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Chain Rules - Online Quiz
Following quiz provides Multiple Choice Questions (MCQs) related to Chain Rules. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using Show Answer button. You can use Next Quiz button to check new set of questions in the quiz.
Q 1 - A rope makes 140 rounds of the circumference of a cylinder, the radius of whose base is 14cm . How many times can it go rounds a cylinder with radius 20 cm?
Answer : C
Explanation
Let the required number of round be x. More radius , less round (Indirect) 20: 14: :: 140 : x ⇒ 20x = (14*140 ) ⇒ x = (14 * 140/20 = 98. Required number of round = 98.
Q 2 - The Qutab minar casts a shadow 150m long at the same time when the vikes minar casts a shadow 120m long on the ground . If the height of the Vikes minar is 80m, what is the height of the Qutab Minar?
Answer : A
Explanation
Let the height of the Qutab Minar be x metres. Longer is the shadow , longer is the object (Direct) 120 : 150 :: 80 : x ⇒ 120 * x = (150 *80)⇒ x= (150 *80)/120=100m. Height of Qutab Miinar = 100m.
Q 3 - 10 women can complete a piece of work in 8 days and 10 children take 12 days to complete it. How many days will 6 women and 3 children together take to complete the work?
Answer : C
Explanation
1 women can complete the work in ( 10 * 8) = 80 days. 1 child can complete the work in ( 10 * 12) = 120 days. 1 women 1 days work = 1/80, 1 child 1 days work = 1/120. ( 6women + 3 children) 1 days work = ( 6 * 1/80 + 3 * 1/120) = ( 3 /40 + 1/40 ) = 4/40=1/10. Required time 10 days.
Q 4 - 25 men can reap a filed in 20 days . When should 15 men leave the work if the whole filed is to be reaped in 74/2 days after they leave the work?
Answer : C
Explanation
Let 25 men work for x days. Work done = x/20 , Remaining work = (1- x/20). 25 men 1 days work = 1/20. 1 men 1 days work = (1/20 * 1/25 ) = 1/500. 10 men 1 days work = (1/500 * 10 ) = 1/50. 10 men 75/2 days work = ( 1/50 * 75/2) = 75/100 = 3/4. ∴ (1 - x/20 ) = 3/4 ⇒ x/20 = 1/4 ⇒ x= (20 1/4 )= 5. Hence , 15 men leave after 5 days.
Q 5 - In a barrack of soldiers there was stock of food for 190 days for 4000 soldiers.After 30 days , 800 soldiers left the barrack. For how many days shall the left over Food last for the remaining soldiers?
Answer : B
Explanation
Remaining food was sufficient for 4000 soldiers for 160 days. Remaining soldiers = ( 4000- 800) = 3200. Let the required number of days be x . Less soldiers, more days (Indirect) 3200 : 4000 :: 160 : x⇒ 3200 x = (4000 * 160 ) ⇒ x = (4000 * 160)/3200 = 200 days.
Q 6 - A garrison of 3300 men had provisions for 32 days , when given at the rate of 850 gm per head . At the end of 7 days , a reinforcement arrived and it was found that the provisions would last 17 days more , when given at the rate of 825 gm per head . What was the strength of the reinforcement?
Answer : C
Explanation
Let the required strength of reinforcement be x. 3300 men had provisions for (32-7 ) = 25 days. Less food per head , more persons (Indirect) Less days , more persons (Indirect) Food per days 825 : 850 :: 3300 : (3300 x ) Days 17 : 25 ∴ 825 * 17 * (3300 + x ) = 850 * 25 * 3300 ⇒ ( 3300 + x ) = 850 * 25 * 3300/825 * 17 = 5000 ⇒ x = ( 5000 - 3300) = 1700 men.
Q 7 - 8 men working for 9 hours a day complete a piece of work in 20 days . In how many days can 7 men working for 10 hours a day complete the same piece of work?
Answer : D
Explanation
Let the required number of days be x . Less men , more days ( Indirect) More working hours, less days (Indirect) Men 7: 8 :: 20 : x Working hrs 10 : 9 ∴ ( 7 * 10 * x ) = ( 8 * 9 * 20) ⇒ x = 8 * 9 * 20 / 7 * 10 = 144/7 = 144/7.
Q 8 - If 42 persons consume 144kg of rice in 15 days , then in how many days will 30 persons consume 48kg of rice?
Answer : D
Explanation
Let the required number of days be x. More persons, more days (Direct) Less rice, less days (Indirect) Persons 30 : 42 :: 15 : x Quantity 144 : 48 ∴ ( 30 * 144 * x ) = ( 42 * 48 *15 ) ⇒ x = 42 * 48 * 15/ 30 * 144 = 7 days
Q 9 - If a certain number of workmen can be a piece of work in 25 days , in what time will another set of an equal number of men do a piece of work twice as great , supposing that 2 of the first set can do as much work in an hour as 3 of the second set can do in an hour
Answer : B
Explanation
Let the required time be x days . Ratio of their speed = 1/2 : 1/3. More work , more time (Direct) Less speed, more work (Indirect) Work 1:2 :: 25:x Speed 1/3 : 1/2 ∴ (1* 1/3 *x) = 2 *1/2 *25 ⇒ x=75days.
Q 10 - A contractor undertakes to do a piece of work in 40 days . He engages 100 men at the beginning and 100 more after 35 days and complete the work in stipulated time. If he had not engaged the additional men, how many days behind schedule would it be finished?
Answer : A
Explanation
[(100 *35 ) + (200 *5)] men can finish the work in 1 day. 4500 men can finish it in 1 day ∴ men can finish it in 4500/100 days = 45 days. This is 5 days behind schedule.