The price of an article is first decreased by 40% and then increased by 20%. The net change in the price will be : [#1123]
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Q1. The price of an article is first decreased by 40% and then increased by 20%. The net change in the price will be :
Q1. The price of an article is first decreased by 40% and then increased by 20%. The net change in the price will be :
(A) 20%
(A) 20%
(A) 20%
(B) 40%
(B) 40%
(B) 40%
(C) 24%
(C) 24%
(C) 24%
(D) 28%
(D) 28%
(D) 28%
Answer: (D) 28%
Answer: (D) 28%
Answer: (D) 28%
Net change (-40 + 20 + )%
= (-20 + )%
= (-20 - 8)%
= -28%
= 28%
Net change (-40 + 20 + )% = (-20 + )% = (-20 - 8)% = -28% = 28%
Net change (-40 + 20 + )% = (-20 + )% = (-20 - 8)% = -28% = 28%
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Related MCQ Quizzes
Q1. p, q, r are three numbers such that the LCM of p and q is q and the LCM of q and r is r. The LCM of p, q and r will be
Q1. p, q, r are three numbers such that the LCM of p and q is q and the LCM of q and r is r. The LCM of p, q and r will be
(A) q
(A) q
(A) q
(B) r
(B) r
(B) r
(C) qr
(C) qr
(C) qr
(D) pqr
(D) pqr
(D) pqr
Answer: (B) r
Answer: (B) r
Answer: (B) r
LCM will be r.
px= q and qy = r, hence r = pxy.
LCM will be r. px= q and qy = r, hence r = pxy.
LCM will be r. px= q and qy = r, hence r = pxy.
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Q2. ‘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 :
Q2. ‘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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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. What is the term for a shape with five sides?
Q4. What is the term for a shape with five sides?
(A) Triangle
(A) Triangle
(A) Triangle
(B) Quadrilateral
(B) Quadrilateral
(B) Quadrilateral
(C) Pentagon
(C) Pentagon
(C) Pentagon
(D) Hexagon
(D) Hexagon
(D) Hexagon
Answer: (C) Pentagon
Answer: (C) Pentagon
Answer: (C) Pentagon
A pentagon is a polygon with five sides, like the shape of the Pentagon building in Washington, D.C.
A pentagon is a polygon with five sides, like the shape of the Pentagon building in Washington, D.C.
A pentagon is a polygon with five sides, like the shape of the Pentagon building in Washington, D.C.
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Q5. If '÷' means ‘addition’, ‘+’ means ‘subtraction’, ‘–’ means ‘multiplication’ and ‘×’ means ‘division’, then the value of 18 ÷ 12 × 4 – 5 is
Q5. If '÷' means ‘addition’, ‘+’ means ‘subtraction’, ‘–’ means ‘multiplication’ and ‘×’ means ‘division’, then the value of 18 ÷ 12 × 4 – 5 is
(A) 25
(A) 25
(A) 25
(B) 35
(B) 35
(B) 35
(C) 40
(C) 40
(C) 40
(D) 33
(D) 33
(D) 33
Answer: (D) 33
Answer: (D) 33
Answer: (D) 33
33
33
33
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Q6. 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
Q6. 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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Q7. 5 : 27 :: 9 : ________
Q7. 5 : 27 :: 9 : ________
Fill the blank.
Fill the blank.
Fill the blank.
(A) 83
(A) 83
(A) 83
(B) 81
(B) 81
(B) 81
(C) 36
(C) 36
(C) 36
(D) 18
(D) 18
(D) 18
Answer: (A) 83
Answer: (A) 83
Answer: (A) 83
5 : (52 + 2) = 5 : 27
Hence
9 : (92 + 2) = 9 : 83
5 : (52 + 2) = 5 : 27 Hence 9 : (92 + 2) = 9 : 83
5 : (52 + 2) = 5 : 27 Hence 9 : (92 + 2) = 9 : 83
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Q8. At what speed should a car travel on a highway to reach a destination 12 km away in 15 minutes?
Q8. At what speed should a car travel on a highway to reach a destination 12 km away in 15 minutes?
(A) 64 kmph
(A) 64 kmph
(A) 64 kmph
(B) 84 kmph
(B) 84 kmph
(B) 84 kmph
(C) 48 kmph
(C) 48 kmph
(C) 48 kmph
(D) 36 kmph
(D) 36 kmph
(D) 36 kmph
Answer: (C) 48 kmph
Answer: (C) 48 kmph
Answer: (C) 48 kmph
12
---- x 60 = 48 KMPH
15
12 ---- x 60 = 48 KMPH 15
12 ---- x 60 = 48 KMPH 15
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Q9. The smallest positive integer that is simultaneously divisible by 6, 8 and 12 is
Q9. The smallest positive integer that is simultaneously divisible by 6, 8 and 12 is
(A) 24
(A) 24
(A) 24
(B) 18
(B) 18
(B) 18
(C) 48
(C) 48
(C) 48
(D) 36
(D) 36
(D) 36
Answer: (A) 24
Answer: (A) 24
Answer: (A) 24
6*4 = 24
8*3 = 24
12*2 = 24
6*4 = 24 8*3 = 24 12*2 = 24
6*4 = 24 8*3 = 24 12*2 = 24
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Q10. The ratio of the radii of two circles is 1 : 3. The ratio of their areas is
Q10. 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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Related Questions
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