
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Answer : 2. "315"
Explanation :
Answer: B) 315 Explanation: Parallelograms are formed when any two pairs of parallel lines (where each pair is not parallel to the other pair) intersect. Hence, the given problem can be considered as selecting pairs of lines from the given 2 sets of parallel lines. Therefore, the total number of parallelograms formed = 7C2 x 6C2 = 315
Q: The average age of a couple and their son was 40 years, the son got married and a child was born just two years after their marriage. When child turned to 10 years, then the average age of the family becomes 38 years. What was the age of the daughter in law at the time of marriage ? 2718 05b5cc6c6e4d2b4197774de59
5b5cc6c6e4d2b4197774de59- 112 yearstrue
- 210 yearsfalse
- 314 yearsfalse
- 413 yearsfalse
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Answer : 1. "12 years"
Explanation :
Answer: A) 12 years Explanation: Given the average age of couple and their son = 40 => Sum of age (H +W +S) = 40×3 Let the age of daughter in law at the time of marriage = D years Now after 10 years, (H + S +W) + 3×12 + D +12 + 10 = 38×5 178 + D = 190 D = 190 -178 = 12 years
Q: k, l, m together started a business. k invested Rs.6000 for 5 months l invested Rs.3600 for 6 months and m Rs.7500 for 3 months. If they get a total profit of Rs.7410. Find the share of l ? 2714 05b5cc6dce4d2b4197774e9a6
5b5cc6dce4d2b4197774e9a6- 1Rs. 1640false
- 2Rs. 2500false
- 3Rs. 2160true
- 4Rs. 3000false
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Answer : 3. "Rs. 2160"
Explanation :
Answer: C) Rs. 2160 Explanation: => 60x5 : 36x6 : 75x3 => 100 : 72 : 75 => 72/247 x 7410 = Rs. 2160
Q: A garrison of 2000 men has provisions for 54 days. At the end of 15 days, a reinforcement arrives, and it is now found that the provisions will last only for 20 days more. What is the reinforcement ? 2714 05b5cc6e7e4d2b4197774edb0
5b5cc6e7e4d2b4197774edb0- 11900true
- 22100false
- 31700false
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Answer : 1. "1900"
Explanation :
Answer: A) 1900 Explanation: Given 2000 ---- 54 days The provisions for 2000 men for 39 days can be completed by 'm' men for only 20 days. i.e, 2000 ----- 39 days == m ---- 20 days => m x 20 = 2000 x 39 m = 3900 So total men for 20 days is 3900 => 2000 old and 1900 new reinforcement. Hence, reinforcement = 1900.
Q: Three men, four women and six children can complete a work in 9 days. A women does double the work a man does and a child does half the work a man does. How many women alone can complete this work in 9 days? 2709 05b5cc6b2e4d2b4197774d374
5b5cc6b2e4d2b4197774d374- 17true
- 28false
- 39false
- 46false
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Answer : 1. "7"
Explanation :
Answer: A) 7 Explanation: Given (3 Men + 4 Women + 6 Children) -----> 9 days But W = 2M and C = M/2 Now, convert Men and Children into Women by 3W2 + 4W + 6W4=> 32 + 4 + 32 = 7 Women Therefore, 7 women alone can complete this work in 9 days.
Q: There are several peacocks and deers in a cage (with no other types of animals). There are 72 heads and 200 feet inside the cage. How many peacocks are there, and how many deers ? 2708 05b5cc73de4d2b4197774f95d
5b5cc73de4d2b4197774f95d- 142 & 24false
- 238 & 26false
- 336 & 24false
- 444 & 28true
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Answer : 4. "44 & 28"
Explanation :
Answer: D) 44 & 28 Explanation: Letx = peacocks y = deers 2x + 4y = 200 x + y = 72 y = 72 - x 2x + 288 - 4x = 200 x = 44 peacocks y = 28 deers
Q: Cube root of 729 then square it 2705 05b5cc658e4d2b4197774bdf7
5b5cc658e4d2b4197774bdf7- 19false
- 236false
- 381true
- 4144false
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Answer : 3. "81"
Explanation :
Answer: C) 81 Explanation: 729 = 9 x 9 x 9 => Cube root of 729 = 9 Now, required square of 9 = 9 x 9 = 81.
Q: A group consists of 4 men, 6 women and 5 children. In how many ways can 2 men , 3 women and 1 child selected from the given group ? 2696 05b5cc6dde4d2b4197774e9d3
5b5cc6dde4d2b4197774e9d3- 1600true
- 2610false
- 3609false
- 4599false
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Answer : 1. "600"
Explanation :
Answer: A) 600 Explanation: Two men, three women and one child can be selected in ⁴C₂ x ⁶C₃ x ⁵C₁ ways = 600 ways.