[2 Mark Question Answer] - MATHS STD 7 Questions - Vidyadip

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19 questions · timed · auto-graded

Question 12 Marks

A garrison of 1200 men has provisions for 15 days. How long will the provisions last if the garrison be increased by 600 men?

Answer

$\begin{aligned} & 1200 \text { men has provision for }=15 \text { days } \\ & 1 \text { man will have that provision for }=15 \times 1200 \text { days } \\ & \therefore 1200+600=1800 \text { men will has that provisions for }=\frac{15 \times 1200}{1800} \text { days } \\ & =10 \text { dyas }\end{aligned}$

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Question 22 Marks

The wages of 10 workers for a six days week are Rs, 1,200. What are the one day wages:
(i) of one worker?
(ii) of 4 workers?

Answer

(i) 10 workers can earn in 6 days $=$ Rs. 1200
1 worker will earn in 6 days $=$ Rs. $\frac{1200}{10}$
1 worker will earn in 1 day = Rs. $\frac{1200}{10 \times 6}=$ Rs. 20
(ii) $\therefore 4$ workers will earn in 1 day $=$ Rs. $20 \times 4=$ Rs. 80

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Question 32 Marks

50 men mow 32 hectares of land in 3 days. How many days will 15 men take to mow it?

Answer

Land is same in both the cases.
Now 50 men can mow land in $=3$ days
$\therefore 1$ man will mow it in $=3 \times 50$ days
and 15 men will mow it in $=\frac{3 \times 50}{15}=10$ dyas

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Question 42 Marks

A wrist watch loses 10 sec in every 8 hours; in how much time will it lose 15 sec.?

Answer

$10 \mathrm{sec}$. are lost in $=8$ hours
$\therefore 1 \mathrm{sec}$ will be lost in $=\frac{8}{10}$ hours
and $15 \mathrm{sec}$. will be lost in $=\frac{8}{10} \times 15$ hours $=12$ hours

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Question 52 Marks

$3 \frac{1}{2} \mathrm{~m}$ of cloth costs Rs. 168 ; find the cost of $4 \frac{1}{3}$ of the same cloth.

Answer

Cost of $3 \frac{1}{2}$ i.e. $\frac{7}{2} \mathrm{~m}$ of cloth $=$ Rs. 168
$\therefore$ Cost of $1 \mathrm{~m}$ of cloth $=$ Rs. $168 \times \frac{2}{7}=$ Rs. 48
and cost of $4 \frac{1}{3}=\frac{13}{3} \mathrm{~m}$ cloth $=$ Rs. $48 \times \frac{13}{3}$
$ =\text { Rs. } 208 $

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Question 62 Marks

A camp has provisions for 60 pupil for 18 days. In how many days, the same provisions will finish off if the strength of the camp is increased to 72 pupil?

Answer

60 pupil have provision for $=18$ days 1 pupil will have provision for
$=18 \times 60$ days (less pupils more days)
and 72 pupils will have provision for $=\frac{18 \times 60}{72}$ days
$=15$ days

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Question 72 Marks

The cost of $\frac{3}{5} \mathrm{~kg}$ of ghee is ₹ 96 ; find the cost of:
(i) one kg ghee
(ii) $\frac{5}{8} \mathrm{~kg}$ ghee

Answer

Cost of $\frac{3}{5} \text{kg}$ of ghee is ₹ 96
(i) $\therefore$ Cost of $1 \text{kg}$ ghee = ₹ $96 \times \frac{5}{3}=$ ₹ 160
(ii) and cost of $\frac{5}{8}$ of ghee =₹ $160 \times \frac{5}{8}=$ ₹ 100

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Question 82 Marks

8 men can finish a certain amount of provisions in 40 days. If 2 more men join with them, find for how many days the same amount of provisions be sufficient ?

Answer

8 men can finish a provision in $=40$ days
1 man will finish in $=40 \times 8$ days
$8+2=10$ men will finish in $=\frac{40 \times 8}{10}$ $=32$ days

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Question 92 Marks

200 men have provisions for 30 days. If 50 men left, the same provisions would last for the remaining men, in how many days?

Answer

200 men have provisions for $=30$ days
1 man will have provisions for $=30 \times 200$ days
$200-50=150$ men will have provisions
for $=\frac{30 \times 200}{150}$ days $=40$ days

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Question 102 Marks

400 men have provisions for 23 weeks. They are joined by 60 men. How long will the provisions last?

Answer

400 men have provisions for $=23$ weeks
1 man will have provisions for $=23 \times 400$ weeks
and $400+60=460$ men will have provisions for $=\frac{23 \times 400}{460}$ weeks $=20$ weeks

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Question 112 Marks

Rs. 750 support a family for 15 days. For how many days will Rs. 2,500 support the same family?

Answer

Rs. 750 , can support a family for $=15$ days
Rs. 1 will support for $=\frac{15}{750}$ days
and Rs. 2,500 will support for $=\frac{15}{750} \times 2500$ days $=50$ days

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Question 122 Marks

A typist takes 80 minutes to type 24 pages. How long will he take to type 87 pages?

Answer

For typing 24 pages, time is required $=80$ minutes
For typing 1 page, time is required $=\frac{80}{24}$ minutes
and for typing 87 pages, time is required
$=\frac{80 \times 87}{24}$ minutes $=290$ minutes

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Question 132 Marks

If a 25 cm long candle burns for 45 minutes, how long will another candle of the same material and same thickness but 5 cm longer than the previous one, burn?

Answer

$25 \mathrm{~cm}$ long candle burn in $=45$ minutes
$1 \mathrm{~cm}$ long candle will burn in $=\frac{45}{25}$ minutes
$25+5=30 \mathrm{~cm}$ long candle will burn in
$=\frac{45 \times 30}{25}$ minutes $=54$ minutes

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Question 142 Marks

8 men complete work in 6 hours. In how many hours will 12 men complete the same work?

Answer

8 men can complete a work in $=6$ hours
1 man can complete the work in $=6 \times 8$
hours 12 men can complete the work in $=\frac{6 \times 8}{12}=4$ hours

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Question 152 Marks

6 books weigh 1 .260 kg. How many books will weigh 3.150 kg ?

Answer

$1 \mathrm{~kg} 260 \mathrm{~g}$ or $1.260 \mathrm{~kg}$ no.of.books $=6$
and in $1 \mathrm{~kg}$. no. of books $=\frac{6}{1.260}$
and in $3.150 \mathrm{~kg}$ no.of.books $=\frac{6 \times 3.150}{1.260}$
$=\frac{6 \times 3150}{1260}=\frac{3150}{210}=15$ books

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Question 162 Marks

A man earns ₹5,800 in 10 days. How much will he earn in the month of February of a leap year?

Answer

Leap year has 29 days in the month of February.
Man earns in 10 days ₹ 5800
Man earns in 1 day ₹ $\frac{5800}{10}$
$\therefore$ Man earns in February month of leap year $=\frac{5800 \times 29}{10}=$ ₹16820

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Question 172 Marks

A carpenter prepares a new chair in 3 days, working 8 hours a day. Atleast how many hours per day must he work in order to make the same chair in 4 days?

Answer

A chair is completed in 3 days working per day $=8$ hours
Then their will be completed in 1 day working for $=8 \times 3$ hours per day (less days more hours)
and it will be completed in 4 days working for $=\frac{8 \times 3}{4}=6$ hours per day.

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Question 182 Marks

If 9 lines of a print, in a column of a book contains 36 words. How many words will a column of 51 lines contain?

Answer

In 9 lines of print, words are $=36$
In 1 line of print, words will be $=\frac{36}{9}$
$\therefore$ in 51 lines of print, words will be
$=\frac{36}{9} \times 51=204$

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Question 192 Marks

Weight of 8 identical articles is 4.8 kg. What is the weight of 11 such articles?

Answer

Weight of 8 articles $=4.8 \mathrm{~kg}$
Weight of 1 article $=\frac{4.8}{8} \mathrm{~kg}$
and weight of 11 articles $=\frac{4.8}{8} \times 11 \mathrm{~kg}$
$ =0.6 \times 11=6.6 \mathrm{~kg} $

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[2 Mark Question Answer] - MATHS STD 7 Questions - Vidyadip