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Question 1 of 5
1. Question
For any integer n, what is HCF (22n + 7,33n + 10) equal to?
Correct
Sol.(b)
HCF of (22n + 7, 33n + 10) is always 1.
Illustration: For n = 1, HCF (29, 43) = 1
For n = 2, HCF (51, 76) = 1
For n= 3, HCF (73, 109) = 1
Incorrect
Sol.(b)
HCF of (22n + 7, 33n + 10) is always 1.
Illustration: For n = 1, HCF (29, 43) = 1
For n = 2, HCF (51, 76) = 1
For n= 3, HCF (73, 109) = 1
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Question 2 of 5
2. Question
A, B and C can do a piece of work in 11 days, 20 days and 55 days respectively, working alone. How soon can the work be done if A is assisted by B on odd days and C on even days?
Correct
Ans- Solution: B
o (A+B)’s 1 day’s work = (1/11 + 1/20 ) = 31/220 o (A+C)’s 1 day’s work = ( 1/11 + 1/55) = 6/55 o Work done in 2 days by them= ( 31/220 + 6/55) = 55/220 = ¼. o Therefore, Whole work will be done in 8 days
Incorrect
Ans- Solution: B
o (A+B)’s 1 day’s work = (1/11 + 1/20 ) = 31/220 o (A+C)’s 1 day’s work = ( 1/11 + 1/55) = 6/55 o Work done in 2 days by them= ( 31/220 + 6/55) = 55/220 = ¼. o Therefore, Whole work will be done in 8 days
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Question 3 of 5
3. Question
Amit covers one-fourth of the total distance at 20 kmph, one-fourth at 10 kmph and rest of his journey at 80 kmph. Find Amit’s average speed for the whole distance?
Correct
Solution: A
Average speed = Total Distance/Total Time
Let the total distance = D Km Therefore, Time taken to complete first D/4 distance = t1 =(D/4)km/20kmph Time taken to complete second D/4 distance = t2 = (D/4)/10kmph Time taken to complete remaining distance (that is D/2) = t3 = (D/2)/80kmph Therefore, Average speed = Total Distance/Total Time = (D)/(t1+t2+3) = D/(D/80 + D/40 + D/160) = D/(2D + 4D + D)/160 = 160 D / 7D Therefore, Average Speed = 22. 85 kmph
Incorrect
Solution: A
Average speed = Total Distance/Total Time
Let the total distance = D Km Therefore, Time taken to complete first D/4 distance = t1 =(D/4)km/20kmph Time taken to complete second D/4 distance = t2 = (D/4)/10kmph Time taken to complete remaining distance (that is D/2) = t3 = (D/2)/80kmph Therefore, Average speed = Total Distance/Total Time = (D)/(t1+t2+3) = D/(D/80 + D/40 + D/160) = D/(2D + 4D + D)/160 = 160 D / 7D Therefore, Average Speed = 22. 85 kmph
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Question 4 of 5
4. Question
A zoo consists of only Tigers and Lions. The number of tigers in a zoo is three times the number of lions. Which one of the following numbers cannot represent the total number of animals in the zoo?
Correct
Ans-. Solution: B
• Let the number of lions = x • We know that the number of tigers in a zoo is three times the number of lions. Hence, the number of tigers = 3x. • Then, x + 3x = 4x = total number of animals. • Thus the total number of animals is a number that must be divisible by 4. • Here, 42 is the only number which is not divisible by 4
Incorrect
Ans-. Solution: B
• Let the number of lions = x • We know that the number of tigers in a zoo is three times the number of lions. Hence, the number of tigers = 3x. • Then, x + 3x = 4x = total number of animals. • Thus the total number of animals is a number that must be divisible by 4. • Here, 42 is the only number which is not divisible by 4
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Question 5 of 5
5. Question
There are two containers P and Q. P contains 56 kg of salt and Q contains 36 kg of sugar. From P 24 kg of salt is taken out and poured into Q. Then 20kg of the mixture from Q is taken out and poured into P. Find the ratio of final quantity of salt to sugar in container P
Correct
Ans-Solution: D •
Initially, the Amount of sugar in Q = 36kg • Now, 24kg of salt is poured in Q, • Total quantity in Q becomes = 36kg (sugar) + 24kg (salt) = 60kg (mixture) • The ratio of salt to sugar in Q becomes = 24 : 36 = 2 : 3 • Now, the Amount of salt in P = 56 – 24 = 32 kg • Again, 20kg of the mixture is taken out from Q and poured into P Therefore, quantity of salt and sugar in P becomes = [32 + 20 * (2/5)] kg of salt + 20 * (3/5) kg of sugar = (32 + 8) kg of salt + 12 kg of sugar = 40 kg of salt + 12 kg of sugar Required ratio = (40/12) = (10/3)
Incorrect
Ans-Solution: D •
Initially, the Amount of sugar in Q = 36kg • Now, 24kg of salt is poured in Q, • Total quantity in Q becomes = 36kg (sugar) + 24kg (salt) = 60kg (mixture) • The ratio of salt to sugar in Q becomes = 24 : 36 = 2 : 3 • Now, the Amount of salt in P = 56 – 24 = 32 kg • Again, 20kg of the mixture is taken out from Q and poured into P Therefore, quantity of salt and sugar in P becomes = [32 + 20 * (2/5)] kg of salt + 20 * (3/5) kg of sugar = (32 + 8) kg of salt + 12 kg of sugar = 40 kg of salt + 12 kg of sugar Required ratio = (40/12) = (10/3)
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