EXERCISE 7 (A)
QUESTION 1
Write each of these percentages as a fraction and a decimal.
(i) 15%
Sol :
=15100
= 0.15
(ii) 84%
Sol :
=84100
= 0.84
(iii) 12.5%
Sol :
=12.5100
= 0.125
(iv) 120%
Sol :
=120100
= 1.2
(v) 3313%
Sol :
=1003×1100
=13
= 0.33
Q2 | Ex-7A | Class-8 | Schand Composite Mathematics | Percentage | myhelper
QUESTION 2
Write each of the following as percentages.
(i) 35
Sol :
35×100
= 60 %
(ii) 1920
Sol :
=1920×100
= 95 %
(iii) 223
Sol :
=223×100
=26623%
(iv) 0.95
Sol :
=95100
=95100×100
= 95 %
(v) 2.575
Sol :
=25751000×100
= 257.5 %
Q3 | Ex-7A | Class-8 | Schand Composite Mathematics | Percentage | myhelper
QUESTION 3
Express each of the following ratios as percents.
(i) 1 : 4
Sol :
=14×100
= 25%
(ii) 11 : 20
Sol :
=1120×100
= 55 %
(iii) 21 : 40
Sol :
=2140×100
=5212 %
Q4 | Ex-7A | Class-8 | Schand Composite Mathematics | Percentage | myhelper
QUESTION 4
Express each of the following percents as ratios.
(i) 45%
Sol :
=45100
=920
or 9 : 20
(ii) 1623%
Sol :
=503
=503×100
=16
or 1 : 6
(iii) 130%
Sol :
=130100
=1310
or 13 : 10
Q5 | Ex-7A | Class-8 | Schand Composite Mathematics | Percentage | myhelper
QUESTION 5
Express the first quantity as a percentage of the second quantity.
(i) 10 cm of 1 m
Sol :
1m=100cm
=10cm100cm×100%
= 10%
(ii) 4 kg of 120 kg
Sol :
=4120×100%
=400120=103=313%
(iii) 110 of 25
Sol :
=(11025)×100%
=(110÷25)×100%
=(110×52)×100%
=14×100%
= 25%
Q6 | Ex-7A | Class-8 | Schand Composite Mathematics | Percentage | myhelper
QUESTION 6
Find the following amounts.
(i) 12% of 600
Sol:
=12100×600
=12×6
= 72
(ii) 3313% of 2400 people
Sol :
=(3313100)×2400
=(1003÷1001)×2400
=(1003×1100)×2400
=(13)×2400
= 800 peoples
(iii) 48% of 1 litre
Sol :
1l = 100ml
=48100×1000ml
= 48×10 ml
= 480 ml
(iv) 717 of 3kg 500g
Sol:
1 kg = 1000 g
3 kg = 3000g
3 kg 500 g = 3000 + 500 g
3 kg 500 g = 3500 g
=(717100)×3500g
=(507÷1001)×3500g
=(507×1100)×3500g
=(114)×3500g
= 250 g
Q7 | Ex-7A | Class-8 | Schand Composite Mathematics | Percentage | myhelper
QUESTION 7
A metal bar is 2.4 meters long.After heating its length increases by 2%.What is the new length?
Sol :
Original length 2.4 m
A.T.Q
Length increased by 2%
=2100×2.40
= 0.048 m
New length = Original Length + Length increased by 2%
= 2.4m + 0.048m
= 2.448 m
Hence , new length is equal to 2.448m
Q8 | Ex-7A | Class-8 | Schand Composite Mathematics | Percentage | myhelper
QUESTION 8
A pair of shoes costs ₹ 3000.During sale, its price reduced by 40%.What is the new price after reduction?
Sol :
Original Price = 3000
During sale price reduced by 40% =40100×3000
= 4×300
= 1200
New Price = Original Price - 40%
= 3000 - 1200
= 1800
Q9 | Ex-7A | Class-8 | Schand Composite Mathematics | Percentage | myhelper
QUESTION 9
Piyush's marks in Mathematics were wrongly read as 65 instead of 85.What is the percentage error in the marks?
Sol :
Error = 85 - 65 = 20
Percentage error =error85×100
=2085×100
=200085=23917%
Q10 | Ex-7A | Class-8 | Schand Composite Mathematics | Percentage | myhelper
QUESTION 10
A man gets a 15% hike salary.If his new salary is ₹ 16,100.What was his original salary?
Sol :
Let the original salary be 100
Increase in salary is 15% of 100 =15100×100 = 15
New salary be 115
When new salary are 115 , original salary are 100
When new salary are 16100 , original salary are =100115×16100 = 14000
ALTERNATE METHOD
Amount=Principal(1+rate of interest100)n
16100=Principal(1+15100)1
16100=Principal(100+15100)
16100=Principal(115100)
Principal=16100×100115
Principal=1610000115
Principal = 14000
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