A. 135

B. 130

C. 120

D. 125

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**ANSWER: B. 130Solution:When minute hand is behind the hour hand, the angle between the minute hand and hour hand at M minutes past H hours is given by:=> 30 * (H-M/5) + M/2When minute hand is ahead of hour hand, the formula becomes:=> 30* (M/5-H) – M/2Applying the second formula as here the minutes hand is ahead of the hour hand.=> 30* (40/5-3) – 40/2=> 30* (8-3) – 20=> 30*5 – 20=> 150-20=> 130Thus the angle formed is 130 degrees. **

A. Thursday

B. Sunday

C. Wednesday

D. Friday

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**ANSWER: B. Sunday Solution15th August 2010 can be written as 2009 + days from 1st January 2010 to 15th August 2010.=> Total number of odd days in 400 years = 0Hence, total number of odd days in 2000 years = 0 (as 2000 is a perfect multiple of 400)Odd in days in the period 2001-2009:7 normal years + 2 leap yeas=> (7*1) + (2*2) = 11=> Odd days will be 11- (7*1) = 4Days from January 1 to August 15 in 2010: 31+28+31+30+31+30+31+15= 227 days.= 32 weeks and 3 days, this gives additional 3 odd days.=> Total odd days= 3+4=7=> 7 odd days=1 week= 0 odd days=> 0 odd days= SundayThus, 15th August 2010 was a Sunday.**

A. 4 hours

B. 6 hour

C. 2 hours

D. 1 hours

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**ANSWER: C. 2 hours SolutionSpeed = distance/timeSpeed downstream: 96 / 8 km/hr = 12kmphSpeed of stream = 4kmphEffective speed of boat = (12-4) kmph= 8kmphDistance to be travelled upstream= 8 kmSpeed upstream = boat speed-current speed= 8-4 kmph= 4 kmphTime taken = distance/speed = 8/ 4 hours=2 hoursThus it will take 2 hours to go upstream.**

A. 400

B. 410

C. 390

D. 380

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**ANSWER: A. 400 SolutionUsing the formula:True Discount = (Banker’s Discount * 100 * 100)+ Total Rate=> True Discount = 420*100/100+15= 400Thus the true discount is Rs 100.**

A. 6.25

B. 5.5

C. 7.25

D. 8

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**ANSWER: A. 6.25 Solution:Runs scored in first 10 overs: => Rate*Overs= 3.2*10 = 32 runs=> Required runs = 282-32 = 250=> Left overs= 40Required rate = runs/overs= 250/40= 6.25=> 6.25 runs per over is needed to win the game.**

A. 4.04

B. 4

C. 40..4

D. 4.004

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**ANSWER: A. 4.04 SolutionGiven Error = 2% while measuring the side of a square.If the correct value of the side of square is 100, the measured value:=> 100 + 2% *100= 100+2=102The area of square with edge 100 = side*side=> 100*100=> 10000The area of square with side 102 = 102*102= 10404Error in area calculation = 10404-1000 = 404% error= (404/10000)*100= 4.04%**

A. 6

B. 5

C. 4

D. 3

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**ANSWER: C. 4 SolutionLet the age of youngest child be XAccording to the question, X + (X+3) + (X+6) + (X+9) + (X+12) = 505X + 30= 505X = 50-305X = 20X = 20/5 = 4Age of youngest child is 4 years.**

A. 20.68

B. 20

C. 20.12

D. 20.5

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**ANSWER: A. 20.68 SolutionLet length of other pipe be XAccording to question, 30 = 45/100 X + X30 = 0.45X + X30 = 1.45 XX= 30/1.45X = 20.68mThus the length of the other pipe is 20.68 metres.**

A. 4 hours 35 minutes

B. 4 hours

C. 3 hours

D. 3 hours 45 minutes

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**ANSWER: D. 3 hours 45 minutes SolutionSpeed of man = 4kmphTime for which he cycled = 45 minutes.Distance covered = speed*time=> 4*45/60=> 3kmTotal time to cover 3 km = 45 minutes + 15 minutes = 60 minutes = 1 hourThus 3 km is covered in 1 hourRequired distance to be covered = 12 kmBeginning from origin point, he covers 9 km in 3 hours taking 15 minutes rest after every 45 minutes. To complete 12 kms he goes another 3 km in 45 minutes. Thus, total time he takes is 3 hours 45 minutes.**

A. 225m

B. 120m

C. 220m

D. 160m

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**ANSWER: A. 225m Solution:Let X be the length of the courseaccording to the question, A beats B by 15 metres and C by 29 metresApplying this, If A runs X metres, B runs (X-15) and C runs ( X-29) metres.B and C together run, B runs X metres and C runs (X-15)=> If B runs 1 metre, C runs X-15/X=> B runs X-15 m, C runs (X-15/x) * (X-15)APPLYING THIS, X-29 = (X-15)(X-15) / XSolving,X**

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