Quantitative Aptitude Practice Question and Answer
8Q: In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there ? 2154 05b5cc774e4d2b41977750011
5b5cc774e4d2b41977750011- 1205false
- 2194false
- 3209true
- 4159false
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Answer : 3. "209"
Explanation :
Answer: C) 209 Explanation: We may have (1 boy and 3 girls)or(2boys and 2 girls)or(3 boys and 1 girl)or(4 boys). Required number of ways = (C16× C34) + C26×C24 + (C36× C14) + (C46) = (24+90+80+15) = 209.
Q: A single pipe of diameter x has to be replaced by six pipes of diameters 12 cm each. The pipes are used to covey some liquid in a laboratory. If the speed/flow of the liquid is maintained the same then the value of x is ? 2153 05b5cc73ae4d2b4197774f902
5b5cc73ae4d2b4197774f902- 114.69 cmfalse
- 229.39 cmtrue
- 318.65 cmfalse
- 422.21 cmfalse
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Answer : 2. "29.39 cm"
Explanation :
Answer: B) 29.39 cm Explanation: Volume of water flowing through 1 pipe of diameter x = Volume discharged by 6 pipes of diameter 12 cms. As speed is same, area of cross sections should be same. Area of bigger pipe of diameter x = Total area of 6 smaller pipes of diameter 12 i.e πR2 = 6πR12 Here R = R and R1 = 12/2 = 6 ⇒R2 = 6×6×6 R = 14.696 => D = X = 14.696 * 2 = 29.3938 cm.
Q: The Value of logtan10+logtan20+⋯⋯+logtan890 is 2152 15b5cc7d0e4d2b41977751204
5b5cc7d0e4d2b41977751204- 1-1false
- 20true
- 31/2false
- 41false
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Answer : 2. "0"
Explanation :
Answer: B) 0 Explanation: = log tan10+log tan890 + log tan20+ log tan880+⋯⋯+log tan450 = log [tan10 × tan890] + log [tan20 × tan880 ] +⋯⋯+log1 ∵ tan(90-θ)=cotθ and tan 450=1 = log 1 + log 1 +.....+log 1 = 0.
Q: A 400-metre-long train is running at the speed of 60 km/hour. A man is running in the opposite direction of train at the speed of 12 km/hour. In how much time the train will cross this man? 2150 15d15fbab44a3a75f13296b82
5d15fbab44a3a75f13296b82- 150 secondfalse
- 230 secondfalse
- 320 secondtrue
- 440 secondfalse
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Answer : 3. "20 second"
Q: Find the Odd One Out? 3, 4, 14, 48, 576, 27648 2145 05b5cc6b0e4d2b4197774d2c6
5b5cc6b0e4d2b4197774d2c6- 14false
- 214true
- 348false
- 427648false
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Answer : 2. "14"
Explanation :
Answer: B) 14 Explanation: Given Number series is : 3 4 14 48 576 27648 It follows a pattern that, 3 4 4 x 3 = 12 but not 14 12 x 4 = 48 48 x 12 = 576 576 x 48 = 27648 Hence, the odd one in the given Number Series is 14.
Q: Express 3/8 and 1/10 as a Decimals? 2145 05b5cc6b3e4d2b4197774d437
5b5cc6b3e4d2b4197774d437- 11.375 & 1false
- 20.375 & 1false
- 313.75 & 1.1false
- 40.375 & 0.1true
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Answer : 4. "0.375 & 0.1"
Explanation :
Answer: D) 0.375 & 0.1 Explanation: 3/8 as a Decimal : In this we should convert the denominator into 10, 100, 1000, 10000, 100000,... to make it simple. Now to make 8 into 1000, we multiply numerator and denominator with 125 => 3 x 1258 x 125 = 3751000 = 0.375 1/10 as a Decimal : 1 divided into 10 parts of each 0.1. Therefore, 3/8 = 0.375 and 1/10 = 0.1 in decimals.
Q: If y/(x - z) = (y + x)/z = x/y then find z : y : x ? 2144 05b5cc761e4d2b4197774fd4e
5b5cc761e4d2b4197774fd4e- 11:2:3false
- 23:2:4true
- 34:3:2false
- 42:3:4false
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Answer : 2. "3:2:4"
Explanation :
Answer: B) 3:2:4 Explanation: Given, y/(x - z) = (y + x)/z = x/yyz = xy + x2 - yz - xz ....(1)Also xy = yx-z ⇒x2 -xz = y2 ....(2)Using (1) and (2), we get yz = xy - yz + y22yz = xy + y22z = x + y Only option (B) satisfies the equation.
Q: If the labour cost 20% of the cost of production and raw material cost 10% of the cost of production and the price on which article is sold is 20% above the cost of production. If the price of labour is increased by 40% and the price of raw material increased by 20% and rest other expenditure of cost remain constant. The industry thus decide to increase the selling price by 10%. Find the new profit percent ? 2140 05b5cc6c7e4d2b4197774de78
5b5cc6c7e4d2b4197774de78- 118%false
- 220%true
- 322%false
- 424%false
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Answer : 2. "20%"
Explanation :
Answer: B) 20% Explanation: Let the Cost of Production of the article = 100 Then, Labour Cost = 20 Raw Material = 10 Other Expenditure = 100 - 10 – 20 = 70 Selling Price of the article = 120 After increasing Labour and Raw material cost by 40% & 20% respectively, New Labour cost = 28 New Raw material cost = 12 New Cost of Production = 70 + 28 + 12 = 110 Then, New SP = 110% of 120 = 120 x 110/100 = 132 => New Gain = 132 - 110 = 22 => New Profit % = 22 x 100/110 = 20%

