Three numbers are in the ratio 3:4:5. The sum of the largest and the smallest equals the sum of the third and 52. The smallest number is [#899]
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Q1. Three numbers are in the ratio 3:4:5. The sum of the largest and the smallest equals the sum of the third and 52. The smallest number is
Q1. Three numbers are in the ratio 3:4:5. The sum of the largest and the smallest equals the sum of the third and 52. The smallest number is
(A) 48
(A) 48
(A) 48
(B) 36
(B) 36
(B) 36
(C) 39
(C) 39
(C) 39
(D) 30
(D) 30
(D) 30
Answer: (C) 39
Answer: (C) 39
Answer: (C) 39
39
39
39
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Related MCQ Quizzes
Q1. ‘A’ starts his journey at 1:00 p.m. from a location P with a speed of 1 m/sec. ‘B’ starts his journey from the same location P and along the same direction at 1:10 p.m. with a speed of 2 m/sec. If ‘B’ meets ‘A’ at the location Q, then the distance PQ is :
Q1. ‘A’ starts his journey at 1:00 p.m. from a location P with a speed of 1 m/sec. ‘B’ starts his journey from the same location P and along the same direction at 1:10 p.m. with a speed of 2 m/sec. If ‘B’ meets ‘A’ at the location Q, then the distance PQ is :
(A) 1.5 km
(A) 1.5 km
(A) 1.5 km
(B) 1.75 km
(B) 1.75 km
(B) 1.75 km
(C) 1.2 km
(C) 1.2 km
(C) 1.2 km
(D) 1.25 km
(D) 1.25 km
(D) 1.25 km
Answer: (C) 1.2 km
Answer: (C) 1.2 km
Answer: (C) 1.2 km
A starts journey before B at a speed of 1 m/s. Hence A will be ahead of B
(10*60)s * 1m/s = 600m
Let A covers a distance of X after starting of B,
Then X + 600m = 2X
=> 2X - X = 600m
=> X = 600m
Hence B will cover a distance of 2X = 2 * 600m = 1200m = 1.2 km
A starts journey before B at a speed of 1 m/s. Hence A will be ahead of B (10*60)s * 1m/s = 600m Let A covers a distance of X after starting of B, Then X + 600m = 2X => 2X - X = 600m => X = 600m Hence B will cover a distance of 2X = 2 * 600m = 1200m = 1.2 km
A starts journey before B at a speed of 1 m/s. Hence A will be ahead of B (10*60)s * 1m/s = 600m Let A covers a distance of X after starting of B, Then X + 600m = 2X => 2X - X = 600m => X = 600m Hence B will cover a distance of 2X = 2 * 600m = 1200m = 1.2 km
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Q2. A number was increased by 40% and thereafter decreased by 40%. The net change in the number in percentage is
Q2. A number was increased by 40% and thereafter decreased by 40%. The net change in the number in percentage is
(A) 16% increase
(A) 16% increase
(A) 16% increase
(B) 16% decrease
(B) 16% decrease
(B) 16% decrease
(C) 32% decrease
(C) 32% decrease
(C) 32% decrease
(D) No change
(D) No change
(D) No change
Answer: (B) 16% decrease
Answer: (B) 16% decrease
Answer: (B) 16% decrease
16% decrease
= 40-40-%
= -%
= -16%
16% decrease = 40-40-% = -% = -16%
16% decrease = 40-40-% = -% = -16%
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Q3. The difference of two numbers is 5 and the difference of their square is 135. Then the sum of the numbers is
Q3. The difference of two numbers is 5 and the difference of their square is 135. Then the sum of the numbers is
(A) 25
(A) 25
(A) 25
(B) 45
(B) 45
(B) 45
(C) 32
(C) 32
(C) 32
(D) 27
(D) 27
(D) 27
Answer: (D) 27
Answer: (D) 27
Answer: (D) 27
27
27
27
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Q4. 100% of 100 when added to 200% of 200 would result
Q4. 100% of 100 when added to 200% of 200 would result
(A) 300
(A) 300
(A) 300
(B) 400
(B) 400
(B) 400
(C) 500
(C) 500
(C) 500
(D) 600
(D) 600
(D) 600
Answer: (C) 500
Answer: (C) 500
Answer: (C) 500
100 * 100% + 200 * 200%
=
=
= 100 + 400
= 500
100 * 100% + 200 * 200% = = = 100 + 400 = 500
100 * 100% + 200 * 200% = = = 100 + 400 = 500
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Q5. A man is 3 years older than his wife and four times as old as his son. If the son becomes 15 years old after 3 years, the present age of his wife is :
Q5. A man is 3 years older than his wife and four times as old as his son. If the son becomes 15 years old after 3 years, the present age of his wife is :
(A) 60 years
(A) 60 years
(A) 60 years
(B) 51 years
(B) 51 years
(B) 51 years
(C) 48 years
(C) 48 years
(C) 48 years
(D) 45 years
(D) 45 years
(D) 45 years
Answer: (D) 45 years
Answer: (D) 45 years
Answer: (D) 45 years
The son becomes 15 years old after 3 years.
Hence son's present age = 15 years - 3 years = 12 years
Man is four times as old as his son.
Hence Man's present age = 12 years * 4 = 48 years
Man is 3 years older than his wife.
Hence present age of his wife = 48 years - 3 years = 45 years.
The son becomes 15 years old after 3 years. Hence son's present age = 15 years - 3 years = 12 years Man is four times as old as his son. Hence Man's present age = 12 years * 4 = 48 years Man is 3 years older than his wife. Hence present age of his wife = 48 years - 3 years = 45 years.
The son becomes 15 years old after 3 years. Hence son's present age = 15 years - 3 years = 12 years Man is four times as old as his son. Hence Man's present age = 12 years * 4 = 48 years Man is 3 years older than his wife. Hence present age of his wife = 48 years - 3 years = 45 years.
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Q6. If 2x = 32, then the value of x is
Q6. If 2x = 32, then the value of x is
(A) 4
(A) 4
(A) 4
(B) 5
(B) 5
(B) 5
(C) 6
(C) 6
(C) 6
(D) 7
(D) 7
(D) 7
Answer: (B) 5
Answer: (B) 5
Answer: (B) 5
25 = 2*2*2*2*2 = 32
25 = 2*2*2*2*2 = 32
25 = 2*2*2*2*2 = 32
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Q7. Three numbers are in the ratio 3:4:6 and their product is 1944. The largest of the numbers is.
Q7. Three numbers are in the ratio 3:4:6 and their product is 1944. The largest of the numbers is.
(A) 21
(A) 21
(A) 21
(B) 12
(B) 12
(B) 12
(C) 18
(C) 18
(C) 18
(D) 9
(D) 9
(D) 9
Answer: (C) 18
Answer: (C) 18
Answer: (C) 18
18
18
18
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Q8. If A : B = 3, then 144A : 36B is
Q8. If A : B = 3, then 144A : 36B is
(A) 4
(A) 4
(A) 4
(B) 12
(B) 12
(B) 12
(C) 3
(C) 3
(C) 3
(D) 16
(D) 16
(D) 16
Answer: (B) 12
Answer: (B) 12
Answer: (B) 12
A : B = 3
=> = 3
=
= x 3
= 4 x 3
= 12
A : B = 3 => = 3 = = x 3 = 4 x 3 = 12
A : B = 3 => = 3 = = x 3 = 4 x 3 = 12
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Q9. The ratio of the radii of two circles is 1 : 3. The ratio of their areas is
Q9. The ratio of the radii of two circles is 1 : 3. The ratio of their areas is
(A) 1:6
(A) 1:6
(A) 1:6
(B) 2:9
(B) 2:9
(B) 2:9
(C) 1:9
(C) 1:9
(C) 1:9
(D) 6:9
(D) 6:9
(D) 6:9
Answer: (C) 1:9
Answer: (C) 1:9
Answer: (C) 1:9
Area = πr2
π12 : π32
= π1 : π9
= 1:9
Area = πr2 π12 : π32 = π1 : π9 = 1:9
Area = πr2 π12 : π32 = π1 : π9 = 1:9
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Q10. A hostel has 120 students and food supplies are for 45 days. If 30 more students joined the hostel, then how many days the hostel will run with the existing food?
Q10. A hostel has 120 students and food supplies are for 45 days. If 30 more students joined the hostel, then how many days the hostel will run with the existing food?
(A) 40 days
(A) 40 days
(A) 40 days
(B) 38 days
(B) 38 days
(B) 38 days
(C) 36 days
(C) 36 days
(C) 36 days
(D) 32 days
(D) 32 days
(D) 32 days
Answer: (C) 36 days
Answer: (C) 36 days
Answer: (C) 36 days
36 days
36 days
36 days
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