ANSWER KEY - Practice Test -1
Class- 8 Subject - Maths Marks : 40 marks
Chapters - Profit and Loss, Time and work
SECTION A each carries 1 marks
Shalini bought an air cooler for Rs. 3300 including a tax of 10%. The price of the air cooler before VAT was added is:-----
Explanation:10% VAT on Rs.100 will make it Rs.110
So, for price including VAT Rs.110, the original price is Rs.100
Then, Price including VAT Rs. 3300, the original price = Rs. (100/110) x 3300 = Rs. 3000.
A shopkeeper purchased 300 bulbs for Rs 10 each. However 10 bulbs were fused and had to be thrown away. The remaining were sold at Rs 12 each. Find the gain or loss % —---
Answer: C 16%
A shopkeeper purchased 500 pieces for Rs 20 each. However 50 pieces were spoiled in the way and had to be thrown away. The remaining were sold at Rs 25 each. Find the gain or loss %. —-
Answer: 12.5%
A trader marks his goods at 40% above the cost price and allows a discount of 25%. What is his gain?
Answer: Let the cost price be Rs. 100.
Then, the marked price =100+40100×100 = Rs. 140
Discount = 25% of 140=25100×140=1404 = Rs. 35
Selling price = marked price – discount = 140 –35 = Rs. 105
Profit percentage =Profit /C.P×100=105−100100×100=5%
A dealer purchased a washing machine for Rs. 7660. He allows a discount of 12% on its marked price and still gains 10%. Find the marked price of the machine.
Answer:
profit = 7660x 10/100=766
7660+766=8426
M.p discount s.p
100 - 12 = 88
? 8426
x= (100 x 8426)/88 = 9575
3 men or 5 women can do a work in 12 days. How long will 6 men and 5 women take to do it?
Answer:
3 men = 5 women
The work done by 1 man = 5/3 women
For 6 men = 5/3 x 6 = 10 women
So 10+5 = 15 women
If 5 women take 12 days then 15 women
5/15 x 12 = 4 days
A man can do a piece of work in 5 days, but with the help of his son, he can do in 3 days. In what time can the son do it alone?
Answer :
In day = ⅕
In 3 days = ⅕ 3 = ⅗
Remaining work = ⅗ - 1/5 = ⅖
Total work in 1 day ⅓ x ⅖ = 2/ 15 = 7 ½ days
A can do 1/3 of a work in 5 days and B can do 2/5 of the work in 10 days. In how many days can both A and B together do the work?
Answer:
A's rate in Work days: =(1/3∗1/5)=1/15
B's rate, in Work days: =(2/ 5 * 1/10)=2/ 50=1 / 25
LCM : A 1 / 15 , and B, 1 / 25 is 1/ 50
combine rate of A and B:(10/ 150+6/ 150)=16 / 150= 8/ 75
Work / rate=time⟹1/ 8/ 75
time=75/8= 9 ⅜ days
One tap can fill a water tank four times as fast as another tap. If together the two taps can fill the water tank in 30 minutes then the slower tap alone will be able to fill the water tank in _______
Answer: Hint. Let the slower pipe alone fill the water tank in t minutes.
Then, faster pipe will fill it in t/4 minutes.
Therefore, 1/t + 4/t = 1/30
⇒ 5/t = 1/30
Now solve for t.\7
Answer: 150 min
45 men can finish a job in 16 days. 30 more men joined working, 6 days after they initiated working. How many days more will they now take to finish the remaining work?
Answer: 6 days
Section B each carries 3 marks
By selling a book for Rs 258, a bookseller gains 20%. For how much should he sell it to gain 30%?
Solution:
Given details are,
Selling price of book is = Rs 258
The man’s gain percent is = 20% of 100 = 20/100
So, let us consider the cost price of book be Rs x
By solving,
x + x×20/100 = 258
x + x/5 = 258
(5x+x)/5 = 258
By cross multiplying
6x= 5×258
x = 1290/6
= 215
Now, the cost price of book is = Rs 215
For a gain of 30% the man should sell the book at = 215 + 215×30/100
= 215 + 64.5
= 279.50
∴ To gain 30% the man should sell the book at Rs 279.50
If the selling price of 10 pens is equal to cost price of 14 pens, find the gain percent.
Solution:
S.P of 10 pens = cost price of 14 pens
let C.P of 1 pen be Rs x
S.P of 10 pens = Rs 14x
S.P of 1 pen =Rs 14x/10
Gain = S.P – C.P
= 14x/10 – x
= 4x/10
Gain % = (gain/cost price) × 100
= (4x/10)/x × 100
= 2/5 × 100
= 40%
By selling 90 ball pens for Rs 160 a person loses 20%. How many ball pens should be sold for Rs 96 so as to have profit of 20%?
Solution:
FOR 90 PENS
C.P - S.P = LOSS
100 - 80 = 20
200 - 160 = 40
S.P for 90 ball pens is = Rs 160
S.P of 1 ball pen = Rs 160/90 = Rs 16/9
The loss percent is = 20% of 100 = 20/100
Let C.P of 1 pen be Rs x
x – x×20/100 = 16/9
x – x/5 = 16/9
(5x-x)/5 = 16/9
4x/5 = 16/9
4x×9 = 16×5
36x = 80
x = 80/36
= Rs 20/9
C.P of 1 ball pen = Rs 20/9
To get a profit of 20%
Let number of pens be ‘x’
S.P of ‘x’ pens is = Rs 96
Selling price of 1 pen is = Rs 96/x
Gain % = (gain/cost price) × 100
20% = [(96/x) – (20/9)] / (20/9) × 100
20/100 = [(96/x) – (20/9)] / (20/9) × 100
(20/100 × 200/9) + 200/90 = 96/x
4/9 + 200/90 = 96/x
(40+200)/90 = 96/x
240/90 = 96/x
24/9 = 96/x
24x = 96×9
x = 864/24 = 36
∴ 36 ball pens can be sold at a price of Rs 96
A dishonest shopkeeper professes to sell pulses at his cost price but uses a false weight of 950 gm for each kilogram. Find his gain percent.
Solution:
Let us consider the cost price of 1000gm pulses be Rs x
Selling price of 950 gm pulses is also = Rs x
Selling price of 1000 gm pulses = x/950 × 1000
So, Gain = SP – CP
Gain = 1000x/950 – x
= (1000x – 950x)/950
= 50x/950
Gain % = (gain/cost price) × 100
= (50x/950)/x × 100
= 50x/950x × 100
= 5/95 × 100
= 100/19
= 5 5/19 %
∴ The Shopkeeper’s gain percent is
5 5/ 19%
The difference between two selling prices of a shirt at profits of 4% and 5% is Rs 6. Find (i) C.P. of the shirt (ii) The two selling prices of the shirt.
Solution:
(i) Let the CP of shirt be = Rs x
Profit (4%) = 4/100 of CP
= 4/100 × x
= 4x/100
Selling Price = C.P + Profit
= x + 4x/100
= (100x + 4x)/100
= 104x/100
(ii) Let the CP of shirt be = Rs x
Profit (5%) = 5/100 of CP
= 5/100 × x
= 5x/100
Selling Price = C.P + Profit
= x + 5x/100
= (100x + 5x)/100
= 105x/100
difference between the two selling price is Rs 6
105x/100 – 104x/100 = 6
(105x-104x)/100 = 6
x/100 = 6
x = 600
∴ Now, C.P of the shirt is = Rs 600
Selling Price of one shirt = 104x/100 = (104×600)/100 = Rs 624
Selling Price of other shirt = 105x/100 = (105×600)/100 = Rs 630
A cistern can be filled by a tap in 4 hours and emptied by an outlet pipe in 6 hours. How long will it take to fill the cistern if both the tap and the pipe are opened together?
Solution:
Inlet tap can fill a cistern in = 4 hours
Inlet tap can fill a cistern in 1 hour = 1/4
Outlet tap can empty the cistern in = 6 hours
Outlet tap can empty the cistern in 1 hour = 1/6
Work done by both pipe in 1 hour = (1/4 – 1/6)
= (3-2)/12
= 1/12
∴ When both tap and pipe are opened together the cistern can be filled in = 1/(1/12) = 12 hours.
If 25 men earn Rs 1000 in 10 days, how much will 15 men earn in 15 days?
Solution:
The given details are,
In 10 days 25 men can earn = Rs.1000
In 1 day 25 men can earn = 1000/10 = Rs 100
In 1 day 1 man can earn = 100/25 = Rs 4
In 15 days 1 man can earn = 15 × 4 = Rs 60
∴ In 15 days 15 men can earn = 60 × 15 = Rs 900
Rani can finish typing a 100 page document in 9 hours, Raji in 6 hours and Rama in 12 hours. How long will they take to type a 100 page document if they work together?
Solution:
The given details are,
Work done by Rani in 1 hour = 1/9
Work done by Raji in 1 hour = 1/6
Work done by Rama in 1 hour = 1/12
Work done by Rani, Raji and Rama together in 1 hour = 1/9 + 1/6 + 1/12
= (4+6+3)/36 (by taking LCM for 9, 6 and 12 which is 36)
= 13/36
∴ Time taken by all three together to complete the work = 1/(13/36) = 36/13 hours.
A and B can do a piece of work in 12 days; B and C in 15 days; C and A in 20 days. How much time will A alone take to finish the work?
Solution:
The given details are,
A and B can do a piece of work in = 12 days
Work done by A and B in 1 day = 1/12
B and C can do a piece of work in = 15 days
Work done by B and C in 1 day = 1/15
A and C can do a piece of work in = 20 days
Work done by A and C in 1 day = 1/20
By adding A, B and C we get,
2(A+B+C)’s one day work = 1/12 + 1/15 + 1/20
= (5+4+3)/60 (by taking LCM for 12, 15 and 20 which is 60)
= 12/60
= 1/5
A+B+C one day work = 1/(5×2) = 1/10
We know that,
A’s 1 day work = (A+B+C)’s 1 day work – (B+C)’s 1 day work
= 1/10 – 1/15
= (3-2)/30 (by taking LCM for 10 and 15 which is 30)
= 1/30
∴ A alone can finish the work in = 1/(1/30) = 30days.
Kiran can paint his doll in 20 minutes and his sister Karthi can do so in 25 minutes. They paint the doll together for five minutes. At this juncture, they have a quarrel and Karthi withdraws from painting. In how many minutes will Kiran finish the painting of the remaining doll?
Solution:
The given details are,
Kiran can paint his doll in = 20 minutes
Kiran can paint his doll in 1 minute = 1/20
karthi can paint the same doll in = 25 minutes
Karthii can paint the same doll in 1 minute = 1/25
Together they both can paint the doll in 1 minute = 1/20 + 1/25
= (5+4)/100 (by taking LCM for 20 and 25 which is 100)
= 9/100
Work done by them in 5 minute = 5 × 9/100
= 9/20
Remaining work = 1 – 9/20
= (20-9)/20
= 11/20
∴ Kiran can paint the remaining doll in = (11/20)/(1/20)
= 11/20 × 20
= 11minutes
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