第九章一个变量中的线性方程–练习9.4 |套装1
问题14.我现在的年龄是儿子的5倍。在6年的时间里,我的年龄将是他那时的三倍。我们现在几岁了?
解决方案:
Let assume that present son’s age will be x years
Present father’s age will be 5x years
Son’s age after 6 years will be (x + 6) years
Fathers’ age after 6 years will be (5x + 6) years
5x + 6 = 3(x + 6)
5x + 6 = 3x + 18
5x – 3x = 18 – 6
2x = 12
x = 12/2= 6
Noe ,present son’s age will be x = 6years
And Present father’s age will be 5x = 5(6) = 30years
问题15.我有1000卢比的十卢比和五卢比的钞票。如果我的十卢比钞票的数量比五卢比钞票的数量多十,那么每个面额有多少张钞票?
解决方案:
Let us assume that the number of five rupee notes be x
Number of ten rupee notes will be (x + 10)
Amount due to five rupee notes will be = 5 × x = 5x
Amount due to ten rupee notes will be = 10 (x + 10) = 10x + 100
The total amount = Rs 1000
5x + 10x +100 = 1000
15x = 900
x = 900/15 = 60
The number of five rupee notes are x = 60
Number of ten rupee notes are x + 10 = 60+10 = 70
问题16.在聚会上,向客人提供了可乐,南瓜和果汁。四分之一的客人喝可乐,三分之一的南瓜,五分之二的人喝果汁,只有三人什么都不喝。总共有几位客人?
解决方案:
Let assume that number of guests will be x
Let The given details are Number of guests who drank colas are x/4
Let Number of guests who drank squash are x/3
Let Number of guests who drank fruit juice are 2x/5
Let Number of guests who did not drink anything are 3
x/4 + x/3 + 2x/5 + 3 = x
By take LCM for 4, 3 and 5 is 60
(15x+20x+24x-60x)/60 = -3
By doing cross-multiply,
(15x+20x+24x-60x) = -3(60)
-x = -180
x = 180
The total number of guests in all were 180.
问题17.测验中有180个多项选择题。如果候选人的每个正确答案和每个未尝试或错误回答的问题均获得4分,则从正确答案的总分中扣除1分。如果考生在考试中获得450分,他正确回答了多少个问题?
解决方案:
Let assume that number of correct answers be x
Number of questions answered wrong are (180 – x) (let)
Total score when answered right = 4x (let)
Marks deducted when answered wrong = 1(180 – x) = 180 – x
4x – (180 – x) = 450
4x – 180 + x = 450
5x = 450 + 180
5x = 630
x = 630/5 = 126
问题18.一个工人工作20天,条件是他每天会收到60卢比,他会工作,每天缺勤会被罚款5卢比。如果他总共缺席多少天,共收到745卢比,他仍然缺席吗?
解决方案:
Let us assume that number of absent days as x
Let the number of present days is (20 – x)
Wage for one day work will be Rs 60
Fine for absent day will be Rs 5
60(20 – x) – 5x = 745
1200 – 60x – 5x = 744
-65x = 744-1200
-65x = -456
x = -456/-65 = 7
问题19:Ravish有三个盒子,总重量为60½Kg。盒子B的重量比盒子A重3.5公斤,盒子C的重量比盒子B重5 1/3公斤。找到盒子A的重量。
解决方案:
The given details are total weight of three boxes is 60 ½ kg = 121/2 kg
Let assume that the weight of box A is x kg
Weight of box B is x + 7/2 kg
Weight of box C is x + 7/2 + 16/3 kg
x + x + 7/2 + x + 7/2 + 16/3 = 121/2
3x = 121/2 – 7/2 – 7/2 – 16/3
Now take LCM for 2 and 3 is 6
3x = (363 – 21 – 21 – 32)/6
3x = 289/6
x = 289/18
问题20.有理数的分子比分母小3。如果分母增加5,分子增加2,我们将得到有理数1/2。找到有理数。
解决方案:
Let assume that denominator be x
The numerator be (x – 3)
We know that formula
Fraction = numerator/denominator
= (x – 3)/x
when Numerator is increased by 2
Denominator is increased by 5, then fraction is ½
(x – 3 + 2)/(x + 5) = 1/2
(x – 1)/(x + 5) = 1/2
By doing cross-multiplication
2(x – 1) = x + 5
2x – 2 = x + 5
2x – x = 2 + 5
x = 7
Denominator is x = 7, numerator is (x – 3) = 7 – 3 = 4
And the fraction is numerator/denominator = 4/7
问题21.在一个有理数中,分子的两倍比分母多2;如果将分子和分母加3,则分母增加2。新的分数是2/3。查找原始号码。
解决方案:
Let us assume the numerator be x
The denominator be (2x – 2)
According formula
Fraction = numerator/denominator
= x / (2x – 2)
So, the numerator and denominator are increased by 3, then fraction is 2/3
(x + 3)/(2x – 2 + 3) = 2/3
(x + 3)/(2x + 1) = 2/3
By doing cross-multiplying
3(x + 3) = 2(2x + 1)
3x + 9 = 4x + 2
3x – 4x = 2 – 9
-x = -7
x = 7
The numerator is x = 7, denominator is (2x – 2) = (2(7) – 2) = 14-2 = 12
And the fraction is numerator/denominator = 7/12
问题22.两个站之间的距离是340公里。两列火车从这些车站的平行轨道上同时开始,相互交叉。其中一个的速度比另一个速度快5 km / hr。如果两列火车在启动2小时后的距离为30公里,请找出每列火车的速度。
解决方案:
Let assume the speed of one train be x km/hr.
Let Speed of other train be (x + 5) km/hr.
Total distance between two stations = 340 km
As we know the formula
Distance = speed × time
Distance covered by one train in 2 hrs. = x × 2 = 2x km
Distance covered by other train in 2 hrs. = 2(x + 5) = (2x + 10) km
Distance between the trains is 30 km
2x + 2x + 10 + 30 = 340
4x + 40 = 340
4x = 340 – 40
4x = 300
x = 300/4 = 75
The speed of one train is 75 km/hr.
Speed of other train is (x + 5) = 75 + 5 = 80 km/hr.
问题23:蒸笼在9个小时内从一个点流到另一个点。它在10小时内覆盖了上游相同的距离。如果水流的速度为1 km / hr。,则求出蒸锅在静止水中的速度以及端口之间的距离。
解决方案:
Let us assume the speed of steamer be x km/hr.
Speed of stream = 1 km/hr. (According to question)
Let Downstream speed = (x + 1) km/hr.
Let Upstream speed = (x – 1) km/hr.
According to formula
Distance = speed × time
= (x + 1) × 9
= (x – 1) × 10
9x + 9 = 10x – 10
9x – 10x = -10 -9
-x = -19
x = 19 km/hr.
The speed of the steamer in still water 19 km/hr.
Distance between the ports is 9(x + 1) = 9(19+1) = 9(20) = 180 km.
问题24. Bhagwanti继承了12000.00卢比。她将其中一部分投资为10%,其余为12%。她从这些投资中获得的年收入为1280.00卢比。
解决方案:
At rate of 10% Let the investment be Rs x (let)
At the rate of 12% the investment will be Rs (12000 – x) (let)
At 10% of rate the annual income will be x × (10/100) = 10x/100 (let)
At 12% of rate the annual income will be (12000 – x) × 12/100 = (144000 – 12x)/100
Total investment = 1280
10x/100 + (144000 – 12x)/100 = 1280
(10x + 144000 – 12x)/100 = 1280
(144000 – 2x)/100 = 1280
By doing cross-multiply
144000 – 2x = 1280(100)
-2x = 128000 – 144000
-2x = -16000
x = -16000/-2= 8000
At 10% of rate she invested Rs 8000 and at 12% of rate she invested Rs (12000 – x) = Rs (12000 – 8000) = Rs 4000
问题25.矩形的长度超出其宽度9厘米。如果长度和宽度分别增加3厘米,则新矩形的面积将比给定矩形的面积多84平方厘米。找到给定矩形的长度和宽度。
解决方案:
Let assume that breadth of the rectangle be x meter
Length of the rectangle be (x + 9) meter (let)
Area of the rectangle length × breadth = x(x +9) m2 (As we know the formula)
When length and breadth increased by 3cm
New length will be x + 9 + 3 = x + 12
New breadth will be x + 3
Area =(x + 12) (x + 3) = x (x + 9) + 84
x2 + 15x + 36 = x2 + 9x + 84
15x – 9x = 84 – 36
6x = 48
x = 48/6 = 8
Length of the rectangle (x + 9) = (8 + 9) = 17cm
breadth of the rectangle is 8cm.
问题26.阿努普和他父亲的年龄之和为100。如果阿努普现在与父亲一样大,他的年龄将是其儿子阿努吉现在的五倍。当Anup和他父亲一样大时,Anuj将比现在的Anup大八岁。他们现在几岁了?
解决方案:
Let assume that age of Anup be x years
Age of Anup’s father will be (100 – x) years
The age of Anuj will be (100-x)/5 years
Anup is as old as his father after (100 – 2x) years
Anuj’s age = present age of his father (Anup) + 8
Present age of Anuj + 100 – 2x = Present age of Anup + 8
(100 – x)/5 + (100 – 2x) = x + 8
(100-x)/5 – 3x = 8 – 100
(100 – x – 15x)/5 = -92
By doing cross-multiplication,
100 – 16x = -460
-16x = -460 – 100
-16x = -560
x = -560/-16 = 35
Present age of Anup is 35 years then, Age of Anup’s father will be (100-x) = 100-35 = 65 years
The age of Anuj is (100-x)/5 = (100 – 35)/5 = 65/5 = 13 years
问题27.一位女士去购物,花了一半的钱用来购买手帕,并给在商店外等候的乞讨者卢比兑换了卢比。她把剩下的一半都用在了午餐上,然后付了两卢比的小费。她把剩下的一半花在书本上,花了三卢比买公交车。到家后,她发现只剩一卢比。她开始花了多少钱?
解决方案:
Let assume that amount lady had be Rs x
Amount spent for hankies and given to begger is x/2 + 1 (let)
Remaining amount is x – (x/2 + 1) = x/2 – 1 = (x-2)/2 (let)
Amount spent for lunch (x-2)/2×1/2 = (x-2)/4
Amount given as tip is Rs 2
Remaining amount after lunch = (x-2)/2 – (x-2)/4 – 2 = (2x – 4 – x + 2 – 8)/4 = (x – 10)/4
Amounts spent for books =1/2 × (x-10)/4 = (x-10)/8
Bus fare is Rs 3
Amount left = (x-10)/4 – (x-10)/8 – 3 = (2x – 20 – x + 10 – 24)/8 = (x-34)/8
According to the question we know that the amount left = Rs 1
(x-34)/8 = 1
By cross-multiplication
x – 34 = 8
x = 8 + 34 = 42
The lady started with Rs. 42