问题13:一所大学的旅馆每月收取的费用是固定的,其余的取决于人们在烂摊子里吃饭的天数。当学生A吃饭20天时,他必须支付₹1000的旅馆费用,而学生B如果学生吃饭26天,则要支付₹1180的旅馆费用。查找固定收费和每天的食物成本。 [CBSE 2000]
解决方案:
Let’s assume that the fixed charge of the college hostel ₹ x
and daily charges be ₹ y
According to given condition,
x + 20y = 1000 —————(i)
x + 26y =1180 ——————(ii)
Subtracting (i) from (ii) and we get,
6y = 180
= y = 30
Substituting the value of y in (i) and we get,
x + 20 × 30 = 1000
= x + 600 = 1000
= x = 1000 – 600 = 400
Hence, Fixed charges ₹ 400 and daily charges ₹ 30.
问题14.长度为4 m的花园的一半长,其宽度为36 m。找到花园的尺寸。
解决方案:
Let’s assume that the length of the garden be x m and width y m.
According to given condition,
x – y = 4 —————–(i)
and x + y = 36 —————–(ii)
Adding both the eqn. we get,
2x = 40
x = 20
put the value if x in eqn. (i) and we get,
2y = 32
y = 16
Length of the garden = 20 m and width = 16 m
问题15:两个补充角中的较大者会比较小者增加18度。找到他们
解决方案:
In two supplementary angles,
Let larger angle = x and smaller angle = y
x – y = 18°
But x + y = 180°
Adding we get,
2x = 198°
= x = 9°
put the value if x in eqn. (i) and we get,
2y = 162°
= y = 81°
Angles are 99° and 81°
问题16:2名女性和5名男性可以在4天内一起完成一件刺绣,而3名女性和6名男性可以在3天内完成一件刺绣。找出一个女人独自完成刺绣所花费的时间,以及一个男人独自完成该刺绣所花费的时间。
解决方案:
Let one woman can do the work in = x days and one man can do the same work = y days
1 women’s 1 days work = 1/x
and 1 men 1 day work = 1/y
2 women and 5 men 1 day work = 2/x + 5/y
then 4 days work = 4(2/x + 5/y) = 1
similarly 3 women and 6 men’s 1 day work = 3/x + 6/y
then 3 day’s work = 3(3/x + 6/y) = 1
= 8/x + 20/y = 1 ——————(i)
= 9/x + 18/y = 1 —————–(ii)
Multiply eqn. (i) by9 and (ii) by 10 we get,
x = 18
put the value of x in eqn. (i) and we get,
y = 36
Hence one women can do the work in 36 days and one men can do the same work in 18 days.
问题17.梅娜去银行提取了2000卢比。她要求收银员只给她50卢比和100卢比的钞票。 Meena总共收到25张音符。找出她收到了多少张纸币₹50和₹100。
解决方案:
Total amount with drawn = ₹ 2000
Let’s assume that the number of ₹ 50 notes = x and of ₹ 100 = y
According to given condition,
x + y = 25 ——————(i)
and 50x + 100y = 2000
= x + 2y = 40 ——————(ii) (Dividing by 50)
Subtracting eqn. (i) from (ii) and we get,
y = 15
Substituting the value of y in (i) and we get.
x + y = 25
x + 15 = 25
x = 25 – 15 = 10
Hence Number of 50 rupees notes 10 and number of 100 rupee notes 15.
问题18:有两个考试室A和B。如果从A到B派出10名考生,则每个房间中的学生人数是相同的。如果将20位候选人从B发送到A,则A中的学生人数是B中学生人数的两倍。找到每个房间的学生人数。
解决方案:
Let A examination room has students x and B room has students y.
According to given condition
x – 10 = y + 10
= x – y = 10 + 10
= x – y = 20 ——————–(i)
and x + 20 = 2 (y – 20)
= x + 20 = 2y – 40
= x – 2y = – 40 – 20 = – 60 ———————(ii)
Subtracting (ii) from (i) and we get,
y = 80
Substituting the value of y in (i) and we get,
x – 80 = 20
x = 20 + 80 = 100
In examination room A, the students are 100 and in B room 80.
问题19:一张铁路半票的价格是全价的一半,半票的预订费与全票的预订费相同。从孟买到艾哈迈达巴德的一张头等舱预订票价为216卢比,一张半价的头等舱预订票价为327卢比。基本头等舱全价票是多少,预订费是多少?
解决方案:
Let the rate of fare of full ticket ₹ x and rate of reservation ₹ y.
According to given condition,
x + y = 216 ——————(i)
and 3/2 x + 2y = 327 ————–(ii)
Multiply eqn. (i) by 4 we get,
x = 210
put the value of x in (i) and we get,
y = 6
Hence rate of fare of full ticket ₹210 and reservation charges per ticket is ₹6.
问题20:一个在神秘和神奇药物中具有神秘力量的巫师看到一只公鸡,继续战斗,私下与两只公鸡的主人交谈。他对一个人说:如果您的鸟获胜,那么您就给我您的赌注,但是如果您没有获胜,我将给您三分之二。在另一方,他以同样的方式答应给予四分之三。从他们两个人那里,他的收益将只有12金币。找出每个公鸡拥有者的钱本金。
解决方案:
Let the first owner of cock has state money x gold coins and second owner has y gold coins
According to given condition,
x – 3/4 y = 12 —————(i)
and y – 2/3 x = 12 —————-(ii)
Adding eqn. (i) and (ii) we get,
x = 42
put the value of x in eqn. (i) and we get,
y = 40
Hence the first owner has 42 gold coin and second owner has 40 gold coin.
问题21:班上的学生排成一排。如果连续多出3个学生,则少1个学生。如果连续少3个学生,则多2个行。找到班上的学生人数。
解决方案:
Let number of rows x and number of students in each row y,
According to given condition,
Number of total students = x * y
(y + 3)(x – 1) = xy
= xy – y + 3x – 3 = xy
= 3x – y = 3 …(i)
and (y – 3) (x + 2) = xy
= xy + 2y – 3x – 6 = xy
= 2y – 3x = 6 …(ii)
Adding (i) and (ii) and we get,
y = 9
Substituting the value of y in (i) and we get,
3x – 9 = 3
= 3x = 3 + 9 = 12
= x = 4
Number of total students = x * y = 4 x 9 = 36
问题22:一个人说:“给我一百个朋友!然后,我将成为你的两倍。”另一个回答是:“如果给我十分,我的财富将是您的六倍”。告诉我他们各自的资本额是多少?
解决方案:
Let’s first person has amount of money ₹x and second person has ₹y
According to given condition,
x + 100 = 2(y – 100)
= x + 100 = 2y – 200
= x – 2y = – 200 – 100
= x – 2y = -300 ………(i)
and 6(x – 10) = (y + 10)
= 6x – 60 = y + 10
= 6x – y = 10 + 60
= 6x – y = 70 ……….(ii)
From (i) we get,
x = 2y – 300
Substituting the value of x in (ii) and we get,
6(2y – 300) – y = 70
= 12y – 1800 – y = 70
= 11y = 70 + 1800 = 1870
y = 170
x = 2y – 300 = 2 x 170 -300 = 340 – 300 = 40
Hence first person has money ₹ 40 and second person has ₹ 170.
问题23:一个店主以8%的利润出售一件纱丽,以10%的折扣出售一件毛衣,因此获得₹1008。如果她以10%的利润出售这套纱丽,以8%的折扣出售一件毛衣,她将得到₹1028。找到纱丽的成本价和毛衣的标价(折扣前的价格)。 [NCERT示例]
解决方案:
Let the cost price of the saree and the list price of the sweater be ₹ x and ₹ y, respectively.
Case 1: Sells a saree at 8% profit + Sells a sweater at 10% discount = ₹1008
= (100+8)% of x + (100-10)% of y = 1008
= 1.08x + 0.9y = 1008 —————-(i)
Case 2: Sold the saree at 10% profit + Sold the sweater at 8% discount = ₹1028
= (100+10)% of x + (100 – 8)% of y=1028
= 1.1x + 0.92y =1028 —————–(ii)
put the value of y in eqn. (i) we get,
x = 600
put the value of x in eqn. (i) and we get,
y = 400
Hence the cost price of the saree and the list price of sweater are ₹600 and ₹400 respectively.
问题24.在竞争性考试中,每个正确答案将获得一个分数,而每个错误答案将被扣除12个分数。 Jayanti回答了120个问题,并获得90分。她正确回答了多少个问题。 [NCERT示例]
解决方案:
Let x be the number of correct answer of the questions in a competitive examination,
then (120 – x) be the number of wrong answers of the questions.
According to given condition,
= x – (120 – x) * 1/2 = 90
= 3x / 2 = 150
= x = 100
Hence, Jayanti answered correctly 100 questions.
问题25.一位店主给书店租书以供阅读。她在前两天收取固定费用,此后每天收取一笔额外费用。 Latika为保留6天的书支付₹22,而Anand为保留四天的书支付₹16。查找固定费用,然后每增加一天收费一次。 [NCERT示例]
解决方案:
Let Latika takes a fixed charge for the first two day is ₹ x and additional charge for each day thereafter is ₹ y.
According to given condition,
Latika paid ₹ 22 for a book kept for six days
i.e., x + 4y = 22 ——————-(i)
and by second condition,
Anand paid ₹ 16 for a book kept for four days
i.e., x + 2y = 16 ———————-(ii)
Now, subtracting Eq. (ii) from Eq. (i), we get
2y = 6 ⇒ y = 3
On putting the value of y in Eq. (ii), we get
x + 2 x 3 = 16
x = 16 – 6 = 10
Hence, the fixed charge ₹ 10 and the charge for each extra day ₹ 3.