arithmetic aptitude problems · web view7. in kiran's opinion, his weight is greater than 65 kg but...
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Arithmetic Aptitude Problems
Arithmetic Aptitude Problems
SAMBHRAM INSTITUTE OF TECHNOLOGY
“APTITUDE PROBLEMS AND SOLUTION”
Department of Master of Computer Applications
Prepared by Namratha K Asst. Professor
Contents:
1. Problems on Age2
2. Problem on Average20
3. Problems on Banker Discount Problem38
4. Problems on Partnerships60
5. Problems on Percentage79
6. Problems on Pipes and Cistern 106
7. Problems on Simple Interest 150
8. Problems on Time and Distance 162
9. Problems on Time and Work 184
10.Group Discussion208
Chapter 1: Problems on Age -
1. Ten years ago, P was half of Q's age. If the ratio of their present ages is 3:43:4, what will be the total of their present ages?
A. 4545
B. 4040
C. 3535
D. 3030
Answer: Option C
Explanation:
Let present age of P and Q be 3x3x and 4x4x respectively.
Ten years ago, P was half of Q's age⇒(3x−10)=12(4x−10)⇒6x−20=4x−10⇒2x=10⇒x=5⇒(3x−10)=12(4x−10)⇒6x−20=4x−10⇒2x=10⇒x=5
Total of their present ages=3x+4x=7x=7×5=35=3x+4x=7x=7×5=35
2. Father is aged three times more than his son Sunil. After 88 years, he would be two and a half times of Sunil's age. After further 88 years, how many times would he be of Sunil's age?
A. 44 times
B. 55 times
C. 22 times
D. 33 times
Answer: Option C
Explanation:
Assume that Sunil's present age =x=x.Then, father's present age =3x+x=4x=3x+x=4x
After 88 years, father's age =212=212 times of Sunils' age⇒(4x+8)=212(x+8)⇒4x+8=52(x+8)⇒8x+16=5x+40⇒3x=40−16=24⇒x=243=8⇒(4x+8)=212(x+8)⇒4x+8=52(x+8)⇒8x+16=5x+40⇒3x=40−16=24⇒x=243=8
After further 88 years,Sunil's age =x+8+8=8+8+8=24=x+8+8=8+8+8=24Father's age =4x+8+8=4×8+8+8=48=4x+8+8=4×8+8+8=48
Father's age/Sunil's age =4824=2=4824=2
3. A man's age is 125%125% of what it was 1010 years ago, but 8313%8313% of what it will be after 1010years. What is his present age?
A. 7070
B. 6060
C. 5050
D. 4040
Answer: Option C
Explanation:
Let the age before 1010 years =x=x. Then,
125x100=x+10⇒125x=100x+1000⇒x=100025=40125x100=x+10⇒125x=100x+1000⇒x=100025=40
Present age =x+10=40+10=50=x+10=40+10=50
4. A man is 2424 years older than his son. In two years, his age will be twice the age of his son. What is the present age of his son?
A. 2323 years
B. 2222 years
C. 2121 years
D. 2020 years
Answer: Option B
Explanation:
Let present age of the son =x=x yearsThen, present age the man =(x+24)=(x+24) years
Given that, in 22 years, man's age will be twice the age of his son⇒(x+24)+2=2(x+2)⇒x=22⇒(x+24)+2=2(x+2)⇒x=22
5. Present ages of Kiran and Syam are in the ratio of 5:45:4 respectively. Three years hence, the ratio of their ages will become 11:911:9 respectively. What is Syam's present age in years?
A. 2828
B. 2727
C. 2626
D. 2424
Answer: Option D
Explanation:
Ratio of the present age of Kiran and Syam =5:4=5:4
Let present age of Kiran =5x=5xPresent age of Syam =4x=4x
After 33 years, ratio of their ages =11:9=11:9⇒(5x+3):(4x+3)=11:9⇒9(5x+3)=11(4x+3)⇒45x+27=44x+33⇒x=33−27=6⇒(5x+3):(4x+3)=11:9⇒9(5x+3)=11(4x+3)⇒45x+27=44x+33⇒x=33−27=6
Syam's present age =4x=4×6=24
6. The sum of ages of 55 children born at the intervals of 33 years each is 5050 years. Find out the age of the youngest child?
A. 66 years
B. 55 years
C. 44 years
D. 33 years
Answer: Option C
Explanation:
Let the age of the youngest child =x=x
Then, the ages of 55 children can be written as x,(x+3),(x+6),(x+9)x,(x+3),(x+6),(x+9) and (x+12)(x+12)
x+(x+3)+(x+6)+(x+9)x+(x+3)+(x+6)+(x+9) +(x+12)=50+(x+12)=50⇒5x+30=50⇒5x=20⇒x=205=4⇒5x+30=50⇒5x=20⇒x=205=4
7. A is two years older than B who is twice as old as C. The total of the ages of A, B and C is2727. How old is B?
A. 1010
B. 99
C. 88
D. 77
Answer: Option A
Explanation:
Let age of C =x=x. Then,Age of B =2x=2xAge of A =2+2x=2+2x
Total age of A,B and C =27=27⇒(2+2x)+2x+x=27⇒5x=25⇒x=255=5⇒(2+2x)+2x+x=27⇒5x=25⇒x=255=5
B's age =2x=2×5=10=2x=2×5=10
8. The average age of a class of 2222 students is 2121 years. The average increased by 11 when the teacher's age also included. What is the age of the teacher?
A. 4848
B. 4545
C. 4343
D. 4444
Answer: Option D
Explanation:
Solution 1
Total age of all students =22×21=22×21
Total age of all students + Age of the teacher =23×22=23×22
Age of the teacher=23×22−22×21=22(23−21)=22×2=44=23×22−22×21=22(23−21)=22×2=44
Solution 2
If age of the teacher was 2121, average would not have changed.
Since average increased by 11,Age of the teacher =21+23×1=44=21+23×1=44
9. A father said to his son, "I was as old as you are at the present at the time of your birth". If the father's age is 3838 years now, what was the son's age five years back?
A. 2020 years
B. 1818 years
C. 1414 years
D. 2222 years
Answer: Option C
Explanation:
Let son's present age be xx years. Then,
(38−x)=x⇒2x=38⇒x=382=19(38−x)=x⇒2x=38⇒x=382=19
Son's age 55 years back =19−5=1419−5=14
10. Ayisha's age is 1/6th1/6th of her father's age. Ayisha's father's age will be twice Shankar's age after 1010 years. If Shankar's eight birthdays was celebrated two years before, then what is Ayisha's present age.
A. 1010 years
B. 1212 years
C. 88 years
D. 55 years
Answer: Option D
Explanation:
Solution 1
Let Ayisha's present age =x=xThen, her father's age =6x=6x
Given that Ayisha's father's age will be twice Shankar's age after 1010 years.Therefore, Shankar's age after 1010 years =12(6x+10)=3x+5=12(6x+10)=3x+5
Also given that Shankar's eight birthdays was celebrated two years before.Therefore, Shankar's age after 1010 years =8+12=20=8+12=20
3x+5=20⇒x=153=53x+5=20⇒x=153=5
Therefore, Ayisha's present age =5=5 years
Solution 2
Shankar's present age =8+2=10=8+2=10
Ayisha's father's age after 1010 years =2(10+10)=40=2(10+10)=40Ayisha's father's present age =40−10=30=40−10=30
Ayisha's present age =16×30=5
11. The sum of the present ages of a son and his father is 6060 years. Six years ago, father's age was five times the age of the son. After 66 years, what will be son's age?
A. 2323 years
B. 2222 years
C. 2121 years
D. 2020 years
Answer: Option D
Explanation:
Let the present age of the son =x=xThen, present age of the father =(60−x)=(60−x)
Six years ago father's age was 55 times the age of the son⇒(60−x)−6=5(x−6)⇒84=6x⇒x=846=14⇒(60−x)−6=5(x−6)⇒84=6x⇒x=846=14
Son's age after 66 years=x+6=14+6=20=x+6=14+6=20
12. Kiran is younger than Bineesh by 77 years and their ages are in the respective ratio of 7:9.7:9. How old is Kiran?
A. 2525
B. 24.5
C. 2424
D. 23.523.5
Answer: Option B
Explanation:
Solution 1
Let age of Kiran and Bineesh be 7x7x and 9x9x respectively
7x=9x−7x=72=3.57x=9x−7x=72=3.5
Kiran's age =7x=7×3.5=24.5=7x=7×3.5=24.5
Solution 2
Kiran's age : Bineesh's age =7:9=7:9
9−7=29−7=222 units =7 =7 (∵ difference in age is 77)⇒7⇒7 units =7×72=24.5=7×72=24.5
i.e., Kiran's age =24.5=24.5
13. Six years ago, the ratio of the ages of Vimal and Saroj was 6:56:5. Four years hence, the ratio of their ages will be 11:1011:10. What is Saroj's age at present?
A. 1818
B. 1717
C. 1616
D. 1515
Answer: Option C
Explanation:
Given that, six years ago, the ratio of the ages of Vimal and Saroj =6:5=6:5
Hence we can assume thatage of Vimal six years ago =6x=6xage of Saroj six years ago =5x=5x
After 44 years, the ratio of their ages =11:10=11:10⇒(6x+10):(5x+10)=11:10⇒10(6x+10)=11(5x+10)⇒5x=10⇒x=105=2⇒(6x+10):(5x+10)=11:10⇒10(6x+10)=11(5x+10)⇒5x=10⇒x=105=2
Saroj's present age=(5x+6)=5×2+6=16=(5x+6)=5×2+6=16
14. At present, the ratio between the ages of Shekhar and Shobha is 4:34:3. After 66 years, Shekhar's age will be 2626 years. Find out the age of Shobha at present?
A. 1515 years
B. 1414 years
C. 1313 years
D. 1212 years
Answer: Option A
Explanation:
After 66 years, Shekhar's age will be 2626 yearsTherefore, Present age of Shekhar =26−6=20=26−6=20
Let present age of Shobha =x=x
Then,20x=43x=20×34=1520x=43x=20×34=15
15. My brother is 33 years elder to me. My father was 2828 years of age when my sister was born while my mother was 2626 years of age when I was born. If my sister was 44 years of age when my brother was born, then what was the age of my father when my brother was born?
A. 3535 years
B. 3434 years
C. 3333 years
D. 3232 years
Answer: Option D
Explanation:
Solution 1
Let my age =x=x
ThenMy brother's age =x+3=x+3My mother's age =x+26=x+26My sister's age =(x+3)+4=x+7=(x+3)+4=x+7My father's age =(x+7)+28=x+35=(x+7)+28=x+35
Age of my father when my brother was born=x+35−(x+3)=32=x+35−(x+3)=32
Solution 2
Father's age was 2828 years when my sister was born.
My sister's age was 44 years when my brother was born.
Therefore, father's age was 28+4=3228+4=32 years when my brother was born.
16. The present ages of A,B and C are in proportions 4:7:94:7:9. Eight years ago, the sum of their ages was 5656. What are their present ages (in years)?
A. Insufficient data
B. 16,30,4016,30,40
C. 16,28,4016,28,40
D. 16,28,3616,28,36
Answer: Option D
Explanation:
Solution 1
Let present age of A,B and C be 4x,7x4x,7x and 9x9x respectively.
(4x−8)+(7x−8)+(9x−8)=56⇒20x=80⇒x=4(4x−8)+(7x−8)+(9x−8)=56⇒20x=80⇒x=4
Hence present age of A, B and C are4×4, 7×44×4, 7×4 and 9×49×4 respectively.i.e., 16,2816,28 and 3636 respectively.
Solution 2
Sum of their present ages =56+3×8=80=56+3×8=80
Present age of A =80×420=16=80×420=16
Present age of B =80×720=28=80×720=28
Present age of C =80×920=36=80×920=36
17. A person's present age is two-fifth of the age of his mother. After 88 years, he will be one-half of the age of his mother. What is the present age of the mother?
A. 6060
B. 5050
C. 4040
D. 3030
Answer: Option C
Explanation:
Solution 1
Let present age of the person =x=xThen, present age of the mother =5x2=5x2
Given that, after 88 years, the person will be one-half of the age of his mother.⇒(x+8)=12(5x2+8)⇒2x+16=5x2+8⇒x2=8⇒x=16⇒(x+8)=12(5x2+8)⇒2x+16=5x2+8⇒x2=8⇒x=16
Present age of the mother=5x2=5×162=40=5x2=5×162=40
Solution 2
Let present age of the mother =5x=5xThen, present age of the person =2x=2x
5x+8=2(2x+8)5x+8=4x+16x=85x+8=2(2x+8)5x+8=4x+16x=8present age of the mother =5x=40=5x=40
18. A is as much younger than B and he is older than C. If the sum of the ages of B and C is 5050 years, what is definitely the difference between B and A's age?
A. Data inadequate
B. 33 years
C. 22 years
D. 55 years
Answer: Option A
Explanation:
Age of C << Age of A << Age of B
Given that sum of the ages of B and C is 5050 years.
Now we need to find out (B's age - A's age). But this cannot be determined with the given data.
19. Sobha's father was 3838 years of age when she was born while her mother was 3636 years old when her brother four years younger to her was born. What is the difference between the ages of her parents?
A. 66 years
B. 55 years
C. 44 years
D. 33 years
Answer: Option A
Explanation:
Solution 1
Let Sobha's age =x=x andher brother's age =x−4=x−4
Sobha's father's age =x+38=x+38Sobha's mother's age =(x−4)+36=x+32=(x−4)+36=x+32
Sobha's father's age - Sobha's mother's age=(x+38)−(x+32)=6=(x+38)−(x+32)=6
Solution 2
Age of Sobha's father when Sobha was born =38=38
Age of Sobha's mother when Sobha was born =36−4=32=36−4=32
Required difference of age =38−32=6=38−32=6
20. The age of father 1010 years ago was thrice the age of his son. Ten years hence, father's age will be twice that of his son. What is the ratio of their present ages?
A. 7:37:3
B. 3:73:7
C. 9:49:4
D. 4:94:9
Answer: Option A
Explanation:
Let age of the son before 1010 years =x=x andage of the father before 1010 years =3x=3x
(3x+20)=2(x+20)⇒x=20(3x+20)=2(x+20)⇒x=20
Age of the son at present =x+10=20+10=30=x+10=20+10=30Age of the father at present =3x+10=3×20+10=70=3x+10=3×20+10=70
Required ratio =70:30=7:3
21. The ages of two persons differ by 1616 years. 66 years ago, the elder one was 33 times as old as the younger one. What are their present ages of the elder person?
A. 1010
B. 2020
C. 3030
D. 4040
Answer: Option C
Explanation:
Let present age of the elder person =x=x andpresent age of the younger person =x−16=x−16
(x−6)=3(x−16−6)⇒x−6=3x−66⇒2x=60⇒x=602=30(x−6)=3(x−16−6)⇒x−6=3x−66⇒2x=60⇒x=602=30
22. Present age of a father is 33 years more than three times the age of his son. Three years hence, father's age will be 1010 years more than twice the age of the son. What is father's present age?
A. 3030 years
B. 3131 years
C. 3232 yeas
D. 3333 years
Answer: Option D
Explanation:
Let the present age the son =x=xThen, present age of the father =3x+3=3x+3
Given that, three years hence, father's age will be 1010 years more than twice the age of the son⇒(3x+3+3)=2(x+3)+10⇒x=10⇒(3x+3+3)=2(x+3)+10⇒x=10
Father's present age=3x+3=3×10+3=33=3x+3=3×10+3=33
23. Kamal was 44 times as old as his son 88 years ago. After 88 years, Kamal will be twice as old as his son. Find out the present age of Kamal.
A. 4040 years
B. 3838 years
C. 4242 years
D. 3636 years
Answer: Option A
Explanation:
Let age of the son before 88 years =x=xThen, age of Kamal before 88 years ago =4x=4x
After 88 years, Kamal will be twice as old as his son⇒4x+16=2(x+16)⇒x=8⇒4x+16=2(x+16)⇒x=8
Present age of Kamal=4x+8=4×8+8=40=4x+8=4×8+8=40
24. If 66 years are subtracted from the present age of Ajay and the remainder is divided by 1818, then the present age of Rahul is obtained. If Rahul is 22 years younger to Denis whose age is 55 years, then what is Ajay's present age?
A. 5050 years
B. 6060 years
C. 5555 years
D. 6262 years
Answer: Option B
Explanation:
Present age of Denis =5=5 yearsPresent age of Rahul =5−2=3=5−2=3
Let present age of Ajay =x=xThen, present age of Rahul =x−618=x−618
x−618=3⇒x−6=3×18=54⇒x=54+6=60x−618=3⇒x−6=3×18=54⇒x=54+6=60
25. The ratio of the age of a man and his wife is 4:34:3. At the time of marriage the ratio was5:35:3 and After 44 years this ratio will become 9:79:7. How many years ago were they married?
A. 88 years
B. 1010 years
C. 1111 years
D. 1212 years
Answer: Option D
Explanation:
Let the present age of the man and his wife be 4x4x and 3x3x respectively.
After 44 years this ratio will become 9:79:7⇒(4x+4):(3x+4)=9:7⇒7(4x+4)=9(3x+4)⇒28x+28=27x+36⇒x=8⇒(4x+4):(3x+4)=9:7⇒7(4x+4)=9(3x+4)⇒28x+28=27x+36⇒x=8
Present age of the man =4x=4×8=32=4x=4×8=32Present age of his wife =3x=3×8=24=3x=3×8=24
Assume that they got married before tt years. Then,(32−t):(24−t)=5:3⇒3(32−t)=5(24−t)⇒96−3t=120−5t⇒2t=24⇒t=242=12
26. The product of the ages of Syam and Sunil is 240240. If twice the age of Sunil is more than Syam's age by 44 years, what is Sunil's age?
A. 1616
B. 1414
C. 1212
D. 1010
Answer: Option C
Explanation:
Let age of Sunil =x=x andage of Syam =y=y
xy=240 ⋯(1)xy=240 ⋯(1)
2x=y+4⇒y=2x−4⇒y=2(x−2) ⋯(2)2x=y+4⇒y=2x−4⇒y=2(x−2) ⋯(2)
Substituting equation (2)(2) in equation (1)(1). We getx×2(x−2)=240⇒x(x−2)=2402⇒x(x−2)=120 ⋯(3)x×2(x−2)=240⇒x(x−2)=2402⇒x(x−2)=120 ⋯(3)
We got a quadratic equation to solve.
Always time is precious and objective tests measure not only how accurate you are but also how fast you are. We can solve this quadratic equation in the traditional way. But it is more easy to substitute the values given in the choices in the quadratic equation (equation 33) and see which choice satisfy the equation.
Here, option A is 1010. If we substitute that value in the quadratic equation, x(x−2)=10×8x(x−2)=10×8 which is not equal to 120120
Now try option B which is 1212. If we substitute that value in the quadratic equation, x(x−2)=12×10=120x(x−2)=12×10=120. See, we got that x=12x=12
Hence Sunil's age =12=12
(Or else, we can solve the quadratic equation by factorization as,x(x−2)=120⇒x2−2x−120=0⇒(x−12)(x+10)=0⇒x=12 or −10x(x−2)=120⇒x2−2x−120=0⇒(x−12)(x+10)=0⇒x=12 or −10Since xx is age and cannot be negative, x=12x=12
Or by using quadratic formula asx=−b±√b2−4ac2a=2±√(−2)2−4×1×(−120)2×1=2±√4+4802=2±√4842=2±222=12 or −10x=−b±b2−4ac2a=2±(−2)2−4×1×(−120)2×1=2±4+4802=2±4842=2±222=12 or −10
Since age is positive, x=12x=12
27. One year ago, the ratio of Sooraj's and Vimal's age was 6:76:7 respectively. Four years hence, this ratio would become 7:87:8. How old is Vimal?
A. 3232
B. 3434
C. 3636
D. 3838
Answer: Option C
Explanation:
Let the age of Sooraj and Vimal, 11 year ago, be 6x6x and 7x7x respectively.
Given that, four years hence, this ratio would become 7:87:8⇒(6x+5):(7x+5)=7:8⇒8(6x+5)=7(7x+5)⇒48x+40=49x+35⇒x=5⇒(6x+5):(7x+5)=7:8⇒8(6x+5)=7(7x+5)⇒48x+40=49x+35⇒x=5
Vimal's present age=7x+1=7×5+1=36=7x+1=7×5+1=36
28. The total age of A and B is 1212 years more than the total age of B and C. C is how many year younger than A?
A. 1010
B. 1111
C. 1212
D. 1313
Answer: Option C
Explanation:
Solution 1
Given that, A + B =12+=12+ B + C⇒ A - C =12=12Therefore, C is younger than A by 1212 years
Solution 2
Total age of A and B is 1212 years more than total age of B and C.
Since B is common in both sides, C is 1212 years younger than A.
29. Sachin's age after 1515 years will be 55 times his age 55 years back. Find out the present age of Sachin?
A. 1010 years
B. 1111 years
C. 1212 years
D. 1313 years
Answer: Option A
Explanation:
Let present age of Sachin =x.=x. Then,
(x+15)=5(x−5)⇒4x=40⇒x=10(x+15)=5(x−5)⇒4x=40⇒x=10
30. Sandeep's age after six years will be three-seventh of his father's age. Ten years ago the ratio of their ages was 1:51:5. What is Sandeep's father's age at present?
A. 3030 years
B. 4040 years
C. 5050 years
D. 6060 years
Answer: Option C
Explanation:
Let the age of Sandeep and his father before 1010 years be xx and 5x5x respectively.
Given that Sandeep's age after six years will be three-seventh of his father's age⇒x+16=37(5x+16)⇒7x+112=15x+48⇒8x=64⇒x=8⇒x+16=37(5x+16)⇒7x+112=15x+48⇒8x=64⇒x=8
Sandeep's father's present age=5x+10=5×8+10=50
Chapter 2: Problem on Average
1. In the first 10 overs of a cricket game, the run rate was only 3.2. What should be the run rate in the remaining 40 overs to reach the target of 282 runs?
A. 6.25
B. 5.5
C. 7.4
D. 5
Answer: Option A
Explanation:
Runs scored in the first 10 overs = 10 × 3.2 = 32Total runs = 282
Remaining runs to be scored = 282 - 32 = 250Remaining overs = 40
Run rate needed = 25040=6.2525040=6.25
2. A grocer has a sale of Rs. 6435, Rs. 6927, Rs. 6855, Rs. 7230 and Rs. 6562 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Rs. 6500?
A. 4800
B. 4991
C. 5004
D. 5000
Answer: Option B
Explanation:
Let the sale in the sixth month =x=x
Then 6435+6927+6855+7230+6562+x66435+6927+6855+7230+6562+x6 =6500=6500
=> 6435+6927+6855+7230+6562+x6435+6927+6855+7230+6562+x =6×6500=6×6500
=> 34009+x=3900034009+x=39000=> x=39000−34009=4991x=39000−34009=4991
3. The average of 20 numbers is zero. Of them, How many of them may be greater than zero, at the most?
A. 1
B. 20
C. 0
D. 19
Answer: Option D
Explanation:
Average of 20 numbers = 0
=> Sum of 20 numbers20=0Sum of 20 numbers20=0
=> Sum of 20 numbers = 0
Hence at the most, there can be 19 positive numbers. (Such that if the sum of these 19 positive numbers is x, 20th number will be -x)
4. The captain of a cricket team of 11 members is 26 years old and the wicket keeper is 3 years older. If the ages of these two are excluded, the average age of the remaining players is one year less than the average age of the whole team. Find out the average age of the team.
A. 23 years
B. 20 years
C. 24 years
D. 21 years
Answer: Option A
Explanation:
Number of members in the team = 11
Let the average age of of the team = xx
=> Sum of ages of all 11 members11=xSum of ages of all 11 members11=x
=> Sum of the ages of all 11 members = 11x11x
Age of the captain = 26Age of the wicket keeper = 26 + 3 = 29
Sum of the ages of 9 members of the team excluding captain and wicket keeper =11x−26−29=11x−55=11x−26−29=11x−55
Average age of 9 members of the team excluding captain and wicket keeper=11x−559=11x−559
Given that 11x−559=(x−1)11x−559=(x−1)
⇒11x−55=9(x−1)⇒11x−55=9x−9⇒2x=46⇒x=462=23 years⇒11x−55=9(x−1)⇒11x−55=9x−9⇒2x=46⇒x=462=23 years
5. The average monthly income of A and B is Rs. 5050. The average monthly income of B and C is Rs. 6250 and the average monthly income of A and C is Rs. 5200. What is the monthly income of A?
A. 2000
B. 3000
C. 4000
D. 5000
Answer: Option C
Explanation:
Let monthly income of A = amonthly income of B = b monthly income of C = c
a + b = 2 × 5050 .... (Equation 1)b + c = 2 × 6250 .... (Equation 2)a + c = 2 × 5200 .... (Equation 3)
(Equation 1) + (Equation 3) - (Equation 2)=> a + b + a + c - (b + c) = (2 × 5050) + (2 × 5200) - (2 × 6250)=> 2a = 2(5050 + 5200 - 6250)=> a = 4000
i.e., Monthly income of A = 4000
6. A car owner buys diesel at Rs.7.50, Rs. 8 and Rs. 8.50 per litre for three successive years. What approximately is the average cost per litre of diesel if he spends Rs. 4000 each year?
A. Rs. 8
B. Rs. 7.98
C. Rs. 6.2
D. Rs. 8.1
Answer: Option B
Explanation:
Total Cost = 4000 × 3
Total diesel used = 40007.5+40008+40008.540007.5+40008+40008.5
Average cost per litre of diesel=4000×3(40007.5+40008+40008.5)=3(17.5+18+18.5)=4000×3(40007.5+40008+40008.5)=3(17.5+18+18.5)
It is important how you proceed from this stage. Remember time is very important and if we solve this in the normal method, it may take lot of time. Instead, we can find out the approximate value easily and select the right answer from the given choices.
In this case, answer =3(17.5+18+18.5)≈3(18+18+18)≈3(38)≈8=3(17.5+18+18.5)≈3(18+18+18)≈3(38)≈8
We got that answer as approximately equal to 8. From the given choices, the answer can be 8 or 7.98 or 8.1 . But which one from these?
It is easy to figure out. We approximated the denominator,(17.5+18+18.5)(17.5+18+18.5) to 3838. However17.5+18.5=18−0.5+18+0.5=8+0.5+8−0.5(8−0.5)(8+0.5)=16(82−0.52) [∵(a−b)(a+b)=a2−b2]=1664−0.2517.5+18.5=18−0.5+18+0.5=8+0.5+8−0.5(8−0.5)(8+0.5)=16(82−0.52) [∵(a−b)(a+b)=a2−b2]=1664−0.25
i.e., 17.5+18.5=16(64−0.25)17.5+18.5=16(64−0.25)We know that 18+18=14=166418+18=14=1664
=> 17.5+18.5>18+1817.5+18.5>18+18
Early we had approximated the denominator to 3838. However, from the above mentioned equations, now we know that actual denominator is slightly greater than 3838. It means answer is slightly lesser than 8. Hence we can pick the choice 7.98 as the answer
Try to remember such relations between numbers which can save lot of time in calculations.
7. In Kiran's opinion, his weight is greater than 65 kg but less than 72 kg. His brother does not agree with Kiran and he thinks that Kiran's weight is greater than 60 kg but less than 70 kg. His mother's view is that his weight cannot be greater than 68 kg. If all are them are correct in their estimation, what is the average of different probable weights of Kiran?
A. 70 kg
B. 69 kg
C. 61 kg
D. 67 kg
Answer: Option D
Explanation:
Let Kiran's weight = x. Then
According to Kiran, 65 < x < 72 ----(equation 1)
According to brother, 60 < x < 70 ----(equation 2)
According to mother, x ≤≤ 68 ----(equation 3)
Given that equation 1,equation 2 and equation 3 are correct. By combining these equations,we can write as
65
8. The average weight of 16 boys in a class is 50.25 kg and that of the remaining 8 boys is 45.15 kg. Find the average weights of all the boys in the class.
A. 48.55
B. 42.25
C. 50
D. 51.25
Answer: Option A
Explanation:
Average weight of 16 boys = 50.25Total weight of 16 boys = 50.25 × 16
Average weight of remaining 8 boys = 45.15Total weight of remaining 8 boys = 45.15 × 8
Total weight of all boys in the class = (50.25 × 16)+ (45.15 × 8)Total boys = 16 + 8 = 24
Average weight of all the boys = (50.25×16)+(45.15×8)24(50.25×16)+(45.15×8)24=(50.25×2)+(45.15×1)3=(16.75×2)+15.05=33.5+15.05=48.55=(50.25×2)+(45.15×1)3=(16.75×2)+15.05=33.5+15.05=48.55
9. A library has an average of 510 visitors on Sundays and 240 on other days. What is the average number of visitors per day in a month of 30 days beginning with a Sunday?
A. 290
B. 304
C. 285
D. 270
Answer: Option C
Explanation:
In a month of 30 days beginning with a Sunday, there will be 4 complete weeks and another two days which will be Sunday and Monday.
Hence there will be 5 Sundays and 25 other days in a month of 30 days beginning with a Sunday
Average visitors on Sundays = 510Total visitors of 5 Sundays = 510 × 5
Average visitors on other days = 240Total visitors of other 25 days = 240 × 25
Total visitors = (510 × 5) + (240 × 25)Total days = 30
Average number of visitors per day =(510×5)+(240×25)30=(51×5)+(24×25)3=(17×5)+(8×25)=85+200=285=(510×5)+(240×25)30=(51×5)+(24×25)3=(17×5)+(8×25)=85+200=285
10. A student's mark was wrongly entered as 83 instead of 63. Due to that the average marks for the class got increased by 1212. What is the number of students in the class?
A. 45
B. 40
C. 35
D. 30
Answer: Option B
Explanation:
Let the total number of students = xx
The average marks increased by 1212 due to an increase of 83 - 63 = 20 marks.
But total increase in the marks = 12×x=x212×x=x2
Hence we can write asx2=20⇒x=20×2=40
11. A family consists of two grandparents, two parents and three grandchildren. The average age of the grandparents is 67 years, that of the parents is 35 years and that of the grandchildren is 6 years. The average age of the family is
A. 32273227 years
B. 31573157 years
C. 28172817 years
D. 30573057 years
Answer: Option B
Explanation:
Total age of the grandparents = 67 × 2Total age of the parents = 35 × 2Total age of the grandchildren = 6 × 3
Average age of the family =(67×2)+(35×2)+(6×3)7=134+70+187=2227=3157=(67×2)+(35×2)+(6×3)7=134+70+187=2227=3157
12. The average weight of A, B and C is 45 kg. If the average weight of A and B be 40 kg and that of B and C be 43 kg, what is the weight of B?
A. 3131 kg
B. 28122812 kg
C. 3232 kg
D. 30123012 kg
Answer: Option A
Explanation:
Let the weight of A, B and C are a,b and c respectively.
Average weight of A,B and C = 45a + b + c = 45 × 3 = 135 --- equation(1)
Average weight of A and B = 40a + b = 40 × 2 = 80 --- equation(2)
Average weight of B and C = 43b + c = 43 × 2 = 86 --- equation(3)
equation(2) + equation(3) - equation(1) => a + b + b + c - (a + b + c) = 80 + 86 - 135=> b = 80 + 86 -135 = 166 - 135 = 31
Weight of B = 31 Kg
13. If the average marks of three batches of 55, 60 and 45 students respectively is 50, 55, 60, what is the average marks of all the students?
A. 53.23
B. 54.68
C. 51.33
D. 50
Answer: Option B
Explanation:
Average marks of batch1 = 50Students in batch1 = 55Total marks of batch1 = 55 × 50
Average marks of batch2 = 55Students in batch2 = 60Total marks of batch2 = 60 × 55
Average marks of batch3 = 60Students in batch3 = 45Total marks of batch3 = 45 × 60
Total students = 55 + 60 + 45 = 160
Average marks of all the students =(55×50)+(60×55)+(45×60)160=275+330+27016=87516=54.68=(55×50)+(60×55)+(45×60)160=275+330+27016=87516=54.68
14. The average age of husband, wife and their child 3 years ago was 27 years and that of wife and the child 5 years ago was 20 years. What is the present age of the husband?
A. 40
B. 32
C. 28
D. 30
Answer: Option A
Explanation:
Let the present age of the husband = hPresent age of the wife = wPresent age of the child = c
3 years ago, average age of husband, wife and their child = 27=> Sum of age of husband, wife and their child before 3 years = 3 × 27 = 81=> (h-3) + (w-3) + (c-3) = 81=> h + w + c = 81 + 9 = 90 --- equation(1)
5 years ago, average age of wife and child = 20=> Sum of age of wife and child before 5 years = 2 × 20 = 40=> (w-5) + (c-5) = 40=> w + c = 40 + 10 = 50 --- equation(2)
Substituting equation(2) in equation(1)=> h + 50 = 90=> h = 90 - 50 = 40
i.e., Present age of the husband = 40
15. The average weight of 8 person's increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg. What is the weight of the new person?
A. 75 Kg
B. 50 Kg
C. 85 Kg
D. 80 Kg
Answer: Option C
Explanation:
Total increase in weight = 8 × 2.5 = 20
If xx is the weight of the new person, total increase in weight = x−65x−65=> 20 = xx - 65=> xx = 20 + 65 = 85
16. There are two divisions A and B of a class, consisting of 36 and 44 students respectively. If the average weight of divisions A is 40 kg and that of division b is 35 kg. What is the average weight of the whole class?
A. 38.25
B. 37.25
C. 38.5
D. 37
Answer: Option B
Explanation:
Total weight of students in division A = 36 × 40Total weight of students in division B = 44 × 35
Total students = 36 + 44 = 80
Average weight of the whole class =(36×40)+(44×35)80=(9×40)+(11×35)20=(9×8)+(11×7)4=72+774=1494=37.25=(36×40)+(44×35)80=(9×40)+(11×35)20=(9×8)+(11×7)4=72+774=1494=37.25
17. A batsman makes a score of 87 runs in the 17th inning and thus increases his averages by 3. What is his average after 17th inning?
A. 39
B. 35
C. 42
D. 40.5
Answer: Option A
Explanation:
Let the average after 17 innings = xTotal runs scored in 17 innings = 17x
Average after 16 innings = (x-3)Total runs scored in 16 innings = 16(x-3)
Total runs scored in 16 innings + 87 = Total runs scored in 17 innings=> 16(x-3) + 87 = 17x=> 16x - 48 + 87 = 17x=> x = 39
18. A student needed to find the arithmetic mean of the numbers 3, 11, 7, 9, 15, 13, 8, 19, 17, 21, 14 and x. He found the mean to be 12. What is the value of x?
A. 12
B. 5
C. 7
D. 9
Answer: Option C
Explanation:
3+11+7+9+15+13+8+19+17+21+14+x123+11+7+9+15+13+8+19+17+21+14+x12 =12=12
⇒137+x12=12 ⇒137+x=144 ⇒x=144−137=7⇒137+x12=12 ⇒137+x=144 ⇒x=144−137=7
19. Arun obtained 76, 65, 82, 67 and 85 marks (out in 100) in English, Mathematics, Chemistry, Biology and Physics. What is his average mark?
A. 53
B. 54
C. 72
D. 75
Answer: Option D
Explanation:
Average mark = 76+65+82+67+855=3755=7576+65+82+67+855=3755=75
20. Distance between two stations A and B is 778 km. A train covers the journey from A to B at 84 km per hour and returns back to A with a uniform speed of 56km per hour. Find the average speed of the train during the whole journey?
A. 69.0 km /hr
B. 69.2 km /hr
C. 67.2 km /hr
D. 67.0 km /hr
Answer: Option C
Explanation:
-------------------------------------------Solution 1-------------------------------------------If a car covers a certain distance at xx kmph and an equal distance at yy kmph. Then,average speed of the whole journey = 2xyx+y2xyx+y kmph.
By using the same formula, we can find out the average speed quickly.
Average speed =2×84×5684+56=2×84×56140=2×21×5635=2×3×565=3365=67.2=2×84×5684+56=2×84×56140=2×21×5635=2×3×565=3365=67.2
-------------------------------------------Solution 2 -------------------------------------------Though it is a good idea to solve the problems quickly using formulas, you should know the fundamentals too. Let's see how we can solve this problems using basics.
Let the distance between A and B = xx
Train travels from A to B at 84 km per hour Total time taken for traveling from A to B = distancespeed=x84distancespeed=x84
Train travels from B to A at 56 km per hour Total time taken for traveling from B to A = distancespeed=x56distancespeed=x56
Total distance travelled = x+x=2xx+x=2xTotal time taken = x84+x56x84+x56
Average speed =Total distance traveledTotal time taken=Total distance traveledTotal time taken
=2xx84+x56=2184+156 =2×84×5656+84=2×84×56140 =2×21×5635=2×3×565=3365=67.2=2xx84+x56=2184+156 =2×84×5656+84=2×84×56140 =2×21×5635=2×3×565=3365=67.2
21. The average age of boys in a class is 16 years and that of the girls is 15 years. What is the average age for the whole class?
A. 15
B. 16
C. 15.5
D. Insufficient Data
Answer: Option D
Explanation:
We do not have the number of boys and girls. Hence we cannot find out the answer.
22. The average age of 36 students in a group is 14 years. When teacher's age is included to it, the average increases by one. Find out the teacher's age in years?
A. 51 years
B. 49 years
C. 53 years
D. 50 years
Answer: Option A
Explanation:
average age of 36 students in a group is 14Sum of the ages of 36 students = 36 × 14
When teacher's age is included to it, the average increases by one=> average = 15Sum of the ages of 36 students and the teacher = 37 × 15
Hence teachers age= 37 × 15 - 36 × 14= 37 × 15 - 14(37 - 1)= 37 × 15 - 37 × 14 + 14= 37(15 - 14) + 14= 37 + 14= 51
23. The average of five numbers id 27. If one number is excluded, the average becomes 25. What is the excluded number?
A. 30
B. 40
C. 32.5
D. 35
Answer: Option D
Explanation:
Sum of 5 numbers = 5 × 27Sum of 4 numbers after excluding one number = 4 × 25
Excluded number= 5 × 27 - 4 × 25= 135 - 100 = 35
24. The batting average for 40 innings of a cricket player is 50 runs. His highest score exceeds his lowest score by 172 runs. If these two innings are excluded, the average of the remaining 38 innings is 48 runs. Find out the highest score of the player.
A. 150
B. 174
C. 180
D. 166
Answer: Option B
Explanation:
Total runs scored by the player in 40 innings = 40 × 50Total runs scored by the player in 38 innings after excluding two innings = 38 × 48Sum of the scores of the excluded innings = 40 × 50 - 38 × 48 = 2000 - 1824 = 176
Given that the scores of the excluded innings differ by 172. Hence let's takethe highest score as x + 172 and lowest score as x
Now x + 172 + x = 176=> 2x = 4=> x =42=42 = 2
Highest score = x + 172 = 2 + 172 = 174
25. The average score of a cricketer for ten matches is 38.9 runs. If the average for the first six matches is 42, what is the average for the last four matches?
A. 34.25
B. 36.4
C. 40.2
D. 32.25
Answer: Option A
Explanation:
Total runs scored in 10 matches = 10 × 38.9
Total runs scored in first 6 matches = 6 × 42
Total runs scored in the last 4 matches = 10 × 38.9 - 6 × 42
Average of the runs scored in the last 4 matches = 10×38.9−6×42410×38.9−6×424=389−2524=1374=34.25=389−2524=1374=34.25
26. The average of six numbers is x and the average of three of these is y. If the average of the remaining three is z, then
A. None of these
B. x = y + z
C. 2x = y + z
D. x = 2y + 2z
Answer: Option C
Explanation:
Average of 6 numbers = x=> Sum of 6 numbers = 6x
Average of the 3 numbers = y=> Sum of these 3 numbers = 3y
Average of the remaining 3 numbers = z=> Sum of the remaining 3 numbers = 3z
Now we know that 6x = 3y + 3z=> 2x = y + z
27. Suresh drives his car to a place 150 km away at an average speed of 50 km/hr and returns at 30 km/hr. What is his average speed for the whole journey ?
A. 32.5 km/hr.
B. 35 km/hr.
C. 37.5 km/hr
D. 40 km/hr
Answer: Option C
Explanation:
-------------------------------------------Solution 1--------------------------------------------If a car covers a certain distance at x kmph and an equal distance at y kmph. Then, average speed of the whole journey = 2xyx+y2xyx+y kmph.
Therefore, average speed=2×50×3050+30=2×50×3080=2×50×38=50×34=25×32=752=37.5=2×50×3050+30=2×50×3080=2×50×38=50×34=25×32=752=37.5-------------------------------------------Solution 2--------------------------------------------Though it is a good idea to solve the problems quickly using formulas, you should know the fundamentals too. Let's see how we can solve this problems using basics.
Total time taken for traveling one side = distancespeed=15050distancespeed=15050Total time taken for return journey = distancespeed=15030distancespeed=15030Total time taken = 15050+1503015050+15030
Total distance travelled =150+150=2×150=150+150=2×150
Average speed = Total distance traveled Total time taken Total distance traveled Total time taken
=2×15015050+15030=2150+130 =2×50×3030+50=2×50×3080=2×50×38=50×34=25×32=752=37.5=2×15015050+15030=2150+130 =2×50×3030+50=2×50×3080=2×50×38=50×34=25×32=752=37.5
28. The average age of a husband and his wife was 23 years at the time of their marriage. After five years they have a one year old child. What is the average age of the family?
A. 21 years
B. 20 years
C. 18 years
D. 19 years
Answer: Option D
Explanation:
Total age of husband and wife (at the time of their marriage) = 2 × 23 = 46
Total age of husband and wife after 5 years + Age of the 1 year old child = 46 + 5 + 5 + 1 = 57
Average age of the family = 573573 = 19
29. In an examination, a student's average marks were 63. If he had obtained 20 more marks for his Geography and 2 more marks for his history, his average would have been 65. How many subjects were there in the examination?
A. 12
B. 11
C. 13
D. 14
Answer: Option B
Explanation:
Let the number of subjects = xThen, total marks he scored for all subjects = 63x
If he had obtained 20 more marks for his Geography and 2 more marks for his history, his average would have been 65=> Total marks he would have scored for all subjects = 65x
Now we can form the equation as 65x - 63x = additional marks of the student = 20 + 2 = 22=> 2x = 22=> x = 222222 = 11
30. The average salary of all the workers in a workshop is Rs.8000. The average salary of 7 technicians is Rs.12000 and the average salary of the rest is Rs.6000. How many workers are there in the workshop?
A. 21
B. 22
C. 23
D. 24
Answer: Option A
Explanation:
Let the number of workers = x
Given that average salary of all the workers = Rs.8000Then, total salary of all workers = 8000x
Given that average salary of 7 technicians is Rs.12000=> Total salary of 7 technicians = 7 × 12000 = 84000
Count of the rest of the employees = (x - 7)Average salary of the rest of the employees = Rs.6000Total salary of the rest of the employees = (x - 7)(6000)
8000x = 84000 + (x - 7)(6000)=> 8x = 84 + (x - 7)(6)=> 8x = 84 + 6x - 42=> 2x = 42=> x = 422422 = 21
Chapter 3: Problems on Banker Discount
Banker's DiscountA. Important Concepts - Banker's Discount
Assume that a merchant A purchases goods worth, say Rs.1000 from another merchant B at a credit of say 4 months.
Then B prepares a bill called bill of exchange (also called Hundi). On receipts of goods, A gives an agreement by signing on the bill allowing B to withdraw the money from A’s bank exactly after 4 months of the date of the bill.
The date exactly after 4 months is known as nominally due date. Three more days (called grace days) are added to this date to get a date known as legally due date.
The amount given on the bill is called the Face Value (F) which is Rs.1000 in this case.
Assume that B needs this money before the legally due date. He can approach a banker or broker who pays him the money against the bill, but somewhat less than the face value. The banker deducts the simple interest on the face value for the unexpired time. This deduction is known as Bankers Discount (BD). In another words, Bank Discount (BD) is the simple interest on the face value for the period from the date on which the bill was discounted and the legally due date.
The present value is the amount which, if placed at a particular rate for a specified period will amount to that sum of money at the end of the specified period. The interest on the present value is called the True Discount (TD). If the banker deducts the true discount on the face value for the unexpired time, he will not gain anything.
Banker’s Gain (BG) is the difference between banker’s discount and the true discount for the unexpired time.
Note: When the date of bill is not given, grace days are not to be added.
B. Important Formulas - Banker's Discount
Let F = Face Value of the Bill, TD = True Discount, BD = Bankers Discount, BG = Banker’s Gain, R = Rate of Interest, PW = True Present Worth and T = Time in Years
BD = Simple Interest on the face value of the bill for unexpired time = FTR100FTR100
PW = F1+T(R100)PW = F1+T(R100)
TD = Simple Interest on the present value for unexpired time =PW ×TR100=FTR100+TR=PW ×TR100=FTR100+TR
TD = BD×100100+TRBD×100100+TR
PW = F - TD
F = BD×TDBD – TDBD×TDBD – TD
BG = BD – TD = Simple Interest on TD = (TD)2PW(TD)2PW
TD = √PW×BGPW×BG
TD = BG×100TRBG×100TR
C. A Simple Example to understand the Basic Concepts of Banker's Discount
What is the present value, true discount, banker's discount and banker's gain on a bill of Rs.104500 due in 9 months at 6% per annum?
F = Rs. 104500
T = 9 months = 912912 years = 3434 years
R = 6%
Banker's Discount, BD =FTR100=FTR100=104500×34×6100=1045×34×6= Rs. 4702.50=104500×34×6100=1045×34×6= Rs. 4702.50
Present value, PW =F1+T(R100)=F1+T(R100)=1045001+(34)(6100)=Rs. 100000=1045001+(34)(6100)=Rs. 100000
True Discount, TD =PW ×TR100=PW ×TR100=100000×34×6100=1000×34×6=Rs. 4500=100000×34×6100=1000×34×6=Rs. 4500
Banker's Gain, BG = BD – TD = 4702.50 - 4500 = Rs.202.50
Problems on Banker's Discount - Solved Examples
1. The banker's discount on a bill due 4 months hence at 15% is Rs. 420. What is the true discount?
A. Rs. 410
B. Rs. 400
C. Rs. 390
D. Rs. 380
Answer: Option B
Explanation:
TD = BD ×100100+ TR=420×100100+(412×15) =420×100100+(13×15)=420×100100+5 =420×100105=84×10021 =4×100=400TD = BD ×100100+ TR=420×100100+(412×15) =420×100100+(13×15)=420×100100+5 =420×100105=84×10021 =4×100=400
2. The banker's discount on a certain amount due 2 years hence is 11101110 of the true discount. What is the rate percent?
A. 1%
B. 5%
C. 10%
D. 12%
Answer: Option B
Explanation:
Let TD = Rs. 1
Then BD = 1110×1= Rs. 11101110×1= Rs. 1110
T = 2
R = ?
F = BD ×TD(BD – TD) =(1110×1)(1110−1)=1110110= Rs. 11F = BD ×TD(BD – TD) =(1110×1)(1110−1)=1110110= Rs. 11
BD = FTR100FTR100
⇒1110=11×2×R100 ⇒110=22R ⇒R=11022=5%⇒1110=11×2×R100 ⇒110=22R ⇒R=11022=5%
3. The present worth of a sum due sometimes hence is Rs.5760 and the banker's gain is Rs.10. What is the true discount?
A. Rs. 480
B. Rs. 420
C. Rs. 120
D. Rs. 240
Answer: Option D
Explanation:
TD = √PW × BG=√5760×10=√57600 = Rs. 240TD = PW × BG=5760×10=57600 = Rs. 240
4. What is the banker's discount if the true discount on a bill of Rs.540 is Rs.90 ?
A. Rs. 108
B. Rs. 120
C. Rs. 102
D. Rs. 106
Answer: Option A
Explanation:
Present Worth, PW = F - TD = 540 - 90 = Rs. 450
Simple Interest on the Present Worth = True DiscountHence Simple Interest on 450 = 90 ------(Equation 1)
Simple Interest on the face value = Bankers Discount=> Simple Interest on 540 = Bankers Discount
From Equation 1, Simple Interest on 450 = 90 Hence, Simple Interest on 540 = 90450×540=540590450×540=5405 = Rs. 108=> Bankers Discount = Rs. 108
5. A bill for Rs. 3000 is drawn on 14th July at 5 months. It is discounted on 5th October at 10%. What is the Banker's Discount?
A. Rs. 60
B. Rs. 82
C. Rs. 90
D. Rs. 120
Answer: Option A
Explanation:
F = Rs. 3000
R = 10%
Date on which the bill is drawn = 14th July at 5 months
Nominally Due Date = 14th December
Legally Due Date = 14th December + 3 days = 17th December
Date on which the bill is discounted = 5th October
Unexpired Time = [6th to 31st of October] + [30 Days in November] + [1st to 17th of December]= 26 + 30 + 17 = 73 Days =73365 year=15 year=73365 year=15 year
BD = Simple Interest on the face value of the bill for unexpired time=FTR100=3000×15×10100 =30×15×10 = Rs. 60=FTR100=3000×15×10100 =30×15×10 = Rs. 60
6. The bankers discount and the true discount of a sum at 10% per annum simple interest for the same time are Rs.100 and Rs.80 respectively. What is the sum and the time?
A. Sum = Rs.400 and Time = 5 years
B. Sum = Rs.200 and Time = 2.5 years
C. Sum = Rs.400 and Time = 2.5 years
D. Sum = Rs.200 and Time = 5 years
Answer: Option C
Explanation:
BD = Rs.100
TD = Rs.80
R = 10%
F = BD ×TD(BD – TD)BD ×TD(BD – TD)
=100×80(100–80)=100×8020=Rs. 400=100×80(100–80)=100×8020=Rs. 400
BD = Simple Interest on the face value of the bill for unexpired time =FTR100=FTR100⇒100=400×T×10100 ⇒100=4×T×10 ⇒10=4×T ⇒T=104=2.5 years⇒100=400×T×10100 ⇒100=4×T×10 ⇒10=4×T ⇒T=104=2.5 years
7. The banker's gain on a sum due 6 years hence at 12% per annum is Rs. 540. What is the banker's discount?
A. 1240
B. 1120
C. 1190
D. 1290
Answer: Option D
Explanation:
TD = BG ×100TRTD = BG ×100TR
=540×1006×12=90×10012 =15×1002 = Rs. 750=540×1006×12=90×10012 =15×1002 = Rs. 750
BG = BD – TD=> 540 = BD - 750=> BD = 540 + 750 = 1290
8. The present worth of a certain bill due sometime hence is Rs. 1296 and the true discount is Rs. 72. What is the banker's discount?
A. Rs. 76
B. Rs. 72
C. Rs. 74
D. Rs. 4
Answer: Option A
Explanation:
BG = (TD)2PW=7221296 =72×721296=12×1236=123= Rs. 4BG = (TD)2PW=7221296 =72×721296=12×1236=123= Rs. 4
BG = BD – TD=> 4 = BD - 72=> BD = 72 + 4 = Rs. 76
9. The banker's discount of a certain sum of money is Rs. 36 and the true discount on the same sum for the same time is Rs. 30. What is the sum due?
A. Rs. 180
B. Rs. 120
C. Rs. 220
D. Rs. 200
Answer: Option A
Explanation:
F = BD ×TD(BD – TD)=36×30(36–30) =36×306=36×5 = Rs. 180F = BD ×TD(BD – TD)=36×30(36–30) =36×306=36×5 = Rs. 180
10. The banker's gain on a bill due 1 year hence at 10% per annum is Rs. 20. What is the true discount?
A. Rs. 200
B. Rs. 100
C. Rs. 150
D. Rs. 250
Answer: Option A
Explanation:
TD = BG ×100TR =20×1001×10 = Rs. 200TD = BG ×100TR =20×1001×10 = Rs. 200
11. The banker's gain of a certain sum due 3 years hence at 10% per annum is Rs. 36. What is the present worth ?
A. Rs. 400
B. Rs. 300
C. Rs. 500
D. Rs. 350
Answer: Option A
Explanation:
T = 3 years
R = 10%
TD = BG ×100TR =36×1003×10=12×10=Rs. 120TD = BG ×100TR =36×1003×10=12×10=Rs. 120
TD = PW × TR100⇒120=PW ×3×10100⇒1200=PW ×3PW =12003 = Rs. 400TD = PW × TR100⇒120=PW ×3×10100⇒1200=PW ×3PW =12003 = Rs. 400
12. The present worth of a certain sum due sometime hence is Rs. 3400 and the true discount is Rs. 340. The banker's gain is:
A. Rs. 21
B. Rs. 17
C. Rs. 18
D. Rs. 34
Answer: Option D
Explanation:
BG = (TD)2PW=(340)23400 =340×3403400=34010=Rs. 34BG = (TD)2PW=(340)23400 =340×3403400=34010=Rs. 34
13. The banker's discount on Rs. 1600 at 15% per annum is the same as true discount on Rs. 1680 for the same time and at the same rate. What is the time?
A. 3 months
B. 4 months
C. 5 months
D. 6 months
Answer: Option B
Explanation:
Bankers Discount, BD = Simple Interest on the face value of the bill for unexpired time.
True Discount, TD = Simple Interest on the present value for unexpired time.
Simple Interest on Rs. 1600 = True Discount on Rs.1680
=> Rs. 1600 is the Present Worth (PW) of Rs. 1680
=> Rs. 80 is the Simple Interest of Rs.1600 at 15%
⇒80=1600×T×15100⇒80=16×T×15⇒5=T×15⇒1=T×3⇒T=13 year=123 months=4 months⇒80=1600×T×15100⇒80=16×T×15⇒5=T×15⇒1=T×3⇒T=13 year=123 months=4 months
14. The banker's gain on a certain sum due 212212 years hence is 925925 of the banker's discount. What is the rate percent?
A. 1813%1813%
B. 1812%1812%
C. 2413%2413%
D. 2212%2212%
Answer: Option D
Explanation:
T = 212212 years = 5252 years
Let the banker's discount, BD = Rs. 1
Then, banker's gain, BG = 925×1 = Rs. 925925×1 = Rs. 925
BG = BD – TD⇒925=1 - TD⇒TD=1−925=1625⇒925=1 - TD⇒TD=1−925=1625
F = BD ×TDBD – TD =1×16251–1625=1625925= Rs. 169F = BD ×TDBD – TD =1×16251–1625=1625925= Rs. 169
BD = Simple Interest on the face value of the bill for unexpired time = FTR100FTR100⇒1=169×52×R100 ⇒100=169×52× R ⇒100=16×5×R9×2 ⇒100=8×5×R9 ⇒R = 100×98×5 =100×940=5×92=452=2212%⇒1=169×52×R100 ⇒100=169×52× R ⇒100=16×5×R9×2 ⇒100=8×5×R9 ⇒R = 100×98×5 =100×940=5×92=452=2212%
15. The banker's gain on a sum due 3 years hence at 12% per annum is Rs. 360. The banker's discount is:
A. Rs. 1360
B. Rs. 1000
C. Rs. 360
D. Rs. 640
Answer: Option A
Explanation:
BG = Rs. 360T = 3 yearsR = 12%
TD = BG ×100TR =360×1003×12=Rs. 1000TD = BG ×100TR =360×1003×12=Rs. 1000
BG = BD – TD=> BD = BG + TD = 360 + 1000 = Rs. 1360
16. The true discount on a certain sum due 6 months hence at 15% is Rs. 240. What is the banker's discount on the same sum for the same time at the same rate?
A. None of these
B. Rs. 278
C. Rs. 228
D. Rs. 258
Answer: Option D
Explanation:
TD = Rs. 240T = 6 months = 1212 yearR = 15%
TD = BG ×100TR⇒240=BG×100(12×15)⇒BG =240×15100×2=120×15100=Rs. 18TD = BG ×100TR⇒240=BG×100(12×15)⇒BG =240×15100×2=120×15100=Rs. 18
BG = BD – TD=> 18 = BD - 240=> BD = 18 + 240 = Rs. 258
17. A bill is discounted at 10% per annum. If banker's discount is allowed, at what rate percent should the proceeds be invested so that nothing will be lost?
A. 1019%1019%
B. 1119%1119%
C. 1111
D. 1029%1029%
Answer: Option B
Explanation:
Let the amount = Rs.100
Then BD = Rs.10 (∵ banker's discount, BD is the simple Interest on the face value of the bill for unexpired time and bill is discounted at 10% per annum)
Proceeds = Rs.100 – Rs.10 = Rs.90
Hence we should get Rs.10 as the interest of Rs.90 for 1 year so that nothing will be lost⇒10=90×1×R100⇒R=10×10090=1009=1119%⇒10=90×1×R100⇒R=10×10090=1009=1119%
18. A banker paid Rs.5767.20 for a bill of Rs.5840, drawn of Apr 4 at 6 months. If the rate of interest was 7%, what was the day on which the bill was discounted?
A. 3rd March
B. 3rd September
C. 3rd October
D. 3rd August
Answer: Option D
Explanation:
F = Rs.5840R = 7%BD = 5840 - 5767.20 = Rs.72.8
BD = FTR100⇒72.8=5840×T×7100⇒T=72.8×1007×5840=10.4×1005840=10405840=104584=1373 years=13×36573 days=65 daysBD = FTR100⇒72.8=5840×T×7100⇒T=72.8×1007×5840=10.4×1005840=10405840=104584=1373 years=13×36573 days=65 days
=> Unexpired Time = 65 days
Given that Date of Draw of the bill = 4th April at 6 months=> Nominally Due Date = 4th October=> Legally Due Date = (4th October + 3 days) = 7th October
Hence, The date on which the bill was discounted= (7th October - 65 days)= (7th October - 7 days in October - 30 days in September - 28 days in August)= 3rd August
19. The banker's discount on a sum of money for 3 years is Rs. 1116. The true discount on the same sum for 4 years is Rs. 1200. What is the rate percent?
A. 8%
B. 12%
C. 10%
D. 6%
Answer: Option D
Explanation:
BD for 3 years = Rs. 1116BD for 4 years = 11163×4 = Rs. 148811163×4 = Rs. 1488
TD for 4 years = Rs. 1200
F = BD ×TD(BD – TD)=1488×1200(1488–1200) =1488×1200288=124×120024 =124×1002=62×100=Rs. 6200F = BD ×TD(BD – TD)=1488×1200(1488–1200) =1488×1200288=124×120024 =124×1002=62×100=Rs. 6200
=> Rs.1488 is the simple interest on Rs. 6200 for 4 years
⇒1488=6200×4×R100⇒R=1488×1006200×4=372×1006200 =37262=6%⇒1488=6200×4×R100⇒R=1488×1006200×4=372×1006200 =37262=6%
20. The true discount on a bill of Rs. 2160 is Rs. 360. What is the banker's discount?
A. Rs. 432
B. Rs. 422
C. Rs. 412
D. Rs. 442
Answer: Option A
Explanation:
F = Rs. 2160TD = Rs. 360PW = F - TD = 2160 - 360 = Rs. 1800
True Discount is the Simple Interest on the present value for unexpired time=>Simple Interest on Rs. 1800 for unexpired time = Rs. 360
Banker's Discount is the Simple Interest on the face value of the bill for unexpired time= Simple Interest on Rs. 2160 for unexpired time=3601800×2160=15×2160=Rs. 432=3601800×2160=15×2160=Rs. 432
21. The banker's gain of a certain sum due 2 years hence at 10% per annum is Rs. 24. What is the present worth?
A. Rs. 600
B. Rs. 500
C. Rs. 400
D. Rs. 300
Answer: Option A
Explanation:
T = 2 yearsR = 10%
TD = BG ×100TR=24×1002×10 =12×10=Rs. 120TD = BG ×100TR=24×1002×10 =12×10=Rs. 120
TD = PW × TR100⇒120=PW ×2×10100⇒1200=PW ×2PW =12002 = Rs. 600TD = PW × TR100⇒120=PW ×2×10100⇒1200=PW ×2PW =12002 = Rs. 600
22. The true discount on a bill for Rs. 2520 due 6 months hence at 10% per annum is
A. Rs. 180
B. Rs. 140
C. Rs. 80
D. Rs. 120
Answer: Option D
Explanation:
F= Rs. 2520T = 6 months = 1212 yearR = 10%
TD = FTR100+TR=2520×12×10100+(12×10)=1260×10100+5=12600105=252021=Rs. 120TD = FTR100+TR=2520×12×10100+(12×10)=1260×10100+5=12600105=252021=Rs. 120
23. What is the present worth of a bill of Rs.1764 due 2 years hence at 5% compound interest is
A. Rs. 1600
B. Rs. 1200
C. Rs. 1800
D. Rs. 1400
Answer: Option A
Explanation:
Since the compound interest is taken here,PW (1+5100)2=1764PW (1+120)2=1764PW (2120)2=1764PW ×441400=1764⇒PW =1764×400441=4×400=Rs. 1600PW (1+5100)2=1764PW (1+120)2=1764PW (2120)2=1764PW ×441400=1764⇒PW =1764×400441=4×400=Rs. 1600
24. If the discount on Rs. 498 at 5% simple interest is Rs.18, when is the sum due?
A. 8 months
B. 11 months
C. 10 months
D. 9 months
Answer: Option D
Explanation:
F = Rs. 498TD = Rs. 18PW = F - TD = 498 - 18 = Rs. 480R = 5%
TD = PW × TR100⇒18=480× T ×5100⇒18=24× T ⇒ T =1824=34 years=12×34 months = 9 monthsTD = PW × TR100⇒18=480× T ×5100⇒18=24× T ⇒ T =1824=34 years=12×34 months = 9 months
25. What is the difference between the banker's discount and the true discount on Rs.8100 for 3 months at 5%
A. Rs. 2
B. Rs. 1.25
C. Rs. 2.25
D. Rs. 0.5
Answer: Option B
Explanation:
F = Rs. 8100R = 5%T = 3 months = 1414 years
BD = FTR100=8100×14×5100 =202520=4054= Rs. 101.25BD = FTR100=8100×14×5100 =202520=4054= Rs. 101.25
TD = FTR100+TR=8100×14×5100+(14×5) =2025×5100+54=2025×5×4400+5 =2025×5×4405=405×5×481 =45×5×49=5×5×4 = Rs. 100TD = FTR100+TR=8100×14×5100+(14×5) =2025×5100+54=2025×5×4400+5 =2025×5×4405=405×5×481 =45×5×49=5×5×4 = Rs. 100
BD - TD = Rs. 101.25 - Rs. 100 = Rs. 1.25
26. The B.G. on a certain sum 4 years hence at 5% is Rs. 200. What is the present worth?
A. Rs. 4500
B. Rs. 6000
C. Rs. 5000
D. Rs. 4000
Answer: Option C
Explanation:
T = 4 yearsR = 5%Banker's Gain, BG = Rs.200
TD = BG ×100TR=200×1004×5 = Rs. 1000TD = BG ×100TR=200×1004×5 = Rs. 1000
TD = √PW × BG1000=√PW ×2001000000=PW ×200PW =1000000200 = Rs. 5000TD = PW × BG1000=PW ×2001000000=PW ×200PW =1000000200 = Rs. 5000
27. The B.D. and T.D. on a certain sum is Rs.200 and Rs.100 respectively. Find out the sum.
A. Rs. 400
B. Rs. 300
C. Rs. 100
D. Rs. 200
Answer: Option D
Explanation:
F = BD×TDBD – TD=200×100200−100 =200×100100= Rs. 200F = BD×TDBD – TD=200×100200−100 =200×100100= Rs. 200
28. The banker's discount on a bill due 6 months hence at 6% is Rs. 18.54. What is the true discount?
A. Rs. 24
B. Rs. 12
C. Rs. 36
D. Rs. 18
Answer: Option D
Explanation:
T = 6 months = 1212 yearR = 6%
TD = BD ×100100+TR=18.54×100100+(12×6) =18.54×100103=1854103=Rs. 18TD = BD ×100100+TR=18.54×100100+(12×6) =18.54×100103=1854103=Rs. 18
29. The banker's discount on a sum of money for 112112 years is Rs. 120. The true discount on the same sum for 2 years is Rs.150. What is the rate percent?
A. 313%313%
B. 413%413%
C. 323%323%
D. 423%423%
Answer: Option A
Explanation:
BD for 112112 years = Rs. 120
=> BD for 2 years =120×23×2=120×23×2 = Rs.160
TD for 2 years = Rs. 150
F=BD ×TDBD – TD=160×150160–150 =160×15010=Rs. 2400F=BD ×TDBD – TD=160×150160–150 =160×15010=Rs. 2400
=> Rs.160 is the simple interest on Rs. 2400 for 2 years
⇒160=2400×2×R100⇒R=160×1002400×2=16048=103=313%⇒160=2400×2×R100⇒R=160×1002400×2=16048=103=313%
30. The present worth of a certain bill due sometime hence is Rs. 400 and the true discount is Rs. 20. What is the banker's discount?
A. Rs. 19
B. Rs. 22
C. Rs. 20
D. Rs. 21
Answer: Option D
Explanation:
BG = (TD)2PW=202400= Rs. 1BG = (TD)2PW=202400= Rs. 1
BG = BD – TD=> 1 = BD - 20=> BD = 1 + 20 = Rs. 21
Chapter 4: Problems on Partnerships
1. Two or more people can get together to do business by pooling resources. The deal is known as partnership. All the people who have invested money in the partnership are called partners.
A partner who manages the business is a working partner. A partner who simply invests the money is a sleeping partner.
2. If investments of all the partners are for the same time period, the gain or loss is distributed among the partners in the ratio of their investments.
Assume P and Q invest Rs. a and Rs. b respectively for a year in a partnership, then at the end of the year,(P's share of profit) : (Q's share of profit) = a : b
3. If investments are for different time periods, then equivalent capitals are calculated for a unit of time.
Assume A invests Rs. a for x months and B invests Rs. b for y months then,(A's share of profit) : (B's share of profit)= ax : by.
1. X and Y invest Rs.21000 and Rs.17500 respectively in a business. At the end of the year, they make a profit of Rs.26400. What is the share of X in the profit?
A. Rs.14400
B. Rs.26400
C. Rs.12000
D. Rs.12500
Answer: Option A
Explanation:
Ratio of the investment=21000:17500=210:175=21000:17500=210:175 =42:35=6:5=42:35=6:5
Share of X in the profit=26400×611=2400×6=14400=26400×611=2400×6=14400
2. X starts a business with Rs.45000. Y joins in the business after 3 months with Rs.30000. What will be the ratio in which they should share the profit at the end of the year?
A. 1:2
B. 2:1
C. 1:3
D. 3:1
Answer: Option B
Explanation:
Ratio in which they should share the profit = Ratio of the investments multiplied by the time period=45000×12:30000×9=45×12:30×9=3×12:2×9=2:1=45000×12:30000×9=45×12:30×9=3×12:2×9=2:1
3. Suresh started a business with Rs.20,000. Kiran joined him after 4 months with Rs.30,000. After 2 more months, Suresh withdrew Rs.5,000 of his capital and 2 more months later, Kiran brought in Rs.20,000 more. What should be the ratio in which they should share their profits at the end of the year?
A. 21:32
B. 32:21
C. 12:17
D. 17:12
Answer: Option A
Explanation:
Here capital is not the same.
Suresh invested 20000 for initial 6 months and 15000 for the next 6 months. Hence his term of ratio=(20000×6+15000×6)=(20000×6+15000×6)
Kiran invested Rs.30000 for 4 months and Rs.50000 for next 4 months. Hence his term of ratio=(30000×4+50000×4)=(30000×4+50000×4)
Suresh : Kiran =(20000×6+15000×6)=(20000×6+15000×6) :(30000×4+50000×4):(30000×4+50000×4)=(20×6+15×6):(30×4+50×4)=(20×3+15×3):(30×2:50×2)=105:160=21:32=(20×6+15×6):(30×4+50×4)=(20×3+15×3):(30×2:50×2)=105:160=21:32
4. Kamal started a business with Rs.25000 and after 4 months, Kiran joined him with Rs.60000. Kamal received Rs.58000 including 10% of profit as commission for managing the business. What amount did Kiran receive?
A. 75000
B. 70000
C. 72000
D. 78000
Answer: Option C
Explanation:
Ratio of the profits of Kamal and Kiran=25000×12:60000×8=25×12:60×8=5×3:12×2=5:4×2=5:8=25000×12:60000×8=25×12:60×8=5×3:12×2=5:4×2=5:8
Let the total profit =x=x
Then Kamal received 10x100=x1010x100=x10 as commission for managing the business.
Remaining profit =x−x10=9x10=x−x10=9x10 which is shared in the ratio 5:85:8
Total amount received by Kamal =x10+9x10×513=x10+9x10×513
⇒x10+9x10×513=58000⇒x+9x×513=580000⇒x(1+4513)=580000⇒x×5813=580000⇒x×113=10000⇒x=130000⇒x10+9x10×513=58000⇒x+9x×513=580000⇒x(1+4513)=580000⇒x×5813=580000⇒x×113=10000⇒x=130000
Kiran's share = 130000−58000=72000130000−58000=72000
5. A and B started a partnership business investing Rs. 20,000 and Rs. 15,000 respectively. C joined them with Rs. 20,000 after six months. Calculate B's share in total profit of Rs. 25,000 earned at the end of 2 years from the starting of the business?
A. 7500
B. 8500
C. 9000
D. 8000
Answer: Option A
Explanation:
A : B : C =20000×24:15000×24:20000×18=20×4:15×4:20×3=4×4:3×4:4×3=4:3:3=20000×24:15000×24:20000×18=20×4:15×4:20×3=4×4:3×4:4×3=4:3:3
B's share = 25000×310=750025000×310=7500
6. A starts a business with a capital of Rs. 85,000. B joins in the business with Rs.42500 after some time. For how much period does B join, if the profits at the end of the year are divided in the ratio of 3 : 1?
A. 5 months
B. 6 months
C. 7 months
D. 8 months
Answer: Option D
Explanation:
Let B joined for xx months. Then
85000×12:42500×x=3:1⇒850×12:425x=3:1⇒850×12×1=3×425x⇒850×4=425x⇒x=885000×12:42500×x=3:1⇒850×12:425x=3:1⇒850×12×1=3×425x⇒850×4=425x⇒x=8
7. A starts a business with Rs.40,000. After 2 months, B joined him with Rs.60,000. C joined them after some more time with Rs.1,20,000. At the end of the year, out of a total profit of Rs.3,75,000, C gets Rs.1,50,000 as his share. How many months after B joined the business, did C join?
A. 4 months
B. 5 months
C. 6 months
D. 7 months
Answer: Option A
Explanation:
Assume that C was there in the business for xx months
A:B:C=40000×12:60000×10:120000×x=40×12:60×10:120x=40:5×10:10x=8:10:2x=4:5:x=40000×12:60000×10:120000×x=40×12:60×10:120x=40:5×10:10x=8:10:2x=4:5:x
C's share =375000×x9+x=375000x9+x=375000×x9+x=375000x9+x
⇒375000x9+x=150000⇒375x9+x=150⇒15x=6(9+x)⇒5x=18+2x⇒3x=18⇒x=6⇒375000x9+x=150000⇒375x9+x=150⇒15x=6(9+x)⇒5x=18+2x⇒3x=18⇒x=6
It means C was there in the business for 6 months. Given that B joined the business after 2 months. Hence C joined 4 months after B joined.
8. A and B invest in a business in the ratio 3: 2. Assume that 5% of the total profit goes to charity. If A's share is Rs. 855, what is the total profit?
A. 1400
B. 1500
C. 1600
D. 1200
Answer: Option B
Explanation:
Assume that the total profit is xx
Since 5% goes for charity, 95% of xx will be divided between A and B in the ratio 3:23:2Therefore, A's profit =95x100×35=95x100×35
Given that A's share is Rs. 855. Therefore,95x100×35=855⇒95x100=855×53=285×5=1425 ⇒x=1425×10095=285×10019=150095x100×35=855⇒95x100=855×53=285×5=1425 ⇒x=1425×10095=285×10019=1500
Hence the total profit = 1500
9. A, B and C invest in a partnership in the ratio: 72,43,6572,43,65. After 4 months, A increases his share 50%. If the total profit at the end of one year is Rs.21,600, then what is B's share in the profit?
A. Rs. 2000
B. Rs. 3000
C. Rs. 4000
D. Rs. 5000
Answer: Option C
Explanation:
Ratio of the initial investment=72:43:65=105:40:36=72:43:65=105:40:36
Therefore, let the initial investments of A, B and C be 105x,40x105x,40x and 36x36x respectively
A increases his share 50% after 4 months. Hence the ratio of their investments=(105x×4)+(105x×150100×8)=(105x×4)+(105x×150100×8) :40x×12:36x×12:40x×12:36x×12=105+(105×32×2)=105+(105×32×2) :40×3:36×3:40×3:36×3=105×4:40×3:36×3=35×4:40:36=35:10:9=105×4:40×3:36×3=35×4:40:36=35:10:9
B's share = total profit × 10541054=21600×1054=4000=21600×1054=4000
10. A, B and C jointly thought of engaging themselves in a business venture. It was agreed that A would invest Rs. 6500 for 6 months, B, Rs. 8400 for 5 months and C, Rs. 10,000 for 3 months. A wants to be the working member for which, he was to receive 5% of the profits. The profit earned was Rs. 7400. What is the share of B in the profit.
A. 2660
B. 1000
C. 2300
D. 4000
Answer: Option A
Explanation:
A is a working member and for that, he receives 5% of the profit = 5% of 7400 =5×7400100=370=5×7400100=370
Remaining amount =7400−370=7030=7400−370=7030
Ratio of their investments=6500×6:8400×5:10000×3=65×6:84×5:100×3=13×6:84:20×3=13×2:28:20=13:14:10=6500×6:8400×5:10000×3=65×6:84×5:100×3=13×6:84:20×3=13×2:28:20=13:14:10
Share of B in the profit=7030×1437=190×14=2660=7030×1437=190×14=2660
11. A, B, C subscribe Rs. 50,000 for a business. If A subscribes Rs. 4000 more than B and B Rs. 5000 more than C, out of a total profit of Rs. 35,000, what will be the amount A receives?
A. 14200
B. 14700
C. 14800
D. 14500
Answer: Option B
Explanation:
Total amount invested = 50000
Assume that investment of C =x=x.Then investment of B =5000+x=5000+x,Investment of A =4000+5000+x=9000+x=4000+5000+x=9000+x
x+5000+x+9000+x=50000⇒3x+14000=50000⇒3x=50000–14000=36000⇒x=360003=12000x+5000+x+9000+x=50000⇒3x+14000=50000⇒3x=50000–14000=36000⇒x=360003=12000
Investment of C =x=12000=x=12000Investment of B =5000+x=17000=5000+x=17000Investment of A =9000+x=21000=9000+x=21000
Ratio of the investment of A, B and C=21000:17000:12000=21:17:12=21000:17000:12000=21:17:12
Share of A = Total profit ×2150×2150=35000×2150=700×21=14700=35000×2150=700×21=14700
12. A, B and C rent a pasture. If A puts 10 oxen for 7 months, B puts 12 oxen for 5 months and C puts 15 oxen for 3 months for grazing and the rent of the pasture is Rs. 175, then how much amount should C pay as his share of rent?
A. 45
B. 35
C. 55
D. 60
Answer: Option A
Explanation:
A : B : C=10×7:12×5:15×3=2×7:12×1:3×3=14:12:9=10×7:12×5:15×3=2×7:12×1:3×3=14:12:9
Amount that C should pay=175×935=5×9=45=175×935=5×9=45
13. A and B entered into a partnership with capitals in the ratio 4 : 5. After 3 months, A withdrew 1414 of his capital and B withdrew 1515 of his capital. At the end of 10 months, the gain was Rs.760. What is A's share in the profit?
A. 310
B. 330
C. 370
D. 350
Answer: Option B
Explanation:
Ratio of the initial capital of A and B = 4 : 5
Hence we can take the initial capitals of A and B as 4x4x and 5x5x respectively
Ratio in which profit will be divided =(4x×3)+34×4x×7=(4x×3)+34×4x×7 :(5x×3)+45×5x×7:(5x×3)+45×5x×7=(12+21):(15+28)=33:43=(12+21):(15+28)=33:43
A's share =760×3376=330=760×3376=330
14. A starts a business with Rs. 3500. After 5 months, B joins with A as his partner. After a year, the profit is divided in the ratio 2 : 3. B's contribution in the capital is
A. 7000
B. 8000
C. 9000
D. 10000
Answer: Option C
Explanation:
Ratio in which profit is divided =2:3=2:3
Assume that B's contribution to the capital =b=b
3500×12:b×7=2:3 ⇒3500×12×3=2×b×7 ⇒b=3500×12×32×7 =500×6×3=90003500×12:b×7=2:3 ⇒3500×12×3=2×b×7 ⇒b=3500×12×32×7 =500×6×3=9000
15. A, B and C shared the profit in a business in the ratio 5 : 7 : 8. They had partnered for 14 months, 8 months and 7 months respectively. What was the ratio of their investments?
A. 10 : 12 : 14
B. 12 : 24 : 28
C. 20 : 22 : 12
D. 20 : 49 : 64
Answer: Option D
Explanation:
Ratio in which A,B and C shared the profit =5:7:8=5:7:8
Assume that the ratio of their investments =a:b:c=a:b:c⇒14a:8b:7c=5:7:8⋯(A)⇒14a:8b:7c=5:7:8⋯(A)
From (A),14a:8b=5:7⇒14a×7=8b×5⇒7a×7=4b×5⇒b=49a20⋯(1)14a:8b=5:7⇒14a×7=8b×5⇒7a×7=4b×5⇒b=49a20⋯(1)
From (A),14a:7c=5:8⇒14a×8=7c×5⇒2a×8=c×5⇒c=16a5⋯(2)14a:7c=5:8⇒14a×8=7c×5⇒2a×8=c×5⇒c=16a5⋯(2)
a:b:c=a:49a20:16a5=1:4920:165=20:49:64a:b:c=a:49a20:16a5=1:4920:165=20:49:64
16. A and B started a partnership business investing capital in the ratio of 3 : 5. C joined in the partnership after six months with an amount equal to that of B. At the end of one year, the profit should be distributed among A, B and C in --- proportion.
A. 10 : 5 : 4
B. 5 : 3 : 4
C. 3 : 4: 5
D. 6 : 10 : 5
Answer: Option D
Explanation:
Initial investment capital ratio of A and B = 3 : 5
Let initial capital of A and B be 3x3x and 5x5x respectively.
Amount that C invested after 6 months =5x=5x (since it is equal to B's investment)
Ratio in which profit should be distributed after 1 year=3x×12:5x×12:5x×6=3×12:5×12:5×6=3×2:5×2:5=6:10:5=3x×12:5x×12:5x×6=3×12:5×12:5×6=3×2:5×2:5=6:10:5
17. A and B partner in a business. A contribute 1414 of the capital for 15 months and B received 2323 of the profit. For how long B's money was used?
A. 12 months
B. 10 months
C. 14 months
D. 16 months
Answer: Option B
Explanation:
B received 2323 of the profit.Then A received remaining 1313 of the profit.=> A : B =13:23=1:2=13:23=1:2
Let the total capital =x=xThen A's capital =x4=x4B's capital =x–x4=3x4=x–x4=3x4
Assume B's money was used for bb monthsThen A:B =x4×15:3x4×b=x4×15:3x4×b
⇒15x4:3bx4=1:2⇒15:3b=1:2⇒5:b=1:2⇒5×2=1×b⇒b=10⇒15x4:3bx4=1:2⇒15:3b=1:2⇒5:b=1:2⇒5×2=1×b⇒b=10
18. A, B and C started a partnership business by investing Rs.27000, Rs.72000, Rs.81000 respectively. At the end of the year , the profit were distributed among them. If C's share of profit is 36000, What is the total profit?
A. 80000
B. 90000
C. 70000
D. 120000
Answer: Option A
Explanation:
A:B:C =27000:72000:81000=27000:72000:81000=3:8:9=3:8:9
Let total profit =p=p
Then p×920=36000p×920=36000
p=36000×209=80000p=36000×209=80000
19. A and B started a partnership business. A's investment was thrice the investment of B and the period of his investment was two times the period of investments of B. If B received Rs 4000 as profit, what is their total profit?
A. 28000
B. 30000
C. 32000
D. 34000
Answer: Option A
Explanation:
Suppose B's investment =x=x.Then A's investment =3x=3x
Suppose B's period of investment =y=ythen A's period of investment =2y=2y
A : B =3x×2y:xy=6:1=3x×2y:xy=6:1
Total profit ×17=4000×17=4000=> Total profit =4000×7=28000=4000×7=28000
20. P and Q invested in a business. The profit earned was divided in the ratio 2 : 3. If P invested Rs 40000, the amount invested by Q is
A. 40000
B. 50000
C. 60000
D. 70000
Answer: Option C
Explanation:
Let the amount invested by Q = q
40000:q=2:3⇒40000×3=2q⇒q=40000×32=6000040000:q=2:3⇒40000×3=2q⇒q=40000×32=60000
21. A and B start a business jointly. A invests Rs 16000 for 8 month and B remains in the business for 4 months. Out of total profit, B claims 2727 of the profit . How much money was contributed by B?
A. 11200
B. 12000
C. 12400
D. 12800
Answer: Option D
Explanation:
B claims 2727 of the profitA claims remaining 5757 of the profit=> A : B =57:27=5:2=57:27=5:2
Let the money contributed by B =b=bThen A:B =16000×8:b×4=16000×8:b×4
Therefore,16000×8:b×4=5:216000×8×2=b×4×516000×2×2=b×53200×2×2=bb=1280016000×8:b×4=5:216000×8×2=b×4×516000×2×2=b×53200×2×2=bb=12800
22. A and B starts a business investing Rs.85000 and Rs.15000 respectively. Find out the ratio in which the profits should be shared.
A. 10 : 3
B. 17 : 3
C. 3 : 10
D. 3 : 17
Answer: Option B
Explanation:
Here A's and B's capitals are there for equal time. Hence
A : B = 85000 : 15000= 85 : 15= 17 : 3
23. A, B and C start a business with each investing Rs 20,000. After 5 months A withdraws Rs 5000, B withdraws Rs 4000 and C invests Rs 6000 more. At the end of the year, a total profit of Rs 69900 was recorded. Find the share of A.
A. 20600
B. 20700
C. 20500
D. 20400
Answer: Option C
Explanation:
A : B : C
=(20000×5+15000×7):(20000×5+16000×7):(20000×5+26000×7)=(20000×5+15000×7):(20000×5+16000×7):(20000×5+26000×7)
=(20×5+15×7):(20×5+16×7):(20×5+26×7)=(20×5+15×7):(20×5+16×7):(20×5+26×7)
=205:212:282=205:212:282
A's share =69900×205205+212+282=69900×205205+212+282=69900×205699=20500=69900×205699=20500
24. A invested Rs 76000 in a business. After few months, B joined him with Rs 57000. The total profit was divided between them in the ratio 2 : 1 at the end of the year. After how many months did B join?
A. 2
B. 3
C. 4
D. 5
Answer: Option C
Explanation:
Suppose B was there in the business for xx months. ThenA : B =76000×12:57000×x=76000×12:57000×x
Therefore,76000×12:57000×x=2:176×12:57x=2:176×12×1=57x×276×4=19x×24×4=x×2x=876000×12:57000×x=2:176×12:57x=2:176×12×1=57x×276×4=19x×24×4=x×2x=8
Hence B was there in the business for 8 months, or joined after 12-8 = 4 months
25. Three partners A, B and C start a business. B's capital is four times C's capital and twice A's capital is equal to thrice B's capital. If the total profit is Rs 16500 at the end of a year, find out B's share in it.
A. 4000
B. 5000
C. 6000
D. 7000
Answer: Option C
Explanation:
Suppose C's capital =x=x. Then,B's capital =4x=4x (Since B's capital is four times C's capital)A's capital =6x=6x (Since twice A's capital is equal to thrice B's capital)
A : B : C =6x:4x:x=6x:4x:x= 6 : 4 : 1
B's share=16500×411=1500×4=6000=16500×411=1500×4=6000
26. In a business, A and C invested amounts in the ratio 2 : 1 , whereas the ratio between amounts invested by A and B was 3 : 2 . If Rs.157300 was their profit, how much amount did B receive?
A. 48000
B. 48200
C. 48400
D. 48600
Answer: Option C
Explanation:
Assume that investment of C =x=xThen, investment of A =2x=2xInvestment of B =4x3=4x3
A:B:C =2x:4x3:x=2x:4x3:x=2:43:1=6:4:3=2:43:1=6:4:3
B's share =157300×46+4+3=157300×46+4+3=157300×413=12100×4=48400=157300×413=12100×4=48400
27. P, Q and R started a business by investing Rs.120000, Rs.135000 and Rs.150000 respectively. Find the share of each, out of the annual profit of Rs.56700.
A. 16800, 18900, 21000
B. 17850, 18900, 21000
C. 16800, 18900, 22000
D. 17850, 18500, 22000
Answer: Option A
Explanation:
P : Q : R = 120000:135000:150000120000:135000:150000=120:135:150=24:27:30=8:9:10=120:135:150=24:27:30=8:9:10
Share of P=56700×827=2100×8=16800=56700×827=2100×8=16800
Share of Q=56700×927=2100×9=18900=56700×927=2100×9=18900
Share of R=56700×1027=2100×10=21000=56700×1027=2100×10=21000
28. If 4 (P's Capital) = 6(Q's Capital) = 10(R's Capital), then out of the total profit of Rs 4650, R will receive
A. 600
B. 700
C. 800
D. 900
Answer: Option D
Explanation:
Let P's capital be pp,Q's capital be qq,and R's capital be rr
Then4p=6q=10r2p=3q=5r⋯(A)4p=6q=10r2p=3q=5r⋯(A)
From (A),q=2p3 ⋯(1)r=2p5 ⋯(2)q=2p3 ⋯(1)r=2p5 ⋯(2)
p:q:r=p:2p3:2p5p:q:r=p:2p3:2p5=15:10:6=15:10:6
R's share =4650×631=150×6=900=4650×631=150×6=900
29. P, Q, R enter into a partnership. P initially invests 25 lakh and adds another 10 lakh after one year. Q initially invests 35 lakh and withdraws 10 lakh after 2 years. R's investment is Rs 30 lakh. In what ratio should the profit be divided at the end of 3 years?
A. 18:19:19
B. 18:18:19
C. 19:19:18
D. 18:19:18
Answer: Option C
Explanation:
P : Q : R=(25×1+35×2):(35×2+25×1)=(25×1+35×2):(35×2+25×1) :(30×3):(30×3)=95:95:90=19:19:18=95:95:90=19:19:18
30. P, Q, R enter into a partnership and their share are in the ratio 12:13:1412:13:14. After two months, P withdraws half of his capital and after 10 more months, a profit of Rs 378 is divided among them. What is Q's share?
A. 114
B. 120
C. 134
D. 144
Answer: Option D
Explanation:
The ratio of their initial investment=12:13:14=6:4:3=12:13:14=6:4:3Let the initial investment of P, Q and R be 6x,4x6x,4x and 3x3x respectively.A:B:C=(6x×2+3x×10):4x×12:3x×12=(6x×2+3x×10):4x×12:3x×12=(12+30):4×12:3×12=(4+10):4×4:12=14:16:12=7:8:6=(12+30):4×12:3×12=(4+10):4×4:12=14:16:12=7:8:6
B's share =378×821=18×8=144=378×821=18×8=144
Chapter 5: Problem on Percentage
Important Concepts and Formulas (Part 1) - Percentage
Percent means for every 100100. So, when percent is calculated for any value, it means that we calculate the value for every 100100 of the reference value.
percent is denoted by the symbol %%.
Example:xx percent is denoted by x%x%
x%=x100x%=x100
example25%=25100=1425%=25100=14
xy=(xy×100)%xy=(xy×100)%
example14=(14×100)%=25%14=(14×100)%=25%
Values to memorize for quick calculations.
1=100%1=100%
16=1623%≈16.67%16=1623%≈16.67%
12=50%12=50%
17=1427%≈14.29%17=1427%≈14.29%
13=3313%≈33.33% 23=6623%≈66.67%13=3313%≈33.33% 23=6623%≈66.67%
18=1212%=12.5%18=1212%=12.5%
14=25% 34=75%14=25% 34=75%
19=1119%≈11.11%19=1119%≈11.11%
15=20%15=20%
110=10%110=10%
If xx is increased by y%y%, then,new number =x+xy100=x(100+y)100=x+xy100=x(100+y)100
If xx is decreased by y%y%, then,new number =x−xy100=x(100−y)100=x−xy100=x(100−y)100
Examples:(1)(1) If 150150 is increased by 10%10%,new number =150+150×10100=165=150+150×10100=165ornew number =150(100+10)100=165=150(100+10)100=165ornew number =150+150×110=165 =150+150×110=165 (since 10%=11010%=110)
(2)(2) If 150150 is decreased by 10%10%,new number =150−150×10100=135=150−150×10100=135ornew number =150(100−10)100=135=150(100−10)100=135ornew number =150−150×110=135 =150−150×110=135 (since 10%=11010%=110)
If the price of a commodity increases by x%x%, the reduction in consumption so as not to increase the expenditure=(x100+x×100)%=(x100+x×100)%
If the price of a commodity decreases by x%x%, the increase in consumption so as not to decrease the expenditure=(x100−x×100)%=(x100−x×100)%
If population of a town =P=P and it increases at the rate of R%R% per annum, then,
population after nn years=P(1+R100)n=P(1+R100)n
population before n years=P(1+R100)n=P(1+R100)n
If the present value of a machine =P=P and it depreciates at the rate of R%R% per annum, then,
value of the machine after nn years=P(1−R100)n=P(1−R100)n
value of the machine before n years=P(1−R100)n
1. Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56% of the sum of their marks. What are the marks obtained by them?
A. 42, 33
B. 42, 36
C. 44, 33
D. 44, 36
Answer: Option A
Explanation:
Let the marks secured by them be xx and (x+9)(x+9)
sum of their marks =x+(x+9)=2x+9=x+(x+9)=2x+9
Given that (x+9)(x+9) was 56% of the sum of their marks.⇒(x+9)=56100(2x+9)⇒(x+9)=1425(2x+9)⇒25x+225=28x+126⇒3x=99⇒x=33⇒(x+9)=56100(2x+9)⇒(x+9)=1425(2x+9)⇒25x+225=28x+126⇒3x=99⇒x=33
Then (x+9)=33+9=42(x+9)=33+9=42
Hence their marks are 33 and 42
2. If A =x%=x% of yy and B =y%=y% of xx, then which of the following is true?
A. None of these
B. A is smaller than B.
C. Relationship between A and B cannot be determined.
D. If xx is smaller than yy, then A is greater than B.
E. A is greater than B.
Answer: Option A
Explanation:
A =x100×y=xy100 ⋯(1)=x100×y=xy100 ⋯(1)
B = y100×x=yx100 ⋯(2)y100×x=yx100 ⋯(2)
Therefore A = B
3. If 20% of a = b, then b% of 20 is the same as:
A. None of these
B. 10% of a
C. 4% of a
D. 20% of a
Answer: Option C
Explanation:
20% of a=ba=b⇒b=20a100=a5⇒b=20a100=a5
b%b% of 20=20×b10020=20×b100=b5=a5×15=a25=4a100=4% of a=b5=a5×15=a25=4a100=4% of a
4. Two numbers A and B are such that the sum of 5% of A and 4% of B is two-third of the sum of 6% of A and 8% of B. Find the ratio of A : B.
A. 2 : 1
B. 1 : 2
C. 1 : 1
D. 4 : 3
Answer: Option D
Explanation:
5% of A + 4% of B =23=23(6% of A + 8% of B)
=> 5A100+4B100=23(6A100+8B100)5A100+4B100=23(6A100+8B100)=> 5A + 4B =23=23(6A+ 8B)=> 15A + 12B = 12A+ 16B=> 3A = 4B=> AB=43AB=43=> A : B = 4 : 3
5. Two employees X and Y are paid a total of Rs. 550 per week by their employer. If X is paid 120 percent of the sum paid to Y, how much is Y paid per week?
A. Rs. 150
B. Rs. 300
C. Rs. 250
D. Rs. 200
Answer: Option C
Explanation:
Solution 1
Let the amount paid to X per week =x=xand the amount paid to Y per week =y=yThen x+y=550 ⋯(1)x+y=550 ⋯(1)
But x=120%x=120% of yy =120y100=12y10 ⋯(2)=120y100=12y10 ⋯(2)
From (1) and (2)12y10+y=550⇒22y10=550⇒y=550022=5002=25012y10+y=550⇒22y10=550⇒y=550022=5002=250
Solution 2
Let the amount paid to X per week =x=xand the amount paid to Y per week =y=yThen x+y=550 ⋯(1)x+y=550 ⋯(1)
But x=120%x=120% of y ⋯(2)y ⋯(2)
From (1) and (2),120% of yy + 100% of yy = 550220% of yy = 550y=550×100220=250y=550×100220=250
Solution 3
Let the amount paid to Y per week =y=yand the amount paid to X per week =120y100=6y5=120y100=6y5
y+6y5=55011y5=550y=250y+6y5=55011y5=550y=250
6. Rahul went to a shop and bought things worth Rs. 25, out of which 30 paise went on sales tax on taxable purchases. If the tax rate was 6%, then what was the cost of the tax free items?
A. Rs. 15
B. Rs. 12.10
C. Rs. 19.70
D. Rs. 16.80